Structural Systems

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Structural systems: behaviour and design

Structural systems: behaviour and design L. T. Stavridis

Published by Thomas Telford Limited, 40 Marsh Wall, London E14 9TP. www.thomastelford.com

Distributors for Thomas Telford books are USA: Publishers Storage and Shipping Corp., 46 Development Road, Fitchburg, MA 01420 Australia: DA Books and Journals, 648 Whitehorse Road, Mitcham 3132, Victoria

First published 2010 Also available from Thomas Telford Limited Structural systems: behaviour and design. Volume 1: Plane structural systems. Stavridis L.T. ISBN: 978-0-7277-4106-6 Structural systems: behaviour and design. Volume 2: Spatial structural systems, foundations and dynamics. Stavridis L.T. ISBN: 978-0-7277-4107-3 Structural analysis with finite elements. Rugarli P. ISBN: 978-0-7277-4093-9 Steel—concrete composite buildings: Designing with eurocodes. Collings D. ISBN: 978-0-7277-4089-2 Designers guide to eurocode 1: Actions on bridges. EN 1991-2, EN 1991-1-1, -1-3 to -1-7 and EN 1990 Annex A2. Calgaro J.-A., Tschumi M., Gulvanessian H. ISBN: 978-0-7277-3158-6 www.icevirtuallibrary.com A catalogue record for this book is available from the British Library ISBN: 978-0-7277-4105-9 # Thomas Telford Limited 2010 All rights, including translation, reserved. Except as permitted by the Copyright, Designs and Patents Act 1988, no part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying or otherwise, without the prior written permission of the Publisher, Thomas Telford Limited, 40 Marsh Wall, London E14 9TP. This book is published on the understanding that the author is solely responsible for the statements made and opinions expressed in it and that its publication does not necessarily imply that such statements and/or opinions are or reflect the views or opinions of the publishers. While every effort has been made to ensure that the statements made and the opinions expressed in this publication provide a safe and accurate guide, no liability or responsibility can be accepted in this respect by the authors or publishers.

Typeset by Academic þ Technical, Bristol Index created by Indexing Specialists (UK) Ltd, Hove, East Sussex Printed and bound in Great Britain by Antony Rowe Limited, Chippenham

‘To the memory of my parents’

Contents Preface

xi

Foreword

xiv

1

Introductory concepts 1.1 Loads 1.2 The structural behaviour of basic materials 1.3 Behaviour of a reinforced concrete member under tension 1.4 Behaviour of a prestressed concrete member under tension 1.5 Numerical examples 1.6 The design process and its control Reference

1 1 5 13 18 22 26 28

2

The use of equilibrium in finding the state of stress and deformation (statically determinate structures) 2.1 Introductory concepts 2.2 The handling of internal forces 2.3 Determining the deformations 2.4 Symmetric plane structures 2.5 Grid structures

29 29 41 65 84 87

3

4

The handling of deformations for determining the stress state in framed structures (statically indeterminate structures) 3.1 Introduction 3.2 The force method 3.3 The deformation method

93 93 94 114

Simply supported beams 4.1 Steel beams (reference material) 4.2 Reinforced concrete beams 4.3 Prestressed concrete beams 4.4 Cantilever beams 4.5 External prestressing

139 139 150 163 177 181 vii


4.6 Design control References

184 185

5

Continuous beams 5.1 Introduction 5.2 Steel beams 5.3 Reinforced concrete beams 5.4 Prestressed concrete beams 5.5 Creep effects 5.6 Composite beams References

187 187 188 195 195 199 209 221

6

Frames 6.1 Overview 6.2 Single-storey, single-bay frames 6.3 One-storey multibay frames 6.4 Multi-storey frames 6.5 Design of sections 6.6 Plastic analysis and design 6.7 Design and check of joints References

223 223 223 235 240 251 257 266 269

7

The 7.1 7.2. 7.3

influence of deformations on the state of stress — elastic stability Overview Buckling of bars The influence of deformation on the response of beams (second-order theory) 7.4 The influence of deformation on the response of frames 7.5 Lateral buckling of beams 7.6 Plastic analysis References

271 271 271

8

Arches 8.1 Basic characteristics of structural behaviour 8.2 Elastic stability — second-order theory 8.3 The girder-stiffened arch system 8.4 The tied-arch system References

297 297 304 308 311 313

9

Cable structures 9.1 Overview 9.2 Cable—beam structures 9.3 The freely suspended cable 9.4 Prestressed cable nets 9.5 Suspension bridges — the suspended girder

315 315 317 322 325 328

viii

277 282 290 293 296

Contents

9.6 Stiffening the suspension cable 9.7 Cable-stayed bridges References

335 339 349

10 Grids 10.1 Overview 10.2 Main characteristics of the structural behaviour of grids 10.3 Layout and structural action of skew bridges

351 351 351 358

11 Plates 11.1 The plate equation as a consequence of its load-bearing action 11.2 Orthogonal plates 11.3 Circular plates 11.4 Skew plates 11.5 Flat slabs 11.6 Folded plates References

361 361 369 385 386 389 399 402

12 Shells 12.1 Introduction 12.2 The membrane action as a basic design concept 12.3 Cylindrical shells 12.4 Dome shells 12.5 Hyperbolic paraboloid shells 12.6 Conoidal shells References

403 403 403 409 423 436 451 453

13 Thin-walled beams 13.1 General characteristics 13.2 The basic assumption of a non-deformable cross-section 13.3 Shear centre 13.4 Warping of thin-walled beams and the stress state due to its prevention 13.5 The bimoment concept 13.6 Two theorems of the bimoment 13.7 Warping shear stresses 13.8 The governing equation for torsion and its practical treatment 13.9 Examples References

455 455 460 461

14 Box girders 14.1 General 14.2 Rectilinear girders 14.3 Curved girders References

481 481 481 495 516

462 470 472 473 474 477 479

ix


15 Lateral response of multi-storey systems 15.1 Introduction 15.2 Formation of the system 15.3 Lateral response 15.4 Temperature effect References

517 517 517 519 528 532

16 Dynamic behaviour of discrete mass systems 16.1 Introduction 16.2 Single-degree-of-freedom systems 16.3 Multi-degree systems 16.4 Approximate treatment of continuous systems 16.5 Design for avoiding annoying vibrations References

533 533 535 555 577 580 586

17 Supporting the structure on the ground 17.1 Overview 17.2 General mechanical characteristics of soils 17.3 Shallow foundations 17.4 Pile foundations References

587 587 587 592 617 621

Bibliography

623

Index

625

x

‘‘PepaideumÝnon gar estin epi tosout! to akribÝ& epizhtein óson h tou prÜgmato& ’ýsi& epidÝcetai’’ AristotÝlh& ‘‘Because it is the essence of education to seek as much accuracy as the nature of things allows’’ Aristoteles

Preface What a technically educated person (engineer, architect or constructor) understands today by the term ‘structural design’ is practically the same as what his fellow man meant 500 — or even more — years ago, namely a procedure based on the application of a particular item of knowledge, and because of which a structure will ‘stand up’ and not ‘fall down’, under whatever actions it may be exposed to during its life span. However, what changed over the years and took the so-called ‘structural theory’ out of the realm of empiricism and intuition was the introduction of analysis in the assessment of structural behaviour and its elevation to an applied scientific field. Of course, the evolution of computational methods together with the wide availability of computer facilities played decisive roles in this. Structural mechanics is now a highly demanding subject, not only from the point of view of its analytical treatment regarding structural behaviour but also with regard to its evaluation and practical application in structural design. Both of these directions have quite distinct characteristics. The analytical approach always poses the question: Given a structural configuration and a certain loading, what is the response and its corresponding deformations? This is a problem governed by strict analytical conditions and requirements, establishing in this way the scientific character of the subject matter, but sometimes creating the illusion that the process of analysis is an end in itself. Of course, the solution to this problem is nowadays ensured, due to powerful numerical methods and the wide availability of personal computers. On the other hand, the practical aspects focus on the application of load-carrying behaviour in the conceptual and working design of structures, by posing the essential question: Given the physical environment and the prevailing service requirements, what structural system made of the appropriate materials will meet the necessary load-bearing requirements in an economical and aesthetically satisfactory way? This problem does indeed constitute an end in itself. It is the first approach which has become established, over the years, in education for many reasons, all of them hinging on the fact that it is this approach which ensures the appropriate ‘scientific’ profile. The second approach, although more realistic, seldom attracts the attention it deserves in the educational civil engineering curricula, a fact that the student unfortunately becomes aware of at a rather later stage, to his or her disappointment. xi


Indeed, the almost exclusive concentration on the computational aspect of structural mechanics dramatically deprives the young and inexperienced engineer from the physical perception of the load-carrying characteristics of a structural system, something which is not due solely to his or her lack of experience. The student concerned, being used to the study of complicated computational scenarios, tends to lack the mental clarity that permits the direct structural perception which is so often required either by the collaborating architect or the constructor on the building site. In this book, an approach to the understanding of load-carrying mechanisms and the behaviour of a wide range of structural systems is presented, with subsequent application to relevant design decisions, which, relying in principle on physical comprehension, is carried out through simple analytical reasoning. However, despite the prevailing non-computer-oriented philosophy, it is recognised throughout this book that the computational procedure in the design office practice needs in every case to be underpinned by the appropriate computer software. A necessary condition, though, for the successful use of such software is the ability of the user to derive, even approximately, results based on a well-cultivated structural perception regarding the load-carrying action and the subsequent preliminary design, which should in any case precede any computer modelling and the corresponding computation: this is exactly what the present book aims at. Thus, through the study of basic load-bearing actions and the behaviour of typical systems, such as simply supported and continuous beams, frames, arches, cable structures of any type, grids, plates, shells, rectilinear and curved thin-walled beams, and multi-storey systems, an attempt has been made to bring insights into these load-carrying characteristics that are necessary for their design and are usually overshadowed by a strictly analytical examination. In this respect, particular attention has been given to the use of reinforced and prestressed concrete, as well as to composite structures, in addition to the ‘traditional’ consideration of steel as the most ‘convenient’ material. Due attention has also been given to the plastic analysis and design of skeletal structures as well as to second-order theory and stability effects, with an emphasis on their practical and efficient use. However, it has been deemed an absolute necessity to consider in advance and in some detail the handling of plane skeletal statically determinate and indeterminate structures in two respective chapters. Moreover, the dynamic behaviour of discretemass structures has been examined in the penultimate chapter, where not only the response of multi-storey systems to earthquake but also that of beams and plates to human-induced dynamic actions as well as to machine operation is discussed. Of course, complete consideration of structural design must also include the foundation problem. Thus, the final chapter is dedicated to this issue, dealing primarily with the behaviour and design of shallow foundations, but also with soil—structure interaction and, to a lesser extent, the pile foundations. It is hoped that, through the systematic examination of the above subjects, the basic elements of structural perception are emphasised which permit the safe preliminary design and dimensioning of any structure, such as a bridge, building, roofing of a large space etc. These elements constitute the basis of not only appropriate computer xii

Preface

modelling but also of the deliberate acceptance — or not — of the numerical results produced by software, which is a very important issue. This book follows a strictly progressive path in the presentation of various subjects. Thus, later chapter build only on previously examined concepts, and some basic knowledge of elementary mechanics is considered a prerequisite. As is well known — since the first century BC, from the Roman architect Vitruvius — a successful structural concept requires the satisfaction of four characteristic properties, namely technical safety, functionality, economy and aesthetic quality. Technical safety means that the available strength should be greater than the resulting response; functionality means, structurally, the control of different deformational characteristics, annoying vibrations included; economy means the successful selection of the structural and foundation system, as well as of the appropriate construction method; and aesthetic quality means the achievement of structural elegance. While the first two criteria are the subject of a knowledgeable technical analysis, successfully meeting the last two requires, on the part of the engineer, ingenuity, creativity and aesthetic judgment — properties not acquired by studying the prevailing design characteristics of the structural systems but are nevertheless prerequisites. Thus, although the successful design of a structure is based on the technical and aesthetic talent of the engineer, the knowledge of the structural principles put forward in this book are an essential basis for any such endeavour. This book is aimed at everyone engaged in the study of structural analysis and design, either as a student, a practising engineer, or even as an architect seeking for a more profound structural understanding of bridge or building engineering. I hope that this books proves useful and achieves its goals. Finally, I would like to express my gratitude to Thomas Telford Ltd for undertaking the publication of this edition of this book and the excellent collaboration with the editorial team, whose authoritative and knowledgeable direction under Matthew Lane has led to an entirely satisfactory result. L.T. Stavridis Athens, June 2010

xiii

Foreword In the process of designing, engineers have to consider the different alternatives for structures, aiming generally towards an optimum of functional performance, resistance, durability and also towards a minimum of costs. In the dialogue with architects, the overall design as well as the suitable structural elements are looked for. Hence, the engineer has to master the art of structural design, has to understand the behaviour and the functions of beams, frames, plates, shells etc. Engineering knowledge is more than the ability to analyse a structure. It enables engineers to propose structural systems and also to determine a first estimation of the overall and sectional dimensions needed. A computer aided numerical analysis can be applied only after the design is chosen. It is a complementary tool, especially helpful to obtain results for dimensioning the structure and for parameter variations. For checking and validating computer results it is needed to apply engineering judgement, plausibility considerations and e.g. simple equilibrium checks. An engineer is a good engineer when he or she — even on vacation — does not only admire structures like bridges, towers, buildings, but when he or she is visualising the flow of forces, the structural elements for tension and those for compression, perhaps even imaging moment diagrams. This is rather easy where ‘form follows function’ (L.H. Sullivan) as for instance in suspension bridges, yet difficult for more complex structures such as thin shell structures. The book of Professor Stavridis Structural systems: behaviour and design follows the above issues very consistently, making a remarkable contribution to this rather rarely encountered book genre. It is an excellent textbook, especially for Civil Engineering students and young professionals. It covers the principal approaches to determine states of equilibrium and to calculate the deformations. For statically indeterminate structures, alternative methods are demonstrated using either forces or deformations as unknowns. In explaining the structural behaviour of beams, frames, plates, shells, thin-walled elements, pre-stressed concrete, foundation elements and dynamically incited systems, basic concepts and directly applicable approaches are presented that allow an estimation of crucial quantities for the design of a very wide range of structures, including bridges, buildings, roofing of larger spaces and foundations. Many graphic figures help to visualise the underlying mechanics. Thus, the main objective is very well achieved: to promote structural understanding as a complementary tool to computer analyses, to make students and readers fit for the art of designing. Heinz Duddeck Professor emeritus Dr.Ing., Dr.Ing. E.h. Institute of Structural Analysis, Technical University of Braunschweig xiv

Foreword Methods of structural analysis have experienced an explosive growth during the last 40 years. But it was the advent of powerful personal computers, along with the evolution of numerical tools (based mainly on the finite element method) and the parallel development of numerous reliable, comprehensive, commercially available computer software, that have enabled the engineer to tackle very complex structural systems. As a consequence, in today’s design offices, analysis of even some rather simple systems is performed (especially by the younger generation of engineers) with the use of such computer codes. Classical as well modern methods of structural analysis (based on the principles of virtual work, compatibility of deformations, matrix analysis) are rather rarely invoked in everyday practice. Yet, these theoretical tools often constitute the major (if not the only) part of the curriculum in civil engineering schools. Several problems may arise from this state of affairs. First, the danger of the ‘black-box syndrome’: when a sophisticated code is used without the analyst having the ability to check if the results are indeed reasonable and to spot any errors in the physical meaning of his/her implicit assumptions and on how these assumptions are materialised in the model. Second, there is little if any training to help the young engineer develop a deeper understanding of how structural systems behave, let alone to sharpen his/her physical intuition; such understanding and intuition are necessary especially in the conception and preliminary design stages. Indeed, conceptual clarity and physical insight are rarely mentioned as key objectives of structural analysis courses. The book by Professor Leonidas Stavridis offers a much needed addition to classical computational structural analysis: a physical approach is developed in which a structural system is decomposed into elements whose behaviour to the applied loads is easily computed ‘from the basics’. Starting in the first chapters with fundamental concepts and applications, the step-by-step exposition becomes progressively more advanced. Structural analysis blends naturally with mechanics of materials — the latter include reinforced and pre-stressed concrete, steel and composites. The in-depth analysis of standard structural systems (such as simply and multi-supported beams, frames, arches, cabled beams) is followed by the exposition of some more advanced topics such as buckling, plates and shells, thin-walled and box girders, grids and curved beams, laterally-loaded multi-story frames and shear walls. It is amazing how the analysis of such complex systems is made so simple, clearly understandable even to a non-specialist civil engineer, as the present writer. This is accomplished to a large extent thanks to the numerous illustrative figures (sketches) xv


which go far beyond the usual ‘formalistic’ figures of most available textbooks: they are imaginative, vivid, self-explanatory. What a difference they make when trying to comprehend difficult topics! For instance, the chapter on ‘Shells’ contains 51 elaborate figures, most of which comprise several sketches while a few of them are a whole page long. The 3D nature of cylindrical, spherical, paraboloid, and conical shells is elucidated with the help of ingeniously-selected isometric views and numerous cross-sections, so that the reader feels that this is a rather simple subject. As an engineer with special interest in soil—foundation—structure interaction I was particularly happy with the comprehensive treatment of foundations. Viewed mainly from a structural engineer’s viewpoint, the pertinent chapter deals not only with some classical deformation-settlement and stress-distribution problems, but also with the interplay between foundation stiffness and structure distress. I believe this book will prove invaluable to both students and active engineers in helping them not only to absorb a huge volume of material but (more significantly) to cultivate ‘engineering intuition’ and develop insight into the physics of structural analysis. For students, in particular, all this will offer the motivation for further study and the desire to later apply in real-life projects both the material and the methodology developed in the book. George Gazetas Professor of Soil Mechanics and Foundation Engineering National Technical University of Athens

xvi

1 Introductory concepts 1.1

Loads

Loads constitute the raison d’eˆtre of structural systems, and therefore their examination precedes anything else. Structures are built and designed in such a way as to carry, safely and in a functionally satisfactory way, only certain loads, and it is the responsibility of the designer to prescribe the loads which the structure is logically expected to be exposed to during its life. The determination of loads raises a difficult problem. Because of the obvious need to design structures by loading them in a manner agreed in advance for each category, structural loads are subject to regulations which typically differ from country to country. Thus, for example, the same bridge would be designed for different loads in the USA than in the UK or in Japan. Before examining the causes of the existence of loads, a distinction between static and dynamic loads should be made. A load P is considered to act statically when the time t1 needed for its full development is definitely longer than the fundamental period T of the structure. This corresponds practically to the time it takes for a structure to perform a complete oscillation, when an arbitrary deflection from its equilibrium situation is imposed on the structure and then left free to oscillate (Figure 1.1). Thus a wind gust increasing from zero to its maximum in 3 s represents a static force for a short, stiff building having a fundamental period of 0.5 s, whereas for a tall, flexible building having a fundamental period of 6 s the wind gust must be considered as a dynamic loading. It is clear that the way in which a dynamic loading is handled differs radically from that of a static one, for the simple reason that, because of the induced permanent motion, inertia forces are developed, which depend — at each moment — on the varying corresponding displacement of the structure. It is because of this complexity that it is always preferable, when feasible, to replace a dynamic loading with an equivalent static one. Loads, from consideration of their natural origin, can be categorised as below. (1) Gravity loads (g ¼ 9.81 m/s2) Loads which consist of the weight of the permanent components of a structure are characterised as dead loads. They are determined according to the specific weight of each material (e.g. reinforced concrete ¼ 25.0 kN/m3, steel ¼ 78.5 kN/m3). These loads are clearly static ones. Structural systems: behaviour and design 978-0-7277-4105-9

Copyright Thomas Telford Limited # 2010 All rights reserved

Structural systems: behaviour and design P

P P

When the system is left free it will vibrate with the fundamental period T t1 t1 > T t1 < T

t

Static loading Dynamic loading

Figure 1.1 Characterisation of load as a static or a dynamic load

Loads due to the occupancy or the use, in general, of mainly horizontal surfaces are characterised as live loads. This use may arise from human activities (rooms in residences or offices with equipment, people on pedestrian bridges or standing in stadiums, etc.) or from equipment installed in industrial spaces, or it may represent traffic loads due to moving vehicles on road bridges, as well as trains on railway bridges. These loads are expressed either as surface loads (kN/m2), or as concentrated loads (kN). They are termed ‘live’ because they may change their position in the structure permanently. This fact always poses the question of the configuration of these live loads, which then leads to consideration of the most unfavourable response in a structure. Live loads — especially moving traffic — are usually considered to act in a static manner, even if, from their very nature, they act dynamically. This is accounted for by an additional percentage of the acting vehicle weight, known as the impact factor. However, these traffic loads also always induce horizontal braking forces, which usually constitute about 5% of their corresponding weight. Normal human activities on specifically formed surfaces, as occur in rooms used for gymnastics and in dance halls, as well as on pedestrian bridges, also have a dynamic influence. (2) Loads arising from the soil that surrounds and/or supports the structure . Earth pressure is exerted on (mainly) vertical surfaces of the structure that come into contact with the surrounding soil. Earth pressure basically arises from the fact that a soil volume resting freely on a horizontal level usually shows a more or less formed slope, the free sides of which are standing in a state of equilibrium (Figure 1.2). This physical slope angle generally depends on the nature of the soil; for example, for sands this angle is roughly equal to 308. Thus, when this sloped surface is bound, through a structural surface, to be retained at a different angle to the physical one — usually vertically — the soil exerts on that surface a force that is stronger the greater the divergence from the physical slope angle. . This situation is in fact similar to that of the hydrostatic pressure, where the corresponding ‘physical slope angle’ obviously equals zero. This is why the earth pressure 2

Introductory concepts

e fac

Retaining wall

ur es

Earth pressure

p

e Fre

slo

Slope angle

Underlying soil

Figure 1.2 Earth pressure loading

(kN/m2) is determined as the product of the height of soil mass, times its specific weight, multiplied by a factor less than that applied for water, i.e. less than unity. Such additional pressures also act on a retaining wall surface, when live loads have to be considered on the free horizontal soil surface (see Chapter 16). . The soil, on which the structure is supported, may exhibit settlement due to causes which are either unrelated to the structure (e.g. groundwater lowering) or related to the deformability of the structure under loads acting on it. In the latter case, there is a substantial interaction between the foundation soil and the structure. Whatever the case, soil settlement definitely constitutes a loading on the structure, causing internal forces within it, unless the structure is statically determinate (as will be examined in Chapters 2 and 3). . During an earthquake the foundation soil exhibits horizontal and/or vertical vibrations, which are imposed on the foundations of the structure. As a result of this dynamic loading, the whole structure is subjected to strong internal forces. As will be examined in Chapter 16, it is the imposed accelerations on the foundation that actually cause this state of stress in the structure; the ensuing soil settlement may be an additional factor, as explained above. (3) Loads arising from the aquatic environment The acting water may be either free, as in the case of a marine structure or a dam, or contained in the surrounding soil. The main effect of water is the hydrostatic pressure, which acts on every surface embedded within a water-containing environment. In the case of strong water motion relative to the structure, additional hydrodynamic pressures come into play. If the structure is confined with its lower part in soil containing groundwater, buoyancy forces develop, which are merely the upward hydrostatic pressure on the lower horizontal surface of the structure (Figure 1.3). Although the buoyancy force decreases the weight of the structure acting on the soil, the lower surface of the structure is acted on by the unreduced weight in the upward direction. (4) Loads arising from climatic conditions . Wind forces represent basically a horizontal loading that is dynamic in character, as explained above. The wind force experienced by a structure is directly related to 3


Water table

Water pressure

Water table

Water pressure

Pressure on the soil The soil is acted on by a reduced pressure because of buoyancy

Pressure on the slab The slab is acted on by the full (unreduced) weight of the structure

Figure 1.3 Loads in an aquatic environment

the form of the structure and its height above the earth’s surface. A thorough determination of wind forces is particularly difficult, as the forces may interact with the ensuing structural deformations (e.g. in the case of cable-suspended bridges), leading to complex aerodynamic calculations. However, in most cases wind pressures can be determined from the appropriate structural codes. . Snow — in fact a gravity loading — is taken into account by means of a prescribed pressure on a roof, according to the appropriate structural codes. The prescribed pressure values take into account the geographic region as well as the altitude of the specific site. . Temperature variation may be uniform (an increase or reduction) or linearly distributed across the thickness of a structural element. It should be noted that the unavoidable shrinkage of concrete during its hardening may be simulated by a uniform temperature fall of 208C. This causes a state of stress in the structure, which may not be negligible (see Chapter 15). A non-uniform temperature change always appears in structural elements exposed to the atmosphere (space covering roofs, bridges, etc.). It should be noted that, as in the case of soil settlement, a temperature variation in a statically determinate structure does not give rise to any internal forces at all (see Chapter 2). (5) Special impact loads Impact loads are dynamic in nature and are due to events such as an explosion, a collision of a body with the structure (e.g. a ship colliding with a bridge pier), etc. 4

Introductory concepts F F A σ = F/A

A

ε = δ/L

δ

σ L L

Deformed specimen

Figure 1.4 The concept of stress

1.2

The structural behaviour of basic materials

The structural strength of a material is determined on the basis of the relationship between the stress and the strain " experienced by an appropriate bar specimen subjected to pure tension or pure compression. Consider a specimen with a length L and a section A, subjected to a force F acting along its longitudinal axis. Assuming a uniform distribution of the force on the section, the concept of stress is introduced as ¼ F/A (kN/m2) In this way the stress is uniformly distributed over the whole section A of the specimen (Figure 1.4). The force F concerns the whole specimen externally, while the stress characterises the internal state of stress of its longitudinal fibres. It may be considered that the application of the force F and the development of the stress over the whole section A represent equivalent factors, given that F ¼ A. The result of the application of the force F is an elongation or shortening of the specimen, depending on whether F is applied as a tensile or as a compressive force. Thus the concept of strain " may be introduced as " ¼ /L (non-dimensional number) The relationship between with " is obtained experimentally, by loading the specimen progressively and at a constant rate, starting from a null loading level and continuously recording the pairs of magnitudes and " on a diagram over two orthogonal axes, as they are calculated on the basis of the measured values of F and .

1.2.1

Steel

Figure 1.5(a) shows the experimentally obtained diagram for the behaviour of steel. The diagram represents a test done on steel in tension, but applies equally to steel in compression. It is clear that, for each pair of values (, "), the force F and the 5

Structural systems: behaviour and design σ (4)

(3) (2)

fy

(1)

Unloading

E εy

εy

0.10

εtot= εy+ εpl

εpl

ε

(a)

εtot σ

fy

2.0 εy

5.0

8.0

ε: ‰

Idealised σ–ε diagram for steel (b)

Figure 1.5 General characteristics of the behaviour of steel

corresponding deformation can be determined as F ¼ A and ¼ " L (a) It should be noted that the test is made by imposing on the specimen a specific deformation " rather than a stress . The rate of application d"/dt of " is approximately 5 105/s. It can be seen from Figure 1.5(a) that during the test four distinct behaviours are observed. In region (1) the stress is proportional to the strain ", which means — as will be shown later — that the corresponding force F is also proportional to the deformation . This fundamental statement, widely known as Hooke’s law, can be written as ¼"E The magnitude of E is derived geometrically as the slope of the straight segment (1). It represents the most basic quantitative characteristic of the load-carrying behaviour of the material and is called the modulus of elasticity. It can be thought of as a measure of the resistance (stiffness) offered by the material against its unit elongation or shortening. In the case of steel, the value of E is approximately 2.1 108 kN/m2. Substituting the above two equations (a) into the preceding equation gives the relationship between the external force F and the deformation of the specimen: F ¼ (E A/L) The expression (E A/L) is called the axial stiffness of the specimen (kN/m). It gives the force required (kN) to deform (either elongate or shorten) the specimen by 1 m; in other 6


words, it gives the resistance (stiffness) (kN) that the specimen offers when subjected to an axial deformation of 1 m. The last equation may also be written in the form ¼ F (L/E A) which shows how the axial deformation of the specimen can be calculated from the externally applied force F. The expression (L/E A) is called the axial flexibility (or deformability) (m/kN). It gives the axial deformation of the specimen (m) when an axial force of 1 kN is applied. It is clear that these two concepts (and values) of ‘stiffness’ and ‘flexibility’ are inversely related. The behaviour of the steel in region (1) is called linear behaviour. This is a very convenient material property because it is governed by an absolute proportionality between the force F (or stress ) and the deformation (or strain "). Thus if the force F (or the stress ) is multiplied by a factor k, the deformation (or the strain ") is also multiplied by the same factor. However, this property is valid only up to a certain stress level fy, at which point region (2) of the behaviour of the steel begins (see Figure 1.5(a)). Up to the point where the stress reaches the value fy, if the load is removed from the specimen the deformation (as well the strain ") will return to zero. In other words, up to this point the deformation is reversible, and for this reason the region up to fy is called the elastic region. For bars used in concrete reinforcement the value of fy may be considered equal to 420 N/mm2, and the corresponding "y, i.e. ( fy/Es), is about 2%. Representative values of fy for the two basic quality types of structural steel are 240 N/mm2 and 360 N/mm2. Once the stress has reached the value fy, the specimen can be deformed further without an increase in the applied force (i.e. stress). The steel exhibits zero stiffness (the slope of the graph is equal to zero), which means it ‘yields’, and so the stress fy is called the yield stress. This state prevails up to a value of " of roughly 10%, where the steel acquires a little stiffness again, as can be seen from the small slope at the beginning of region (3). To obtain a further increase in the stress ", a small increase in (i.e. in the force F) is required, and this property of the steel is called hardening. After region (1), the steel no longer exhibits ‘elastic behaviour’. Removing the loading from the specimen at any time will lead to a course parallel to the ‘elastic region’, and so will not lead to zero deformation but to a ‘remaining’ deformation, represented by the strain "pl. Thus the specimen cannot revert to its initial length and will be either a bit longer or a bit shorter, according to whether the stress imposed was a tensile or compressive one. Therefore, it is said that the material has undergone a plastic deformation, i.e. a plastic strain "pl. It is clear that the total deformation of the specimen is represented by the total strain "tot (see Figure 1.5(a)), which is given by "tot ¼ "y þ "pl The material property of yielding, i.e. of deformability under the constant stress fy, is a very important structural property of steel, called the ductility, which confers beneficial effects on the load-carrying capacity of structures. This holds not only for steel structures but also for reinforced or prestressed concrete structures, as will be explained later with 7


respect to the plastic behaviour of these materials. Now, regarding the very small slope of region (3), this minimal stiffness is considered to be practically zero, exactly as in the yield state, and consequently the very small increase in stress fy in this region is simply ignored. So, the state of yielding can be considered to extend up to values of " of about 10%, but for practical design purposes the value is taken as 8%. However, for the detailed control requirements in an investigational study, the aforementioned hardening property of steel must be taken into account. It can be see from region (4) in Figure 1.5(a) that to increase the deformation (i.e. of strain ") further requires increasingly less force F (i.e. stress ), until the point where the specimen fails. This region is of no practical interest for the design of structures. From the foregoing it can be concluded that, for practical purposes, the maximum level of stress on the steel must not exceed the value fy. Thus in practice the behaviour of steel can be taken as the behaviour of a homogeneous, ideal, elastic—plastic material (see Figure 1.5(b)). Such behaviour characteristics are used in the development of the classic methods of structural analysis. This idealised material behaviour is generally used in this book, but special attention will be given to those peculiarities of concrete structures which do not necessarily conform to this ideal behaviour. The stress—strain diagrams in Figure 1.6(a) are for naturally hard steel, the strength of which is due entirely to its chemical composition. This quality of steel is used for both structural elements and reinforcing bars. However, when prestressed concrete is used, a prestressing steel of very high strength is needed (for reasons that will be explained later), and this is obtained through a cold forming procedure. The stress—strain diagram for high-strength steel is shown in Figure 1.6(c). Comparing this with the graph for natural steel, it is immediately observed that there is no ‘yield plateau’, which means that there is not a clearly recognisable yield stress in the way that has been previously established. However, a yield stress fPy may be considered, which corresponds to the point where removing the load from the specimen will lead to a remaining plastic strain "pl of 2%. This yield stress may be considered as constant after the value "Py 8% has been reached, as shown in the idealised diagram in Figure 1.6(b). For practical purposes the actual ultimate tensile strength of the prestressing steel beyond fPy can thus be ignored.

1.2.2

Concrete

Steel exhibits practically the same stress—strain diagram whether the force applied is a tension or compression force. However, the behaviour of concrete under tension is totally different from its behaviour under compression, as the compressive strength of concrete is much higher than its tensile strength. The stress—strain diagram for a specimen of concrete under compression is shown in Figure 1.7. It is seen that up to a loading level roughly equal to one-third of the maximum compressive strength fc the concrete exhibits linear elastic behaviour. The constant slope in this region represents the value of the modulus of elasticity of concrete Ec, which is roughly equal to 25 000 kN/m2. 8


σ: N/mm2

Reinforcing steel ~400

Structural steel

~240 E 2‰

ε

8‰ (a) σ: N/mm2

Prestressing steel ~1600

Unloading σ: N/mm2

~1600

fPy

8.0 εPy

10.0

ε: ‰ ~E

Idealised σ–ε diagram for prestressing steel

ε

2‰

(b)

(c)

Figure 1.6 Characteristics of the behaviour of reinforcing and prestressing steel

σ: N/mm2 ~20 to ~45

fc

~fc /3

1.5–2.5‰

ε

Linear behaviour

Figure 1.7 Stress—strain diagram for concrete under compression

9


After the point fc/3 and up to the highest sustainable stress fc the relationship between and " is no longer linear, and consequently Hooke’s law does not apply. The strain corresponding to the highest stress level fc lies roughly between 1.5% and 2.5%. The stress fc represents the compressive strength of the specimen. Any further deformation of the concrete requires an applied stress of less than fc. This fact makes concrete in this last region of the stress—strain diagram, from a design point of view, effectively useless. An interesting fact observed experimentally is that increasing the speed of imposing the deformation increases the maximum sustainable stress (i.e. strength) fc. It is found that if a stress " is imposed at a speed of 102 to 101/s — which corresponds to the rate during an earthquake — there is an increase in the compressive strength fc of 50%. It should be noted that no such increase in strength is observed in the case of steel. As has been already pointed out, the tensile strength of concrete fct is much lower than its compressive strength. The tensile strength can generally be estimated using the empirical relationship fct ¼ 0:45

pffiffiffi fc

ðN=mm2 Þ

and is of the order of 2 N/mm2. Obviously, the concept of a yield stress for concrete simply does not exist. Exceeding the tensile strength of concrete leads automatically to cracking. In most concrete structures the load under which cracking occurs is only a part of the loading that the structure has to support under service conditions, i.e. under conditions of everyday use. Thus it is necessary to place reinforcing bars in the concrete in order to overcome the tensile stresses within the structure. Despite this measure, the concrete can still crack, and this poses different requirements on the design from those for a steel-only structure. The placement of reinforcing bars actually creates a new composite material, reinforced concrete, which is able to carry not only compressive but also tensile internal forces. The particularities of the behaviour of reinforced concrete under axial tension are dealt with in Section 1.3.

1.2.2.1 Creep If a compressive stress 0 within the elastic range is applied to a concrete specimen and is maintained at a constant level for a long time, it will be seen that, after an initial elastic shortening, the specimen will continue to shorten as time progresses. This phenomenon is called creep and is an unavoidable property of concrete (Figure 1.8). This deformation will continue for a period of more than 1 year, and will finally take a value that is larger than the initial elastic deformation. This extra deformation due to creep depends on the initial elastic deformation "el and the creep factor ’. The value of ’ depends on the age of the specimen at the time when the stress 0 was applied (accounted for by a factor k ), the relative humidity of the environment (a dry environment enhances creep), and the ratio of water to cement weight. 10

Introductory concepts σ0 σ0 εel = σ0/E Immediate deformation

εcr = φ∞ · εel

σ0

Gradual (slow) deformation

εtot = εel + εcr

εtot = σ0 · (1 + φ∞)/E = σ0/Einit

The compressive stress of concrete remains permanently constant

Figure 1.8 The creep behaviour of concrete

The final creep strain "cr, i.e. the creep strain after an infinite time, can be written as (see Figure 1.8) "cr ¼ ’1 "el ¼ ’1 0 Ec where ’1 ¼ ’ k Depending on the age of the concrete at the time when the stress 0 is applied, the factor k takes different values, being 1.8, 1 and 0.02 for ages of 1 day, 28 days and 5 years, respectively. It must be pointed out that this deformation is not reversible because, after the unloading of the specimen, i.e. after the removal of the stress 0 that caused "el, a remaining deformation is observed. Thus, the total strain "tot of the specimen at an infinite time is "tot ¼ "el þ "cr ¼ (1 þ ’1) "el The value of ’1 normally lies between 2.0 and 3.0. g The basic problem with regard to creep is to determine the final strain "1 when an initial stress 0 is applied at time 0 (the ‘age’ of the concrete) and then varies (say, increases) gradually over time until its final value 1 (Figure 1.9). Systematic research over the years has led to a practical and very functional conclusion known as Trost’s proposal (Menn, 1990) which, assuming that the principle of superposition of creep deformations is generally valid, can be expressed as (see Figure 1.9) 0 ð1 þ ’1 Þ "1 ¼ 0 ð1 þ ’1 Þ þ 1 Ec Ec where the value of the coefficient (which depends on 0) is of the order of 0.85. The last relationship may be written in the form 0 "1 ¼ 0 þ 1 Einit Edif 11

Structural systems: behaviour and design σ σ∞

Gradual (slow) additional application of stress

Additional deformation Edif = E/(1 + µ · ϕ∞)

ε = ∆σ/Edif ε∞

+

∆σ

σ0

Influence of time duration

Einit = E/(1 + ϕ∞)

ε = σ0/Einit τ0

σ0 t∞

Instantaneous application of stress t

Variation of the initially applied stress (Trost’s proposal)

Figure 1.9 The creep behaviour of concrete with a variation in applied stress

where Einit ¼

Ec 1 þ ’1

and

Edif ¼

Ec 1 þ ’1

This expression of "1 is of particular importance. First, it shows that the effect of creep is generally equivalent to a reduction in the modulus of elasticity Ec of concrete. For the initially applied stress 0 a value of the modulus equal to Einit is considered, whereas for any further gradual increment in the stress — assumed as (1 — 0) — the value Edif applies. In other words, the concrete responds instantaneously to an initially applied stress through the value Ec, but over the course of time responds with the value Einit, on the condition that the applied stress remains constant. For any further response due to an additional gradual variation in the initially imposed stress, the value Edif must be applied. In this way, an instantaneous elastic deformation w due to an initial stress results in the value w1 ¼ w (1 þ ’1) by applying the modulus Einit, whereas the gradual application of an additional stress corresponding to an ‘instantaneously produced’ elastic deformation w results in the value w,1 ¼ w (1 þ ’1) by applying the modulus Edif . Of course the total deformation that actually occurs is w1 þ w,1. g When looking at the deformation properties of concrete, shrinkage must also be considered. During its curing and drying there is a shrinking of the concrete mass, and this is expressed as the corresponding shortening strain "s. For practical purposes the value of "s is taken as 20 105. This is equivalent to the effect of a uniform temperature fall of 208C, assuming a thermal coefficient of 105/8C, a value that is also valid for steel.

1.2.2.2 Relaxation The property of concrete relaxation has characteristics opposite to those of creep. More precisely, if a shortening strain 0/Ec is imposed on a concrete specimen through an applied compressive stress 0 and care is taken such that this deformation is kept 12

Introductory concepts σ0 σ∞ < σ0

Why the stress decreases

εel σ0 Immediate deformation

Blocked deformation

Free deformation

Figure 1.10 The relaxation of concrete

constant (i.e. is blocked), then as time progresses there is a decrease in the concrete stress (Figure 1.10). In other words, the stress required for the initially imposed specimen shortening to remain constant decreases continuously over time. This reduction in stress under constant strain is called relaxation and is due to the creep property of concrete. It can be explained as follows: if the initial compressive stress 0 is held constant, the specimen should shorten further over time, according to its creep behaviour. It is obvious that, in this hypothetical configuration, in order to keep the initial shortening intact, an additional tensile stress has to be applied, which of course leads to a reduction in the initial compressive stress (see Figure 1.10). The amount of relaxation is easily determined on the basis of Trost’s relation (see Section 1.2.2.1). It must simply hold that 0 ð1 þ ’1 Þ ¼ 0 "1 ¼ 0 ð1 þ ’1 Þ þ 1 Ec Ec Ec from which it is obtained that ’1 1 ¼ 0 1 1 þ ’1 In this way, for ’1 ¼ 2.0 and with ¼ 0.85, the final value 1 amounts to only 26% of the initially applied stress 0. This last equation plays a particular role in evaluating the stress in statically indeterminate concrete structures under an imposed deformation, as will be explained in Section 5.5.1. g It should be noted that the phenomenon of relaxation occurs also in prestressing steel, wherein there is a percentage loss of the initially applied stress under constant strain. This loss depends on the bar diameter and is roughly in the region of 5%.

1.3

Behaviour of a reinforced concrete member under tension

The peculiarity of concrete as a structural material, as compared with steel, does not lie so much in the fact that the —" diagram is not linear after a certain stress level, but in 13

Structural systems: behaviour and design Bond stresses As, Es

Z

Ib

Ac, Ec ZC

Z

ZS

Z = Zs + Zc Zs < Z

τS

Bond stresses

Z Ib

ZS

Z = Zs + Zτ

ZC

Zτ = Zc

Bond stresses Zτ Steel stress

Equivalent section

σs

Concrete tensile steel

σc

Z

σc (fct) Ib

Ac,i, Ec

Figure 1.11 The distribution of stresses due to bond forces

its inability to afford a significant tensile resistance, a fact that makes the placement of reinforcing steel bars necessary, as already noted above. Thus, an understanding of the behaviour of a simple reinforced concrete member under axial tension is of basic importance. Consider a straight concrete member (Figure 1.11) with a reinforcing bar placed at the centre of its square section and sticking out a little at both ends of the member so that a self-equilibrating pair of tensile forces Z can be applied at its extremities. Let Ac and As be the cross-sectional areas of the concrete and the steel member, respectively, and Ec and Es the corresponding moduli of elasticity. Under the action of the forces Z the steel bar becomes elongated. It is of absolutely basic importance that this steel elongation is imposed on the concrete too. This is achieved through the adhesion or bond forces between the steel and the concrete, which are more efficient if the surface of the steel bar is roughly formed (e.g. ribbed). The bond between the two materials ensures that all along the reinforcing bar "s ¼ "c. Obviously, if there is a total absence of a bond then "c ¼ 0, which means that the concrete would be completely uninvolved in carrying the tensile load — something that is undesirable, as will become clear. Before the material stresses near the ends of the member are examined, a section near the middle of the specimen is considered, in order to determine the tensile stress of the concrete c and the steel s at this point. If Zc and Zs are the tensile forces carried by the concrete and the steel, respectively, then considering the equilibrium of, say, the left-hand part as a free body (see Figure 1.11): Z ¼ Zc þ Zs ¼ Ac c þ As s ¼ Ac ("c Ec) þ As ("s Es) 14


Given the identical strains of the two materials ("s ¼ "c), then s ¼ c (Es/Ec) and, more specifically, on the basis of the last equation: s ¼ "s Es ¼

Z

E As þ Ac c Es

(a)

and c ¼ "c Ec ¼

Z

Z ¼ Es Ac;i Ac þ As Ec

(b)

From the above equation it may be deduced that the concrete behaves as if its section were homogeneous and equal to Es Ac;i ¼ Ac þ As Ec This cross-section Ab,i, which is somewhat larger than Ac, is called the equivalent crosssection (see Figure 1.11). The concept of the idealised cross-section, which will also be used in other cases in this book, allows the material — in this case the concrete — to be considered as ‘homogenous’, by considering a cross-section that is larger than its real dimensions. If the initial magnitude of the force Z is small enough, producing a stress c in the concrete that is lower than its tensile strength fct — thus leaving the concrete uncracked — it is immediately concluded from equation (a) that s < Z/As and consequently, according to Figure 1.11, the tensile force on the reinforcing bar Zs ¼ As s at the considered point is always smaller than the external force Z. However, in order to explain the longitudinal equilibrium of the reinforcing bar considered as a free body, it must be concluded, and indeed should be intuitively clear, that friction (or adhesion) stresses are developed along the bar on its lateral surface. These are better known as bond stresses s, and contribute to the overall equilibrium (see Figure 1.11). The distribution of these bond stresses along the reinforcing bar is shown qualitatively in Figure 1.11 (Menn, 1990). It can be seen that these stresses develop over a restricted length lb and then disappear. This length lb, known as the bond or anchorage length, allows the introduction of the external force Z into the concrete in order to produce a uniformly distributed state of stress. The anchorage length depends on the layout of the reinforcement and the quality of the concrete. The bond stresses act on the concrete in the opposite sense to the way they act on the steel reinforcement, and they explain its equilibrium as a free body, given that no force is acting at its left-hand end, while in the location of the reinforcing section the tensile force Zc ¼ Ac c is acting. The integral of the bond stresses acting over the length lb gives the bond force Z which, on the basis of equilibrium of the reinforcing bar, is Z ¼ Z — As s, while from the longitudinal equilibrium of the concrete part Z ¼ Zc (see Figure 1.11). 15

Structural systems: behaviour and design σc (fct)

Ib

σc (fct)

Ib

Ib

Concrete tensile stress

Ib

Crack Bond stresses Zr

Zr σs

Crack

Steel stress

Crack fct

Zr

Ib

Crack Zr

Next crack location Ib

Zr

Crack

σc < fct

Zr

1.0) is agreed in advance, the structure should be safe against collapse for every load P þ P less than P, while for a load P (or greater) the structure may reach a state of collapse. 26


According to the ‘old’ method, the so-called working stress design, if the load P would cause stresses equal to the magnitude of (ultimate strength/), it was assumed that a load P would lead to the ‘ultimate strength’ (i.e. to the collapse of the structure) and, moreover, that a load P þ P < P should produce stresses that would be lower than this ‘ultimate strength’. Of course, according to the structural knowledge now available, both these assumptions are false. The concept of the factor of safety for a designed structure can be applied through the theory of plastic analysis, which will be examined later in this book. However, the factor of safety can be practically ensured by means of the following consideration (see Chapters 5 and 6). Let R represent the ultimate resistance (strength) of a designed section in some part of the structure, and let S represent a sectional force of that section corresponding to the load P, under the condition that the equilibrium requirements are satisfied. The ‘sectional forces’ S account for the whole state of stress of the section under consideration, including its axial, bending, shearing and torsional stresses (this will be clarified in Chapter 2). However, until now, the examination of the state of stress has been restricted to the axial stress, and for the time being the axial force can be considered as being representative of the ‘sectional force’ concept. If the structure is designed in such a way that the relation S R (a) is satisfied everywhere, then it can be proved that the structure will collapse at a load greater than or equal to P, which means that the existing factor of safety is greater than or equal to . In recent structural codes, the above relationship is given in the following form: S S R/ R

(with S R ¼ )

The factor S is intended to cover the uncertainty regarding the application of loads, as well as all kinds of inaccuracy or inconsistency in the design of the structural model. Clearly, therefore, smaller factors are applied for permanent loads than for live loads (see e.g. Eurocode 1.3 for permanent loads and 1.5 for live loads). The factor R expresses the uncertainty with regard to the accepted strength of the materials used (being greater for concrete than for steel). Thus, the control of technical safety lies in the verification that the ‘magnified’ internal forces during the ‘state of service’ are smaller than the reduced strengths of the designed sections. However, it should be emphasised that the dominant condition for safety remains the initially formulated one, according to which a load of greater than or equal to P (but not less) should lead to collapse. This is noted in particular because, in cases where the structural safety is connected to the so-called buckling load — which is examined in detail in Chapter 7 — the mere satisfaction of the condition (a) above under simple equilibrium requirements is not relevant. Of course, the control of a designed structure at the ultimate load level alone does not suffice, as it does not ensure satisfactory behaviour of the structure in everyday use, a 27


factor that is equally important. Thus, there must be additional control of the response of the structure to the service loads ‘P’, taking into consideration excessive deformations, excessive crack widths in the case of concrete, or perhaps annoying vibrations induced by people using the structure or by machines, etc. Finally, it should be noted that in some special cases the design of a structure could be based on some ‘allowable stresses’, which should not be exceeded.

