Dynamics of Structures Volume 2 - Applications Chowdhury

Dynamics of Structure and Foundation – A Unified Approach

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


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)


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?


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?


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


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


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


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


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


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


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


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.


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.


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.



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.



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


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.



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.



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


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


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.


B


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


[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.



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



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.



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.



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


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.



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.



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.



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.



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


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.



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



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.



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


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.


M


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


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


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


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.



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


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.



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


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)


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.



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.



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.



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.



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


mφ32 0.47475407 0.07529536 0.12271058 0.67276001


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


for the third mode.


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


0.4905 0.34335

0.6867 0.2943

0.7848 0.245



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

Mode 1

Mode 2

Mode 3


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


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.



• • • •

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


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.



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.



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.



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.



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



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.



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


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 .



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


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


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


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.



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


(1.7.6)


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


BH4


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


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


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.



$ 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 =



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


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”.


(1.8.2)


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


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


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


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


• • •

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


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.



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.



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.


11

12

13 14

15


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


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.



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


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.



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.



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



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.


8

9

10

Damping Ratio(%)


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.



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


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


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.



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.


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


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.


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.



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


• •

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


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


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



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.



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)


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


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.



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.


(2.2.18)


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


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.



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 =


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


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)


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


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. . ..



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)


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˙


Based on the above the equation of motion becomes

x¨ 200 + −300 φ¨

50 0

0 100

+

3000 −4500


−4500 11750


x˙ φ˙

−300 988


•

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



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


200 −300 0.1188173 = 1.86308 −300 988 0.0542329


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


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)


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


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


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%.


0.20273 −0.08399


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



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



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.



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.



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.



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.



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} =



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



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


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.



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


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


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


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.



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.


Figure 2.5.4 Mathematical model.


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


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



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


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


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



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


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.


(2.5.11)


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



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



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



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



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



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.



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 =


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


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


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.



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


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


(2.5.27)


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



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


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


X2 =

GS1 L EAβ − ; 4β 4L

X3 =

Gb r0 Cb 2 (2.5.42)


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


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


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.


(2.5.61)


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.



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)


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


1 β2 2 4 − + β 7 33 675

¯ b cos2 β (2.5.73) + r 0 ρb G b C


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


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


−Gf Sf Df G f Sf D f

(2.5.79)


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+


sin 2β 2β

!

and A = Kgroup + Gf Sf Df ; B = Gf Sf Df .

(2.5.84)


√ √ 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


K12 K22 K32

⎤ K13 K23 ⎦ K33

(2.5.86)


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



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


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.


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


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


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


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


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



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 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


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


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;



• •

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



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



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


(2.5.103) (2.5.104)


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.



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


X Y 2βz 2βz + cos + η sin dz 2 2 L L

(2.5.114)


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)


− 34 η

which can be further simplified to

(2.5.118)


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


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


Table 2.5.16 Suggested value of Sx1 for Poisson’s ratio of soil = 0.50. Poisson’s ratio 0.50


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




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


(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


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


(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


Sx2 0.8652a0 a0 + 0.00874 41.59a0 0.83a0 + a0 + 3.90 56.559a0 0.96a0 + a0 + 4.68 7.334a0 +


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 +


(2.5.132)

1 . βe

(2.5.133)


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)


(2.5.138) (2.5.139)


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


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


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


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


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)


(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


(2.5.153) Mz =

E p I p P0 K (1 − r2 )2 + (2Dr)2

2β 2 − βz e L L2

sin


(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


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


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)


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.


1

9 0.

8 0.

7 0.

6 0.

5 0.

4 0.

3 0.

2 0.

1 0.

0

0


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


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



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


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.



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.



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


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


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)


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


(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


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


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.



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.



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




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


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


(2.5.210)


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


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


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


0

(2.5.224)


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


(2.5.229)


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


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


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)


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


(2.5.258)


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


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


➔ Jx =

(2.5.267) Mp r20 [I1 + 4λ2 I2 ] 4

(2.5.268)


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


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.



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:



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


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


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. . . . . .



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.



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


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.



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. . . . . .



•

•

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


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.



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


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


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



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



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 .


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


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 )


Loading location


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


−Cy Zc Cφy + Cy Zc2 − WZc

−Ky Zc y 0 = 0 Kφy + Ky Zc2 − WZc φ


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


and −2269540φ21 − 5319678φ22 = 0


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.



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


0 900611 −2269540 + 0.0203 18588.4 −2269540 18194648 −46071 . 318233


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


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


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 )


Kh = m


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


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


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


+ (2Dr)2

= 0.332

where r =

236 = 9.7 24.34


➔ 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


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


Time Steps

Displacement in Y direction Angular Rotation


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.


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


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.



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,



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


m . K


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.


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)


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.



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.


(2.7.1)


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


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


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.


(2.7.5)

242 Dynamics of Structure and Foundation: 2. Applications N

{{ξï } + 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


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



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.



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

{{ξï } + 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


(2.7.14)


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


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



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;



• • • • • • • • • • • • • • • • • •

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



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



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


δdyn =

V ωna


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.



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.



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.



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.



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.



• • • •

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.



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.



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


x1


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 =


0.0667 0.0684

and

0.01898 . −0.240


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


1474 6316


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.



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.


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


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


Amplitude(mm)


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.


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.



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.



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.



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.



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


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


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)



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


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)



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


tan φ =

2Dω /ωn 2 1 − ωωn

(2.11.6)


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


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.


6

3


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


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.



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



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.


c


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


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)


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.


(d)


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


(2.13.9) (2.13.10)


(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


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


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.



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.



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.



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.



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


(2.14.4)


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).



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.


k1/2


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).



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.


D

C

Wb

Kb

Kd

Kd


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).



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.



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.



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.


+

∇ 2 π

(r)2

.

(2.14.25)


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



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.


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.



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.



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.



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


(2.15.5)


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)



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


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

ξï + 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.



• • • •

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.



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


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


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.



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


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.



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).



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.



• • •

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.



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.



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


•

•

•

• • • • • • • •

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.



•

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


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).



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


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


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


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


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.



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


(2.18.5)

where Ps = Ks δs

(2.18.6)


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



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


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


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.



• • • • • • •

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.



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


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.



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


Steel quantity


• • • • •

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.


1200 1580 100kN 2238


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.



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


1308

900

3


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.



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 )



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.


965

1308

900


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.



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.



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



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.



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


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


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



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.



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



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.



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



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.



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



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


Khi Xhi2


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



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


and = 0.4


For transverse frame 2, we have 4

δi = 1.9375 × 10−5 m

i=1

For calculation of and

4

i=1 δi


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


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



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



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



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



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.



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



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



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.



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.



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



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.



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


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


Figure 2.18.21 Comparison of transient and operating response.



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


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


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



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.



| −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



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


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


Y φ U θ

802.8 18.43 60.41 157

0.44338 −2.26×10−04 −0.018798 −0.085314





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


Displacement under normal operation

Displacement under 2nd transient Transient frequency = 60.41 rad/sec

Displacement

Frequency


Y φ ϒ θ

802.8 18.43 60.41 157

0.44338 −2.26×10−04 −0.018798 −0.085314





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


Y φ U θ

802.8 18.43 60.41 157

0.44338 −2.26×10−04 −0.018798 −0.085314






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


Transient frequency = 157 rad/sec


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



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.



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


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


Figure 2.18.28 Amplitudes at the second transient frequency @ 60 rad/sec.


U



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


Figure 2.18.30 Amplitudes at third transient frequency @ 157 rad/sec. 2.00E-03


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.


Y1 Y2 Y3


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.



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.


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.


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


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.


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


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.


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.



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.



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.



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


(3.1.4c) (3.1.4d)


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.


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.


(3.1.9)


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


(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


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


1.76−

190 152

= 2.498;

and


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


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.



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.



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


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


(3.1.17)


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


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


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


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


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.


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


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.



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.



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;



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.



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.


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.



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.


(3.2.6a)


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


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


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


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)


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


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.


3000


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 =



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.



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.



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∗


m K,

we have for this generalized case

(3.2.33)


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


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∗



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



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 } =


[ϕn ]T {P(t)} [ϕn ]T [M][ϕ]

(3.2.38)


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.


(3.2.46)


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.



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



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


⎤ 40

⎦ 20


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



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


and

⎧ ⎫ ⎨0.078⎬ {Sv } = 0.036 m/sec ⎩ ⎭ 0.025


•

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


⎧ ⎫ ⎫ ⎤⎧ 9.81 ⎬ 0 ⎨−0.5⎬ ⎨ −20 0 ⎦ −0.5 ×27.386×0.036 = 9.81 kN ⎩ ⎭ ⎩ ⎭ −9.81 20 1.0 n=2 40


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


(3.2.50)


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


(3.2.54)


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



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



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, {äg }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.



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]{äg }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]


(3.3.3)


•

Calculate modified load at time t + t ˙ t + α3 X ¨ t} ˆ t+t } = [M]{äg }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


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.


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)


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 {ξï=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 {ξï=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 } .


(3.3.10)


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.



• • •

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 {ξï } + 2Di ωi {ξ˙i } + ω2 {ξi } = {äg } we have the three equations as: ξ¨1 + 0.281ξ˙1 + 1.640961ξ1 = a¨ g ; ξ¨3 + 0.4135ξ˙3 + 17.098ξ3 = a¨ g .


ξ¨2 + 0.3162ξ˙2 + 9.99824ξ2 = 0;


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


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.



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


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)


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


dz2

dz

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

(3.3.26)


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


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).


(3.3.37)


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


Time period (secs) ' WH3 T1 = 1.787 EIg ' WH3 T2 = 0.285 EIg ' WH3 T2 = 0.102 EIg

(3.3.39)


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.



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


(3.3.47)


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


Sa2 g

Sa3 g

[f3 (ξ )],

[f3 (ξ )]

and

(3.3.52)


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


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



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



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.



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 =


(3.3.55)


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.



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


w3 (z, t) = κ3


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)


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


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]



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



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



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.


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


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


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 .



• • • •

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.



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.



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


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.



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.



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)


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


(3.4.13)


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 =


(3.4.20) (3.4.21)


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.



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


Time period (secs) 2 γc T1 = 4.68 HB Eg 2 γc T2 = 1.36 HB Eg 2 γc T3 = 0.455 HB Eg


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.



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


(3.4.30)


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


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,


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.


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.


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


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.


(3.4.49)


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


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)


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


, a constant number.

(3.4.62)


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 some simplification we have, H2 T = 6.19 B

'

γc CFS Ec g

(3.4.64)

where, CFS is fluid-structure interaction coefficient given by CFS = The amplitude of displacement is given by w(z)max = κi β where κi =

κi =

Sa f1 (ξ )max ω2

i=1 mi φi /

at z = H.

2 i=1 mi φi ,

I2 χm + I3 . I 2 χs + I 1

(3.4.65)

for this particular case, the expression becomes

γw Hw2 1 γc A 0 H 1 0 (1 + ψξ )f (ξ )dξ + 2g 0 f (ξ )dξ g 2 γw H w 1 γc A 0 H 1 2 2 0 (1 + ψξ ) f (ξ ) + 2g 0 f (ξ ) g

=

I4 + χm I5 I 3 + χ m I2

(3.4.66)

Based on this the maximum displacement becomes w(z)max = 1.01175κ i β

γc CFS B2 E c

Sa g

H4

(3.4.67)

The moment and shear are thus given by M = −EI Hence,

d2w dz2

and V = EI

d3w dz3

Mz = −0.2964 (1 + ψξ )3 γc κi βC FS BH 2 (Sa /g)f (ξ )

Vz = −0.557 (1 + ψξ )3 γc κi βC FS BH(Sa /g)f (ξ )

(3.4.68) (3.4.69) (3.4.70)

The values of f (ξ ) and f (ξ )are given Table 3.4.1. The integral values I1 to I5 are as given hereafter for various Bt /B ratio are given in Table 3.4.2. © 2009 Taylor & Francis Group, London, UK

Analytical and design concepts for earthquake engineering 501 Table 3.4.1 Z/H

Moment coefficients

Shear coefficient

0 0.2 0.4 0.6 0.8 1

1 0.725493 0.461163 0.229918 0.063898 0

1 0.988851 0.917369 0.742928 0.440703 0

Table 3.4.2 Values of integral constants for various Bt /B ratio. Bt/B

I1

I2

I3

I4

I5

0.05 0.1 0.15 0.2 0.25 0.3

1.102 1.13 1.159 1.189 1.221 1.254

1.855 1.855 1.855 1.855 1.855 1.855

0.434 0.509 0.583 0.658 0.733 0.808

0.33 0.369 0.408 0.447 0.485 0.524

1.067 1.067 1.067 1.067 1.067 1.067

Table 3.4.3 Values stiffness (χs) and mass (χ m) parameter for dam. H/B

1.5

1.4

1.3

1.2

1.1

1

0.9

0.8

0.7

0.6

0.5

χs χm

0.300 0.293

0.244 0.273

0.195 0.254

0.154 0.234

0.118 0.215

0.089 0.195

0.065 0.176

0.045 0.156

0.030 0.137

0.019 0.117

0.011 0.098

Table 3.4.4 Values of fluid structure influence coefficient CFS . Bt /B

1.5

1.4

1.3

1.2

1.1

1

0.9

0.8

0.7

0.6

0.5

0.05 0.1 0.15 0.2 0.25 0.3

0.589 0.624 0.656 0.688 0.718 0.746

0.605 0.642 0.676 0.709 0.741 0.770

0.618 0.656 0.692 0.727 0.760 0.791

0.626 0.667 0.704 0.741 0.775 0.807

0.630 0.672 0.712 0.750 0.785 0.818

0.628 0.672 0.714 0.753 0.790 0.824

0.621 0.668 0.710 0.751 0.789 0.825

0.610 0.657 0.702 0.744 0.783 0.820

0.593 0.642 0.688 0.732 0.772 0.810

0.572 0.623 0.670 0.715 0.756 0.795

0.548 0.600 0.648 0.694 0.736 0.776

Values of χs and χm are given in Table 3.4.3 for various H/B ratio. The coefficient CFS the fluid structure coefficient is given in Table 3.4.4 for varying Bt /B and H/B ratio. The modal mass participation κi factors are as given in Table 3.4.5 for various H/B and Bt /B ratio. With the above tables now available it becomes quite simple to analyze the dam for fluid structure interaction under earthquake force. Intermediate values can be linearly interpolated without any significant error as the variations as observed are not significant. The analysis can well be carried out by a simple calculator or can very well be programmed in a spread sheet. © 2009 Taylor & Francis Group, London, UK

502 Dynamics of Structure and Foundation: 2. Applications Table 3.4.5 Values of modal mass participation factor κi . Bt /B

1.5

1.4

1.3

1.2

1.1

1

0.9

0.8

0.7

0.6

0.5

0.05 0.1 0.15 0.2 0.25 0.3

0.657 0.648 0.640 0.632 0.625 0.619

0.661 0.650 0.642 0.634 0.626 0.620

0.664 0.653 0.644 0.636 0.628 0.622

0.668 0.656 0.647 0.638 0.630 0.623

0.672 0.659 0.649 0.640 0.631 0.624

0.676 0.663 0.652 0.642 0.633 0.626

0.681 0.667 0.655 0.645 0.635 0.627

0.686 0.671 0.658 0.648 0.637 0.629

0.692 0.675 0.662 0.650 0.639 0.631

0.699 0.680 0.666 0.654 0.642 0.633

0.706 0.686 0.670 0.657 0.645 0.635

A word regarding damping ratio of the system based on which Sa /g is to be chosen. As per ICOLD (International Committee Of Large Dams) the damping values of the concrete in a dam is low and would be around 2–5%.While based on study by Newmark and Roesenbleuth and also by Zangar (1953) water is usually found to have a damping in the range of 15%. Thus as conservative estimate a value of 8–10% may be chosen while selecting the value of Sa /g for a coupled model. 3.4.2.8

Lumped mass and stiffness approach for F luid-Structure interaction

We now present herein a lumped mass model for the same analysis where the fluid and the dam are coupled together. The reasons for choosing this model are as mentioned hereunder: • • •

Engineers working with structural dynamics are quite conversant with this approach. The model is simple and easy to work with in practical design situation. It is particularly advantageous to use this model when over and above the fluid the foundation interaction is also to be considered (soil-fluid-structure interaction).

For large dams it has been particularly observed that considering fluid as an incompressible medium and performing the analysis separately (as suggested by codes of many countries including IS) the forces at times are in considerable variation to the case when soil is considered in the analysis. As such at least for economic reasons ICOLD recommends to use soil-fluid structure interaction wherever deemed feasible or perform a comprehensive Finite Element analysis. As a prelude to such analysis the present model would give a very good estimate on how the loads are affecting the dam and whether it is worth such elaborate analysis. In many of the cases the present analysis would suffice for a comprehensive interaction analysis where further sophistication may not be necessary. We have already established previously that for a fluid medium extending to infinity in horizontal direction of height Hw the equivalent spring stiffness and lumped mass can be 2 /2g, respectively. To couple this with the expressed as Kf = π 2 Bm /8 and Mf = γw Hw dam structure the most intuitive model would be to couple directly fluid stiffness to the dam stiffness by static coupling which we are so conversant with. If we however do this we would actually be grossly wrong!! © 2009 Taylor & Francis Group, London, UK


A

Ra=2W/3

B

L/3

2L/3

Rb=W/3

W

Figure 3.4.8 A simple supported beam supporting triangular load.

uf

mf =

wH w 6g 2

Kf =

us

ms =

Ks =

2

B m

8

BH g EB

3

12 H

I3 4 1 3 I1

Figure 3.4.9 Mathematical model for fluid structure interaction.

Housner et al. have shown that when an infinite fluid vibrates, a part of the fluid mass get locked with the wall and vibrates in same phase as the wall (called convective mass and the balance fluid vibrates on its own called the sloshing mass. Now let us look at the mass expression it is obvious that the loading is triangular (hydrostatic) in nature. For a beam of span L having triangular load at as shown in Figure 3.4.8, it is elementary to show that equivalent lumped mass at A is mL/3 and that at B is mL/6. Here m is the mass per unit length and L the span. Since for the dam the base is considered fixed the term ml/3 goes to the foundation and is not effective and the active fluid participation is only ml/6. Thus based on this principle the mathematical model for the dam and the fluid is expressed as in Figure 3.4.9. 3.4.2.9

Model for f luid-structure interaction of the dam and reservoir

In the above mathematical it is thus assumed that two third of the weight of water gets transfered to the dam foundation and one third of the water vibrates on its own © 2009 Taylor & Francis Group, London, UK


(sloshing mode). The free vibration equation of the coupled fluid structure system can now be expressed as ms 0

2 3 K + Kf u¨ s + s mf u ¨ −Kf f 3

−Kf Kf

0

2 3 us =0 uf

(3.4.71)

where, Ks = stiffness of the dam structure; Kf = stiffness of the fluid, Ms and Mf = mass of the dam and the equivalent fluid respectively. The eigen-solution of the above problem can now be carried out by the standard expression

Ks + Kf − ms ω2 −Kf

−Kf m Kf − 3f ω2

2 3 us =0 uf

(3.4.72)

It is evident from above that the frequency vis-à-vis the time period of the system will get modified. Neither the fluid mode will have a frequency @ 4Hw /c nor the 2 fundamental time period of the structure will remain T1 = 4.68 HB γc /gE for Bt /B = 0.1, say. The dynamic pressure and the response of the dam will now have to be obtained based on this coupled frequency as derived above. Let ω1 ≤ ω2 be the two modified natural frequencies ofthe fluid and structure and φ11 φ12 ; let the corresponding eigen vectors be expressed as [ϕ] = φ21 φ22 The displacement of the dam is then given by ui = ϕi Here

Lni Mni

Sai

(3.4.73)

ωi2

Mni = [ϕi ]T [M] [ϕi ]

and Lni = [ϕi ]T [M] [I]

(3.4.74)

I is an identity matrix. Thus, the maximum displacement for a particular mode is given by ui = ϕi

Lni Mni

Sai ωi2

f (ξ )

at ξ → 1.0

(3.4.75)

Once displacement are observed the maximum moment and shear for the two modes are obtained from the expression M = −EI Thus,

d2u dz2

and V = −EI

Mmax = −

d3u

Ec I0 (1 + ψξ )3 μ21 umax H2


(3.4.76)

dz3 and

Vmax = −

Ec I0 (1 + ψξ )3 μ31 umax H3 (3.4.77)


Displacement(mm)

Comparison of displacement Rigourous Vs Lumped Mass approach 15 10

Disp. Analytic solution

5

Disp. Lumped Mass

0 0

0.2

0.4

0.6

0.8

1

z/H

Figure 3.4.10 Comparison of displacement analytical versus lumped mass approach.

The final design moments and shear are then

MR =

M12 + M22

and VR =

V12 + V22

(3.4.78)

Once the moment and shear are known they may be multiplied by the coefficients given in the Table 3.4.1 to get the values along the height of the dam. You might wonder how correct would be the result? Just for comparison a 100 m high dam 70 m wide with Bt /B = 0.1 was analysed by the rigorous semi-analytic method and the lumped mass approach, the displacement plot is compared in Figure 3.4.10. It can be observed that the values are in excellent agreement and gave almost same results. The method derived here is simple and practical. It does not need an elaborate expensive analysis and can well be carried out in a spread sheet. This would give a far more realistic result than what is suggested in IS-1893 (1984). One major advantage now is that coupling the soil stiffness to consider the foundation interaction effect becomes relatively simple.

3.4.2.10

Consideration of f luid- structure-foundation interaction

Having established the validity of the lumped mass model proposed above, the model can be further extended to couple the soil stiffness as shown in Figure 3.4.11. © 2009 Taylor & Francis Group, London, UK


Mf =

uf

2 wH w 6g

Kf =

Ms =

us

Ks =

h

2 Bm 8

BH g EB

3

12 H

Kx

I3 4 1 3

I1

K

Figure 3.4.11 Lumped mass stiffness model of dam.

3.4.2.11

Lumped mass stiffness model of dam with soil-f luid-structure interaction

We had shown in Chapter 1 (Vol. 2) that the equivalent stiffness of the soil structure system is given by 1 1 1 h¯ 2 = + + Ke Ks Kx Kθ

(3.4.79)

where, Ke = equivalent stiffness of the soil structure system; Ks = stiffness of dam as derived earlier; Kx = lateral stiffness of the dam foundation; Kθ = rocking stiffness of foundation, and, h¯ = center of gravity of the dam from bottom. Based on above the equation of motion of free vibration gets modified to ms 0

0 mf 3

2 3 K + Kf u¨ s + e u¨ f −Kf

−Kf Kf

2 3 us =0 uf

(3.4.80)

The rest of the steps now remain same as previous. This shall be elaborated latter by a suitable numerical example. 3.4.2.12

F inite element analysis of concrete dam

Finite element stress analysis of concrete dams is the most usual practice in industry. As stated in Chapter 2 (Vol. 1) the first application of FEM in civil engineering was carried out for concrete dam (Norfolk Dam analysis by Clough). Since then it has been a common practice that the stress analysis for such dams are invariably carried out by © 2009 Taylor & Francis Group, London, UK


applying FEM software. This is quite justified considering most of the concrete dams are massive in nature and, secondly in many cases a number of cooling water pipes run through its cross-sections to control the heat of hydration during the setting of concrete. The heat generated is significant in such cases and if not properly controlled can result in thermal cracks. In many cases tunnels run through the sections that are used to carry water pipes and also used for maintenance. Often one is interested to know the stress concentration around such openings and ensure no cracks generate around them. Figure 3.4.12 shows a conceptual finite element model for a concrete dam. For static analysis the problem is quite straightforward and does not pose any problem. For uncoupled vibration analysis also, the solution is not complicated. One does the eigen value and modal analysis for the dam only and applies the hydrodynamic pressure as point loads on the surface of the dam and combines the stresses generated by these load cases to finally arrive at the dynamic stresses induced in the dam body. However, for coupled fluid structure interaction, the analysis poses some difficulties that are discussed as hereunder: •

• •

The reservoir in question is unbounded. As such to ensure that no waves are reflected back from the finite element fluid boundary one has to take the boundary significantly away from the dam. This greatly increases the model size and at times can make the analysis quite laborious and expensive. Fluid Poisson’s ratio is = 0.5, this at times makes the matrix singular and poses severe difficulties in eigen value solutions in certain cases. Majority of the Finite Element software that are used commercially for structural analysis (except ANSYS and NASTRAN) do not have fluid elements in their library as such trying to simulate 2D fluid elements by using normal structural elements (mostly by trying to use a Poisson’s ratio of 0.49) does not often work.

Thus if one has not developed a special purpose source code circumventing the above mentioned problem the task is indeed difficult to carry out a complete fluid structure interaction analysis based on normal finite element structural software that are available in the market. Free surface of water

Figure 3.4.12 Conceptual finite element model of a concrete dam.



Free surface of water

Fluid mass @ M /3 f

Fluid spring @

2

Bm /8

Figure 3.4.13 Conceptual finite element model of a concrete dam with fluid modeled as spring and lumped mass.

We now propose a simple procedure based on which the above problem can well be avoided and show how we can do such analysis while carrying out a FEA. It was shown earlier that the reservoir fluid can effectively be represented by equivalent spring and lumped mass as furnished in Equation (3.4.51). These are in effect stiffness and mass contribution per unit area. Thus if we multiply this value by the effective contributing area for each node at wall-water surface we get an effective spring contribution for each node along the depth of the dam. It is apparent that sum of all such springs at each node must be equal to (π 2 Bm )/8. For the mass distribution as explained earlier 2Mf /3 of the mass goes to foundation can be ignored for dynamic analysis of the dam and the balance lumped mass Mf /3 can be connected to the top of the dam. Based on above argument Figure 3.4.12 gets modified to Figure 3.4.13. The problem has now simplified considerably. We do not have to worry about the infinite fluid domain or about the numerical difficulties encountered in trying to model a mock fluid in 2D. The spring and lumped mass adequately models the fluid and coupled analysis for the dam can now be easily carried out in any of the commercially available software like SAP 2000, GTSTRUDL, etc.

Example 3.4.1 A concrete dam of height 100 m, base width 70 m has top width of 7 m. The maximum height of water it contains is 97.5 m. The dam is built on site of hard rock and is in zone IV as per IS-1893 (2002). The grade of concrete used is M25. Consider unit weight of water as 10 kN/m3 and that of concrete as 25 kN/m3 . Bulk modulus of water (Bm ) = 2.11 × 106 kN/m2 . Shear wave velocity of soil = 750 m/sec. Unit weight of soil = 19 kN/m3 . Perform Earthquake analysis based on © 2009 Taylor & Francis Group, London, UK


• • • •

The IS-code method. Proposed dynamic analysis. Dynamic analysis with fluid structure interaction. Dynamic analysis with coupled soil fluid structure interaction considering 20% overall damping.

Solution: For the dam being in zone IV: As per IS-code Z = 0.24, I = 3.0, R = 1.5 (Assuming the dam to be un-reinforced). ZI This gives β = 2R = 0.24. IS-code method 2

The time period is given by, T = 5.55 HB

γc gEs .

Here, Ec = 28500000 kN/m2 , H = 100 m B = 70 m, γc = 25 kN/m3 g = 9.81 m/sec2 . This gives, T = 0.237 sec for which Sa /g for 5% damping = 2.5. The weight of the dam, W = 70+7 2 × 100 × 1 × 25 = 96250 kN per meter width Thus base shear is given by, VB = 0.6αh W → VB = 0.6 × β × (Sa /g) × W = 34650 kN. The base moment is given by, MB = 0.9αh W h¯ Here h¯ = 36.36 m from base; thus, Mb = 0.9 × β × (Sa /g) × W × h¯ = 1889811 kN · m. The Moment and shear force at various depth are given as Z/H

Cm

Cs

Moment (kN · m)

Shear (kN)

0 0.2 0.4 0.6 0.8 1

1 0.675 0.35 0.15 0.05 0

1 0.875 0.65 0.4 0.18 0

1889811 1275622 661433.9 283471.7 94490.55 0

34650.00 30318.75 22522.5 13860 6237 0

The hydrodynamic pressure is given by, p = Cs αh γw h Here as suggested in IS-1893, we take same αh as that of dam i.e Sa /g = 2.5 for T = 0.237 sec. The hydrodynamic pressure is given hereafter in a tabulated form as Z/H

Cs

pdyn (kN/m2 )

0 0.2 0.4 0.6 0.8 1

0 0.3528 0.5292 0.645519 0.712875 0.735

0 206.388 309.582 377.6288 417.0319 429.975



Considering, Vz = 0.726p · z and Mz = 0.299 p· z2 . The moments and shears are calculated at various depth as

Z/H

Moment

Shear

0 0.2 0.4 0.6 0.8 1

1.23 × 10+06 9.51 × 10+05 6.46 × 10+05 3.53 × 10+05 1.18 × 10+05 0.00 × 10+00

3.04 × 10+04 2.36 × 10+04 1.60 × 10+04 8.77 × 10+03 2.92 × 10+03 0.00 × 10+00

Thus total design moment (due to dam itself and hydrodynamic pressure) as per IS code is Z/H

Moment (IS-code)

Shear (IS-code)

0 0.2 0.4 0.6 0.8 1

3.12 × 10+06 2.23 × 10+06 1.31 × 10+06 6.37 × 10+05 2.12 × 10+05 0.00 × 10+00

6.51 × 10+04 5.39 × 10+04 3.86 × 10+04 2.26 × 10+04 9.16 × 10+03 0.00 × 10+00

Proposed dynamic analysis 2

Here Bt /B = 0.1, this gives, T = 4.68 HB

γc gEs .