Reference Menn C. (1990) Prestressed Concrete Bridges. Basel: Birkhaüser Verlag.

28

2 The use of equilibrium in finding the state of stress and deformation (statically determinate structures) 2.1


Load-bearing structures consisting of longish members of, usually, constant crosssection, the largest dimension of which does not exceed about one-fifth of the member length, are called skeletal structures. These members, which are usually rectilinear but may also be curved, are assumed to be made of one of the materials examined in Chapter 1. The skeletal structures consisting of these elements must be able to take up all the loads for which they are designed and transfer them safely to the ground through appropriately designed supports, which together are called the foundation. Any such element is by its very nature a three-dimensional one but, in order to be studied, it is idealised as the centroidal line of cross-sections all along the member (Figure 2.1), which is supposed to fulfil all its relevant bearing properties. In this chapter, all these lines that substitute as the elements of a structure are considered as belonging to a certain plane, thus forming a plane skeletal system. This system, if appropriately supported, may be loaded either in its own plane, in which case it is called a plane structure, or perpendicularly to it, in which case it is called a grid structure. This chapter deals mostly with plane structures.

2.1.1

Forces

The concept of force is of basic importance. Each force is a vector quantity and is fully determined when the following items are known: (a) its application point, (b) its line of action (direction) with the respective sense and (c) its magnitude. It is understandable that with the application point already known, the force is also determined if its projections Px and Py on two arbitrary axes x and y (not necessarily orthogonal) are known. The unit of force is the kilonewton (kN). The loads on plane structures are considered either as forces concentrated on specific points or as distributed forces along a skeletal member (Figure 2.2). The latter are measured in kN/m, expressing how many kN are uniformly acting on a length of 1 m along the element. Usually, when designing structures uniformly distributed loads with a constant value of kN/m along a member are considered. Structural systems: behaviour and design 978-0-7277-4105-9


Structural systems: behaviour and design Centroidal axis

Members are simulated by lines (model)

Real member layout Support Foundation

Support

Foundation

Figure 2.1 Idealisation of skeletal members

2.1.2

Conditions of equilibrium

Equilibrium constitutes the most basic condition that must exist and prevail in a structure. It concerns the forces acting on the structure, as well as the forces that act on any part arbitrarily cut out of it. It should be made clear that the following three conditions are necessary and sufficient for the equilibrium of any forces acting on a plane formation. This means that if equilibrium exists in a plane formation, the satisfaction of these three conditions with regard to the acting forces is implied; and, conversely, the fulfilment of these three conditions will ensure the equilibrium of any plane formation undergoing these forces. The three conditions are (Figure 2.3): (1) The sum of the vectorial projections of all the acting forces on an arbitrary axis should be zero, i.e. the forces should be mutually cancelled. This axis may be an arbitrarily chosen line with an arbitrary positive sense. The vectorial projections are signed with respect to the selected positive sense of the axis. (2) The sum of the vectorial projections of all the acting forces on another arbitrary axis should also be zero. This second axis should not be parallel to the first. All other previous remarks remain the same. (3) The sum of all the moments of all the forces with respect to an arbitrary point of their plane should be zero. This presupposes for the plane considered that an arbitrary rotational positive sense should be selected (either clockwise or anticlockwise). Thus the moment of each force with respect to the selected point may be analogously signed.

3.0 kN/m 4.0 kN/m

2.0 kN/m

Px

∆s

Py

3.0 · ∆s Py Px

Figure 2.2 External forces on a member

30

Stress and deformation (statically determinate structures)

Reference point of moments

O

x

y

Positive sense

Positive sense

Equilibrium ΣPx = 0 ΣPy = 0 ΣMo = 0

Arbitrary selection of axes

Figure 2.3 The three equilibrium conditions of a free formation

The need to satisfy the above three conditions, which are expressed by the three equations shown in Figure 2.3, has the following basic consequence. If a plane formation is in equilibrium under various forces and moments containing, at most, three unknown pieces of information, then these unknown data can be determined directly using the three equations of equilibrium. Furthermore, the above equilibrium conditions lead to the following conclusions (Figure 2.4): . When a plane formation is in equilibrium under three concentrated forces, these three forces pass through the same point. This is explained by the fact that, otherwise, P1

P2 P3 P1 P2 Equilibrium

Triangle of forces

P3

The three forces must pass through the same point

B A

The two self-equilibrating forces lie on the direction AB

The acting forces at the member ends lie on their axes

Figure 2.4 The consequences of the three equilibrium conditions

31


the moment of any one force with respect to the point of intersection of the other two cannot be balanced. It should be noted that in the case when one force is completely known, but for the other two only their directions are known, then both their sense and magnitude may be qualitatively (and quantitatively) determined through the so-called triangle of forces, as shown in Figure 2.4. In the triangle of forces, all forces must follow the same rotating sense. Thus the triangle can be used to quickly obtain an idea of the relative magnitude and sense of the two forces, in contrast to the more time-consuming process of undertaking a purely analytical manipulation of the equilibrium equations. . When a plane formation is in equilibrium under only two concentrated forces, these two forces must act on the line connecting their points of application, and the two forces must be equal and opposite in magnitude. This has a direct consequence with regard to rectilinear members that are embedded in a plane structure in such a way that the only forces they can receive are concentrated at their ends and without a moment. Such (hinged) members develop an exclusively axial response (tensile or compressive) (see Figure 2.4).

2.1.3

Principle of action and reaction

This is a fundamental principle of statics. When a body I comes into contact in any way with a body II, then at the common point of contact two forces are developed: (a) body I receives from body II the force PI, and (b) body II receives from body I the force PII. According to the action and reaction principle, PI ¼ —PII.

2.1.4

Support conditions

There are three possible ways of supporting a plane skeletal structure at any point — simple, hinged and fixed support. Each way consists of a certain number of restrained and freely developing displacements with regard to the supported point of the structure. The restrained displacements give rise to the development of forces that act as reactions on the structure. . Simple support. In this case the displacement of a supported point is restrained (nullified) in a specific direction a. This point may be displaced freely in the perpendicular direction, and can rotate freely (Figure 2.5). Consequently, at the supported point the

a

Symbol

a Simple support

Figure 2.5 Simple support

32

Reaction in a specific direction a


or Symbol Hinge

Force with an arbitrary direction R

R

Figure 2.6 Hinged support or connection

structure undergoes a reaction of only one concentrated force in the direction a, and no force in a perpendicular direction and no moment can be developed there. . Hinged support. In this case the displacement of the supported point is restrained (nullified) in any direction, but the point may rotate freely (Figure 2.6). Consequently, at the supported point the structure may undergo as a reaction any force in any direction, while any moment is excluded. The notion of ‘hinged’ may also be used for the connection of two elements or skeletal formations if, through this connection only, a force may be mutually transferred (according to the principle of action and reaction). Obviously, this force may have an arbitrary direction (see Figure 2.6). . Fixed support. In this case the supported point is restrained in any direction as well as against rotation (Figure 2.7). As a direct consequence the structure may be acted upon at this point by a force of any direction as well as by a moment. The notion of ‘fixed’ is also used for the connection of two members or even skeletal formations, when, between them, a force of any direction as well as a moment may be mutually transmitted. g

Symbol Fixed support

Force with an arbitrary direction and moment R M M R

Figure 2.7 Fixed support or connection

33


Equilibrium

Reactions on the structure

Actions on the ground

Figure 2.8 The equilibrium of a structure

As pointed out previously, a structure is in equilibrium under the acting external loads (forces and/or moments), as well as the reactions it receives from its supports. The system of all these forces must satisfy the three equilibrium conditions listed above, thus being a self-equilibrated system (Figure 2.8). Furthermore, the supporting ground receives the equal and opposite forces of the reactions acting on the structure. How these forces are carried by the soil mass itself is something which, although it may relate to the stress state of the structure, is examined as a soil—structure interaction (this is covered mainly in Chapter 17).

2.1.5

Basic assumptions

At this point it is appropriate to mention the basic assumptions that govern the analysis of skeletal structures, as considered in the present chapter and Chapter 3. (1) As has been already mentioned, the length of the members is more than about five times the maximum transverse direction of their cross-section. (2) The linear relationship between stress and strain is valid: ¼ E " (see Section 1.2). (3) The deformations of the structure are so small that the equilibrium, which obviously exists at the deformed configuration, can be examined in the undeformed condition. This important assumption, which is illustrated in Figure 2.9 and constitutes the so-called first-order theory, is generally valid for the majority of skeletal structures designed. However, for some structure types and in many cases of skeletal structures, where under certain conditions this assumption has to be abandoned, a different approach must be taken to their analysis. These conditions are examined extensively in Chapter 7 (see also Chapters 8 and 9). A direct consequence of the two last assumptions is the so-called superposition principle. According to this principle, a structure receiving the total of loadings A and B develops a state of stress and deformation, consisting of the sum of stress states and deformations due to the loadings A and B separately. Thus if a state of stress or deformation due to 34


Undeformed

Deformed Undeformed

Deformed

According to the first-order theory, the equilibrium is examined in the undeformed configuration

Figure 2.9 The influence of deformation on the equilibrium state

the loading A is considered, those states due to k A are obtained by multiplying the former by the factor k.

2.1.6

Statically determinate formation

If, in a rigid skeletal formation under an arbitrary system of self-equilibrated forces, it is possible to determine the forces and moments acting on the ends of an arbitrarily cut out part of the formation on the basis only of the three equilibrium conditions, then the considered formation is called statically determinate. The most usual type of a statically determinate formation is the so-called ‘open’ or ‘tree’ skeletal system (Figure 2.10). Indeed, if the formation is cut at an arbitrary place, then the equilibrium of either of the two pieces, which are now free bodies, will allow the determination of the three unknowns, and thus the specification of the force and the moment at the point of the cut. It is clear that each cut out piece is considered ‘fixed’ to its adjacent one. A more complex type of statically determinate formation is the so-called three-hinged connection between three formations (I, II and III in Figure 2.11). By considering the mutually transferred forces (the action—reaction principle) at each hinge, three unknown forces appear, namely RI—II, RII—III, RI—III. Each of these forces represents two unknowns, e.g. the projections on two orthogonal axes x and y. The total of the

Figure 2.10 Statically determinate formation of the ‘open’ or ‘tree’ type

35

Structural systems: behaviour and design Three-hinged formation Loading of the formation must be self-equilibrating RI–II I

I II

II RI–II RI–III RI–III

RII–III III RII–III

III The equilibrium of I and II ensures also the equilibrium of III

Figure 2.11 The three-hinged formation

six unknowns is determined by setting out the system of three equilibrium equations for any two formations, say I and II. The unknowns thus determined ensure automatically the equilibrium of the third formation also. A special case of the three-hinged formation is the triangular truss formation of rectilinear bars, which receives (self-equilibrated) forces (no moments) only at its nodes (Figure 2.12). According to the above, the members of the truss formation will develop exclusively axial forces, and consequently there are three resulting unknowns. The equilibrium of each of the three nodes leads to the consecutive determination of all three axial forces, by requiring each time the vanishing of the sum of the

Equilibrium of projections on the selected axis determines an unknown force each time

(1)

Self-equilibrating forces Self-equilibrating forces

(2)

Figure 2.12 The truss formation

36

(3)

Stress and deformation (statically determinate structures) Statically indeterminate formation

Three unknowns Two unknowns

Self-equilibrating loading

The three equations of equilibrium are not sufficient for the determination of five unknowns

Figure 2.13 A statically indeterminate formation

projections of all the acting forces on an axis perpendicular to either of the unknown forces (see Figure 2.12). The addition of a further node linked to two other nodes of the existing formation, and so on, creates broader statically determinate formations, which can be analysed in the manner described above. If the formation is not a statically determinate one, it is called statically indeterminate. For example, the closed system shown in Figure 2.13 is subjected to a self-equilibrating loading. Cutting the formation into two parts reveals five unknowns which cannot be ‘covered’ by the three equilibrium equations for either part.

2.1.7

Statically determinate support

There are two ways of supporting a plane formation (whether statically determinate or statically indeterminate) such that the developed reactions can be determined using only the equilibrium conditions (equations): the simply supported manner and the fixed supported manner. In the simply supported case (Figure 2.14), one point of the formation has a hinged support while another has a simple support, restraining the displacement with respect to a direction a. Direction a is not allowed to pass through the hinged support; if it does so the structure will not be rigid, being a movable mechanism. Clearly, the

Arbitrary direction: two unknowns

Specified direction: one unknown

Figure 2.14 A simply supported formation

37


One unknown Arbitrary direction: two unknowns

Figure 2.15 A fixed formation

unknown reaction of the simple support may be determined directly if the moment equilibrium of all the acting forces is considered with respect to the hinged support, the corresponding moment of the hinge reaction obviously being null. The components of this reaction about two arbitrary axes are also determined directly by considering the equilibrium of the projections with respect to the perpendicular directions on either of these two axes. In the fixed support case, any point of the formation may be fixed (Figure 2.15). The three unknowns of the reactions are determined by applying the three equilibrium conditions applied to the formation, which is considered as a free body, under the acting loads and reactions. When there are two plane formations, there is another manner of statically determinate support: three-hinged support (Figure 2.16). The two plane formations are mutually connected at any one point with a hinge, while each of them is supported on the ground also through a hinge. The three hinges are not allowed to be on the same line, otherwise the structure is not rigid. In such a system there are three unknown forces — the two reaction forces and the mutually transferred force through the common hinge — with unknown direction, which means there are six unknowns in total. As can be seen from Figure 2.16, these six unknowns can be determined through the two groups of three equilibrium equations valid for each one of the two formations.

RG G

RG

A B RA

RA RB

Figure 2.16 Three-hinged support

38

RB


Statically determinate support of statically determinate formations leads to statically determinate structures

Figure 2.17 The transition from the statically determinate formation to a structure

At this point, the difference between formation and structure should be made clear. By ‘formation’ is meant a free body that can be in equilibrium only under the action of a selfequilibrating system of forces. A ‘structure’ may be subjected to any forces and moments (i.e. loads) lying in its own plane, and it is the role of the supports to offer the appropriate reactions in order to establish the equilibrium. If a statically determinate formation is supported in one of the three ways described above, the structure so created is called statically determinate (Figure 2.17). Clearly, the addition of further supports increases the number of unknowns, making it impossible to determine them through the use of the equilibrium conditions alone. The structure formed in this case is called statically indeterminate (Figure 2.18). Moreover, it is obvious that supporting a statically indeterminate formation with a statically determinate support leads also to a statically indeterminate structure. The distinction between statically determinate and statically indeterminate structures is examined more thoroughly in the following section.

2.1.8

Compound structures

The three types of statically determinate structure examined in the previous section are considered to be fundamental. The formation of any plane structure, however complex, is practically based on them. Indeed, by initially setting any one of the three types of structure, any point on it may be considered as the ‘ground’ to be used as the support for another of the three structural types previously examined. Each point of the new

A statically determinate formation supported indeterminately constitutes a statically indeterminate structure

Figure 2.18 Statically indeterminate support of a statically determinate formation

39


Synthesis of structure

II III

I

I: Fixed supported

II: Simply supported

III: Three-hinged supported R3

Analysis of structure Opposite path

R2 III

R3

R2

II R1

I

R1 Each time a simple structure is analysed

All actions are known

Figure 2.19 The formation and analysis of a composite statically determinate structure

formed system may be regarded as the ‘ground’ for supporting a further structure, and so forth. The final structure formed in this way is a statically determinate one, which means that all its internal forces can be determined on the basis of the three equilibrium conditions alone. The way to ensure that this determination can be done directly is illustrated using the example complex structure shown in Figure 2.19. First, the individual parts of the structure are sought. These may be supported successively in any one of the three ways described in Section 2.1.7. Thus, part I is selected first, then part II is supported on it (in a statically determinate manner), and finally part III is supported, again in a statically determinate way. Now if the ‘analysis’ of the structure follows the opposite path to that of ‘synthesis’, it will proceed unimpeded. Indeed, the analysis of the last part (III) will directly yield some reactions, on the basis of which part II may be analysed. Finally, by applying to part I the equal and opposite reactions of parts III and II (which are already known), the whole analysis can be completed in a direct manner. It should at first be examined whether any composite structure can be formed in the above manner. If it can, this means that the structure is statically determinate and rigid. If no composite structure can be formed, then it should be examined whether it is possible for the structure to become statically determinate by appropriate modifications to the continuity of some members or to the supports. 40


Three times statically indeterminate Nine times statically indeterminate

(–3) (–1) (–3) (–1) (–3)

Statically determinate structure (–1)

Statically determinate structure

Figure 2.20 The degree of statical indeterminacy

Cutting the continuity at a point where two parts are monolithically (‘fixed’) connected, means that three magnitudes (i.e. unknowns) are removed. Of course, cutting off a hinged bar means the removal of only one magnitude, namely the axial force (Figure 2.20). The ‘modification’ of a support also leads to the removal of some effective magnitudes. For example, modifying a fixed support to become a simple support means the removal of two magnitudes, whereas modifying it to a hinge support means the removal of one magnitude only. Moreover, removing a fixed, a hinged or a simple support leads to the removal of three, two and one magnitude, respectively. The total number of magnitudes removed in the above way, so that the remaining structure becomes statically determinate, gives the so-called degree of statical indeterminacy of the initial plane structure considered (see Figure 2.20). The overwhelming majority of structures encountered in practice are statically indeterminate. However, becoming familiar with handling statically determinate structures is absolutely necessary for the clarification of the concepts that are essential for understanding the load-carrying action of structures.

2.2

The handling of internal forces

2.2.1

Sectional forces

In order to ensure that any member of a plane skeletal structure has equilibrium as a free body, there must be a suitable force and moment acting at its ends. Under the loading, the skeletal structure ACBD shown in Figure 2.21 develops the vertical reactions RA ¼ 24.0 kN and RB ¼ 36.0 kN. The reaction RA can be determined 41

Structural systems: behaviour and design 5 kN/m

C

m

B

D 2.0

RB = 36 kN

A RA = 24 kN 3.0

3.0

4.0

2.0 m

5 kN/m

V

M = –49.50 kN m N = –2.85 kN V = –8.54 kN

N m

M 5 kN/m

24 kN 49.50 kN m 36 kN

m 2.85 kN

8.54 kN

5 kN/m 5 kN/m 49.50 kN m m 9.00 kN 24 kN

49.50 kN m m

36 kN

9.00 kN The section is acted upon by one force and one moment

Figure 2.21 The internal forces required for the equilibrium

by expressing the moment equilibrium of all external forces (uniform load, RA, RB), with respect to point B (whereby the contribution of RB is null), while RB is obtained from the equilibrium of the projections of the external forces on a vertical axis after RA has been determined. Considering point m at a horizontal distance of 3.0 m from A and examining the equilibrium of part (Am) — chosen because it is ‘simpler’ than the part mCBD — the three magnitudes, which represent the fixed connection between the parts Am and mCBD, are sought. What are specifically sought are the two components N and V of the force acting on section m, corresponding to the member axis and the perpendicular direction to it, respectively, together with the acting moment M. The positive senses of the axes N and V as well as of the moment M are chosen arbitrarily. The determination of N, V and M follows the three equilibrium conditions given in Section 2.1.2: 42


(1) Considering the moment equilibrium with respect to point m, the contributions of Nand V vanish, and the moment M ¼ 49.50 kN m is deduced directly, the negative sign expressing the fact that it is applied in the opposite sense than has been assumed. (2) The equilibrium of the projections on direction N — the contribution of V being zero — allows direct determination of the component N ¼ 2.85 kN, indicating thereby that it is also acting in the opposite sense. (3) Finally, by considering the sum of the projections on direction V, the contribution of N automatically vanishes and the component V ¼ 8.54 kN may be determined directly. Its sense is again the opposite to what has been assumed. It is clear that the equilibrium of part mCBD requires at the point m the application of the equal and opposite forces M, V and N (see Figure 2.21). These three magnitudes M, V and N are called the bending moment, the shear force and the axial force, respectively, and constitute the sectional forces at point m. However, it must be made clear that section m receives, apart from the bending moment, a unique force, actually the resultant F of V and N: pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi F ¼ 2:852 þ 8:542 ¼ 9:00 kN This force, for the case examined, happens to be vertical, as the vertical equilibrium of the part Am implies (see Figure 2.21), and it is emphasised that the magnitudes N and V simply represent the components of this unique force F with respect to two arbitrarily chosen axes. However, these specific axes have been purposefully chosen along the direction of the corresponding normal and shear stresses, as will be explained below. g The action of a bending moment on a cross-section induces normal stresses , which should have such a distribution that they offer to the section precisely the acting moment M (Figure 2.22). The resulting distribution is a linear one, has a zero value at the centroidal axis of the section, and exhibits compressive and tensile stresses on either side of it, thus offering a resultant compressive force on one part and a resultant tensile force on the other part of the section. It is understood that these two forces should be equal and with opposite senses in order to offer the moment M when coupled.

Compression

D

y

z M

Centroidal axis

Z Tension

D=Z M=D·z

Diagram of normal stresses

Figure 2.22 The development of bending stresses

43

Structural systems: behaviour and design Compression

Compression

Compression

DM

N

D M

z

e1 e2

Centroidal axis

ZM

Z Tension

Tension DM = ZM M = D · zM = D · e1 + Z · e2

D–Z=N

Figure 2.23 The stresses caused by the combined bending and axial force

It is found that the normal stresses developed at each fibre on the cross-section are proportional to its distance y of the fibre from the centroidal axis, as well as to the acting moment M, while remaining constant across the width of the section. Moreover, these stresses are inversely proportional to the moment of inertia I of the cross-section, and thus they obey the relation: M y I As discussed in Chapter 1, the axial force N produces uniformly distributed normal stresses (either compressive or tensile) according to the relation ¼ N/A, where A represents the cross-sectional area. The coexistence of a bending moment and an axial force leads to a superposition of the normal stresses (compressive and tensile) due to bending and those (compressive or tensile) due to the axial force (Figure 2.23): ¼

N M þ y A I Moreover, acting of a shear force V on a section causes shearing stresses of the same sense, but not evenly distributed over the height of the section, which naturally sum over the whole section to this force exactly (Figure 2.24). The developed shearing stress over the height of the section, which remains constant across the width b of the section at the point considered, is given at a ¼

τ

V

Centroidal axis

Distribution of shearing stresses τ (rectangular section) max τ = 1.50 · (V/A)

y

Figure 2.24 The shearing stresses caused by the shear force

44

Stress and deformation (statically determinate structures) m N

m

N N

N Positive axial force

Figure 2.25 The convention for the positive axial force

distance y from its centroidal axis by VS bI where S represents the so-called first moment of area of the portion of the section, lying beyond the place of the measured width b, with respect to the centroidal axis. g ¼

It is appropriate now to examine how the positive sign of the above sectional forces is defined. . Axial force. The axial force N is considered ‘positive’ if it causes tensile stresses (Figure 2.25). In the example examined previously (see Figure 2.22), the axial force at point m, which is equal to 2.85 kN, causes compressive stresses and is, consequently, ‘negative’. . Shear force. The shear force V at a point m is considered positive if, together with the shear force of a neighbouring section, which for equilibrium reasons must act in the opposite sense, it constitutes a clockwise couple (Figure 2.26). Thus, in the previous example, the shear force of 8.54 kN at point m is positive (see Figure 2.21). . Bending moment. For the bending moment, signing is not physically important. However, it is of primary importance to determine on each occasion on which side of the section the tensile stresses (or compressive stresses) are developed (Figure 2.27). It is concluded that in the above example the bending moment of 49.5 kN m causes tension at the bottom fibres of section m. While the signing of the axial and shear force does have a physical meaning, because it suggests the actual sense of the axial and shearing stresses, respectively, on the section, the signing of the bending moment has no such meaning. However, if a so-called positive border is arbitrarily established on the member, then the bending moment may be considered ‘positive’ if the tensile stresses it produces are on this positive side; otherwise it is ‘negative’ (see Figure 2.27).

m

Clockwise couple m

V V

V

V Positive shear force

Figure 2.26 The convention for the positive shear force

45

Structural systems: behaviour and design 49.50 kN m

49.50 kN m m

m Tensioned fibres

Tensioned fibres Tensioned fibres (Moment = +49.50)

m Positive border

m

Positive border (Moment = –49.50) The selection of the positive border is arbitrary The tensioned side is independent of the positive border

Figure 2.27 The bending moment and tensioned fibres

2.2.2

Determination of sectional forces

The explanation given in the previous section allows for the systematic determination of the sectional forces at every point on a skeletal model. Using the same example as above, in order to find the sectional forces at the characteristic positions A, Cl, Cr, Bl, Br and D, the structure is cut at the respective position each time, and that part is considered which seems more convenient from the aspect of equilibrium examination. The positive sectional forces N and V are then placed on the section, the selected sense of M being indifferent (Figure 2.28). . Position A. The tiny cut out part is shown in Figure 2.28, together with the known external force and the unknown sectional forces. Following the determination of each force through the equilibrium of the projections on its own axis, it is found that N ¼ 7.59 kN, V ¼ þ22.57 kN and M ¼ 0.0 kN m. . Position Cl. The selected part ACl is shown in Figure 2.28 together with its external forces and the unknown sectional forces in their positive sense. It is found that N ¼ þ1.90 kN, V ¼ 5.69 kN and M ¼ þ54.00 kN m. . Position Cr. The selected part ACr is shown in the Figure 2.28. It is found that N ¼ 0.0 kN, V ¼ 6.0 kN and M ¼ þ54.00 kN m. Note that the values of N and V have been differentiated with respect to position Cl, whereas this is not the case for the bending moment M. . Position Bl. As the more convenient part for the examination of the equilibrium, part BlD (see Figure 2.28) is selected. It is found that N ¼ 0.0 kN, V ¼ 26.0 kN and M ¼ 10.0 kN m. . Position Br. The examined part is now BrD (see Figure 2.28). According to the positive senses of the sectional forces, it is found that N ¼ 0.0 kN, V ¼ þ10.0 kN and M ¼ 10.0 kN m. . Position D. The equilibrium of the tiny cut out part (see Figure 2.28) leads to the conclusion that all the sectional forces there are null. g 46

Stress and deformation (statically determinate structures) 5 kN/m

l

l

r C

r D

B

2.0

RB = 36 kN

A RA = 24 kN 3.0

3.0

4.0

2.0 m

M N

N = –7.59 kN V = +22.57 kN M = 0.0 kN m

A V

24 kN

5 kN/m

5 kN/m M l

r

N

V

V N = +1.90 kN V = –5.69 kN M = +54.00 kN m

24 kN

A 24 kN

5 kN/m V M l B 36 kN

N = 0.0 kN V = –6.00 kN M = +54.00 kN m

5 kN/m

M N

N

C

C

A

M

D

N = 0.0 kN V = –26.00 kN M = –10.0 kN m

N

V r B

V M D

N = 0.0 kN V = +10.00 kN M = –10.0 kN m

N

D

N = 0.0 kN V = 0.0 kN M = 0.0 kN m

Figure 2.28 The systematic determination of sectional forces

The ultimate objective of this analysis is to construct three diagrams, giving for each point of the member the value of the bending moment, the shear and the axial force. The diagrams are ‘read’ in the direction perpendicular to the member axis. With regard to the axial force and the shear force diagrams, it does not matter on which side of the member axis the positive values appear, but it is absolutely necessary that these sectional forces are signed. However, for the bending moments the convention must be followed that they must always appear on the tensioned side of the section. The values of M, V and N obtained in the example considered above are shown in their appropriate positions in Figure 2.29. The procedure that must be followed in 47


l

l

r

r

C

B

D

RB = 36 kN

A RA = 24 kN

7.59 1.90 26.0

[N]

5.69

6.00

10.0

[V ]

22.77

10.0

54.00

54.00 [M]

Figure 2.29 Diagrams showing the calculated sectional forces

order to round out these values to a corresponding continuous diagram is elucidated in Section 2.2.5.

2.2.3

Relationships between loading and sectional forces

In a skeletal structure an element of length s is cut out and its equilibrium is considered under the transverse load p(s) and the longitudinal load h(s). The parameter s is referred to the longitudinal sense of the element. This element is acted upon by the actions M, V and N at one end and by the (generally) modified actions M þ M, V þ V and N þ N at the other (Figure 2.30). Applying the equilibrium conditions for this element, the following differential relations are readily obtained: dV ¼ pðsÞ ds 48

dM ¼ VðsÞ ds

dN ¼ hðsÞ ds

Stress and deformation (statically determinate structures) p(s) M + ∆M

M h(s)

N + ∆N

N

V

∆s

V + ∆V

Figure 2.30 The equilibrium of an elementary rectilinear part

On the basis of the first two, an equally important relationship is deduced: d2 M ¼ pðsÞ ds2 From the first relation it may be seen that the ‘bearing mechanism’ of the transverse load p(s) on a beam is expressed through the variability (dV/ds) of the shear force. Moreover, the last relationship rewritten as d dM ¼ pðsÞ ds ds expresses the same fact. The above equations are directly related to the diagrams of the sectional forces M, V and N — as they are themselves actually functions referred to the axis of each member — and allow the following conclusions to be drawn: . The slope of the tangent at every point on the diagram of V is identical to the value of the transverse distributed load acting at that point. . The slope of the tangent at every point on the diagram of M is identical to the value of the shear force V at that point. Consequently, wherever the shear force V is null, the diagram of M exhibits either a maximum or a minimum value. . The slope of the tangent at every point on the diagram of N is identical to the value of the longitudinal distributed load at this point. Moreover, on the basis of the above, the following may also be concluded: . A region of a rectilinear member which is not acted upon by a transverse distributed load has a constant V. In the same region M varies linearly. . In a region of a rectilinear member having a constant transverse distributed load, the shear force V varies linearly. In the same region M varies according to a parabolic law. . A change in value in the transverse distributed load means a change in the slope in the diagram for V (a higher transverse distributed load leads to a greater slope). . A null longitudinal distributed load means constant N. . A constant longitudinal distributed load means that the diagram for N varies linearly. A concentrated transverse force at some point will cause a discontinuity in the V diagram, due to a different value of V on each side of the load (Figure 2.31), whereas 49

Structural systems: behaviour and design M

P

Vl

Vr

M

∆s Equilibrium: V r = V I – P

Figure 2.31 The influence of a concentrated load

the value of bending moment M will be unaffected. However, the slope of tangent of the moment diagram M will be different on the two sides of the load, adapting to the corresponding values of the shear force. Finally, the application of a concentrated moment at a certain point will cause a discontinuity in the bending diagram, with different values of M on either side; however, the corresponding values of V will remain the same.

2.2.4

The simply supported beam

Knowing how the bending moment and shear force diagrams for a simply supported beam are drawn for some typical loadings is essential for the treatment of any plane skeletal structure involving sectional force diagrams.

2.2.4.1 A beam under a uniform load Both reactions are vertical and equal to half of the total vertical load (Figure 2.32). The diagrams for V and M are shown in Figure 2.32. q

P A

q · L/2

L

q · L/2

P · b/L

B a

P · a/L

b L

q · L/2 P · a/L q · L/2

P · b/L

[V ]

[V ] P Equilibrium

q · L2/8 [M ]

P · b/L

P · a/L

q · L2/8 [M]

Tangent

Figure 2.32 Sectional force diagrams for a simply supported beam

50

P · a · b/L


The following may be observed: the shear force diagram has a constant slope and becomes zero at the midspan, where the maximum value of M occurs. The slope of the parabolic diagram for M follows the value of the shear force. The maximum bending moment has the value q L2/8 and causes tension of the lower fibres, as in fact occurs over the entire length of the beam. Moreover, it is useful to note the way in which the parabolic diagram for M is drawn, as shown in Figure 2.32.

2.2.4.2 A beam under a concentrated load The reactions and the diagrams for V and M are obtained as shown in Figure 2.32. The different shear forces on each side of the concentrated load satisfy the equilibrium of the tiny cut out segment as shown in Figure 2.32, and correspond to the different slopes of the bending moment diagram.

2.2.4.3 An oblique simply supported beam under a uniform load The inclination angle of the beam axis is equal to a (Figure 2.33). The uniform load refers to the horizontal projection L. The beam receives thereby a constant transverse and longitudinal distributed load. The specific values of these loads is not of interest. Both reactions are vertical and equal to the half the total vertical load (q L), i.e. they are identical to those of the ‘horizontal’ beam. Diagrams V and N . Position A. From the equilibrium of the tiny cut out element (see Figure 2.33) it is found that VA ¼ þ(q L/2) cos and NA ¼ (q L/2) sin . . Position B. From the equilibrium of the tiny cut out element it is found that VB ¼ (q L/2) cos and NB ¼ þ(q L/2) sin . The diagrams for V and N are linear due to the constant transverse and longitudinal distributed loads, respectively. Bending diagram M Midspan m. The equilibrium of the cut out part Am (see Figure 2.33) yields a bending moment equal to (q L2/8), tensioning the bottom fibres, i.e. exactly the same as for the previous ‘horizontal’ beam. This means that at each point of the oblique beam the developing bending moment is equal to that of the projected point on the ‘horizontal’ beam. The parabolic diagram M referred to the oblique axis is drawn in the same manner as shown previously (see Figure 2.33).

2.2.5

Drawing the bending moment diagram of a member

Following the procedure given in the preceding two sections, the moment diagram for any straight member, the bending moments at its ends having been already determined, may be drawn directly. Specifically, it is assumed that the end bending moments M1 and 51

Structural systems: behaviour and design q q′ B

q′

q · L/2 α

A

L′ L

q · L/2 Total load: q · L = q ′ · L′

(Determination of the load components is not necessary)

(q · L/2) cos α

V

N A

B V N

q · L/2

q · L/2

[V] V: positive N: compressive

V: negative N: tensile

(q · L/2) sin α

(q · L/2) cos α

The moment in the middle is independent of the beam length

Tangent

[N ] q · L2/8

[M ]

Tangent

(q · L/2) sin α q

q · L2/8 m A

L/2

q · L2/8

L The two moment diagrams are absolutely equivalent

q · L/2

Figure 2.33 The load carrying response of an oblique simply supported beam

M2 of the skeletal member 1—2, which is subjected to a transverse loading in an arbitrary direction, as shown in Figure 2.34, have already been determined. These moments are drawn accordingly on a diagram [M] (see Figure 2.34). The transverse loading is represented indicatively by a distributed load. The length of the projection 10 —20 of the member in a direction perpendicular to the load is L. As noted in the previous section, the moment diagram M0 of the simply supported beam 1—2 under the considered load is identical to that of the simply supported beam 10 —20 . 52

Stress and deformation (statically determinate structures) q

M2 2

N2

V2 V1 q 1 N1

M1

L 2

2 V2

2

N2

M2 q · L2/8

1

[V]

1

[N]

M1 1

N1

M0

V1

L

Tangent

1′

2′ M0

q · L2/8

The diagram may be ‘suspended’ from any axis

Figure 2.34 Sectional force diagrams for a member with known end actions

The question of how the bending diagram for the member 1—2 is drawn between the values M1— and M2 can now be answered simply. It is drawn by ‘hanging’ the diagram M0 from the straight line connecting the two end values M1— and M2, to that side where the tensioned fibres of the beam 10 —20 are. It is understood that diagram for M0, like values M1 and M2, is drawn perpendicularly to direction 1—2. Note that the shear forces at the ends of a straight member can always readily be determined from the equilibrium requirements, once the end bending moments of the member are known.

2.2.5.1 Application to a cantilever The cantilever, being a member that is fixed at one end, generally constitutes a basic element in the formation of plane skeletal structures, and the understanding of its state of stress, as expressed through the corresponding sectional forces diagrams, has its own importance. The loading of a horizontal cantilever of length L by a vertical uniformly distributed load q causes at its fixed end the bending moment (q L2/2), which imposes tension in the top fibres (Figure 2.35). Between this value and the zero value at the free end, the diagram of the simply supported beam is ‘hung’ with the characteristic value (q L2/8) in 53

Structural systems: behaviour and design q q·L

P

q · L2/2

P·L

L

P

L

P

q·L [V]

[V]

q · L2/8

q · L2/2

P·L

[M ]

[M ]

q · L2/8 q

L [M] q · L2/2 q·L q · L2/2

The fixed-end moment is independent of the oblique cantilever length

Figure 2.35 Sectional force diagrams for a cantilever

the middle. Clearly, the resulting diagram should exhibit a horizontal tangent at the free end, given that the corresponding shear force is null. The same cantilever under a vertical concentrated force at its free end develops at its fixed end the bending moment (P L), causing tension in the upper fibres. Because of the absence of a transverse load, the bending moment values at the two ends are connected by a straight line (see Section 2.2.3). Moreover, Figure 2.35 shows how the bending moment diagram for an ‘oblique’ cantilever is drawn, according to the above.

2.2.6

The three-hinged frame

The preceding sections were essentially devoted to the examination of beams. However, the formation of a skeletal structure that will ‘cover’ or ‘close’ a space leads to the concept of a frame. In its simplest form a frame consists of a beam (called a girder) as the ‘horizontal’ element and two columns in the ‘vertical’ or ‘oblique’ sense which are rigidly connected to it. Clearly, in simply supported frames, like those shown in Figure 2.36, the bending response of the girder does not differ at all from 54


The vertical reactions lead to bending moments identical to those of a simply supported beam

Bending moment of simply supported beam

Three-hinged frame

Figure 2.36 The consequences of the three-hinged formation

that of the simply supported beam. If, however, the structure is formed as a three-hinged one, as described in Section 2.1.7, this situation changes. The reactions are no longer vertical and their horizontal components cause bending moments in the columns, as well as at the ends of the girders, resulting in significantly reduced midspan moments (see Figure 2.36). Considering the three-hinged arch subjected to a vertical loading shown in Figure 2.37, it can be concluded from the ‘horizontal’ equilibrium requirement that the horizontal components H of the reactions of the supports A and B are equal in magnitude and of opposite sense. The moment equilibrium with respect to support B yields a relationship between the components VA and H. One more relationship between these two components may be obtained by considering the moment equilibrium of part AG with respect to point G (see Figure 2.37). The remaining components VB, VG and HG can then be readily determined. In the case when the external loading acts in a perpendicular direction to the line connecting the supports A and B (Figure 2.38), some interesting conclusions may be drawn. First, on the basis of the previously described procedure, the vertical reactions VA and VB work out to be identical to those of the simply supported beam AB. 55


G

H

A

B

VA

H

VB The developed reactions at A and B are a single force The appearing components correspond to arbitrarily chosen directions

H G

G

VG A

H

H

B

VA Moment equilibrium with respect to B

VB

H

A

VA Moment equilibrium with respect to G

Determination of the components of the reaction at A

Figure 2.37 Analysis of a three-hinged arch

The horizontal component H (thrust), may be expressed as H¼

MG f

where MG is the bending moment of the simply supported beam AB at the corresponding point G on the hinge (see Figure 2.38). The last equation has particular importance. It shows that the higher the frame, the less its horizontal thrust, i.e. the more the reaction approaches the vertical. Conversely, if the frame is made lower, the horizontal thrust H increases and the inclination of the reaction becomes smaller. Another conclusion is that the component H is directed inwards inasmuch as the bending moment MG of the simply supported beam at G causes tension in the lower fibres. Otherwise (which is certainly unusual), the component H is directed outwards (see Figure 2.38). g Finally, a graphical or ‘qualitative’ determination of the reactions and internal forces of a three-hinged arch is presented in Figure 2.39. When one of the two parts of the frame is unloaded, it is possible to determine qualitatively the direction and sense of both the reactions and the internal force in G. Given that the part AG is two-hinged and 56


G f H

A

B

H

H = MG/f VA

VB

The vertical components are identical to the reactions of the simply supported beam

A

G

B

MG VA

VB

[M]

G

A

B MG B

A

G

The horizontal thrusts are directed outwards

Figure 2.38 A three-hinged arch with a horizontal base line

unloaded, it is directly concluded (see Section 2.1.2) that the forces at A and G are acting along the line AG. Due to the fact that the whole structure is acted on by three forces only (i.e. the two reactions and the resultant of the uniform load), its equilibrium is governed by the corresponding triangle of forces. The acting forces should run in the same rotating sense (see Section 2.1.2). Thus the sense of direction of all the relevant forces is determined, and then, following the procedure given in Section 2.2.2, the sectional forces diagrams can be set out in a purely qualitative manner (see Figure 2.39).