Substituting the values, we have, T = 0.20 sec and that gives Sa /g = 2.5. Corresponding eigen vectors for the first mode is {φ}T = 0.986

− 0.167

0.020

Hence, M(z) = −0.0335β

Sa g

z 3 γc BH 2 1 + ψ H

× (3.466f1 (z) − 3.679f2 (z) + 1.234f3 (z)) where fi (z) =

μ2i H2

−sin μHi z − sin h μHi z + αi cos μHi z + cos h μHi z and ψ = −0.9.



Similarly,

Sa V(z) = −0.0335 g

z 3 γc BH 1 + ψ (6.5f1 (z) H

− 17.27f2 (z) + 9.693f3 (z)) Substituting the values, we have Dynamic Moments and Shears as follows Z/H

Moment dynamic (kN·m)

Shear Dynamic (kN)

0 0.2 0.4 0.6 0.8 1

1.65 × 10+06 1.04 × 10+06 6.86 × 10+05 2.88 × 10+05 3.30 × 10+04 0.00 × 10+00

7.58 × 10+03 1.25 × 10+04 1.11 × 10+04 −2.38 × 10+03 −2.33 × 10+03 −8.01 × 10−04

We observe here a very interesting thing while the Bending moment is hardly affected by the higher modes, the shear profile is significantly influenced by it. The free field time period of fluid is given by T = 4Hc w where Bm = ρw c2 . Substituting the value, ρw = 10/9.81 = 1.02 kN-sec2 /m, we have T = 0.27 sec. Considering 15% damping for fluid we have Sa /g = 2·5 and taking Response reduction factor32 R = 1·5, we have β = 0.24. 12 S a γw Hw πz p = β sin and considering Vdyn = −0.6455pz × z π(π + 2) g 2H Mdyn = −0.4014 pz × z2 , we have Z/H

Dynamic pressure (kN/m2 )

Moment Proposed Dynamic (kN·m)

Shear Proposed Dynamic (kN)

0 0.2 0.4 0.6 0.8 1

0 134.2987 255.4514 351.5987 413.329 434.5999

1.69 × 10+06 1.29 × 10+06 8.22 × 10+05 3.98 × 10+05 1.05 × 10+05 0.00 × 10+00

2.71 × 10+04 2.06 × 10+04 1.32 × 10+04 6.38 × 10+03 1.68 × 10+03 0.00 × 10+00

Thus the total moments and shears due to proposed dynamic analysis and hydrodynamic pressure is given by Z/H

Moment proposed dynamic (kN · m)

Shear proposed dynamic (kN)

0 0.2 0.4 0.6 0.8 1

3.34 × 10+06 2.33 × 10+06 1.51 × 10+06 6.86 × 10+05 1.38 × 10+05 0.00 × 10+00

3.47 × 10+04 3.31 × 10+04 2.43 × 10+04 4.00 × 10+03 −6.58 × 10+02 −8.01 × 10−04

32 We have assume R = 1.5 considering the dam as unreinforced.



Response considering fluid structure interaction semi analytic approach Though the method is analytic we will hardly use any analysis here, but would refer extensively to the design tables furnished earlier. Based on the tables furnished we have the following design parameters. For H/B = 100/70 = 1.43, Bt /B = 0.1 Hw /B = 1.4 μ1 = 1.875, I1 = 1.13, I2 = 1.855, I3 = 0.509, I4 = 0.369, I5 = 1.067, χm = 0.272, χs = 0.259 and Cfs = 0.629; Now considering ' H 2 γc CFS T = 6.19 we have T = 0.21 secs → Sa /g = 2.5 B Ec g m I5 Considering, κi = II43 +χ +χm I2 , we have κ1 = 0.65 The displacement thus in mm is given by

u(z)max = 1.01175κi β

γc CFS B2 Ec

Sa g

H4

u(max) = 0.012 m Ec × I = 28500000 × (70)3 /12 = 8.146 × 1011 kN/m2 E I (1+ψξ )3 μ2

E I (1+ψξ )3 μ3

1 1 Thus, considering Mmax = − c 0 H 2 umax and Vmax = − c 0 H 3 umax . We have, Mmax = 3.469 × 106 kN/m2 and Vmax = 4.77 × 104 kN, at ξ = 0 Multiplying by the coefficients furnished in the table, the variation with depth is as shown hereunder: renewcommand

Z/H

Moment FSI

Shear FSI

0 0.2 0.4 0.6 0.8 1

3.469 × 10+06 1.387 × 10+06 4.193 × 10+05 7.762 × 10+04 4.865 × 10+03 0.000 × 10+00

4.77 × 10+04 2.60 × 10+04 1.15 × 10+04 3.45 × 10+03 4.62 × 10+02 6.77 × 10−03

Fluid structure interaction lumped mass model In this case for the first mode the dam stiffness is given by Ks =

EB3 μ41 I1 = 1.138 × 107 kN/m 12H 3

The fluid stiffness is given by Kf =

π 2 Bm = 2.603 × 106 kN/m 8



The mass of dam in first mode is given by Ms =

γ BH I3 = 9080 kN-sec2 /m g

The mass of fluid is given by, mf =

2 γw Hw 10 × 97.52 = = 4845 kN-sec2 /m. 2g 2 × 9.81

The free vibration equation is

or

2 3 2 3 K + Kf −Kf us u¨ s 1 + s =0 u¨ f uf −Kf Kf 0 Mf 3 2 3 2 3 u¨ s 1.398 × 107 −2.603 × 106 us 9.08 × 103 0 + =0 uf 0 1615 u¨ f −2.603 × 106 2.603 × 106 Ms

0

On eigen value solution, we have 2 {λ} =

3

2

895.02 2256

⇒ {ω} =

29.917 47.502

3 rad/sec

The corresponding Eigen vectors are given as

0.445 −0.40 [ϕ] = 1 1

The time period of the system33 is thus 2 {T} =

0.21 0.132

3 sec

Here for the first mode Mn1 = [φ1 ]T [Mφ1 ] Here [φ1 ]T = [0.445, 1] and this gives Mn1 = 3411. Ln1 = [φ1 ]T M[I] here I = [1, 1]T and this gives Ln1 = 5653.

33 Observe here that under uncoupled condition the time period of the dam was T = 0.20 sec and that of fluid was T = 0.27 sec. this is now different. While this matches exactly with semianalytic method @0.21 sec.



L2n1 /Mn1 = 9369; thus percentage of modal mass participation (MMP) factor is given by MMP =

9369 × 100 = 87.603% 9080 + 1615

For T = 0.21 sec, Sa1 /g = 2.5, thus design acceleration is given by Sa1 = 2.5 · β · g = 2.5 × 0.24 × 9.81 = 5.886 m/sec2 . Sv1 = Sa1 /ω1 = 0.197 m/sec.

Ln1 Mn1

Sv1 ω1

f (ξ ); at ξ = 1.0 (i.e. z = H), f (ξ ) = 2.7242. 0.013 Substituting the values we have max = m 0.030 Thus it is apparent that for the first mode, displacement of the dam is 13 mm while that of the water is 30 mm34 . Considering M = −EI d 2 u/dz2 and V = −EI d 3 u/dz3 we have μ2 Mmax = EI0 12 max(struct) or Mmax = 28500000 × 28580 × [(1.875)2 / H (70)2 ] × 0.013 = 3.782 × 106 kN · m at base and Vmax = 28500000 × 28580 × [(1.875)3 (70)3 ] × 0.013 = 7.091 × 104 kN. at base Similarly for second mode Thus max = φ1

Mn2 = [φ2T M[φ2 ]. Here [φ2 ]T = [−0.40, 1] and this gives Mn2 = 3068. Ln2 = [φ2 ]T M[I] and this gives Ln2 = −2017. L2n1 /Mn1 = 1326; thus percentage of modal mass participation (MMP) factor is given by MMP =

1326 × 100 = 12.397% 9080 + 1615

For T = 0.132 sec, Sa2 /g = 2.5, thus design acceleration is given by Sa2 = 2.5 · β · g = 2.5 × 0.24 × 9.81 = 5.886 m/sec2 . Sv2 = Sa2 /ω2 = 0.124 m/sec. Thus

max = ϕ2

Ln2 Mn2

Sv2 ω2

f (ξ )

at ξ = 1.0 f (ξ ) = 2.7242.

34 It is to be noted that based on semi-analytic method the displacement obtained for the dam is 12 mm thus variation with lumped mass model is of the order of 7% only.



1.869 × 10−3 m −4.672 − 10−3

Substituting the values we have max =

Thus for the second mode, displacement of the dam is 1.8 mm while that of the water is −4.6 mm. Considering M = −EI d 2 u/dz2 and V = −EI d 3 u/dz3 we have Mmax = EI0

μ21 max(struct) H2

→ Mmax = 28500000 × 28580 × [(1.875)2 /(70)2 ] × 1.869 × 10−3 = 5.351 × 105 kN · m at base and Vmax = 28500000 × 28580 × [(1.875)3 (70)3 ] × 1.869 × 10−3 = 1.003 × 104 kN · m at base The above now can multiplied by the coefficients given in design table to obtain the values as shown hereunder Z/H

Moment (1st mode)

Shear (1st mode)

0 0.2 0.4 0.6 0.8 1

3.78 × 10+06 1.51 × 10+06 4.57 × 10+05 8.46 × 10+04 5.30 × 10+03 0.00 × 10+00

7.09 × 10+04 3.87 × 10+04 1.71 × 10+04 5.13 × 10+03 6.86 × 10+02 1.01 × 10−02

Soil – fluid-structure interaction In this case as calculated previously Ks =

EB3 μ41 I1 = 1.138 × 107 kN/m 12H 3

and Kf =

π 2 Bm = 2.603 × 106 kN/m. 8

Considering foundation contact area = 70 × 1 = 70 m2 , equivalent circular footing radius is r = 4.72 m. For rocking mode considering base moment of inertia of the dam as I = B3 /12 equivalent radius rθ = 13.812 m. 19 Dynamic shear modulus = × 7502 = 1089450 kN/m2 . 9.81 Poisson’s ratio = 0.45 32G(1 − ν)r 8Gr3θ Kh = = 2.662 × 107 kN/m; Kθ = = 1.392 × 1010 kN/m. 7 − 8ν 3 (1 − ν) h¯ = 36.36 m from the base, 1 1 1 h2 = + + ⇒ Ke = 4.536 × 106 kN/m Ke Ks Kh Kθ



The free vibration equation is

2 3 2 3 K + Kf −Kf us u¨ s 1 + e =0 u¨ f uf −Kf Kf 0 Mf 3 2 3 2 3 u¨ s 9080 0 7.139 × 106 −2.603 × 106 us ⇒ + = 0. uf 0 1651 u¨ f −2.603 × 106 2.603 × 106 Ms

0

On eigen value solution we have 2

3

2

403.76 {λ} = 1994

20.094 ⇒ {ω} = 44.667

3 rad/sec

The corresponding eigen vectors are given as 0.749 [ϕ] = 1

−0.237 1

2

The time period of the

system35

3 0.313 is thus {T} = sec. 0.141

Here for the first mode Mn1 = [φ1 ]T M[φ1 ] Here [φ1 ]T = [0.860, 1] and this gives Mn1 = 6716. Ln1 = [φ1 ]T M[I] here I = [1, 1]T and this gives Ln1 = 8420. L2n1 /Mn1 = 10560; thus percentage of modal mass participation (MMP) factor is given by MMP =

9868 × 100 = 98.719% 9080 + 807.5

For T = 0.313 sec, Sa1 /g = 2.5 thus design acceleration is given by Sa1 = 0.6 · 2.5 · β · g 0.6 × 2.5 × 0.24 × 9.81 = 3.532 m/sec2 considering 20% damping of the soil. Sv1 = Sa1 /ω1 = 0.176 m/sec. Thus

max = ϕ1

Ln1 Mn1

Sv1 ω1

f (ξ ); at ξ = 1.0 (i.e. z = H), f (ξ ) = 2.7242.

35 Observe here that under uncoupled condition the time period of the dam was T = 0.199 sec and of fluid was T = 0.27 sec. this is now different.



0.022 m 0.030

Substituting the values we have max =

Thus for the first mode, displacement of the dam is 22 mm while that of the water is 30 mm. Considering M = −EI d 2 u/dz2 and V = −EI d3 u/dz3 , we have Mmax = EI0

μ21 max (struct) H2

→ Mmax = 28500000 × 28580 × [(1.875)2 /(70)2 ] × 0.022 = 6.413 × 106 kN · m at base and Vmax = 28500000 × 28580 × [(1.875)3 (70)3 ] × 0.022 = 1.202 × 105 kN · at base Similarly for second mode Mn2 = [φ2T [Mφ2 ]. Here [φ2 ]T = [−0.237, 1] and this gives Mn2 = 2126. Ln2 = [φ2 ]T M[I] and this gives Ln2 = −539.81. L2n2 /Mn2 = 137.033; thus percentage of modal mass participation (MMP) factor is given by MMP =

137.033 × 100 = 1.281% 9080 + 807.5

For T = 0.141 sec, Sa2 /g = 2.5 thus design acceleration is given by Sa2 = 0.6·2·5·β · g = 0.6 × 2.5 × 0.24 × 9.81 = 3.532 m/sec2 . Sv2 = Sa2 /ω2 = 0.079 m/sec. Ln2 Sv2 Thus max = ϕ2 f (ξ ) Mn2 ω2

at ξ = 1.0 f (ξ ) = 2.7242.

2.906 × 10−4 m = −1.225 − 10−4

Substituting the values we have max

Thus for the second mode, displacement of the dam is 0.29 mm while that of the water is −1.225 mm. Considering M = −EI d 2 u/dz2 and V = −EI d 3 u/dz3 we have Mmax = EI 0

μ21 max (struct) H2

or Mmax = 28500000×28580×[(1.875)2 /(70)2 ]×2.906×10−4 = 8323 kN · m at base and Vmax = 28500000 × 28580 × [(1.875)3 (70)3 ] × 2.906 × 10−4 = 1561 kN at base. © 2009 Taylor & Francis Group, London, UK


The above now can multiplied by the coefficients given in design table to obtain the values hereunder Z/H

Moment (1st mode)

Shear (1st mode)

0 0.2 0.4 0.6 0.8 1

6.41 × 10+06 2.57 × 10+06 7.75 × 10+05 1.44 × 10+05 9.00 × 10+03 0.00 × 10+00

1.20 × 10+05 6.55 × 10+04 2.89 × 10+04 8.69 × 10+03 1.16 × 10+03 1.70 × 10−02

The values computed above are now compared in Figs. 3.4.14 and 15.

Comparison of Bending Moment along Dam Height 7.00E+06

Moment (ISCode)

Moment9kN.m)

6.00E+06 5.00E+06

Moment dynamic

4.00E+06 3.00E+06

Moment FSI

2.00E+06 1.00E+06

Moment lumped mass

0.00E+00 -1.00E+06

0

0.2

0.4

0.6

0.8

1

Moment SFSI

z/H

Figure 3.4.14 Comparison of bending moment by various methods.

Comparison of Shear force

1.40E+05

Shear kN

1.20E+05 1.00E+05

Shear (IS code)

8.00E+04

Shear dynamic

6.00E+04

Shear FSI

4.00E+04

Shear lumped

2.00E+04

Shear SFSI

0.00E+00 -2.00E+04

0

0.2

0.4

0.6

0.8

1

z/H

Figure 3.4.15 Comparison of shear force by various methods.



3.5 ANALYSIS OF EARTH DAMS AND EMBANKMENTS

3.5.1 Dynamic earthquake analysis of earth dams In countries like India, USA, Australia where a major part of the economy is largely dependent on agriculture and consists of a number of rivers and canals, for the purpose of irrigation and flood control (specially in rural areas) dams made out of natural earth are very popular. The system usually consists of carefully chosen natural soil with a clay core (to control the seepage) and has been successfully utilized for effective utilization of water stored thereby. We shall deal with the earthquake analysis of such dams in this section. Unlike concrete dams, which are far more rigid, the behaviour of earth dams are different under earthquake. Though theoretically concrete dams also do not behave exactly as rigid structures they are usually assumed to behave as rigid without much practical error. In such case it is assumed that the motion at the base is same in all parts of the concrete dam. On the contrary for earth dams the constituent material being much softer it mostly behaves as flexible structure where the acceleration induced within the dam varies with height and could be different at different points of the dam.

3.5.2 Mononobe’s method for analysis of earth dam One of the earliest analyses of such earth dam was proposed by Mononobe (1936) in his classic paper, considering the dam as an isosceles triangle having uniform mass density. As width of the base of the dam is considered as far greater then the height of the dam, it is assumed that the shear deformation is predominant and bending deformation which is secondary in nature may be ignored. With respect to Figure 3.5.1, for a strip dz, the horizontal shear force and the inertial shear is given by ρ(az)dz

∂ 2X ∂t 2

∂s dz = ∂z

→

ρaz

∂ 2X ∂t 2

=

∂s ∂z

(3.5.1)

where, X is the horizontal amplitude of displacement and s is the shear force. The shear modulus36 of the soil medium is given by G=

s/az Shear stress = Shear strain dX/dz

which gives, s = Gaz

Differentiating the above with respect to Z we have ∂X ∂ 2X ∂s = Ga + Gaz 2 ∂z ∂z ∂z

from which we have,

36 This is considered constant over the depth in this case.


dX dz

(3.5.2)


Figure 3.5.1 Shear force in a triangular wedge shaped earth dam.

ρz

∂ 2X ∂X ∂ 2X = Ga + Gaz ∂z ∂t 2 ∂z2

(3.5.3)

which is the basic equation of shear vibration of the triangular wedge. For solution of the above equation considering, z = z H, where H is the height of the dam, and considering, X = x(z ) sin ωt, we have ∂ 2X = −xω2 sin ωt ∂t 2 ∂X 1 ∂x sin ωt = ∂z H ∂z

or

and

∂X ∂x = sin ωt ∂z ∂z ∂ 2X 1 ∂ 2x = 2 sin ωt 2 ∂z H ∂z 2

and

(3.5.4)

Substituting the above in the basic equation of shear vibration, we have

∂ 2x ∂x + z 2 G ∂z ∂z

+ ρH 2 z xω = 0 2

(3.5.5)

The above is the Bessel’s equation, whose solution is given by the expression x = AJ 0

ρ Hω z G

(3.5.6)

where, A is a constant and J0 is the Bessel function of the first kind and zero order. Implementing the boundary condition at the base of the dam (z = 1), at x = 0, we have ρ =0 (3.5.7) J0 Hω G



The above can be expanded as

ρ Hω G

1 0.0507 1 =n− + + · · ·, 4 4n − 1 n π 1 0.0507 vs ωn = π n − + H 4 4n − 1

➔

Considering, T = 2π/ωn we have,

T=

where n = 1, 2, 3. . .

(3.5.8) (3.5.9)

2H

vs n −

1 4

+

0.0507 4n−1

(3.5.10)

Thus, for the first three modes, we have time periods as shown in Table 3.5.1. The IS-code only furnishes the first fundamental mode as mentioned above for the calculation of time period. We started the derivation of the time period equation based on the equilibrium of the elemental strip given by ρ(az)dz

∂ 2X ∂t 2

=

∂s dz ∂z

which gives maximum value of Again considering, ωn = gives,

∂ 2X ∂t 2

vs Hπ

∂2X ∂t 2

= −x(0.767)2 π 2 max

∂ 2X = −xω2 sin ωt ∂t 2

and

as (−xω2 )

n−

1 4

+

0.0507 4n−1

(3.5.11)

vs π , which for n = 1; ωn = 0.767 H

vs2 H2

(3.5.12)

Substituting the value of x derived previously, we have

∂ 2X ∂t 2

max

ρ v2 π G = −AJ0 H · z · 0.767 (0.767)2 π 2 s2 G H ρ H

proportional to

(3.5.13)

vs2 J0 (0.767πz ); H2

the numerical constants being considered within the constant proportionality. Table 3.5.1 Time period of earthen dams for first three modes. Mode

Time period (sec)

1 2 3

2.607 H/vs 1.138 H/vs 0.726 H/vs


522 Dynamics of Structure and Foundation: 2. Applications Table 3.5.2 Variation of acceleration along the height of the dam as per Mononobe. z = z/H

0.0

0.1

0.3

0.5

0.7

0.9

1.0

Maximum acceleration α, J0 (2.41z )

1.00

0.986

0.874

0.668

0.406

0.127

0.00

v2

For a particular dam as the Hs2 is a constant quantity, hence maximum acceleration 2

∂ X is proportional to J0 (2.41z ). ∂t 2 max

The value of J0 (2.41z ) for various values of z are as shown in Table 3.5.2. The above values may be used to determine the inertial force at different heights of the dam. When resonance occurs the deformations would tend to infinite value without any internal friction, however due to internal friction the maximum deflection is restricted to a finite value. Mononobe calculated the ratio of top and bottom deflection for G and ρ and also when the parameters vary linearly with depth. Considering linear variation with depth the acceleration is expressed as α = α0

d0 − d H − z 1+ d H

(3.5.14)

where α is the acceleration at any depth z below top, α0 is the ground level acceleration and d0 is the maximum displacement at the top. The ratio of top and bottom displacement came to 2.5 and 3.5 respectively from which Mononobe concluded that maximum acceleration at the top may be 2.5 to 3.5 times the acceleration at the base.

3.5.3 Gazetas’ method for earth dam analysis Gazetas (1982) developed solutions to the shear beam wave equation for the case where the shear modulus value varies with the depth of the dam given by G(z) = Gav (z/H)α , where Gav is the average shear modulus of material constituting the dam. The nth natural frequency of the dam considering the dam to be triangular in shape (i.e. h/H = 1) is given by ωn =

vs βn (4 + α)(2 − α), H 8

(3.5.15)

where vs = average shear wave velocity of the soil in the dam and βn is a function given below in Table 3.5.3 for the first three modes; h is the height of the dam to it s crest, and H is full height of the triangle. The mode shape for the nth natural frequency is given by ϕn (z) = (z/H)−α/2 Jq [βn (z/H)1−α/2 ]


(3.5.16)

Analytical and design concepts for earthquake engineering 523 Table 3.5.3 Frequency coefficient for earth dam for first three modes as per Gazetas (1982). α

β1

β2

β3

0 1/2 4/7 2/3 1

2.404 2.903 2.999 3.142 3.382

5.520 6.033 6.133 6.283 7.106

8.654 9.171 9.273 9.525 10.174

where Jq = Bessel’s function of the first kind of order q = α/(2 − α) and can be evaluated from the expression ∞

Jq (x) =

k=0

x q+2k (−1)k k! (q + k + 1) 2

(3.5.17)

where (x) is the gamma function and is given by the expression ∞ (x) =

e−x xn−1 dx.

(3.5.18)

0

3.5.4 Makadisi and Seed’s method for analysis of earth dam A simplified procedure proposed by Makadisi and Seed (1977) taking into consideration the strain dependent degradation of shear modulus and damping of soil possibly remains the most popular method for dam analysis in design office and was developed in similar line of Mononobe’s method as explained earlier. Based on the shear beam equation as derived earlier they found the acceleration at any level z as a function of time is given by u(z, ¨ t) =

∞ 2J0 (βn z/H) n=1

β n J1 β n

ωn Vn (t)

(3.5.19)

where, J0 , J1 = Bessel functionof first kind of order zero and one; βn = the zero value of frequency equation, J0 (ωH ρ/G) = 0; ωn = natural frequency of the nth mode = βn Vs /H. t Vn (t) =

u¨ g e−λn ωn (t−τ ) sin[ωdn (t − τ )]dτ

0


(3.5.20)


√ where, ωdn = ωn 1 − λ2 the damped natural frequency and λn = critical damping ratio. Thus,

u(z, ¨ t) =

∞

ϕn (z)ωn Vn (t)

(3.5.21)

n=1

where, ϕn (z) =

∞ 2J0 (βn z/H) n=1

βn J1 βn

= modal participation factor.

Considering the first three modes of vibration, the corresponding values of βn are always β1 = 2.4, 5.52 and β3 = 8.65 which gives the first three natural three natural frequencies as shown in Table 3.5.4. At the crest of the dam (z = 0) the corresponding values of mode participation factors φn (0) for the first three modes are given in Table 3.5.5. Thus the crest acceleration at each mode is given by u(0, ¨ t) =

∞

ϕn (0)ωn Vn (t)

(3.5.22)

n=1

when based on response spectrum analysis the maximum expected acceleration is given by u(0, ¨ t) = ϕn (0)San

(3.5.23)

Thus for the first three modes the maximum acceleration is given by u¨ 1 max = 1.6Sa1 ;

u¨ 2 max = 1.06Sa2 ,

and

u¨ 3 max = 0.86Sa3

Table 3.5.4 Natural frequencies for first three modes for earth dams as per Makadis and Seed (1977). Mode

Frequency

1 2 3

ω1 = 2.4vs /H ω2 = 5.52vs /H ω3 = 8.65vs /H

Table 3.5.5 Modal participation factor for first three modes for earth dams as per Makdisi and Seed (1977). Mode

Modal participation factor

1 2 3

φ1 = 1.6 φ2 = 1.06 φ3 = 0.86



The maximum SRSS values of the acceleration is thus given by the expression 3

u¨ n max =

(u¨ n max)2

(3.5.24)

n=1

To estimate the strain compatible material properties an expression for the average shear strain over the entire section should be determined. From the shear slice theory, the expression for shear strain at any level in the dam is given by γ (z, t) =

∞ 2J1 (βn z/H) n=1

Hωn J1 (βn )

Vn (t) =

∞ H 2J1 (βn z/H) ωn Vn (t) vs2 βn2 J1 (βn ) n=1

∞ H = 2 ϕ¯n (z)ωn Vn (t) vs

(3.5.25)

n=1

2J (βn Hz ) where ϕ¯n (z) = β 12 J1 (β = shear mode participation factor. n) n It is recommended that the contribution of higher modes being small it is sufficient to consider the contribution of the first mode only over the entire depth of the dam for calculation of the average shear strain. Thus, the maximum average shear strain is obtained as

γmax (z) =

H ϕ¯1 Sa1 vs2

(3.5.26)

It has been shown that the average value of the first factor is given by, ϕ¯1avg = 0.3 Assuming an equivalent cyclic shear strain as approximately 65% of the average shear strain γavg (max) = 0.65 × 0.3 ×

H Sa1 vs2

(3.5.27)

Having obtained a new value for the average shear strain a new set of modulus and damping value can obtained from the expressions37 G=

Gmax (1 +

ψ ψr )

and

ψ/ψr Dc = Dm 1 + (ψ/ψr )

(3.5.28)

The iterations are carried out till the values become constant with respect to the earlier cycle. It has been observed that the system generally converges by 3 cycles.

37 For further explanation of how to modify the damping and shear modulus with respect to shear strain refer Chapter 1 (Vol. 2) under the section titled Geo-technical Consideration for Dynamic Soil Structure Interaction.



Water Line W Failure line

y

W

Figure 3.5.2 Slope failure of dam under earthquake force.

3.5.5 Calculation of seismic force in dam and its stability Once the modal acceleration and its SRSS values is obtained, for assessing the damage incurred in a dam shown in Figure 3.5.2, the usual design office practice is to undertake a pseudo-static analysis. For major damage observed in a dam is due to stability failure specially on the upstream side. To asses this, a slope stability analysis is usually carried out to evaluate the factor of safety which should be more than unity under the seismic force in addition to all other forces working on the dam. For stability analysis, any standard method like Slip-Circle, Fellenius’ or Bishop’s method38 may be applied which needs to be modified to cater for the applied seismic force. For instance the factor of safety (F) based on slip circle method is modified to: n F=

i=1

N − U − αy T tan ϕ + ni=1 cS n i=1 T + αy N

(3.5.29)

where, F = factor of safety; N = normal component of the weight of the soil; T = tangential component of the weight of the soil; φ = friction angle resistance of soil; c = cohesion of soil; S = width of segments of slices considered for analysis; αy = seismic acceleration coefficient (Sa /g) in the horizontal direction obtained from code based on the fundamental time period as derived earlier, and U = force due to internal Pore pressure within the dam. 3.6 ANALYSIS OF EARTH RETAINING STRUCTURES

3.6.1 Earthquake analysis of earth retaining structures Retaining walls supporting earth make an important component in infra-structure works like highways, roads, ports, bridge abutment etc. It has not been uncommon that during post earthquake relief operations39 , relief could not be sent in time, for damage to retaining wall itself has cut off the link

38 For details of these methods refer any standard text book on soil mechanics. 39 Specially in mountainous region.



road or tilting of the same has resulted in secondary collapse of bridge girders thus making some part of relief area inaccessible thus compounding the problems of the relief workers who have anyway a tough job to execute. Thus, understanding the behaviour of such structures under earthquake load is of paramount importance to civil engineers undertaking such tasks. It is important at times that such structures remain functional even after a severe earthquake maintaining the vital link among two places. One of the earliest methods for earthquake analysis of such retaining wall has been proposed by Mononobe (1926) and Okabe (1926) which still remain the backbone of analysis in almost all design office and code of practice.

3.6.2 Mononobe’s method of analysis of retaining wall Mononobe developed a pseudo static analysis for estimation of earth pressure on a gravity type retaining wall during an earthquake. He basically extended the original Coulomb’s theory to develop the soil pressure behind a retaining wall (Figure 3.6.1) catering to the horizontal and vertical component of the ground acceleration. Coulomb (1776) derived the equation of active and passive earth pressure as

Pa = KA =

1 KA γ H 2 2

and

Pp =

1 KP γ H 2 , 2

where

(3.6.1)

cos2 (ϕ − β) 1 2 sin(δ+ϕ) sin(ϕ−i) 2 2 cos β cos(δ + β) 1 + cos(δ+β) cos(β−i)

= coefficient of active earth pressure, and KP =

cos2 (ϕ + β) 1 2 sin(δ+ϕ) sin(ϕ+i) 2 cos2 β cos(δ − β) 1 − cos(δ−β) cos(i−β)

= coefficient of passive earth pressure, in which, γ = unit weight of soil; H = height of the retaining wall; φ = friction angle of the soil; δ = angle of friction between the wall and soil; β = angle subtended with the back of the wall with vertical, and i = slope of soil with respect to the horizontal. Mononobe and Okabe modified the above coefficient of earth pressure considering the horizontal and vertical seismic coefficient earthquake force αh and αv to

KAE =

cos2 (ϕ − β − θ) 1 2 sin(δ+ϕ) sin(ϕ−θ−i) 2 cos θ cos2 β cos(δ + β + θ ) 1 + cos(δ+β+θ) cos(β−i)


(3.6.2)


C

A

i

h

W

v

W

W H N Pa Fr F

B

Figure 3.6.1 Mononobe’s force diagram for gravity type retaining wall under earthquake.

αh where, θ = tan−1 1−α similarly the coefficient of passive earth pressure under v earthquake is given by

KPE =

cos2 (ϕ + β − θ) 1 2 sin(δ+ϕ) sin(ϕ−θ +i) 2 2 cos θ cos β cos(δ − β + θ ) 1 − cos(δ−β+θ) cos(i−β)

(3.6.3)

Mononobe’s solution was based on the following assumptions: • • • • •

The failure in soil takes place along the plane BC The movement of the wall is sufficient to produce active plane At failure full shear strength along the failure plane is mobilized The soil mass behind the wall behaves a rigid body αh and αv are maximum peak acceleration in horizontal and vertical direction.

It will be observed that in the above calculation no reference to time period is made, which clearly shows the analysis to be pseudo static in nature. The αh and αv values are considered as peak acceleration is quite justified in this case because of the following two reasons. © 2009 Taylor & Francis Group, London, UK


2

8 1.

6 1.

4 1.

2 1.

1

8 0.

6 0.

4 0.

0.

2

1.200 1.000 0.800 0.600 0.400 0.200 0.000

0

Sa/g

Response spectrum as per Japanese Code for Medium soil

Time period in second

Figure 3.6.2 Response spectrum for medium type soil as per Japanese code.

• •

At the time when Mononobe worked out the solution, major retaining walls were gravity type wall being massive was considered to be infinitely stiff i.e. have time period T → 0. This invariably makes the structure stiff attracting more force to it. As per Japanese code (Figure 3.6.2), for time period up to nearly 0.8 second, it is a practice to consider maximum acceleration on the structure.

Mononobe considered the dynamic pressure distribution as parabolic in nature and for active case it was assumed that the force is acting at a height H/2 from the bottom of the wall. While for passive case again considering a parabolic distribution the force was assumed to be acting at a height of 2/3 H from the base of the wall. The above procedure has almost universally dominated the design office practice in almost all countries40 though characteristics of retaining wall have undergone significant change with advent of Reinforced Concrete structures. While in the early thirties when Mononobe developed this theory almost all retaining walls were either made of Masonry or plain concrete and were usually massive in nature. It sustained its stability by its own weight. However with advancement in design techniques in RCC, they have progressively become much thinner and fit for the purpose, globally. Presently, the common practice is to go for a cantilever retaining wall for a height up to 6.0 meter beyond which engineers normally provide counterfort retaining wall which would provide overall economy in design. The typical sketches of such walls are as shown in Figure 3.6.3. It is obvious from the above figure that present day retaining walls have become much more flexible as such considering maximum acceleration as inducing the design pressure may not be true in all cases. Moreover, considering its flexibility the time period will surely play a dominant role in dynamic response of the system where a pseudo static analysis only could become insufficient. In-spite of the above limitation Mononobe’s pseudo static method is in extensive use irrespective of whether the wall is a cantilever or a counterfort or if it is thin or thick.

40 Including India where IS code follows the same procedure.



Cantil ever retaining wall

Counter fort retaining wall

Figure 3.6.3 Present day cantilever and counterfort retaining walls made of RCC.

Before we try tackle the above limitations as posed in Mononobe’s method we describe herein other methods which are also in practice for evaluation of pressure due to earthquake force.

3.6.3 Seed and Whitman’s method Seed and Whitman (1970) developed an expression which may also be used to determine the horizontal pseudo static force acting on a retaining wall. As per their calculations the active earth pressure may be expressed as

PAE =

3 amax γs H 2 8 g

(3.6.4)

According to this work the location pseudo static force is assumed to act at height of 0.6 H above the base of the structure.

3.6.4 Arango’s method Arango41 (1969) developed this method which is deemed practical and simple by extending the Coulomb’s theory of static earth pressure. As mentioned previously the

41 Dr. Ignacio Arango is Head of the Geotechnical division of Bechtel San Francisco office. A Bechtel fellow, who has contributed significantly in many areas of Geotechnical engineering specially in the area of Liquefaction Potential of soil.



static pressure is expressed as 1 KA γ H 2 2

Pa =

where,

(3.6.5)

cos2 (ϕ − β) 1 2 sin(δ+ϕ) sin(ϕ−i) 2 cos2 β cos(δ + β) 1 + cos(δ+β) cos(β−i)

KA =

The active earth pressure can be expressed in the form Pa =

1 1 γ H 2 Ac , 2 cos2 β

where, Ac =

(3.6.6)

cos2 (ϕ − β) = KA cos2 β 1 2 2 sin(δ+ϕ) sin(ϕ−i) cos(δ + β) 1 + cos(δ+β) cos(β−i)

Mononobe’s expression on the other hand is given by 1 γ H 2 (1 − αv )KAE 2 1 1 = γ H 2 (1 − αv ) Am 2 cos θ cos2 β

PAE =

where,

(3.6.7)

Am = KAE cos θ cos2 β =

cos2 (ϕ − β − θ ) 1 2 sin(δ+ϕ) sin(ϕ−θ −i) 2 cos(δ + β + θ ) 1 + cos(δ+β+θ) cos(β−i)

Comparing Am and Ac it shows that Am can be determined from the solution for Ac by redefining the slope of the back of wall as βˆ where, βˆ = β + θ and î = i + θ ˆ î) = KA (β, ˆ î) cos2 βˆ and Thus, Am = Ac (β, PAE = =

1 1 ˆ î) cos2 βˆ γ H 2 (1 − αv ) KA (β, 2 cos θ cos2 β 1 ˆ î)F γ H 2 (1 − αv )KA (β, 2

where, F =

cos2 βˆ . cos θ cos2 β

(3.6.8)

So far we have explained the various pseudo-static methods available for analysis for retaining wall. The first attempt to explain dynamic characteristics of such wall was proposed by Steedman and Zeng (1990) based on a pseudo-dynamic method. © 2009 Taylor & Francis Group, London, UK


3.6.5 Steedman and Zeng’s method For the fixed base cantilever as shown in Figure 3.6.4, if subjected to ground acceleration ah the acceleration at depth z is given by H−z a(z, t) = ah sin ω t − vs

(3.6.9)

Considering the pressure on the wall is resulting from the triangular wedge only being at a state of incipient failure, mass of a thin strip of depth dz within the soil wedge is given by m(z, t) =

γ (H − z) dz g tan α

where, γ is the unit weight of soil.

(3.6.10)

The total inertial force Ph acting on the wall can thus be expressed as H Ph (t) =

m(z)a(z, t) = 0

λγ ah [2π H cos ωζ + λ (sin ωζ − sin ωt)] 4π 2 g tan α

(3.6.11)

is the vertically propagating shear wave length and ζ = t − H in which, λ = v2π vs . sω For a special case when the failed wedge act as a rigid block (i.e. vs → ∞), we have [Ph (max)]lim vs →∞ =

γ H 2 ah a = h W = αy W 2g tan α g

(3.6.12)

The above can be stated to be equivalent force as assumed in the Mononobe’s method.

Ph

z

W

H

Ps = 45 ±

/2

Figure 3.6.4 Force diagram of cantilever retaining wall as per Steedman and Zeng’s (1990) method.



The total static plus dynamic thrust can then be obtained by resolving the force in the wedge and is given by PAE =

Ph (t) cos(α − ϕ) + W sin(α − ϕ) cos(δ + ϕ − α)

(3.6.13)

The total earth pressure is obtained by differentiating above expression with respect to z thus pAE

∂PAE γz αh γ z cos (α − ϕ) sin(α − ϕ) z = = + sin ω t − ∂z tan α cos(δ + ϕ − α) tan α cos (δ + ϕ − α) vs (3.6.14)

The first term in the above equation represents the static pressure acting at a height of H/3 from base while the second term represents the dynamic pressure where the thrust point is found to be varying with time and is given by hd = H −

2π 2 H 2 cos ωζ + 2πλH sin ωζ − λ2 (cos ωζ − cos ωt) 2πH cos ωζ + πλ (sin ωζ − sin ωt)

(3.6.15)

For very low frequency motion the dynamic thrust is found to at H/3 (this is when H/λ is small and the backfill moves in the same phase). For higher frequency motions the point of application is found to move higher up on the wall.

3.6.6 Dynamic analysis of RCC retaining wall So far we have explained the various pseudo static and pseudo dynamic methods available for analysis of retaining walls under earthquake force. We have also explained briefly some of the limitations of the pseudo-static method which is perhaps more appropriate for gravity type of wall. We now present herein a procedure for dynamic response of cantilever and counter fort type of retaining wall. While studying the dynamic response of structure previously we had seen that the acceleration it would be subjected to depends on the time period of the structure, then considering a suitable damping ratio we can find out the earthquake response based on code prescribed response spectrum. Based on the above, we proceed to explain the investigation which has been carried out by the present authors.

3.6.7 Dynamic analysis of cantilever and counterfort retaining wall This method applies standard modal response technique based on Rayleigh Ritz Method to evaluate the dynamic response of the soil wall system (Chowdhury and Dasgupta 2004). Though the response herein has been obtained based on IS-code © 2009 Taylor & Francis Group, London, UK


A

B

Possible Failure Line H

Wall

= 45 ± / 2 D X

Figure 3.6.5 Cantilever retaining wall with soil behind it.

(IS-1893 2002 to be specific) can very well be adapted to any international code provided the response history of the site in question is available either directly or as prescribed in the national code of practice (like UBC, Eurocode etc), We start with a simple case of retaining wall (Figure 3.6.5) with soil profile as shown above. It is assumed here like Mononobe’s case that • • • •

the soil profile under active case is at incipient failure when the failure line makes angle α = tan(45 + φ/2) as shown in the above figure. Since soil profile is already under failed condition under static load, the soil will not induce any stiffness in the overall dynamic response but will only contribute to the inertial effect. Since the cantilever wall is relatively thin the mass contribution of the wall itself may be ignored compared to that of the soil. The wall thus contributes only to the stiffness of the overall soil-structure system. The retaining wall is fixed at the base and foundation compliance has been ignored for the time being.

It will be observed that the assumptions made are identical to what Mononobe or Steedman and Zang has assumed in their analysis. Based on the above assumption the mass distribution of soil along the height of the wall is as shown hereafter. As shown in Figure 3.6.6, is the mass distribution of the failed wedge ABD. For an elemental strip dz in vertical direction mass distribution is given by

m(z) =

γ zdz g tan α


(3.6.16)


Z A

B

m (z ) =

D

zdz g tan α

; α = 45 +

φ 2

α Y

Figure 3.6.6 Mass distribution of the failed soil wedge under active soil pressure.

For the analysis of time period as a first step, we develop the stiffness and equivalent mass contributing to the dynamic response of the system. For this we use the Rayleigh Ritz method to obtain the stiffness and mass of the wall-soil system. We had already shown earlier while deriving the time period of chimneys that for a flexural beam the stiffness and mass matrix may be obtained from the expressions ⎡ L ⎤ mij = ⎣ m(z)ϕ i (z)ϕ j (z)dz⎦

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

and,

0

L kij =

EI(z) 0


dz2

dz

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

(3.6.17)

where m(z) is as defined above; E = Young’s modulus of the wall material, and I(z) = Average moment of inertia of the wall. Considering the wall as a cantilever flexural member the displacement y(z, t) can represented trigonometrically by

Ym = sin

μm z μm z μm z μm z

− sin h − αm cos − cos h H H H H

where, m = number of modes 1, 2, 3, . . . ; μm = 1.875, 4.694,7.855, m = 1, 2, 3 . . . m, and αm = © 2009 Taylor & Francis Group, London, UK

sin μm +sin hμm . cos μm +cos hμm

(3.6.18)

2m−1 2 π,

for


Thus for failure wedge of the soil, the mass contribution for an element dz for the first mode is given by H [M]ij = 0

γz γ ϕi (z)ϕ j (z)dz = g tan α g tan α

H zφii (z)φj (z)dz

(3.6.19)

0

Considering, the shape function as μi z μi z μi z μi z

− sin h − αi cos − cos h and 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

(3.6.20)

The double derivative of the above is given by μ2i μi z μi z μi z μi z −sin cos − sin h + α + cos h i H H H H H2 2 μj μj z μj z μj z μj z − sin h + αj cos + cos h ϕj = 2 −sin H H H H H ϕi =

and (3.6.21)

Before performing the integration we change the above to generalized co-ordinate by considering, ξ = Hz when, dξ = dz H and as z → 0, ξ → 0 and as z → H, ξ → 1. Based on the above, we can now express the double derivative as μ2i −sin μi ξ − sin hμi ξ + αi cos μi ξ + cos hμi ξ H2 μ2j f (ξ )j = 2 −sin μj ξ − sin hμj ξ + αj cos μj ξ + cos hμj ξ H f (ξ )i =

and (3.6.22)

Thus stiffness of the system can now be expressed as

kij =

1 EIμ2i μ2j

H3

f (ξ )i f (ξ )j dξ

(3.6.23)

0

and, mass of the system is given by γ H2 mij = g tan α

1 ξ f (ξ )i f (ξ )j dξ

where i, j = 1, 2, 3, . . . . . . . . .m.

(3.6.24)

0

where γ = unit weight of soil; g = acceleration due to gravity; E = Young’s modulus of the wall material, and, I = moment of inertia of the wall. © 2009 Taylor & Francis Group, London, UK


Thus for the first three modes the stiffness matrix42 is given by ⎡ μ41

1

⎤ f (ξ )21 dξ

⎢ ⎢ 0 ⎢ ⎢ 1 EI ⎢ [K]ij = 3 ⎢μ22 μ21 f (ξ )2 f (ξ )1 dξ H ⎢ 0 ⎢ ⎢ ⎣ 2 2 1 μ3 μ1 f (ξ )3 f (ξ )1 dξ 0

μ42 μ23 μ22

1

1 0

f2 (ξ )2 dξ

f (ξ )3 f (ξ )2 dξ

0

μ43

1 0

f (ξ )23 dξ

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

and the mass matrix is given by ⎡

1

⎤ ξ f (ξ )21 dξ

⎢ ⎢ 0 ⎢1 γ H2 ⎢ ⎢ ξ f (ξ )2 f (ξ )1 dξ [M]ij = g tan α ⎢ ⎢0 ⎢1 ⎣ ξ f (ξ )3 f (ξ )1 dξ 0

1

ξ f2 (ξ )2 dξ

0 1

ξ f (ξ )3 f (ξ )2 dξ

0

1 0

ξ f (ξ )23 dξ

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

(3.6.25)

The above integrals can very easily be solved based on Numerical analysis between the limits 1 to 0 when we have

[K]3×3

⎡ 22.936 EI ⎣ = 3 −0.002 H 0.006

−0.002 468.044 −0.11

⎤ 0.006 −0.11 ⎦ 3812.81

and the mass matrix is given by

[M]3×3

⎡ 1.496 −0.205 γ H2 ⎣ −0.205 0.573 = g tan α 0.028 −0.188

⎤ 0.028 −0.188⎦ 0.533

(3.6.26)

Converting the above into standard eigen-value form of Aϕ = λϕ and applying the generalized Jacobi technique43 we have ⎡ 15.317 EIg tan α ⎣ 0 [λ] = γ H5 0

0 846.507 0

⎤ 0 0 ⎦ 8235

42 The matrix is symmetric about is diagonal. 43 The technique has been worked in detail in Chapter 5 (Vol. 1).


(3.6.27)


and the corresponding eigen vectors are given as ⎫ ⎤⎧ 0.139 −0.031 ⎨f1 (ξ )⎬ 0.989 −0.353⎦ f2 (ξ ) ⎭ ⎩ f3 (ξ ) −0.046 −0.935

⎡

1 [ϕ] = ⎣−6.833 × 10−3 1.161 × 10−4 since [λ] = ω2 and T = ⎡ 1.6049 [T] = ⎣ 0 0

2π ω

we have

0 0.2198 0

⎤ 0 γ H5 0 ⎦ EIg tan α 0.0743

Considering moment of inertia of the wall as I = can modify the fundamental time period as ⎡ 5.556 0 0.761 [T] = ⎣ 0 0 0

(3.6.28)

(3.6.29)

1 3 12 t

having one meter width we

⎤ 0 γ H5 0 ⎦ Et 3 g tan α 0.257

(3.6.30)

Calculation of amplitude In terms of response spectrum analysis maximum displacement amplitude Sd is given by Sd = ωSa2 , expressing the above in terms of codal formulation we may express it as Sd = κi

ZI Sa 2R ω2

(3.6.31)

where, κi = modal participation factor and is given by ⎛H ⎞ ⎛ 1 ⎞ *1 *H mi φi2 dφ ⎠ = ⎝ ξ fi (ξ )dξ ξ fi (ξ )2 dξ ⎠ κi = ⎝ mi φi dφ 0

0

0

(3.6.32)

0

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 the values as shown in Table 3.6.1. ZI Now considering, β = 2R an IS-code factor we can write the time dependent function of displacement as, Sd = κi β

Sa . ω2

(3.6.33)

Thus, for the first mode, we have Sd = 0.619β

Sa1 γ H 5 λ1 EIg tan α


(3.6.34)

Analytical and design concepts for earthquake engineering 539 Table 3.6.1 Modal participation factor for cantilever walls for the first three modes. Mode number

Mass participation factor (ki )

1 2 3

0.619 0.224 0.09

Substituting the value of λ = 15.317, calculated earlier, we have Sd = 0.485β

Sa1 γ H 5

(3.6.35)

Et 3 g tan α

where t = the average thickness of the wall. The complete displacement function is thus given by w(z, t) = φ(z) · q(t) and for first mode w(z, t) = 0.5387β

Sa1 γ H 5 × [f1 (ξ ) − 6.833 × 10−3 f2 (ξ ) + 1.161 × 10−4 f3 (ξ )] Et 3 g tan α (3.6.36)

Ignoring the effect of second and third mode which is found very small w(z, t) = 0.485β

Sa1 γ H 5 Et 3 g tan α

f1 (ξ )

(3.6.37)

2

Considering, EI ddzw2 = −Mz , we get M(z, t) = −0.0404β

Sa1 γ H 3 2 [μ f ξ )]; g tan α 1 1

V(z, t) = −0.0404β

Sa1 γ H 2 3 [μ f (ξ )] g tan α 1 1 (3.6.38)

The above can now be represented as w(z, t) = Coeffdi × β

Sa1 γ H 5 Et 3 g tan α

V(z, t) = −Coeffvi × β

Sa1 γ H 2 g tan α

;

M(z, t) = −Coeffmi × β

Sa1 γ H 3 g tan α (3.6.39)

where, Coeffdi , Coeffmi and Coeffvi are the dynamic amplitude, moment and shear coefficients for the first three mode (i = 1, 2, 3). © 2009 Taylor & Francis Group, London, UK

540 Dynamics of Structure and Foundation: 2. Applications Table 3.6.2 Factors of dynamic amplitude moments and shears for cantilever retaining wall. Z/H

coeff d1

coeff d2

coeff d3

Coeff m1

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 0.00567 0.02216 0.04866 0.08437 0.12851 0.18029 0.23893 0.30367 0.37376 0.44849 0.52717 0.60913 0.69376 0.78050 0.86883 0.95829 1.04849 1.13913 1.22997 1.32086

0.00000 0.00016 0.00058 0.00117 0.00187 0.00260 0.00327 0.00384 0.00425 0.00446 0.00444 0.00418 0.00367 0.00293 0.00198 0.00085 −0.00043 −0.00180 −0.00324 −0.00472 −0.00620

0.00000 0.38681 0.00002 0.36018 0.00006 0.33358 0.00011 0.30703 0.00016 0.28061 0.00019 0.25441 0.00020 0.22854 0.00018 0.20313 0.00014 0.17835 0.00008 0.15434 0.00001 0.13130 −0.00006 0.10941 −0.00012 0.08888 −0.00016 0.06992 −0.00017 0.05274 −0.00015 0.03758 −0.00010 0.02464 −0.00002 0.01418 0.00008 0.00641 0.00019 0.00158 0.00030 −0.00010

Coeff m2

Coeff m3

Coeff v1

Coeff v2

Coeff v3

0.01143 0.00870 0.00599 0.00333 0.00080 −0.00154 −0.00362 −0.00537 −0.00673 −0.00767 −0.00815 −0.00819 −0.00780 −0.00705 −0.00600 −0.00476 −0.00342 −0.00214 −0.00103 −0.00026 0.00004

0.00135 0.00082 0.00031 −0.00016 −0.00053 −0.00079 −0.00089 −0.00083 −0.00064 −0.00033 0.00003 0.00040 0.00072 0.00094 0.00104 0.00101 0.00086 0.00063 0.00040 0.00022 0.00020

−0.53200 −0.53190 −0.53123 −0.52945 −0.52607 −0.52063 −0.51272 −0.50197 −0.48805 −0.47065 −0.44953 −0.42447 −0.39527 −0.36178 −0.32389 −0.28149 −0.23452 −0.18293 −0.12669 −0.06578 −0.00021

−0.05400 −0.05389 −0.05320 −0.05147 −0.04843 −0.04396 −0.03807 −0.03093 −0.02280 −0.01405 −0.00510 0.00356 0.01145 0.01809 0.02303 0.02589 0.02633 0.02411 0.01908 0.01112 0.00020

−0.01062 −0.01052 −0.00993 −0.00857 −0.00640 −0.00360 −0.00048 0.00258 0.00514 0.00688 0.00754 0.00707 0.00556 0.00328 0.00063 −0.00193 −0.00390 −0.00484 −0.00440 −0.00232 0.00153

It will however be shown based on a numerical problem subsequently that moments and shears developed in the wall for the higher modes (i = 2, 3, etc.) have insignificant contribution i.e. their SRSS values almost same as the moment, shear and amplitude values for the first mode. Table 3.6.2 showing values of these coefficients are furnished for reference. It is to be understood that once the time period and subsequent acceleration value Sa /g is known from the code one has to only multiply the displacement, moment, and shear expressions by the values in Table 3.6.2 to arrive at the dynamic design data without resorting to any elaborate dynamic analysis. Analysis for passive case For passive case the analysis remains the same except the fact that the angle α becomes tan 45 − φ/2 in lieu of what has been shown earlier. We now explain the above theory based on a suitable numerical problem.

Example 3.6.1 A retaining wall of height 5.8 m has top thickness of wall as 200 mm and bottom thickness as 500 mm. The unit weight of soil it retains has a value of 22 kN/m3 . The angle of friction of soil is 28◦ . the unit weight of the concrete wall is considered as 25 kN/m3 . Consider the wall is in zone IV as per IS-code © 2009 Taylor & Francis Group, London, UK


(1893, 2002) and is resting on Hard soil. Determine the earthquake force acting on the wall. Solution: Based on the above theory, Average thickness of wall = 200+500 = 350 mm; Unit weight of concrete = 2 25 kN/m3 ; Thus, Young’s modulus of concrete = 5700 fck × 1000 kN/m2 = 2.85 × 107 kN/m2 . Angle α (active case) = 45 + 21 × 28 = 59◦ ; Angle α (passive case) = 45 − 12 × 28 = 31◦ . As per IS-code consider Importance factor as 1.0 and ductility factor as R = 2.0. ZI Considering β = 2R we have, β = 0.24×1.0 = 0.06 2×2 Now considering ⎡

5.556 [T] = ⎣0 0 ⎡

5.556 [T] = ⎣0 0

0 0.761 0

⎤ 0 γ H5 ⎦ 0 , we have 3 g tan α Et 0.238

0 0.724 0

⎤ 0 22 × 5.85 ⎦ 0 which gives 7 2.85 × 10 × 0.353 × 9.81 tan 59 0.238

⎧ ⎫ ⎨0.473⎬ {T} = 0.065 secs, for active case for the first three modes. ⎩0.022⎭

Similarly for passive case considering α = 31 degree, we have ⎡

5.567 [T] = ⎣0 0

0 0.724

⎤ 0 22 × 5.85 ⎦ 0 , which gives 7 2.85 × 10 × 0.353 × 9.81 tan 31 0.238

⎧ ⎫ ⎨0.787⎬ {T} = 0.108 sec for passive case ⎩ ⎭ 0.036

Corresponding to the time periods for the first three modes the Sa /g values for active and passive case are shown hereafter. The values are obtained considering 7% damping for the RCC wall i.e. codal values scaled by a factor 0.9.

Mode

Sa /g (active case)

Sa /g (passive case) m/sec 2

1 2 3

1.9 1.773 1.173

1.144 2.25 1.355



The displacement shear and moment factors are as shown below: w(z, t) = β

Sa γ H 5 , Et 3 g tan α

M(z, t) = −β

Sa1 γ H 3 g tan α

and

V(z, t) = −β

Sa1 γ H 2 g tan α

For the first three modes, we have

Mode

Amplitude coefficient (active case)

Amplitude coefficient (passive case)

1 2 3

0.00809 0.00755 0.00499

0.013499 0.02655 0.015992

Mode

Moment coefficient (active case)

Moment coefficient (passive case)

1 2 3

−294.46 −274.61 −185.02

−490.06 −964.421 −596.64

Mode

Shear coefficient (active case)

Shear coefficient (passive case)

1 2 3

−50.768 −47.345 −31.9

−84.493 −166.279 −102.87

Multiplying the above factors by the coefficients as furnished earlier in Table 3.6.2, the dynamic displacement moment and shear for active and passive case are obtained as shown below Dynamic amplitude (mm), moment (kN · m) and shear (kN) for active case Z/H

D1

D2

D3

0.000 0.050 0.100 0.150 0.200 0.250 0.300 0.350 0.400 0.450 0.500 0.550 0.600

0.000 0.046 0.179 0.394 0.683 1.040 1.459 1.933 2.457 3.024 3.628 4.265 4.928

0.000 0.000 −113.899 −3.138 −0.250 −27.009 −2.557 −0.339 0.001 0.000 −106.060 −2.388 −0.152 −27.004 −2.552 −0.336 0.004 0.000 −98.225 −1.644 −0.057 −26.970 −2.519 −0.317 0.009 0.001 −90.408 −0.916 0.029 −26.879 −2.437 −0.273 0.014 0.001 −82.628 −0.220 0.099 −26.708 −2.293 −0.204 0.020 0.001 −74.913 0.423 0.145 −26.431 −2.081 −0.115 0.025 0.001 −67.295 0.994 0.164 −26.030 −1.803 −0.015 0.029 0.001 −59.815 1.475 0.154 −25.484 −1.464 0.082 0.032 0.001 −52.516 1.849 0.118 −24.777 −1.080 0.164 0.034 0.000 −45.447 2.105 0.062 −23.894 −0.665 0.219 0.034 0.000 −38.662 2.238 −0.006 −22.822 −0.242 0.241 0.032 0.000 −32.217 2.248 −0.074 −21.549 0.169 0.225 0.028 −0.001 −26.172 2.143 −0.133 −20.067 0.542 0.177 (continued)


M1

M2

M3

V1

V2

V3

Analytical and design concepts for earthquake engineering 543 Z/H

D1

D2

0.650 0.750 0.800 0.850 0.900 0.950 1.000

5.613 7.029 7.753 8.482 9.216 9.950 10.686

0.022 −0.001 −20.589 1.936 0.006 −0.001 −11.065 1.306 −0.003 0.000 −7.257 0.940 −0.014 0.000 −4.175 0.587 −0.024 0.000 −1.888 0.284 −0.036 0.001 −0.464 0.070 −0.047 0.002 0.028 −0.012

D3

M1

M2

M3

V1

V2

V3

−0.175 −18.367 0.857 0.105 −0.186 −14.291 1.226 −0.062 −0.159 −11.906 1.247 −0.124 −0.117 −9.287 1.142 −0.154 −0.074 −6.432 0.903 −0.140 −0.041 −3.340 0.526 −0.074 −0.036 −0.010 0.010 0.049

Dynamic amplitude (mm), moment (kN · m) and shear (kN) for passive case Z/H

D1

D2

D3

M1

M2

M3

V1

V2

V3

0.000 0.050 0.100 0.150 0.200 0.250 0.300 0.350 0.400 0.450 0.500 0.550 0.600 0.650 0.700 0.750 0.800 0.850 0.900 0.950 1.000

0.000 0.077 0.299 0.657 1.139 1.735 2.434 3.226 4.100 5.046 6.055 7.117 8.223 9.366 10.537 11.729 12.937 14.155 15.378 16.605 17.832

0.000 0.004 0.016 0.032 0.051 0.070 0.088 0.104 0.115 0.121 0.120 0.113 0.099 0.079 0.053 0.023 −0.012 −0.049 −0.088 −0.127 −0.167

0.000 0.000 0.001 0.002 0.003 0.003 0.003 0.003 0.002 0.001 0.000 −0.001 −0.002 −0.003 −0.003 −0.002 −0.002 0.000 0.001 0.003 0.005

−189.559 −11.020 −0.807 44.954 8.980 1.092 −176.512 −8.388 −0.491 44.946 8.962 1.082 −163.473 −5.773 −0.185 44.889 8.846 1.021 −150.463 −3.215 0.093 44.739 8.559 0.881 −137.515 −0.773 0.318 44.453 8.054 0.659 −124.675 1.486 0.469 43.993 7.310 0.371 −111.997 3.491 0.530 43.325 6.332 0.049 −99.548 5.179 0.498 42.417 5.144 −0.265 −87.401 6.493 0.381 41.240 3.792 −0.529 −75.637 7.393 0.200 39.770 2.336 −0.707 −64.345 7.860 −0.018 37.986 0.849 −0.776 −53.618 7.896 −0.239 35.867 −0.592 −0.727 −43.557 7.525 −0.430 33.400 −1.904 −0.572 −34.265 6.799 −0.563 30.570 −3.009 −0.337 −25.847 5.788 −0.621 27.368 −3.830 −0.064 −18.414 4.586 −0.601 23.786 −4.305 0.199 −12.077 3.302 −0.512 19.817 −4.379 0.401 −6.948 2.061 −0.379 15.458 −4.010 0.498 −3.142 0.996 −0.237 10.705 −3.172 0.452 −0.772 0.248 −0.133 5.559 −1.849 0.239 0.047 −0.041 −0.116 0.017 −0.034 −0.158

If we compare the SRSS value with that of the first mode the values compared as given hereafter.