2.2.7

The funicular structure

The development of bending moments in a skeletal structure leads to a linear distribution of stresses in a section, which means that the bearing capacity of all the longitudinal fibres of the member cannot be fully exploited. Obviously, the optimum utilisation of the material used is accomplished only when the fibres of each section are stressed uniformly, which clearly means the development of exclusively axial forces. Thus the question is raised about the form that a structure should have in order to carry a specific vertical loading by developing exclusively axial stress, being itself supported by two hinges at its ends, spanning a distance L (Figure 2.40). 57


G P Resultant of distributed load

A RA

P

RB

B

RA Triangle of forces

RB

RG G [M]

[V]

[N]

[M]

[V]

[N]

A RA RG

G

B RB

Figure 2.39 Qualitative determination of the sectional forces in a three-hinged frame

This question may be answered directly from a physical point of view, if an absolutely flexible medium, such as a cable, is fixed at the given points and is subjected to the loading under consideration. The cable will take such a form that only tensile forces will be developed within it. In the case of a material that can sustain compression rather than tension, the absolutely equivalent mirror image of the ‘cable system’ may be considered, and the loading sense may be reversed, thus causing exclusively compressive stresses (see Figure 2.40). In this way the familiar arch form may be created. This form is called a funicular structure. It is understandable that at every point the externally acting load is in equilibrium with the internal axial (compressive) forces. In other words, the loads are transferred to the supports exclusively through axial (compressive) forces (see Figure 2.40). With regard to the geometrical form y(x) of such an arch, it is understandable (see Section 2.2.6 and Figure 2.38) that, as the vertical components of the reactions are 58


L

The cable develops exclusively axial tensile forces balancing the loads at each node

Static configuration identical to the previous one

Reversing the loads leads to a reversal of the response Development of exclusively compressive forces

Figure 2.40 Conception of the funicular structure

equal to the reactions of the corresponding simply supported beam, the bending moment of the arch at a point with an ordinate y(x) will result from the corresponding bending moment M0(x) of the beam, reduced by the corresponding moment of the thrust H (Figure 2.41). Thus M(x) ¼ M0(x) H y(x). The requirement M(x) ¼ 0 leads to the equation M0 ðxÞ H which states that the sought-after form is affine to the bending diagram of the corresponding simply supported beam, by the undefined factor 1/H. This factor is fixed only when a third point of the arch is also defined, as for example the crown yðxÞ ¼

59


y(x) H

A

B

x

VB

VA Equal

H

M = M0 – H · y = 0 Infinite number of solutions: y(x) = M0(x)/H

VA

Equal

VB

M0(x0) M0(x)

Once the ‘third’ point is specified the solution is ascertained

y0 (x = x0)

y(x) H

A

B

x x0

VA

H

VB

Pressure line: y(x) = M0(x) · [y0/M0(x0)] H = M0(x0)/y0

Figure 2.41 Formation of the funicular structure

height at a distance x0 from the support A. Then, with the term y(x0) already specified, H¼

M0 ðx0 Þ y0 ðx0 Þ

and finally yðxÞ ¼

yðx0 Þ M0 ðxÞ M0 ðx0 Þ

(see Figure 2.41)

The curve y(x) is also called the pressure line. For a constant uniformly distributed load the curve is a parabola (Figure 2.42). However, in order to take up a group of concentrated loads, and according to the bending moment diagram M0, a polygonal funicular arch form is the appropriate one. At this point it should be noted that if the constant uniform load q stays perpendicular to the axis of a very flexible medium such as a cable, then the cable takes the form of a circular segment with radius R, requiring a certain force P to be applied at its ends. This 60

Stress and deformation (statically determinate structures) q q q·R

f

H

H

q·R

Constant tensile force Circular arch

L q · L/2

q

q · L/2 S

R ≈ L2/8 · f

H

H=q·R

Variable axial force S with constant component H

q·R

q·R Constant compressive force

Figure 2.42 Typical forms of funicular arches

force goes through the whole length of the cable as the only existing axial tensile force, and is given by P ¼ q R (see Figure 2.42). Obviously, the same funicular form acts as a compression path for the same loading in the opposite sense. It should be pointed out that, while in this funicular form the axial force is kept constant, in the examined case of a constant vertical uniform loading q the axial force in the funicular parabolic form keeps change constantly over the entire arch length, as shown in Figure 2.42. It is clear that the specific line pressure y(x) leads to an exclusively axial stress state only for the specific loading under consideration. Any other loading will cause bending. Moreover, it is obvious that the slightest deviation from the funicular form under the given loading will equally lead to bending. Of course, the greater this deviation is, the greater the degree of bending developed (see Section 6.2).

2.2.8

Trusses

In contrast to monolithic structures, which have rigid joints and develop bending and shear, trusses are load-bearing structures consisting of rectilinear two-hinged bars. They are loaded by concentrated forces at their nodes only, and consequently exclusively axial forces develop in their members. The truss loading is applied to the nodes by means of small beams supported on each node. It is these small beams that actually receive the incoming loads, and then transmit them as concentrated forces to the truss joints (Figure 2.43). Despite the fact that the truss members are never connected by a pin joint, this idealisation leads to safe results. Moreover, although the self-weight of the bars does not actually act directly on the nodes, it may, nevertheless, be safely considered as directly acting on them. g A statically determinate truss is analysed by first determining its reactions and then examining the equilibrium of each node as a free body, under the corresponding loading and the unknown axial forces acting on the cut out bars. Using the two equations of 61


Figure 2.43 Indirect loading of a truss through joint loads

equilibrium for each node, a linear system of equations is obtained, which will lead to the determination of all the unknown axial forces. For the overwhelming majority of trusses it is also possible, by examining the equilibrium at each node through use of the triangle of forces, to determine qualitatively the two unknown axial forces at each node, starting from the support joints where the reactions are already known. It is useful to note that in an unloaded node with three concurring bars, two of which are on the same line, the third bar is under no stress. Furthermore, if at a node there are four concurrent forces lying in only two different directions, then the two forces in each direction must be equal and opposite. Although the above described joint equilibrium method always leads to the full analysis of a truss, it does not help one in understanding the bearing action of the truss itself, and the determination of the axial force in specific bars is rather cumbersome. g A truss consisting of an upper and a bottom chord, with vertical and diagonal bars between them, can instead be thought of as a monolithic structure, because both its bending and shear action can be determined through its axial stress state, as described below. If at a region of a simply supported beam, for example, a part is removed and substituted by three two-hinged bars, as shown in Figure 2.44, the resulting structure will continue to be statically determinate and rigid (see Sections 2.1.6 and 2.1.7). Cutting these three bars through a vertical section and considering, for example, the equilibrium of the left-hand part, it is understandable that the forces C and Z may be used as substitutes for the compressive and tensile stresses of the section, respectively, and thus simultaneously give the corresponding bending moment at the position considered. Moreover, by examining the vertical equilibrium of the same part it is clear that the diagonal force D must provide the required positive shear force at the same place. This specific shear force may be offered only through a tensile force D, whereas for the cross-symmetric position of the diagonal bar the positive shear force can be offered only through a compressive force D (see Figure 2.44). Considering in this manner the region of interest, the stress state in a truss in that region may always be determined using the sectional force diagrams for the corresponding monolithic structure. g For the truss shown in Figure 2.45, the axial forces O, U and D in the bars can be determined directly using the free-body equilibrium of the cut out left-hand part of 62


The three connecting bars can offer all the sectional forces C D Z The inclination and the direction of the diagonal determine its response

C D Z

Figure 2.44 The sectional forces occurring due to the truss action

the truss. To determine the axial force O, the moment equilibrium with respect to point m2, where the forces U and D intersect, is considered, in order to get rid of their contribution. It is found that O ¼ Mm2 =hm2

(compressive)

To determine the axial force U, the moment equilibrium with respect to point m1 (the intersection of forces O and D) is considered, in order to eliminate their involvement, as above: U ¼ Mm1 =hm1

(tensile)

Finally, the axial force D is determined through the requirement for vertical equilibrium, the contribution of force U being thereby eliminated: D ¼ Vm/sin m þ O/sin ’ Thus the compressive axial force O ‘helps’ the tensile force D to ‘offer’ to the considered cut out left-hand part the required positive shear force Vm (see Figure 2.45). Consequently, the value of D is lower than the value that would obtained if the upper chord was horizontal. 63


O ϕ

m1

hm2 hm1 D αm U

m2 O O = Mm2/hm2 U = Mm1/hm1 D = (Vm – O sin ϕ)/sin αm

D U

The inclined compressive force of the upper chord contributes to take up the shear, therefore relieving the diagonal

Mm1

Mm2

Vm

Figure 2.45 Analysis of a truss considered as a monolithic beam

Finally, the axial force of the vertical bar at node m1 may be obtained from the vertical equilibrium of the cut out node. g As has been shown, an appropriately formed truss may be used as a substitute for any monolithic structure, such as a beam, frame or arch, as it can provide all the necessary sectional bending and shear forces (Figure 2.46). The main structural advantage of trusses

Three-hinged frame

The axial forces of the truss offer all the sectional forces of the corresponding monolithic structure

Figure 2.46 A truss used as a substitute for a monolithic structure

64


is that, due to the exclusively axial response of their members, the cross-sectional area of the truss members can be exploited fully, and this naturally leads to material economy.

2.3

Determining the deformations

2.3.1

Overview

Deformations constitute the ‘visible’ part of the behaviour of structures, in the sense that the stresses and the sectional forces are ‘derived’ quantities, not being directly measurable. Deformations may be ‘external’, such as the displacements and rotations developed at any point of a plane skeletal structure, or ‘internal’, the latter being better known as ‘strains’ (Figure 2.47). When a plane skeletal structure is acted upon by a coplanar loading, and possibly also a temperature variation, each point on the structure undergoes a displacement, which is represented by a vector and a rotation, either clockwise or anticlockwise, in the same plane as the displacement. It should be emphasised that, during the deformation, the angles between any monolithically connected parts (joints) — strictly speaking the angles between the tangents of the deformed lines at the joints — remain unchanged, meaning that the rotation angles at the ends of all concurring members at a joint are equal. This does not happen when two members are connected by a hinge (see Figure 2.47). It should also be noted that, while the strains are deduced from the sectional forces and from the temperature variation, this is not the case for the external deformations. g The determination and general consideration of the external deformations, which will hereafter be referred to simply as ‘deformations’, has particular importance in the study of the behaviour of structures. The development of deformations in a structure, as a result of the incoming sectional forces in general, is inevitable and must be contained within prescribed limits. Excessive deformations, whether permanent or not, will weaken a structure, and should always be controlled and, if necessary, appropriately settled. The determination of deformations is absolutely necessary in order to calculate and understand the response of statically indeterminate systems, which account for the overwhelming majority of structures. Rotation

Displacement

The angle at each joint remains unchanged

The change in the relative angle at the hinge

The relative rotation of neighbouring sections (‘internal deformation’)

Figure 2.47 Deformations of plane structures

65


2.3.2

Causes of deformation

Before dealing with the procedure for determining any deformation that occurs in a skeletal plane structure, it is appropriate to examine the kind of strains that are caused by the sectional forces and by temperature variations.

2.3.2.1 Bending moment A self-equilibrating pair of bending moments M acting on the opposite, initially parallel, sides of a skeletal element of length s causes a curving with a radius of curvature r and, consequently, a relative rotation ’ of the two parallel sides (Figure 2.48). Such curving must occur because, otherwise, the shortening and the elongation of the extreme fibres would change in an unacceptable way the right angles of the edges of the element. There is a basic ‘constitutive’ relationship between the curvature 1/r and the bending moment M: 1 M ¼ r EI Given that: 1/r ¼ ’/s, then ’ ¼

M s EI

Obviously the curving of a straight member will cause displacements transversely to its axis, and for skeletal structures this is the main reason for the development of deformations. The diagram of the bending moments allows it to be concluded directly how each region is curved, given that the curvature always occurs on the tensioned side of the member (Figure 2.49). In areas where the moment is null, the corresponding region has no curvature. To obtain a qualitative picture of the deformation for the whole structure (see Figure 2.49), each member is first deformed separately, according

∆ϕ ∆s r Shortening M

M

M

M

Elongation The change in the right angle is unacceptable

Figure 2.48 Bending deformation

66

(1/r) = M/EI ∆ϕ = (M/EI) · ∆s


[M]

[M]

At each point the curvature is proportional to the bending moment

Figure 2.49 A qualitative estimate of deformation on the basis of the bending moment diagram

to its moment diagram. The deformed members are then connected together, retaining the initial angles at each joint and considering the allowable deformations at the supports.

2.3.2.2 Axial force If the opposite sides of an element are acted upon by two self-equilibrating axial forces, an elongation or a shortening l of the element will occur, depending on whether the element is tensioned or compressed respectively (Figure 2.50). The relationship between the change in length and the axial force (Hooke’s law) is already known from Section 1.2.1: N s EA Clearly, in structures bearing their loads exclusively through axial forces, such as trusses or structures having a funicular form, the finally developed deformation is due exclusively to the elongation or shortening of their elements. l ¼

∆I/2

∆I/2

N

∆I/2

N ∆s Elongation

∆I/2

N

N ∆s

∆I = (N/EI) · ∆s

Shortening

Figure 2.50 Axial deformation

67

Structural systems: behaviour and design ∆h/2 ∆h/2 V ∆h/2

V

∆s

h

∆h/2

L

∆h = (V/GA) · ∆s

L/h > 3 The contribution of the shear forces is neglected

Figure 2.51 Deformation due to the shearing force

2.3.2.3 Shear force If the opposite sides of the element are acted upon by a pair of shearing forces V, a certain mutual ‘sliding’ h of the two sides occurs. This results to the development of an ‘angle of shearing deformation’ , or shearing strain, which is given by ¼ h/s (Figure 2.51). This may be written as V s GA where A is the cross-sectional area and G is the so-called shear modulus, which for concrete has roughly the half value of its modulus of elasticity and for structural steel is 0.8 108 kN/m2. The contribution of the shearing deformation to the overall deformation of a skeletal structure is generally negligible. It begins to play an appreciable role only in the case of so-called deep beams, where the ratio of the member length to the section depth is less than 3. As stated at the beginning of this chapter, skeletal structures in general do not belong to this category, and therefore the influence of shear forces on their deformation may be disregarded. h ¼

2.3.2.4 Temperature variation If an element is subjected to a uniform temperature T, which is different from the temperature T0 that it was created under, then its length undergoes a change lT according to the equation lT ¼ Ts aT s where Ts ¼ TT0 and aT is the temperature coefficient, which has the same value for steel and concrete up to 10 —5/8C. An increase or decrease in the initial temperature of the element by Ts leads to an elongation or shortening of the element, respectively. This change in length is not accompanied by any axial stress at all (Figure 2.52). Usually the temperature change is not uniform across the depth h of the element, which means that the the fibres at the two extremes of the section are at different temperatures. Thus, for example, in the cross-section shown in Figure 2.53, the 68

Stress and deformation (statically determinate structures) Uniform temperature difference ∆Ts ∆IT/2

∆IT/2

∆IT/2

∆IT/2

∆s

∆s

Temperature increase (elongation)

Temperature fall (shortening) ∆IT = ∆Ts · αT · ∆s

Figure 2.52 Deformation due to a uniform change in temperature

temperature of the top fibre is 108C and that of the bottom fibre is þ408C, while the initial temperature T0 is assumed to be þ58C. The temperature variation across the region between the extreme fibres is assumed to be linear. Thus the middle fibre has a temperature Tm ¼ þ158C, and consequently it undergoes a temperature increase of Ts ¼ 15 5 ¼ 108C and the corresponding elongation. Any non-uniform temperature change can always be considered as a superposition of a uniform temperature equal to the existing temperature of the middle fibre Tm and a non-uniform change T, which accounts for equal but opposite temperatures at the top and bottom fibres, having the value (T/2). In the present case the value is T ¼ 40 þ 10 ¼ 508C (see Figure 2.53). An element that undergoes such a temperature variation (i.e. an equal shortening and elongation of the two extreme fibres) develops a curvature with radius r, in the same Initial temperature: +5°C –10

+15

–50/2

(∆T/2)

h +15 +40

∆s

+15

+50/2

∆T

(∆T/2)

∆T

Temperature increase of the middle fibre: (15 – 5) = 10°C ∆φ Shortening ∆φ = (αT · ∆T/h) · ∆s

Due to ∆T

h

Elongation

Curvature in order to preserve the right angles

Figure 2.53 Deformation due to a non-uniform temperature change

69


manner as occurs in bending. However, without the presence of stress, the deformation is actually purely geometrical in nature. The consequent relative rotation ’ of the two opposite sides of the element is ’ ¼

T T s h

Of course it is always valid that 1/r ¼ ’/s. In conclusion, a non-uniform temperature change in an element, causes: (1) a uniform elongation or shortening, depending on the temperature change Ts in the middle fibre of the element; and (2) a curvature that is determined by the temperature difference T between the fibres located at the two extremes of the element. It should be noted that the imposition of any temperature variation on a statically determinate structure does not cause reactions, and consequently does not affect the sectional force diagrams at all — if these in fact exist due to other loads. However, it does cause deformations.

2.3.3

The principle of virtual work

This principle allows the determination of any deformation in a skeletal structure subjected to a certain loading, if the sectional forces diagrams (mainly M and N) are already known. The principle of virtual work applies to any skeletal structural system, and links a freely selected (arbitrary), self-equilibrating loading, which consists of concentrated forces or/and moments, the so-called virtual loading, with those deformations of the system under the real loading, which correspond to the specific loads of the virtual loading on the one hand, and the sectional forces diagrams due to the virtual and real loading on the other. Although this principle applies to any structure, whether determinate or indeterminate, and to any part of it, for simplicity it is illustrated below for a simple structure. When subjected to a uniform loading, the fixed structure shown in Figure 2.54 develops a state of stress, represented by the diagrams M and N, and a deformation due to this stress state. This situation is called real loading. It is clear that at any point on the structure certain curvatures and strains are developed, and these are represented by the quantities ’real and lreal, respectively (see Section 2.3.2). For the same structure, another arbitrary (‘imaginary’) loading may be considered that is totally independent of the real one. This imaginary loading may consist of, for example, the force P (2.0 kN) and the moment M (10.0 kN m), as shown in Figure 2.54, which, together with the induced reactions RP (2.0 kN) and RM (4.0 kN m), constitute a self-equilibrating system of forces, called the virtual loading, as also mentioned previously. In order to clarify how the above quantities are interconnected, first the structure is considered under the virtual loading only, i.e. the self-equilibrating force system Pvirt that gives rise to the bending moment and the axial force diagrams Mvirt and Nvirt, respectively. The real load is then superposed on the virtual loading, and the structure deforms further. Obviously this additional deformation is identical to the 70

Stress and deformation (statically determinate structures) 8.0 kN/m

36.0

9.0

36.0 δ

ϕ

[M real]

5.0 m

[N real]

24.0 3.0 m Real loading P = 2.0 kN

M = 10.0 kN m

6.0 4.0

2.0

[M virt]

[N virt]

RP = 2.0 kN RM = 4.0 kN m Virtual loading Principle of virtual work Work of external forces of virtual loading on the corresponding real deformations: (2.0 · δ) + (10.0 · ϕ) + (4.0 · 0) + (2.0 · 0) Equal to the work of internal forces of virtual loading on the corresponding real deformations: Work of bending moments: –1.0 · 36.0 · 4.0 · 5.0/EI + [(1/3) · 36.0 · 6.0 · 3.0/EI – (1/3) · 6.0 · 9.0 · 3.0/EI] = –558.0/EI Work of axial forces: +1.0 · 24.0 · 2.0 · 5.0/EA = +240.0/EA (negligible)

Figure 2.54 The principle of virtual work

one caused by the real loading itself, represented by the quantities ’real and lreal (see above). During this additional loading both the external (self-equilibrating) forces of the virtual loading and the corresponding internal (sectional) forces remain constant and produce some work on the imposed (real) deformations (i.e. displacements and rotations). Thus the external forces Pvirt of the virtual loading (reactions included) produce the work X virt resp;real P on those real displacements (and rotations) that correspond to them and, moreover, the internal (sectional) forces Mvirt and Nvirt produce work on the corresponding real deformations ’real and lreal, which can of course be expressed in terms of the real 71


sectional forces (see above). This work is given by ð ð ð ð Mreal Nreal ds þ Nvirt ds Mvirt ’real þ Nvirt lreal ¼ Mvirt EI EA The principle of virtual work states that the above two works are equal, and consequently ð X virt resp;real ð virt Mreal Nreal ¼ M P ds þ Nvirt ds EI EA For the case under consideration the left-hand side of the above equation can be written as X virt resp;real ¼ P þ M ’ þ RP 0 þ RM 0 P The deformations and ’ are shown in Figure 2.54, whereas the null terms represent the null displacement and rotation of the fixed support in the real situation. In the above equation all the terms are known quantities except resp,real, i.e. the real deformations ( and ’) involved. Thus, if on the left-hand side of the equation the sought-after deformation is the only unknown, it can be determined directly. However, for this to be possible the virtual loading must be chosen appropriately such that the work produced by it is restricted to this specific deformation only. It is thus obvious that, in order to determine a particular displacement, the virtual loading should consist of a corresponding force for that displacement plus the other forces (reactions) required for equilibrium, whereas the determination of a rotation requires the virtual loading to have a corresponding moment to this rotation, plus the other forces that ensure equilibrium (Figure 2.55). It is, therefore, clearly convenient if the ‘corresponding’ concentrated virtual action is chosen to be a unit force or moment. If the calculated deformation has a negative sign, its sense is opposite to that of the virtual load chosen. Thus, referring to Figure 2.55, the sought-after deformations are determined from the equations ð Mreal 1:0 þ 3:0 0 þ 1:0 0 ¼ Mvirt EI and ð

1:0 ’ þ 1:0 0 ¼ Mvirt

Mreal EI

These integrals account for each separate member of the skeletal system; their evaluation is facilitated by using the values given in Table 2.1. The values given in this table are for the integrals of the products of those functions that are commonly occurring variations of the sectional forces diagrams M and N, referred over a length L, with constant EI. It is clear that for the evaluation of the above integrals the bending moment diagrams M must be appropriately signed, as indicated in Section 2.2.1. It should also be noted 72

Stress and deformation (statically determinate structures) 8.0 kN/m

8.0 kN/m

δ

ϕ

Sought-after deformation 5.0 m

Real loading

5.0 m

3.0 m

3.0 m 1.0 kN

[M virt] Determination of δ

1.0 kN

Sought-after deformation

3.0 kN m Virtual loading

1.0 kN m

[M virt] Determination of ϕ

1.0 kN m

Figure 2.55 The choice of the virtual loading

that when a skeletal structure exhibits a bending response the contribution of the axial forces to its deformations is practically negligible compared with that of the bending moments. Thus usually only the bending moments need be taken into account in the integrals. g When, in addition to the above considered deformations there is a variation in temperature, the internal forces due to the virtual loading will produce additional work on the corresponding real deformations ’real and lreal (see Section 2.3.2): ð ð ð ð T ds þ Nvirt T Ts ds Mvirt ’real þ Nvirt lreal ¼ Mvirt T h

Table 2.1 Characteristic integral values for the moment diagram products M2

M1 M1 M1

M2

M2

1 M1 M2 L/EI

1 2 M1 M2 L/EI

2 3 M1 M2 L/EI

1 2 M1 M2 L/EI

1 3 M1 M2 L/EI

1 3 M1 M2 L/EI

1 2 M1 M2 L/EI

1 6 M1 M2 L/EI

1 3 M1 M2 L/EI

73


Because the above integrals contain a number of constant terms, the calculation, based on the appropriate areas of the diagrams for Mvirt and Nvirt, is relatively simple. However, due attention must be paid to the sign of the work. The sign of the first integral is positive or negative, depending on whether or not the curvature due to Mvirt is of the same sense as the curvature caused by the existing T (see Section 2.3.2). The sign of the second integral (which is not negligible) is positive or negative depending on whether or not the axial deformation due to Nvirt corresponds to that of the existing Ts. It is clear that if the above integrals are signed in the suggested manner, then all the quantities involved must be used with their absolute value. Thus the principle of virtual work may be written in its complete form as ð X virt resp;real ð virt Mreal Nreal ¼ M ds þ Nvirt ds P EI EA ð ð T ds þ Nvirt T Ts ds þ Mvirt T h It may be recalled here that the term Mreal/EI expresses the curvature 1/r at every point (see Section 2.3.2) and, consequently, the first term on the right-hand side may also be written as ð Mvirt ð1=rÞ ds g In the case of a truss the above relationship is simplified, and can be written as: X virt Nreal lþ N T Ts l EA Finally, it is noted that if the shearing deformation has to be taken into account on the basis of the existing shear forces, then the following term should be added to the righthand side of the above equation: ð V real ds Vvirt GA X

2.3.4

Pvirt resp;real ¼

X

Nvirt

A wider use of the principle of virtual work

The conditions under which the principle of virtual work may be applied were described in the previous section. However, it is also useful to examine the problem of finding the deformations in statically indeterminate structures. As an example, the statically indeterminate frame shown in Figure 2.56 is considered. Under the action of some loading, this frame develops the bending moment diagram shown in the figure. The horizontal displacement of the upper left-hand joint is to be determined. This deformation can, in principle, be found by considering a virtual loading consisting of a unit horizontal force applied at this joint, and the ensuing reactions, and determining the corresponding bending moment diagram. The procedure for determining the virtual work is then applied as set out in Section 2.3.3. 74

Stress and deformation (statically determinate structures) δ

7.37

1.0

7.37 6.0 m 13.58 8.98

8.98 4.60

4.60

13.58 6.0 m

15.05

15.05

Inappropriate way of determining δ

8.0 m δ

δo

1.0

7.37

8.98

7.37

o

u δu

4.60

4.60

ϕu Real loading

1.0

6.0 1.0

6.0 Virtual loading

The principle of virtual work is always valid 12.0 15.05

1.0

12.0

Real loading Virtual loading Direct determination of δ

Figure 2.56 A wider application of the principle of virtual work

However, although this procedure is correct, it is practically impossible to carry out by simple computational means. The self-equilibrating force system of virtual loading can be chosen much more simply if the left column of the frame is thoroughly cut out and subjected to the same virtual load at the top, accompanied by the actions required for equilibrium at its base (i.e. an equal and opposite force and an appropriate moment). The bending diagram so obtained corresponds to that for a vertical cantilever (see Figure 2.56). The principle of virtual work may now be applied directly: ð Mreal ds 1 þ 1 0 þ 12:0 0 ¼ Mvirt EI It is understood that the integral refers only to the selected members. g In order to point out the absolute freedom to use the principle of virtual work, the left upper column of the frame is considered alone (see Figure 2.56). The unit horizontal 75


force at the top of this column is appropriately equilibrated by a force and a moment at its base. The system of these three forces is a self-equilibrating one and clearly constitutes a virtual loading, and consequently the principle may be applied directly: ð Mreal ds 1 o 1 u 6:0 ’u ¼ Mvirt EI where o, u and ’u are the corresponding real deformations to the virtual forces considered, i.e. the horizontal displacements at the points o and u, and the rotation at point u. The integral obviously concerns only the selected member. However, although the above equation is absolutely correct, it is of no help at all in the determination of o, as this is not the only unknown in the equation.

2.3.5

The Betti—Maxwell theorem

This theorem is of particular importance in structural theory and is valid for any type of structure. Consider, for example, the structure shown in Figure 2.57. Now consider two separate loadings: a concentrated load P in state I and a moment M in state II. Each of these forces acts at a different point. In state I the structure develops a rotation ’ at the point where the moment is acting in state II, whereas in state II the structure develops a displacement at the point where the concentrated force is acting in state I. According to Betti’s theorem the (hypothetical) work produced by the force P (remaining constant) on the ‘path’ is equal to the work produced by the moment M (remaining constant) on the ‘path’ ’. That is, P¼M’ Of course, instead of the moment M, there may be any other concentrated force in its place, i.e. instead of the pair (M, ’) another one (P, ) may be considered. Obviously the deformations considered in each instance should correspond to the type of forces producing work on their ‘path’. Thus it may be generally written: (force I) (corresponding deformation II) ¼ (force II) (corresponding deformation I) ϕ

M

δ

P I

II

Betti: P·δ=M·ϕ Maxwell: if P = 1 and M = 1 then δ = ϕ

Figure 2.57 The Betti—Maxwell theorem

76


The relationship above has particular importance if the forces P and M have unit value. Then it is: ¼’ Of course this relationship must be interpreted as an equality of work (1 ¼ 1 ’) rather than as an absurd equality between two quantities having different dimensions. Nevertheless, for unit forces it may be written as (respective deformation to II) ¼ (respective deformation to I) The above relationship is also known as the reciprocal theorem or Maxwell’s theorem.

2.3.6

The beam equation and finding the deflection curve

The principle of virtual work can be used to determine the deformations at specific points of a structure. However, in order to find the deflection curve for one or more members of a skeletal structure, it is necessary to return to the basic equation for the beam, governing its deflection w(x) under an applied distributed load p(x). As the curvature 1/r at a point of a beam with abscissa x is expressed to a very good approximation by the value (dw2/dx2), its relationship to the bending moment (see Section 2.2.3) may be written as dw2 M ¼ ðaÞ EI dx2 Moreover, the equation for equilibrium between the bending moment and the load (see Section 2.2.3) is

d2 M ¼ pðxÞ ðbÞ dx2 By twice differentiating equation (a), the classical beam equation is obtained: dw4 ¼ pðxÞ dx4 The physical meaning of this equation is that, if the beam is ‘compelled’ to deform by w(x), it will offer a resistance qi equal to EI

dw4 dx4 This resistance, which is directed in the opposite sense to w, is of course in equilibrium with the applied load p(x) causing this deformation, so that qi ¼ p(x). Solving the above equation mathematically in order to determine the function w(x) is cumbersome, and does not provide the required physical picture. Thus, in order to find the deflection curve, instead of the purely mathematical approach, the following procedure (Mohr’s theorem) is preferred (Figure 2.58). It is clear that in the same way that equation (b) can be used to find the diagram for M from the load p, qi ¼ EI

77


1

2 [M] δ

Developed deformation

M/EI 1

2 Fictitious loading of a simply supported beam M/EI [M]

1

1

Deflection curve

2

Bending moment diagram M due to fictitious loading: deflection curve for unyielding ends

2 δ The diagram M drawn between the displaced ends leads to the final deflection curve

Figure 2.58 Finding the deflection curve using the bending moment diagram

equation (a) can be used to determine w on the basis of M, i.e. as the bending moment diagram that results from application of the load M/EI. Thus, in any rectilinear member of a skeletal structure having a diagram for M, the deflection curve transverse to its axis and between its displaced ends may be drawn as the moment diagram of the corresponding simply supported beam loaded with M/EI. Note that this loading must always be directed towards the tensioned fibres of the part of the member under consideration (see Figure 2.58). It is also clear that the transverse displacements at the ends of the member may be determined by appropriate application of the principle of virtual work (see Section 2.3.3). A further remark to be made is that equation (a) written in the form d dw M ¼ dx dx EI suggests an analogy with the equilibrium relation dV/dx ¼ p, and consequently the slope of the deflection curve may be found from the loading M/EI in the same way the shear force can be found from the loading p. Thus the slopes of the non-displaced ends of a beam with bending diagram [M] may be represented by the reactions of a simply supported beam loaded by the diagram for M/EI, directed as explained above. 78


2.3.7

Elastic support of structures

The load-bearing structures are supported on the ground. Thus, at the points of support the ground takes up forces and moments that are equal and opposite to the reactions, and consequently the ground deforms. Obviously the role of the ground as a means of support may be undertaken by any other appropriately formed structure. If the action applied by a structure on the body on which it is supported causes an elastic deformation exclusively in the sense of the action itself, then the corresponding support offered to the structure is called elastic simple support or elastic clamped support, depending on whether the action is a force or a moment.

2.3.7.1 Elastic simple support Here it is assumed that the ground at the point of application of a vertical concentrated force P deforms by , according to the relationship ¼ fs P The coefficient fs is called the coefficient of elastic support, and represents the settlement of the ground under a unit concentrated force. The elastic support is provided by a translational spring. Clearly the settlement is directed in the same sense as P (Figure 2.59). The elastic simple support may be represented by any elastic body which, under a concentrated load P, develops only a corresponding displacement , and no rotation, as would occur, for example, in an axially loaded bar or a simply supported beam loaded at its middle.

2.3.7.2 Elastic clamped support Here it is assumed that the ground at the point of application of a moment M rotates by ’ according to the relationship ’ ¼ f’ M The coefficient f’ is called the coefficient of elastic rotation, and represents the rotation of the ground under a unit moment. The elastic clamped support is provided by a rotational spring. Clearly the rotation ’ has the same sense as M (Figure 2.60). The elastic clamped support may be represented by any elastic body that under a moment M develops only a P

P

P P

δ

Ground δ

δ fs

P

δ = fs · P fs

L

δ L

fs = L/(EA)

fs = L3/(48 · EI)

Figure 2.59 Elastic simple support

79

Structural systems: behaviour and design M Ground

M

M

ϕ fϕ

fϕ

ϕ

M ϕ = fϕ · M

ϕ L fϕ = L/(3 · EI) M ϕ fϕ = L/(6 · EI)

L

L

Figure 2.60 Elastic clamped support

rotation ’ and no displacement, as would occur, for example, at the simply supported end of a beam or at the joint of an unmoveable frame. g At this point it should be emphasised that a statically determinate structure, which is supported elastically in the above sense, develops exactly the same state of stress as if it were unyieldingly supported, although it develops a different deformation. The cantilever shown in Figure 2.61, subjected to a uniform load p and elastically clamped with a coefficient f’ , develops an identical bending moment diagram to

p

p · L2/2

fϕ L

φ

Same diagram M as for unyielding support

fϕ Ground

p · L2/2

p · L2/2

Opposite sense than the rotation caused P=1

1.0 · L

1.0 · L Virtual loading 1.0

Figure 2.61 Elastically clamped structure

80

Stress and deformation (statically determinate structures) p

p

fs

δ

p · L/2

L p · L/2

Ground

fs

δ

Figure 2.62 A elastically simply supported structure

the diagram produced for a fixed support. However, its deflection at the free end differs. According to the principle of virtual work the bending moment may be calculated as ! # ð " L2 Mreal ds f’ Mvirt 1 ð1 LÞ p 2 EI Note that, the moment 1 L, as an external force (reaction) of the chosen ‘virtual loading’, produces work on the corresponding ‘real’ rotation due to the action ( p L2/2) of the ‘real loading’ on the ground, but the two moments have the opposite sense. The fact that the reaction ( p L2/2) has an opposite sense to that of the developed rotation at the support is important. Moreover, referring to the elastically supported beam shown in Figure 2.62, it is similarly noted that the reaction ( p L/2) acting on the beam has an opposite sense to the corresponding support settlement at this point. It is understood that the latter remarks are generally valid, regardless of whether or not the structure is statically determinate.

2.3.8

The concepts of flexibility and stiffness

If in a structure a deformation caused by a certain force is expressed through the quantity f by the relationship Deformation ¼ Force f, then it is obvious that, at the same point on the structure, the force required in order to produce a similar deformation as the above may be expressed through the quantity k by the relationship Force ¼ Deformation k, where k ¼ 1/f. The above remarks relate directly to the concepts of flexibility and stiffness, respectively, which are merely a consequence of the deformability of structures. Thus the term flexibility f means the deformation at some specific point of a structure caused by a unit force. The term stiffness k means the opposite; namely, the force that must be applied at a certain point on a structure in order for it to develop a unit deformation. According to the above discussion of elastic simple and clamped supports, the quantities fs and f’ introduced in Section 2.3.7 represent flexibility quantities, 81


δ

δ

L δ = P · L3/(48 · EI)

L

Offered resistance = stiffness

Flexibility

P = stiffness

P = δ · (48 · EI/L)

3

fs = L3/(48 · EI)

δ=1

Stiffness P

δ = P · L/(EA)

δ

Flexibility L

P = δ · (EA)/L

fs = L/(EA)

Stiffness

Figure 2.63 Flexibility and stiffness with respect to displacement

whereas the values 1/fs and 1/f’ represent stiffness quantities. Specifically, these last values represent the force and moment required to act on the springs in order to produce a unit displacement and rotation, respectively. The simply supported beam shown in Figure 2.63, under the concentrated load at the middle, develops at this point a deflection equal to ¼

L3 P 48 EI

The quantity (L3/48 EI) represents the flexibility of the beam with respect to the vertical deflection of its middle, which is the deformation itself as the force is unity (P ¼ 1). If a structure is supported at the middle of the beam (see Figure 2.63), it can be considered as elastically supported with a coefficient of elastic support fs equal to (L3/48 EI). Furthermore, given that P ¼ (48 EI/L3) , the quantity (48 EI/L3) represents the stiffness of the beam at its middle with respect to the vertical deflection, and is the force required to produce the unit displacement ¼ 1. This specific stiffness of the beam is proportional to EI of its cross-section, and it decreases rapidly as the length of the beam increases. Moreover, in the axially loaded member (see Figure 2.63) the quantity EA/L represents its axial stiffness, and is the compressive (or tensile) force required to produce a unit shortening (or elongation). g 82

Stress and deformation (statically determinate structures) M ϕ

ϕ L φ = M · L/(3 · EI)

Offered resistance = stiffness

Flexibility

M = stiffness

M = φ · (3 · EI/L) ϕ=1

Stiffness

L

fφ = L/(3 · EI)

Figure 2.64 Flexibility and stiffness with respect to rotation

The same beam as above loaded with the concentrated moment M at its ends develops a rotation angle ’ (Figure 2.64): L M 3 EI The quantity (L/3 EI) represents the flexibility of the beam with respect to its end rotation, being the rotation due to the unit moment M ¼ 1. If a structure is monolithically connected to the end of the beam (see Figure 2.64), the structure may be considered as elastically clamped with a coefficient of elastic clamped support f’ equal to (L/3 EI). Furthermore, given that M ¼ (3 EI/L) ’, the quantity (3 EI/L) represents the stiffness of the beam with respect to the end rotation, and is the moment required to produce the unit rotation ’ ¼ 1. This specific stiffness of the beam is proportional to EI and inversely proportional to its length. g ’¼

From the two cases described above it can be seen that the stiffness represents the resistance (force) offered by the structure against a deformation (typically of unit value). Clearly the concept of stiffness is quite general, and it is not restricted to the case of a simple beam. For example, the horizontal force that must be applied to the node of the frame shown in Figure 2.65 in order to produce a unit horizontal displacement at that point is taken as the ‘lateral stiffness of the frame’ at that point. P = stiffness

δ=1

Figure 2.65 Lateral stiffness

83


Finally, it should be noted that, while a change in the sectional properties EI of a member in a skeletal structure clearly affects its deformability, the state of stress of the statically determinate structure remains totally unaffected.

2.4

Symmetric plane structures

In many cases structures are designed such that they possess an axis of symmetry with regard to their geometric and elastic properties (Figure 2.66). When such a structure is subjected to a symmetric or antisymmetric loading, its sectional forces, as well as deformations, at the points of intersection with this axis must obey certain rules. An antisymmetric loading is one which, when the sense of all loads on each side of the symmetry axis is inverted, results in a symmetric loading. An arbitrary loading of a symmetric structure may always be considered as a superposition of a symmetric and an antisymmetric part, as shown in Figure 2.67. The properties of both symmetric and antisymmetric loading are always very useful, and are described below.

2.4.1

Symmetric loading

A symmetrical loading with respect to a (vertical) axis of symmetry is visually directly obvious. Symmetrical loading also includes any concentrated load P0 lying on this axis. All the sectional forces and the deformations are developed symmetrically to the axis (Figure 2.68). Based on this fact the following conclusions may be deduced, and are valid for all points A lying on the intersection of the structure with the axis of symmetry. (a) Deformations . The rotations and displacements perpendicular to the axis are null. . Displacements are developed only along the axis of symmetry. (b) Internal forces . The (vertical) component along the axis of symmetry of the internal force acting on the section is equal to P0/2, as may be concluded from the vertical equilibrium of the cut out joint around the axis of symmetry. In the case when P0 ¼ 0, this component p

p

p

P

P

q

q

M

M Symmetric loading

Figure 2.66 Symmetric and antisymmetric loading

84

p

P

P

q

q

M

M Antisymmetric loading

Stress and deformation (statically determinate structures) p

p/2

P

p/2

p/2

P/2

p/2

P/2

P/2

P/2

q/2

q/2

q/2

q/2

q

Arbitrary loading

Symmetric

Antisymmetric

Figure 2.67 Splitting of an arbitrary loading into a symmetric and an antisymmetric part

obviously vanishes. Note that this component does not generally represent the shearing force of the relevant section. . Both the bending moment and the (horizontal) component perpendicular to the axis of symmetry of the internal force are normally developed. The above stress and deformation state at point A allows the structure to be cut out at each such point and to be provided there by a movable fixed support, as shown in Figure 2.68. This kind of support permits movement parallel to the axis of symmetry, as well as the development of both a bending moment and a horizontal reaction. However, the development of rotation, of displacement perpendicular to the axis and of a reaction along the axis is not possible. In the case of a concentrated force P0 acting on the structure, an applied force at point A equal to P0/2 must be considered. Thus it is concluded that the whole structure works exactly the same as the above ‘half-model’.

Rotation = 0 Displacement = 0 A

P0/2

P0 P0

P0/2

P0/2 A

δ

Symmetric stress state ϕ

P

δ

Always existing

ϕ

Symmetric deformation

P/2

P A

A Equivalent model

Figure 2.68 Properties arising from a symmetric loading

85


2.4.2

Antisymmetric loading

A concentrated force H0 may also count as an antisymmetric loading, perpendicular to the axis, as may a concentrated moment M0, both acting at the intersection points of the structure with the axis. All sectional forces and deformations are developed antisymmetrically to the axis (Figure 2.69). Based on this fact the following conclusions may be deduced, and these are valid for all points A lying on the intersection of the structure with the symmetry axis. (a) Deformations . The displacements perpendicular to the axis are null. . The rotations and the displacements perpendicular to the axis are normally developed. (b) Internal forces . The moment equilibrium of a joint cut out around the axis implies that the bending moment is M0/2, and in the case when M0 ¼ 0 it obviously vanishes. . The (horizontal) component perpendicular to the axis of symmetry of the internal force acting on the section is H0/2, as may be concluded from the horizontal equilibrium of the cut out joint around the axis of symmetry. In the case when H0 ¼ 0, this component obviously vanishes. Note that this component does not generally represent the axial force of the section under consideration. . The (vertical) component along the axis of symmetry of the internal force is normally developed.

Displacement = 0 A M0

M0 H0 δ

δ

Antisymmetric stress ϕ

A

H0/2

M0/2 H0

M0/2

Always existing

H0/2 Equivalent model

ϕ

Antisymmetric deformation

A A (EI) N = 0

Figure 2.69 Properties arising from an antisymmetric loading

86

H0/2

(EI/2)


The above stress and deformation state at point A allows the structure to be cut out at each such point and to be provided there with a simple support, as shown in Figure 2.69. This kind of support satisfies all the above requirements at point A. The presence of a concentrated force or moment at joint A of the structure implies the action on the supported point of the forces H0/2 and/or M0/2. Thus it is concluded that the whole structure works in exactly the same way as the above ‘half-model’. It should be noted that in the case where members of the structure lie on the axis of symmetry, in the ‘half model’ these members have to be considered to have EI/2. However, their bending and shear forces in the actual structure have a value double that in the ‘half model’. Moreover, the axial force in all such members vanishes. This may be explained if a value for the axial force of such a member is assumed and then the structure is ‘seen’ from the ‘back’, i.e. the same structure with an identical loading but in the opposite sense. The fact that for the same member and the same loading two opposite values of the axial force may be concluded proves that this axial force simply cannot exist.

2.5

Grid structures

2.5.1

General

The plane structures examined so far have been loaded and deformed within their own plane. In this section the basic concepts concerning the stressing and deformation of grid structures will be presented as they were defined in Section 2.1, i.e. the concepts as they apply to plane structures that are loaded perpendicularly to their own plane (Figure 2.70). In this respect it is appropriate for the grid to be referred to a horizontal plane defined by two arbitrary axes OX and OY. For any cut out part of a grid in equilibrium the following conditions should be satisfied: X ! X! X ! My ¼ 0; Pz ¼ 0 Mx ¼ 0;

Y Z 2

2 3

1

3

1

O

4 X

Equilibrium Σ MOX = 0 Σ MOY = 0 Σ MOZ = 0

4 The reactions at 1 and 4 together with the external loading satisfy the equilibrium conditions

Figure 2.70 Equilibrium conditions for a loading perpendicular to a plane structure

87


This means that the sum of the moment vectors of all the forces acting on the examined part, with respect to both axis OX and axis OY, must be null. Furthermore, the sum of all acting vertical forces must also be zero.

2.5.2

Sectional forces and deformations

For the better description of the sectional forces and deformations at a point on a grid member, account should be taken of the vertical plane to the member at the point considered. The intersection of this plane with the member reveals two free edges, both of which exhibit identical deformation vectors but equal and opposite vectors for their sectional forces. These magnitudes are referred to a local orthogonal system (x, y, z) that has its origin at the centroid of the section (Figure 2.71). Its x axis coincides with the axis of the member, while the two other axes y and z lie on the edge plane, in the horizontal Y Z

O

X

The ‘screw’ convention

Sectional forces

Deformations

z

Local system

Vz (V ) My

y

ϕy

(MB) O Mx

x

w

(MT) Plane Oxy coincides with the plane OXY

Loading vertical to the plane OXY (grid) V

MT

MB

Positive sectional forces

MT MB

V

Figure 2.71 Sectional forces and the deformations developed

88

ϕx


and the vertical senses, respectively. It is assumed that the section is symmetrical about the horizontal y axis. The vectors with double arrows in Figure 2.71 are moments and rotations that obey the ‘screw rule’, i.e. they express the sense in which a screw must be rotated in order to proceed in the direction indicated by the arrow. In a grid where each member is loaded within its corresponding plane (x, z) the following internal forces are developed (see Figure 2.71): . shear force Vz (i.e. V) . bending moment My (i.e. MB) . torsional moment Mx (i.e. MT). The bending moment MB together with the shear force V induce bending within the vertical plane (x, z) of each member. Moreover, at each point on the grid, the deformations developed consist of a vertical deflection w (with respect to the z axis) and a rotation represented by a horizontal vector. The projection ’y of this vector onto the y axis (i.e. perpendicular to the member axis) refers to the bending of the member (bending rotation), whereas the projection ’x onto the x axis refers to its torsion (twisting angle). The above sectional forces are signed as follows (see Figure 2.71): . The shear force is positive if, together with the shear force acting on the next cut out section, it forms a clockwise couple. . The bending moment MB is positive if it causes tension at the bottom fibres. . The torsional moment MT is positive if its vector acts in a ‘tensile sense’ on the section. These three sectional forces can always be determined through the three previously stated equilibrium conditions for a cut out part of a grid, provided that all the other forces acting on it are known (Figure 2.72). For the determination of any deformation, the principle of virtual work is always applicable, which, in order to allow for the contribution of torsional moments in

Y

Equilibrium of cut out part

Z

O

(Grid)

X For the determination of sectional forces the number of unknown magnitudes must not exceed 3

Figure 2.72 Determination of the sectional forces on a grid

89


carrying the vertical loads, is now written in the form: ð X virt resp;real ð virt Mreal Mreal T ¼ MB B ds þ Mvirt ds P T EI GIT

2.5.3

Torsion

It is already clear that the stressing factor that differentiates grids from plane structures is the torsion. The basic property of the torsional behaviour of a member having a length L is the relationship between the acting torsional moment MT at its free end with the corresponding twisting angle ’T of the end section of this member (Figure 2.73): ’T ¼ MT

L GIT

where IT represents the torsional moment of inertia and G is the shear modulus of the material of which the member is composed (see Section 2.3.2). From the above equation it is concluded that the magnitude GIT/L expresses the torsional stiffness of the member, being, according to Section 2.3.8, that torsional moment required to produce a unit twisting angle, or, in other words, the torsional resistance offered by the member if subjected to a unit twisting angle ’T ¼ 1. For an orthogonal section having dimensions b/t (b > t), IT is calculated according to the expression IT ¼ b t3/, the coefficient varying from 3.0 for oblong sections to 7.0 for square sections. A torsional moment MT causes a rotating flow of shearing stresses on a full orthogonal cross-section, diminishing towards the centre and having a resultant moment vector with respect to the centre of the section identical to MT (Figure 2.73). In the case of orthogonal sections, the maximum value max of these shearing stresses lies at the middle of the longest side and is proportional to the stress magnitude MT/(b t2) by a ϕT MT

b>t t

τmax

L

MT ϕT

ϕT = MT · (L/GIT)

b τmax = (3 ~ 5) · MT/(b · t 2) 3 for b/t > 10 5 for b/t = 1

Figure 2.73 The torsional response of a beam having an orthogonal full section

90


t

V = MT · b/(2 · Fk)

h

τ (constant)

MT

MT ϕT

Fk V = MT · h/(2 · Fk)

t

b

The shear force of each wall is proportional to its height

Figure 2.74 The torsional response of a beam with a hollow box section

factor of 3.0 for oblong sections and 5.0 for square sections. Note the inverse relationship between max and the square of the thickness of the section. A commonly used type of section is the hollow box section (Figure 2.74). This section has a constant wall thickness t, and takes up the torsional moment MT through a rotating peripheral constant shearing stress , which, according to the Bredt formula, is given by ¼

MT 2 Fk t

where Fk is the area enclosed by the middle line of the walls, i.e. slightly less than the area of the full section (see Figure 2.74). It may be concluded that each wall of the examined section of the rectilinear member is subjected to a constant peripheral force flow vl ¼ t ¼ MT/(2 Fk) acting per unit length (kN/m), which thus causes a total shear force V over each wall (see Figure 2.74). It is found that this shear flow v has a constant value for all walls, even if their thicknesses are different. Of course the relationship between the twisting angle ’T and the torsional moment MT is also valid here, the torsional moment of inertia IT now being equal to 4 t F2k =L , where L is the perimeter of the middle line of the section. Note that the quantity IT, and consequently the torsional stiffness GIT/L, is, for the same cross-sectional area, much greater in a hollow than in a full section.

2.5.4

Types of support

The two following ways of supporting a point on a grid structure may practically be distinguished (Figure 2.75). . Simple support. This type of support simply prohibits the vertical movement. The deformations ’x and ’y are freely developed. The only reaction developed is a vertical force R. 91

Structural systems: behaviour and design Simple support

Fixed support

ϕy My

ϕx R

Mx

R

ϕx = 0 ϕy = 0 w=0

Figure 2.75 Two ways of supporting a grid structure

. Fixed support. With this type of support all possible deformations are constrained. The developed reactions are a vertical force R and the two moments Mx and My. If the whole grid is supported in such a way that only three reaction forces are developed, the reactions can be determined using the three equilibrium conditions mentioned previously. If more than three reaction forces exist, the equilibrium conditions are insufficient for the determination of these forces.

92

3 The handling of deformations for determining the stress state in framed structures (statically indeterminate structures) 3.1

Introduction

As pointed out in the previous chapter, the vast majority of plane frames are statically indeterminate structures. This means that it is not possible to determine their stress state (i.e. the diagrams M, V and N) by using only the equilibrium equations, even if their reactions can be determined, as shown in Figure 3.1. In such a structure, one can always perform suitable modifications consisting of either removing support elements or altering the continuity of the structure wherever needed so that the system is changed to a statically determinate one (see Section 2.1.8). Each modification corresponds to a specific number of forces that are automatically released (a maximum of three), which means that their development in the new system is excluded. The statically determinate structure that results in this way is called the primary structure, and henceforth it constitutes the base on which the forces suppressed by the modifications will be determined. The number N of these forces is the degree of redundancy of the initial structure (Figure 3.2). These forces are called statically redundant forces and may be considered as (unknown) external actions applied to the primary (statically determinate) structure. It is clear that at the points where the continuity in the statically redundant structure has been released, the external actions may be considered to be acting on the two ends of the corresponding section with equal and opposite sign (see Figure 3.2). It is obvious that for any arbitrary value of these forces and the given external loads, the primary structure is in a state of equilibrium and has a specific deformation. Thus in a statically indeterminate structure, which after suitable modifications is transformed to a statically determinate system, an unlimited number of equilibrium states under the specific loading can be considered. This fact automatically raises the question of which stress state among all the possible stress states will really be developed in the structure. Assuming an elastic behaviour, one unique solution exists for this problem. This solution can be found on the basis of the deformability of the primary structure, as the deformations which are developed at the modification points are not compatible with those of the real structure (see Figure 3.2). Structural systems: behaviour and design 978-0-7277-4105-9



The three equilibrium equations are not sufficient for the determination of the sectional forces

Hinge removed

Statically determinate

Figure 3.1 The inadequacy of the equilibrium equations for determining the stress state

However, the compatibility is restored by the application of concrete external statically redundant forces on the primary structure. Thus, if continuity is broken at some point in a monolithic member, both section ends should not only come in contact after the final application of the statically redundant forces, but their relative rotation should be zero, unless the break in continuity takes place at an internal hinge. It is clear that when a support is removed the developing shift (or rotation) should have the value imposed by the real conditions. Clearly the number of statically redundant forces will always coincide with the number of deformation conditions that should be satisfied in this way, a fact that always leads to a unique solution. This way of dealing with the problem, which is described in further detail below, is known as the force method.

3.2

The force method

3.2.1

Physical overview of the force method

The procedure that may be applied for any statically indeterminate system involves the three steps described below. Step 1 Specific modifications are made to the redundant system in order to transform it into in a statically determinate (and rigid) one, i.e. the primary structure. All considerations hereafter will concern this system only. 94

Stress state in framed structures (statically indeterminate structures)

Hyperstatic structure

Releasing of three redundant forces

There are various possibilities for selecting the primary system

Redundant forces

Development of: • relative rotation • relative horizontal displacement • relative vertical displacement

Requirement for restoring the continuity of deflection curve

Primary system

Primary system

Releasing of three redundant forces

Redundant forces Development of displacement Development of rotation

Requirement for zero rotation Requirement for zero displacement

Primary system

Primary system

Figure 3.2 The selection of the primary structure and the corresponding requirements

The modifications have the following two consequences for the primary system: . they exclude the development of some specific (redundant) forces X1, X2, . . . , etc. . they allow the development of deformations at the specific points, corresponding to the forces X1, X2, . . . , etc. Therefore, the following deformations are calculated in the primary structure under the given external loads: . deformation along the direction of X1 . deformation along the direction of X2. Step 2 The same deformations as above are calculated for the imposition of the external loads X1, X2, . . . , etc. only. Each of these deformations results as a linear combination of X1, X2, . . . , etc. 95


Step 3 The total deformations of the primary structure are expressed as a superposition of those determined in Steps 1 and 2. It only remains, therefore, to enforce these deformations to take the values they must actually have. This automatically implies the determination of a system of as many equations as there are unknowns X1, X2, . . . , etc. Thus the role of the statically redundant forces becomes explicit: the forces X1, X2, . . . , etc. should take such values that if applied as external forces to the primary structure together with the external loads they cause the specific deformations that occur in the real redundant system (compatibility of deformations). It is clear that the force method consists exclusively in dealing with the deformations that are developed in the primary structure. It should be noted that the satisfaction of the two criteria of equilibrium and compatibility of deformations is an absolutely necessary condition that a stress state in a statically indeterminate structure has to fulfil, within the premises of an assumed elastic behaviour. g The procedure described above is now applied in the following two examples. Example 1 Consider the two-hinged frame shown in Figure 3.3. Of the various modifications that can be made to make this structure statically determinate, the interruption of the continuity of the girder by placing a hinge at node G is selected. The primary structure thus created is a three-hinged frame. Step 1. With this modification, the possibility of the development of a bending moment X1 at node G automatically disappears, while the deformation corresponding to X1 is developed, i.e. the relative rotation of the two ends at node G. The magnitude of X1 is therefore considered to be the statically redundant force. Step 2. Now X1 is considered as an external individual loading on the primary structure, due to which a corresponding deformation along the direction of X1 is developed. Step 3. Superposition of the deformations obtained in Steps 1 and 2 gives the total relative rotation at the ends of the hinge due to the external load X1, which will obviously depend on the value of X1. It is clear that the only acceptable value for X1 is the one that causes a deformation identical to the deformation of the real indeterminate system, i.e. a zero relative rotation. Example 2 Consider the suspended beam shown in Figure 3.4. Of the various modifications that can be made to make this structure statically determinate, the interruption of the continuity of bar CD and changing the fixed support A to a hinged one are selected. Step 1. In the primary structure thus created: . The development of an axial force X1 in the bar CD and of a bending moment X2 at node A, is excluded. 96


G

Redundant structure

X1 G

Primary structure

X1

Development of relative rotation Primary structure

Development of relative rotation Primary structure

Requirement for elimination of relative rotation

Figure 3.3 A redundant force as an external action

. Deformations due to the external loads are developed in the directions of X1 and X2, i.e. a relative shift of the cut points in the bar CD and a rotation at end A respectively. The magnitudes of X1 and X2 are the statically redundant forces. Step 2. The imposition of X1 and X2 only leads to a corresponding deformation in the direction of X1 (a relative shift of the cut points) and a corresponding deformation in the direction of X2 (a rotation of end A) will develop. Obviously these deformations will depend on the values of X1 and X2. Step 3. Superposition of the deformations obtained in Steps 1 and 2 gives the relative shift and rotation that will result from the simultaneous action of the external loads and any arbitrary values of the statically redundant forces X1 and X2. However, it is obvious that only those values of X1 and X2 are sought after that will lead to the deformations that actually occur in the real redundant system, i.e. a zero relative shift and a zero rotation.

3.2.2

The analytical application of the force method

An analytical application of the force method is now presented using Example 2 in Section 3.2.1, as illustrated in Figure 3.5. As already explained, for the primary structure 97

Structural systems: behaviour and design C

A

D Redundant structure

B

C X1

X1 A

X2

Relative displacement

D Primary structure

Relative displacement X1

Rotation

B

Relative displacement

X1

Rotation

X2 Rotation

Requirement for null relative displacement Requirement for null rotation

Figure 3.4 Redundant forces as external actions

two deformations have to be determined when the given external loads as well as the loads X1 and X2 are applied: deformation in the direction of X1 and deformation in the direction of X2. The first deformation is denoted by s1 and the second one by s2 . The deformation s1 is considered positive when, corresponding to the deformation effect of the arbitrary direction of X1, the two section ends are approaching each other. Deformation s2 is considered positive when, corresponding to the deformation effect of the arbitrary direction of X2, the end is rotating clockwise. The following loading states are considered: . State 0: loading due to external loads. . State 1: loading due to a unit force X1 ¼ 1. . State 2: loading due to a unit moment X2 ¼ 1. It is obvious that, according to the superposition principle, any deformation in the final state corresponding in the total action of the external loads and the individual actions X1 and X2, is equal to the deformation value in State 0, plus X1 times its value in State 1, plus X2 times its value in State 2. Thus, for the relative shift s1 of the section ends 1s ¼ (relative shift in State 0) þ X1 (relative shift in State 1) þ X2 (relative shift in State 2) 98

Stress state in framed structures (statically indeterminate structures) 20 kN/m 5.0 m 10.0

5.0 m

A = 0.002 m2 I = 0.00029 m4 E (homogeneous)

Redundant structure ∆s1 X1

∆s2

Relative displacement F10 (virtual loading: [1]) 20 kN/m

X2

X1

20 kN/m

Primary structure

Relative displacement F11 1 kN

Relative displacement F12

· X1

1 kN

· X2 1 kN

Rotation F20 [M0]

(virtual loading: [2]) [0]

Rotation F21 [M1]

[1] 1.49

562.50 F10 = –5121.9/EI F20 = 2812.5/EI

Rotation F22

[2]

[M2] 1.0

F11 = 11.1/EI + 11.2/EA F21 = –4.97/EI

Step 1 Dependent on the external loading ∆s1 = F10 + X1 · F11 + X2 · F12 ∆s2 = F20 + X1 · F21 + X2 · F22 Requirement by redundant structure: ∆s1 = 0 Requirement by redundant structure: ∆s2 = 0

(F12 = F21)

F12= –4.97/EI F22= 5.0/EI

Step 2 Independent of the external loading

Step 3

X1 = 299.70 kN X2 = –264.65 kN m

Figure 3.5 The analytical application of the method of forces

while for the rotation s2 of end A it will be s2 ¼ (rotation in State 0) þ X1 (rotation in State 1) þ X2 (rotation in State 2) At this point, a more appropriate notation is introduced for the deformations in order to systemise their computation. Fij is used to denote the deformation of the (primary) structure, and this deformation corresponds to the redundant force Xi when the structure is loaded only with Xj ¼ 1, in other words the deformation corresponding to Xi in state j. As Fij corresponds to a deformation due to a unit load, it is called the flexibility coefficient (see Section 2.3.8). Fi0 is used to denote the deformation of the primary structure, corresponding to the 99


redundant force Xi when the structure is loaded only with the given external loading (State 0). Using the above defined notation and according to the aforementioned superposition, it may be written for s1 and s2 : s1 ¼ F10 þ X1 F11 þ X2 F12 s2 ¼ F20 þ X1 F21 þ X2 F22

(a)

The physical significance of the factors Fij is as follows: . F10, the relative shift due to the external loads; F11, the relative shift due to X1 ¼ 1; F12, the relative shift due to X2 ¼ 1. (A positive sign corresponds to the deformation due to X1.) . F20, the rotation due to the external loads; F21, the rotation due to X1 ¼ 1; F22, the rotation due to X2 ¼ 1. (A positive sign corresponds to the deformation due to X2.) All the above deformations can be computed through the principle of virtual work. For the magnitudes F10, F11 and F12 the loading X1 ¼ 1 has to be used as the virtual one (State 1), whereas for the magnitudes F20, F21 and F22 the loading X2 ¼ 1 is used as the virtual one (State 2). Thus: ð M 5121:88 (kN m) 1 F10 ¼ M1 0 ds ¼ EI EI ð ð M1 N 11:10 11:18 þ (kN m) ds þ N1 1 ds ¼ 1 F11 ¼ M1 EI EA EI EA ð M 4:97 (kN m) 1 F12 ¼ M1 2 ds ¼ EI EI ð M 2812:5 1 F20 M2 0 ds ¼ (kN m) EI EI ð M 4:97 (kN m) 1 F21 ¼ M2 1 ds ¼ EI EI ð M 5:0 1 F22 ¼ M2 2 ds ¼ (kN m) EI EI It is observed that F12 ¼ F21, and this is justified according to the Betti—Maxwell theorem (see Section 2.3.5). Thus, in general, Fij ¼ Fji. Note that the above equations (a) allow the calculation of s1 , s2 for any specific values of X1 and X2. The third step of the method of forces is now applied, i.e. the satisfaction of the compatibility of the deformations. This means that s1 and s2 are required to take the values they have in the real structure, i.e. zero and zero, respectively. If the 100

Stress state in framed structures (statically indeterminate structures) 299.70 kN 20 kN/m

20 kN/m 5.0 m 10.0 m

5.0 m

A = 0.002 m I = 0.00029 m E (homogeneous)

264.65 kN m Primary structure

Redundant structure 264.65 34.99

250.0

[M ]

[V]

77.03

268.06 43.0

[N]

57.0

62.50 122.97

Figure 3.6 Sectional forces diagrams drawn on the basis of the determined redundant forces

left-hand side of equations (a) is known, then X1 and X2 can be determined from the resulting equations. From the solution of equations (a) it results that X1 ¼ 299.70 kN and X2 ¼ 264.65 kN m. Thus if X1 and X2 are replaced in the primary structure with their actual senses and values then, together with the external loading, the sectional forces diagrams can be readily obtained, being obviously identical to those of the actual redundant structure (Figure 3.6).

3.2.3

The effect of temperature change

The imposition of a temperature change in a statically indeterminate structure with respect to an initial temperature (either uniform across all fibres of the cross-section, or linearly variable along the depth of the section) causes a stress state; unlike in a statically determinate structure, where only deformations are caused and there is no stress at all (see Section 2.3.2). This is because under the applied temperature change the selected primary structure is deformed (without being stressed), developing specific deformations at the modified points of the structure. Therefore, specific redundant forces must be applied at the relevant points in order to re-establish the required deformation values, i.e. those occurring in the statically indeterminate structure (usually zero). The stresses developed will be due to these redundant forces only. The redundant forces are determined in the same way as described previously. It is clear that temperature effects occur only in State 0, i.e. in the deformations Fi0. For the structure under consideration and for the state of temperature depicted in Figure 3.7, it is obtained (see Section 2.3.3): ð ð T ds 1 F10 ¼ N1 ðT Ts Þ ds þ M1 T h 1 105 40 ¼ 1:11:18 105 20 1:49 15:0 ¼ 0:011 kN m 2 0:50 101

Structural systems: behaviour and design Tinit = +20°C T = 0°C

5.0 m

A = 0.002 m2 I = 0.00029 m4 E = 2.1 · 108 kN/m2

T = +40°C 10.0 m 5.0 m Redundant structure Shortening

∆s1

X1

X1

Tinit = +20°C

T = 0°C ∆s2

X2

Curvature due to ∆T No axial temperature variation

T = +40°C Primary system

Figure 3.7 The application of the force method in the case of a temperature change

ð

ð

1 F20 ¼ N2 ðT Ts Þ ds þ M2

T T ds h

1 105 40 ¼ 0 þ 1:0 15:0 ¼ 0:006 kN m 2 0:50 In the linear system of equations (a) given in the previous section, all the flexibility coefficients remain the same and, by requiring that s1 ¼ s2 ¼ 0, their solution yields X1 ¼ 40.32 kN and X2 ¼ 32.0 kN m.

3.2.4

The effect of support settlement

The imposition of specific shifts or rotations on the supports of indeterminate structures also creates stresses, unlike the case in statically determinate structures in which only deformations are caused, and no stresses (see Section 2.3.7). This can be explained as follows (Figure 3.8).

First case

Second case

δ

X1 Development of δ required

δ

X1 Requirement for null rotation

The redundant force causes stress

Figure 3.8 The stress state due to support settlement

102

δ


The consideration of a primary structure will show the imposed deformation either to coincide with a redundant force or not. In the first case, it is obvious that the imposition of redundant forces on the primary structure will be required, these forces having such values that the actual deformation state at the points of the supports is produced. The stress state of the indeterminate structure is caused by these forces (see Figure 3.8). In the second case, where the imposed deformation does not coincide with some selected redundant force, the primary structure under the imposition of the deformation considered at its supports will develop at the modified points certain deformations corresponding to the redundant forces, without of course being stressed. It is obvious that the imposition of redundant forces as an external loading will be required, so that the resulting deformations at these points are identical to the real ones. The stress state of the redundant structure is caused by these forces (see Figure 3.8). Two examples, using the same structure considered above, are now given to demonstrate the procedure used for the case where there are support settlements (Figure 3.9). In the first example (see Figure 3.9(a)), in addition to the uniform loading, the structure is subjected to an imposed rotation of the fixed support A by 0.01 rad in the clockwise sense. The coefficients ‘F’ in the previously established system of equations (a) (see Section 3.2.2) obviously remain unaltered: s1 ¼ F10 þ X1 F11 þ X2 F12 s2 ¼ F20 þ X1 F21 þ X2 F22 The redundants X1 and X2 result from the requirement to create through equations (a) such values s1 and s2 that are identical to the actually occurring ones, i.e. s1 ¼ 0 and s2 ¼ þ0:01. It is: 1 ¼ 376.50 kN, 2 ¼ 68.21 kN/m. In the second example, the same structure under the external loading is subjected to both to an imposed rotation at point A of 0.01 rad clockwise and to a support settlement at point B of 0.02 m downwards (see Figure 3.9(b)). The same superposition is now (a)

(b)

C

C 20 kN/m

20 kN/m 5.0 m

A

0.01 rad 10.0 m

B

D

∆s1 X1

A ∆s2

X2

X1

0.01 rad 10.0 m

5.0 m A = 0.002 m2 I = 0.00029 m4 E = 2.1 · 108 kN/m2

Redundant structure C

A

C X1

B

5.0 m

Redundant structure

20 kN/m

D Primary system

B

D

∆s1

X1

0.02 m

20 kN/m B

A ∆s2

X2

D Primary system 0.02 m

The imposed rotation is not included in the primary system

Figure 3.9 The application of the force method in the case of support settlements

103


made, by considering in State 0 the imposed settlement of support B of 0.02 m downwards, as it does not correspond to a statically redundant size. The role of X1 and X2 now is to cause a zero relative displacement at the cut point of bar CD and a þ0.01 rad rotation of end A, when applied to the primary structure subjected to the external loading and to the settlement of support B. Equations (a) again allow the calculation of s1 and s2 for any value of X1 and X2, but the coefficients F10 and F20 are obviously different from those in Section 3.2.2, although all the other coefficients remain the same. According to the principle of virtual work: ð

M 5121:88 1 F10 þ 0:298 0:02 ¼ M1 0 ds ¼ ðkN mÞ EI EI ð 1 M 2812:5 1 F20 0:02 ¼ M2 0 ds ¼ ðkN mÞ 15 EI EI The system of equations (a) leads to the determination of values for X1 and X2 that will cause s1 ¼ 0 and s2 ¼ þ0:01 rad, i.e. X1 ¼ 412.34 kN and X2 ¼ 48.61 kN m.

3.2.5

The influence of EI in the stress state

In the case where a statically redundant system is subjected only to external loads, the deformations of the primary structure, F10, F20, . . . , etc., as well as Fij are inversely proportional to the value EI for the members and hence, as examination of equations (a) in Section 3.2.2 shows, the stress state, being determined by the redundant forces X1, X2, . . . , etc., definitely depends on the ratio of the individual values of EI to each other and not on their absolute values. Thus, the frames (a) and (b) in Figure 3.10 have exactly the same bending moment diagram, while frame (c) has a different one. However, in the case of a temperature change or an imposed support settlement, the developing deformations in the fundamental structure are geometrical ones and do not depend on EI. As a result — see equations (a) in Section 3.2.2 — the final stress state (i.e. X1, X2, . . . , etc.) depends on the absolute value of EI. Thus, in Figure 3.10, the higher EI values in frame (b) cause greater stresses, due to some imposed temperature change or some support settlement, than those developed in frame (a). It is obvious that the presence of external loads simultaneously with either a temperature change or an imposed settlement, or even both, does not change the fact that X1, X2, . . . , etc., ultimately depend on the absolute value of EI.

EI

3EI EI

9EI EI

(a)

3EI

3EI

(b)

Figure 3.10 The influence of member stiffness on the stress state

104

EI

EI

(c)


3.2.6

Checking the results

Given a statically indeterminate structure with an applied loading and the resulting sectional diagrams, there is always the question of whether or not these diagrams are correct. If suitable modifications are performed to the supports and/or inside the structure, so that it becomes a statically determinate one, then the stress (and deformation) state of the redundant structure may be determined on the basis of the actual loading and the externally imposed relevant redundant quantities, which may be taken immediately from the corresponding diagrams [M], [V] and [N]. Now, the corresponding deformations at the points of modification in the primary structure may be determined through the principle of virtual work, and the control of the validity of the sectional diagrams consists of checking whether the calculated deformations are identical to the real ones. If this is not the case, then the sectional diagrams are erroneous. Of course an equilibrium check must have been undertaken, i.e. a check to ensure that the diagrams [M], [V] and [N], are in ‘equilibrium’ with the external loads (see Section 2.2.5).

3.2.7

Analysis of elastically supported redundant structures

Elastically supported redundant structures clearly develop a different stress state than the one developed by a redundant structure having immovable supports, as support settlements occur, and with a non-prescribed value. As discussed in Section 2.3.7, the settlement of an elastic support (displacement or rotation) is due to the corresponding action on the ground, having always its direction, i.e. the direction opposite to that of the reaction on the structure. Thus the analysis of elastically supported redundant structures does not present any peculiarity, and follows the normal course of the analysis of statically indeterminate structures with support settlement (see Section 3.2.4). It need only be taken into account whether or not the selected redundant forces correspond to the elastic supports (Figure 3.11). In the case where the redundant size X1, for example, corresponds to an elastic support, then the corresponding s1 actually takes the value (which will be required to be developed in the primary structure): s1 ¼ fs X1 It is clear that the modifications that will lead to the fundamental structure can also leave the flexible supports intact, and therefore the primary structure is simply a statically determined system with elastic supports, and can be handled as described in Section 2.3.7 (see Figure 3.11).

3.2.8

Qualitative handling of the method of forces

In many cases the qualitative determination of the reactions and the sectional diagrams of statically redundant structures is possible without the need for any 105


EI = 42 000 kN/m2 3.0 m

20 kN/m

fs

X1

Hyperstatic structure EI

Primary system X1

fϕ = 0.001 rad/kN m fs = 0.001 m/kN

fϕ

fϕ

Ground fs

∆s1

6.0 m The settlement is not included 20 kN/m

1 kN

[0] fϕ

Displacement F10 (virtual loading is state [1])

[1]

Displacement F11 (virtual loading is state [1])

· X1

fϕ ∆s1 = F10 + X1 · F11 Requirement from redundant structure: ∆s1 = –X1 · fs

X1 = 57.9 kN

Figure 3.11 The application of the force method in the case of elastic supports

particular numerical calculations. This is always desirable, not only in order to have a direct sense of the response of a structure, but also in order to be able to test any computed results. The strategy for making a qualitative analysis of a statically indeterminate structure according to the force method does not differ in any way from the one used for its quantitative analysis (see Figures 3.13 and 3.14). After the choice of suitable modifications, the primary structure is considered subjected to the external loading. The redundant forces are already located. The primary structure is deformed under the external loads. What is sought are the senses of the corresponding deformations at the points of action, and these can be found once the whole deformation configuration of the primary structure has been drawn. The drawing of the deformation configuration is accomplished on the basis of the corresponding bending moment diagram, as detailed in Section 2.3.2. The bending of any part of a bar always occurs toward the side of tension fibres. It should not be overlooked that in any rigid joint the angles between the connected members are maintained, regardless of any joint deformation. Only in hinged joints do adjacent members develop different rotations. As the specific deformations found in this way will not generally comply with the actual ones of the redundant structure, the suitable senses of the redundant forces may be easily concluded in 106


Redundant structure X1

Redundant structure

X1

Primary structure

X1

Primary structure Redundant force

Redundant force

Redundant force

[M0]

[M0]

Deduced sense of X1

X1

Deduced deformation

X1

X1

Deduced sense of X1

Deduced deformation

X1

X1 [M] [M]

X1

Ttop > Tbottom Redundant structure

Redundant structure Ttop > Tbottom

X1 Redundant force

X1

Primary structure

X2

Redundant force

X1

Primary structure Redundant force

Redundant force [M0] Not developed

Deduced deformation

[M0] X1

Deduced deformation

X2

X1

X1 Deduced senses of X1

Deduced senses of X1 and X2

X1

X1 [M]

X2

X1 [M]

Figure 3.12 Qualitative plot of the bending moment diagram

order to restore these real deformations. Plotting the sectional diagrams is then possible. Figure 3.12 shows the application of the above technique for some typical structures, such as the fixed-end beam, the continuous beam and the two-hinged frame. 107


3.2.9

Consequences of the elastic support concept

The beam AB in Figure 3.13 has both its ends A and B elastically clamped and a coefficient of elastic rotation f’ (see Section 2.3.7.2). The beam is subjected to a uniform load q. The determination of the redundant reactions X1 based on the primary structure of the simply supported beam leads to the relationship f’ X1 ¼ s1 ¼ F10 þ F11 X1 from which it results that X1 ¼

q L2 =12 1 þ 2 f’ EI=L

It is observed at first that for f’ ¼ 0, i.e. for a fully fixed support, the value of the redundant reaction is independent of EI and is equal to q L2/12, while as the term f’ increases the reaction decreases, depending always on EI. For a very large f’ the value of X1 approaches zero and the beam behaves as if it were simply supported.

q

fϕ

fϕ EI

A

B

L q

X1

X1

qL2/8

X1

X1

Increasing of spring flexibility leads to decrease of end moment and vice versa q L1/(3E1I1) = fϕ E1I1

L2/(3E2I2) = fϕ A

L1

L1

B

L

E2I2 L2

q

L1/(4E1I1) = fϕ E1I1

EI

A

EI L

L2/(4E2I2) = fϕ B

E2I2 L2

The loaded beam develops the same response in all three cases

Figure 3.13 The elastic support concept in a continuous beam

108


As explained in Chapter 2 (see Section 2.3.7), the existence of an elastic support can be modelled by a connection to a simply supported beam with length L and rigidity (EI) such that f’ ¼ L /3 (EI) . Thus, the elastically clamped beam, behaves in precisely the same way as the intermediate span of the continuous beam shown in Figure 3.13. It is also understood (see Section 2.3.7) that an increase or a decrease in the flexibility f’ means a decrease or increase, respectively, in the stiffness of the end beams, which is of course expressed as 3 (EI) /L , i.e. as the inverse of the flexibility f’. Therefore, according to the above behaviour of X1 with respect to the flexibility f’ and using the more usual term of stiffness, it is concluded that the increase in the length L leading to a reduction in stiffness causes a reduction in the support moment MA, which is obviously identical to the redundant size X1 of the isolated beam. Of course, an increase in the EI of the end beams leads to an increase in stiffness, and consequently an increase in the end moments MA. Moreover, it is concluded that an increase in the EI of the beam AB leads, according to the above analysis, to a reduction in the support moment MA (see Figure 3.16). It is clear that an increase in, for example, the moment MA leads to a reduction in the moment in span AB, as the constant moment diagram for the simply supported beam is always hanging by the points that are associated with the moment MA. Clearly the existence of an elastic support can be related to the connection with a simply supported beam or the connection with a fixed-hinged beam. In this case the rotation deformability of the free end is smaller than the corresponding one of the simply supported beam, according to the relationship ’¼

L M 4 ðEIÞ

and in this case f’ is computed from the expression L /4 (EI) (see Figure 3.13). It can be concluded from the above that the increase in stiffness of members monolithically connected to the loaded beam always leads to an increase in the moment at the connection node, whereas a reduction in the stiffness of members neighbouring the loaded beam always leads to a reduction in the support moment of the beam and, accordingly, to an increase in its span moment, as previously explained (Figure 3.14). Hence: . A reduction in the length of the adjacent beams (i.e. an increase in their stiffness with respect to the loaded beam AB), leads to an increase in the support moments MA, MB (with a simultaneous reduction in the span moment). . An increase in the stiffness of the loaded beam AB with respect to the adjacent beams causes a reduction in the support moments MA, MB and, consequently, to an increase in the span moment. . A very long length (very low stiffness) of the adjacent beams causes the beam AB to behave as a simply supported one, whereas a very short length (very high stiffness) of the adjacent beams causes the beam AB to behave as a fixed-end one. It is obvious that the continuous beam shown in Figure 3.15 develops precisely the same bending response as the corresponding frame. (This happens because the frame does not 109

Structural systems: behaviour and design Relative increase (or decrease) in the stiffness of the adjacent members causes an increase (or decrease) in the end moments q

A

B q

Increase in end moments

q

q

Decrease in end moments

q

Figure 3.14 The effect of a variation in the stiffness of continuous beams

develop any shift at its nodes, like the continuous beam.) Accordingly, all the previous observations regarding the continuous beam are also applicable to frames. Thus, an increase in, for example, the height of the column leads to a reduction in the joint moment MA, and vice versa. Because in the case of a frame the moment MA is due exclusively to the horizontal reaction H, it is concluded that an increase in the height of the column leads to a reduction in the horizontal reaction H, and vice versa (see Figure 3.15). Finally, a reduction in the EI of the girder leads to an increase in the moment MA and, accordingly, to a reduction in its span moment. Clearly, stiffening of the cross-section of the columns leads also to an increase in the moments MA, MB .

3.2.10 The influence of stiffness on the stress distribution It is clear from the discussion in the previous section that, when a loaded member is connected to other unloaded members, an increase in the stiffness of the unloaded members with respect to the loaded one leads to an increase in the end moments of the loaded member, and vice versa. This conclusion can be extended, in that a relative change in member stiffness (bending or axial) in a structure results in an increase in the stresses in the strengthened members and to a reduction in the stresses in the less stiff members (Figure 3.16). Further to this property, which is attributed only to the static redundancy (the stress state of statically determinate structures remains unaffected by any changes in the 110

Stress state in framed structures (statically indeterminate structures) Relative increase (or decrease) in the stiffness of the adjacent members causes an increase (or decrease) in the end moments q

A

B q q

q

A

B End moments decrease Span moment increases Decrease in H

H

H

End moments increase Span moment decreases Increase in H

q q

q

A

B End moments increase Span moment decreases Increase in H

End moments decrease Span moment increases Decrease in H H

H

Figure 3.15 The influence of a variation in the stiffness of frames

rigidity of their members), it may be concluded that, when a load (force or moment) is transferred to the supports through members that are compelled to develop common deformations, the most rigid members carry the largest portion of the bending response and the least rigid ones the smallest proportion. g As a first example, a system of two beams connected by a vertical bar at their midpoints is examined (Figure 3.17). Assume that the axial rigidity EA of the bar is very high, which ensures the same (vertical) shift at the midpoint of each beam. Considering as redundant the axial force X1 of the binding bar, it is found that X1 ¼

P 3 L2 ðEIÞ1 1þ L1 ðEIÞ2

Expressing the stiffness of the two beams as k1 ¼ 48 (EI)1/L31 and k2 ¼ 48 (EI)2/L32 , then X1 ¼

P k 1þ 1 k2 111


EI

EI EI

Increasing EI causes a decrease in the tie force and consequently a decrease in the joint moments

Increasing EI of the beam causes an increase in its moments and a decrease in the girder moments

Strengthening the support sections causes an increase in the support moments and a decrease in the span moments

Figure 3.16 The influence of the rigidity of the members on the bending moment distribution

This result has a particular importance as it shows the distribution of the load P on the two beams: The bottom beam is loaded with P2 ¼ X1, while the top beam is loaded with the force P1 ¼ PX1. It results that P1 ¼ P

k1 k1 þ k2

and P2 ¼ P

k2 k1 þ k2

If one of the two beams is turned by 908 in the horizontal plane, the above results and conclusions remain valid. Of course they concern the transfer of a vertical load P in two horizontal directions, a subject that has a significant practical importance (see Chapter 10). If the actual axial stiffness ks ¼ EA/L of the binding bar is taken into account, the above results are written as X1 ¼ 112

P 1þ

Stress state in framed structures (statically indeterminate structures) P

P

(1)

(1) X1

(2)

(2)

If stiffness (2) > stiffness (1) then beam (2) is loaded by >P/2

If stiffness (2) < stiffness (1) then beam (2) is loaded by M1 > M3 Beam (2) takes the largest part of the load N

MT N2

N1 a

b L

M1

M2 a

b L

Figure 3.17 The distribution of a nodal load in statically redundant systems

with P1 ¼ P

1 1þ

and P2 ¼ P where

1þ

1 1 ¼ k2 þ k1 ks

As a second example, the joint between the connected bars in Figure 3.17 is considered. The moment M is applied to the joint; the joint is immovable. It is obvious that the sum of the moments M1, M2 and M3 should, for equilibrium reasons, be equal to M. Moreover, given that stiffness (2) > stiffness (1) > stiffness (3), then M2 > M1 > M3 (see Section 3.3.7). Finally, a fixed beam subjected to either a concentrated load N or a concentrated torsional moment MT is examined. Both cases are characterised by the requirement 113


that at the point of application of the load the axial displacement or the twist angle, respectively, must be common to the two parts of the beam. Thus, according to the foregoing, and to Sections 2.3.8 and 2.5.3, it is obtained that, respectively (see Figure 3.17): b N1 ¼ N ; L

N2 ¼ N

a L

and b M1 ¼ MT ; L

M2 ¼ MT

a L

Note that the above results come out the same whether N or MT is thought of as the transverse load P acting on a simply supported beam, with ‘N’ and ‘M’ being the reaction of the beam, respectively (see Figure 2.32).

3.3

The deformation method

3.3.1

Introduction

On a plane structure subjected to a given loading, it is always possible to apply suitable external actions on its nodes such that all shifts and rotations developed at the nodes are eliminated. These actions (forces and moments) may be directly determined as follows. As each node, cut out as a free body, will receive at the member ends bending moments, shear forces and possibly axial forces corresponding to the fixed state of each member, these external actions merely have to satisfy the equilibrium of the node, and so are directly determinable. This is illustrated by the two examples shown in Figure 3.18. It is now understood that the stress state of the examined structure results by superposing the already known stress state of the ‘fixed’ structure onto the one resulting from a ‘nodal loading’, which consists merely of the equal and opposite forces to the above actions. It is then obvious that the nodal deformations of the structure are due exclusively to this last nodal loading. Thus, the problem of the initial structure is substantially reduced from the analysis point of view to one of a specific nodal loading, as the stress state of the ‘fixed’ structure may be considered as ‘trivial’. g The deformation method is based on the fact that the stress state of any member of a structure loaded only at its nodes is determined by the developing displacements and rotations at its two ends. Of course, if the member is statically determinate, i.e. it is either simply supported or a cantilever, it remains unaffected by the displacements or rotations of its supports. It should also be pointed out (something that will be commented on later as well) that the rotations that develop at the hinged ends of members do not cause stresses. Given the above, the ‘active’ members in a structure (subjected to the determined nodal loading according to the precedents), will fall into one of the following two categories: 114

Stress state in framed structures (statically indeterminate structures) 6.0 kN/m 6.0 kN/m

12.5 kN m

15.0 kN

15.0 kN

15.0 kN 6.0 · 5.02/12 6.0 · 5.0/2

12.5 kN m

12.5 kN m

The nodal actions eliminate the deformations of the nodes

5.0 m

15.0 kN

12.5 kN m Joint deformations

‘Fixed state structure’ 6.0 kN/m 6.0 kN/m

4.0 kN/m

18.75 kN 11.25 kN (5/8) · 6.0 · 5.0 10.42 kN m

5.0 m

10.0 kN

6.0 · 5.02/8 4.0 · 5.0/2 (3/8) · 6.0 · 5.0 4.0 · 5.02/12

18.75 kN 11.25 kN 10.0 kN 10.42 kN m Joint deformations

4.0 kN/m The nodal actions eliminate the deformations of the nodes ‘Fixed state structure’

Figure 3.18 The concept of the fixed structure and the imposition of nodal actions

. monolithically connected at one end and hinged at the other (simply fixed) . monolithically connected at both ends (fixed—fixed). The deformation method consists of the following three steps: (1) the introduction of specific rotations and vectors of nodal shifts of the structure as unknowns (2) the expression of the sectional forces in the members on the basis of their nodal deformations (3) the determination of the unknown deformations from the equilibrium requirements at each node. It is obvious that the resulting stress state must be superposed on that of the fixed structure. It is clear from step (2) the application of the deformation method is based on the force method, which is essentially the only way to deal properly with the stress state of a statically indeterminate structure. However, thanks to standardised results (as will be explained later), it is possible to overcome this and apply the deformation method, which may eventually lead to a smaller number of unknowns than are encountered in the force method. Despite the fact that step (1) is the most critical, it is appropriate at this stage to examine the simply fixed beam and the fixed—fixed beam, in order to obtain the standardised results required in step (2). 115


3.3.2

The simply fixed beam

3.3.2.1 The fixed state The redundant reaction X1 is determined for some external load, using the compatibility requirement for the primary structure of a simply supported beam (Figure 3.19): rotation of i ¼ 0 ¼ rotation of i due to the external load þ rotation of i due to X1 For a vertical load downwards, X1 causes tension in the top fibres, as may also be deduced from the developed curvature of the elastic line. A uniform load q causes the moment: X1 ¼ q L2/8. The bending moment diagram is shown in Figure 3.19. The reactions at A and B are equal to 5 q L/8 and 3 q L/8, where the dimension L refers to the length of the horizontal projection of the beam. g Regarding now a temperature difference T between the top and bottom fibres (Figure 3.20), the above compatibility requirement leads to the value X1 ¼ 3 (EI T T)/2h. It can be seen that the bending moment is independent of the length of the beam, but the shear force is not.

q

q

k

k

k

α i

i

X1

i

L q

q 5 · q · L/8

5 · q · L/8 3 · q · L/8 Elastic line

q · L2/8

L

q · L2/8 q · L2/8

q · L /8 2

[M]

[M] The reactions and bending response are independent of beam inclination

Figure 3.19 A simply fixed beam under a uniform load

116

3 · q · L/8


Ttop > Tbottom 3 · (EI · αT · ∆T)/(2h) [M ]

3 · (EI · αT · ∆T) · cos α/(2hL)

3 · (EI · αT · ∆T)/(2h) The bending response is independent of the beam length

Figure 3.20 A simply fixed beam under a temperature difference

3.3.2.2 Imposed rotation of the fixed end End i is subjected to an anticlockwise rotation ’, as shown in Figure 3.21. The redundant reaction X1 is determined from the compatibility requirement: rotation of i ¼ ’ ¼ rotation of i due to X1 The result is that X11 ¼ (3EI/L) ’, causing tension at the top fibres. L is the length of the beam. The sense of this reaction may also be ascertained qualitatively by considering the curvature of the elastic line at end i. The bending moment diagram is shown in Figure 3.21. Clearly the reaction X1 is the moment required to rotate end i of the simply supported beam ik by ’. Hence, the stiffness of the simply supported beam ik with respect to end rotation is 3EI/L. k

i

k

L i

ϕ

X1

Elastic line (3 · EI/L ) · ϕ 2

(3 · EI/L2) · ϕ

(3 · EI/L) · ϕ ϕ

(3 · EI/L) · ϕ

Required moment to produce rotation [M]

Figure 3.21 A simply fixed beam subjected to a fixed end rotation

117


3.3.2.3 Imposed relative displacement of the beam ends End i is subjected to a transverse displacement downwards, as shown in Figure 3.22. The redundant reaction X1 is determined using the compatibility requirement: rotation of i ¼ 0 ¼ rotation of i due to displacement þ rotation of i due to X1 It results that X11 ¼ (3 EI/L2) , causing tension at the bottom fibres, as may also be deduced qualitatively from the corresponding curvature of the elastic line. The bending moment diagram is shown in Figure 3.22. From this qualitative approach it is obvious that the bending moments (and shear forces) of the member will be the same as above if, instead of end i, end k is subjected to a transverse displacement upwards, or, more generally, if a relative shift is imposed k

k

k

X1 i L

i

i ∆

∆

(3 · EI/L3) · ∆ ∆ Elastic line (3 · EI/L2) · ∆ (3 · EI/L ) · ∆ 3

The bending response is due only to the relative transverse shift

Elastic line (3 · EI/L3) · ∆

(3 · EI/L3) · ∆

(3 · EI/L2) · ∆

k

i

Elastic line

[M]

(3 · EI/L3) · ∆ ∆

Required force to produce displacement

Figure 3.22 A simply fixed beam under an imposed displacement of the supports

118


transversely to the axis of the beam, provided, however, that the resulting elastic line causes with its curvature at end i a tension on the same (bottom) fibres. It is concluded that only the relative shift plays a role in the bending response, and not the individual shifts of the two member ends. It is also clear that the shear force at the support k is the transverse force required to shift end k of the cantilever by . Hence, the cantilever stiffness with respect to the transverse shift of its free end is 3 EI/L3.