Z/H

Moment (1st mode)

Moment (SRSS)

Shear first mode

Shear SRSS

0.000 0.050 0.100 0.150 0.200 0.250 0.300 0.350

−189.559 −176.512 −163.473 −150.463 −137.515 −124.675 −111.997 −99.548

189.881 176.712 163.575 150.497 137.518 124.685 112.053 99.684

44.954 44.946 44.889 44.739 44.453 43.993 43.325 42.417

45.855 45.843 45.764 45.559 45.181 44.598 43.785 42.728 (continued)


544 Dynamics of Structure and Foundation: 2. Applications Z/H 0.400 0.450 0.500 0.550 0.600 0.650 0.700 0.750 0.800 0.850 0.900 0.950 1.000

Moment (1st mode) −87.401 −75.637 −64.345 −53.618 −43.557 −34.265 −25.847 −18.414 −12.077 −6.948 −3.142 −0.772 0.047

Moment (SRSS) 87.642 75.997 64.823 54.197 44.205 34.937 26.495 18.986 12.531 7.258 3.305 0.822 0.132

Shear first mode

Shear SRSS

41.240 39.770 37.986 35.867 33.400 30.570 27.368 23.786 19.817 15.458 10.705 5.559 0.017

41.417 39.845 38.003 35.880 33.459 30.720 27.635 24.173 20.299 15.977 11.175 5.863 0.162

It is observed that the difference in the values are insignificant as such for these type of structure performing a first mode analysis based on the fundamental time period only should suffice for practical designs.

3.6.8 Some discussions on the above method The method as shown above is an approximate method for evaluation of the time period of the structure44 , however is far more realistic then considering a maximum value of 0.9 × 2.5 = 2.25 m/sec2 , which would result in significant over design of the retaining wall thus making it a more expensive proposal. Another major advantage is that unlike the pseudo static or dynamic case one need not estimate the point of application of this load, the moments and shear expression are directly derived from the amplitude expression. Based on the above numerical problem it is seen that fundamental time period is the most significant mode. Thus arriving at a closed form expressions for more complex cases like soil sloped at an angle or a counter fort retaining wall where the wall acts like a plate with appropriate boundary conditions become much simpler45 .

3.6.9 Extension to the generic case of soil at a slope i behind the wall In this case, shown in Figure 3.6.7, we need to take the additional soil mass of the triangle ABE converted it into a form so that it fits into the integration scheme we had proposed. Thus for having same mass as triangle ABE we take en equivalent trapezium

44 We should remember that the integrations performed where based on numerical analysis and not a closed form one as such errors due to truncation is surely to be expected. 45 This we will see subsequently as we take them up subsequently.



C

E O

A B

H

D

Figure 3.6.7 Retaining wall with soil sloped at an angle behind the wall.

ABEC equal to triangular area ABE, when equating the area of ABEC and triangle ABE it can be proved that cos α sin i AC = H( 1 + ζ − 1) where ζ = sin(α − i) √ Now considering H = H + AC = H 1 + ζ H [M]ij = 0

γz γ φi (z)φj (z)dz = g tan α g tan α

γ H 2 mij = g tan α

H zφii (z)φj (z)dz 0

1 ξ f (ξ )i f (ξ )j dξ

where i, j = 1, 2, 3, . . . . . . . . . m

0

in natural co-ordinate γ H 2 (1 + ζ )2 = g tan α

1 ξ f (ξ )i f (ξ )j dξ 0


where i, j = 1, 2, 3, . . . . . . . . .m.

(3.6.40)


Since we had seen earlier that the first mode analysis suffice for these type of structure restricting the above expansion to the first mode, we have 1

γ H 2 (1 + ζ )2 m1 = g tan α

ξ f (ξ )21 dξ

and

EIμ41 k1 = H3

0

f (ξ )2i dξ

(3.6.41)

0 2π ω ,

Considering λ = ω2 and T = Fundamental time period as [T] = 5.556(1 + ζ )

1

we have

γ H5

(3.6.42)

Et 3 g tan α

Based on above we can find out the fundamental time period of the system and subsequently the expression Sa /g from the code for an appropriate damping of the system. It will be observed that for ζ = 0 the time period value reduces to the fundamental time period for the case where the soil plane is considered parallel to the base. The modal mass participation is given by the expression H κ1 = 0

γz ϕ1 (z)dz/ g tan α

1 κ1 =

H 0

γz ϕ 2 dz g tan α 1

(3.6.43)

1 ξ f12 dz

ξ f1 (ξ )dξ z/ 0

and remains the same as in the earlier case.

0

H (1+ζ ) Thus, for the fundamental mode, Sd = 0.619β Sa1λ1γEIg tan α . Substituting the value of λ we have 5

Sd = 0.485β

2

Sa1 γ H 5 (1 + ζ )2 Et 3 g tan α

(3.6.44)

the complete displacement function is thus given by w(z, t) = φ(z) · q(t)

which gives, w(z, t) = 0.485β

M(z, t) = −0.0404β

Sa1 γ H 3 (1 + ζ )2 2 μ1 F1 (ξ ) g tan α

V(z, t) = −0.0404β

Sa1 γ H 2 (1 + ζ )2 3 μ1 F1 (ξ ) g tan α

Sa1 γ H 5 (1 + ζ )2 Et 3 g tan α

F1 (ξ )

where, the value μ1 = 1.875 (3.6.45)



Plan view of counter fort retaining wall

Side view of the wall

Figure 3.6.8 Plan view and elevation of counterfort retaining wall.

The coefficients are as furnished in Table 3.6.2.

3.6.10 Dynamic analysis of counterfort retaining wall When retaining walls are more then 6.0 meter high normally for economic reason counterforts are provided to reduce the moment in the supporting wall. In such cases unlike cantilever, the wall does not behave as pure flexural member but behaves as plate having three sides fixed and one side free. To the best of our knowledge there exists no solution to this problem in terms of earthquake induced dynamic force. Present state of art is to develop the additional pressure based on Mononobe’s coefficient add it too the static pressure (hydrostatic in nature) and solve for the moments and shears based on coefficients as furnished in IS-3370 Part IV or similar literature. In other words neither the time period nor the spatial distribution of the load is accounted for, considering its mode shape which again is a function of its boundary condition like free, fixed etc. For such walls when subjected to force due to an earthquake solutions have to be sought based on plate vibration in lieu of beam vibration as shown previously and the solution becomes downright tricky. However, since we have seen earlier that it is fundamental mode which basically governs the design with a little bit of intelligent mathematical manipulation it is not too difficult to arrive at the time period vis-a-vis dynamic amplitude shear and moments for such type of retaining wall. A typical counterfort retaining wall is shown in Figure 3.6.8. It is evident that the wall between the buttress in this case will behave as plate instead of a beam having boundary condition of three sides continuous and one side free. Shown in Figure 3.6.9, is the wall of a counterfort retaining wall of height H and width b. Also, b is the dimension between the two consecutive counterforts. The failed soil mass under active and passive pressure is shown by the triangular wedge which contributes its inertial effect on the wall.



H Fixed edge(typ.)

b

Figure 3.6.9 Wall with three edges fixed with soil mass under earthquake.

Here the displacement function of the wall can be represented by w(z, y, t) =

n

φi (z, y)qi (t)

(3.6.46)

i=1

The kinetic energy of the system is given by 1 T= 2

H b 0 0

∂w(z, y, t) m(z, y) ∂t

2

Substituting the displacement function in above equation we have 1 T= 2

H b m(z, y)

- n i=1

0 0

=

n n

⎤ .⎡ n φi (z, y)q˙ i (t) ⎣ φj (z, t)q˙ j (t)⎦

⎡

⎢ q˙ i (t)q˙ j (t) ⎣

i=1 j=1

H b

j=1

⎤

⎥ m(z, y)φi (z, y)φj (z, y)⎦

0 0

From above we deduce that the mass coefficient is given by ⎡ ⎤ H b ⎢ ⎥ m(z, y)φi (z, y)φj (z, y)⎦ dy · dz mij = ⎣ 0 0


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

(3.6.47)


The strain energy equation of the plate is given by (Timoshenko and WoinowskyKrieger 1958)

1 V= 2

H b 0 0

⎡

∂ 2w ∂ 2w D(z, y) ⎣ + ∂z2 ∂y2

/

∂ 2w ∂ 2w − 2(1 − ν) − ∂z2 ∂y2

∂ 2w ∂z∂y

2 0⎤ ⎦

(3.6.48) where, w(z, y, t) =

n

φi (z, y)qi (t).

i=1

It can be proved based on Lagrange’s equation that on differentiation of the above equation with respect to qj (t) the stiffness coefficient of an isotropic plate is given by -

H b Kij =

D 0 0

∂ 2 φj ∂ 2 φii + ∂z2 ∂y2

∂ 2 φj ∂ 2 φi + ∂y2 ∂z2

∂ 2 φi ∂ 2 φ j −(1 − ν) ∂z2 ∂y2

. ∂ 2 φi ∂ 2 φ j −2 dy · dz ∂z∂y ∂z∂y

(3.6.49)

3

Et where, D = 12(1−ν 2 ) the flexural stiffness of thin plate; E = Young’s modulus of concrete; t = thickness of the wall; ν = Poisson’s Ratio of concrete, and φi , φj = mode shapes at different modes whose values have to be chosen for the appropriate boundary condition and shall be a function of both z and y. To select the appropriate mode shape let us imagine a thin strip of element in vertical and horizontal direction and try to visualize how they will deflect under a dynamic loading. It is obvious that while vertical strip behaves as a cantilever beam, the horizontal strip shall behave as a beam whose ends are fixed. Based on the above assumption the shape function for the mode are considered as

φ(z, y) = F(z) · F(y), where

(3.6.50)

μz μz μz μz − sin h − αz cos − cos h H H H H 3π y 3πy 3π y 3π y F(y) = sin − sin h − αy cos − cos h 2b 2b 2b 2b F(z) = sin

For first mode μ = 1.875 αz =

sin μ + sin hμ cos μ + cos hμ


and

αy =

3π sin 3π 2 − sin h 2 3π cos 3π 2 − cos h 2

(3.6.51)


3.6.10.1

Calculation of mass coefficient

For first mode analysis as φi (z, y) = φj (z, y), we have H b m= 0 0

γz φ 2 dy · dz = g tan α

H b 0 0

γz F(z)2 F(y)2 dy · dz, g tan α

(3.6.52)

where F(z) and F(y) are as defined above. For ease of analysis we convert the integration basis from local to natural co-ordinate when H b (ξ + 1) and y = (η + 1) which implies 2 2 b H dz = dξ and dy = dη 2 2 μ(ξ + 1) μ(ξ + 1) μ(ξ + 1) μ(ξ + 1) Thus, F(ξ ) = sin − sin h − αz cos − cos h 2 2 2 2 3π(η + 1) 3π(η+1) 3π(η + 1) 3π(η+1) and F(η) = sin − sin h − αy cos − cos h . 4 4 4 4 z=

Substituting the above functions in mass equation, we have 1 1 m= −1 −1

γ H2b F(ξ )2 F(η)2 dηdξ 8g tan α

(3.6.53)

the above can be further simplified to the expression m=

γ H2b C1 8g tan α

(3.6.54)

1 1 F(ξ )2 F(η)2 dηdξ .

where, C1 = −1 −1

3.6.10.2

Derivation of stiffness coefficient

For derivation of stiffness coefficient for the first mode as φi (z, y) = φj (z, y), we have from the stiffness expression furnished above as H b K= 0 0

⎡

∂ 2φ ∂ 2φ D⎣ + 2 2 ∂z ∂y

2

⎧ 2 ⎫⎤ ⎬ ⎨ ∂ 2φ ∂ 2φ ∂ 2φ ⎦dy · dz − (1 − ν) 2 2 2 − 2 ⎩ ∂z ∂y ∂z∂y ⎭ (3.6.55)



Using, φ(z, y) = F(z) · F(y), we have ∂ 2φ ∂ 2φ ¨ ¨ = F(z)F(y) = F(y)F(z) ∂z2 ∂y2

and

∂ 2φ ˙ F(z). ˙ = F(y) ∂y∂z

Substituting the above in the stiffness equation and on expansion, we have H b K=

1 22 1 22 1 2 ¨ ¨ ¨ F(z) ¨ D F(y)F(z) + F(z)F(y) + 2ν F(y)F(z)F(y)

0 0

1 22 ˙ F(z) ˙ + 2(1 − ν) F(y)

(3.6.56)

where F(z) and F(y) are as defined previously. For further derivation of the stiffness matrix it is essential to derive the first and second derivatives of F(z) and F(y) and are as shown hereafter μz μz μz μz

− sin h − αz cos − cos h H H H H μ μz μz μz μz ˙ F(z) = cos − cos h + αz sin + sin h H H H H H

F(z) = sin

μ2 μz μz

μz μz ¨ F(z) = 2 αz cos + cos h − sin − sin h . H H H H H Transforming the above in natural co-ordinate, we have

F(ξ ) = sin

μ(ξ + 1) μ(ξ + 1) μ(ξ + 1) μ(ξ + 1) − sin h − αz cos − cos h 2 2 2 2

= f (ξ ). say ˙ ) = μ cos μ(ξ + 1) − cos h μ(ξ + 1) + αz sin μ(ξ + 1) + sin h μ(ξ + 1) F(ξ H 2 2 2 2 μ = f (ξ ), say H 2 ¨F(ξ ) = μ αz cos μ(1 + ξ ) + cos h μ(1 + ξ ) − sin μ(1 + ξ ) − sin h μ(1 + ξ ) 2 2 2 2 H2 =

μ2 f (ξ ), say. H2



Similarly for y direction on differentiation and transformation to natural co-ordinate we have F(η) = sin

3π(η + 1) 3π(η + 1) 3π(η + 1) 3π(η + 1) − sin h − αy cos − cos h = f (η) 4 4 4 4

3π 3π(η + 1) 3π(η + 1) 3π(η + 1) 3π(η + 1) 3π ˙ F(η) = cos − cos h + αy sin + sin h = f (η) 2b 4 4 4 4 2b 9π 2 3π(1 + η) 3π(1 + η) 3π(1 + η) 3π(1 + η) 9π 2 ¨ F(η) = α cos f (η) + cos h − sin − sin h = y 2 4 4 4 4 4b 4b2

Substituting the above expressions in the stiffness equations, we have ⎤ ⎡ 1 1 1 1 4 1 2 2 1 DHb ⎣ μ4 81π 2 2 K= f (η)f (ξ ) dηdξ + f (η)f (ξ ) dηdξ ⎦ 4 H4 16b4 ⎡ +

−1 −1

−1 −1

DHb ⎣ 9μ2 π 2 ν 4 2b2 H 2

1

1

1

2 f (η)f (ξ )f (η)f (ξ ) dηdξ

−1 −1

9(1 − ν)μ2 π 2 + 2b2 H 2

1 1

22

1

⎤

f (η)f (ξ ) ⎦

(3.6.57)

−1 −1

Considering, 1 1 C2 =

1 1

2

[f (η)f (ξ )] dηdξ ;

C3 =

−1 −1

1

−1 −1

1

C4 =

[f (ξ )f (η)]2 dηdξ

[f (η)f (ξ )f (η)f (ξ )]dηdξ −1 −1

1 1 and

C5 =

[f (η)f (ξ )]2 dηdξ .

−1 −1

(3.6.58) We have, . 81π 4 9μ2 π 2 ν 9(1 − ν)μ2 π 2 μ4 C2 + C3 + C4 + C5 H4 16b4 2b2 H 2 2H 2 b2 . μ4 b 81π 4 H 9π 2 μ2 ν 9(1 − ν) ➔ K=D C4 + C5 C2 + C3 + 8Hb 8Hb 4H 3 64b3

DHb K= 4

-


(3.6.59)


3.6.10.3

Derivation of the time period Considering the time period as T = 2π m K , substituting the value of m and k obtained above, we have 3 4 4 T = 2π 5

8Dg tan α

γ H 2 bC1 μ4 b C 4H 3 2

+

81π 4 H C3 64b3

+

9π 2 μ2 ν C4 8Hb

+

9(1−ν)π2 μ2 C5 8Hb

Considering the value of D as given previously and considering H/b = r as the aspect ratio the above on some manipulation and simplification may be expressed as T = 2π

12γ H 5 (1 − ν 2 ) 8Et 3 g tan α(X1 r4 + X2 r2 + X3 νr2 + X4 )

(3.6.60)

where X1 , X2 , X3 , X4 are constants whose values are mentioned below. It will be observed that here that the time period expression is very similar to the what we got for beam except the fact that for plate element the aspect ratio (r) factor in the denominator affect the equation including the Poisson’s ratio (ν). 3.6.10.4

The integration constants

The integration constants for C1 to C5 are solved numerically based on Simpson’s method and are furnished in Table 3.6.3 for reference. Similarly the constants X1 , X2 , X3 , X4 is given by X1 = 75.3121451,

X2 = −3.2405562,

X3 = 17.688041,

Table 3.6.3 Integration constants for time period calculation. Coefficients

Integral function 1

C1

−1 −1 1

C2

C5

1

−1 −1 1

C4

1

−1 −1 1

C3

1

1

−1 −1 1 1 −1 −1

F(ξ )2 F(η)2 dηdξ

Value 12.444

[f (η)f (ξ )]2 dηdξ

7.716

[f (ξ )f (η)]2 dηdξ

7.60

[f (η)f (ξ )f (η)f (ξ )]dηdξ [f (η)f (ξ )]2 dηdξ


−1.033 5.638

X4 = 1.91589362


3.6.10.5

Calculation of nodal forces under dynamic load

Once the time period is calculated and the response acceleration is obtained from IS-1893, the shear force/or the nodal force is obtained from the expression ZI mi ϕi · Sa 2R n

Vi = κi

where

(3.6.61)

i=1

κ1 =

n

mi φi /

i=1

n

mi φi2 is the modal mass participation factor

i=1

for the present plate problem it can be represented by 1

κ1 =

1

−1 −1 1

1

−1 −1

(1 + ξ )f (ξ )f (η)dξ dη (3.6.62)

(1 + ξ )f 2 (η)f 2 (ξ )dηdξ

The above on integration gives a unique value of κ1 = 0.423. Substituting this we have value in the above equation, ZI Vi = 0.423β ·Sa ni=1 mi ϕi , where β = 2R and the equation can be further transformed into 0.423βSa γ H 2 b Vi = 8g tan α

1 1 (1 + ξ )f (ξ )f (η)dξ dη.

(3.6.63)

−1 −1

In the above expression the term Sa /g is a function of the time period as derived above. The above on integration will give the dynamic shear force. However this is not required and the nodal forces can be found out for various values of ξ and η for boundary +1 to −1 to obtain the nodal force coefficient. Once these coefficients are known they can multiplied by the constant term to obtain dynamic force imposed by the soil at various points of the plate. The summation of all these force over the surface of the wall will give the total shear induced on the wall due to the earthquake. Thus,

Vi (z, y) =

0.423βSa γ H 2 b ψ(ξ , η) 8g tan α

(3.6.64)

The values of ψ(ξ , η) are as given in Figure 3.6.10, as coefficients on the plate.

aγ H b These coefficients, when multiplied by the term 0.423βS , will give the nodal 8g tan α force at various points of the plate. It will be interesting to note from the surface envelope generated below that as the load coefficients are dependent on the assumed shape function there generic curve 2



-1 1 0

-0.8 -0.6 -0.4 -0.2 0 .44 1.44 2.56 3.40

0.0 3.72

0.2 3.42

0.4 2.60

0.6 1.48

0.47

0.8

1.0 0.0

2.89

2.65

2.02

1.153

0.365

0

1.98 8

1.50 7

0.863

0.27 3

0

0.8 0 0.342

1.12

1.98

2.64

0 0.255

0.84

1.48 5

1.98

0.547

1.06

1.407

1.54

1.41 7

1.074

0.614

0.194

0

0.6

0.4

0 0.18 2

2.16

0.2 0

0.12 1

0.399 0.70 7

0.941

1.03

0.94 8

0.71 9

0.411

0.13 0

0

0.0 0

0.074

0.24 5

0.43 4

0.578

0.632

0.58 1

0.441

0.253

0.07 9

0

-0.2 0

0.041

0.13 3

0.23 5

0.313

0.342

0.31 5

0.238

0.136

0.043

0

-0.4 0

0.02 0

0.060 0.10 4

0.134

0.152

0.14 0

0.10 6

0.061

0.01 9

0

-0.6 0

0.00 5

0.020

0.032

0.043

0.047

0.04 3

0.03 3

0.018

0.006

0

0.003 0.00 4

0.006

0.006

0.006

0.004

0.003

0.0007

0

0

0

0

0

0

- 0.8 0 0.000 7

-1.0 0

0

0

0

0

0

Figure 3.6.10 Load coefficients (ξ , η) for plates with three edge fixed and one edge free under.

follows the same shape as that of the assumed mode that is, it varies like a cantilever in vertical direction and a beam fixed at edge in horizontal direction. This is surely logical and is in variance to equivalent static load where hydrostatic force is profile is assumed. The variation is more profound in horizontal direction for in normal analysis this is considered as constant (like a rectangular shape) while in reality it is hyperbolic in nature with zero at the edge and maximum at the centre. Thus if the wall is spanning in one direction (i.e. a one way slab) when major load spans horizontally along the shorter span, present state of art of arriving at the Shear Force and Bending Moment could be significantly in variation to the reality. Once the force are known it can be put directly as an input in a FEM analysis of a plate subdivided up into meshes as shown in Figure 3.6.11 and applying the forces as nodal loads we get the shear and bending moment in the wall. © 2009 Taylor & Francis Group, London, UK


Nodal Load function

4 3.5 3 2.5 f(P)

2 1.5 1 S11

0.5 S6

Width

S1 -1

- 0.6

- 0.2

0.2

0.6

1

0

Height

Figure 3.6.11 Surface envelope of nodal loads of counterfort wall under first mode.

Else the amplitude, dynamic moments and shear can be obtained by the method as shown here after. 3.6.10.6

Calculation of dynamic amplitude

As shown at the outset the displacement is defined by w(z, y, t) =

n

φi (z, y)qi (t)

(3.6.65)

i=1

For structural systems under earthquake the dynamic amplitude can be calculated from the expression w(z, y) = κi φ(zi yi )Sd

(3.6.66)

where, Sd = ωSa2 ; Sa = acceleration spectrum and is a function of time period, and ω2 = square of the natural frequency. Based on the codal provision the displacement spectrum can be expressed as Sd =

βSa ω2


(3.6.67)


Considering ω2 = ω2 =

4π 2 , T2

substituting equation of time period we have

8Dg tan α(X1 r4 + X2 r2 + vX3 r2 + X4 ) γ H5

(3.6.68)

where D = flexural stiffness as expressed earlier. ➔ Sd =

βSa γ H 5 8Dg tan α(X1 r4 + X2 r2 + νX3 r2 + X4 )

(3.6.69)

κi βSa γ H 5 F(z)F(y) 8Dg tan α(X1 r4 + X2 r2 + νX3 r2 + X4 )

(3.6.70)

which gives, w(z, y) =

where, F(z) and F(y) are as defined in Equation (3.6.51). Considering (r) = X1 r4 + X2 r2 + νX3 r2 + X4 , a function of aspect ratio above expression can be modified and written as w(z, y) =

κi βSa γ H 5 F(z)F(y) 8Dg tan α (r)

(3.6.71)

Displacement function f(d) for r=2

4.000 3.000 f(d) 2.000 1.000 S8 0.000 -0.8

Height

-0.2

0.4

1

Width S1

Figure 3.6.12 Amplitude function of wall for aspect ratio H/b = 2.


558 Dynamics of Structure and Foundation: 2. Applications κi βγ Sa Now considering, A = 8g tan α , a constant for a system for a particular mode. Expression w(z, y) can now be expressed as

w(z, y) =

A F(z)F(y) 5 H D (r)

(3.6.72)

It is thus observed that the dynamic amplitude w(z, y) is function of the flexural stiffness of the plate, (r) the aspect ratio of the plate, and the shape function F(z) and F(y). Shown in Figure 3.6.12 is the displacement envelope of the plate with typical aspect ratio of r = 2 and Poisson’s ratio of 0.25. The coefficients are scaled to 1000. The values when multiplied by the term A (and divided by 1000) will give the dynamic amplitude for the first mode. 3.6.10.7

Calculation of dynamic moments and shear

The bending moment of a thin plate is given the expression . . ∂ 2w ∂ 2w ∂ 2w ∂ 2w Mz = D + ν 2 ; My = D +ν 2 ; ∂z2 ∂y ∂y2 ∂z . . ∂ ∂ 2w ∂ 2w ∂ ∂ 2w ∂ 2w Qz = D ; Qy = D + + ∂z ∂z2 ∂y ∂z2 ∂y2 ∂y2 -

(3.6.73)

A F(z)F(y) 5 H . D (r) Substituting the value of w(z, y) above we have

where, w(z, y) =

A Mz = (r)

-

. μ2 9π 2 ν F (z)F(y) + F(z)F (y) H 5 H2 4b2

where F (z) and F (y) are as defined in previously in local and natural co-ordinates. The above expression of Mz can be further modified to . A 9π 2 νr2 2 Mz = μ F (z)F(y) + F(z)F (y) H 3 . (r) 4

(3.6.74)

From the above the expression we observe that moment varies as the cube of the height and is a function of the basic shape function and their derivative, Poisson’s ratio and the aspect ratio. Shown in Figure 3.6.13 is a typical envelope for aspect ratio H/b = 2 and Poisson’s ratio = 0.25. © 2009 Taylor & Francis Group, London, UK


Function f(Mz) for r=2

0.15 0.1 0.05 f(Mz) 0 -0.05 S7

-0.1 0.8

0.2

-0.4

-1

Width S1

Height

Figure 3.6.13 Variation of dynamic moment in vertical (Z) direction for H/b = 2.

Similarly for y direction we have . A 9π 2 r2 2 My = νμ F (z)F(y) + F(z)F (y) H 3 (r) 4

(3.6.75)

where, substituting the values of z in either local and nodal co-ordinates moment coefficient and envelope can be plotted. We show in Figure 3.6.14 a typical envelope of Moment (My ) for aspect ratio r(H/b) = 2 and Poisson’s ratio = 0.25. Similarly for shear force in Z and Y direction can be expressed as . A 9π 2 μr2 3 Qz = μ F (z)F(y) + F (z)F (y) H 2 and (r) 4 . A 3μ2 πr 27π 3 r3 Qy = F (z)F (y) + F(z)F (y) H 2 (r) 2 8

(3.6.76)

Looking at above expressions we see that the shear force varies as the square of the height and is a function of the aspect ratio and the basic shape function. The above procedure gives a comprehensive solution for walls with three side fixed and one side free subjected to dynamic earth pressure under earthquake load where the moments, shears and amplitude are dependent on the time period, geometry of the wall, its boundary condition as well as the material and engineering property of the soil. © 2009 Taylor & Francis Group, London, UK


Function f(My) for r=2

0.6 0.4 0.2 f(My) 0 -0.2 S8 S1

Width

0.8

0.2

-0.4

-1

-0.4

Height

Figure 3.6.14 Variation of dynamic moment in vertical (Y) direction for H/b = 2.

3.6.11 Soil sloped at an angle i with horizontal

√ In this case like we did in cantilever wall let H = H + x = H 1 + ζ , where α sin i 46 ζ = cos . sin(α−i) Thus the mass coefficient can now be considered as varying between, H to 0 H b m= 0 0

γz F(z)2 F(y)2 dy · dz g tan α

(3.6.77)

The above on transformation to natural co ordinate can be represented as 1 1 m= −1 −1

γ H 2b γ H 2 b(1 + ζ ) F(ξ )2 F(η)2 dηdξ = C1 8g tan α 8g tan α

(3.6.78)

The time period is thus given by T = 2π

12γ H 5 (1 + ζ )(1 − ν 2 ) 8Et 3 g tan α(X1 r4 + X2 r2 + X3 νr2 + X4 )

46 Refer to the figure we have drawn for the cantilever retaining wall with sloped soil surface.


(3.6.79)


It will be observed that for i = 0, ζ = 0 the above equation converges to equation (for time period with wall having soil parallel to the ground. The constants X1 , X2 , X3 , X4 remains same as mentioned earlier. The modal mass participation factor shall remain same as earlier i.e.,

κ1 =

1 1 −1 −1 (1 + ξ )f (ξ )f (η)dξ dη 1 1 2 2 −1 −1 (1 + ξ )f (η)f (ξ )dηdξ

= 0.423

(3.6.80)

Thus the nodal load can now be expressed as

0.423βSa γ H 2b Vi = 8g tan α =

1 1 (1 + ξ )f (ξ )f (η)dξ dη −1 −1

0.423βSa γ H 2 (1 + ζ )b 8g tan α

1 1 (1 + ξ )f (ξ )f (η)dξ dη

(3.6.81)

−1 −1

Thus we see that the constant term is multiplied by an additional factor (1 + ζ ) and the integration constants ψ(ξ , η) remain same as expressed earlier. It may again be noted that for i = 0, ζ = 0 when the above expression converges to expression shear equation for the soil parallel to the ground. From above we can safely deduce from mathematical similarity that for this case. The term A as expressed in earlier can be now be expressed as A=

κi βγ Sa (1 + ζ ) 8g tan α

(3.6.82)

The rest of the steps remain same as explained earlier. 3.6.11.1

Design moments and shear coefficients

Observing the values of moments and shear as derived above it can be inferred that they can be expressed in the form M(z) = Coeffz ×

0.423βγ Sa (1 + ζ )H 3 ; 8g tan α

M(y) = Coeffy ×

0.423βγ Sa (1 + ζ )H 3 8g tan α

Q(z) = Coeffz ×

0.423βγ Sa (1 + ζ )H 2 ; 8g tan α

Q(y) = Coeffz ×

0.423βγ Sa (1 + ζ )H 2 . 8g tan α

The coefficients are furnished in Table 3.6.4 to 3.6.6 © 2009 Taylor & Francis Group, London, UK

(3.6.83)

Table 3.6.4 Moment coefficients in vertical direction (Mz ). Aspect ratio

H/b

−1.0

−0.8

−0.6

−0.4

−0.2

0.0

0.2

0.4

0.6

0.8

1.0

2.0

−1.0 −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 0.8 1.0

0.1016 0.0877 0.0737 0.0601 0.0469 0.0345 0.0234 0.0139 0.0065 0.0017 0.0000

0.0548 0.0473 0.0398 0.0326 0.0256 0.0191 0.0133 0.0084 0.0046 0.0022 0.0015

0.0102 0.0089 0.0077 0.0067 0.0059 0.0052 0.0046 0.0043 0.0042 0.0044 0.0050

−0.0273 −0.0234 −0.0192 −0.0149 −0.0106 −0.0063 −0.0022 0.0015 0.0046 0.0071 0.0088

−0.0526 −0.0452 −0.0374 −0.0295 −0.0216 −0.0139 −0.0067 −0.0003 0.0051 0.0092 0.0117

−0.0618 −0.0531 −0.0440 −0.0348 −0.0255 −0.0166 −0.0083 −0.0009 0.0053 0.0100 0.0128

−0.0533 −0.0457 −0.0379 −0.0303 −0.0234 −0.0176 −0.0122 −0.0073 −0.0080 −0.0005 0.0118

−0.0285 −0.0244 −0.0201 −0.0156 −0.0111 −0.0066 −0.0024 0.0014 0.0046 0.0072 0.0089

0.0087 0.0076 0.0066 0.0058 0.0052 0.0047 0.0043 0.0042 0.0043 0.0045 0.0051

0.0530 0.0458 0.0386 0.0316 0.0248 0.0186 0.0129 0.0082 0.0046 0.0023 0.0016

0.0999 0.0861 0.0724 0.0590 0.0460 0.0339 0.0230 0.0136 0.0064 0.0017 0.0000

2.25

−1.0 −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 0.8 1.0

0.0804 0.0694 0.0583 0.0475 0.0371 0.0273 0.0185 0.0110 0.0051 0.0013 0.0000

0.0433 0.0374 0.0315 0.0257 0.0202 0.0150 0.0104 0.0065 0.0035 0.0015 0.0009

0.0081 0.0070 0.0061 0.0052 0.0045 0.0038 0.0033 0.0029 0.0028 0.0028 0.0031

−0.0216 −0.0185 −0.0153 −0.0120 −0.0087 −0.0055 −0.0024 0.0003 0.0026 0.0044 0.0055

−0.0416 −0.0358 −0.0297 −0.0236 −0.0175 −0.0117 −0.0062 −0.0014 0.0026 0.0056 0.0073

−0.0489 −0.0420 −0.0350 −0.0278 −0.0207 −0.0139 −0.0076 −0.0020 0.0027 0.0061 0.0080

−0.0421 −0.0362 −0.0301 −0.0242 −0.0187 −0.0140 −0.0097 −0.0058 −0.0056 −0.0005 0.0073

−0.0225 −0.0193 −0.0160 −0.0126 −0.0091 −0.0058 −0.0026 0.0002 0.0026 0.0044 0.0056

0.0069 0.0060 0.0052 0.0045 0.0039 0.0034 0.0030 0.0028 0.0028 0.0029 0.0032

0.0420 0.0362 0.0305 0.0249 0.0196 0.0146 0.0101 0.0063 0.0034 0.0016 0.0010

0.0790 0.0681 0.0573 0.0467 0.0364 0.0268 0.0182 0.0108 0.0050 0.0013 0.0000

−1.0 −0.8 −0.6 −0.4 −0.2

0.0652 0.0562 0.0473 0.0385 0.0301

0.0351 0.0303 0.0255 0.0208 0.0163

0.0066 0.0057 0.0049 0.0042 0.0035

−0.0175 −0.0150 −0.0125 −0.0099 −0.0072

−0.0338 −0.0290 −0.0242 −0.0193 −0.0145

−0.0396 −0.0341 −0.0284 −0.0227 −0.0171

−0.0342 −0.0294 −0.0245 −0.0197 −0.0153

−0.0183 −0.0157 −0.0130 −0.0103 −0.0076

0.0056 0.0048 0.0042 0.0036 0.0030

0.0340 0.0293 0.0247 0.0202 0.0158

0.0641 0.0552 0.0465 0.0379 0.0295


0.0 0.2 0.4 0.6 0.8 1.0

0.0221 0.0150 0.0089 0.0042 0.0011 0.0000

0.0121 0.0084 0.0052 0.0027 0.0011 0.0006

0.0029 0.0024 0.0021 0.0019 0.0019 0.0020

−0.0047 −0.0024 −0.0003 0.0015 0.0028 0.0036

−0.0098 −0.0055 −0.0018 0.0013 0.0036 0.0048

−0.0117 −0.0067 −0.0023 0.0013 0.0039 0.0052

−0.0114 −0.0079 −0.0047 −0.0041 −0.0005 0.0048

−0.0050 −0.0025 −0.0003 0.0015 0.0028 0.0037

0.0026 0.0022 0.0020 0.0019 0.0019 0.0021

0.0118 0.0081 0.0050 0.0027 0.0011 0.0007

0.0217 0.0147 0.0087 0.0041 0.0011 0.0000

2.75

−1.0 −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 0.8 1.0

0.0539 0.0465 0.0391 0.0319 0.0249 0.0183 0.0124 0.0074 0.0034 0.0009 0.0000

0.0291 0.0251 0.0211 0.0172 0.0135 0.0100 0.0069 0.0042 0.0022 0.0009 0.0004

0.0054 0.0047 0.0040 0.0034 0.0028 0.0023 0.0019 0.0016 0.0014 0.0013 0.0014

−0.0145 −0.0124 −0.0103 −0.0082 −0.0061 −0.0041 −0.0022 −0.0005 0.0009 0.0019 0.0025

−0.0279 −0.0240 −0.0200 −0.0160 −0.0121 −0.0084 −0.0049 −0.0019 0.0006 0.0024 0.0033

−0.0328 −0.0282 −0.0235 −0.0189 −0.0143 −0.0099 −0.0059 −0.0024 0.0005 0.0025 0.0036

−0.0283 −0.0243 −0.0203 −0.0164 −0.0127 −0.0095 −0.0065 −0.0039 −0.0031 −0.0004 0.0033

−0.0151 −0.0130 −0.0108 −0.0086 −0.0064 −0.0043 −0.0023 −0.0006 0.0008 0.0019 0.0025

0.0046 0.0040 0.0034 0.0029 0.0025 0.0020 0.0017 0.0015 0.0013 0.0013 0.0014

0.0281 0.0243 0.0204 0.0167 0.0131 0.0097 0.0067 0.0041 0.0021 0.0009 0.0005

0.0530 0.0457 0.0384 0.0313 0.0244 0.0180 0.0122 0.0072 0.0034 0.0009 0.0000

3.0

−1.0 −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 0.8 1.0

0.0453 0.0391 0.0329 0.0268 0.0209 0.0154 0.0104 0.0062 0.0029 0.0008 0.0000

0.0244 0.0211 0.0177 0.0145 0.0113 0.0084 0.0058 0.0035 0.0018 0.0007 0.0003

0.0046 0.0040 0.0034 0.0028 0.0023 0.0019 0.0015 0.0012 0.0010 0.0009 0.0010

−0.0122 −0.0105 −0.0087 −0.0070 −0.0052 −0.0035 −0.0020 −0.0006 0.0005 0.0013 0.0017

−0.0235 −0.0202 −0.0169 −0.0135 −0.0103 −0.0072 −0.0043 −0.0018 0.0002 0.0016 0.0023

−0.0276 −0.0237 −0.0198 −0.0159 −0.0121 −0.0085 −0.0052 −0.0023 0.0001 0.0017 0.0025

−0.0238 −0.0204 −0.0171 −0.0138 −0.0107 −0.0080 −0.0055 −0.0032 −0.0024 −0.0004 0.0023

−0.0127 −0.0109 −0.0091 −0.0073 −0.0055 −0.0037 −0.0021 −0.0007 0.0005 0.0013 0.0018

0.0039 0.0034 0.0029 0.0024 0.0020 0.0017 0.0014 0.0011 0.0010 0.0009 0.0010

0.0236 0.0204 0.0172 0.0140 0.0110 0.0081 0.0056 0.0034 0.0017 0.0007 0.0003

0.0445 0.0384 0.0323 0.0263 0.0205 0.0151 0.0102 0.0061 0.0028 0.0007 0.0000

2.5


Table 3.6.5 Moment coefficients in horizontal direction (My ). Aspect ratio

H/b

−1.0

−0.8

−0.6

2.0

−1.0 −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 0.8 1.0

0.407 0.351 0.295 0.240 0.187 0.138 0.093 0.055 0.026 0.007 0.000

0.219 0.189 0.159 0.129 0.101 0.075 0.051 0.030 0.014 0.004 0.000

0.041 0.035 0.030 0.024 0.019 0.014 0.010 0.006 0.004 0.002 0.001

−0.109 −0.094 −0.079 −0.064 −0.050 −0.036 −0.024 −0.014 −0.005 0.000 0.002

−0.210 −0.181 −0.152 −0.124 −0.096 −0.070 −0.047 −0.027 −0.011 −0.001 0.003

−0.247 −0.213 −0.179 −0.146 −0.113 −0.083 −0.055 −0.032 −0.013 −0.001 0.003

−0.213 −0.184 −0.154 −0.125 −0.098 −0.071 −0.048 −0.027 −0.011 −0.001 0.003

2.25

−1.0 −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 0.8 1.0

0.3217 0.2774 0.2334 0.1901 0.1483 0.1092 0.0739 0.0439 0.0205 0.0054 0.0000

0.1733 0.1495 0.1258 0.1025 0.0800 0.0589 0.0400 0.0238 0.0112 0.0031 0.0002

0.0324 0.0280 0.0236 0.0193 0.0151 0.0113 0.0078 0.0049 0.0026 0.0012 0.0008

−0.0864 −0.0744 −0.0626 −0.0508 −0.0395 −0.0289 −0.0192 −0.0110 −0.0045 −0.0003 0.0014

−0.1665 −0.1436 −0.1207 −0.0981 −0.0764 −0.0559 −0.0374 −0.0216 −0.0093 −0.0012 0.0018

−0.1955 −0.1686 −0.1417 −0.1153 −0.0897 −0.0657 −0.0440 −0.0255 −0.0110 −0.0016 0.0020

−1.0 −0.8 −0.6 −0.4 −0.2

0.2608 0.2249 0.1892 0.1541 0.1203

0.1405 0.1212 0.1020 0.0831 0.0648

0.0263 0.0227 0.0191 0.0156 0.0122

−0.0700 −0.0604 −0.0507 −0.0412 −0.0321

−0.1350 −0.1164 −0.0979 −0.0796 −0.0620

−0.1585 −0.1367 −0.1149 −0.0935 −0.0728


−0.4

−0.2

0.0

0.2

0.4

0.6

0.8

1.0

−0.114 −0.098 −0.083 −0.067 −0.052 −0.038 −0.025 −0.014 −0.006 0.000 0.002

0.035 0.030 0.025 0.021 0.016 0.012 0.009 0.005 0.003 0.002 0.001

0.212 0.183 0.154 0.125 0.098 0.072 0.049 0.029 0.014 0.004 0.000

0.399 0.344 0.290 0.236 0.184 0.136 0.092 0.055 0.026 0.007 0.000

−0.1686 −0.1454 −0.1222 −0.0994 −0.0773 −0.0566 −0.0379 −0.0219 −0.0094 −0.0012 0.0018

−0.0902 −0.0777 −0.0653 −0.0531 −0.0413 −0.0301 −0.0201 −0.0115 −0.0047 −0.0003 0.0014

0.0275 0.0237 0.0200 0.0163 0.0128 0.0096 0.0067 0.0042 0.0023 0.0011 0.0008

0.1678 0.1447 0.1218 0.0992 0.0774 0.0571 0.0387 0.0231 0.0109 0.0030 0.0003

0.3160 0.2726 0.2293 0.1867 0.1457 0.1073 0.0727 0.0431 0.0202 0.0053 0.0000

−0.1367 −0.1179 −0.0991 −0.0806 −0.0628

−0.0731 −0.0630 −0.0530 −0.0431 −0.0335

0.0223 0.0192 0.0162 0.0132 0.0104

0.1361 0.1173 0.0987 0.0804 0.0628

0.2562 0.2210 0.1859 0.1514 0.1182

0.0 0.2 0.4 0.6 0.8 1.0

0.0885 0.0600 0.0356 0.0167 0.0044 0.0000

0.0478 0.0324 0.0193 0.0091 0.0025 0.0002

0.0091 0.0063 0.0039 0.0020 0.0009 0.0005

−0.0235 −0.0157 −0.0090 −0.0038 −0.0004 0.0009

−0.0454 −0.0305 −0.0177 −0.0078 −0.0012 0.0012

−0.0534 −0.0358 −0.0209 −0.0092 −0.0015 0.0013

−0.0460 −0.0309 −0.0179 −0.0079 −0.0013 0.0012

−0.0245 −0.0164 −0.0094 −0.0040 −0.0004 0.0009

0.0077 0.0054 0.0033 0.0018 0.0008 0.0005

0.0463 0.0314 0.0187 0.0088 0.0024 0.0002

0.0870 0.0589 0.0350 0.0164 0.0043 0.0000

2.75

−1.0 −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 0.8 1.0

0.2157 0.1860 0.1565 0.1274 0.0995 0.0732 0.0496 0.0294 0.0138 0.0036 0.0000

0.1162 0.1002 0.0843 0.0687 0.0536 0.0395 0.0268 0.0159 0.0075 0.0020 0.0001

0.0217 0.0188 0.0158 0.0129 0.0101 0.0075 0.0052 0.0032 0.0016 0.0007 0.0003

−0.0579 −0.0499 −0.0420 −0.0341 −0.0266 −0.0194 −0.0130 −0.0075 −0.0033 −0.0004 0.0006

−0.1116 −0.0963 −0.0809 −0.0659 −0.0513 −0.0376 −0.0253 −0.0148 −0.0065 −0.0012 0.0008

−0.1311 −0.1130 −0.0950 −0.0773 −0.0602 −0.0442 −0.0297 −0.0174 −0.0077 −0.0014 0.0009

−0.1130 −0.0975 −0.0820 −0.0667 −0.0519 −0.0381 −0.0256 −0.0149 −0.0066 −0.0012 0.0008

−0.0604 −0.0521 −0.0438 −0.0356 −0.0277 −0.0203 −0.0136 −0.0079 −0.0034 −0.0005 0.0006

0.0184 0.0159 0.0134 0.0109 0.0086 0.0064 0.0044 0.0027 0.0014 0.0006 0.0004

0.1125 0.0970 0.0816 0.0665 0.0519 0.0382 0.0259 0.0154 0.0073 0.0020 0.0001

0.2119 0.1827 0.1537 0.1252 0.0977 0.0719 0.0487 0.0289 0.0135 0.0036 0.0000

3.0

−1.0 −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 0.8 1.0

0.1813 0.1564 0.1315 0.1071 0.0836 0.0616 0.0417 0.0247 0.0116 0.0030 0.0000

0.0977 0.0843 0.0709 0.0577 0.0451 0.0332 0.0225 0.0134 0.0063 0.0017 0.0001

0.0183 0.0158 0.0133 0.0108 0.0085 0.0063 0.0043 0.0026 0.0013 0.0005 0.0002

−0.0487 −0.0420 −0.0353 −0.0287 −0.0223 −0.0164 −0.0110 −0.0064 −0.0028 −0.0004 0.0004

−0.0939 −0.0809 −0.0681 −0.0554 −0.0431 −0.0317 −0.0213 −0.0125 −0.0056 −0.0011 0.0006

−0.1102 −0.0950 −0.0799 −0.0650 −0.0507 −0.0372 −0.0250 −0.0147 −0.0066 −0.0013 0.0006

−0.0950 −0.0819 −0.0689 −0.0561 −0.0437 −0.0321 −0.0216 −0.0126 −0.0056 −0.0011 0.0006

−0.0508 −0.0438 −0.0368 −0.0300 −0.0233 −0.0171 −0.0115 −0.0067 −0.0029 −0.0005 0.0004

0.0155 0.0134 0.0112 0.0092 0.0072 0.0053 0.0037 0.0023 0.0012 0.0005 0.0003

0.0946 0.0816 0.0686 0.0559 0.0436 0.0321 0.0218 0.0130 0.0061 0.0017 0.0001

0.1781 0.1536 0.1292 0.1053 0.0821 0.0605 0.0410 0.0243 0.0114 0.0030 0.0000

2.5


Table 3.6.6 Shear coefficients in vertical direction (Qz ). Aspect ratio

H/b

−1.0

−0.8

−0.6

−0.4

−0.2

0.0

0.2

0.4

0.6

0.8

1.0

2.0

−1.0 −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 0.8 1.0

0.5596 0.5588 0.5534 0.5393 0.5134 0.4728 0.4157 0.3406 0.2466 0.1331 0.0000

0.3016 0.3006 0.2973 0.2894 0.2751 0.2530 0.2221 0.1815 0.1308 0.0696 −0.0021

0.0564 0.0547 0.0528 0.0502 0.0467 0.0419 0.0357 0.0278 0.0181 0.0066 −0.0068

−0.1502 −0.1529 −0.1539 −0.1521 −0.1468 −0.1371 −0.1227 −0.1031 −0.0781 −0.0478 −0.0121

−0.2897 −0.2931 −0.2935 −0.2890 −0.2777 −0.2583 −0.2300 −0.1918 −0.1435 −0.0850 −0.0161

−0.3401 −0.3438 −0.3441 −0.3385 −0.3251 −0.3023 −0.2688 −0.2240 −0.1673 −0.0985 −0.0176

−0.2933 −0.2967 −0.2972 −0.2925 −0.2811 −0.2615 −0.2327 −0.1941 −0.1452 −0.0859 −0.0162

−0.1569 −0.1595 −0.1605 −0.1586 −0.1530 −0.1429 −0.1278 −0.1073 −0.0812 −0.0496 −0.0123

0.0478 0.0460 0.0441 0.0418 0.0386 0.0344 0.0290 0.0223 0.0141 0.0044 −0.0070

0.2920 0.2910 0.2877 0.2800 0.2662 0.2448 0.2149 0.1756 0.1264 0.0672 −0.0022

0.5498 0.5490 0.5437 0.5299 0.5044 0.4646 0.4085 0.3347 0.2423 0.1308 0.0000

2.25

−1.0 −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 0.8 1.0

0.4428 0.4422 0.4379 0.4267 0.4062 0.3741 0.3289 0.2695 0.1951 0.1053 0.0000

0.2386 0.2380 0.2354 0.2292 0.2179 0.2005 0.1761 0.1440 0.1038 0.0555 −0.0013

0.0446 0.0436 0.0423 0.0404 0.0378 0.0341 0.0292 0.0231 0.0155 0.0064 −0.0043

−0.1189 −0.1205 −0.1209 −0.1192 −0.1147 −0.1068 −0.0952 −0.0796 −0.0598 −0.0358 −0.0076

−0.2292 −0.2313 −0.2311 −0.2270 −0.2177 −0.2022 −0.1795 −0.1492 −0.1109 −0.0646 −0.0100

−0.2691 −0.2714 −0.2710 −0.2661 −0.2551 −0.2367 −0.2100 −0.1744 −0.1295 −0.0750 −0.0110

−0.2321 −0.2341 −0.2339 −0.2298 −0.2204 −0.2046 −0.1817 −0.1510 −0.1123 −0.0653 −0.0101

−0.1241 −0.1258 −0.1261 −0.1243 −0.1195 −0.1113 −0.0992 −0.0829 −0.0623 −0.0372 −0.0077

0.0378 0.0367 0.0354 0.0338 0.0314 0.0282 0.0241 0.0188 0.0123 0.0046 −0.0044

0.2310 0.2304 0.2278 0.2218 0.2109 0.1940 0.1703 0.1393 0.1004 0.0536 −0.0014

0.4351 0.4344 0.4302 0.4193 0.3991 0.3676 0.3232 0.2648 0.1917 0.1035 0.0000

−1.0 −0.8 −0.6 −0.4 −0.2

0.3590 0.3585 0.3550 0.3460 0.3293

0.1935 0.1930 0.1909 0.1859 0.1768

0.0362 0.0355 0.0346 0.0332 0.0311

−0.0964 −0.0974 −0.0975 −0.0959 −0.0921

−0.1858 −0.1871 −0.1867 −0.1831 −0.1754

−0.2182 −0.2196 −0.2189 −0.2147 −0.2055

−0.1882 −0.1895 −0.1890 −0.1854 −0.1775

−0.1006 −0.1017 −0.1017 −0.1000 −0.0960

0.0306 0.0299 0.0290 0.0278 0.0260

0.1873 0.1868 0.1848 0.1800 0.1711

0.3527 0.3522 0.3488 0.3399 0.3236


0.0 0.2 0.4 0.6 0.8 1.0

0.3033 0.2667 0.2185 0.1582 0.0854 0.0000

0.1627 0.1429 0.1169 0.0844 0.0452 −0.0009

0.0282 0.0243 0.0193 0.0132 0.0058 −0.0028

−0.0856 −0.0761 −0.0634 −0.0474 −0.0279 −0.0050

−0.1626 −0.1441 −0.1195 −0.0884 −0.0508 −0.0066

−0.1905 −0.1687 −0.1398 −0.1033 −0.0591 −0.0072

−0.1646 −0.1459 −0.1209 −0.0895 −0.0514 −0.0066

−0.0893 −0.0794 −0.0661 −0.0493 −0.0290 −0.0050

0.0235 0.0201 0.0159 0.0107 0.0044 −0.0029

0.1575 0.1383 0.1131 0.0816 0.0436 −0.0009

0.2980 0.2620 0.2147 0.1554 0.0839 0.0000

2.75

−1.0 −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 0.8 1.0

0.2969 0.2965 0.2936 0.2861 0.2723 0.2508 0.2205 0.1807 0.1308 0.0706 0.0000

0.1600 0.1596 0.1579 0.1538 0.1463 0.1347 0.1183 0.0968 0.0699 0.0375 −0.0006

0.0299 0.0294 0.0288 0.0277 0.0260 0.0237 0.0205 0.0164 0.0113 0.0052 −0.0019

−0.0797 −0.0804 −0.0803 −0.0789 −0.0756 −0.0702 −0.0623 −0.0518 −0.0385 −0.0223 −0.0034

−0.1537 −0.1545 −0.1540 −0.1509 −0.1443 −0.1337 −0.1183 −0.0979 −0.0722 −0.0411 −0.0045

−0.1804 −0.1814 −0.1806 −0.1769 −0.1692 −0.1566 −0.1386 −0.1146 −0.0844 −0.0478 −0.0049

−0.1556 −0.1565 −0.1559 −0.1527 −0.1461 −0.1353 −0.1198 −0.0991 −0.0730 −0.0415 −0.0045

−0.0832 −0.0839 −0.0838 −0.0823 −0.0789 −0.0732 −0.0650 −0.0540 −0.0401 −0.0232 −0.0034

0.0253 0.0248 0.0242 0.0232 0.0218 0.0197 0.0170 0.0135 0.0092 0.0041 −0.0020

0.1549 0.1545 0.1529 0.1489 0.1416 0.1303 0.1145 0.0937 0.0676 0.0362 −0.0006

0.2917 0.2913 0.2884 0.2811 0.2676 0.2465 0.2167 0.1775 0.1285 0.0694 0.0000

3.0

−1.0 −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 0.8 1.0

0.2496 0.2492 0.2468 0.2405 0.2289 0.2109 0.1854 0.1519 0.1100 0.0594 0.0000

0.1345 0.1342 0.1328 0.1294 0.1231 0.1133 0.0995 0.0815 0.0588 0.0316 −0.0004

0.0252 0.0248 0.0243 0.0234 0.0221 0.0201 0.0175 0.0140 0.0097 0.0046 −0.0014

−0.0670 −0.0675 −0.0673 −0.0660 −0.0632 −0.0586 −0.0520 −0.0431 −0.0319 −0.0183 −0.0024

−0.1292 −0.1298 −0.1292 −0.1265 −0.1209 −0.1119 −0.0989 −0.0817 −0.0601 −0.0339 −0.0032

−0.1517 −0.1523 −0.1515 −0.1483 −0.1417 −0.1311 −0.1159 −0.0957 −0.0703 −0.0396 −0.0035

−0.1308 −0.1314 −0.1308 −0.1280 −0.1224 −0.1132 −0.1001 −0.0827 −0.0608 −0.0343 −0.0032

−0.0700 −0.0704 −0.0702 −0.0689 −0.0660 −0.0612 −0.0542 −0.0449 −0.0332 −0.0191 −0.0024

0.0213 0.0209 0.0205 0.0197 0.0185 0.0168 0.0145 0.0116 0.0080 0.0037 −0.0014

0.1302 0.1299 0.1286 0.1252 0.1191 0.1096 0.0963 0.0788 0.0569 0.0305 −0.0004

0.2452 0.2449 0.2425 0.2363 0.2249 0.2072 0.1822 0.1492 0.1080 0.0583 0.0000

2.5


Table 3.6.7 Shear coefficients in horizontal direction (Qy ). Aspect ratio

H/b

−1.0

−0.8

−0.6

−0.4

−0.2

0.0

0.2

0.4

0.6

0.8

1.0

2.0

−1.0 −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 0.8 1.0

−3.7646 −3.2466 −2.7311 −2.2244 −1.7359 −1.2781 −0.8654 −0.5138 −0.2404 −0.0631 0.0000

−3.7055 −3.1947 −2.6847 −2.1820 −1.6961 −1.2394 −0.8265 −0.4733 −0.1968 −0.0148 0.0550

−3.3533 −2.8905 −2.4278 −1.9708 −1.5285 −1.1122 −0.7351 −0.4118 −0.1579 0.0106 0.0776

−2.5797 −2.2235 −1.8670 −1.5146 −1.1732 −0.8517 −0.5603 −0.3101 −0.1133 0.0180 0.0710