3.3.3

The fixed-end beam

3.3.3.1 The fixed state The redundant reactions X1 and X2 are determined from the compatibility requirement in the primary structure of the simply supported beam (Figure 3.23): rotation of i ¼ 0 ¼ rotation of i due to external load þ rotation of i due to X1 þ rotation of i due to X2 q

q

k

k

k

X2 α i

i

X1

i

L q

q q · L2/12

q · L/2 q · L/2

q · L2/12

L

q · L2/12

Elastic line 2

q · L /12 q · L2/12

[M]

q · L2/12

[V ]

q · L/2

q · L/2

[M] (q · L/2) · cos α [V ]

(q · L/2) · cos α

The reactions and bending response are independent of the beam inclination

Figure 3.23 A fixed-end beam under uniform load

119


Ttop > Tbottom

[M]

No shear developed

(EI · αT · ∆T )/h

(EI · αT · ∆T )/h

The bending response is independent of the beam length

Figure 3.24 A fixed-end beam under a temperature difference

rotation of k ¼ 0 ¼ rotation of k due to external load þ rotation of k due to X1 þ rotation of k due to X2 For an external load in a downward direction it is obvious, from the curvature of the elastic line, that X1 and X2 are causing tension in the top fibres. A uniform load q corresponds to the values X1 ¼ X2 ¼ q L2/12. Note that the dimension L refers to the horizontal projection length of the beam. g Regarding now a temperature difference T between the top and bottom fibres, the above compatibility requirements lead to the values (Figure 3.24) X1 ¼ X2 ¼ EI T T/h It can be seen that the value of the constant bending moment along the beam is independent of the beam length, and no shear forces develop.

3.3.3.2 Imposed end rotation End i is subjected to a anticlockwise rotation ’ (Figure 3.25). The redundant reactions X1 and X2 are determined from the compatibility requirements: rotation of i ¼ ’ ¼ rotation of i due to X1 þ rotation of i due to X2 rotation of k ¼ 0 ¼ rotation of k due to X1 þ rotation of k due to X2 It results that X1 ¼

4 EI ’ L

and 2 EI ’ L as shown in Figure 3.25. The length of the beam is denoted by L. The resulting curvatures at the ends of the elastic line of the beam automatically indicate the side of the tensioned fibres and, accordingly, the sense of the bending moments developed at the X2 ¼

120

Stress state in framed structures (statically indeterminate structures) X2

k

k

k

i

L

ϕ

i

Elastic line

(6 · EI/L2) · φ

X1

i

(2 · EI/L) · φ

(6 · EI/L2) · φ (4 · EI/L) · φ

(4 · EI/L) · φ

ϕ

[M]

Required moment to produce rotation

Figure 3.25 A fixed—fixed beam under an imposed end rotation

ends. It can be seen that the redundant reactions have the same sense, i.e. the one corresponding to the rotating end being double that of the other end. From the opposite curvatures at the two ends, it is concluded that there will exist a point in the beam where the curvature, and thus the bending moment also, will be zero (inflection point). It is clear that the reaction X1 is the moment required to rotate the freely supported end i of the simply fixed beam by ’, with a resulting moment diagram identical to the one above. Hence, the stiffness of the simply fixed beam ik with respect to rotation of its freely supported end is 4 EI/L.

3.3.3.3 Imposed relative displacement of the beam ends End i is subjected to a transverse downward displacement (Figure 3.26). The redundant reactions X1 and X2 are determined from the compatibility requirement: rotation of i ¼ 0 ¼ rotation of i due to displacement þ rotation of i due to X1 þ rotation of i due to X2 rotation of k ¼ 0 ¼ rotation of k due to displacement þ rotation of k due to X1 þ rotation of k due to X2 It results that 6 EI L2 The resulting curvatures at the ends of the beam automatically indicate the side of the tensioned fibres and, accordingly, the sense of the bending moments developed at the X1 ¼ X2 ¼

121

Structural systems: behaviour and design k

k

k

k

X1

X2

i L

i

i

i

∆

∆

(12 · EI/L3) · ∆

(6 · EI/L2) · ∆ ∆

(6 · EI/L2) · ∆ Elastic line

(12 · EI/L3) · ∆

(6 · EI/L2) · ∆ The bending response is due only to the relative transverse shift

Elastic line

(12 · EI/L3) · ∆

(12 · EI/L3) · ∆ (6 · EI/L2) · ∆

k Elastic line

i

[M]

(12 · EI/L3) · ∆ ∆

[V]

Required force for producing displacement

Figure 3.26 A fixed-end beam under an imposed end displacement

ends. It is observed that the redundant reactions have the same sense and are equal in magnitude. As the curvatures at the two ends are opposite, it is concluded that at the midpoint of the beam the curvature, and hence the moment, will be zero (inflection point). This qualitative analysis shows that the bending moments will be the same as those above if, instead of end i, end k is subjected to a transverse displacement upwards, or, more generally, if a relative shift is imposed transversely to the axis of the beam, provided, however, that the resulting elastic line causes, with its curvature at end i, tension in the same (bottom) fibres. It is concluded that only the relative shift plays a role in the bending response, and not the individual shifts of the two member ends. 122

Stress state in framed structures (statically indeterminate structures) k

m

m

k

k

Joint rotations leave the hinged bar unaffected Unknown rotations are referred only to developing end moments

Figure 3.27 Determination of the unknown nodal rotations

It is also clear that the shear force at the support k is the transverse force required to shift the fixed end k of the beam by (without rotation). Hence, the fixed-end beam stiffness with respect to a relative transverse shift of its ends is equal to: 12 EI/L3.

3.3.4

Recognition of the unknown nodal deformations

3.3.4.1 Rotations and displacements At each rigid joint (and also at every point) of any plane structure a displacement vector and rotation are developed. With respect to the deformation method, the rotation of a node can be considered as an unknown magnitude if a bending moment can be developed in at least one of its adjacent members (Figure 3.27). The nodes k of the structure shown in Figure 3.27 do not justify an unknown rotation, although the members develop a rotation at their ends. The nodes m develop rotations that do not affect the stress state of the two-hinged bar. With regard to the displacement vectors, these may be considered as unknowns at each node, as they can be freely developed. However, in order to restrict their number, the following justified assumption is made. The lengths of all bending members (i.e. those that do not develop exclusively axial forces) are assumed to remain unchanged. This means that the projections of the displacement vectors of the two ends on the beam axis are identical, and thus a relative shift of the two ends along the bar is excluded (Figure 3.28). Of course, it should not be concluded on the

2

3

The displacement vectors have equal projections on the direction 2–3

1

4

Figure 3.28 Displacement vectors ensuring the length of the bars is unchanged

123

Structural systems: behaviour and design ∆6

∆8 ∆5

∆2

∆7

∆2

∆3

∆1

∆4 ∆1

Assumption of unchanged lengths

Figure 3.29 Restriction of the number of unknown nodal displacements

basis of this assumption that, according to Hooke’s law, the axial forces should be considered zero, as these are simply determined from the equilibrium of the nodes. In the vast majority of framed structures that carry their loading through a bending mechanism, the effects of this assumption on the accuracy of the results are insignificant. Nevertheless, if a member develops exclusively axial forces, as a two-hinged bar without intermediate loading, then its stress state is determined on the basis of its change in length, according to Hooke’s law (see Section 1.2.1). Thus, the number of unknown nodal displacements is considerably reduced. For example, without assuming an unchanged length of the bars, the displacement components 1 to 8 of the frame nodes shown in Figure 3.29 are all independent of each other (eight unknowns), while with the above assumption the two displacements 1 and 2 are sufficient to describe any deformed configuration of the nodes.

3.3.4.2 Basic consideration of the kinematics of members For the final determination of the unknown nodal displacements, according to the criteria presented in the next section, it is essential first to become familiar with the kinematics of the hinged connection between three arbitrary points (Figure 3.30). The system of bars ABC shown in Figure 3.30 is immovable. Node B cannot be displaced because this would mean a change in length of some bar. If node A is now c going to point C0 , the displaced by ~ a going to point A0 and node C is displaced by ~ question arises of how node B will be displaced such that from its new position B0 the lengths B0 A0 and B0 C0 are equal to the lengths BA and BC, respectively. Based on the fundamental assumption of small deformations governing the linear behaviour of structures, the replacement of circular arches by their tangent is allowed (see Figure 3.30). Therefore, two segments, equal and parallel to AB and CB, are drawn from the points A0 and C0 , respectively. The intersection of the vertical lines at the ends of these two segments will fix the new position B0 of point B. 124

Stress state in framed structures (statically indeterminate structures) D

Replaces circular segment with centre C′

E

DE = DE′

B′ Replaces circular segment with centre A′

Displacement of point B

E′ Replaces circular segment

B

C′ →

A

a

A′ →

c

B′A′ = BA B′C′ = BC C

Figure 3.30 Kinematic handling of a system with two hinged solid bars

3.3.4.3 Selection of the unknown nodal displacements In order to determine the unknown node displacements in a plane skeletal structure, which may represent any arbitrary displacement configuration of the system, all rigid joints are first changed to hinged ones, thus converting the whole structure to a mechanism. A group of shift vectors is selected that satisfies the following two conditions (Figure 3.31): (1) If all the proposed possibilities of shifting (displacement vectors) are blocked, the shift of any node should be impossible. (2) Releasing one by one and consecutively every displacement vector of the proposed group, while keeping all the others blocked, the displacement vector of each node should be expressed kinematically in terms of the displacement of the released one only, according to the previous section (see Figure 3.31). The process of recognising the unknown displacement vectors is a heuristic one. This means that it is simply checked whether or not a proposed group ‘’ fulfils the criteria (1) and (2). If it does, then it is the right one. If not, then another group should be sought.

3.3.5

The procedure for the deformation method

The basic steps of analysis using the deformation method are described below, and applied to the example structure shown in Figure 3.32. 125


All nodes unshifted

Unshifted 5

5

One unknown displacement

3

4

3

4

Based on 3–4–2 1

2

Fulfilment of the first condition All nodes unshifted

1

2

1

2

Fulfilment of the second condition

3

1

4

Based on 2–5–3 2 3

1

Based on 1–2–5 5 Fulfilment of the first condition All nodes unshifted

One unknown displacement

4

Based on 5–3–4 5

Fulfilment of the second condition

Figure 3.31 The method of determining the unknown nodal displacements

Step 1: Recognition of the unknown rotations. All rotations of solid nodes are a priori independent of each other (see Section 3.3.4.1) (see Figure 3.32). Step 2: Recognition of the unknown displacements. The structure is transformed into a mechanism by replacing the rigid nodes with hinges and searching for a displacement group that satisfies the two criteria set out in Section 3.3.4.3 (see Figure 3.32). Step 3: Determination of the stress state in the fixed structure and then of the required external actions in order to ensure the nodal equilibrium. Automatic determination of the active nodal loads on the structure with equal and opposite actions. The nodal deformations of the structure are exclusively due to this final nodal loading (see Figure 3.32). Step 4: Determination of the developing moments and shear forces at the ends of each bar due to the imposed rotations and shifts of the ends that have been considered as unknowns, based on the assumed arbitrary senses of the unknown rotations and displacements (Figure 3.33). In order to facilitate the determination of the physical sense of the bending moments, the corresponding elastic line resulting for each member is drawn. After the physical 126

Stress state in framed structures (statically indeterminate structures) 20.0 kN/m

5

4

3

20.0 · 3.02/8

3.0 m

20.0 · 3.0 · 5/8 20.0 kN/m 1

20.0 kN/m

5

2 3.0 m

20.0 · 3.0

3

5.0 m

3 20.0 kN/m

ϕ3

ϕ4

1 20.0 · 3.0 · 3/8 3

20.0 · 3.02/2 20.0 · 5.02/12 4

4

3

20.0 · 5.0/2

20.0 · 5.0/2

‘Fixed structure’ Step 1

1

2

41.67

∆4

∆4

Step 3

90.0

41.67

22.5

4 Based on 1–3–4 ∆4/sin α 3

[M ]

α 1

Step 2

2

20.0 kN/m

37.5 kN

110.0 kN 70.83 kN m

110.0 kN 70.83 kN m

50.0 kN

50.0 kN 37.5 kN

41.67 kN m

41.67 kN m

20.0 kN/m This loading ‘fixes’ the structure

The node deformations are due to this loading

Figure 3.32 Steps 1 to 3 in the analysis of a framed structure

sense of the end moments and shear forces has been determined, the set up of the signed expressions of the moments and shear forces at the ends of the members follows, on the basis of superposition. The adoption of a specific code of positive sign moments is, of course, necessary. In the example examined, the convention based on a predefined positive border is followed. With regard to the axial forces, as pointed out in Section 3.3.4.1, while their development is recognised, they cannot be expressed through the displacement vectors of nodes, because of the assumption of unchanged lengths. They are determined through the equilibrium of nodes. 127

Structural systems: behaviour and design Step 4 Member 1–3 (3EI/4.24) · ϕ3

(3EI/4.242) · ∆4/sin α

Physical forces

ϕ3

3

3

∆4/sin α Conventional expressions M31 = –(3EI/4.24) · ϕ3 + (3EI/4.242) · ∆4/sin α V31 = –(3EI/4.242) · ϕ3 + (3EI/4.243) · ∆4/sin α

1 (3EI/4.243) · ∆4/sin α

1 (3EI/4.242) · ϕ3 Member 3–4 (4EI/5.0) · ϕ3

(2EI/5.0) · ϕ3 4

3 ϕ3

3

4

∆4

∆4/sin α

Physical forces (6EI/5.02) · ϕ3

(2EI/5.0) · ϕ4

(4EI/5.0) · ϕ4

ϕ4

(6EI/5.02) · ∆4/tan α 4

3 3

4 ∆4/tan α (6EI/5.02) · ϕ4

(6EI/5.02) · ∆4/tan α

Conventional expressions M34 = +(4EI/5.0) · ϕ3 + (2EI/5.0) · ϕ4 + (6EI/5.02) · ∆4/tan α M43 = –(2EI/5.0) · ϕ3 + (2EI/5.0) · ϕ4 + (6EI/5.02) · ∆4/tan α V34 = –(6EI/5.02) · ϕ3 – (6EI/5.02) · ϕ4 – (12EI/5.03) · ∆4/tan α V43 = V34

(12EI/5.03) · ∆4/tan α

Member 4–2 (6EI/3.02) · ∆4

(4EI/3.0) · ϕ4 4

Physical forces

∆4

4

ϕ4 (6EI/3.02) · ϕ4

2

(2EI/3.0) · ϕ4

(12EI/3.03) · ∆4

2

Conventional expressions M42 = +(4EI/3.0) · ϕ4 – (6EI/3.02) · ∆4 M24 = –(2EI/3.0) · ϕ4 + (6EI/3.02) · ∆4 V42 = –(6EI/3.02) · ϕ4 + (12EI/3.03) · ∆4

(6EI/3.02) · ∆4

Figure 3.33 Step 4 in the analysis of a framed structure

Step 5: Equilibrium equations. After the determination of signed end moments and shear forces at each member, in terms of the unknown shifts and rotations, the three equilibrium equations for each node of the structure are formulated (Figure 3.34). The equation of equilibrium for the moments does not present a problem, but the axial 128

Stress state in framed structures (statically indeterminate structures) Step 5 110.0 kN 70.83 kN m

50.0 kN

+V43

+V34

41.67 kN m 37.5 kN

+N34 3

+M31

4

+M43

+M34

+V42

N43 = N34

+V31

+N31

+N43 +M42

+N42 Equilibrium of node 3 (three equations)

Equilibrium of node 4 (three equations)

Unknowns involved: ϕ3, ϕ4, ∆4, N31, N34, N42


forces of bars may be additional unknowns in the equations for the equilibrium of the forces (see Figure 3.34). However, in many cases the determination of the unknowns can be done more rapidly if, after formulating the equilibrium equations for the moments at all nodes, suitable cut out parts of the structure are considered, in order to supplement through their global equilibrium the number of missing equations needed (Figure 3.35). Step 6: Determination of unknowns and calculation of final end moments and forces. After the analysis of the system and the determination of the unknowns, the end moments of the members are calculated from the expressions found in Step 4. These moments are superposed on those of the fixed state according to Step 3, so that the final end moments of the structure are determined (Figure 3.36). The shear forces are then calculated from the equilibrium of each member. The axial forces are determined directly through the equilibrium of the nodes.

O

Moment equilibrium of part 3–4 w.r.t. O: M31 + V31 · 7.07 – 70.83 – 110.0 · 5.0 – 37.5 · 5.0 110.0 kN 70.83 kN m

41.67 kN m

37.5 kN 3

4

+V42 +N42

(2)

Moment equilibrium of node 4: +M42

+V31

(1)

Moment equilibrium of node 3: M31 – M34 – 70.83 = 0

+M31 +N31

– 41.67 – M42 + V42 · 5.0 = 0

50.0 kN

M43 – M42 – 41.67 = 0

(3)

Unknowns involved: ϕ3, ϕ4, ∆4 Solution: ϕ3 = –59.6/EI, ϕ4 = +44.0/EI, ∆4 = +266.8/EI


129

Structural systems: behaviour and design Step 6 According to ϕ and ∆: M31 = +106.82 kN m M34 = +35.99 kN m M42 = –119.1 kN m M43 = –77.43 kN m M24 = +148.5 kN m 90.0 77.43

41.67

41.67

22.5

119.1 4

3 35.99 106.82

[M]

[M]

1

According to Step 4 (ϕ, ∆)

2

148.5

According to Step 3 (Fixed structure) 119.1

90.0 5.68

119.1 4

3 84.32 [M] 1

2

148.5 124.8

72.7

89.2

60.0 1.3 27.3 [V]

[N] 82.4 89.2

72.7

41.1


3.3.6

The effect of an elastic support

Within the framework of the deformation method, it is appropriate to recognise the influence of an elastic support on the end reactions of a fixed beam subjected to a uniform load (Figure 3.37). Elastic support through a spring with rigidity ks In order, at first, to eliminate the settlement of end A (see Section 3.3.1), the external vertical force (3 p L/8) must be applied upwards, leaving the spring unstressed. It is clear then that the opposite force applied at end A downwards, will cause the developing 130

Stress state in framed structures (statically indeterminate structures) p

(a)

A

B

A

kϕ

EI

ks

p

(b)

L

L

p

p

p · L2/12

3 · p · L/8 ∆=0 ks

kϕ

Stressless spring

ϕ=0

Stressless spring p · L2/12

ϕ

3 · p · L/8 ∆

B EI

kϕ

ϕ

ks p · L2/12

3 · p · L/8 A

A

kϕ

(3 · EI/L3) · ∆

(4 · EI/L) · ϕ

ks ks · ∆ = RA Equilibrium RA < 3 · p · L/8

kϕ · ϕ = MA Reaction at A

Equilibrium MA < p · L2/12

Figure 3.37 The consequences of an elastic support of a beam

settlement . The real stress state obviously constitutes a superposition of these two states. Equilibrium of node A can be written as 3 EI 3pL þ ks ¼ 3 8 L and, given that the reaction RA of the beam at A is equal to the spring force, it results that RA ¼ ks ¼

3 pL=8 R0A ¼ 3 EI 3 EI 1þ 3 1þ 3 L ks L ks

Thus the presence of the spring generally induces a reduction in the reaction R0A that is developed at the unyielding support of the beam under any load, according to the above equation. Flexible support through a rotational spring with rigidity ku In order to eliminate the end rotation at A, the application of an external moment ( p L2/12) is required, according to Section 3.3.3. The rotational spring is then unstressed. 131


Application of an equal and opposite moment at end A will cause the development of a rotation ’ at this point. The real stress state constitutes a superposition of these two states. Moment equilibrium in node A requires that 4 EI p L2 ’ þ k’ ’ ¼ L 12 and, given that the reaction MA at support A is identical to the moment of the spring, MA ¼ k’ ’ ¼

p L2 =12 M0A ¼ 4 EI 4 EI 1þ 1þ L k’ L k’

Similarly, it is concluded that the rotational compliance of the support expressed by the rotational spring induces a reduction in the reaction M0A developed in the fixed support of the beam under any load.

3.3.7

The distribution of a nodal action

The determination of how a moment applied to a node is distributed to the adjacent members, assuming that no shift but only rotation is possible, is particularly useful. The qualitative conclusion drawn in Section 3.2.10 is now quantified using the deformation method. Equilibrium of the examined node under the moment M shows that the developing moments at the ends of the members have the opposite sense to the moment above (Figure 3.38). Moreover, each Mi will be proportional to the rotation ’ of the node, according to the expression Mi ¼ ki ’ where ki is the rotational stiffness of the corresponding member, which is equal to 4 EI/Li or 3 EI/Li, depending on whether the opposite end of the member is fixed or hinged, M Node equilibrium ϕ

ϕ ϕ

Stressless member k1 · ϕ

k1 = 3 · EI/L1

k2 · ϕ k3 · ϕ

k3 = 4 · EI/L3

(1) (2) k2 = 4 · EI/L2

M

(3)

1 1 2 2

3 Mi =

ki ∑k

M

Figure 3.38 The distribution of a nodal moment to the adjacent bars

132

3


respectively. It is obvious that if a bar with a hinged end is connected to the node, then no moment is developed, and hence no end rotation in the member is developed. According to the above, nodal equilibrium for the applied moments requires: X ki M ¼ M1 þ M2 þ M3 þ . . . ¼ k1 ’ þ k2 ’ þ k3 ’ þ . . . ¼ ’ which results in M ’¼P k Thus, the moment developing in each member will be k Mi ¼ Pi M k The moment transmitted to the opposite fixed end is, of course, Mi/2, with the same sense as that resulting from the elastic line of the member. Thus, according to the last equation, the conclusion of Section 3.2.10 is justified, i.e. the distribution of the nodal moment depends on the stiffness of the corresponding member. This conclusion is also valid in cases where certain members are assigned under the imposition of a force P to a common shift: The force P is distributed to the members depending on their rigidity. In the frame shown in Figure 3.39, due to excessive stiffness of the girder, the nodes are displaced by , but are not rotated. The force P is distributed to the three columns as a shear force, proportionally to their shifting rigidity ks, as determined in Sections 3.3.2.3 and 3.3.3.3. Similarly to the above equation, the following is obtained: kðiÞ Vi ¼ Ps P ks The joints are not rotated

P δ

δ

ks

H

2ks

δ

ks/4

H/2

P

V

V/4

2V

P = V + V/4 + 2V P M

M/2 2M [M]

[V ]

2V

M V = P/3.25

V/4

Figure 3.39 The distribution of a horizontal force on vertical columns

133


3.3.8

Qualitative handling of the deformation method

The qualitative examination of a structure, according to the deformation method, does not differ substantially from its quantitative treatment, as described in the previous sections. The determination of the fixed structure and of the nodal loading that causes the rotations and shifts at the nodes can be done as explained in Section 3.3.1. In a case where the structure does not develop nodal displacements, the senses of the nodal rotations are always identical to those of the nodal moments, thus permitting the elastic line for each member to be drawn directly, and the corresponding qualitative bending moment diagram plotted (Figure 3.40). The nodes that do not have a nodal loading are first considered fixed and the sense of their rotations deduced from the corresponding acting moment transferred to them by the neighbouring rotating

3

4

3

5

4

5

[M] fixed structure 1

2

1

External actions in order to fix the structure

2

Opposite actions cause the node deformations

3

1

4 [M] due to node rotations

Elastic line due to node external actions

[M]

2

Fixed moment 3–4 decreases Fixed moment 4–3 increases Fixed moment 4–5 decreases

[V]

Figure 3.40 Qualitative determination of the moment diagram for immovable nodes

134

5


nodes. It should be remembered that the distribution of a nodal moment to the adjacent members occurs according to their rigidity (see Sections 3.3.7 and 3.2.10). The final moment diagram M will naturally result from superposing this last moment diagram on the one for the fixed state. For the loaded members, the final diagram is essentially a ‘modified’ fixed-state one, which is usually not reversed by the superposed end moments. For the unloaded members the final diagram is deduced directly from the last diagram, due to the incoming rotations. g If the structure does develop a displacement, the sense of that displacement must be determined first. This can be done as follows (Figure 3.41). In a first stage, a suitable additional support is added to the structure in order to prevent the development of displacement, and the qualitative diagram of bending moments in the modified structure

? Shift development

Shift constraint

External actions for node fixation

[M0] Fixed structure

[M1] due to rotations

Development of node rotations (Stronger moment at left-hand column) Equilibrium

[M0] + [M1]

[V]

Shifting goes to the right

(First stage) Stronger shear force at left-hand column

Figure 3.41 Qualitative determination of the displacement sense at a node

135


is then plotted, according to the precedents. The acting shear forces on the nodes are obtained from this diagram of bending moments. Next, examination of the equilibrium of a suitably cut out part of the structure containing the imported support allows the determination of the sense of the reaction developed and, accordingly, of the developing displacement, as this must have the opposite sense to that of the reactions. After the determination of the displacement sense, it is clear that the final bending state results by superposing the ‘first stage’ of the structure under the full loading with immovable nodes on the ‘second stage’, where only the known developing displacement is imposed on the structure (Figure 3.42). This second stage is treated exactly as for a loaded structure with immovable joints (first stage), the only difference being that the fixed state due to the loads is now due to the ‘known’ displacement. The procedure for the examined frame is shown in Figure 3.42. The bending state of the second stage qualitatively modifies the one in the first stage only if the loading of the structure is along the direction of the developing displacement. Thus, for the frame shown in Figure 3.41 the form of moment diagram [M] after superposition with the one in Figure 3.42 remains qualitatively unchanged, except for the fixed support of the column where the existing moment is assumed to cause tension

Second stage

[M0] Nodes fixed Imposition of shift Nodes not rotated

Node equilibrium

[M1]

External node actions for release of rotations

[M0] + [M1]

Bending moments of second stage

Figure 3.42 Qualitative determination of the bending state in the second stage

136


[M]

Shift prevention Fixed structure

Shift development First stage

[M]

Nodal loading for rotation fixity

Rotation development

[M]

Second stage

Imposition of shift. Nodes fixed

Bending state with fixed nodes

[M]

Equilibrium of fixed nodes

Nodal actions for rotation release

Final moment diagram

[M]

Figure 3.43 Qualitative determination of the moment diagram for displaced nodes

137


on the external fibres, as concluded from the elastic line in Figure 3.42. However, in the example shown in Figure 3.43, the bending state of the second stage is predominant. It is obvious that the above qualitative examination of redundant structures, through the consideration of the deformations of their nodes, leads to a clear result only for simple structures that have zero or, at most, one degree of displacement freedom. Nevertheless, such an examination can also be useful for more complex structures, in that numerical results (particularly if obtained using a computer) can be predicted qualitatively or judged a posteriori.

138

4 Simply supported beams 4.1

Steel beams (reference material)

4.1.1

Service conditions (elastic behaviour)

The simply supported beam as a structure transferring loads to its two hinge supports has the biggest possible deviation from the corresponding funicular structure for the same transverse loading (see Section 2.2.7) and, consequently, it develops for that reason the maximum bending response for that loading. Nevertheless, this ‘unfavourable’ result is counterbalanced by the structural advantages arising from the geometric linearity of the structure. Loads acting transversely to the beam direction are carried to the supports through the axial direction, as shown in Figure 4.1, in a cross-section at the middle of the beam. The forces D and Z represent the resultants of the compressive and the tensile stresses, respectively, and, due to equilibrium requirements, they are obviously equal in magnitude and opposite in direction. These forces produce the bending moment M ¼ D z at each specific point along the beam. As the distance z between the lines of action of the two forces is definitely smaller than the relatively small height h of the cross-section, it follows that the forces D and Z are going to be considerable, for example significantly larger than the reactions. According to the well-known relationship given in Section 2.2.1 between stresses and moments, the normal stress distribution reveals that a significant portion of the cross-section in the neighbourhood of its centre is barely stressed, and therefore the section is not adequately exploited. This is the reason why bending is generally considered ‘undesirable’ or ‘unfavourable’. The extreme values of the stresses appear at the top and bottom of the section, and are given by the expression max ¼ M ( ymax/I) ¼ M/W where W ¼ I/ymax is the elastic section modulus for the appropriate top and bottom values of y (see Figure 4.1). It should be noted that the above equation is based on the Bernoulli assumption (Euler—Bernoulli beam theory), which states that transverse lines perpendicular to the mid-surface before deformation remain perpendicular to the deformed midsurface. In other words, the strain distribution " through the thickness of the beam is linear (Figure 4.2). In practice, this means that, if a contour is drawn for an arbitrary cross-section before the deformation of the beam (application of the loads) in such a Structural systems: behaviour and design 978-0-7277-4105-9



σ = M · y/l

Compressive stresses D

y

M z Z Tensile stresses

Neutral axis

Figure 4.1 Bending of a simply supported beam

way that its plane is perpendicular to the beam axis, this contour will maintain both its flatness and its orthogonality to the beam axis also in the deformed configuration (see Figure 4.2). g In general, it is desirable to design a cross-section which can carry the largest possible bending moment M for a given area and to preserve, of course, the values of the end stresses within acceptable limits. This means that the moment of inertia for the section under consideration should be increased by amassing as much material as possible towards the top and bottom of the section, and thus increasing to a maximum the distance between the internal forces D and Z (Figure 4.3). A question, however, arises: How thin may the part of the section become which connects the compression and the tension zones of the cross-section? To answer this, equilibrium in the longitudinal direction is considered separately for the upper and lower halves of the left-hand side of the beam. The left end of either part is not subjected to any horizontal forces, whereas at the right end the applied forces are the compressive force D for the upper part and the tensile force Z for the lower part (Figure 4.4). It becomes apparent that, in order to satisfy the equilibrium condition in the horizontal direction, it is required to have longitudinal forces T (kN/m) along the length of the σ

ε

h

Undeformed beam

Figure 4.2 Plane sections remain plane after deformation

140

Deformed beam

Simply supported beams D D M

Thickness?

Z

Z Moving material towards the outer fibres decreases the compressive and tensile forces Greater bending-moment-carrying capacity

Figure 4.3 Moment-carrying mechanism for a cross-section

two pieces with directions to the right for the upper half and to the left for the lower one. Development of such forces, i.e. shearing stresses, also implies the presence of equal shear stresses acting along the side surface in the vertical direction (see Figure 4.4). It is easily seen that these shear stresses are due to the variation of the bending moments, i.e. due to the variation of forces D and Z, because of which the axial equilibrium of a specific part of the beam requires the development of horizontal and vertical shear stresses, as shown in Figure 4.4. This can be considered as the physical interpretation of the well-known equation dM/dx ¼ V (see Section 2.2.3). One should keep in mind that the presence of shear stresses in the vertical direction may also be deduced from equilibrium in the vertical direction of the left half of the beam, because their integration over the thickness yields the shear force V.

D

D + ∆D

D

h/2 τ

τ τ

T=τ·b D·z=M T=τ·b

Shear stress generation τ

h/2 Z

Z

Z + ∆Z

(Section width b)

[T ]

Identical longitudinal force T and shear force V distribution

h Shear stress distribution (orthogonal section)

τmax = 1.50 · (V/A)

Figure 4.4 Mechanism for the development of shear stresses in a beam

141


These shear stresses are determined from the expression (see Section 2.2.1) ¼ V S/b I The distribution of shear stresses along the interface between the upper and lower halves of the beam and, consequently, of their horizontal resultant forces T follows the distribution of the vertical shear forces V (see Figure 4.4). In addition, the shear stress distribution in the transverse section (or the depth of the beam) vanishes at the top and bottom, and it appears to have a maximum value at the centre of the cross-section. Note that pffiffiffi this maximum shear stress is bounded by the limit of failure, which is the value fy/ 3. Consequently, in order to design a beam with a pair of flanges, one being under tension and the other under compression, so as to produce a bending moment, an appropriate thickness for the connecting section (web) is required for the safe development of horizontal shear stresses between them. This conclusion plays a very important role in the design of beams, as will be explained later. The fact that the area around the mid-height of a beam is underworked compared with its top and bottom fibres has led to the subtraction of this material ‘surplus’ and the creation of beams such as the one shown in Figure 4.5, aiming at a reduction in weight. The flanges must be able to withstand the force M/z both in tension and in compression. This, however, is not enough, because the same force must also be carried in shear. That is, the resultant longitudinal shear force M/z must be held by all parts of the beam located between contiguous openings and crossed by the horizontal section. In fact, this force varies along the length proportionally to the distribution of the

Z = D = M/z D

D

z

T Z

Z

Identical T and V distribution

Figure 4.5 Shear transfer mechanism in a beam with openings

142

T

Simply supported beams

vertical shear force diagram (see Figure 4.5). This means that, in the case of a uniform distribution for the openings, it is sufficient to check the regions closest to the two supports. g If the openings are quite wide, so that the beam can be modelled using the indeterminate structure shown in Figure 4.6 (Vierendeel beam), then it should be taken into account that the constituent parts of the beam, namely the two-column frames, are particularly deformable due to the shear forces developed by the beam (see Figure 4.6). This deformation of the frames is attributed only to bending, as shown in the figure. Thus, frames adjacent to the supports deform more than the rest. All of them, however, present the typical behaviour of a two-column frame subjected to a force parallel to its girder, for which the moments vanish at the middle of the girder and close to the middle of the two columns (see Sections 2.4.2 and 6.2.1). Placement of hinges at these points removes the difficulties of static indeterminacy and permits the direct evaluation of internal forces carried by the members in the model (see Figure 4.6). At this point, a remark is necessary. The deformation of the solid beam having a ratio of span length to depth greater than 3, as has been discussed in Section 2.3.2, is mostly caused by the deformation of its elements due to bending moments, assuming that the

z

Constituent frame

1

2

m D1

D2

V1/2

V2/2

V1/2

T Z1

Z2 m

D1 = M1/z

V2/2 T

D2 = M2/z

V1/2

m

V2/2

V1/2

V2/2 m

T = D2 · D1

Addition of diagonal bars suppresses the shearing action The deformations are drastically reduced

Figure 4.6 Modelling of a beam with rectangular openings (Vierendeel beam)

143


corresponding shear forces make a, relatively, insignificant contribution. In the present case of a Vierendeel beam, shear deformation plays the primary role, as has already been pointed out, and the resulting deformations are comparatively large. How much larger can be determined by drastically restricting the shear deformability, which can be easily accomplished by adding diagonal members, as illustrated in Figure 4.6. Thus, the two-column frame with a diagonal bar becomes much less deformable, resulting in a vertical deflection at the middle of the new simply supported system which is almost 10 times smaller than before. This last model has, of course, nothing to do with the beam with openings under consideration. It is obvious that a Vierendeel beam may also be designed to bridge a long span, with a relatively large available depth. g It is common in practice to construct a composite beam consisting of two separate horizontal sections which are simply laid one onto the other without any adhesives or connecting devices (unilateral contact). Assuming that between the two parts there is no friction, a comparison between such a composite beam and a homogeneous beam of the same length and with the same vertical load is unfavourable for the composite beam (Figure 4.7). If, for example, the two ‘half-beams’ had the same thickness, it would be wrong to consider that the upper one provides the compressive force D and the lower one the tensile force Z for the formation of the bending moment, as in the corresponding case of the homogeneous beam. The reason for this is that there is no shear interaction between these two forces. Thus, the developed normal stresses within each beam have the distribution depicted in Figure 4.7, and every half-beam is stressed and deformed independently of the other, with an effective moment of inertia corresponding to half the thickness of the homogeneous beam, e.g. for a rectangular cross-section the effective moment of inertia is equal to I/8, where I applies to the entire section. In order, therefore, to restore this shear interaction, there must be an appropriate structural formation so that each of the two halves can develop the horizontal shear force previously mentioned and, indeed, in the correct direction (see Figure 4.7). Only then can the system of the two beams can be considered equivalent to the single

h/2 h/2

σ

T

T

σ

The two beams work like a homogeneous one only if shear forces T are offered

h Wrong structural pattern: impossible for forces T to develop Correct structural pattern: shear force T development

Figure 4.7 Coupling of two half-beams in bending requires their interaction in shear

144


homogeneous beam. In other words, only then can the moment of inertia I of the homogeneous beam also be used for the composite beam. The structural formation in question requires a jagged interface between the upper and lower halves, which is very common for wooden beams, but for composite structures made out of steel and concrete it is realised in a different way, as will be studied extensively in the next chapter. g In the discussion above, reference is always made to either the horizontal or vertical section of a beam, for which the expressions of normal and shear stresses refer to a Cartesian system of axes conveniently applied in the longitudinal direction of the beam. This choice of reference axes is arbitrary. Instead of considering a transverse section at the middle of the beam, it may as well be a slant section for which the equilibrium requirements yield completely different normal (perpendicular to the plane of the section) and shear (in the plane of the section) stresses. To confirm this, let us consider an element with a rectangular area on the neutral axis of the beam whose sides, as is already known, are subjected only to shear stresses (Figure 4.8). It should be pointed out that the directions of the shear stresses follow the equilibrium requirements of Cauchy’s relation. If this rectangle is cut by an arbitrary straight line, two triangular shaped areas will be formed, and each of them will also have to be in equilibrium. On the inclined side, the developed normal and shear stress components will have values determined directly from the values of the shear stresses on the sides of the rectangular element.

Principal stresses (τ = 0) σ1 Tension

(τ = 0) σ2 Compression

τ

τ

n

io

ss

σ

e pr

m

Co

Te

ns

io

n

Figure 4.8 Principal stresses and principal directions

145


A question can be posed: Is every single one of these numerous pairs of stresses (, ) revealed by the randomly drawn sections realised in the structure? The answer is ‘definitely no’. The structure is stressed only in the direction of a normal stress (without any shearing), which is the only one realised by the structure, and this by itself is able to maintain the equilibrium of the triangular element. It can be proved that at every point in the structure there are only two such orthogonal directions, called principal directions (see Figure 4.8). The corresponding stresses (compressive and tensile) are called principal stresses. Hence, the structure ‘recognises’ only the principal stresses, and although the (, ) pairs do not represent reality, they nevertheless legitimately represent the principal stresses, and are used as suitable quantities for the various structural checks. Principal stresses have different values and directions at each point in the structure. It can be ascertained that a (homogeneous) simply supported beam subjected to a uniformly distributed load develops the stress trajectories shown in Figure 4.9. It is noted again that the principal directions are mutually orthogonal at every point. Solid lines indicate tensile stresses, while dashed lines indicate compressive stresses. It is observed that at the end fibres the evaluated normal stresses coincide with the principal ones, having compression at the top and tension at the bottom. In addition, because at the neutral axis of the cross-section only shear stresses are computed, the predominant state there is characterised by tensile and compressive principal stresses at a 458 angle. This may also be deduced by inspecting the deformation of a rectangular element which is in equilibrium according to Cauchy’s relation (see Figure 4.8). Elongated diagonals identify the direction of the principal tensile stresses, while shortened diagonals denote the principal direction of compression. Understanding this approach for direct determination of the principal directions of tension and compression is important because it is used in many cases. g

q

R

Figure 4.9 Principal stress trajectories in a simply supported beam

146


r (1/r) = (∆ε/h) = M/EI

εupper M h

εlower ∆ε

Figure 4.10 Correlation between curvature and a bending moment

In the case of steel beams, it should be noted that their deformability may play a more important role in their design than the allowable state of stress. The deflection of a steel beam should generally not exceed 1/500 of its span. At this point, it should be recalled that deformation due to bending, which in most cases is solely responsible for the formation of the elastic curve of the beam (compared with the shear deformation), as was previously explained, is directly related to the curvature (1/r), as this is used in the principle of virtual work (see Section 2.3.3). The curvature, which varies with position along the length of the beam, may, in any case and independently of the material in use, be expressed as shown in Figure 4.10, with the only precondition being the preservation of the plane nature of the section: 1 "upper þ "lower " ¼ ¼ r h h The above equation yields, in the case of elastically behaving material, the well-known expression M/EI (see Section 2.3.2). Finally, it can be shown that the deflection of a uniformly loaded simply supported beam of length l is 5 q l4/384 EI.

4.1.2

State of failure (plastic behaviour)

According to the elastic behaviour consideration, the maximum allowable moment My in a beam is that which causes at any of the two end fibres (top or bottom) the yield stress fy (Figure 4.11). This fibre will be the one located furthest away from the neutral axis, at a distance ymax. Thus, My ¼ fy (I/ymax) and, in the case of a section exhibiting double symmetry, My ¼ fy [I/(h/2)] 147

Structural systems: behaviour and design py Dy y A fy

L

Zy

My = py · L2/8

My = fy · (1/ymax) pu fy

A/2

Dpl z

A/2 Zpl Plastic neutral axis

α = 1.17

Mpl = pu · L2/8

Mpl = fy · (A/2) · z

α = Mpl/My α = 1.5

fy

α = 1.7 My = 2 · fy · (l/h)

h

Mpl = fy · (A/2) · z Sections of double symmetry

Figure 4.11 Plastic behaviour of steel beams

In contrast to the case of a bar subjected to uniform tension, for which the development of the stress fy obviously means that all cross-sectional fibres yield simultaneously and, inevitably, failure of the bar under the axial force Ny ¼ A fy, in the case of beam bending, it is possible to apply a greater moment, i.e. higher values than My, because not all the fibres of the cross-section will have yielded. Increasing the loading and requiring the cross-sections to remain plain during the process — as can be experimentally verified — the rest of the fibres gradually yield either in tension or in compression. The final yield state is characterised by yielding of all the fibres in the location of the maximum bending moment. The distribution of stresses is uniform in the region of tension as well as in the region of compression. The fact that the tensile force should be equal to the compressive force due to equilibrium requirements in the horizontal direction helps to determine the plastic neutral axis at that location where the section is divided into portions of equal areas (above and below the line). Now, the moment Mpl experienced by the beam is created by the couple of the two resultant forces, and is equal to Mpl ¼ fy (A/2) z It is obvious that for members possessing two planes of symmetry, the plastic neutral axis bisects the depth of the beam (axis of symmetry). 148


According to the previous discussion, the ultimate bending moment, which causes failure of the beam, can be higher than the value My if plastic analysis is adopted, or, in other words, the beam can withstand a greater bending moment before failing. A measure of this increase is the ratio a ¼ Mpl/My, which is called the shape factor, and takes values between 1.14 and 1.70, as shown for the cases of sections with two planes of symmetry in Figure 4.11. g By the time all the fibres in the critical section are ready to yield, two regions of total length ly will have already developed on either side of the section, in which the bending moments Mpl begin to diminish at the critical position, reaching a minimum value of My at the position where only the upper and lower fibres yield (Figure 4.12). The rest of the beam continues to deform elastically, because there are no other sections with their extreme fibres in a yield state. The plastic region of length ly is characterised by a progressive increase of the curvature from its ends inwards and up to the critical section, as implied by the equation 1 "upper þ "lower ¼ r h

qu

Elastic range

Elastic range

My

Mpl ly

My

Plastic range qu

Mpl

2

Mpl = qu · L /8

Elastic range

Plastic hinge in simply supported beam: impossible to carry further loading

Elastic range

q

Plastic hinge in a fixed-hinged beam: capacity for further increase of the load

Mpl = q · L2/8 Elastic range

Figure 4.12 The plastic hinge

149


In this region, although the stresses remain constant and equal to fy, the deformations due to yielding of the end fibres of the section increase continuously, resulting in considerably higher values for the curvature (1/r). In the case of a simply supported beam, formation of such region is equivalent to failure (collapse). However, because the behaviour along the plastic region is quite complicated, and in order to simplify the static analysis, the so-called plastic hinge was devised and implemented, as will be described next. It is assumed that as long as the bending moment Mpl does not appear in the section with the maximum moment, the whole structure behaves elastically, and the curvature is expressed at any point of the beam as M/EI. Once the loading reaches the value qu (see Figure 4.12) and the moment Mpl has developed in the critical section, at that exact point a hinge is considered which carries on its sides a couple of external moments Mpl that are equal in magnitude and opposite in direction. The structure is thus changed, as it has acquired a hinge at that point and, although in the case of simply supported beams it becomes a mechanism and thus cannot take any further loading, it remains in equilibrium under the load qu. This is justified by the presence of the external moments Mpl at the hinge, which would have also been there internally in the absence of the hinge. A characteristic property of the plastic hinge as it appears in the case of indeterminate structures is the ability to develop a relative change in slope between the two members on either side of the hinge with a further increase in the load, offering continuously a constant resistance Mpl which acts as an external action (see Figure 4.12). The reason for this is that, because the cross-section is in the yield state, the strains can increase freely, as was previously explained. In this way, the curvature at that location increases accordingly (see the equation above), without any corresponding rise in the stress values and hence in the resulting moment Mpl. Of course, in the case of the simply supported beam, the modified structural system with the plastic hinge, which was previously described, fails to equilibrate even with a load slightly larger than the qu, and it inevitably fails. Therefore, if the simply supported beam is loaded with a uniformly distributed load q (see Figure 4.12), the failure load qu may be evaluated by requiring qu l2/8 ¼ Mpl, as qu ¼ Mpl/(l2/8). It should be pointed out that the above concept assumes pffiffiffi that the beam does not fail due to excessive shear stresses, whose limit value is fy/ 3, as previously mentioned.