−1.4184 −1.2225 −1.0263 −0.8323 −0.6444 −0.4673 −0.3067 −0.1687 −0.0601 0.0125 0.0421

−0.0269 −0.0232 −0.0195 −0.0158 −0.0122 −0.0089 −0.0058 −0.0032 −0.0011 0.0002 0.0008

1.3690 1.1799 0.9906 0.8033 0.6219 0.4510 0.2960 0.1628 0.0579 −0.0121 −0.0407

2.5424 2.1913 1.8399 1.4926 1.1562 0.8393 0.5520 0.3055 0.1114 −0.0180 −0.0703

3.3319 2.8721 2.4122 1.9581 1.5185 1.1048 0.7301 0.4088 0.1564 −0.0112 −0.0778

3.6985 3.1886 2.6796 2.1776 1.6925 1.2366 0.8242 0.4715 0.1953 0.0134 −0.0564

3.7640 3.2460 2.7305 2.2236 1.7351 1.2770 0.8640 0.5121 0.2385 0.0608 −0.0027

2.25

−1.0 −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 0.8 1.0

−3.3510 −2.8899 −2.4311 −1.9800 −1.5452 −1.1377 −0.7703 −0.4573 −0.2140 −0.0562 0.0000

−3.2984 −2.8439 −2.3904 −1.9437 −1.5121 −1.1067 −0.7404 −0.4273 −0.1826 −0.0220 0.0386

−2.9849 −2.5733 −2.1620 −1.7562 −1.3639 −0.9949 −0.6610 −0.3751 −0.1511 −0.0030 0.0545

−2.2963 −1.9795 −1.6627 −1.3500 −1.0474 −0.7627 −0.5048 −0.2839 −0.1104 0.0045 0.0499

−1.2625 −1.0883 −0.9141 −0.7420 −0.5754 −0.4186 −0.2766 −0.1548 −0.0592 0.0043 0.0296

−0.0239 −0.0206 −0.0173 −0.0141 −0.0109 −0.0079 −0.0052 −0.0029 −0.0011 0.0001 0.0006

1.2186 1.0504 0.8822 0.7161 0.5554 0.4040 0.2669 0.1494 0.0571 −0.0042 −0.0286

2.2631 1.9508 1.6386 1.3304 1.0322 0.7516 0.4974 0.2797 0.1087 −0.0047 −0.0494

2.9659 2.5568 2.1482 1.7450 1.3551 0.9884 0.6566 0.3724 0.1497 0.0025 −0.0547

3.2922 2.8385 2.3859 1.9399 1.5090 1.1043 0.7385 0.4259 0.1815 0.0210 −0.0397

3.3505 2.8894 2.4306 1.9794 1.5446 1.1369 0.7693 0.4562 0.2126 0.0546 −0.0019

−1.0 −0.8 −0.6 −0.4 −0.2

−3.0187 −2.6033 −2.1900 −1.7837 −1.3920

−2.9713 −2.5620 −2.1538 −1.7518 −1.3637

−2.6889 −2.3182 −1.9482 −1.5834 −1.2308

−2.0686 −1.7833 −1.4984 −1.2173 −0.9455

−1.1373 −0.9805 −0.8237 −0.6691 −0.5195

−0.0216 −0.0186 −0.0156 −0.0127 −0.0099

1.0978 0.9464 0.7951 0.6458 0.5014

2.0387 1.7575 1.4767 1.1997 0.9318

2.6718 2.3035 1.9357 1.5732 1.2228

2.9658 2.5572 2.1497 1.7484 1.3609

3.0182 2.6029 2.1896 1.7832 1.3915


0.0 0.2 0.4 0.6 0.8 1.0

−1.0249 −0.6939 −0.4120 −0.1928 −0.0506 0.0000

−0.9992 −0.6700 −0.3889 −0.1693 −0.0255 0.0282

−0.8994 −0.5998 −0.3435 −0.1429 −0.0108 0.0398

−0.6900 −0.4587 −0.2608 −0.1057 −0.0033 0.0364

−0.3788 −0.2515 −0.1425 −0.0570 −0.0005 0.0216

−0.0072 −0.0048 −0.0027 −0.0011 0.0000 0.0004

0.3656 0.2427 0.1375 0.0550 0.0004 −0.0209

0.6799 0.4520 0.2569 0.1040 0.0031 −0.0361

0.8936 0.5958 0.3410 0.1417 0.0104 −0.0399

0.9971 0.6684 0.3877 0.1684 0.0248 −0.0289

1.0243 0.6932 0.4111 0.1918 0.0494 −0.0014

2.75

−1.0 −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 0.8 1.0

−2.7460 −2.3682 −1.9922 −1.6226 −1.2663 −0.9323 −0.6313 −0.3748 −0.1754 −0.0461 0.0000

−2.7029 −2.3306 −1.9596 −1.5942 −1.2415 −0.9105 −0.6116 −0.3564 −0.1573 −0.0271 0.0212

−2.4460 −2.1089 −1.7726 −1.4412 −1.1211 −0.8203 −0.5485 −0.3162 −0.1345 −0.0152 0.0299

−1.8817 −1.6223 −1.3634 −1.1081 −0.8614 −0.6296 −0.4199 −0.2406 −0.1003 −0.0079 0.0274

−1.0346 −0.8920 −0.7495 −0.6091 −0.4734 −0.3458 −0.2304 −0.1316 −0.0543 −0.0034 0.0162

−0.0196 −0.0169 −0.0142 −0.0116 −0.0090 −0.0066 −0.0044 −0.0025 −0.0010 −0.0001 0.0003

0.9986 0.8609 0.7235 0.5879 0.4569 0.3337 0.2223 0.1270 0.0524 0.0032 −0.0157

1.8545 1.5989 1.3437 1.0921 0.8489 0.6204 0.4138 0.2371 0.0988 0.0077 −0.0271

2.4304 2.0955 1.7613 1.4320 1.1138 0.8150 0.5449 0.3140 0.1334 0.0149 −0.0300

2.6979 2.3263 1.9558 1.5911 1.2391 0.9086 0.6102 0.3553 0.1565 0.0265 −0.0218

2.7456 2.3678 1.9918 1.6222 1.2658 0.9318 0.6307 0.3741 0.1746 0.0452 −0.0010

3.0

−1.0 −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 0.8 1.0

−2.5184 −2.1718 −1.8270 −1.4880 −1.1613 −0.8550 −0.5789 −0.3437 −0.1608 −0.0422 0.0000

−2.4788 −2.1375 −1.7973 −1.4624 −1.1393 −0.8361 −0.5623 −0.3287 −0.1465 −0.0275 0.0163

−2.2432 −1.9342 −1.6259 −1.3223 −1.0291 −0.7538 −0.5050 −0.2925 −0.1265 −0.0177 0.0231

−1.7257 −1.4879 −1.2506 −1.0168 −0.7909 −0.5787 −0.3870 −0.2231 −0.0949 −0.0107 0.0211

−0.9488 −0.8181 −0.6876 −0.5589 −0.4347 −0.3179 −0.2124 −0.1221 −0.0515 −0.0051 0.0125

−0.0180 −0.0155 −0.0130 −0.0106 −0.0082 −0.0060 −0.0040 −0.0023 −0.0010 −0.0001 0.0002

0.9158 0.7896 0.6636 0.5395 0.4195 0.3068 0.2050 0.1178 0.0497 0.0049 −0.0121

1.7008 1.4664 1.2325 1.0021 0.7795 0.5703 0.3813 0.2198 0.0935 0.0105 −0.0209

2.2289 1.9218 1.6155 1.3138 1.0225 0.7489 0.5017 0.2905 0.1256 0.0174 −0.0231

2.4742 2.1334 1.7939 1.4596 1.1371 0.8343 0.5610 0.3278 0.1459 0.0270 −0.0168

2.5180 2.1715 1.8267 1.4877 1.1609 0.8546 0.5785 0.3432 0.1602 0.0415 −0.0008

2.5



It will be observed from above that these coefficients are a function of the aspect ratio of the slab. Normally the ratio of H/b for counterforts varies between 2 to 3. We furnish below the coefficients for moments and shears for various values between 2 to 3 in increment of 0.2547 . We now explain the above with a typical numerical problem.

Example 3.6.2 A counter fort retaining wall of height 7.5 m has counter forts spaced at 3.0 m. The average thickness of the wall is 300 m having RCC grade as M25. The wall is resting on hard soil in zone IV as per IS-1893-2002. Unit weight of the backfill soil is 22 kN/m3 having friction angle of 28◦ . The soil is sloped to the horizontal plane at angle I = 15o . Consider, Econc = 2.85 × 107 kN/m2 and Poisson’s ratio of concrete as 0.3. Determine the time period of the wall and find out the horizontal dynamic moments and shears in the wall under active earth pressure? Solution: Here we have, φ = 28◦ ; i = 15◦ ➔ α = 45 + 0.5 × 28 = 59◦ . α sin i cos 59 sin 15 Considering, ζ = cos = 0.192, for the problem. sin(α−i) = sin 44 T = 2π = 2π

12γ H 5 (1 + ζ )(1 − ν 2 ) + X2 r2 + X3 νr2 + X4 )

8Et 3 g tan α(X1 r4

12 × 22 × 7.55 (1 + 0.192)(1 − 0.32 ) 8 × 2.85 × 107 × 0.33 × 9.81 tan 59(75.3 × 2.54 − 3.25 × 2.52 + 17.68 × 0.3 × 2.52 + 1.72)

= 0.03 sec

Thus, Sga = 1 + 15T = 1.45, for hard soil; here, β = In the horizontal direction

M(y) = Coeffy ×

0.423βγ Sa (1 + ζ )H 3 8g tan α

ZI 2R

and Q(y) = Coeffy ×

=

0.24×1 4

= 0.06.

0.423βγ Sa (1 + ζ )H 2 8g tan α

M(y) =

0.423 × 0.06 × 22 × 1.45 × 7.53 × 1.192 Coeff(y) = 30.58 × Coeff(y) 8 × 9.81 × tan 59

Q(y) =

0.423 × 0.06 × 22 × 1.45 × 7.52 × 1.192 Coeff(y) = 4.077 × Coeff(y) 8 × 9.81 × tan 59

47 Intermediate values may be linearly interpolated.



Thus, multiplying by the appropriate coefficient, we can have My and Qy as given hereunder Horizontal Moment M y H/b

−1.0

−0.8

−0.6

−0.4

−0.2

0.0

0.2

0.4

0.6

0.8

1.0

−1.0 −0.8 −0.6 −0.4 −0.2 0.0 0.2 0.4 0.6 0.8 1.0

7.975 6.878 5.786 4.712 3.678 2.708 1.833 1.088 0.509 0.134 0.000

4.298 3.706 3.118 2.540 1.983 1.461 0.990 0.589 0.278 0.076 0.005

0.804 0.694 0.584 0.477 0.374 0.278 0.192 0.119 0.063 0.027 0.016

−2.141 −1.846 −1.551 −1.261 −0.981 −0.718 −0.479 −0.276 −0.117 −0.012 0.028

−4.128 −3.560 −2.993 −2.434 −1.895 −1.389 −0.932 −0.542 −0.237 −0.038 0.037

−4.847 −4.180 −3.514 −2.859 −2.226 −1.632 −1.096 −0.638 −0.281 −0.047 0.040

−4.180 −3.604 −3.030 −2.465 −1.919 −1.407 −0.944 −0.549 −0.240 −0.038 0.037

−2.235 −1.927 −1.620 −1.317 −1.024 −0.749 −0.501 −0.289 −0.122 −0.013 0.028

0.681 0.587 0.495 0.404 0.318 0.237 0.164 0.102 0.055 0.025 0.016

4.161 3.588 3.019 2.459 1.920 1.414 0.959 0.571 0.269 0.074 0.005

7.836 6.757 5.685 4.630 3.613 2.660 1.801 1.069 0.500 0.131 0.000

Horizontal Shear Qy H/b −1.0

−0.8

−0.6

−0.4

−0.2

0.0

0.2

0.4

0.6

0.8

1.0

−1.0 −12.307 −12.114 −10.963 −8.434 −4.637 −0.088 4.476 8.312 10.893 12.091 12.305 −0.8 −10.614 −10.445 −9.451 −7.271 −3.997 −0.076 3.858 7.165 9.391 10.426 10.612 −0.6 −8.929 −8.781 −7.943 −6.109 −3.358 −0.064 3.241 6.020 7.892 8.764 8.927 −0.4 −7.272 −7.142 −6.455 −4.963 −2.728 −0.052 2.633 4.891 6.414 7.128 7.270 −0.2 −5.675 −5.560 −5.018 −3.855 −2.118 −0.040 2.044 3.799 4.986 5.549 5.673 0.0 −4.179 −4.074 −3.667 −2.813 −1.544 −0.029 1.491 2.772 3.643 4.065 4.176 0.2 −2.829 −2.732 −2.445 −1.870 −1.025 −0.019 0.990 1.843 2.429 2.725 2.826 0.4 −1.680 −1.585 −1.400 −1.063 −0.581 −0.011 0.561 1.047 1.390 1.581 1.676 0.6 −0.786 −0.690 −0.582 −0.431 −0.232 −0.004 0.224 0.424 0.578 0.687 0.782 0.8 −0.206 −0.104 −0.044 −0.013 −0.002 0.000 0.002 0.013 0.042 0.101 0.202 1.0 0.000 0.115 0.162 0.149 0.088 0.002 −0.085 −0.147 −0.163 −0.118 −0.006

3.7 UNYIELDING EARTH RETAINING STRUCTURES

3.7.1 Earthquake Analysis of rigid walls when the soil does not yield In many cases when a retaining wall is sufficiently rigid, or a basement wall of a building or an underground tank is restrained at top by a rigid slab, the wall becomes unyielding as such the triangular profile of failure which usually generates during the static case does not take place. During such case (under static loading) we design it for a condition of earth pressure at rest. Now we have a catch! Since the soil is not under incipient failed condition, the assumption made for dynamic analysis of cantilever and counterfort retaining wall as shown previously is not valid in this case (where we ignored the stiffness of the soil) and a completely different approach has to be adopted to obtain its dynamic response. The major problem it boils down to is – what will be the dynamic pressure induced on the wall in such case? © 2009 Taylor & Francis Group, London, UK


Building a

H

z

Basement BedRock x

Figure 3.7.1 Schematic sketch of building with basement wall.

A procedure is proposed herein for evaluation of the same (Chowdhury and Dasgupta 2007). Shown in Figure 3.7.1, is a typical basement of a tall building (usually used for car parking and sundry building services). The objective is to find the dynamic pressure on the wall of the basement due to earthquake waves propagating through the soil medium. Normally for deep basement wall there will be built in intermediate floor slabs, which make the wall sufficiently rigid due to which the soil in contact with the wall does not yield and a plain strain condition prevails. The soil medium and the basement is assumed to be resting on stiff soil considered as the bedrock – the level from which the ground acceleration propagates. Let the depth of the basement is H and the soil medium in horizontal direction has been considered having a finite dimension a, where lim a → ∞, is a reality. Having defined the problem with basic conditions we can argue that for shear waves propagating through the soil medium the wave propagation equation can be represented by ∂ 2 u(x, z, t) ∂ 2 u(x, z, t) 1 ∂ 2 u(x, z, t) + = ∂x2 ∂z2 Vs2 ∂t 2

(3.7.1)

where Vs = shear wave velocity of the soil medium; u(x, z, t) = the displacement function and can be considered as u = H(x)Q(z)G(t); three independent functions. Without getting into the details of theory of partial differential equation it can be shown that Equation (3.7.1) can be broken up into three ordinary differential equations of second order, given by (Kreyszig 2001) d2G + λ2 G = 0, dt 2

where, λ = vs · i


where i is a constant.

(3.7.2)


d 2 H(x) + k2 H(x) = 0, where, k is another constant. dx2 d 2 Q(z) + p2 Q(z) = 0, where, p2 = i2 − k2 . dz2

(3.7.3) (3.7.4)

The general solution of Eqns. (3.7.3) and (3.7.4) are given by H(x) = A cos kx + B sin kx : Q(z) = C cos pz + D sin pz

(3.7.5)

We now impose the following boundary condition: At x = 0, u = 0, or H(x) = 0, which implies A = 0. At x = a (where a is very large), u = 0 or H(a) = 0 which implies, H(a) = B sin ka = 0, implying, k = mπ/a. Thus,

Hm (x) = sin

(3.7.6) mπ x a

(3.7.7)

At the free surface i.e. the superstructure base interface we have, dQ(z) At z = 0, shear strain ∂u ∂z = 0 or dz = 0, which implies D = 0. At z = H, displacement, u = 0, i.e., Q(H) = 0 ➔ C cos pH = 0 cos pH = 0, which implies, p=

(2n − 1)π . 2H Q(z) = cos

and hence,

(3.7.8) (2n − 1)πz 2H

(3.7.9)

Thus, the eigenvector of the problem can be established as φ(x, z) = H(x)Q(z) = sin

mπ x (2n − 1)π z cos a 2H

where, m, n = 1, 2, 3 . . .

For calculation of eigen values, we have, p2 = i2 − k2 and λ = vs · i thus,

p2 =

λ2 − k2 Vs2

or λ = vs p2 + k2 .

Substituting the value of p and k from Eqns. (3.7.8) and (3.7.6) we have λ = vs π

m2 (2n − 1)2 + . a2 4H 2


(3.7.10)


For the fundamental mode, considering m, n = 1 and, lim a → ∞, we have λ = ω = vs π 0 +

1 vs π = . 2 2H 4H

(3.7.11)

4H Considering, T = 2π ω , we have, T = vs which is basically the free field time period in one dimension for the site. For lim a → ∞, we drop the first term of eigen function (in the x direction) in Equation (3.7.10) to determine the displacement and pressure at wall face and consider the eigen function as only

ϕ(z) = cos

(2n − 1)π z 2H

(3.7.12)

Based on modal response technique the maximum amplitude function can be defined by Sd = Sa /ω2 where Sd = maximum displacement, and Sa = acceleration which is the function of time period, 4H/vs , and can be read off from the normalized response given in the code. ZI Considering, β = 2R , the code factor, we can write, u(z) = κi β ωSa2 φ(z). Now substituting the value of ω from Equation (3.7.10), we have

u(z) =

4 Sa H 2 πz 4 S a γs H 2 πz κ β cos = κ β cos i i 2 2 2 2H Gg 2H π vs π

(3.7.13)

where G = ρvs2 ; γs = unit weight g = acceleration due to gravity, and κi = modal of soil; mass participation factor = mi φi / mi φi2 . The modal participation factor can thus be considered as κi =

mi φi mi φi2

H =

πz γ z cos dz 2H

0

6 H γ z cos2 0

The above, on integration by parts gives, κi = © 2009 Taylor & Francis Group, London, UK

πz dz 2H

8 π+2 .

(3.7.14)


The strain within the soil body is given by εxx =

∂u = 0; ∂x

➔ εzz = −

εzz =

∂u ∂z

πz 2 S a γs H which gives, εzz = − κi β sin π Gg 2H

16 S a γs H πz β sin . π(π + 2) Gg 2H

(3.7.15)

The constitutive stress-strain relationship under plain strain condition is given by ⎧ ⎫ ⎡ 1−ν ⎨σxx ⎬ 2G ⎣ ν σzz = ⎩ ⎭ (1 − 2ν) τxz 0 Thus,

σxx =

ν 1−ν 0

⎤⎧ ⎫ 0 ⎨εxx ⎬ 0 ⎦ εzz 1−2ν ⎩γ ⎭ xz 2

2G(1 − ν) 2Gν εxx + εzz 1 − 2ν 1 − 2ν

As εxx = 0, in this case we have, σxx = on the wall as pdyn = −

(3.7.16)

2Gν 1−2ν εzz ,

which gives the dynamic pressure

16 S a γs H πz 2ν β sin π(π + 2) 1 − 2ν g 2H

(3.7.17)

Where, negative sign indicates that the pressure is acting in the direction of the wall. ➔

pdyn = −

32 S a γs H πz ψν β sin π(π + 2) g 2H

where ψv =

ν 1 − 2ν

(3.7.18)

The above can be further simplified to pdyn (z) = −Coeffν β

Sa γ s H , g

(3.7.19)

where the coefficients may be read off for different values of Poisson’s ratio (0.25, 0.3 and 0.4) as shown in Figure 3.7.2.

3.7.2 Ostadan’s method Ostadan (2004) conducted extensive study on the subject based on analysis carried out in SASSI 2000 (Lysmer et al. 1999) with various soil properties and Poisson’s ratio value and came with a normalized dynamic pressure coefficient curve given by the expression p(z) = −0.0015 + 5.05z − 15.84z2 + 28.25z3 − 24.59z4 + 8.14z5 .

(3.7.20)

The above normalized pressure coefficient when compared with the analytical solution proposed in Equation 3.7.19 gives quite closely matched value as shown in Figure 3.7.3. © 2009 Taylor & Francis Group, London, UK


Variation of dynamic coefficient of pressure with depth

Pressure coefficient

1.4 1.2 1 n=0.25

0.8

n=0.3 0.6

n=0.4

0.4 0.2 0 0

0.1 0.2

0.3 0.4 0.5 0.6

0.7 0.8

0.9

1

Z/H

Figure 3.7.2 Variation of dynamic pressure coefficient with depth.

1 0.8

Closed form Solution

0.6 0.4 0.2

1

8 0.

6 0.

4 0.

0.

2

0 -0.2

0

Pressure coefficient

1.2

Ostadan's curve

Z/H

Figure 3.7.3 Comparison of normalized pressure on the wall. (Proposed & Ostadan)

Based on the above equation, Ostadan proposed a simplified method for determination of dynamic pressure on such unyielding walls whose steps are as mentioned hereunder: •

Perform free field soil column analysis and obtain the ground response at the depth corresponding to the base of the wall in the free field. The response motion in terms of acceleration response spectrum at 30% damping should be obtained. The free field soil column analysis may be performed using Computer program like SHAKE with input motion specified either at the ground surface or at the depth of foundation base-mat. The choice for location of control motion is an important


Dynamic Pressure (psf)


2000 1500

Closed form

1000 500 0 -500

Simplified Method 0

3

6

9

12

15

18

21

24

27

30

Height in feet

Figure 3.7.4 Comparison of dynamic pressure for the 30 ft basement wall with Vs = 1000 ft/sec.

• • • •

decision that needs to be made consistent with development of the design motion. The location of input motion may significantly affect the dynamic response of the building and seismic soil pressure amplitude. Obtain total mass of the soil body, m = 0.5 × ρ × H 2 × ψv , for the present case ψv = 2/[(1 − ν)(2 − ν)]0.5 , here ρ is mass density of the soil and ν is Poisson’s ratio. Obtain the lateral seismic force from the product of the total mass obtained above and the acceleration value of the free field response at the soil column obtained at the depth of the bottom of the wall. Obtain the maximum seismic soil pressure at the ground surface level by dividing the lateral force obtained above by a factor 0.744 H (which actually the area under curve for the equation furnished by him as mentioned above). Obtain the pressure profile by multiplying the peak pressure from the above step by the pressure distribution relationship as furnished earlier.

Ostadan’s method has been compared with analytical method proposed here48 for a basement wall which is 30 feet deep having shear wave velocity of soil as 1000 ft/sec. The density of soil considered as 125 lbs/ft3 . The wall is considered to be in a zone subjected to severe earthquake where Z = 0.24, I = 1.2 and R = 2.0. The out come of the results are shown in Figure 3.7.4. It will be observed that variation is not too wide and well within the acceptable limit of civil engineering design. 3.8 EARTHQUAKE ANALYSIS OF WATER TANKS

3.8.1 Analysis of water tanks under earthquake force Water tanks resting on ground, underground or on staging at a height form an important part of infrastructure and township development. This is shown in Figure 3.8.1.

48 Personal communications with Dr. Ignacio Arango and Dr. Farhang Ostadan April 2005.



Figure 3.8.1 Typical overhead water tank modeled as an inverted pendulum.

Even in industrial sector like power, petrochemical industry it forms an important ingredient of process engineering or fire fighting. As such in post earthquake scenario many of the tanks storing water it becomes essential that they remain functional with nominal damage. In India water tanks resting on ground were usually given the reprieve of any earthquake analysis not many years ago49 . While overhead tanks were mostly treated as an inverted pendulum where the time period is derived from the expression T = 2π

stat g

where, stat =

W K

(3.8.1)

where, W = weight of water in tank + weight of tank + 1/3rd wt of staging frame, and, K = lateral stiffness of the frame. And water tanks both resting on ground as well as overhead have found to have undergone damage during earthquakes in India and abroad. In India the damage is possibly due to the fact that the IS-code recommendation on the pressure induced on the tank wall is incomplete. IS-1893-1984 for instance only suggests the pressure coefficients for impulsive force and does not cater to the convective or hydro-dynamic sloshing force, whose effect could be significant, especially if the tank is shallow. Nor does the code have any provision for calculation of the sloshing height for which if sufficient free board is not kept could result in damage of the roof slab.

49 For most of the case it was observed that impulsive force suggested by IS-1893 1984 provided with forces of small magnitude unless and until the tank was situated in a place where earthquake has a severe intensity.



Sloshed liquid Sloshing Mass

h Impulsive Mass Sa

D

Figure 3.8.2 Typical cylindrical tank with liquid divided into impulsive and sloshing mass.

Connected by flexible spring to wall

hsl

Connected by rigid link to wall hi

Figure 3.8.3 Typical mathematical model of water tank with liquid impulsive and sloshing mass.

It is unfortunate that code committee did not update this previously though superior mathematical model for analysis of such liquid retaining vessel has been in existence since 1957 (Housner 1957). Before we delve into the details of such mathematical model for analysis of such water tanks, let us see how the fluid behaves, when acceleration is induced at the base of such tanks. Let us consider a cylindrical vessel of diameter D containing liquid of height h (Figure 3.8.2). When the vessel is subjected to an acceleration Sa at its base a part of the liquid (called the impulsive mass) moves along with the container as a rigid body at the bottom of the tank. Balance mass at the top of the tank acts in more flexible manner and induces a convective or a sloshing force on the tank wall. What part of the mass would act as an impulsive mass and sloshing mass depends50 upon the aspect ratio h/D. In simplistic mechanical analogy the above can be represented as shown in Figure 3.8.3. Haroun and Housner (1981) derived the values of sloshing and impulsive mass considering the wall of the tank as rigid and later derived the same for flexible wall.

50 The total impulsive and sloshing mass constitutes the full mass of liquid in the tank.



We will not go into details of the derivation of the same from the fundamentals, which are given elsewhere (Housner 1963) but will deal with final results only, both for circular and rectangular tanks which can be directly used for computation of pressure in a tank either manually or through a computer. In Table 3.8.1, Table 3.8.1 Design parameters for dynamic analysis of tanks with fluids. Sl. No.

1

2

3

4

5

6

7

8

9 10

Rectangular tank √ L tan h 3 Wi H = √ L W 3 H 3 hi = H (Excluding base pressure) 8 √ L 3 1 hi − H = √ L H 8 2 tan h 3 H

Wi = W

√ R 3 H √ R 3 H

tan h

3 H (Excluding base pressure) 8 √ R 3 hi 1 H − = √ R H 8 2 tan h 3 H

hi =

(Including base pressure) Wi Pi = Sa0 g L H Wsl = 0.527 tan h 1.58 W H L H cos h 1.58 −1 hsl L =1− H H H 1.58 sin h 1.58 L L (Excluding base pressure) H cos h 1.58 −2 hsl L =1− H H H 1.58 sin h 1.58 L L (Including base pressure) H 1.58 KS = tan h 1.58 Wsl L L H Sa θh = 1.58 2 tan h 1.58 ω L L

(Including base pressure) Wi Pi = Sa0 g Wsl R H = 0.318 tan h 1.84 W H R H cos h 1.84 −1 hsl R =1− H H H 1.84 sin h 1.84 R R (Excluding base pressure) H cos h 1.84 − 2.01 hsl R =1− H H H 1.84 sin h 1.84 R R (Including base pressure) H 1.84 tan h 1.84 KS = Wsl R R H Sa θh = 1.534 2 tan h 1.84 ω R R

Psl = Wsl θh sin ωt

Psl = 1.2Wsl θh sin ωt

0.527L cot h 1.58 11

Circular tank

dmax =


g −1 ω2 θh L

H L

dmax

H 0.408R cot h 1.84 R = g −1 ω2 θh R


W = Total weight of liquid in the tank; Wi = Weight of impulsive fluid assumed to fastened rigidly with the tank wall at a height hi above the tank bottom; hi = Height from the bottom of the tank where the impulsive force acts on the wall; Pi = Impulsive force which acts on the tank wall; Sao = Acceleration at zero time period to be considered = g (9.81 m/sec2 ) as per IS-1893 2002; Wsl = Sloshing weight of the fluid oscillating at the top part of the tank; hsl = Height from the bottom of the tank where the sloshing force acts on the wall; Ks = Equivalent spring stiffness of the sloshing fluid; θsl = Angular amplitude of free oscillation at the free fluid surface; Psl = Sloshing force acting on the tank wall; dmax = Sloshing Height of the liquid at free surface; L = Half width of rectangular tank of width 2 L; R = Radius of circular tank; H = Height of fluid in the tank; ω = natural frequency of the sloshing mass =

Ks g Wsl .

The total moment acting on the base of the tank (when base pressure is included) and on the wall (when base pressure is excluded) is given by M = Pi × hi + Psl × hsl

(3.8.2)

Above is the original form in which Housner presented a solution to the problem and has been the backbone of further research on this topic for next 30 years. The basic assumption in Housner’s hypothesis which may not be always true (though a conservative estimate) was the impulsive time period considered to be zero in his analysis. The assumption was justified in the above case for Housner assumed the tank to be infinitely rigid but in reality the tank could also be flexible when the time period may have a finite value (albeit low compared to the sloshing time period). Further researches by Veletsos & Young (1977) and Veletsos (1984) have defined the impulsive time period when the wall is not rigid.

3.8.2 Impulsive time period for non rigid walls Codes from different countries use different formulas for derivation of the time period. We present here the most practical one which is easy to apply and amenable to electronic computation. Eurocode 8 (Part IV, 1998) suggests the formula proposed by Veletsos for circular tank, which is given by 2 2R ρH H H − 0.15 where, Ci = 0.01675 + 0.46 (3.8.3) Ti = Ci E w t R R Here R = radius of the tank; H = height of the fluid in tank; ρ = mass density of the fluid; Ew = Young’s modulus of the wall, and t = thickness of the wall. [See Figure 3.8.4].



we

P

Hw hˆ

Figure 3.8.4 Load deflection diagram of tank wall.

For rectangular tank, formula proposed in Eurocode 8 is given by Ti = 2π d/g

(3.8.4)

where d is deflection of the wall due to an equivalent uniformly distributed load of we =

0.5W i + Ww BH

(3.8.5)

where, Wi = weight of the impulsive fluid; Ww = weight of wall perpendicular to the direction of the earthquake force; B = inside width of the tank; H = height of fluid, and we = equivalent udl acting on the wall. To calculate d, following steps may be considered, c.g. of the u.d.l., we may be calculated from the expression 0.5W i hi + Ww Hw hˆ = 0.5W i + Ww

(3.8.6)

In which, Hw = height of the wall, An equivalent concentrated load P may be calculated from the expression, P = we H, where H = height of fluid in the tank. The deflection at the level of fluid surface is thus given by

δ=

ˆ 3 P(h) 3EI

where, I =


t3 12

(3.8.7)


It should be noted that the above is valid for tank walls which are free at top and have an aspect ratio L/H > 2.0 i.e. it behaves as a one way slab. This may not be valid for walls with other boundary conditions but is what is in vogue at present.

3.8.3 Sloshing time period of the vibrating f luid The sloshing time period of the vibrating fluid for a circular tank may be obtained from the expression

Tsl = 2π

D/g

3.68 tan h 3.68H D

(3.8.8)

where, D = diameter of the tank. Similarly for rectangular tank,

Tsl = 2π

2L/g

3.16 tan h 3.16H 2L

(3.8.9)

where, 2L = inside length of tank parallel to the direction of the earthquake.

3.8.4 Calculation of horizontal seismic force for tank resting on ground Once the time-period for the impulsive and sloshing modes is known the corresponding accelerations may be obtained from the chart as furnished in the IS-code. The seismic coefficient may then be obtained from expression Ahi =

ZISai 2Rg

(3.8.10)

for impulsive force, and [For RCC tank damping factor considered is usually 5% and for steel tank this is considered as 2%, as such corresponding damping factor from the code should be read]. Ahsl =

ZISasl , 2Rg

(3.8.11)

for sloshing force where damping considered for fluid is normally 0.5%. Here, Z = zone factor as explained earlier; I = importance factor, and R = ductility factor. The new IS-1893 (2002) is yet to arrive at the importance factor to be recommended for liquid retaining structures; in absence of such data recommendations as followed in UBC 97 may be followed: For non important tanks consider, I = 1.0; © 2009 Taylor & Francis Group, London, UK


For tanks supplying water to public community or meant for fire fighting in important industry like power plant or petrochemical plants or containing liquid having nominal hazard I = 1.25; For tanks containing hazardous or toxic liquid higher importance factor between I = 1.5 to 1.75 may be considered. The assessment of ductility factor is far more complex as the basis of Sa /g as given in UBC and that IS-1893 does not correspond one to one and a comparison of base shear for similar structures has to be seen, which is obviously a topic of research. In absence of such data presently, a conservative value of R between 1.5 and 2 may be used for tanks resting on the ground. For overhead tanks similarly, a value of R = 1.2–1.3 may be used for non ductile detailing and R = 2–2.2 may be used for tank frames with ductile detailing.

3.8.5 Calculation of base shear for tanks resting on ground Once the seismic coefficients are known, the impulsive and sloshing shear force may be obtained from Vi = Ahi (W i + Ww + W r )

Vsl = Ahsl Wsl

and

(3.8.12)

where, Wi = weight of sloshing fluid; Ww = weight of wall, and Wr = weight of roof coming on the wall. The resultant shear may either be considered as V = Vi + Vsl (Absolute sum)52

➔

V=

Vi2 + Vsl2 (SRSS value).

(3.8.13)

3.8.6 Calculation of bending moment on the tank wall resting on the ground The bending moment for impulsive and sloshing force at the base of the wall is given by Mi = Ahi (Wi hi + Ww Hw + W r Hr )

and

Msl = Ahsl Wsl hsl

(3.8.14)

where, hi = height of impulsive mass from the bottom of the tank as explained earlier; Hw = height of the c.g. of the wall mass; Hr = height of the c.g. of the roof mass, and Hsl = height of the sloshing mass from the bottom of the tank. In this case the height hi and hsl shall be calculated for the case “Excluding base pressure”. For moment in the base of the tank (i.e. the pressure induced in the soil) the same expression as above may be used with exception that the height hi and hsl shall be calculated for the case “Including base pressure”.

52 This is the recommendation of Euro code 8.



The resultant Moment can now be obtained based on expression M = Mi + Msl (Absolute sum) ➔

M=

Mi2 + Msl2 (SRSS value)

(3.8.15)

3.8.7 Calculation of sloshing height In absence of any recommendation form IS code presently sloshing height may be calculated based on Euro-code model δ = 0.84 × Ahsl × L for rectangular tank, δ = 0.84 × Ahsl × radius of tank for circular tank.

(3.8.16)

We now further elaborate the problem based on a suitable numerical example.

Example 3.8.1 A rectangular RCC fire water tank (Figure 3.8.5) is resting on ground having a size of 7.5 m × 7.5 m × 6.5 m is constructed in a refinery site which is classified as zone IV as per IS-1893 2002. The average thickness of wall is considered to be 450 mm. Grade of concrete used for constructing the tank is M30. The tank is covered by a roof slab which is simply supported on the four walls having thickness of 200 mm. Nature of ground on which it is resting is considered hard. Calculate the seismic force at the wall base and on the foundation.

500

6000

Figure 3.8.5 Elevation and plan view of the water tank with typical wall slab detail resting on ground.



Solution: Weight of water in tank = 7.5 × 7.5 × 6 × 10 = 3375 kN; Wt. of roof slab = 8 × 8 × 0.2 × 25 = 320 kN, and Wt. of one wall = 7.5 × 6.5 × 0.45 × 25 = 548 kN. Here L = 0.5 × 7.5 = 3.75 m, and H = 6.0 m. Based on Housner’s expressions, impulsive mass is given by, √

L tan h 3 H Wi = W, √ L 3H

√

tan h 3 3.75 6 = × 3375 = 2475.8 kN √ 3.75 3 6

⎡ ⎤ √ L 3 1 3 H √ − ⎦ = 3.342 m (IBP) hi = H = 2.25 m (EBP)53 = H ⎣ 8 8 2 tanh 3L H

Sloshing mass is given by L H Wsl = 0.527 tan h 1.58 W = 1097.6 kN. H L

cos h 1.58 H L −1

H = 3.977 m (EBP); hsl = 1 − H 1.58 H sin h 1.58 L L

cos h 1.58 H L −2

H = 4.358 m (IBP) hsl = 1 − H 1.58 H sin h 1.58 L L Equivalent weight, we = C.G. of the load, hˆ =

0.5Wi +Ww BH

0.5Wi hi +Ww Hw 0.5Wi +Ww

= =

0.5×2476+548 7.5×6

= 39.7 kN/m2 .

0.5×2476×2.25+548×3.25 0.5×2476 + 548

= 2.56 m.

Equivalent concentrated force P, P = we × H × 1 = 39.7 × 6 × 1 = 238 kN.

I=

t3 1 × (0.45)3 = = 7.59375 × 10−3 m4 ; Econc = 3.122 × 107 kN/m2 . 12 12

d=

ˆ 3 P(h) 238 × (2.56)3 = 5.614 × 10−3 m. = 3EI 3 × 3.122 × 107 × 7.5938 × 10−3

53 Here EBP means excluding base pressure and IBP = Including base pressure.



Calculation of impulsive time period

Ti = 2π

d = 2π g

5.614 × 10−3 = 0.150 sec 9.81

For T = 0.150 sec, Sa /g = 2.5 as per IS-1893. Calculation of sloshing time period Considering, 2L = 7.5 m; H = 6 m; Tsl = 2π √

2L g

3.16 tan h( 3.16H ) 2L

= 3.110 secs

which gives, Sa /g = 0.321, for the sloshing mode with 5% damping and it needs to be multiplied by a factor of 2.98 to convert it into an equivalent acceleration with 0.5% damping for the fluid. For zone IV consider Z = 0.24 and I = 1.25, thus Ahi =

ZISai , 2Rg

for impulsive force

0.24 × 1.25 × 2.5 = 0.1875, 2×2 ZISasl = , for sloshing mode 2Rg =

Ahsl

➔ Ahsl =

and

0.24 × 1.25 × 0.321 × 2.98 = 0.072 2×2

(Mulitpiled by a factor 2.98 to cater to 0.5% of damping for fluid). Calculation of Base Shear Weight of roof = 320 kN, so assume 1/4th of the weight coming on each wall. Hence, Vi = Ahi (Wi + Ww + Wr ) = 0.1875 × (2476 + 548 +

1 × 320) = 582 kN, 4

for impulsive force. Vsl = Ahsl Wsl = 0.072 × 1098 = 79 kN

for sloshing force.

Thus resultant shear is given by, V = Vi2 + Vsl2 = 587 kN. Calculation of Bending Moment at base of wall Mi = Ahi (Wi hi + Ww Hw + W r Hr ) = 0.1875 × (2476 × 2.25 + 548 × 3.25 + 80 × 6.6) = 1478 kN · m, for impulsive mode Msl = Ahsl Wsl hsl = 0.072 × 1098 × 3.977 = 314.40 kN · m, for sloshing mode. Resultant Moment, M = Mi2 + Msl2 = 1511 kN · m. Calculation of Bending Moment at foundation © 2009 Taylor & Francis Group, London, UK


Mi = Ahi (Wi hi + Ww Hw + Wr Hr ) = 0.1875 × (2476 × 3.342 + 548 × 3.25 + 80 × 6.6) = 1984 kN · m, for the impulsive mode. Msl = Ahsl Wsl hsl = 0.072 × 1098 × 4.358 = 344.52 kN · m, for the sloshing mode. Resultant Moment, M = Mi2 + Msl2 = 2013 kN · m Calculation of sloshing height δ = 0.84 × Ahsl × L = 0.84 × 0.072 × 3.75 × 1000 = 227 mm, less than the free board of 500 mm provided.

3.9 MATHEMATICAL MODEL FOR OVERHEAD TANKS UNDER EARTHQUAKE

3.9.1 Earthquake Analysis for overhead tanks Water served to communities for daily use and even for industrial purpose in many cases is stored in overhead tanks, so that it can be distributed under adequate pressure. Thus in a post earthquake scenario it is essential that they remain serviceable to serve the community and also mitigate secondary damages like fire. Overhead water tanks come in different shapes like rectangular, Intze type, conoids etc supported on frames constituting of beam columns or single circular shaft etc. as shown Figure 3.9.1.

Figure 3.9.1 Typical overhead water tank with its staging system modeled as two mass system.



In this case also the liquid in the tank may be considered as a two mass lumped system like the case of tank resting on ground constituting of impulsive and sloshing mass. The impulsive and the sloshing mode may be treated as two uncoupled system where the impulsive mass of the fluid as obtained by the Housner’s expression may be added to the tank mass and 1/3rd of that of the staging and whose dynamic response may be obtained from the expression T = 2π

Wi + Wt + 13 Ws gKs

(3.9.1)

where, Wi = weight of impulsive fluid; Wt = weight of the tank; Ws = weight of the staging, and Ks = stiffness of the staging frame. The sloshing time period may be obtained from the expression T = 2π

Wsl gKsl

(3.9.2)

where, Wsl is the sloshing mass as per Housner’s expression and Ksl is the fluid stiffness, given by Ksl H H = 0.83266 tan h2 1.58 W L

(3.9.3)

else they can be obtained from the graph as shown in Figure 3.9.2.

Impulsive,Sloshing mass and stiffness Ratio

1.2

Design Ratio

1 0.8

Wi/W

0.6

Ws/W

0.4

KsH/W

0.2

8 2.

5 2.

2 2.

9 1.

6 1.

3 1.

1

7 0.

4 0.

0.

1

0

L/H

Figure 3.9.2 Impulsive, sloshing mass and stiffness parameters for rectangular tank with different L/H.



In the above analysis everything is fine except the fact that the staging stiffness needs to be evaluated. The easiest way it can be done is by modeling the frame in a computer analysis program like STAAD-Pro, SAP 2000, GTSTRUDL, etc. and apply a unit load at center of mass of the tank and the water and find out the deflection at the base of the tank (top of staging). Knowing the deflection the stiffness value Ks may be obtained from the relationship, P = Ks · d

(3.9.4a)

where, d is the deflection and P is the applied load. Else, for a regular frames the stiffness may be obtained from the formula Ks =

12nEI

(3.9.4b)

jL3

where n is the number of coulumns in the frame and j is the number of storey and L is the height of column per storey. Once the time periods are established as per Equation (3.9.1) the calculation becomes quite straight forward. The shear force at the top of the frame, shown in Figure 3.9.3, is given by Vi = Ahi (W i + WT +

1 W ) 3 fr

and Vsl = Ahsl Wsl

(3.9.5)

where, Wi = weight of impulsive fluid; WT = weight of the tank; Wfr = weight of the frame, and Wsl = weight of sloshing fluid. Unit Load acting at the cg of tank+water

Rigid Links(Typical) Nodes(typical) Beam elements(typical)

Figure 3.9.3 Typical computer model for staging for determing the deflection/stifness of the frame.



And, A=

ZI Sa 2R g

(3.9.6)

In which, Sa = the accelerations due to impulsive and sloshing mode time periods as calculated above. The resultantant shear at the top of the frame is given by V=

Vi2 + Vsl2

(3.9.7)

The overturning moment in impulsive mode at the base of staging is thus given by

1 Wi hi + Hst + WT + Wst Hcg 3

Mi = Ahi

(3.9.8)

where, hi = impulsive height of the fluid as per Housner’s expression considering the “Including base pressure case”, Hst = height of the staging frame, and Hcg = height from the base of staging to the c.g. of the tank + fluid. The overturning moment in sloshing mode at the base of staging is thus given by Msl = Ahsl Wsl (hsl + Hst )

(3.9.9)

where, hsl = sloshing height of the fluid as per Housner’s expression considering the “Including base pressure case”.

Impulsive, sloshing and stiffness ratio for circular tank

1.2

Design Ratio

1 0.8

Ws/W

0.6

KsH/W

0.4

Wi/W

0.2

8 2.

5 2.

9 1.

2. 2

6 1.

3 1.

1

7 0.

4 0.

0.

1

0

Radius/Height

Figure 3.9.4 Impulsive, sloshing mass and stiffness parameters for circular tank with different R/H.



For circular tank the steps remain exactly same except the stiffness value which gets modified to H Ksl H 2 = 0.58512 tan h 1.84 . (3.9.10) W R The sloshing design parameters can also be obtained form the graph as furnished in Figure 3.9.4.

3.9.2 Hydrodynamic pressure on tank wall and base Other then knowing the overturning moement at the base of the tank and the maximum moment in the wall for big tanks, it is essential that we know the hydrodynamic pressure on the wall and the base of the tank, without which curtailment of reinforcement in the wall is not possible. The hydrodynamic distribution of pressure in the tank is given in the following.

3.9.3 Hydrodynamic pressure for circular tank The impulsive pressure is given by phi (dyn) = ψi (z)Ahi ρgH cos φ

(3.9.11)

2 D tanh 0.866 H , in which, ρ = mass density of the where, ψi = 0.866 1 − Hz liquid; φ = circumferential angle, and z = vertical distance of a point from the bottom of the tank wall. Impulsive hydrodynamic pressure in vertical dirfection on a strip of length ‘b’ is given by sin h 0.866 xh

pvi (dyn) = 0.866Ahi ρgH cos h 0.866 bh

(3.9.12)

The sloshing pressure for circular tank wall is given by 1 psi (dyn) = ψs (z)Asi ρgD 1 − cos2 ϕ cos ϕ 3

(3.9.13)

cos h(3.674 z ) where, ψs = 0.5625 cos h 3.674 HD . ( D) Sloshing pressure in vertical direction on the base slab is given by

pvs (dyn) = ψsv Ahs ρgD where, ψsv = 1.125

x D

−

4 3

(3.9.14) x 3 D

sec h 3.674 H D .

where, x = horizontal distance of a point on base of tank in the direction of seismic force from the center of tank. © 2009 Taylor & Francis Group, London, UK


3.9.4 Hydrodynamic pressure for rectangular tank For rectangular tank the impulsive pressure is given by phi (dyn) = ψi (z)Ahi ρgH

(3.9.15)

2 where, ψi = 0.866 1 − Hz tan h 0.866 2L H , in which Impulsive hydrodynamic pressure in vertical dirfection on the base slab is given by pvi (dyn) = 0.866Ahi ρgH

sin h(0.866 xh ) cos h(0.866 2L ) h

(3.9.16)

The sloshing pressure on tank wall is given by psi (dyn) = ψs (z)Asi ρg (2L)

(3.9.17)

cos h(3.162 z ) where, ψs = 0.4165 cos h 3.674 2L H . ( 2L ) Sloshing pressure in vertical direction on the base slab is given by

pvs (dyn) = ψsv Ahs ρg (2L) where, ψsv = 1.25

x 2L

−

4 3

x 3 2L

(3.9.18)

H sec h 3.162 2L .

3.9.5 Effect of vertical ground acceleration Vertical gorund acceleration increases the effective weight of liquid which induces additional pressure on the tank wall. The distribution of this pressure is taken as similar to that of hydrostatic force. Hydrodynamic pressure on tank wall due to vertical acceleration may be taken as pvw =

2 A ρgH (1 − z/H) 3 h

where, Ah =

1.25ZI R

(3.9.19)

as per IS-code

3.9.6 Pressure due to inertia of the wall Pressure due to inertia of the wall itsdelf may be taken as pmw (dyn) = Ahi tρ w g

(3.9.20)

where, t = average thickness of the wall, and ρw = mass density of the tank wall. © 2009 Taylor & Francis Group, London, UK


3.9.7 Maximum design dynamic pressure The maximum dynamic design pressure may be obtained by taking the SRSS value of the above pressures and be expressed as pdes (dyn) =

(phi + pmw )2 + p2si + p2vw

(3.9.21)

Example 3.9.1 Shown in Figure 3.9.5 is an elevated rectangular water tank of capacity 500 m3 resting on medium soil which is classified as falling in a zone of IV as per IS Code. Grade of concrete used is M25. Based on dynamic analysis find out the overturning moment on the top of the foundation. Shear at top of staging, and hydrodynamic pressure on the tank wall and base. Code to be used is IS 18932002. Col size 600 × 600, beam size 500 × 750.

9000

250(typ)

6350

750(typ) 4250(typ)

4450

4450

4450 20,000 4450

Plan view of column

Figure 3.9.5



Solution: Capacity of water tank = 500 m3 ; Weight of water in tank = 500 × 10 = 5000 kN, and Height of water = 500/81 = 6.17 = 6.2 m (say). Calculation of weight of tank Weight of roof slab (150 mm thick) = 9.5 × 9.5× 0.15 × 25 = 338.4 kN54 Weight of side wall = 9.25 × 4 × 6.35 × 0.25 × 25 = 1468.4 kN Weight of bottom slab(350 mm thk) = 9.5 × 9.5 × 0.35 × 25 = 789.7 kN. Thus weight of empty tank = 338.4 + 1468.4 + 789.7 = 2596.5 kN. Calculation of weight of staging Height of staging = 20 m. Size of column = 600 × 600. Thus weight of 9# of column = 20 × 0.6 × 0.6 × 9 × 25 = 1620 kN. Length of peripherial beam = 35.6 = 36 meter(app). Size of beam = 500 × 750 mm2 . Weight of peripherial beams = 36 × 0.5 × 0.75 × 25 × 4 (levels) = 1350 kN. Length of internal beams = 18 m (app). Weight of internal beams = 18 × 4 (levels) × 0.5 × 0.75 × 25 = 675 kN. Total weight of staging = 1620 + 1350 + 675 = 3645 kN. Calculation of impulsive weight of water. As per Housner √

L tan h 3H Wi = √ L W 3H Here, L = 4.5 m, H = 6.2 m, W = 5000 kN; thus, Wi = 5000 × 3381 kN. Calcualtion of sloshing weight of water Wsl L H = 0.527 tan h 1.58 W H L

➔

4.5 6.2 Wsl = 5000 × 0.527 × tan h 1.58 = 1864 kN 6.2 4.5

54 Unite Weight of concrete considered as 25 kN/m3 .


0.8503 1.257

=


Calculation of column stiffness Kcol =

12nEI jL3

Here, Ec = 2.85 × 107 kN/mm2 ; I = (1/12) × 0.6 × 0.63 = 0.0108 m4 ➔ EI = 307800 kN/m2 . Here, n = 9, and j = 4 and height of column, L = 5 m, and we have, Kcol =

12nEI 12 × 9 × 307800 = = 66485 kN/m. 3 jL 4 × 53

Calculation of impulsive time period T = 2π

Wi + Wt + 13 Ws = 2π gKs

3381 + 2597 + 13 × 3645 = 0.6598 sec. 9.81 × 66485

For soil of medium strength, we have Sa /g = 1.36/T = 1.36/0.66 = 2.06 m/sec2 for 5% damping. Considering 7% damping for RCC design Sa /g is multiplied by a factor 0.9, as per IS-1893 2002 ➔ Sa/g = 2.06 × 0.9 = 1.855 m/sec2 . For zone IV, Z = 0.24; Importance factor = 1.5 (say), and R = 1.5, for non ductile frame. ➔

Ahi =

ZI Sa 0.24 × 1.5 = × 1.855 = 0.2226 2R g 2 × 1.5

Thus, impulsive shear at base slab level of the tank Vhi = 0.2226 × (3381 + 2597 + 1/3 × 3645) = 1626 kN. Calculation of sloshing time period Tsl = 2π

2L g

3.16 tan h

3.16H 2L

and substituting the value of H = 6.2 m and L = 4.5 m as stated above, we have, Tsl = 3.43 secs. © 2009 Taylor & Francis Group, London, UK


Considering,

Sa g s

=

1.36 T

= 0.3965 m/sec2 for 5% damping.

Considering, the fluid damping as 0.5% as per IS-1893–2002, the above value gets modified to

Sa g

= 0.3965 × 2.98 = 1.18 m/sec2 . s

ZI Sa 0.24×1.5 Thus, Ahs = 2R g = 2×1.5 × 1.18 = 0.1416 Thus sloshing shear at base slab level of the tank

Vhi = 0.1416 × 1864 = 264 kN · (Wsl = 1864 kN, the sloshing weight of water) Thus, total shear acting at bottom of the tanks (top of staging) = (1626)2 + (264)2 = 1647 kN. Overturning moment at the base of the staging hi = H

√ L √ 3H L 1 tan h 3 − ; 2 H 8

√ 4.5 3 6.1 x6.2 6.2 √

− hi = = 3.80 m; 8 2 tan h 3 4.5 6.1

2597 × 23.30 + 3645 3 × 10 = 19.06 m. 2597 + 1215 1 Mi = Ahi Wi hi + Hst + WT + Wst Hcg 3 Hcg =

➔ Mi = 0.2226 × (3381 × (3.80 + 20) + (2597 +

1 × 3645) × 19.06) 3

= 34086 kN · m For the sloshing mode

➔

H cos h 1.58 L −2 hsl

=1− H H 1.58 H L sin h 1.58 L

Substituting the values H = 6.2 m and L = 4.5 m, we have, hsl = 4.586 m Thus, Msl = Ahsl Wsl (hsl + Hst ) = 0.1416 × 1864 × (4.586 + 20) = 6489 kN · m. Thus, the resultant overturning moment is given by M=

340862 + 64892 = 34698 kN . m



Calculation of hydrodynamic pressure Based on the Equation 3.9.21 hydrodynamic pressure on wall for various cases are furnished hereafter z/H

Impulsive pressure on wall

Sloshing pressure wall

Presure due to vertical acceleration

Pressure due wall inertia

Resultant design pressure(kN/m2 )

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

10.21798 10.1158 9.809262 9.298363 8.583104 7.663486 6.539508 5.21117 3.678473 1.941416 2.27E-15

1.02172 × 10−18 3.91873 × 10−17 3.00498 × 10−15 2.30469 × 10−13 1.7676 × 10−11 1.35567 × 10−09 1.03974 × 10−07 7.97436 × 10−06 0.000611599 0.046906968 3.5975604

9.76 8.784 7.808 6.832 5.856 4.88 3.904 2.928 1.952 0.976 1.08E-15

1.4125 1.4125 1.4125 1.4125 1.4125 1.4125 1.4125 1.4125 1.4125 1.4125 1.4125

15.18307 14.49346 13.67087 12.70428 11.58468 10.30475 8.858648 7.241974 5.452368 3.493355 3.864919

Similarly in vertical direction hydrodynamic pressure on base slab is

X/2L

Sloshing pressure in vertical direction

Impulsive pressure in vertical direction

Design pressure (kN/m2 )

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

0 0.183688 0.36369 0.53632 0.697892 0.84472 0.973117 1.079398 1.159877 1.210868 1.228683

0 0.58238833 1.167154385 1.756685596 2.35338885 2.959700315 3.578095387 4.211098795 4.861294911 5.531338298 6.223964552

0 0.61067 1.222506 1.836732 2.454688 3.077885 3.708062 4.347235 4.99775 5.662323 6.344084

3.10 PRACTICAL ASPECTS OF EARTHQUAKE ENGINEERING We are almost through with this chapter and take this opportunity to pacify those hardened professionals whom we had perhaps bored to stupor with double integrals, partial differentials and hyperbolic trigonometric functions. Before we start with this topic, we would like to point out that – perhaps we could make the reader realize that it is not a very easy subject to grasp. It requires competence in multifaceted subject like engineering geology, soil dynamics, structural dynamics, fluid dynamics and finally applied mathematics which is not a very easy thing to achieve in a nutshell. It is for this, specialists from different field convene together to pool their knowledge and experience to develop a unified design policy which is otherwise known as the code of practice. 55 55 And violations of the same, in the name of economy or financial budget is not uncommon.



A professional engineer undertaking a design is usually guided by three things • • •

Cost involved Engineering and construction schedule Sociological importance of the structure in hand.

To what sophistication an analysis should be carried out depends a lot on the budget the client has the engineering and construction schedule he has to meet and social outcry it would create in case the structure undergoes damage during an earthquake. For instance a commercial building or a hotel sustaining damage during an earthquake without collapse would cause a much lesser furor then a reactor building or a heavy water container undergoing even a minor crack during an earthquake. For radiation effects emanating from those cracks could have a catastrophic effect on the surrounding and would possibly result in complete shut down of the plant till such cracks are rectified. This would possibly result in power shortage in an area for months and could result in a huge revenue loss for industries dependent on such power. Chemical plants storing toxic and hazardous material if undergoes damage can again have deadly consequence on the surrounding and can ravage the ecological balance so badly that it could take years to restore the same. Public building like hospitals, town halls, schools where people mostly take refuge in the post earthquake scenario must remain functional for relief work to be effectively carried out. People could surely argue that – what’s the big deal? As the code suggests we take a higher importance factor and design it for a higher force. It is indubitable a fact that the argument do have some substance in it. But it has been seen in many cases that though the force induced in the structure was possibly lesser then expected, structures have undergone a spectacular failure while there are structures which was subjected to a far higher force then it was designed for and yet it has survived the shock with only minor damages. Reasons attributing to such spectacular failures have been very simple. • •

The structures were inherently planned poorly making them generically weak under earthquake force56 And last but not the least improper detailing causing improper stress dissipation path resulting in considerable damage to the structure.

We would like to re-emphasise at this point that irrespective of the most sophisticated analysis one undertakes the most advanced software one may use if the same is not followed up with well conceived structural arrangement and proper detailing can still result in collapse. On the other hand, analysis of structures based on simple seismic coefficient method and plane frame analysis carried out by simple portal method, but detailed properly and having robust structural configuration has been found to survive severest of the shock.