4.2

Reinforced concrete beams

4.2.1

Service conditions

4.2.1.1 Bending In a concrete beam the transverse loads are also carried to the supports in the axial direction, through a compressive and a tensile force. The problem, however, is that, although the compressive force D can be supported by the concrete, the tensile force Z may be sustained only up to a limiting value determined by the low and unreliable 150

Simply supported beams D0 Unreinforced concrete Bending moment capacity M clearly limited Z0 MD) is equal to M ¼ ZP zP þ Zs zs (b) The magnitude and position of the force D as the resultant of the compressive stress block of the concrete may be determined from the shortening strain "c and the height x of the uncracked concrete region (see Figure 4.32). It is clear that, due to the valid assumption of a plane section, the magnitudes P and "s may also be expressed in terms of "c and x. If "c is still less than 2% (Bieger et al., 1993), "c ð6 "c Þ b x fc 12 8 "c a¼ x 4 ð6 "c Þ D¼

On the basis of the prestressing force P and the cross-sectional areas of prestressing and reinforcing steel AP and As respectively, "c and x can be readily determined for each bending moment M, from equations (a) and (b). Subsequently, the reinforcement steel and tendon stresses are determined directly. An increase in the bending moment leads, firstly, to the yielding of the steel reinforcement and, with a further increase in the moment M, under the constant stress fsy of the mild reinforcement, the tendon will, in turn, also reach the yielding stress fPy. At this stage, the concrete develops the maximum compressive stress fc at the top of the section, whereas its shortening strain "c, which is now greater than 2%, cannot exceed 3.5%. The section will then develop its maximum bending resistance Mpl: Mpl ¼ AP fPy zP þ As fsy zs At this stage, the magnitude and the location of the compressive concrete force may be estimated reliably by assuming a fictitious compressive stress block having a constant uniform stress of fc and extending from the top edge up to a depth of 0.80 x (see Figure 4.32). In this way, the magnitudes D, zP and zs may be expressed as D ¼ fc b (0.80 x) 175


zP ¼ d cP 0.40 x zs ¼ d cs 0.40 x In the present state, the known forces ZP and Zs allow the determination of the compressive concrete force D, the depth x of the compressed zone, and consequently the lever arms zP and zs may be readily deduced. It should be noted that the depth x of the compressive zone represents the distance of the neutral axis in the strain diagram up to the top of the section, and this must not be confused with the height of the fictitious compressive stress diagram, which, of course, equals 0.80 x. g It is clear that the previous consideration of the ultimate state concerns both full and partial prestressing, and gives a unified treatment of the limit state of reinforced concrete sections in general, either with mild or prestressed reinforcement. Actually, the only difference between them lies in the fact that, in the case of prestressing, a certain pre-elongation is stored in the high-strength prestressing steel, owing to which an additional tensile force is available, which in turn means the development of an additional equal compressive force on the section. This leads to an increase in the bending resistance, according to the corresponding moment of the resulting couple of forces. It can be seen, then, that in the ultimate state the presence of prestressing is not connected to the development of a bending moment as in the case of service conditions, but is reflected simply in the formation of the bending resistance Mpl of the section (Menn, 1990). However, as will be illustrated in the next chapter, this holds only for statically determinate structures. g According to the previous discussion, it is clear that the increase in the tensile force in the prestressing steel from its initial value P — after the various losses have occurred — up to the value corresponding to the yielding stress fPy is feasible only because of the existing bond, which allows the cable to ‘keep pace’ with the ‘evolution’ of the linear strain diagram " over the height of the beam. If the bond of the tendon is not established by grouting, as mentioned in Section 4.3.1.2, the increase in the cable force will depend only on the global deformation of the beam, which is difficult to analyse. And given that the prestressing steel does not generally reach the yielding point in the design, it may be assumed, for safety reasons, that the tendon force will not be subjected to any increase beyond its initial value, thus resulting in a decreased bending resistance Mpl. In many cases, the use of unbonded tendons may be — apart from avoiding the grouting — preferable to bonded ones from a constructional point of view, because of their easy exchangeability and the possibility of re-stressing at a later time. These constructional advantages may take precedence, in the design phase, over the drawback of the clearly reduced ultimate bending resistance in comparison with bonded tendons. Although the first prestressed concrete structures of earlier times used unbonded tendons, which were soon abandoned because of the clear structural advantages of bonded ones, the last two decades have seen the frequent use of unbonded tendons (primarily in bridges), placed either ‘externally’ or ‘internally’ with 176


respect to the beam section. In such cases, however, according to what has been previously discussed, the decompression moment MD, which is independent of the existence of a bond, may represent a safe assessment of the ultimate bending resistance of the section. g As pointed out previously, when partial prestressing is adopted, particular attention has to be paid to the service condition, where the cracking of the concrete must be taken into account. An acceptable cracking width is ensured through the restriction of the additional stress increase in both the tendons and the reinforcing steel to within 200 N/mm2. The determination of these tensile stresses, and also of the compressive stresses of the concrete — which also have to be held within logical limits — may be carried out through the equations previously considered. Finally, aspects that affect the necessary control of deflection under service conditions can be assessed — according to Section 4.2.2.1 — on the basis of the relevant expression of the curvature 1/r in Section 4.1.1 and by taking into account the previously obtained top and bottom strains of the section.

4.3.2.2 Shear Everything that has been mentioned in Section 4.2.2.2 regarding the shear ultimate state in reinforced concrete is obviously also valid in the case of prestressing, where, according to Section 4.3.1.3, the struts are inclined at about 308. The relieving influence of the cable prestressing force on the effective shear has to be also taken into account in the ultimate state, as emphasised in this section. It is clear that, for safety reasons, the inclined compressive prestressing force after all losses should be taken into account rather than the corresponding yielding force.

4.4

Cantilever beams

The response of the cantilever beam to gravity loads is characterised by the fact that the bending moments as well as the shearing forces increase towards the fixed end. The cantilever beam is a structural system that is particularly sensitive because it has a single support where the maximum bending and shearing responses occur simultaneously, and therefore a possible failure at this point will cause total collapse. In addition, the cantilever beam exhibits particularly strong deformability. It is clear that the layout of internal forces due to gravity loads is the inverse of that for the simply supported beam. The tensile stresses are developed in the top region, whereas the bottom region is compressed. The steadily increasing bending response towards the fixed end suggests a continuous decrease in the structural height towards the free end, in order to achieve a better exploitation of the various sections in bending and save material in this way, given also the favourable distribution of weight decreasing towards the free end. This structural form has a beneficial effect on the shearing response of the beam, as is now explained. 177


V

V

Approximately constant shearing stress

Maximum shearing stress varies like V

(I-section)

M

DV

z

D

The contribution of the vertical component of the compressive force to the required V allows the beam shearing stresses to cover only the remaining force

∆z/∆x X

Figure 4.33 Favourable influence of variable beam height on the shearing response

The inclined compressive force in the lower region of the section contributes substantially, with its vertical component, to the total shearing force required for the equilibrium (Figure 4.33). Consequently, the developed shearing stresses on the vertical section of the cantilever are less than those that would be developed if the section height remained constant along the beam, as in this case the vertical component of the horizontal compressive force is simply missing. As a result, for a parabolic decrease in the height of the beam towards the free ends, an approximately constant shear along the beam is achieved. If an I-section is considered, then the inclined compressive force D equals M/z, and its vertical component is (M/z) (z/x). In the case of a constant height section, the shearing stress is ¼

V b z w

In the case of a variable section, the above stress is reduced to 1 M z ¼ w V b z z x The resulting constant shearing response allows the adoption of a constant web thickness bw along the complete length of a cantilever beam. g With regard to reinforced concrete beams, apart from the above introductory remarks, neither the bending nor the shearing load-carrying actions differ from what has been previously discussed for simply supported beams. The statically determinate truss model of the beam takes the form shown in Figure 4.34, and consequently all the 178


Figure 4.34 Truss model for a reinforced concrete cantilever beam

relevant results in Section 4.2.1.2 are also valid in this case with respect to the state of stress of either the inclined struts or the stirrups. In the case of a variable beam height, it is clear that the previously discussed approximately constant response in shear allows the use of stirrups of constant crosssectional area. g With regard to additional prestressed concrete beams, the following remarks apply: (1) The deviation forces and the anchor forces of the selected tendon profile should produce bending and shear moment diagrams which are of opposite sense to those caused by the gravity loads, thus contributing to the effective reduction of these loads. (2) In the case of a beam with constant cross-section, the layouts (a) and (b) in Figure 4.35 may be examined. Each layout produces quite different bending moments and shear diagrams, but nevertheless have the same bending values at their ends (see Figure 4.35). It is, of course, obvious that even the inclined straight cable tendon profile (c) will cause a bending and shear response, which reduces that due to gravity loads. It is clear that, in the considered cantilever system, the forces acting on the beam due to the cable prestressing consist of the compressive force at the free end of the beam (inclined or not), as well as the deviation force of the tendon (see Figure 4.35). In each case, the static action of the prestressed tendon on a cross-section is equivalent to a compressive force equal to P applied at the centroid of the tendon with the appropriate inclination. In contrast to the simply supported beam, in the cantilever system the prestressed tendon produces tensile bending stresses at the bottom fibres, whereas the gravity loads produce tension at the top. Assuming, now, that the beam is under the prestressing actions only, with the compressive force P acting on the corresponding trace of the tendon, a bending moment M due to a gravity load applied afterwards will cause a shift a of the force P downwards equal to a ¼ M/P (Figure 4.36). It is also clear that the behaviour of the beam in response to a further increase in load will follow the same stages as described in previous sections, by simply inverting the tensile and compressed regions. (3) For beams with variable height, in addition to the above mentioned shearing response, all that has been discussed in Section 4.3.1.1 regarding the handling of 179


f f

(a)

(b)

Stronger shear relief due to prestressing

Higher bending resistances due to prestressing

u

u P P

VP

VP

MP

MP

(c)

Figure 4.35 Influence of the cable profile on the response of a cantilever beam

the cable profile applies (see Figure 4.26). Thus, in the case, for example, where the centroidal axis of the beam has a parabolic track, the straight cable shown in Figure 4.37 is equivalent to a fictitious parabolic one. This causes uniform upward deviation forces which oppose the gravity loads. Of course, the statically most favourable result is achieved when the cable profile, having exhausted all the available eccentricity at the fixed end, follows a parabolic trace towards the centroid of the free end cross-section, as shown in layouts (a) and (b).

Initial location of prestressing force P a M P

a = M/P

Figure 4.36 Downward shifting of the compressive force due to the bending action of gravity loads

180

Simply supported beams Fictitious cable profile

Although the tendon is straight upward deviation forces are developed

(a)

(b)

VP P

Higher shearing force

Higher bending resistances

Figure 4.37 Influence of the cable profile in a cantilevered beam of variable height

While in both layouts (a) and (b) the required prestressing force, in order to overcome the bending moment at fixed end, is the same, layout (b) offers, for a given prestressing force, greater ultimate bending resistance throughout the length of the cantilever, while the shear relief offered at the fixed end by layout (a) is greater. However, in the case of cantilever-constructed girder bridges using the so-called ‘free cantilever method’, layout (b) is imposed for constructional reasons (see Section 5.5.2). g Considering now the control of shear, it is clear that the effective shearing force Veff, according to the previous discussion, is equal to (see Figure 4.33) Veff ¼ V

M z VP z x

where VP is the vertical component of the cable action on the examined section (Figure 4.37). The actual shearing stress of the cross-section will then be ¼

4.5

Veff bw z

External prestressing

In a beam, it is possible, through an appropriate external prestressing layout, to offer concentrated loads opposing those due to gravity, according to the principle given in Section 4.3.1.1 181


Compression

Compression

Tension

h Tension

Tension L/3

L/3

L/3

The statically indeterminate structure develops deflections at the locations of the struts

R

R

R

R S

The appropriate prestressing force S ensures the exertion of force R The deflections at the locations of the struts are eliminated and the bending response becomes identical to that of the continuous beam

Figure 4.38 The purpose of external prestressing

The layout of the structural system shown in Figure 4.38 offers the simply supported beam of span length L upwards concentrated forces through the two vertical struts, clearly reducing the bending response (g L2/8) of the beam if it were in isolation. The structural action of this statically indeterminate system under the permanent load g leads inevitably to a settlement at the strut joints, which may be eliminated through appropriate prestressing of the three straight cables. More simply, the prestressing of the cables should be such that the developed strut forces are equal to the reactions of a three-span continuous beam with rigid supports. Thus, the bending response of the simply supported beam for the given loads will be identical to the much more favourable one of a continuous beam (see Figure 4.38). If the struts are ordered at the third points of the span, given that the reaction R at the interior support of a continuous beam with three equal spans is R ¼ 1.10 g L, then the prestressing force S, which must be introduced to the horizontal cable, can readily be obtained from the equilibrium at the nodes (Figure 4.39): g L2 S ¼ 0:978 8h 182

Simply supported beams g

0.10 · g ·(L/3)2 0.025 · g ·(L/3)2 L/3

L/3

L/3 γ·q

g Service condition

Ultimate state Mpl

0.025 · g · (L/3)2 h

2

g · L /8

h

(γ · q) · L2/8

S g · L/2 L/2

(γ · q) · L/2

L/2

AP fPy

Figure 4.39 Structural action of the system

It is appropriate to confirm this last result by also remarking that the overall bending moment (g L2/8) of the system may consist of the bending moment of the continuous beam at its middle section (0.025 q (L/3)2) and the corresponding moment (S h) exerted by the internal force S of the horizontal cable (see Figure 4.39): g L2/8 ¼ 0.00278 g L2 þ S h Prestressing of the horizontal cable with the above force ensures, as pointed out, the function of the simply supported beam as a continuous beam over four rigid supports. An additional live load p will cause an increase of the axial force of all three cables according to the behaviour of the actual statically indeterminate system, which depends on both the rigidity of the beam EI and the cross-sectional area As of the cable. g In the ultimate limit state, under the load q, where q is the total service load of the structure, it may be considered that the ultimate bending moment (( q) L2/8) is undertaken by the bending resistance of the beam Mpl and the moment which the yielding cable force contributes with respect to the centroid of the cross-section of the beam (see Figure 4.39): ( q) L2/8 ¼ Mpl þ As fPy h This equation may be used for the assessment of the cross-sectional area of the cable in the preliminary design. g The beam can be made of steel, reinforced concrete or prestressed concrete. The third choice is made when the resulting ‘continuous’ concrete beam has a relatively long span (i.e. of the order of 15 m). Thus by providing the beam with a stronger ultimate bending resistance Mpl, the required prestressing steel cross-sectional area As of the cable is 183


reduced, according to the last equilibrium equation. However, the choice of a highstrength steel for the cable is crucial. It should be noted here that due to the presence of a significant compressive force in the beam, a more realistic calculation of Mpl may rely on the factors discussed in Sections 6.5.1 and 6.5.2, regarding the steel and reinforced concrete sections, respectively. Finally, it should be pointed out that the optimal distance between the vertical struts, with regard the required prestressing force S of the cable in order to counterbalance the permanent load of the beam, equals 0.44 L, instead of L/3. Thus, S ¼ 0:92

4.6

g L2 8h

Design control

As has been pointed out in Section 1.6, the design control of a structure includes, in addition to the control in service conditions of deformations, cracking, annoying vibrations, etc., the control of ultimate strength. This is no longer the control of the stresses developed under service loads, but concerns checking that the specified working loads q multiplied by a safety factor (depending on the code followed), which are approximately equal 1.80, are less than (or equal to) the corresponding ultimate load which causes the failure of the structure. According to the so-called ‘static theorem of plasticity theory’, which will be examined in particular in Chapters 5 and 6, this condition is ensured — excluding failure by loss of elastic stability, or the presence of second-order effects (see Chapter 7) — if the following equation, given previously in Section 1.6, is satisfied: S R ( ¼ S R) The sense of the factors S and R, has been clarified in Section 1.6. The above equation applies to any cross-section which can be regarded as critical and relates to the sectional forces of the structure examined. The term S represents the sectional force of the cross-section examined, and results from the corresponding sectional force diagram, which should be in equilibrium with the external loads q. This, with regard to the bending moments of a beam or frame structure, can be ensured by two things. First, the moment values at the joints must be in equilibrium, on the one hand, and the support reactions must ensure the global equilibrium of the structure, on the other hand. Second, the bending moment diagram, which is drawn between the end values of a straight member, must be identical to the bending moment diagram of the corresponding simply supported beam (see Section 2.2.5). While such a bending moment diagram is compatible with the prevailing equilibrium of the structure, it may generally present a deviation from the corresponding diagram of the elastic solution for the loads ( q), as stated in particular in Chapters 5 and 6. The shear force diagram may obviously be determined from the adopted bending moment diagram. The term R represents the corresponding ultimate resistance of the cross-section, e.g. if bending is considered it will be equal to Mpl. 184


Thus, while for the statically determinate structures the compatible moment diagram with the overall equilibrium is unique, for statically indeterminate structures, where an unlimited number of adoptable bending moment diagrams can satisfy the equilibrium requirements, the design of cross-sections has a large number of options, depending on decisions made by the engineer. Finally, the remark made in Section 4.3.2.1 is repeated here, namely that in the case of a prestressed statically determinate structure, such as a simply supported beam or a cantilever beam where prestressing does not cause any support reactions, the presence of prestressing is reflected only in the ‘resistance term’ R, i.e. in the corresponding determination of Mpl.

References Bieger K.W., Lierse J., Roth J. (1993) Stahlbeton— und Spannbetontragwerke. Berlin: Springer Verlag. Menn C. (1990) Prestressed Concrete Bridges. Basel: Birkaüser Verlag. Walther R., Miehlbradt M. (1990) Dimensionnement des Structures en Be´ton. Lausanne: Presses polytechniques et universitaires romandes.

185

5 Continuous beams 5.1

Introduction

The form of the bending moment diagram of a continuous beam in which the tension is alternated between the top and bottom fibres between the regions of supports and spans, respectively (see Figure 3.14), suggests that such a beam behaves as consisting of intermediate simply supported beams resting on cantilevers, extending from the internal supports up to the points where the bending moments become null. The length of these cantilevers constitutes a percentage of the corresponding span, which, for internal spans having a constant depth, is of the order of 20%, depending on the type of loading, as well as the span ratios (Figure 5.1). For beams of large spans, as are usually applied in bridges, it is statically purposeful to follow a parabolic change in depth in all spans, with the maximum depth at the supports and the minimum depth at the midspan sections. In this way there is an increase in the moments at the supports (see Section 3.2.10) and, according to the equilibrium requirement (suspension of the moment diagram for the simply supported beam between the values at the supports), a reduction in the midspan moments. Moreover, the cantilever lengths are increased compared with those of the corresponding beam with constant depth (see Figure 5.1). Thus, in the case of a constant depth, the bending moment diagram is independent of the moment of inertia of the beam, as explained in Section 3.2.5. The aforementioned change in depth leads to a proportional distribution of selfweight that, for a support-to-midspan depth ratio of 3 (a common value for bridges with large spans), results in a significant increase (up to 35%) in the lengths of the considered ‘cantilevers’ of the internal spans. Therefore, the ‘simply supported beams’ are limited to 30% of the span lengths, with far smaller bending moments than those of the ‘cantilevers’ (see Figure 5.1). The fact that the thus resulting support moments do not differ substantially from those of the cantilevers extending up to the midspan gave rise to the construction of concrete bridges according to the so-called balanced cantilever method, namely the construction of progressively smaller prestressed cantilever segments from both sides of the two free-standing supports (piers) until the midspan, and thereafter the restoration of the beam continuity (see Section 5.5.2). This depth variation in the ‘cantilevers’ contributes favourably to the shear response, as explained in Section 4.4.


~0.20 · L

~0.60 · L

~0.20 · L

L

~0.30 · L ~0.35 · L

~0.35 · L L

Figure 5.1 Load-bearing action of a continuous beam

5.2

Steel beams

As, mainly for reasons of economy, steel beams are nowadays designed using the ‘plastic theory’ rather than the ‘elastic’ one, it is advisable to be familiar with this means of design, as applied to a statically indeterminate continuous beam, by considering first the behaviour of a fixed-ended beam and then that of a fixed simply supported beam. The bending (and shear) response at any span can be described using the basic characteristics for either beam type.

5.2.1

Fixed-ended beam

A fixed-ended beam under a uniform load q develops the bending moment diagram shown in Figure 5.2. From the figure and the discussion above it is obvious that there must be two cantilevers with length 0.21 L. The moment of the support q L2/12 is 188

Continuous beams q

q · L2/12 2

q · L /8

0.21 · L

0.58 · L

0.21 · L

Figure 5.2 Load-bearing action of a fixed-ended beam

double the maximum moment of the midspan, which is of course the span moment of the ‘fictitious’ simply supported beam of length 0.58 L. It is assumed that the crosssection of the beam is symmetrical with respect to the horizontal axis of bending, and accordingly it has the same bending resistance Mpl for moments causing tension either at the top fibres or at the bottom ones. For some load q1 the bending moment will first reach the plastic resistance value at the supports. Thus q1 L2/12 ¼ Mpl and q1 ¼ 12 Mpl/L2. For this load a plastic hinge will be developed at both ends of the beam, as discussed in Section 4.1.2 (Figure 5.3). Obviously, the beam does not collapse under this load, and is still in a position to accept more loading. During the imposition of any additional load, the plastic hinges at the two ends rotate under the constant moment Mpl, which can be considered as being externally applied. q1

Mpl

Mpl qu qu = 16 · Mpl/L2

Mpl

Mpl qu · L2/8 Mpl

1

Virtual loading for the determination of rotation at the plastic hinge

1

1

1

Figure 5.3 Beam behaviour due to plastic hinge formation

189


The diagram of the bending moment corresponding to each load q is obtained by hanging the diagram for the simply supported beam q L2/8 from the constant points that correspond to Mpl. Clearly the beam will collapse when the load q takes the value qu at which a bending moment Mpl is developed at the midspan of the beam. In this case, an additional plastic hinge will be created at the midspan (first-degree mechanism) and no further increase in load will be possible, as discussed in Section 4.1.2 for a simply supported beam. Thus it may be written that qu L2/8 Mpl ¼ Mpl and, consequently, qu ¼ 16 Mpl/ L2 (see Figure 5.3). The problem now of designing a beam under a certain service load qser is that Mpl should be defined such that the beam may sustain, or even better collapse at, a load qu ¼ qser (see Section 4.6). According to the last equation the value sought for Mpl is given by Mpl ¼ ( qser) L2/16 It should be pointed out that designing a beam for the ‘collapse’ load qser by following the elastic moment diagram means that a value of ( qser) L2/12 must be selected for Mpl, which is of course higher than above value, and consequently represents an uneconomical solution. Precisely for this reason it was pointed out in Section 4.6 that in the equation S R (a) the design value S does not necessarily correspond to the elastic solution for the load qser. Thus, in order to design a fixed-ended beam with length L, so that the ultimate bending resistance is exploited at the ends as well as at the midspan, one should consider the bending moment diagram for the simply supported beam that ensures balance with the applied load qser and then suitably shift its base until equal parts Mpl are cut at both ends as well as the midspan (see Figure 5.3). Thus, it is required that ( qser) L2/8 ¼ Mpl þ Mpl This result coincides with the preceding one. It is important though to point out that, according to the static theorem of plastic analysis (see Section 6.6.2), the satisfaction of the above inequality (a) at any point on the beam ensures that the value qser is less than or equal to the ultimate load, provided that the value S belongs to a bending moment diagram which is in equilibrium with the external load qser. The confirmation that ( qser) < qu, can be made when selecting both Mpl ¼ ( qser) L2/16 and Mpl ¼ ( qser) L2/12. It is, of course, clear that while in the first case the ultimate load qu coincides with qser, in the second case, according to the above equation, it results in qu ¼ 16 Mpl/L2 ¼ 1.33 ( qser) It is purposeful at this point to find also the corresponding Mpl that would result if the design was made on the basis of the allowable stress fy/. As the section at the beam end is the most unfavourable one, according to Section 4.1.2 My/ ¼ qser L2/12. As Mpl ¼ a My, it follows that the selected section on which the load qser causes the maximum allowable stress fy/ should have a bending strength equal to 190

Continuous beams

Mpl ¼ a ( qser) L2/12. This value of the moment resistance leads to a collapse load equal to qu ¼ 16 Mpl/L2 ¼ a 1.33 ( qser). For a ¼ 1.15, which is appropriate for I-shaped cross-sections, qu ¼ 1.53 ( qser). This result shows that the safety factor determined according to the method of permissible stresses is practically equal to 1.53 . Finally, another point of interest is the rotation developed in the plastic hinge at the instant of collapse. This rotation, which must be developed in order to realise the aforementioned moment diagram, is limited by the plastic deformability (ductility) of the material, which for steel (as experimentally shown) is of the order of 0.1 rad. This rotation can be determined directly from the existing moment diagram for the instant of collapse by applying the principle of virtual work. A self-equilibrating system of unit moments, applied at the ends of the simply supported beam, is selected as the virtual load. According to the principle of virtual work, and taking into account the above (see Figure 5.3): ð Mpl L 2 a fy L M ds ¼ (see Section 4.1.2) ¼ 2 1 ’ ¼ M EI 3Eh 3 EI For the values a ¼ 1.15 and fy ¼ 4.2 105 kN/m2, it results that ’ ¼ 0.77 10 —3 (L/h). It is concluded that for practical values of the ratio L/h, i.e. values of 25—30, the resulting value of ’ is feasible. The resulting value of the angle ’ for the elastic diagram of bending moments must be equal to zero, as the diagram under consideration is supposed to satisfy, besides the equilibrium, the compatibility of the deformations. Such a requirement cannot, of course, be satisfied for the adopted design diagram, as elastic behaviour has been exceeded and plastic behaviour has already been established.

5.2.2

Fixed simply supported beam

A fixed simply supported beam under a uniform load q develops the bending moment diagram shown in Figure 5.4. It can be seen that there is a cantilever part of length q

q · L2/8

q · L2/8 max M 0.375 · L

0.25 · L

0.75 · L

Figure 5.4 Load-bearing action of a fixed simply supported beam

191

Structural systems: behaviour and design q1

Mpl

L qu

qu = 12 · Mpl /L2

Mpl qu · L2/8 Mpl

5 · L/12

(γ · qser) · L2/8

M

M

5 · L/12 1

Virtual loading for the determination of rotation at the plastic hinge

1

Figure 5.5 Ultimate strength and design of a fixed-ended beam

0.25 L, with a fixed end moment equal to q L2/8. The maximum span moment corresponding to the bending moment of a ‘fictitious’ simply supported beam of length 0.75 L occurs at a distance 3 L/8 from the free end, and is equal to q L2/14.2. The beam at first develops a plastic hinge at the fixed end under the load q1, so that Mpl ¼ q1 L2/8, and it can be loaded further, up to the collapse under the total load qu. Under this load, and with Mpl applied externally at the henceforth ‘hinged’ end, the beam develops its maximum span moment, which is also equal to Mpl (Figure 5.5). It is found that the load qu that leads to collapse is qu ¼ 12 Mpl/L2. The same design procedure used for the fixed-ended beam can be equally applied to the fixed simply supported beam also. On the basis of the bending moment diagram of the simply supported beam under an external load qser, a position of the rotating baseline around the simply supported end is sought at which the cut maximum value of the span is equal to the one corresponding to the fixed end (see Figure 5.5). These equal values are obviously the plastic moment Mpl, and according to the above this results in Mpl ¼ ( qser) L2/12. 192

Continuous beams

Again, the design of such a beam for a load capacity qser, made on the basis of the corresponding maximum value of the elastic diagram involving Mpl ¼ ( qser) L2/8, leads to a less economic solution, as according to the above expression the corresponding ultimate load is qu ¼ 12 Mpl/L2 ¼ 1.50 ( qser) Moreover, a design done using the method of permissible stresses under a ‘safety factor’ 0 imposes, according to Section 4.1.2, My/ ¼ qser L2/8, and consequently Mpl ¼ a ( qser) L2/8. This selected bending resistance (a ¼ 1.15) leads to an ultimate load qu ¼ 1.72 ( qser). The rotation developed in the plastic hinge is again found, according to the principle of virtual work, on the basis of the adopted bending moment diagram (see Figure 5.5): ð Mpl L M ds ¼ 1 ’ ¼ M EI 6 EI a value which is identical to the one for the fixed-ended beam (see Section 5.2.1).

5.2.3

Continuous beam

The case of a continuous beam can now be examined. In general, given the load to be l carried qser, the plastic moments Mpl at midspan and at the supports (Mm pl , and Mpl t and Mpl , respectively) should be so determined that the resulting ‘suspended’ moment diagram M0 of the corresponding simply supported beam between any two successive supports exhibits values that are less than or equal to the corresponding value of Mpl (Figure 5.6). This condition is ensured if the following equation is satisfied at each span (Menn, 1990): 0

M ð qser Þ

Mm pl

Mlpl þ Mrpl þ 2

ðaÞ

(γ · qser)

M lpl

M rpl Mm pl M rpl

M lpl

(γ · qser) · L2/8

Mm pl

Figure 5.6 Design of a continuous beam based on plastic behaviour

193

Structural systems: behaviour and design (γ · qser)

(γ · qser) · L22/8 M1

(γ · qser) ·

L12/8

M2

L1

L2 1

1

1 1 Virtual loading for the determination of the relative rotation at the plastic hinge

Figure 5.7 The required check of the relative rotation at the plastic hinge

For a beam of constant depth, it is purposeful to select a cross-section with double symmetry and having the same value of Mpl for the moments causing tension at the top and at the bottom. The satisfaction of equation (a) for beams with roughly equal spans (min l/max l 0.8) and uniform load qser is ensured if for the external spans a value Mpl ¼ ( qser) L2/11 and for the internal spans a value Mpl ¼ ( qser) L2/16 are selected, as explained previously. To reiterate, according to the static theorem of plastic analysis, the design of the beam (i.e. the determination of Mpl), based on the above equation of equilibrium (a), ensures that the load qser will be less than or equal to the ultimate load of the beam. It should also be emphasised that the adopted bending moment diagram, based on equation (a), is only acceptable if the relative rotations of the plastic hinges in the internal supports are within the permissible limits, according to the above. Their determination is always made by direct use of the principle of virtual work, as shown in Figure 5.7. Assuming a uniformly distributed load qser, the developed (relative) rotation in the plastic hinge on any internal support is ’¼

ð qser Þ ðL31 þ L32 Þ ðM1 þ 2 MÞ L1 ðM2 þ 2 MÞ L2 24 EI 6 EI 6 EI

Note again that, based on the elastic moment diagram, the aforementioned rotation is zero, given that the elastic solution is supposed to satisfy the compatibility requirements on both sides of the considered hinge at each support, and therefore requires the nullification of the relative rotation, due to continuity of the deflection line. However, in cases of intensely different spans or rigidities, the relative rotation of the plastic hinge at an internal support, determined on the basis of the corresponding adopted moment diagram, may be rather prohibitive, and the structure must be designed following the elastic solution, although always according to the strength condition S R. 194

Continuous beams

5.3

Reinforced concrete beams

Although it has always been common practice to calculate the bending moments of continuous beams for a specific loading on the basis of elastic analysis, it should be recognised that the variation in the moments along a continuous beam will cause cracking in those parts where the cracking moment is exceeded, while all other regions will remain uncracked. The cracking moment is the bending moment that causes tensile stresses on the homogeneous section equal to the tensile strength of concrete, which is of the order of 200 kN/m2. However, given that cracking of the section causes a reduction in the moment of inertia, the variable rigidity along each span will obviously lead to a moment diagram that is different from the one resulting for the uncracked sections over the entire beam length. At the same time, factors irrelevant to the considered loading, such as shrinkage, creep or possible thermal variation due to environmental effects, may lead to the tensile strength of the concrete being exceeded, and hence to cracking, before the anticipated loading to reach this point. For all these reasons a precise follow-up of the stress state of the beam becomes practically impossible, and therefore a plastic analysis will be more useful in such a case. This objective of the analysis is to design and reinforce the beam such that the load qser is less than or equal to the ultimate load of the beam. This can be achieved, as stated previously, using the static theorem of plastic analysis (see Section 6.6.2), and by adopting such a bending moment diagram that equation (a) in Section 5.2.3 holds true for each span. However, it should be emphasised that attention must be paid to ensure that the deviation of the adopted moment diagram from the elastic solution does not exceed a ‘reasonable’ limit. The strict criterion to determine whether the adopted moment diagram leads to permissible results for design is the magnitude of the (relative) rotation that each plastic hinge is supposed to develop on the basis of the diagram. The maximum ‘allowable’ rotation value lies in the range 0.02—0.05 rad, depending on the cross-sectional area of the existing reinforcement. This condition is generally satisfied without special calculation if the deviation of the moment values at the supports from the corresponding values in the elastic solution is not more than 20%. Of course, it is understood (always according to the static theorem of plastic analysis) that the adoption of the elastic moment diagram for the load qser and the satisfaction at every point of the relationship S(qser) R ensures the above objective, although not in the most economical way. Everything examined in Section 5.2 with regard to steel beams has an analogous application in the case of reinforced concrete beams. Moreover, whatever applies to the design of a simply supported beam and a cantilever with regard to reinforced concrete has a direct application in the determination of the reinforcements for a continuous beam, as such a beam is composed of such analogous parts.

5.4

Prestressed concrete beams

5.4.1

Tendon design and structural performance

In the case of prestressed concrete, the layout of tendons is also based on the synthesis of cantilevers and simply supported beams, as explained previously. Basically, the tendon 195

Structural systems: behaviour and design uS uF

uF

Self-equilibrating loading LF

LS uS

uF

uF · LF = uS · LS

Common tangent Keeping the minimum radius of curvature

LF

LS

Figure 5.8 Geometric requirements for the layout of tendons

layout aims to provide the beam, via the induced deviation forces, with the opposite loads to those due to gravity, which stress the beam. In order for the greatest possible bending resistance to be offered, both in the regions of the supports for moments causing tension in the top fibres and in the spans for moments causing tension in the bottom fibres, the tendons are arranged in such a way as to exhaust the corresponding placement limits. The thus resulting reverse curvatures on the path of the tendon must ensure that the minimum applicable radius of curvature over the support for the specific tendon is not exceeded. It is understood that the equilibrium of the tendon as a free body imposes specific geometrical conditions on the positions of turning points and on the curvatures of the tendons (Figure 5.8). As the continuous beam is a statically indeterminate structure, the bending moment due to prestressing is no longer equal to the product of the prestressing force due to its eccentricity with respect to the centroidal axis. In other words, the compressive force on the concrete section is not applied at the trace of the tendon, as is the case in the simply supported beam or the cantilever. This is explained below using the example of a continuous beam with two equal spans (Figure 5.9). For the analysis of the statically indeterminate beam under prestressing loads only, the force method is applied and, in order to convert the structure into a statically determinate one, the intermediate support is removed by selecting the corresponding reaction as the redundant unknown. The simply supported beam is then deformed upwards under the deviation forces and, in order to restore the compatibility of deformations with respect to the actual system, which requires an immovable intermediate support, a downward applied (redundant) force must be applied. Thus a triangular diagram of bending moments MSP results, causing tension in the bottom fibres, and 196

Continuous beams Self-equilibrating loading

P

P

Deviation forces

Statically determinate beam

Deformed statically determinate beam

Redundant force

[MSP] Redundant prestressing moments due to the required concentrated (redundant) force The compressive force on each cross-section shifts upwards due to the redundant prestressing moments

P

P

Bending moments due to self-equilibrating prestress loading

[MP]

MP = M0 + MSP

Figure 5.9 Analysis of a prestressed statically indeterminate beam

this diagram is then superposed on the one for the ‘statically determinate’ prestressing bending moments, which are the product of the prestressing forces due to the existing eccentricity (see Figure 5.9). As a result, the compressive force on the concrete is shifted (as explained in Chapter 4, see Figures 4.28 and 4.36) upwards relative to the trace of the cable, in the regions of both the midspan and the internal supports, where it is of course located outside the section. In this manner, the finally developed bending moments MP due to prestressing are given by MP ¼ M0 þ MSP 197


where M0 represents the statically determinate moments ( ¼ P e), and MSP are the socalled ‘statically indeterminate’ or ‘constraint’ moments, which are due to the action of the hyperstatic (redundant) force only. It should be noted that the development of the aforementioned redundant force implies also the development of self-equilibrating reactions in the continuous beam, while in the simply supported system loaded with the self-equilibrating prestressing forces no such reactions are developed. It is clear that the moments MSP are caused by this system of self-equilibrating reactions. Of course, the fact that the developing reactions in the continuous beam are self-equilibrating arises immediately also from the fact that they are caused by a self-equilibrating loading such as prestressing. The moments MP may be calculated directly from the statically indeterminate system under the ‘external’ loads imposed by the prestressing, i.e. the deviation forces and the possible moments at the two ends of the continuous beam due to eccentricity of the anchor forces (see Figure 5.9). The statically indeterminate moments will then be MSP ¼ MP M0 It should be noted that as the loading is increased and the time of cracking due to a more intense bending response is approached, the redundant prestressing moments do not change significantly. However, when the ultimate state is approached and the continuous beam becomes in practical terms a number of simply supported parts constituting a statically determinate system, the redundant prestressing moments approach vanishing. In any case, the above moments must be determined because, as will be explained in the next section, they are taken into account in the control of the design. With regard to large-span beams, as occur in bridges, girders with variable depth (following a parabolic law) are normally used, and consequently all that applies is what has been discussed already in Section 5.1. The bending moment diagram has a favourable shape and there is a positive influence on the shearing forces in the regions of the supports, which for practical purposes leads to a constant shear stress design value.

5.4.2

Control of the design

The adoption of a moment diagram that is in equilibrium with the external loads and has been determined on the basis of equation (a) in Section 5.2.3 has already been examined for steel and for concrete beams. In the case of prestressed concrete, the statically redundant moments should be superposed on the moments due to external loads. As explained previously (see Section 4.6), the statically determinate prestressing moments do not count towards the loads, but simply contribute to the bending resistance. This is reasonable, as only the redundant prestressing moments modify the equilibrium state of the system, by changing the reactions arising from the gravity loads. Despite the fact that the structure is designed for a supposed failure state where, as previously stated, the statically redundant moments do not develop, the normal 198

Continuous beams

values of the latter moments are nevertheless generally taken into consideration according to Section 5.4.1. Therefore, the strength relationship M(qser) MR can be written (Menn, 1990) as M(qser) þ MSP MR It should be noted that this relationship is valid regardless of whether or not the depth varies across the cross-section. The required bending resistance MR in various locations of the beam is achieved, as for the simply supported beam, by considering a combination of prestressed tendons and reinforcing steel (see Section 4.3.2.1). It should be noted that the moments MSP are ‘legitimately’ superposed with the values from the diagram M(qser), multiplied by some factor P, due to the static theorem of plasticity theory, as they arise from a self-equilibrating state (see Section 6.6.2). Of course, at the ultimate limit state, where the system becomes statically determinate, these moments may be considered as zero (i.e. P ¼ 0), as it in fact happens. g As noted previously, at a preliminary stage the system is first designed based on ultimate strength requirements, and this system is then checked in the serviceability state. It is obvious that the design principles laid out in Section 4.3 — particularly regarding partial prestress — may be directly applicable in the case of continuous beams. In the service condition, the satisfactory behaviour of the beam is first checked against cracking. If only compressive stresses are developing in the concrete at every location, besides the ultimate strength, it should also be checked that the area of reinforcing steel area is at least the minimum required by the appropriate structural code; this is usually expressed as a percentage of the section. The presence of tensile stresses in the concrete above the limit of 2 N/mm2 will lead to cracking (partial prestressing) and, accordingly, both the stresses in the reinforcing steel and the stress increase in the prestressed tendons must be checked. These stress values must not exceed the limit of 200 N/mm2.

5.5

Creep effects

The presence of creep as a progressive cause of deformation raises immediately the problem of the redistribution of the initial stresses in those systems where the state of stress under service conditions is based on the compatibility of deformations (i.e. in the statically indeterminate structures in general, and in this respect in continuous beams as well). The problem arises of whether the system has been definitively formed or will undergo a change during the construction procedure. Although this problem is complex, thanks to the understanding of the creep behaviour of concrete (see Section 1.2.2) it may be dealt with in a direct manner (Menn, 1990), as described below. 199


5.5.1

Statically redundant systems

What may be said in general about any structure the members of which have the same creep properties (i.e. structures that are made entirely concrete of the same specifications and age) is that, although the creep causes an increase in deformations, it does not affect the state of stress (axial, bending and shear responses) caused by external loading (excluding the cases of support settlement and temperature variation). The reason for this lies simply in the fact that a uniform change in the elasticity modulus of the members of a statically redundant system (more specifically a reduction due to creep) involves exactly what was stated in Section 3.2.5. Thus, if one takes into account that creep implies a reduction in the elasticity modulus (see Section 1.2.2.1) and this reduction is uniform (for members that have the same creep properties), then, according to Section 3.2.5, no change takes place in the state of stress, although the deformation increases. The continuous beam can be used as a representative example of a statically redundant system. For simplicity, a continuous beam of two spans under an arbitrary loading is considered (Figure 5.10). The statically determinate primary system consists of the two adjacent simply supported beams, and thus the moment at the internal support B is the unknown redundant.

A

B

C

F l10 F r10 X1 = 1

F l11

F r11

The bending diagram remains unchanged with time if all members have the same creep properties X1

∆X1

The right beam does not creep at all

X1

Initial diagram of bending response Final (redistributed) bending response due to creep

Figure 5.10 Creep effects on the state of stress of a continuous beam

200

Continuous beams

With the load imposed on the continuous beam, the magnitude X1 is instantaneously developed. This value results from the compatibility condition ðFl10 þ Fr10 Þ þ X1 ðFl11 þ Fr11 Þ ¼ 0 where F1 denotes the rotation at support B of the primary system due to the external load or due to X1 ¼ 1, for both the beam on the left (l) and the beam on the right (r). The quantities (Fl10 þ Fr10 ) and (Fl11 þ Fr11 ) represent the corresponding rotation at end B of the beams due to the external load and X1 ¼ 1, respectively (see Section 3.2.2). It is now assumed that X1 changes because of creep, so that after an adequate time it reaches the value (X1 þ X1). The total value of the relative rotation at support B ( ¼ 0) is obtained by superposing the following magnitudes (see Sections 1.2.2.1 and 3.2.2, where for simplicity ’1 is simply written as ’): . the relative rotation due to external loads Fl10 ð1 þ ’Þ þ Fr10 ð1 þ ’Þ . the relative rotation due to the imposition of X1 X1 ½Fl11 ð1 þ ’Þ þ Fr11 ð1 þ ’Þ . the relative rotation due to the gradual imposition of the change in X1, (i.e. of X1) X1 ½Fl11 ð1 þ ’Þ þ Fr11 ð1 þ ’Þ It is clear that multiplication of the ‘elastic’ deformation values by (1 þ ’) or (1 þ ’) means that Einit or Edif, respectively, is adopted as the effective elasticity modulus of concrete, according to Section 1.2.2.1. Thus, the compatibility condition after some elapsed time takes the form Fl10 ð1 þ ’Þ þ Fr10 ð1 þ ’Þ þ X1 ½Fl11 ð1 þ ’Þ þ Fr11 ð1 þ ’Þ þ X1 ½Fl11 ð1 þ ’Þ þ Fr11 ð1 þ ’Þ ¼ 0 Based on the aforementioned initial compatibility relation, this last equation can be written more simply as X1 ½Fl11 ð1 þ ’Þ þ Fr11 ð1 þ ’Þ ¼ 0 from which it is immediately deduced that X1 ¼ 0. The conclusion is that X1 ¼ 0 is not valid if the creep is not identical in all the members of the structural system. This is obvious in the ‘extreme’ case of the same beam, when its right-hand part, being made of steel, does not creep at all (Figure 5.10). After some time has elapsed, for the primary system that was previously examined the total value of relative rotation at support B (which is obviously also equal to zero) will be composed of superposing the following: . the relative rotation due to the external loads Fl10 ð1 þ ’Þ þ Fr10 201


. the relative rotation due to the imposition of X1 X1 ½Fl11 ð1 þ ’Þ þ Fr11 . the relative rotation due to the gradual imposition of the change in X1 (i.e. X1) X1 ½Fl11 ð1 þ ’Þ þ Fr11 The compatibility condition is now written as Fl10 ð1 þ ’Þ þ Fr10 þ X1 ½Fl11 ð1 þ ’Þ þ Fr11 þ X1 ½Fl11 ð1 þ ’Þ þ Fr11 ¼ 0 and, taking into account the initial condition, which naturally remains valid, Fl10 ’ þ X1 Fl11 ’ þ X1 ½Fl11 ð1 þ ’Þ þ Fr11 ¼ 0 and therefore X1 ¼

’ ðFl10 þ X1 Fl11 Þ Fl11 ð1 þ ’Þ þ Fr11

From the numerator in the above equation it may be observed that if the (absolute) value of the bending moment at the internal support (i.e. X1) is smaller than that for the lefthand fixed simply supported beam, then X1 has the same sign as X1, i.e. the bending moment at the support increases. This becomes immediately clear in the case where the right-hand beam is unloaded. The system will then have an identical bending action to the frame ABC in Figure 5.11, where the column is made of steel. The reduction in the effective elasticity modulus of the loaded beam due to creep causes (see Section 3.2.9) an increase in the bending moment at the support (see Figure 5.11). The last equation may also be applied in any other case where there is cooperation between concrete (C) and steel (S) members (as shown in Section 6.2.1 also). In this case it is more convenient to write the above equation in the form X1 ¼

’ ðFC10 þ X1 FC11 Þ FC11 ð1 þ ’Þ þ FS11

g

∆X1 The vertical member does not creep A

B

The bending behaviour of the frame is identical to that of the corresponding continuous beam C

Figure 5.11 Moment redistribution due to creep

202

Initial diagram of bending response Final (redistributed) bending response due to creep

X1 X1

∆X1

Continuous beams δ

X1: Instantaneously developed reaction δ X1

Deformation due to instantaneous imposition of settlement

Additional deformation due to creep which is eliminated by ∆X1 ∆X1

Reduction in the instantaneously developed reaction

Initial diagram of bending response Final bending response due to creep

Figure 5.12 Relieving creep effect on the state of stress due to an instantaneous settlement

It is known that the imposition of loading on a statically indeterminate structure causes a state of stress that depends only on the ratio of the rigidity EI of its members and not on their absolute values. However, if a deformation is imposed (support settlement or temperature change), the absolute value of EI (i.e. the bending stiffness of each member) plays a predominant role. In such cases it is logical that the redundant forces change over time because of creep, as can be shown using the same example of the two-span continuous beam. In the continuous beam shown in Figure 5.12 a settlement is imposed on the internal support. In the primary system of the simply supported beam, the reaction X1 is considered as the redundant force. The value acquired instantaneously by the reaction X1 (downwards) is determined from the compatibility equation X1 F11 ¼

(i.e. X1 ¼ /F11)

where F11 is the displacement caused by the unit force X1 ¼ 1. This reaction decreases with time, and acquires a final value of (X1 þ X1). The reduction by X1 is determined by satisfying the compatibility condition at the final time phase, as follows. The displacement of the reaction application point at the final time phase results from the superposition of the following two effects: the deformation due to the application over a long time of the force X1; and the deformation due to the gradual imposition over the same time interval of its variation X1. The first deformation is equal to X1 F11(1 þ ’), and the second deformation is equal to X1 F11(1 þ ’). Thus, X1 F11(1 þ ’) þ X1 F11(1 þ ’) ¼ 203


Combining this with the preceding relationship gives ’ X1 ¼ X1 1þ’ and therefore the final value of the internal reaction (definitely decreased) is ’ ðX1 þ X1 Þ ¼ X1 1 1þ’ It is clear that the value in parentheses applies to both the bending moments and the shear forces of the beam. This reduction is very pronounced (see Figure 5.12). For typical values of ’ ¼ 2.5 and ¼ 0.80, the final reaction is 16% of X1. Note here the absolute correspondence with the phenomenon of ‘relaxation’ (see Section 1.2.2.2). Indeed, the above expression is identical to the one derived in Section 1.2.2.2 for the final stress of concrete. It is not unusual for the settlement to be imposed on the structure gradually, starting from a zero value. This occurs in cohesive soils (e.g. clay), in which settlements develop gradually (see Section 17.1.2). It is obvious that, in this case, no initial reaction X1 exists. However, a reaction is being gradually developed, acquiring its final value X1 when the process of settlement growth is completed (Figure 5.13). If accepted that this process follows the time function of creep deformation, then the above compatibility condition may be written as X1 F11(1 þ ’) ¼ Considering again, as previously, the value X1 ¼ /F11, which corresponds to the initial reaction resulting from an instantaneous imposition of a settlement , it follows that X1 ¼ X1

1 1þ’

In this case the developing reaction is 33% of X1.

δ Gradually imposed settlement Gradually developed reaction which finally causes δ on the statically determinate beam ∆X1

Bending response due to instantaneous settlement Developed bending response due to creep

Figure 5.13 Relieving creep effect on the state of stress due to a gradually imposed settlement

204

Continuous beams

The above results are generally valid for any type of imposed deformation and for any degree of redundancy as long as the creep is identical in all the members in the structure. Thus, for an instantaneous imposition of a deformation, a reduction factor applied to the instantaneously developing redundant forces is ’ 1 1þ’ whereas for a gradual imposition of a deformation the ratio of the developing sectional forces to those corresponding to instantaneous imposition is equal to [1/(1 þ ’)]. It may be understandable that in all the above cases, the redistribution of stresses caused by creep under service conditions are statically equivalent to superposing a self-equilibrating state of stress to the initial one and therefore does not affect the ultimate state checks, according to the static theorem of plastic analysis (see Section 6.6.2).

5.5.2

Change of structural system

For the construction of continuous beams in bridge engineering with a certain number of equal spans, it is customary to use precast girders for each span that are first treated as simply supported beams and then treated with their continuity restored in the regions of the supports by means of a monolithic connection (Figure 5.14). It is clear that at the instant of connection the bending moment diagram for the system is composed of the g

g

F10 2 · F10: Initial relative rotation (2 · F10) · ϕ: Increment of relative rotation in the free condition Eliminated by the developed bending moment ∆X1 = 1 F11 ∆X1 2

g · L /8

Initial bending diagram Final bending diagram

Figure 5.14 Redistribution of the bending moments due to the change of structural system

205


diagrams for the simply supported beams. However, as time passes and due to the creep properties of concrete, bending moments are developed at the supports, and these tend to reach the values that correspond to a continuous system initially constructed monolithically, with a consequent corresponding reduction in span moments. The reason for the development of this state of stress, as shown by the example of a continuous beam with two equal spans in Figure 5.14, is that the initial rotation F10 at the two ends of the internal supports will not remain constant over time in the free system, but will increase until the final time phase by the value F10 ’. This increase, however, is prevented by the restored continuity, through the gradually developing moment X1. This moment considered alone causes at each end a rotation with a final value X1 F11(1 þ ’), where F11 is, as previously, the instantaneous rotation of each end due to X1 ¼ 1. The compatibility of deformations requires that the relative rotation at the internal support is zero: 2 F10 ’ þ X1 2 F11(1 þ ’) ¼ 0

(see Figure 5.14)

Introducing the moment value X1 ¼ 2 F10/2 F11 that corresponds to the monolithically constructed system (due to the self-weight), it results that ’ X1 ¼ X1 1þ’ For the values of ’ and already used in previous sections, it is found that the finally developed moment at the support is 83% of the corresponding ‘elastic’ moment. g A similar situation also exists when, in order to construct a bridge of a large span over a difficult natural obstacle (deep canyon, river, etc.) the two half parts of the bridge are purposefully constructed as cantilevers, as mentioned earlier in this chapter, and then connected together (balanced cantilever method). In such a case, which is shown in Figure 5.15, it is clear that, as soon as continuity is restored at the connection, the diagram of bending moments for the system is composed of that for the two cantilevered parts. However, at the point of connection a bending moment develops in time, which brings tension to the bottom fibres and acquires a significant percentage of the moment corresponding to the ‘monolithic’ system. It is obvious that, for the maintenance of equilibrium, the moments at the supports are correspondingly decreased. The analysis is precisely the same as in the last problem (see Figure 5.15). The relative rotation (2 F10) of the two ends in a hypothetical free state, up to the final instant, would increase by (2 F10 ’). This, however, is prevented by the gradual increase in moment X1, which causes a relative rotation X1 (2 F11) (1 þ ’). Thus, the compatibility equation remains exactly the same as previously, and the result is identical. The ratio X1/X1 of the developing bending moment X1 to the span moment X1 that would be developed in a monolithic system is equal to ’/(1 þ ’). Of course, since, until the two free ends approach each other, the concrete has already reached some age, the value of factor ’ is now smaller (see Section. 1.2.2.1). Therefore, a smaller percentage of the span moment of the monolithic system develops, this being of the order of about 60%. It is understandable that such a bending moment should be 206

Continuous beams

F10 2 · F10: Initial relative rotation 2 · F10 · φ: Free increment of relative rotation Eliminated by the deformational contribution of the developed bending moment ∆X · (2 · F11) · (1 + µ · ϕ) F11 ∆X = 1

∆X

∆X: Developed bending moment after the junction ∆X

∆X

∆X

Bending diagram M just after the junction Bending diagram M just after a long time period

Figure 5.15 Redistribution of the bending state of stress in a free cantilever construction

taken up by an appropriate layout of tendons, placed on the bottom side (‘midspan tendons’). g The last problem considered here is again concerned with bridge engineering (Menn, 1990). The structure considered is an a posteriori cast-in-place concrete deck plate on precast prestressed concrete girders, already placed parallel to each other at equal distances (Figure 5.16). Over the course of time, the initially unstressed plate will develop under its total self-weight a stress state due to creep of the beam. Obviously, if the beams are made of steel — i.e. they do not exhibit creep — this state of stress will not be developed. The redistribution of stresses is due to fact that the top fibres of the beam will continue to shorten even after the final casting of the top plate, and, therefore, the stresses will drift, with a proportional shortening — i.e. compression stress state — of the bottom fibres of the top plate as well. The active cross-section consists then of the beam and a certain ‘effective’ width of the top plate. 207

Structural systems: behaviour and design ∆εPl

Apl

N0Pl = 0 a

∆N Pl a

eB AB

N0B

∆εB

M0B

∆N B ∆M

B

The variations in the internal forces are self-equilibrating

Figure 5.16 Redistribution of the internal forces in a concrete composite section

At the moment of restoring of the full section, the total self-weight is taken only by the beam. The beam is, therefore, subjected to the total self-weight and prestressing forces. In the characteristic midspan section, there is a compressive force NB0 on its centroid and a moment MB0 that causes tension in the bottom fibres. As time passes, a compressive force NPl develops in the top plate, as already mentioned, which of course will be accompanied by a corresponding change NB in the compressive force of the beam and a change MB in its bending moment (see Figure 5.16). Due to the much smaller bending stiffness of the top plate compared with that of the beam, it can be assumed that its developing bending MB will, in the course of time, become negligible. These last three unknowns are determined through the compatibility of deformation, and through the equilibrium of the cross-section. Compatibility of deformation means that the variation in the strain "B in the extreme top fibre of the beam is identical to the strain "Pl in the bottom fibre of the plate, while equilibrium of forces in the cross-section imposes that the variations in the sectional forces in the two parts constitute a self-equilibrating system. Thus NB0 MB0 þ eB EB AB EB IB and, given that both "B and "Pl are gradual changes, "B0 ¼

"Pl ¼ B

" ¼

NPl ð1 þ Pl ’Pl Þ EPl APl "B0

NB MB B ’ þ þ e EB AB EB IB B

! ð1 þ B ’B Þ

The differentiated values of and ’ are due to the fact that the two concrete parts have different ages at the time when creep is considered to start. The previously stated conditions for the determination of the three unknowns NPl, NB and MB are written as: "B ¼ "Pl 208

Continuous beams

NPl ¼ NB MB ¼ a NPl where a is the distance between the centroids of the plate and the beam cross-sections. As expected, it is found that the distribution of normal stresses on the full crosssection tends towards the one that would be developed if the total cross-section was uniformly cast (and prestressed) from the beginning.

5.6

Composite beams

5.6.1

Basic shear function

The concept of using a composite beam to transfer a transverse load over two supports lies in the utilisation of a concrete zone of sufficient thickness to carry the compressive force, and of a steel beam to carry the tensile force, both forces arising from the bending action of the so-formed composite beam (Figure 5.17). As examined in Section 4.1.1, this can be accomplished only under the condition that the horizontal — and consequently also the vertical — shearing cooperation between the two parts is ensured. Thus, if the longitudinal equilibrium of the upper part of the beam extending from the free end up to the midlength of the beam span is considered,

M=0

max M

max D [M]

max M

M=0

max Z

Figure 5.17 The equilibrium between internally developed actions

209


a significant compressive force at the right-hand end will be ascertained, which can only be balanced by the development of shearing forces (stresses) on the top flange of the steel beam. This is accomplished by attaching (welding) special steel connectors, in the form of headed studs, to the top flange of the steel beam at the steel—concrete interface. The connectors can transmit the horizontal forces, with sustainable stresses in the concrete flange (see Figure 5.17). Moreover, in the regions of the internal supports of a composite continuous beam, the top concrete flange is under tension, and accordingly an appropriate reinforcement must be placed near the top surface of the flange. It is obvious that, in the same region, the shear connectors located between the support and the point where the moment vanishes are supposed to offer (and take) the tensile force developed by the reinforcement at the support section (see Figure 5.17). The aim of the subsequent elastic analysis is to determine: first, the normal stresses developed in the cross-section; and, secondly, the horizontal shearing forces borne by the shear connectors.

5.6.2

Bending behaviour of composite beams

In order to determine the normal stresses due to bending, it is essential to consider an equivalent section made solely from steel, by dividing the concrete section area Ac by the ratio n ¼ Es/Ec, also known as the ‘modular ratio’ (see Section 4.2.2.1). This is explained by the fact that both a concrete cross-section Ac and a steel cross-section Ac/n develop the same force for the same imposed strain ", as Ac " Ec ¼ (Ac/n) " Es (Figure 5.18). If Ast, As and Ac are the cross-sectional areas of the steel beam, the reinforcement and the concrete, respectively, and if z are the corresponding centroidal distances from the bottom surface of the steel beam, then the distance zm of the centroid of the total crosssection will be zm ¼

1 ½A z þ As zs þ ðAc =nÞ zc Am st st

As

Compression σc

Ac h

Ast z

Ac /n: Statically equivalent steel section n = Es/Ec σst Tension Stresses due to the bending moment

Figure 5.18 Composite cross-section

210

M

Continuous beams

where Am ¼ [Ast þ As þ (Ac/n)] Moreover, if Ii0 is the moment of inertia of each part of the cross-section, by employing Steiner’s theorem the moment of inertia Im of the total cross-section with respect to the centroidal axis will be P P Im ¼ Ii0 þ Ai (zizm)2 It is obvious that the centroidal axis also constitutes the neutral axis of the total crosssection. Then, the stresses will result from the simple bending formula; for the concrete part they should be divided by n. Thus, for a bending moment M causing tension at the bottom fibres st ¼

M z Im m

s ¼

M ðzs zm Þ (compression) Im

(tension)

1 M c ¼ ðh zm Þ n Im

5.6.3

(compression)

Shear transfer

The fact that composite beams are necessarily constructed in two stages is quite important. First, the steel beam is set in place, and then the top concrete plate is cast. In the concrete-casting phase the steel beam is acting alone, simply carrying the weight of the concrete. In this phase it must be ensured that there is no noticeable deformation of the beam under its own weight and that of the concrete, possibly by providing suitable intermediate temporary supports. Only after the concrete has hardened are the intermediate supports removed, and any further loading can then be borne by the composite steel—concrete system. The developing stresses can then be calculated according to the above formulae. Naturally, an essential condition for the operation of the composite system is the appropriate take-over of the horizontal shearing flow. This consists, as previously explained, of the horizontal force vel per unit length being transmitted between the two materials at the steel—concrete interface, and is equal to the product of the considered intersurface width (i.e. the width of the top flange) and the corresponding shearing stress: vel ¼

V Sc Im n

The value Sc is the first moment of inertia of the concrete cross-section with respect to the neutral axis of the composite cross-section previously determined, so that the value (Sc/n) corresponds to the assumed equivalent ‘steel section’. 211


[V]

vel The distribution of the longitudinal shearing forces is identical to that of the transverse ones

Figure 5.19 The transfer of the longitudinal shearing forces

It should be checked, of course, that the above force, which varies along the length of the beam according to the corresponding value of the shear force V (Figure 5.19), can be taken over by the corresponding connectors located in the appropriate positions. In the case where there is no need to use intermediate temporary supports, the aforementioned longitudinal shearing forces will be developed only for those loads applied after the concrete hardening. The resistance that a shear connector can offer is constituted by the shear resistance VRs developed in its shaft and the local resistance VRc of the concrete, which receives by reaction the same force through the connector shaft, mainly through its specially formed head, as previously mentioned. Thus (according to indicative regulatory requirements) VRs ¼ 0.7 fuD AD where fuD is the tensile strength of the connector material and AD is the cross-section of the connector ( ¼ d2/4, where d is the diameter of the connector), and pffiffiffiffiffiffiffiffiffiffiffiffi VRc ¼ 0:35 d2 fc Ec Naturally, the minimum of these two VR values is the effective one. The strength of the stiffeners is checked using the condition that the total longituÐ dinal shear force V1 ¼ vel dx between points with zero and maximum bending response (see Figure 5.17) is safely taken by the locally available number m of connectors: V1 =m VR =R

5.6.4

ðR ¼ 1:1Þ

Construction stages — stress control

Once the shearing cooperation between the two parts of the composite beam has been established, it is necessary to understand the mechanism of the development of stress, 212

Continuous beams

Concrete after hardening remains free of stresses

Figure 5.20 Restoration of the top flange of the beam with concrete

both during the construction phases and thereafter during service. As noted previously, the construction process plays an important role, and must take into account the creep properties of the compressed concrete. When the finally formed composite beam is loaded with a permanent action, the stresses developed due to this loading will change over time. As shown in Section 1.2.2, the elasticity modulus of concrete decreases over time to the value Einit and, accordingly, the ratio n increases (see Section 5.6.2). In order to understand the implication of this, three cases of constructing a simply supported beam are examined. . A steel beam is without any intermediate supports, loaded by its self-weight and the weight of the still fresh concrete (Figure 5.20). Clearly, even after the concrete has hardened and shear cooperation has been established, the concrete bears no compressive stresses and, provided that no additional load is applied, this stress state will not change over time. If, however, an additional permanent load is applied immediately after the hardening of the concrete, this will cause compressive stresses in the concrete, which will decrease over time. After a long time the state of stress will be equivalent to the superposition of the initial one (where only the steel beam was stressed) and the additional one, which is of course calculated on the basis of the ratio n ¼ Es/Einit. . The provisional support of the steel beam at two points. In the concrete-casting phase, the steel beam acts as a three-span continuous beam under its self-weight, and the weight of the fresh concrete, developing reactions at the internal supports equal to A (Figure 5.21). After the concrete has hardened, the effective section of the beam changes, but the reactions A do not change because the stiffness ratio is maintained. The removal, afterwards, of the intermediate supports is equivalent to the application of the forces A, with the opposite sense to the previous ones, on the restored simply

Insertion of temporary supports

A

A

Reactions remain the same after hardening of concrete Removal of temporary supports

A

A

The compressive forces of concrete keep reducing with time

Figure 5.21 Provisional support for the restoration of the function of the top flange

213

Structural systems: behaviour and design Insertion of intermediate supports: imposition of upward displacements Concrete free of stress also after hardening

Reactions unchanged after concrete hardening

A + AP Removal of intermediate supports

A

A

AP

AP

Due to self-weight Due to imposition of displacements A + AP

Additional loading beyond self-weight Additional compressive concrete stresses

Figure 5.22 Prestressing of the top flange via temporary intermediate supports

supported beam. This means that, according to the above, the resulting stresses (compressive in the concrete and tensile in the steel) will change over time, the final values being determined on the basis of the ratio n ¼ Es/Einit. Clearly, each additional permanent load applied after the hardening of the concrete will cause stresses that can be calculated in the same way, whereas a transient live load will cause stresses that can be calculated on the basis of the modular ratio n ¼ Es/Ec. . Providing the concrete with an initial compression force (i.e. prestressing). The steel beam is provided with two temporary supports at the points where the reactions A develop. Just before concreting, an equal upward displacement is imposed on each support. It is obvious that in this way the moments at the supports of the steel beam are increased, as are the corresponding reactions which are increased by AP (Figure 5.22). It is understandable that, after concrete casting and until concrete hardening, the total reactions at the intermediate supports will be exactly the same and equal to (A þ AP), where A includes also the beam reactions under the self-weight of the concrete. No stresses have yet developed in the concrete. Next, the removal of the intermediate supports can be treated as previously, where the permanent forces (A þ AP) are acting with opposite sense on the restored simply supported composite beam (see Figure 5.22). In this way, an additional compressive stress is offered to the concrete, which is irrelevant due to the permanent loads acting on the beam. The thus caused stresses will again change over time and reach final values based on the ratio n ¼ Es/Einit, with final compressive stresses in the concrete that are smaller than the initial ones. As it may be seen, in the last two cases, despite the various construction measures, no additional final sectional forces are introduced, besides those caused by the self-weight g. Therefore, the normal stress diagram in each case is always equivalent to a bending moment (g L2/8) and a zero normal force. 214

Continuous beams

5.6.5

Temperature change

There are two good reasons for considering the state of stress that develops in a composite beam due to a temperature difference of its two constituent parts, i.e. concrete and steel. The first reason is that, due to the smaller thermal conductivity of the concrete, an exterior temperature change causes a short-term differential temperature between the steel and the concrete. The second reason is the concrete shrinkage, which is prevented to some degree by the steel and which, as mentioned in Section 1.2.2.1, is equivalent to an imposition of a fall in the temperature of the concrete. The implications for the response of a composite beam of a temperature difference can be assessed on the basis of equilibrium and compatibility considerations (Figure 5.23). For a fall in temperature of T in the concrete, for example, the concrete is left free to shrink by "c ¼ aT T, and a tensile force NT is then applied to restore the section (Ac/n) of ‘concrete’ to its initial state: NT ¼ ("c Es) (Ac/n) This force causes a corresponding tensile stress in the concrete, while leaving the steel part unaffected. Given, however, that no external force is applied to the section, an equal force with opposite sense (compressive) must be applied at the same point on the total (composite) section. This force, referred to the centre of gravity of the section, is accompanied by a bending moment that causes tension in the bottom fibres, equal to MT ¼ NT ac where ac is the distance between the centre of gravity of the composite section and the concrete part. The superposition of the two states contains the tensile force NT acting on the concrete alone, and the compressive force NT and the moment MT acting on the composite section. The combined effect is the self-equilibrating stress diagram shown in Figure 5.23. ∆T

N∆T

N∆T +

ac N∆T

–

M∆T +

Free shortening

M∆T

ϕ

Equivalent actions

ϕ

Final stresses

M∆T

Simply supported beam: constant bending response, development of curvature

Figure 5.23 The state of stress in a simply supported beam due to a temperature difference

215

Structural systems: behaviour and design Additional actions required to eliminate the rotations M∆T M∆T

M∆T M∆T

Fixed-ended beam: null bending response, null curvature

Figure 5.24 The effect of a temperature change on a fixed-ended beam

This moment (as well as the above stress diagram) is constant along the length of the beam, and causes a constant curvature 1/r ¼ MT/(Es Im), as well as an equal rotation ’ at both beam ends: ’¼

L MT L ¼ 2 r 2 Es Im

(The same result is, of course, also obtainable via the principle of virtual work.) g With regard to a fixed-ended beam — subjected only to this temperature change — the above state of stress of a simply supported beam must have superposed on it two appropriate equal and opposite moments applied at the beam ends, so that the above rotations vanish. These moments are obviously equal to MT and, consequently, they eliminate the curvature of the beam, leaving it straight (Figure 5.24). The state of stress of a fixed-ended beam is thus composed of the superposition of the force NT acting only on the concrete part, and the compressive force NT acting on the composite section. The stresses that result are uniform in both the concrete and the steel section, and are given by, respectively, c ¼

NT N Ast T ¼ T T Es Ac n Am n Am

st ¼

NT Ac ¼ T T Es Am n Am

(tensile)

(compressive)

Thus, with regard to the internal spans of a continuous beam, this last result constitutes a satisfactory approximation. g With regard to a fixed simply supported beam, the support moment, which must cancel the aforementioned rotation ’, is obtained according to the principle of virtual work, and is equal to 1.5 MT. This moment is superposed on the bending moment diagram of the simply supported beam and leads to the diagram shown in Figure 5.25. The stresses that develop in the section are calculated through superposition of the imposed axial forces NT, as above (see Section 5.6.2). Thus, for a moment M causing tension in the bottom fibres, the above stresses of the fixed-ended beam are superposed, for the concrete, on the compressive stress: 1 M c ¼ ðh zm Þ n Im 216

Continuous beams Additional moment for the elimination of rotation M∆T

1.5 · M∆T M∆T

Fixed simply supported beam: variable bending response 0.5 · M∆T

M∆T

Figure 5.25 The effect of a temperature change on a fixed simply supported beam

For the steel they are superposed on the tensile stress: st ¼

M z Im m

This state of stress can approximate to a satisfactory degree the one corresponding to the end spans of a continuous beam.

5.6.6

Continuous beam behaviour — prestressing

As pointed at the beginning of this chapter, in the case of a continuous beam and at the region of internal supports, the concrete is under tension and needs to be reinforced. However, this does not prevent the concrete from cracking. The calculation of the stresses due to a bending moment causing tension in the top fibres is performed, according to Section 5.6.2, by considering the total cross-section without the concrete part, i.e. Ac ¼ 0. It is obvious that in this case the neutral axis will be within the steel profile. The possible case of a cantilever is treated in the same way. Checking of the shear connectors can again be done, as for the case of the simply supported beam, on the basis of the relationship Vl/m VR/ R where in this case the longitudinal shearing force Vl is equal to the maximum force developed in the reinforcement at the corresponding internal support position (Figure 5.26). g Cracking under service conditions can be prevented if an analogous construction procedure to the one described in Section 5.6.4 is followed, in order to offer the concrete an initial compressive force (prestressing). More specifically, in the example of the two-span beam shown in Figure 5.27, an upward shift is imposed on the internal support before concrete casting. As a result, the beam develops, besides the bending moment due to the self-weight of the fresh concrete, an additional triangular bending diagram causing tension in the top fibres, resulting from the ‘statically 217


(M = 0)

(max M)

D

(min M )

(M = 0)

Z = f s · As

Total force Vl (=D)

Total force Vl (=Z)

Figure 5.26 The equilibrium of the internal forces in characteristic positions

determinate’ simply supported beam loaded with an upward force equal to X0. After casting, hardening of concrete and restoration of the composite section, while the concrete remains free of stress, the same shift is imposed downwards on the internal support, so that a uniform level applies for all support points. It is clear that, in this second phase, bending moments opposite to those in the first phase will be developed along the whole length of the beam, causing compression of the concrete plate. These bending moments will be greater than those in the first phase, as the composite section is now strengthened with respect to the previous one, where only the steel part existed (see Section 3.2.5). In fact the bending moments can be so large that the resulting compression cancels the corresponding tension due to the additional permanent and live loads. Obviously, the resulting (triangular) bending moment diagram is due to the downward reaction X1 (being greater than X0), acting on the primary system of the simply supported beam (see Figure 5.27). The development of compressive stresses in the concrete plate will cause creep, which leads directly to an undesired decrease in the plate compression stresses due to the decrease in reaction X1. Note that, while the concrete plate is stressed only because Self-weight of fresh concrete plus steel beam

Concrete free of stress before imposition of downward displacement

δ

δ Bending response due to imposition of displacement

X0

X1

Reaction developed X 1 > X0

Reaction developed Bending moment diagram causes compression of the concrete plate

Reaction due to self-weight

(X1 – X0): Reaction due to imposition of the two consecutive displacements

Figure 5.27 The introduction of prestressing in the top flange of a continuous beam

218

Continuous beams X1 = 1 (t = ∞) F i1n1it

∆X1 = 1 (t = ∞) F1d1if

Figure 5.28 The effect of creep on the development of the final stress state

of X1, which decreases over time, the steel beam is stressed by the superposition of the continuous beam action under the total self-weight and the application of a downward force X1 X0 on the simply supported system, which constitutes the additional reaction of the continuous beam due to the two imposed phases. However, this force decreases over time following the decrease in X1 (see Figure 5.27). This decrease can be estimated using the analytic approach described in Section 5.5.1 via the force method, as this procedure is always adequate for dealing with creep problems due to a redistribution of forces, by following the compatibility of deformations. Thus, the developed force X1 and its variation X1 are applied to the ‘statically determinate’ simply supported beam for the whole time period examined (theoretically infinite) (Figure 5.28). The resultant deformation due to the two actions must obviously be equal to . According to Section 1.2.2, the final shift due to X1 ¼ 1 is denoted by Finit 11 , which is obtained using the ratio n ¼ Es/Einit, while the final shift due to X1 ¼ 1 is denoted by Fdif 11 , and this is obtained using the ratio n ¼ Es/Edif . Thus (see Section 5.5.1) dif ¼ X1 Finit 11 þ X1 F11

It should be noted that the developed reaction X1 is initially caused by the ‘instantaneous’ imposition of downwards, or — in other words — the application of force X1 in the simply supported beam instantaneously causes the shift . The value of X1 is given by X1 ¼ /F11 where F11 is obtained the basis of the modular ratio n ¼ Es/Ec. It is clear that init F11 < Fdif 11 < F11

The two last equations lead to the result that the final value (X1 þ X1) of the reaction X1 is equal to ! Finit 11 F11 (reduction) ðX1 þ X1 Þ ¼ X1 1 Fdif 11 Thus the reduction factor of the initial compressive stress of the plate (due to X1) is equal to Finit F 1 11 dif 11 F11 219


The final bending moment of the steel beam is obtained by superposing on its continuous beam action under the gravity loads a concentrated downward force on the ‘simply supported’ system (see Figure 5.27) that is equal to ! Finit F X1 1 11 dif 11 X0 F11 It is clear that the required prestressing of the concrete part should be calculated for the service condition on the basis of the above reduction factor.

5.6.7

Plastic analysis

Although elastic analysis is commonly used for the design of composite beams, it is useful to consider some basic characteristics of their plastic design. First, it should be pointed out that the plastic design of composite structures is not affected at all by any stress redistribution that may occur under service conditions. g With regard to the simply supported beam, exceeding the bending resistance Mpl of a section means collapse of the structure. The computation of the bending resistance Mpl requires the determination of the neutral axis position, so that the compressive limit force developed in the concrete above the neutral axis is equal to the limit tensile force in the steel below it. At this point the stresses are considered to have reached their maximum limit value in all fibres of the section, that is fc for concrete and fy for steel (Figure 5.29). Assuming that the neutral axis lies within the concrete section at a distance x from the extreme fibre with an effective width beff , the compressive force in the section is equal to D ¼ fc beff x beff D

fc

x a

Mpl

Z Span region

fy fsy

Z fy a D Support region

Figure 5.29 Plastic analysis of a composite section

220

Mpl fy

Continuous beams

The tensile force Z results from the entire steel sectional area As: Z ¼ fy As The distance x is determined through the relationship D ¼ Z, and hence, the position of the neutral axis is also verified. In the uncommon case where the resulting value of x is higher than the thickness of the concrete plate, the procedure is repeated in the same manner. The plastic moment Mpl is given by Mpl ¼ Z a where a is the distance between the points of application of the forces D and Z. With regard to shear, the ultimate shear force Vpl is of course carried only by the crosssection Aw of the steel web and, as described in Section 4.1.1, it is equal to fy Vpl ¼ pffiffiffi Aw 3 For a continuous beam — and analogously the cantilever — the computation of the bending resistance of the section at the internal supports where the concrete is completely cracked requires the determination of the neutral axis position in a purely steel section. In this case, the neutral axis lies within the steel section, so that the sum of the tensile forces in the reinforcement and the corresponding top part of the beam is equal to the compressive force of the bottom part. It is obvious that the limit stress fy prevails along the whole height of the section. The plastic moment Mpl is calculated accordingly (see Figure 5.29). g With regard to the check of the shear connectors, the only difference from the elastic analysis (see Section 5.6.3) lies in the determination of the total longitudinal shear force Vl between points with zero and maximum bending stresses (i.e. at plastic hinges). In regions where the bottom fibres are in tension due to bending, the shear force Vl is equal to the tensile force of the steel beam fy Ast , provided that the neutral axis of the section lies within the concrete section. In the uncommon case where the neutral axis lies even lower, the force Vl is obviously equal to the compressive force of the concrete section fc Ac. Moreover, it is obvious that in regions where the top fibres are in tension due to bending, the force Vl is equal to the reinforcement tensile force fsy As. The corresponding check of the available m shear connectors placed in the examined region is performed through the requirement V1 m VR

References Menn C. (1990) Prestressed Concrete Bridges. Basel: Birkhaüser Verlag.

221

6 Frames 6.1

Overview

A frame is a skeleton structural system consisting of beams and columns designed to carry loads of various nature (gravity as well as horizontal loads) and to transfer them to the ground. In this chapter, plane frames will be considered, which are loaded only in their plane. Frames are either designed to cover or bridge a specific span, characterised as ‘one-storey frames’, where the girder of the frame may not always be a straight member, or may constitute basic elements of a space system carrying superimposed floors in a building, and therefore the girders placed on more than one level are necessarily horizontal and straight. In this last case, the member system bearing the floors is a space system, but in the overwhelming majority of buildings the whole structural arrangement is such that the space system can be considered as a group of multi-storey plane frames connected with each other (see Chapter 15).

6.2

Single-storey, single-bay frames

These usually consist of a horizontal beam (girder) and two columns, which are connected to the two ends of the beam and are based on the ground.

6.2.1

Design for vertical and horizontal loads

The simplest form of such a structure is the post-and-beam frame, where the columns are fixed to the ground and the beam is simply supported on their tops. Thus, for vertical loads the beam acts as a simply supported beam, transferring its vertical reactions to the column tops, without inducing bending moments in them (Figure 6.1). It is obvious that if two cantilevers are set up on the left and the right side, the bending moment of the girder will be decreased, thus resulting in a statically more favourable structure. If, indeed, the length of each cantilever is about 35% of the length of the central span L then the span and support moments for a load q will be equal to q L2/16, which is half of those for the simply supported beam. In the case where a cantilever is placed only on one side, the length of the cantilever must be equal to 41% of the central span in order to have again equal span and support moments with a value of q L2/11.9. For a horizontal load W acting at the level of the girder, it is obvious that the beam remains unstressed in both bending and shearing (see Figure 6.1), while the whole load Structural systems: behaviour and design 978-0-7277-4105-9


Structural systems: behaviour and design q

q

Without bending q · L2/16 h

2

q · L /8

L

0.35 · L

L

0.35 · L

q

W

Without bending

q · L2/11.9 V = +(W/2)

(W/2) · h

V = +(W/2)

(W/2) · h

0.41 · L

L

Figure 6.1 Set-up of post-and-beam frames

is carried equally by the two columns (provided that they have the same stiffness, see Section 3.2.10), thus acting as two cantilevers with a horizontal load equal with W/2 applied at their tops. Columns usually connect monolithically to the girder. Then, the frame is formed connected either to the columns hinged to their foundation or fixed (Figure 6.2). The foundation is considered as a solid massive concrete element which is based directly on the ground, being rigidly connected to it and aiming, through its larger dimensions, to transfer much smaller stresses to the ground than those developing at the column base. The static conditions for their design will be examined in Chapter 17. In the case of a hinged connection at the foundation, the frame is either statically determinate, if it has an additional hinge (usually at the midpoint of the girder), or it has its girder continuous, in which case it is one-time statically indeterminate. In the case where the columns are fixed to the foundation and the girder is continuous, the system is statically three-times indeterminate. g Since it is useful to examine the bending of the above-considered frames, also in correlation with the so-called pressure line, the following should be recalled (see Section 2.2.7). For any vertical load q(x) that must be taken up by a structure transferring the loads only through axial forces, i.e. without bending (and shear), at two points on the ground lying on a horizontal line at a distance L and enclosing its projection, there exist infinite solutions regarding the axis of the structure. All these solutions constitute affine curves of the bending moment diagram M0(x) of a simply supported beam having length L and loaded by q(x) (Figure 6.3). It is obvious that q(x) can also include concentrated loads or 224

Frames

Three-hinged frame

Two-hinged frame

Fixed frame

Figure 6.2 One-bay rigid frames

even be composed only of such loads. Thus, if y(x) represents the curve which the structure should follow, then y(x) ¼ k M(x)0 The choice of a value for k means the choice of a specific curve, i.e. a specific pressure line. This value for k is the inverse value 1/H of the horizontal component H of the reaction R of the system, which will of course have the direction of the tangent at the beginning of the selected curve (see Figure 6.3).

k = 1.20 k = 1.0

y

k = 0.55 A

B

A

B x

M0

y(x) = k · M0

Pressure lines for the given loading

k S2 S1

y0

H

∆x

A RA

H

x0

B H = 1/k

RB

The selection of a third point establishes the pressure line and the value of k The acting compressive force along the pressure line is not constant but its horizontal component H has a constant value

Figure 6.3 Characteristics and determination of a pressure line

225


The pressure line ensures at each point x the equilibrium between a corresponding vertical load q x and the two forces S(x x/2) and S(x þ x/2) that act tangentially at the two ends of the segment x. It is emphasised that the aforementioned axial (compressive) forces of the pressure line do not have a constant value, but they do have a constant horizontal component equal to H (see Figure 6.3). It is clear that with two support points, the determination of one more point y(x0) lying on the curve is required in order to determine the curve completely. Then, k ¼ y(x0)/M0(x0), and y(x) is already defined, so that the structure that follows this curve will develop only axial forces. The structure will develop compressive forces if the point y(x0) lies above the line connecting the supports (AB), and will develop tensile forces if it lies below the line. However, in this consideration the second case is not of much interest (see Figure 6.3). Now, if a hinge is inserted at the position y(x0) and the structure does not follow the specific curve but takes a different form while remaining a three-hinged frame, then it will develop bending moments at each point, which are the product of the distance of this point from the pressure line and the corresponding axial force acting on it (Figure 6.4). The compressive sense of the axial forces of the pressure line will provide a particularly useful result, i.e. it will indicate which fibres of the frame are under tension due to bending (see Figure 6.4). g Regarding the portal frame in Figure 6.5, it can be seen that the placement of a hinge at the midspan of the girder under a uniform load due to gravity causes tension exclusively at the outside fibres. The form of the moment diagram is not necessarily favourable. The girders are bent with the same moment magnitude (q L2/8) that is developed in the simply supported beam, and, moreover, the columns also bend with the same moment.

m

n

em

Rm

en

y0 Rn

RB

x0

A

B

RA

Mm = Rm · em

Mn = Rn · en

m

n em Rm

en Rn RB

A RA

B The bending moment at each point of the frame depends on its distance from the pressure line

Figure 6.4 Correlation between the pressure line and the developed bending

226

Frames

Independent of the height M 0G

W · h/2

W

G h

Pressure line

W/2

H = M 0G/h

W/2

Thrust inversely proportional to height

M 0G/(1 + f /h) G

f

Pressure line

h

H = M 0G/(h + f)

Figure 6.5 Load-carrying action of a three-hinged frame

It is obvious that the form of the pitched frame significantly decreases its bending response since it is adapted more to the pressure line. Note that the horizontal component H of the reaction of the frame is inversely proportional to its height; the moment at the joint, however, is independent of the absolute value of the height (see Figure 6.5). Of course, for a horizontal load W acting on the three-hinged frame (Figure 6.5), without taking into account the fact that the girder also participates in load bearing, the same bending moments are developed as those for the ‘cantilever’ shown in Figure 6.1. The developing moment at the column top is (W/2) h. If the girder is continuous, with regard always to hinged supports (Figure 6.6), the beam develops, for a vertical uniform load q, moments at its ends that cause tension on the top fibres. So, the maximum span moment is clearly decreased in comparison with that of the simply supported beam, since, as is known, this configuration is always ‘suspended’ from the points of the end moments. If the stiffness ratio ¼ (IR/L)/(IS/h) of the girder and columns is introduced, the expression for the exact value of the moment MB at the joint is (Salvadori and Levy, 1967) MB ¼

q L2 3 12 3 þ 2

For common stiffness ratios of the girder and columns (1.0 < < 2.50), the bending moment diagram of the girder takes a zero value (placement of hinges) at distances 227


q Equilibrium ~0.10 · L

Pressure line

0.10 · L

~0.10 · L

q

~0.045 · q · L2 IR h

0.40 · q · L IS

(~Direction of reaction)

[M ] H H

0.045 · q · L2/h L W

q · L/2 W · h/2

[M ]

W/2

W/2

Figure 6.6 Load-carrying action of a two-hinged frame

from the two ends in the range 0.13 L and 0.07 L, respectively. An intermediate value of 0.10 L allows, through the equilibrium of the imaginary two-hinged extreme part, an approximate computation of the horizontal component H ( ¼ 0.045 q L2/h) of the reaction and then of the moment at the top of the column (H h ¼ 0.045 q L2). In any case, the direction of the reaction does not differ substantially from the line connecting the hinges. Note that the pressure line passing from the supports as well as from the points of zero moments in the girder clearly shows the distribution of the bending response, according to the distance of each point of the frame from it, while the prevailing compression force has to increase towards the supports in order to always maintain a constant horizontal component. The bending response, due to a horizontal load at the girder level, does not differ from the corresponding three-hinged frame with a hinge in the middle, given the antisymmetric character of the system (see Figure 6.6). In cases where it is not desirable for the ground to take up the horizontal reaction H, a steel tension member (tie) is inserted in the frame, developing an axial force which depends on the ratio of its axial stiffness (Es As) to the bending stiffness (EI) of the frame (Figure 6.7). The structure has again one degree of statical redundancy. It can be seen that an increase (or reduction) in Es As with respect to EI causes an increase (or reduction) in the tensile force of the member and, consequently, of the bending 228

Frames

EI

EI After creep

Bending response before creep EsAs

P

P=H

Steel tie Increase in the tie axial force due to creep

Prestressing with the thrust force of the hinged frame leaves the bending moments unchanged

Figure 6.7 Frame with a tension member — the creep effect

moment at the column top, as well as of the axial compression of the girder (see Section 3.2.10). In the case where the frame is made of concrete, a redistribution of internal forces will of course take place due to creep. The concrete will present a decreasing E in time and, as a result, the tensile force in the steel member will increase, with the same consequences for the frame as above. The change in the tensile force X1 of the tie is determined through precisely the same expression that gives the change in the state of stress in the heterogeneous frame example given in Section 5.5.1, which is ’ ðFB10 þ X1 FB11 Þ X1 ¼ B F11 ð1 þ ’Þ þ FSt 11 All the underlying deformation magnitudes arise according to the method of forces, introducing the force X1 of the tie as the statically redundant unknown. Of course, if the tie is prestressed with the force corresponding to the horizontal reaction of the two-hinged frame, then no change in the whole state of stress will take place, since the frame will behave as a statically redundant system with hinged supports and uniform creep properties (see Figure 6.7 and Section 5.5.1). g In the case of a fixed frame (Figure 6.8), the moment MB at the ends of the girder is a little higher than the one corresponding to the two-hinged frame, that is (Salvadori and Levy, 1967), q L2 2 12 2 þ The points of zero moments are now at larger distance from the ends, that is between 0.15L and 0.09L, relatively to the values 1.0 < a < 2.50. As an intermediate distance, it may be considered roughly 0.12L. Moreover, since for a uniformly distributed vertical load the column top is simply rotating without shifting, it follows that the moment at the top of the column will be double than the one at the base (see Section 3.3.3), and consequently, the moment diagram of the column is zero at 13 of the height up from the base. MB ¼

229


Equilibrium ~0.12 · L

~0.12 · L

0.12 · L

Pressure line

q ~0.053 · q · L2

IR h

0.38 · q · L

2 · h/3

IS H

H

~0.026 · q · L2

H

(~Direction of reaction)

0.079 · q · L2/h

H

q · L/2

L

3 · W · h/16

W 3 · h/8 W/2

W/2

W/2

5 · W · h/16

W/2

Figure 6.8 Load-carrying action of a fixed frame

From the equilibrium of the extreme imaginary two-hinged part, it can again be realised that the direction of the force transferred through the bottom hinge does not significantly differ from the line connecting the hinges, which has, of course, a slope smaller than that previously encountered in the two-hinged frame. Now, given the constant value (q L/2) of the vertical component of the force transferred through the bottom hinge, it can be concluded that its horizontal component, which is obviously equal to the horizontal component of the reaction of the frame, is higher than that of the two-hinged frame. From the equilibrium of the above extreme two-hinged part, H ¼ 0.079 q L2/h, and, for the moment at the top of the column, H (2 h/3) ¼ 0.053 q L2. The moment at the fixed support is MV ¼ H (h/3) ¼ 0.026 q L2. In this case, the corresponding pressure line can again be considered to be passing through the column hinges, as well as the hinges in the girders, as shown in Figure 6.8. In the case of a fairly deformable soil, the foundation receiving this last moment rotates, and, as a result, the point of zero moments — i.e. the imaginary hinge — shifts downwards, approaching the base of the column. The frame then behaves more like a two-hinged one, causing a moment reduction at the end of the girder and, accordingly, an increase in the span moment. A horizontal load W acting at the level of the girder leads to a moment diagram, which becomes zero at midspan due to antisymmetric conditions, as well as at points nearly three-eighths of the column height from top. Thus, the top part of the structure acts as a three-hinged frame (see Figure 6.8). The horizontal components of the forces transferred 230

Frames

through the column hinges, and consequently the reactions of the frame, are equal to W/2, while the moments at the column base are equal to (W/2) (5 h/8) ¼ 5 W h/16, and the moments at the column top are equal to (W/2) (3 h/8) ¼ 3 W h/16. In the case of a fairly deformable soil, the imaginable column hinges will be shifted downwards, and, as a result, this latter moment at the top will increase, approaching the value W h/2 of the two-hinged frame, while the moments at the fixed base of the column will decrease, approaching the value (W h/2) of the two-hinged frame, while the moments at the fixed column base will decrease.