56 Like irregular geometry in plan causing additional torsion not catered to properly etc.



One of the major limitations a civil engineer faces specially in the building industry is that in many of the major buildings during conceptual stage he has very little control on selection of material and planning of the functional space for this is principally controlled by the architects57 . But how well the structure will behave under earthquake depends a lot on these decisions. If the architect concerned does not have appreciation of the problems earthquake could create, may lead to a situation of impasse–when in extreme case can even result in replacement of the project civil engineer58 . While it is surely not the job of a structural engineer to put spanner in every aspiring wheel an architect could conjure, yet if he sees something that could seriously mar the performance of the building should be pointed out clearly – if possible with comparative numbers enabling an architect to make a quantitative assessment of the issue. The bottom line is that it is necessary to have an unbiased continuous and an open dialogue between the architect and the engineer to arrive at the most optimal shape and configuration which is structurally sound and aesthetically pleasing. During planning stage of structure at a location susceptible to severe earthquake if some fundamental rules are followed and adhered to much of the risk of a collapse can surely be significantly mitigated. We discuss a few of the important ones hereafter •

•

Avoid the fundamental period of the building to be near the free field site of the motion, equating the two as shown in Chapter 1 (Vol. 2) one can arrive at the critical height of the building which an architect could be made aware to suppress the seismic excitation of the building. Avoid irregular geometry in plan these creates additional force. If not properly taken care off can lead to significant damage of the building. We explain this point with an example.

Shown in Figure 3.10.1 is a typical plan view of a school building with a playground59 . It is obvious that shear center of such building will be along the chain dotted line as shown in the figure. Thus an earthquake force acting on the building would act along this line and would invariably create additional torsion in the building which if not properly catered for could result in severe cracking at the junction. Now the point remains is that does this functional concept be rejected at the outset citing it is dangerous? For wearing an architects hat one can envisage a number of functional advantage with this type of configuration as a school building. So, what are the other options a structural engineer is left with? He can surely under take a detailed dynamic analysis of such building and cater to the additional torsion or can simply do the following: The trick is simple – break up the building into three regular modules as shown in Figure 3.10.2 by providing construction joints as shown which surely makes the

57 Not to mention, those mafias (with high level political nexus) who in the name of promoters today control almost everything in building industry and have polluted the complete work ethics of the building industry in India. 58 If the firm concerned is primarily an architectural firm. 59 A common feature one observes in many public schools in England and Germany, playground in front is usually called a quadrangle.



Figure 3.10.1 Typical plan of a school building.

Construction joint (Typ.)

Figure 3.10.2 Plan of a school building with construction joint.

configuration much simpler to handle while the functional configuration conceived by the architect can still be maintained 60 . •

Avoid rapid change in stiffness in vertical direction.

This is a common problem faced in many congested urban area where due to lack of space the ground floor is completely kept open for the cars to park while the top portion constitute of residential or office complex with usual curtain walls as shown in Figure 3.10.3. In such cases, the top portion of the building having higher stiffness would possibly have a low time period which shows the structure is quite stiff. This would thus attract a significant force which when gets transferred to the foundation, suddenly finds a level which has a much poorer stiffness then rest of the building and if the members are not sturdy enough to transfer this shear force would invariably result in a failure at the column beam junction as marked as the weak zone.

60 A similar situation can happen in a pipe rack configuration too where due to process requirement and pipe stress limitations the configuration cannot be changed. Again opting for a separation joint will do the trick in such case.



Weakest link

Figure 3.10.3 Typical office building with open space for car parking in ground floor.

Figure 3.10.4 Office building with open shear wall and access cut-out.

It can well be envisaged that while we cannot reject the option61 , yet try to arrive at a solution which would make the building safe against earthquake. The easiest solution would to provide one bay with a shear wall which would be stiff enough to absorb the load as shown in Figure 3.10.4. We had just cited a few examples to give some idea. All international codes including IS-1893 (2002) has come up with do and don’ts in term of building planning and should be adhered to as much as practicable. Detailing is another aspect – that needs to be given proper attention. Ductile detailing as such is gaining importance more and more to attenuate the effect of earthquake force. IS code has developed a special code, IS-13920 for the same which should be adhered to. Ductility is an important aspect which safe guards a structure by dissipating the energy induced in the body due to seismic force by cracking thus preventing a total collapse. A detailed discussion is beyond the scope of this book and interested reader may refer to a number of excellent reference like Park & Pauley 1975, Dowrick 2002.

61 For doing so might result in a rejection of the building plan itself citing provisions have not been kept for adequate parking space and the client does not have budget to provide a basement parking.



A word or two for the rookies. . . . . . In eagerness to carry out a clever and sophisticated analysis it is not uncommon to see freshmen start with an elaborate computer model at the outset and finally get lost in maze of numbers loosing sight of the basic issue as to how the structure is behaving in general. Whether there is an inherent flaw in the arrangement that generates additional forces in one part of the system or not etc. Study the plan carefully, if required seek help of more experienced people and discuss. Always start with a simple model like a stick model or a simple plane frame model to have a first feel of the system behavior before launching into a more detailed and elaborate analysis – we can assure you this will eliminate a number of erroneous design decisions vis-a-vis re-work at later stage of the project when cost implication in terms of schedule can be more severe.

3.10.1 Epilogue We are almost at the end of the road, for readers seeking more information on the subject we would encourage him to read the literatures mentioned at the end of the chapter. Read them, if you are really interested in this topic we can assure you that you will enjoy them immensely. Considering this is not a handbook our intention is not to work out design problems in completeness enabling one to follow them blindly. The purpose was to provide you with the basic essence of the phenomena and encourage you to understand the fundamental mechanics behind it. Finally a word of apology, to the bridge engineers for not having addressed -such an interesting topic. The reasons were basically the following: Bridge engineering being a topic by itself would significantly increase the volume of this book62 , finally private bridges in India is a rare commodity and most of the bridges and flyovers are controlled based on legislation of IRC63 and guidelines prescribed by MOST64 . Even though detailed dynamic analysis is possible for such bridges however, as per Indian practice, most of the dynamic loads coming on the bridges due to moving vehicles or earthquake are catered for based on dynamic load factor or pseudo static methods. Applying too much sophistication in their analysis may be construed as a rude intrusion in their “all is fine” world of slumber and may not be approved by the legislative body. In USA bridge codes put forward by AASHTO65 have regular provisions for design of bridges under dynamic earthquake loading are not officially recognized in the country. Till such modifications are brought about by IRC and MOST we thought it prudent not to venture in this otherwise a very interesting subject.

62 63 64 65

The intenion has never been to make international weightlifters out of the reader. Indian Road Congress. Ministry of Surface Transport of India. American Authority of state highway official.


References

1 Alpan, I. 1961, ‘Machine foundation and soil resonance’, Geotechnique, Vol. 11. 2 Almuti, A.M. 1976, ‘Large flexible Turbine foundation’, Methods of Structural Analysis, ASCE, NY, pp. 707–719. 3 Ambraseys, N. (1995), The prediction of earthquake peak ground acceleration in Europe. J. Earthq. Eng. & Str. Dyn., Vol. 24, pp. 467–490. 4 Aneja, I. 1975, ‘Dynamic Response of Systems—Turbine generators on Various Foundations’, Proceedings of the American Power conference, Vol. 37, pp. 528–540. 5 Applied Technology Council 1982, ‘Tentative provisions for the development of seismic regulations for buildings’, ATC 3-06, NBS SP-510, NSF 78-8, National Bureau of Standards USA. 6 Archer, J.S. 1963, ‘Consistent Mass Matrix for Distributed Mass Systems’, Jour. of Struct. Division, ASCE, Vol. 89, No. STA4, pp. 161–178. 7 Arya, S.C., O’Neill, M.W. & Pincus, G. 1979, Design of Structures and Foundations for Vibrating Machines, Gulf Publishing Co., Houston, Texas. 8 Arya, A.S. & Drewer, R. 1997, ‘Mathematical modeling and computer simulation of elevated foundations supporting vibrating Machinery’, Tansaction of IMACS, Vol. XIX, No. 4, Dec. 9 Bannerjee, P.K & Sen, R. 1987, ‘Dynamic behavior of axially and laterally loaded piles and pile groups’, Chapter 3, Developments in Soil Mechanics and Foundation Engineering, Vol. 3, ed., Bannerjee, P.K. and Butterfield, R., Elsivier Applied Science, London. 10 Baranov, V.A. 1967, ‘On the calculation of excited vibrations of an embedded foundation’, Voprosy Dinamiki Prochnocti, No. 14, Polytech. Ins. Riga, pp. 195–209. 11 Brooker E.W. & Ireland H.O. 1965, ‘Earth pressure at rest related to Stress History’, Canadian Geotechnical Journal, Volume II, No. 1. 12 Barkan, D.D. 1962, Dynamics of Bases and Foundations, McGraw-Hill Book Co. NY. 13 Blake, T.F., Hollingsworth, R.A. & Stewart, J.P., eds. 2002, Recommended procedures for implementation of DMG, Thomas F Blake (Fugro-West Inc., Ventura, Ca.): Special Publication 117. 14 Bojtsov, G., Postnov, M., Paliy, V., Chuvikovsky, V.S. 1982, ‘Dynamics and Stability of construction’, Vol. 3, Structural Mechanics of Ship Leningrad, Sudostronei (In Russian), p. 317. 15 Broms, B.B. 1965, ‘Design of Laterally Loaded Piles’, J. of Soil Mechanics and Foundation Engng. Div., ASCE, 91, No. SM3, pp. 79–98. 16 Bycroft, G.N. 1956, ‘Forced vibration of rigid circular plate on a semi infinite half space on an elastic stratum’, Philosophical Transactions Royal Society of London, Series 248, pp. 327–368. 17 Chopra, A.K. & Chakrabarti, P. 1973, ‘The Koyna Earthquake and damage to Koyna Dam’, Bull. Seis. Soc. America, 63, No. 2, pp. 381–97.


606 References 18 Chowdhury, I. 1984, Dynamic soil-structure interaction of Turbo-Generator Foundation, M.Tech Dissertation, Department of Civil Engineering, IIT Kharagpur. 19 Chowdhury, I. & Som, P.K. 1993, ‘Dynamic Pile structure interaction of Boiler Feed Pump Frame Foundation’, Indian Geotechnical Conference, Vol. 1, pp. 411–414. 20 Chowdhury, I., Ghosh, B. & Dasgupta, S.P. 2002, ‘Analysis of Hammer foundation considering soil damping’, International Conference on Advancement in Civil Engineering, Vol. 2, Indian Institute of Technology (Kharagpur) India, pp. 1019–1028. 21 Chowdhury, I. & Dasgupta, S.P. 2002, ‘Earthquake Response of Soil-Structure System’, Indian Geotechnical Journal, 32(2), pp. 309–328. 22 Chowdhury, I. & Dasgupta, S.P. 2004, ‘Earthquake response of cantilever and counterfort retaining wall’, Electronic Journal of Geotechnical Engineering, Vol. 8, [Online] Available at: http:/www.ejge.com/ 23 Chowdhury, I. & Dasgupta, S.P. 2006, ‘Dynamic response of piles under vertical loads’, Indian Geotechnical Journal, Vol. 36, No(2), April. 24 Chowdhury, I. & Dasgupta, S.P. 2007, ‘Dynamic Earth pressure on rigid unyielding walls under Earthquake Force’, Indian Geotechnical Journal, Vol. 37(2), pp. 81–93. 25 Chowdhury, I. 2008, ‘Some Myths and Facts about Dynamic Soil Structure Interaction and its application to Building Design’, keynote lecture, Conf. on Challenges and Application of Mathematical Modeling in Building Science and Technology, Central Building Research Institute of India, Roorkee, India. 26 Chowdhury, I & Dasgupta, S.P. 2008, ‘Earthquake Analysis of Concrete Dams considering fluid structure and fluid-structure soil interaction’ (communicated). 27 Chowdhury, I. & Dasgupta, S.P. 2008, ‘Dynamic analysis of piles under lateral loads’, Indian Geotechnical Journal (In Press). 28 Cohen, M. & Jennings, P., ‘Silent Boundary Methods For Transient Analysis’, Computational Method in Transient Analysis—Computational Method in Mechanics, Vol. 1, North Holland. 29 Coulomb, C.A. 1776, Essai sur une Applicaion des Regles de Maximis er Minims a quelques Problemes de Satique, relatifsa l’Archutecture-Mem. Roy des Sciences Paris, Vol. 3, p. 38. 30 Craig R.H. 1981, Structural Dyanmics, John Wiley Publication, NY. 31 Crede, C.E. 1951, Vibration and Shock Isolation, John Wiley & Sons, NY. 32 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. 33 Dasgupta, S.P. & Kameswara Rao, N.S.V.K. 1978, Dynamics of rectangular footings by Finite elements, Journal of GT Division, ASCE, Vol. 104(5). 34 Der Kiureghian, A. 1981, ‘A Response Spectrum Method for Random Vibration Analysis for MDF Systems’, Earthquake Engineering and Structural Dynamics, 9, pp. 419–435. 35 ‘Design Criteria for Turbine Generator Pedestal’, 1970, Journal of Power Division, ASCE, Vol. 96, Jan, pp. 1–22. 36 Dobry, R. & Gazetas, G. 1988, ‘Simple Method for dynamic stiffness and damping of floating piles groups’, Geotechnique, 38, No. 4, pp. 557–574. 37 Dowrick, D.J. 2002, Earthquake Resistant Design of Structures a Manual for Architects and Engineers, John Wiley & Sons, NY. 38 Dowrick, D.J. 2003, Earthquake Risk Management, John Wiley Publication, U.K. 39 Erden, S.M. & Stokoe, K.H. 1975, ‘Effects of embedment on foundation response’, Shock and vibration problems in Geotechnical Engineering, ASCE National Convention, Denver, Colorado. 40 Eurocode-8 1998, Design Provisions for Earthquake resistance of Structures Part 4- Silos, tanks and pipelines. European Committee for Standardization Brussels.


References 607 41 Federation of Emergency Management Agency(FEMA)1997, Building Seismic Safety Council, Washington D.C., USA. 42 Gazetas, G. 1982, ‘Shear vibrations of vertically inhomogeneous earth dams’, Int J. Num and Analytical Methods in Geo-mechanics, Vol. 6, No. 1, pp. 219–241. 43 Gazetas, G. & Tassoulas, A.L. 1987, ‘Horizontal Stiffness of Arbitrarily shaped embedded foundation’, Journal of GT Division, ASCE, Vol. 113, No. 5. 44 Ghosh, D., Chakroborty, D. & Batavyal, H. 1984, ‘Dynamic Response and Static Analysis of RCC Frames supporting high speed centrifugal machines with Soil Structure Interaction’, International conference on case histories in geo-technical engineering, St Louis, University of Missouri Rolla, USA. 45 Ghosh, D.K. & Batavayal, H.N. 1985, ‘Analysis of Structural response to earthquake for 150M high RCC Chimney with soil-structure interaction’, Proc. National Seminar on Tall Chimneys, Vigyan Bhavan, New Delhi. 46 Gupta, B.N. 1972, ‘Effect of foundation embedment on the dynamic behavior of foundation of the foundation—soil system’, Geotechnique, 22(1), pp. 129–137. 47 Hardin, B.O. & Richart, F.E., Jr. 1963, ‘Elastic wave velocities in granular media’, JSMFD, ASCE, Vol. 89, No. SM1. 48 Hardin, B.O. 1965, ‘The nature of damping in sands’, J. Soil Mech. Found. Div., Am. Soc. Civ. Eng. 91(SM-1), Part 1, pp. 63–97. 49 Hardin, B.O. & Drnevich, V.P. 1973, ‘Shear Modulus and damping in soils, Design equations and curves’, JSMFD, ASCE, Vol. 98, No. SM7, July. 50 Haroun, M.A. & Housner, G.W. 1981, ‘Seismic design of liquid storage tank’, Journal of Technical Councils of ASCE, Vol. 107, No. TCI, pp. 191–207. 51 Housner, G.W. 1957, ‘Dynamic Pressure on Accelerated Fluid Containers’, Bulletin of Seismological Society of America, Vol. 47(1). 52 Housner, G.W. 1963, ‘Dynamic Analysis of fluids in containers subjected to acceleration’, Nuclear Reactors and Earthquakes Report No. TID 7024, U.S. Atomic Energy Commission, Washington DC. 53 Hsieh, T.K. 1962, ‘Foundation Vibrations’, Proceedings of Inst. of Civil Engineers, Vol. 22, pp. 211–226. 54 Humar, J.L. 1990, Dynamics of Structure, Prentice-Hall of India, New Delhi. 55 Hurty, W.C. & Rubenstein, M.F. 1967, Dynamics of Structure, Prentice Hall of India Pvt. Ltd., New Delhi. 56 Idriss, I.M. 1999, ‘An update to the Seed-Idriss simplified procedure for evaluating liquefaction potential’, Proc. TRB Workshop on New Approaches to Liquefaction, January, Publication No. FHWA-RD-99-165, Federal Highway Administration. 57 Indian Standards Institution 1984, 2002, ‘Indian Standard Criteria for Earthquake Resistant Design of Structures’, IS: 1893 (Part 1), ISI, New Delhi, India. 58 Indian Standards Institution 1968, ‘Indian Standard Code of Practice for Design and Construction of Machine Foundations’, Part I to Part IV, IS: 2974, ISI, New Delhi, India. 59 Indian Standards Institution 1993, ‘Indian Code of practice for ductile detailing of Reinforced Concrete Structure Subjected forces’, IS: 13920, ISI, New Delhi, India. 60 Ishibashi, I. & Zhang, K. 1993, ‘Unified dynamic shear modulii and damping ratios of sand and clay’, Soils and Foundations, Vol. 33, No. 1, pp. 182–191. 61 Jadi, H. 1999, Prediction of lateral dynamic response of Single Pile embedded in Clay, Ph. D. Thesis, University of Missouri Rolla, Missouri. 62 Kardestuncer, H. ed., 1987, Finite Element Handbook, McGraw-Hill, NY. 63 Karnovsky, I. & Lebed, O. 2001, Formulas for Structural Dynamics, McGraw-Hill, NY. 64 Kaynia, A.M. & Kausel, E. 1982, ‘Dynamic behavior of pile groups’, 2nd Int. Conf. on Numerical. Methods. in Offshore Piling, Austin, Texas.


608 References 65 Lasdon, L.S., Warren, A.D., Jain, A. & Ratner, M. 1978, ‘Design and testing of a generalized reduced gradient code for nonlinear programming’, ACM Trans. Math. Software, 4, pp. 34–50. 66 Lysmer, J. & Richart, F.E. Jr. 1966, ‘Dynamic response of footings to vertical loading’, J. Soil Mech. and Found. Div. Proc., ASCE, Vol. 92, No. SM1, pp. 65–91. 67 Jennings, P.C. & Bielek, J. 1973, ‘Dynamics of building soil interaction’, Bulletin of Seismological society of America, 63,1,9–48. 68 Kameswar Rao, N.S.V. 1977, ‘Dynamic soil structure system—A Brief Review’, J. Struct. Engg., India, Vol. 4. 69 Kameswara Rao, N.S.V. 1998, Vibration Analysis and Foundation Dynamics, Wheeler Publishing, New Delhi. 70 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. 71 Kramer, S. 2004, Geotechnical Earthquake Engineering, Pearson Education Publication. New Delhi, Section 7.5, Page 297. 72 Kreyszig, E. 2001, Advanced Engineering Mathematics, 8th Edition, John Wiley & Sons Inc., NY. 73 Lysmer, J. & Kuhlemeyer, R.L. 1969, ‘Finite Dynamic Model of Infinite Media’, J. EM. Divn, ASCE, EM4. 74 Lysmer, J., Ostadan, F. & Chen, C. 1999, ‘A System for Analysis of Soil Structure Interaction’, SASSI-2000, University of California, Berkeley. 75 Luco, J.E. 1986, ‘Soil Structure interaction effects on seismic response of tall chimneys’, Soil Dynamics and Earthquake Engineering, Vol. 5(3), pp. 170–177. 76 Major, A. 1980, Dynamics in Civil Engineering—Analysis and Design, Vols. I–IV, Akademia Kiado, Budaapest and Collets Holding London. 77 Makadisi, F.I. & Seed, H.B. 1977, A simplified procedure for estimating earthquake induced deformations in dam and embankments, EERC Report -77/19, University of California Berkeley. 78 Mario, Paz. 1987, Structural Dynamics, CBS Publishers Ltd., New Delhi. 79 Meirovitch, L. 1967, Analytical Methods in Vibrations, Macmillan Co., NY. 80 Meirovitch, L. 1975, Elements of Vibration Analysis, McGraw-Hill Book Co. NY. 81 Mononobe, N. & Matsuo, H. 1926, ‘On determination of earth pressure during earthquakes’, Proc. on World Engineering Congress, p. 9. 82 Mononobe, N. 1936, ‘Seismic stability of earth dam’, Proc. Second Congress on Large Dams, Vol. IV. 83 Murray, D.A. 1967, Introductory Course on Differential Equations, Orient Longman (India). 84 Murthy, V.N.S. 1991, Soil Mechanics and Foundation Engineering, Vols. I &II, Sai-Kripa Technical Consultants, Bangalore(India). 85 Naeim, F. 1989, The Seismic Design Handbook, Van Nostrand Reinhold, NY. 86 Navarro, C. 1992, ‘Influence of soil flexibility on seismic behavior of chimneys’, Soil Dynamics and Earthquake Engineering, Vol. 11, pp. 403–409. 87 Newcomb, W.K. 1951, ‘Principles of foundation design for engines and compressors’, Trans. ASME, 73, 307–318. 88 Novak, M. & Beredugo, Y.O. 1972a, ‘Vertical vibration of embedded footings’, J. Soil Mech. Found. Div., ASCE, 12, pp. 1291–1310. 89 Novak, M. & Beredugo, Y.O. 1972b, ‘Coupled horizontal and rocking vibration of embedded footings’, Canadian Geotechnical Journal, 9, pp. 477–497. 90 Novak, M. 1974, ‘Dynamic stiffness and damping of piles’, Canadian Geotechnical Journal, Vol. 11, pp. 574–598.


References 609 91 Novak, M. & El Sharnouby, B. 1983, ‘Stiffness and damping constants for single piles’, J. Geotech. Engng. Div., ASCE, 109, pp. 961–974. 92 Novak, M. & El Hifnawy, L. 1983, ‘Vibration of Hammer Foundation’, Journal. of Soil Dynamics and Earthquake Engineering, 2(1), 43–53. 93 Ohsaki, Y. & Iwasaki, R. 1973, ‘On dynamic shear modulus and Poisson’s ratio of soil deposits’, Soils and Foundations, 13(4), pp. 61–73. 94 Okabe, S. 1926, ‘General theory of earth pressure’, Journal of Japanese Society of Civil Engineering, Vol. 12, No. 1. 95 Ostadan, F. 2004, ‘Seismic Soil Pressure for Building Walls—an Updated Approach’, International Conference on Soil Dynamics and Earthquake Engineering (11th ICSDEE), University of California, Berkeley, January. 96 Park, R. & Pauley, T. 1975, Reinforced Concrete Structures. 1st ed., John Wiley & Sons, NY. 97 Paz, M. 1991, Structural Dynamics: Theory and Computation, Van NostrandReinhold, NY. 98 Peck, R.B., Hanson, W.E. & Thornburn, T.H. 1980, Foundation Engineering, 2nd edn, Wiley Eastern Ltd., New Delhi. 99 Poulos, H.G. & Davis, E.H. 1980, Analysis and design of Pile foundations, John Wiley Publication. 100 Randolph, M.F. & Poulos, H.G. 1982, ‘Estimating Flexibility of Offshore Pile groups’, Proceedings of the 2nd International Conference on Numerical methods in Offshore Piling, University of Texas, Austin, USA. 101 Rausch, E. 1959, ‘Maschinen fundamente und andere dynamisch beanspruchte baukonstructionen’, VDI Verlag, Dusseldorf. 102 Richart, Jr. F.E., Hall, Jr. J.R. & Woods, R.D. 1970, Vibrations of Soils and Foundations, Prentice-Hall, Englewood Cliffs, New Jersey. 103 Rosenblueth, E. 1951, A Basis for Aseismic Design, PhD thesis, Univ. of Illinois, Urbana. 104 Sadhegpour, N. & Chowdhury, I. 2008. Aerodynamic response of Tall Chimneys considering foundation compliance- 12th International Conference in Advancement in computational methods in Geo-mechanics-India. 105 Schmertmann, J.H. 1978, ‘Use the SPT to measure Dynamic Soil Properties?_ Yes, But . . . !’, Dynamic Geotechnical Testing, ASTM, STP 654, ASTM, Philadelphia, Pa, pp. 341–355. 106 Seed, H.B. & Whitman, R.V. 1970, ‘Design of Earth Retaining Structures under Dynamic Loads’, Proc. ASCE Specialty Conference on Lateral Stresses in Ground and Design of Earth Retaining Structures, ASCE, pp. 103–147. 107 Seed, H.B. & Idriss, I.M. 1970, ‘Soil Moduli and Damping Factors for Dynamic Response Analysis’, Report No. 70-1, EERC, Berkeley, California. 108 Seed, H.B., Tokimatsu, K., Harder, L.F. & Chung, R.M. 1984, The Influence of spt. procedures in soil liquefaction resistance evaluations. Report No. UCB/EERC-84/15, University of California at Berkeley, 1984. 109 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. 110 Shames, I.H. & Dym, C.L. 1995, Energy and Finite Element Method in Structural Mechanics, New Age International Publishers Ltd., New Delhi. 111 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. 112 Srinivasulu, P. & Lakshmannan, N. 1978, ‘Dynamic response of turbo-generator pedestal’, ASCE, Spring convention, Pittsburgh, Pensylvania, April, pp. 24–28. 113 Srinivasalu, P. & Vadiyanathan, C.V. 1977, Handbook of Machine Foundations, Tata Mcgraw-Hill, New Delhi.


610 References 114 Steedman, R.S. & Zeng, X. 1990, ‘The Seismic Response of Waterfront Retaining Walls’, Proc Speciality Conference on Design and Performance of Earth Retaining Structures, Speciality and Technical Publication 25, Cornell University, Ithaca, NY. 115 Steven, J. 1978, ‘Prediction of pile response to vibratory loads’, Proc. Xth Offshore Technology Conference, Vol. IV, May, Houston, Texas. 116 Stevens, G.R. 1980, New Zealand Adrift, A.H. & A.W. Reed, Wellington. 117 Task Committee on Turbine Foundations 1987, Design of large steam turbine-generator foundations, ASCE, NY. 118 Terzaghi, K. & Peck, R.B. 1967, Soil Mechnics in Engineering Practice, John Wiley & Sons, NY. 119 Timoshenko, S. 1956, Strength of Material Vol. 2, Van Nostrand & Reinhold Publications, NY. 120 Timoshenko, S.P. & Woinowsky-Krieger (1958), Theory of Plates and Shells, McGrawHill Kogakusha, Ltd., Tokyo. 121 Tschebotarioff, G.P. 1953, ‘Performance records of engine foundation’, ASTM Special Technical Publication, No. 156. 122 Veletsos, A.S. & Meek, J.W. 1974, ‘Dynamic behavior of building foundation system’, Journal of Earthquake Engineering and Structural Dynamics, Vol. 3(2), pp. 121–138. 123 Veletsos, A.S. & Young, 1977, ‘Earthquake response of liquid storage tank’, Proceeding of 2nd Engineering Mechanics specialty conference, ASCE, Raleigh, pp. 1–24. 124 Veletsos, A.S. 1984, ‘Seismic response and design of liquid storage tanks’, Guidelines for seismic design of Oil and gas pipelines system, ASCE, NY, pp. 255–370. 125 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. 126 Waas, G. & Hartmann, H.G. 1981, ‘Piles subjected to dynamic horizontal loads-European simulation meeting on modeling and simulation of large scale Structural Systems’, Capri, p. 17. 127 Weaver, Jr. W. & Gere, J.M. 1986, Matrix Analysis of Framed Structure, CBS Publishers & Distributors, Delhi, India. 130 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. 131 Whitman, R.V. 1970, Soil Structure Interaction—Seismic design for Nuclear power plants, The MIT press, Cambridge, Massachusets. 132 Whitman, R.V. 1972, ‘Analysis of Soil-Structure interaction—A state of the art review’, Soil Publications No-300, Massachusetts Institute of Technology, April. 133 Whitman, R.V. 1973, ‘Analysis of soil structure interaction—A state of the art review’, Soil Publication, No 300, M.I.T. 134 Wilson, J.L. 2003, Aseismic design of tall chimneys, CICIND Research Report, Vol. 13. 135 Wolf, J.P. & VonArx, G.A. 1978, ‘Impedance function of a group of vertical piles’, Proc. ASCE Speciality Conf. on Earthquake Engg. and Soil Dynamics, Pasadena, California, pp. 1024–1041. 136 Wolf, J.P. 1985, Dynamic Soil Structure Interaction, Prenctice-Hall Englewood Cliffs, NJ. 137 Wolf, J.P. 1988, Dynamic Soil Structure Interaction in Time Domain, Prenctice-Hall, Englewood Cliffs, NJ. 138 Wolf, J.P. 1994, Foundation Vibration Analysis: Using Simple Physical Model, PrenticeHall, Englewood-Cliffs, NJ.


Dynamics of Structures Volume 2 - Applications Chowdhury

Recommend Documents