6.2.2

Lateral stiffness

In addition to the developing state of stress in the frame, also of basic importance is the assessment of its lateral stiffness, that is, the resistance that the frame presents to the imposition of a specific (unit) horizontal displacement at the joint, or — to put it another way — of its deformability (or flexibility), that is, of the produced displacement due to a specific (unit) horizontal force at the joint. As explained in Section 2.3.8, stiffness and flexibility are mutually opposing concepts. If represents the displacement under a force P and k is the stiffness, then, always, ¼ P/k. In the case of a frame with pinned supports, which is subjected to a horizontal force P (Figure 6.9), a shift is developed according to the expression ¼

P h3 2 þ 1= 12 EIS

corresponding to the following expression of stiffness k: EI 12 k ¼ 3S h 2 þ 1= It should be noted that in this shift the deformability of the columns plays a more significant role than that of the girder, which enters the expression only via the ratio . The stiffness kS of the columns in the case of a very stiff girder appears as the stiffness of reversed cantilevers, that is, kS ¼ 3 EIS/h3. Thus, the applied force P is opposed by the total stiffness of the columns, that is, 2 kS ¼ 6 EIS/h3, and, therefore, ¼ P/kS ¼ P h3/6 EIS. This result corresponds to an ‘infinite’ value of , and constitutes the maximum limit for k. It is clear that, by the reduction of the girder stiffness, this latter shift will be increased depending on the ratio . A fixed frame under a horizontal force P, according to the above, ‘exhibits’ its imaginary column hinges at roughly five-eighths of the column height (see Figure 6.9). Thus, the horizontal shift of the girder consists of the shift of the two-hinged frame with three-eighths of the height plus that of the top of the ‘cantilever’ bottom parts of the columns with height (5h/8), loaded with a force (P/2). Based on the above, the resulting horizontal shift is roughly one-quarter of the previous one, ffi

P h3 2 þ 1= 48 EIS

indicating a quadruple stiffness compared with the two-hinged frame (see Figure 6.9). 231


δ

δ

IR

P

~δ/4

IS

IS

P/2

Elongation of the diagonal

IR

P

Sh

P/2

~1/4 elongation Four-fold stiffness

IR

S

~δ/4

3 · h/8

Stiffness: k

P

IR

A IS

IS

S A

Stiffness: k + ∆k Reduction in the horizontal displacement (increase in the horizontal stiffness) P

S/(3 ~ 5)

For the same percentage of stiffness increment a four-fold cross-sectional area of tie is required P

Compressive member, ineffective because of buckling Activated by inverting the loading sense

Figure 6.9 Lateral stiffness of a two-column frame and its strengthening

Once again, the deformability of the columns plays the main role. In the case of a very stiff girder, the ends of the columns do not rotate, and hence rigidity is kS ¼ 12 EIS/h3. It follows that ¼ P/2 kS ¼ P h3/24 EIS a value that corresponds to ‘infinite’ , and constitutes the maximum limit for k. The increasing effect of the reduction in the girder stiffness is obvious. It is important to point out that the deformability of the frame, which is relatively high, is, in practice, identified through the increase in the length d of its diagonal, and can be considerably limited by placing a hinged diagonal member. The state of stress of the frame is, of course, mainly due to the horizontal displacement of its nodes and, secondarily, to their rotation (see Figure 6.9). Applying the method of forces, the ‘statically redundant’ value of the tension member force S results from the relative gap F10 of its ends due to the force P, and the relative approach F11 due to the force S ¼ 1: F10 ¼ (L/d) 232

Frames

F11 ¼ (L/d)2 (/P) þ d/ES A S ¼ F10/F11 It is obvious that the horizontal component Sh ¼ S (L/d), being opposite to the loading force P, leads to a reduction in the horizontal displacement as well as in the stress state in the frame. Based on the above relations, it follows that, for the hinged frame, Sh ¼

P 1 þ ðd=LÞ2

kd ES A

while for the fixed frame, the force Sh is computed on the basis of a four-fold higher value for k, i.e. (4 k). The frame displacement thus becomes ¼ (P Sh)/k, and, as a result, the existing stiffness k increases by 2 L E A k ¼ S d d Obviously, this latter frame displacement is ¼ P/(k þ k). The last relation permits the determination of a cross-section A for the truss member, which reduces both the horizontal displacement as well as the stress state of the frame due to a load P to any desirable ratio %. Hence, 2 d d k A¼ 100 L ES while, for the fixed frame, the corresponding value is obviously four times higher. The new stiffness value is 100 g 100 Given that the above change in length of the diagonal is the same when elongated and shortened, it can be seen that a hinged member placed in the opposite direction and acting as a compression member has the same effect as above. However, due to the small section A with respect to the length of the compression member, this bar becomes immediately inactive due to buckling, as will be explained in Chapter 7. Since, therefore, the direction of the loading force P may be alternated (e.g. due to wind or earthquake), it is common practice to place two crossing bars, so that for a specific direction of the force P only the tension member is active, while the other member is considered to be inactive (see Figure 6.9). Masonry embedded in a concrete or steel frame definitely causes a reduction in both the deformation and stress state of the frame under a horizontal loading. The action of the wall can be simulated with a diagonal compression member having the same thickness as the wall and a width equal to approximately 1/4—1/5 of the length of the diagonal (Rosman, 1983). k

233


Thus, for a fixed column frame with geometric characteristics L ¼ 6.0 m, h ¼ 4.0 m, IR ¼ IS ¼ 0.005 m4, E ¼ 30 000 N/mm2 and wall thickness t ¼ 15 cm, assuming that Es ¼ 5000 N/mm2, then — based on the above — the ‘equivalent’ cross-sectional area is A ¼ (d/5) 0.15 ¼ 0.216 m2, which makes the frame stiffness four times stronger. It is assumed, of course, that the stress in the compression member, S Sh ¼ ¼ ð0:20 dÞ t 0:20 L t has a bearable value with regard to its compressive strength.

6.2.3

Frames with inclined legs

Frames with inclined legs are better than orthogonal frames for transferring gravity loads at two support points on the ground, since they correspond more closely to the pressure line (Figure 6.10). These frames develop smaller bending moments, thus leading to more g

[M] Pressure line The small deviations from the pressure line reduce the bending W

W/2

W/2 The vertical component of the reaction reduces the bending of the leg p Symmetric part

p/2

Antisymmetric part p/2

p/2

The symmetric loading causes little bending response, contrary to the antisymmetric one p

The absence of horizontal reaction leads to intense bending p/2

Monolithic extension from both sides

Relatively small bending relief

Figure 6.10 Behaviour of a frame with inclined legs

234

Frames

economic use of materials. In addition, their stressing due to horizontal loads at the girder level is clearly favourable. The horizontal girder may be monolithically extended to both sides, so that a continuous passage for transport loading (i.e. a bridge) is formed between two specific locations (see Figure 6.10). In the case where the girder is extended from both sides through simply supported beams, the sensitivity of the frame should be taken into account in view of bending stresses towards half-loading (with live loads), as shown in Figure 6.10. However, with regard to bigger spans (up to 100 m), the self-weight plays a more significant role in this configuration than the live load and, consequently, the sensitivity is limited through the monolithic connection of girders (see Figure 6.10).

6.3

One-storey multibay frames

6.3.1

Vertical loads

In the case of a hinged connection of the girder to the columns being fixed at their base, the same characteristics appear exactly as in the frame with one span (see Figure 6.1). It is obvious that a lateral load applied at the level of the girder will be taken up by the shear forces V at the top of the columns, distributed proportionally to their stiffness (3 EI/h3) and causing directly at the fixed support of the columns the moments V h (see Section 3.3.7 and Figure 3.42). These frames are usually formed with the girder monolithically connected to the columns, and with spans that do not differ substantially between them. The columns are considered to have the same height (Figure 6.11). In such a frame, a gravity load uniformly distributed over the whole length of the girder does not cause a substantial horizontal displacement of the frame, and hence the bending stresses in the columns depend only on the rotation of their top. This rotation can be neglected in practice for the internal columns, and thus neither bending nor shear appears in them. As a result, the girder of each internal bay acts, basically, as a fixed-ended beam. However, the rotation at the two extreme joints cannot be neglected. This rotation causes a bending moment at the top of the two external columns, as well as at the ends of the corresponding girders, with the same magnitude, of course. The above moment MB, which causes tension of the external fibres of the column and the girder, depends on the stiffness ratio ¼ (IR/L)/(IS/h), and for a uniform load q it is ðhÞ

MB ¼

q L2 1 12 1 þ 1:33

(for a hinged connection)

and q L2 1 (for a fixed support on the ground) 12 1 þ Since the internal joints of the girder are considered not to rotate, it is obvious that the moment of the external joint under consideration will be transmitted with half its ðfÞ

MB ¼

235


q · L2/8 – MB(h)/2

q · L2/12

MB(h)

MB(h)

IR B

B

h

IS Without bending

MB(h)/h L

MB(h)/h b q

q · L2/8 – MB(f)/2

MB(f)

q · L2/12 MB(f)

IR B

B IS

1.5 · MB(f)/h

Without bending

1.5 · MB(f)/h

Figure 6.11 Bending behaviour of a multibay frame under vertical loads

magnitude to the interior joint, this time causing tension at the bottom fibres. This moment is superposed on the fixed simply supported moment q L2/8 of the extreme span, thus resulting in the final bending moment of the first internal joint (see Figure 6.11). The difference between this moment and that from the fixed-ended beam of the neighbouring span should naturally be taken by the first internal column (joint equilibrium); however, it is small in most cases. ðfÞ In the case of a fixed frame, the above moment MB will also be transmitted with half its magnitude to the column base, causing tension at its internal fibres (see Figure 6.11). Of course, at the external columns, the developed reactions have horizontal compoðhÞ ðfÞ nents directed towards the interior of the frame, equal to MB /h or 1.50 (MB /h), while the column compressive forces result from the vertical equilibrium of each joint under the corresponding shear actions of the girders.

6.3.2

Horizontal loads

With respect to the lateral loading of the frame, a concentrated force P at the external joint of the girder is again considered. Any pre-existing different distribution of the horizontal force along the girder will only affect its axial stress state and nothing else (Figure 6.12). Such a loading causes identical horizontal shifts and joint rotations in practice, resulting in zero moments at the midspan of the girders, due to antisymmetrical rotation of their ends. The columns on the other side, which develop bending moments due mainly to their top displacement, in the case of fixed frames also, in 236

Frames P · h/b

P · h/b

n: number of bays

(P/n) · h

P · h/b

P · h/b

P

h

Without N P/2n

P/2n P/n

P/n

P/n

P · h/b P

P · h/b

P/n

P/n

P/n

P/n IR

IR IS/2 P/2n

P/2n

IS/2

P/2n P/n

P/2n P/n

N=0

P h/2

P/2n

P/2n P/n

P/n Fixed frame The bending moments are halved

P/n

Figure 6.12 Bending behaviour of a multibay frame under horizontal loads

practice, exhibit zero moments at their midheight. However, at the external columns, the point of zero moments lies slightly higher than their midheight (compared with at three-eighths in single-bay frames), but this is not particularly important for their preliminary design. Of course, hinges may be inserted at the above zero-moment points, still leaving the frame statically indeterminate (see Figure 6.12). In a frame with columns hinged at their base, it can be considered that the n single-bay frames that make up the structure carry the concentrated force P equally. Thus, each subframe takes a horizontal force P/n. The direct result is that the horizontal reaction (shear) at the hinge of the external columns is P/2 n, while in all the internal columns it is equal to P/n (see Figure 6.12). The above considerations lead to the following results: . As the vertical reactions corresponding to two adjacent subframes cancel each other, it may be concluded that the axial forces of all the internal columns are zero. Thus, the moment P h of the load with respect to the frame base is taken up only by the two vertical reactions of the external columns, which are at a distance b from each other, i.e. by axial forces P h/b (tensile and compressive). . The shear force at the midspan hinge of each girder is P h/b, causing at each span end a moment of half the magnitude of that developing at the top of each internal column. Indeed, the bending moment at the top of the internal columns is P h/n — double the corresponding value for the external columns — and the moment at the ends of each girder is (P h/b) (L/2) ¼ P h/2n. g 237


A frame with fixed columns at its base can be considered, according to the above, as a frame with hinged columns of height h/2, and, therefore, all the previously developed results are, by analogy, valid. Thus, the shear forces in the columns remain the same, while the moments in the girders and the columns are halved with respect to those of hinged frames.

6.3.3

Lateral stiffness

The horizontal displacement of a multibay frame is identical to the shift of one of its single-bay subframes. The columns in each have half the moment of inertia of the actual value IS, while the moment of inertia of the girder is equal to the existing value IR (Figure 6.13). Thus, the displacement results from the expression for in Section 6.2.2 on the basis of a horizontal force P/n, and a moment of inertia of the columns (IS/2): ¼

2 P h3 2 þ 1=ð2 Þ 12 n EIS

¼

IR =L IS =h

(hinged columns)

δ

IR

IR

P

IS

n: number of bays

IS

Hinged columns P

P/n

P/n

IR

δ

P/n

δ IS/2

P/n δ

IR δ

Stiffness = k*

IS/2

Total stiffness: k = n · k* Fixed columns: stiffness = 4 · k*, caused δ/4

P

P – Peff n–1 δ

P – Peff n – 1 IR

Peff Sh

δ

S

S

Stiffness = k* + ∆k (with tie)

P – Peff n – 1 IR

δ IS/2

δ IS/2

Stiffness = k* (without tie)

Total stiffness: k = n · k* + ∆k δ = P/k = (Peff – Sh)/k*

Figure 6.13 Lateral stiffness of a multibay frame and its strengthening

238

Frames

The stiffness k of each subframe is then k ¼

EIS 12 3 2 þ 1=2 2h

while the stiffness k of the whole frame is k ¼ n k . Note that for fixed columns the stiffness k is four times higher (see Figure 6.13), according to Section 6.2.2. Restriction of the horizontal shift can again be attained by inserting hinged diagonal members in the frame. The placement of one such diagonal bar (with length d) in only one bay increases the stiffness k , according to Section 6.2.2, by 2 L E A k ¼ S d d and thus the stiffness of the whole frame becomes equal to n k þ k. This ‘strengthened’ bay will now take up a higher horizontal force Peff (see Section 3.3.7), equal to Peff ¼ P

k þ k n k þ k

and, according to Section 6.2.2, the horizontal component Sh opposing the action Peff applied to the bay being examined will be Sh ¼

Peff 2 d ðk þ kÞ d 1þ L ES A

It should be recalled that the axial force S developing in the tension member is S ¼ Sh

d L

The obviously reduced displacement of the whole frame now becomes ¼ P/(n k þ k) a value which may also result by simply considering the ‘strengthened’ opening according to Section 6.2.2: ¼ (Peff Sh)/k ¼ Peff/(k þ k) Once again, it is possible to determine the required cross-section A of the tension member that reduces the horizontal displacement due to a load P to a desirable percentage %. Thus, 2 d d nk A¼ 100 L ES while for the fixed frame the corresponding value is four times higher. 239


The new stiffness is obviously equal to n k

100 100

g

It can be seen that multibay frames present a higher stiffness than single-bay ones, and this means, of course, that they are more sensitive to the imposition of a constraint such as temperature change or support settlement (see Section 3.2.5). With regard to concrete frames, creep acts as a reducing factor for the stress state due to shrinkage. If such a constraint is applied within a small time interval (e.g. ground settlement of non-cohesive soil) the instantaneously resulting state of stress reduces after some time by a percentage of the order of 100 ’/(1 þ ’)%, whereas, if the constraint is imposed gradually (e.g. shrinkage, a temperature change), the final developed state of stress is only a part (e.g. 100/(1 þ ’)%) of that which corresponds to the final deformation, if applied instantaneously (see Section 5.5.1). However, the impending cracking will reduce the calculated stiffness for the uncracked state, and, accordingly, the ‘elastic’ response of the structure to the imposed constraint is not as intense as initially calculated.

6.4

Multi-storey frames

Multi-storey frames are a constitutive element of multi-storey buildings, offering as plane systems the possibility of taking up linear distributed loads at various levels, acting in their own plane. The considered loads can be either gravity loads or horizontal loads due to wind or earthquake actions. In the case of horizontal loads, these may also be carried through a combined action of the frame and a shear wall, interconnected at various levels. The examined frames are considered as not having particular differentiation with regard to the column sections along the storeys or the girder spans and the floor heights. Such a frame is a structure with a high degree of static redundancy, and its analysis obviously requires the use of a special computer program. However, it is necessary to cultivate a basic perception for the load-bearing action of the frame based on an approximate, yet structurally sound, approach that permits the preliminary design of girders and columns and which constitutes a solid basis for evaluating the computer results.

6.4.1

Vertical loads

The gravity loads on the girders, as for the one-storey multibay frames examined above (see Section 6.3.1), are considered to cause negligible rotations at the joints of the internal columns, and thus these are not stressed in bending but only axially. The rotations are, however, not negligible at the external columns, where they cause bending moments. Thus, each girder may be considered as a continuous beam being subjected at its ends to bending moments causing tension at the top fibres (Figure 6.14). 240

Frames

h

h M 0R

MS

MS h

h

Without bending Without bending

L

0.21 · L

0.21 · L

h MR h/2

IS

MR

MS

IR

MS

IR

h

h/2

MR

LR

IR

MS h

IS

Figure 6.14 Bending of a multi-storey frame under vertical loads

For equal storey heights and column sections, as well as for the same loading of girders in all floors, the rotations of the external nodes can be considered equal, inducing a directly zero moment at the midheight of columns. This allows one to consider each girder to be connected at its external joint with two columns hinged at their ends above and below, having height h/2, with the internal joint of the girder (approximately) fixed (see Figure 6.14). Using the girder to column stiffness ratio ¼ (IR/LR)/(IS/h), as introduced in Section 6.2.1, the moment MS at the bottom 241


end of the top column and at the top of the bottom column, according to this model, is MS ¼

1 ð0Þ MR 2 þ 0:66

while the moment MR of the girder that balances them at the joint is MR ¼

2 ð0Þ MR 2 þ 0:66 ð0Þ

In these relations, MR is the fixed-end moment of the external fixed-ended span of the girder. Note that considering the full height of columns with fixed ends above and below (see Figure 6.14), and taking into account the effect of loading of the two girders, leads to the result MS ¼

1:5 ð0Þ MR 2:5 þ

which is unfavourable in comparison with the previous equation. However, considering reinforced concrete frames, the use of this expression seems more logical, given the expected reduction in girder stiffness relative to the columns, due to girder cracking. The fact that in the external columns at the ground level the column moment at their top is double the moment at the fixed bottom and, accordingly, the moment becomes zero at a distance h/3 from the base, leads to the conclusion that the above results are always on the safe side. Considering now the internal spans of each girder, they may be considered as having roughly fixed supports for the permanent loads, i.e. they may be treated as fixed-ended members with fictitious hinges placed at a distance from the supports equal to 21% of the span length (see Figure 6.14). However, in view of the unfavourable layout of live loads necessary to obtain the highest moments in the spans, the imaginary intermediate ‘simply supported’ beam should have a length of 80% of the span, instead of 58% with regard to the permanent loads. It is also obvious that when calculating the support moments due to live loads the corresponding ‘cantilevers’ must be considered as having a length 21% of the span (see Section 5.1).

6.4.2

Horizontal loads

The horizontal loads Pi are usually applied as concentrated forces at the levels of girders. This way, each floor can be considered as a multibay single-storey frame, which, through the shear forces of the columns of the upper floor, takes up the sum of all horizontal loads acting above the considered girder level. Thus, according to Section 6.3.2, the midpoints of columns and girders have zero moment, and the shear forces of the columns are determined according to the above. Again, only the external columns develop a substantial axial stress, as was also clarified in Section 6.3.2 (Figure 6.15). Each Pfloor i receives, through the column shear forces of the upper floor, a horizontal force Pi — the summation of all the loads above this level — which is distributed to 242

Frames

h ΣPi

Without axial forces

Storey i + 1

P

h

Without axial forces

Storey i

h

Tensile N

Compressive N

n: number of bays

h

ΣPi Storey i ΣPi /2n

ΣPi /n ΣPi + 1/2n

ΣPi /n ΣPi + 1/n

ΣPi /n ΣPi + 1/n

Storey i + 1

P

Storey i ΣPi /2n

ΣPi /n

ΣPi /n

ΣPi /n ΣPi + 1/n

ΣPi + 1/2n

MR MR ΣPi /n

ΣPi /2n

Moments are increasing towards lower storeys

Figure 6.15 Bending of a multi-storey frame under horizontal loads

the columns as explained earlier for single-storey frames. Each column with its P corresponding shear ( Pi/n) considered at the hinge position (i.e. at its midlength) develops a moment Mst,i at the top and at the bottom equal to P Mst,i ¼ ( Pi/n) (h/2) The moments MR at the girder ends on the left and right side of the column are determined by equilibrium at the corresponding node. For equal heights of columns and equal girder spans, the moment of the girder between the ith and (i þ 1)th floor is MR ¼ ðMst;i þ Mst;i þ 1 Þ=2 that is P P MR ¼ ( Pi þ Pi þ 1) (h/4n) It is clear that the bending stresses in the girders increase as long as their level approaches the base (see Figure 6.15). 243


P

h

(P/2) · h/2 h

H

h

‘Shear’ action

(P/2) · H

‘Bending’ action

Rigidly connected girders: column moments practically unchanged

Figure 6.16 Bending and shear of a multi-storey system under horizontal loads

It is important to note that the column moments do not increase up to the foundation with respect to height, i.e. much as would happen in a cantilever, although the multistorey frame seems ‘macroscopically’ a tall structure (cantilever) fixed to the ground. The bending moment in each column depends only on the shear force that the column transfers, as well as on its floor height. Thus, a single load applied at the top level causes in any column (through its antisymmetric part) roughly equal moments in each floor up to its base. This kind of response is obviously due to the monolithic connection between girders and columns, as Figure 6.16 illustrates, and it is a direct consequence of the ‘shear beam’ behaviour that such a frame exhibits, as will be explained in the next section. Indeed, in the system shown in Figure 6.16, where the girders are connected to the columns through hinges, the bending in the columns, due to the antisymmetrical part of loading — for which the simply supported girders are unstressed — makes them act as cantilevers fixed to the ground, developing moments much higher than the corresponding ones in a frame with monolithic joints.

6.4.3

Lateral stiffness

In the case of multi-storey frames that are loaded with horizontal concentrated forces at the level of the girders, the developing horizontal shift at the level of each girder is of great interest. This shift is, of course, common to all the joints of each floor, and is basically produced by the deformation of a typical bay panel taken from the ‘interior’ of the frame, as illustrated in Figure 6.17. This panel comprises the girder and the halflengths of the two columns at its left and right ends, which are hinged at their remote ends, according to the above. P The hinged ends P of the half-columns are acted Pi/n, respectively (see Section on by the corresponding shear forces Pi þ 1/n and 6.4.2). 244

Frames w p

x h

Storey i+1

h

Storey i

x

H

GA

n: number of bays Storey drift proportional to bending moment

ϕR · h/2

[M] (Shear beam)

Shear deformation ΣPi + 1/n

ΣPi + 1/n

Storey i + 1

Ast

h/2 IR

ϕR N

ϕR IS

Storey i

ϕR · h/2 ∆w

N n · LR

h/2

ΣPi /n

Ast

ΣPi /n

Increasing with the ‘shear force’, i.e. downwards

∆w

Figure 6.17 Deformation behaviour of a multi-storey frame under horizontal loads

It can be seen from Figure 6.17 that the relative horizontal shift w of the top and bottom hinged ends of the column places the relative horizontal shift between the two consecutive girder floors at the level under consideration. This shift contributes the rotation ’R of the girder end (the same at both ends, hence P the antisymmetric P Pi þ 1/n and Pi/n stress state) with the magnitude 2 ’R h/2, as well as the forces acting at the hinged ends of the cantilevers above and below with lengths h/2. Thus, P P Pi þ 1 ðh=2Þ3 Pi ðh=2Þ3 MR ðLR =2Þ h w ¼ 2 þ þ 3 EIR 2 n 3 EIS n 3 EIS P P Pi þ Pi þ 1 h2 LR h ¼ þ 2 12 n E IR IS The length LR obviously represents an average value of the girder spans. 245


P P The expression ( Pi þ Pi þ 1)/2 gives the shear force V of a fictitious ‘cantilever’ beam at the position of the considered girder, and the shift w, according to the above relation, is proportional to this value (Franz and Scha¨fer, 1988). It can be seen that the multi-storey frame acts like a beam, the elastic curve ‘w’ of which is developed due to the shear and not the bending deformation of its elements. In such a shear beam, the slope dw/dx at every point is equal to the shear angle , which is expressed on the basis of the shear force V at the point under consideration, and the effective cross-section A as ¼ V/AG, that is dw VðxÞ ¼ dx A G

ðaÞ

(see Section 2.3.2.3 and Figure 2.51)

Note that the deformation of this beam under a transverse load p is not governed by the equation dw4/dx4 ¼ p/EI (see Section 2.3.6) but by that resulting from differentiating the latter relation, that is dw2/dx2 ¼ (dV/dx)/AG ¼ p/AG, conforming to the elastic curve of the cantilever beam shown in Figure 6.17. Staying with the shear beam concept, it can be alternatively stated that the deformations of a shear beam may be obtained according to the following formulation of the principle of virtual work, where only the shear force terms come into play (see Section 2.3.3): X vir resp;real ð vir Vreal ¼ V ds P GA In this way, a shear beam cantilever having a height H and subjected to a horizontal load p(x) develops at its top a shift , which can be determined by considering as virtual loading a unit force at the top. By applying the principle of virtual work, according to the above relation, 1 H 1 p H2 ¼ 1 ¼ 1:0 ð p HÞ 2 GA GA 2

ðbÞ

which means that the deflection (or shift) at the top of a shear cantilever beam is proportional to the bending moment at its base. Returning now to the multi-storey frame being examined, since dw/dx ¼ w/h, an analogous relation to the above (a) can also be established here. According to the above, dw h LR h þ ¼ kV VðxÞ ¼ VðxÞ dx 12 n E IR IS where

h LR h kV ¼ þ 12 n E IR IS

It is clear that kV now plays the role of the term 1/AG in relation (a). From the last relation the following conclusion is drawn: ð wV ¼ kV V dx ¼ kV MðxÞ ðcÞ 246

Frames

that is, the shift of the frame — as well as of any ‘shear beam’ — is proportional to the bending moment, exactly as was deduced previously using the virtual work principle. Of course, the last result re-establishes the above-mentioned role of the factor kV, in correlation also to the above result (b). The effective difference between the two types of beam now becomes clearer. While in the ‘bending beam’ the curvature (d2w/dx2) is proportional to the bending moment, according to the relation d2w/dx2 ¼ kM M(x), where kM ¼ 1/EI (see Section 2.3.6), in the case of the ‘shear beam’ it is the deflection that is proportional to the bending moment. After all the above analysis it should again be emphasised that, although the multi-storey frame deforms under horizontal loads following the bending of its columns and girders, it exhibits a lateral stiffness behaviour, which may be simulated by a shear beam with specific sectional characteristics described by the factor kV. The above relation (c) allows the calculation of the relative shift of each level with respect to the top, which represents the origin for the coordinate x (see Figure 6.17). Thus, the shift wTV at the top can first be calculated based directly on the moment developing at the foundation (x ¼ H). For each inferior level A at a distance x from the top, the relative shift wA ¼ wTV wA V results directly from applying the above A relation (c): w ¼ kV M(x). Thus, the distribution of shifts along the height becomes comprehensible, as shown in Figure 6.17. It is, of course, clear that in the case of hinged girders commented on in Section 6.4.2 (see Figure 6.16), the deformation of the two ‘bending beams’ gives a much higher shift than that resulting from the monolithic frame acting as a ‘shear beam’. g With the increase in the frame total height with respect to its width, the axial forces of the external columns are also increased, so that, in combination with their axial deformation (tensile and compressive, respectively), they cause a horizontal shift, in addition to that caused by the bending of columns and girders. Since the external columns are in tension and compression, through the force NðxÞ ¼

MðxÞ n LR

the elongation or the shortening strain of these columns is "¼

N EAst

Then, assuming the girders to be completely rigid, the frame can be simulated with a bent beam, the developing curvature of which, according to Figure 6.17 (see Section 4.1.1 and Figure 4.10), is 1 2" 2N 2 MðxÞ ¼ ¼ ¼ r n LR n LR E Ast ðn LR Þ2 E Ast The displacement at the top wN is determined according to the principle of virtual work, by considering as a virtual load a concentrated unit force at the top and 247

Structural systems: behaviour and design Increasing with the ‘shear force’, i.e. downwards

S Sh

P

w

Sh S

S

∆w

P

The girder is acted on by the difference of Sh which opposes P LR

∆w

h

∆w · cos ω

h

1/cos ω 1/cos ω

Storey i S=1

(S = 0)

h

h

h

Elongation of the diagonal (S = 0)

h

Shortening of the diagonal due to S = 1

Figure 6.18 Lateral stiffening of a multi-storey frame

assuming a parabolic distribution of MðxÞ (see Section 2.3.3): ð H2 Mðx ¼ HÞ 1 g wN ¼ Mvir dx ¼ r 3 ðn LR Þ2 E Ast A restriction of the frame deformability, and of its stress state accordingly, can be achieved through the insertion of hinged diagonal bars, in the same sense as has already been examined for single-storey multibay frames (Figure 6.18). The developing diagonal tensile force S of a hinged bar with a cross-sectional area A in a frame panel depends on the relative shift w between the two levels. In this case, the tensile force may be expressed as S ¼ F10/F11, through the developed gap F10 between the ends of the cut tension bar and their relative approach F11 due to the force S ¼ 1, according to the procedure of the method of forces followed in Section 3.2.2. As is shown in Figure 6.18, F10 ¼ w cos !, where, according to the above, w ¼ kV VðxÞ h Considering now F11, this can be obtained in the same way as F10, that is F11 ¼ w1 (LR/d) þ d/(ES A) 248

Frames

where w1, according to Figure 6.18 and the previous analysis, may be calculated as w1 ¼ kV

28 h ð28 1Þ h d kV ¼ kV h cos ! cos ! LR

Thus, for the horizontal component Sh ¼ S (LR/d), the following expression is obtained: Sh ðxÞ ¼ VðxÞ

ðLR =dÞ2 d 1þ k V h Es A

From this expression, it is clear that the axial force of the tension bar is proportional to the shear force V(x), as is also the case with w. In this way, the girder at each level receives the resultant of the forces Sh above and below it, a force opposite to the horizontal loads P. The distribution of these ‘relieving’ forces obviously follows that of the shear forces V, having a constant rate of change for equal loads P (see Figure 6.18). The role of masonry should be recalled here, in contributing to such ‘relieving’ actions that lead to an effective increase in the lateral frame stiffness.

6.4.4

Mixed system

A mixed system is considered to be the coupling of a multi-storey ‘shear-acting’ frame with a cantilever ‘bending’ beam (such as a ‘shear wall’) with a large moment of inertia, fixed at its base and connected to the frame at each level with a hinged — and hence axially stressed — bar. As such a system may represent a typical structural constituent of high-rise buildings, its behaviour under horizontal loads applied at each level, generally equivalent to a distributed load ph (Figure 6.19), is of primary interest. The load ph is distributed between the two ‘members’ in such a way that a common horizontal shift w is developed at each level. For the ‘bending-acting’ beam the imposition of a shift w(x) along its height requires the application of a distributed force pb ¼ EI(dw4/dx4). Moreover, imposition of a shift w(x) along the height of the ‘shear-acting’ beam requires the application of a distributed force ps which, according to the previous section, is expressed as ps ¼ (1/kV) (dw2/dx2). The coordinate x increases from the bottom to the top. Thus, the following relation results: ph ¼ pb þ ps ¼ EI

dw4 1 dw2 dx4 kV dx2

The distribution pattern of the horizontal shift along the frame and the ‘bending’ beam due to a uniform lateral load is quite different, as shown in Figure 6.19. The deformability of the ‘bending’ beam decreases in its lower regions and increases in its higher regions, in contrast to what occurs in the shear member. Thus the frame ‘is supported’ in its lower levels on the ‘bending’ beam, while in the higher levels it is 249


ph

ph

Bending deformation

Shearing deformation

Compression in connecting bars (increase towards the top)

Tension in connecting bars (increases towards the bottom)

Approximate consideration ph

F10: Relative approach (negative)

F11: Relative gap (positive)

ph

X=1

H

EI

EI

The force of the connecting bar results in a compressive force

[M]

Figure 6.19 Coupling of a multi-storey frame and a shear wall

the ‘bending’ beam which is supported by the frame. This means that the hinged connecting bars are in tension in the lower levels, whereas in the higher levels they are compressed. Attempting to find an analytical solution to the above equation does not make much sense, given the ability to analyse the system through an appropriate computer program for framed structures. Nevertheless, of basic importance is an understanding of the structural action of such a system. In order, therefore, to assess the developing state of stress, the two members might be considered to be connected by a hinged bar only at their top. The corresponding compressive force X developed is then determined via the method of forces. 250

Frames

If F11 is the relative gap between the top of the bending beam and the frame, due to a mutual compressive force X ¼ 1, then, according to Section 6.4.3, F11 ¼

1 H3 þ kV H 3 EI

(positive)

Compatibility requires that X ¼ F10/F11, where F10 is the relative displacement of the tops of the bending beam and the frame (negative), due to the external horizontal loads (see Figure 6.19). After the determination of the compression force in the bar, an estimate of the stress state of the frame as well as of the cantilever beam can easily be made, according to the above. This estimate is on the safe side.

6.5

Design of sections

The defining characteristic of a stress state of a frame, in contrast to a beam, is obviously the additional presence of axial forces — more so for the columns than for the girders. In each column coexist, besides the shear force, which is basically treated in the same way as for the beams, an axial force and a bending moment that require a particular check. This check — and consequently the member design — refers to the cross-section resistance as well as to the stability of the whole, as will be examined in the next chapter. The cross-section check that will be considered next is performed — as is well known — utilising two aspects. Following the first aspect, the developing longitudinal normal stresses due to the existing pair of values (M, N) under the service condition should not exceed some ‘allowable’ limits. According to the second aspect, the examined section should contribute — under the extreme stress state of its materials — the pair of values (Md, Nd) that correspond to service loads ‘magnified’ by S and will cause the section to ‘fail’: Md ¼ S M Nd ¼ S N

6.5.1

Steel sections

The developing longitudinal stresses at the extreme fibres of a steel cross-section under a compressive force N and a bending moment M, as shown in Figure 6.20, are calculated using the following relation that expresses superposition of axial and bending stresses: ¼ N þ M ¼ ðN=AÞ ðM=WÞ The above stresses should generally be lower than a value of about 160 N/mm2. The stresses in the whole cross-section are distributed linearly between the above two values, and the stress diagram creates the pair of values (M, N) (see Figure 6.20). As has been pointed out previously, the design of a cross-section is nowadays performed at the ultimate state. In this state it is considered that all the fibres of the cross-section develop the yield stress fy (see Figure 4.11), and the section has dimensions 251

Structural systems: behaviour and design NR Npl σN

σM

(= fy · A)

Unsafe region

M Safe region

N

Due to N

σM Due to M Superposition

(0, 0)

Mpl 1.18 · Mpl

MR

Interaction curve

Figure 6.20 Check of a steel section in eccentric compression

such that it can accommodate the pair of values (Md, Nd) — or some other ‘safer’ ones — through a suitable distribution of the stress diagram. In order to check whether this is possible, the following relation must be satisfied: Nd Md þ 1 Npl 1:18 Mpl where Npl is the highest compression force that the cross-section can develop, i.e. fy A, while Mpl is the cross-section resistance in pure bending, as determined in Section 4.1.2. The above relation, considered as an equality, yields the moment MR, which, along with a given axial force NR, causes cross-sectional failure, and vice versa. This is depicted by the interaction diagram in Figure 6.20. If the above equation is applied for a specific cross-section, the axial force Nd is not necessarily the limit value for the moment Md under consideration, but, generally, some other force N > Nd may be. The same procedure is applied if Nd is considered as the basic sectional force, instead of Md.

6.5.2

Reinforced concrete sections

The presence of axial forces together with the bending moment requires the placement of reinforcement on both ‘active’ sides of the section. The reason for this is that column bending results from, in particular, the lateral loads, which can be cyclical. This imposes a symmetrical layout on the reinforcement in the section. The determination of stresses in the concrete and the reinforcement for a given pair of values (M, N) lies basically in finding an appropriate (linear) diagram of strains " in the cross-section. This means the determination of the following two values: the shortening strain "c of the concrete and the elongation (or even shortening) strain "s of the reinforcement (Figure 6.21). From these values, the stresses in the concrete and the reinforcement of the cross-section can be determined directly. Both "c and "s should have such values that the computed tensile and compressive resultant forces are statically equivalent to the pair of forces (M, N) in the cross-section. 252

Frames Statically equivalent to (N, M) M Ds

Dc

Zs N εs

εc As

As

Cracked region Statically equivalent to (NR, MR) MR Zs

NR (compressive)

NR Npl Unsafe region

εs

Dc

Ds

2.0‰ Safe region

εc = 3.5‰ As

As (0, 0)

εc, εs: Determine D and Z

Mpl

MR

Interaction curve

Figure 6.21 Concrete section symmetrically reinforced under axial compression and bending

Furthermore, in the case of reinforced concrete, the design is preferably implemented for the ultimate state of the cross-section. This means that the concrete cross-sectional area Ac and the reinforcement cross-sectional area As are designed so that the pair of values (Md, Nd) correspond to the highest possible stressing of concrete and steel reinforcement. For the concrete, the highest response in failure level is the development of a compressive strain equal to 3.5%, corresponding to a compressive resistance fc, depending on its quality, while for the steel it is the yield stress fy, roughly equal to 420 N/mm2, that corresponds to a strain of 2.0%. Since, in the ultimate state being considered, the section remains plane too (Bernoulli), a linear distribution of strains over the cross-section is sought, so that the resulting compressive and tensile forces ‘cover’ the above pair of sectional forces (Md, Nd), on the basis of the corresponding stress—strain diagrams for concrete and steel. Keeping a constant strain of 3.5% at the outer fibre of the compression side, for each selected strain line trimming out at the level of tensioned reinforcement, a value of "s between 2% and 10% (see Figure 6.21), corresponds to the ‘ultimate state’ pair (MR, 253


NR), so that for the given cross-section a particular curve can be drawn. This curve, called the interaction curve (between MR and NR), constitutes the means by which each reinforced concrete section can be identified. Thus, on the basis of the interaction curve, a given value of the moment Md corresponds to an ‘ultimate’ axial force NR, which should be greater than Nd (see Figure 6.21). However, for preliminary design purposes and for a given pair of internal forces (Md, Nd) corresponding to ‘factored’ service loads by S, the following consideration may be followed instead, offering a more practical procedure. The internal forces are referred to the centroid of the section. Nevertheless the pair (Md, Nd) can be substituted by a single compressive force Nd, applied at a distance from the centroid equal to c ¼ Md/Nd. Two cases will be examined. In the first case, if the shifted force Nd lies between the two extreme reinforcement positions having a distance z, so that c < z/2, then, although the compressive force Nd may lie outside the core of the section, mainly compressive stresses develop in the section. In this case it may be assumed, in practice, that the cross-section reaches its ultimate state by superposition of the following two states: (a) a uniform distribution of ultimate compression stresses fc over the whole area affected by the compressive force, and (b) the development of a compressive and tensile yield stress fy at both reinforcement layers which are acted on by the bending moment (Figure 6.22). Thus, for a specific section Ac with symmetric reinforcement As, that axial force NR is sought, which, for a given c (meaning MR ¼ c NR), causes the section to fail. Equivalency of the ‘internal’ forces of the section with the ‘external’ forces (MR, NR) requires, as shown in Figure 6.22, the moment resultants about the tension reinforcement to be equal, i.e. fy As z þ fc Ac (z/2) ¼ NR (z/2) þ MR Setting as Npl the resistance of the cross-section in pure compression, Npl ¼ fy (As þ As) þ fc Ac

MR = c · NR NR

M

N

fc · Ac

fy · As

NR fc

fc

c As

As z

fy

fy

As

As

z

Figure 6.22 Practical assessment of ultimate resistance for small eccentricity

254

fy · As

z

Frames

and, consequently, NR ¼ Npl

1 c 2 þ1 z

Thus, for the case where c < z/2, a practical check for a section is the examination of whether, for the given value of c ( ¼ Md/Nd) the calculated value NR is equal to or higher than the design axial force Nd. In the second case, where the shifted force Nd lies outside the cross-section, i.e. when c > z/2, the neutral axis clearly limits the compression area. The section exhausts its resistance with the development of yield stresses in the tension reinforcement and a suitable development of compressive force in the compression reinforcement and in the compression area. Note that the total of these compression forces is higher than the force of the yielded tension reinforcement, in order for the corresponding ‘external’ compression force of the section (Figure 6.23) to be offered. Again, as above, an axial force NR is sought which, acting on a given section with symmetric reinforcement As under a given eccentricity c (meaning combined with the moment MR ¼ c NR), causes the section to fail. Equivalency of the ‘internal’ forces of the section with the ‘external’ forces (MR, NR) applied to the centroid of the section requires, as shown in Figure 6.23, the moment resultants about the compression reinforcement to be equal. Considering the distance between the concrete and steel compressive forces to be essentially negligible (see Figure 6.23), fy As z ¼ MR NR (z/2) and, hence, NR ¼

As f y c 1 z 2 MR = c · NR

MR

NR

NR εs

c fy · As

NR

εc

Dc fy · A s

fy

fy

As

As z

As

z

As

Figure 6.23 Practical assessment of ultimate resistance for large eccentricity

255


This relation can be used either for checking that for a given c based on the design values (c ¼ Md/Nd) the calculated NR is equal to or higher than the design axial force Nd, or for the direct determination of the reinforcement area As based on the values Nd ( ¼ NR) and c ¼ Md/Nd.

6.5.3

Prestressed concrete

The application of prestressing in frames concerns mainly the girders in order to counter gravity loads. It also concerns, with regard to single-storey simply supported portal frames, the insertion of an externally prestressed tension member between the column bases in order to avoid an outward thrust on the foundation, in case the soil is inadequate to offer the required resistance to this thrust (see Figure 6.7). Girder prestressing, basically, follows the logic of a simply supported prestressed beam (or of a continuous beam), with a possible eccentricity of the anchored cable at the extreme joints, in order to compensate for the presence of end moments there (Figure 6.24). The issue of the active compression force in the girder should be noted. Due to the stiffness of the columns, external application of a horizontal compression force P at

The deviation forces remain unchanged

P

P

A decrease in girder compressive force

P

P

More intense decrease in girder compressive force They induce M0

Statically indeterminate

Statically determinate

X1

X1 X2

Resulting bending response: MP MP = M0 + MSP

(X1, X2) Redundant forces: induce MSP

Figure 6.24 Prestressing action in the statically redundant frame

256

Frames

the extreme nodes of the frame does not imply the development of the same compressive force in the girder as well. As long as the columns are stiffer, the portion of the compression force transferred to the girder is decreased. Therefore, although prestressing the cable with a force P causes undiminished deviation forces on the girder based on P, the corresponding axial force in the girder resulting from the statically redundant frame action is, nevertheless, clearly smaller than P (see Figure 6.24). The treatment of the statically ‘determinate’ and ‘redundant’ prestressing moments takes place in the same way as examined for the continuous beam (see Section 5.4). More specifically, the frame under the external anchor forces, as well as the deviation forces, both calculated directly based on the force P, develops as a statically redundant system the bending moments MP, where (see Figure 6.24) MP ¼ M0 þ MSP The statically determined moments M0 expressed with respect to P ( ¼ P e) correspond to the statically determined primary system resulting through appropriate interventions under the support conditions (see Figure 6.24), since in this case the girder receives the entire compression force P. The redundant prestressing moments MSP result from the above relation, and are treated as sectional forces in the strength-checking relation, as examined in Section 5.4.2. As explained in Section 5.4, the moments MSP are caused by the statically redundant forces applied to the primary system under consideration, in order to restore the actual deformation conditions at the support points (seee Figure 6.24). As was also pointed out in Section 5.4, these moments are essentially unaffected by the progressive cracking due to load increment, since they depend on the relative stiffness of the frame members rather than on their absolute values. Of course, in the phase of collapse where the frame has become a statically determined system (see Section 6.6), these moments vanish. However, they may generally be considered to be multiplied by a varying factor that is normally equal to unity.

6.6

Plastic analysis and design

As in the design of beams, the main objective for frames and for any type of structure is to design the structure in such a way that the service loads factored by S will either lead directly to collapse or to a non-collapse state which, however, is as close as possible to collapse.

6.6.1

General overview and plastic behaviour

Any type of loading that is applied to a frame structure consists of distributed and concentrated loads that can always be expressed in the form ‘ P’ associated with a single parameter for each load (Figure 6.25). If the cross-sections of the structural members are already designed with an ultimate resistance pair of values (MR, NR), then the parameter can always be determined so that the resulting loading induces collapse of the structure. 257

Structural systems: behaviour and design λ·P λ · (P/2)

2.50 · EI 2

5.0 m

EI

3

4

2 · Mpl

Mpl

Mpl Mpl = 60.0 kN m

1

5.0 m

EI

5

5.0 m

λ · 1.0

λ · (1.0/2)

60.0

λ = 45.43 78.63 Joint 4 49.21 λ · 1.0 16.26

λ · (1.0/2)

60.0

λ = 54.85

60.0

60.0 140.95 62.76

66.59

102.82 2.8

40.93

Joint 5 60.0

25.65

λ · 1.0 λ · (1.0/2)

60.0 50.53 60.0

λ = 61.00

78.95

116.35 320.0 λ = 64.00

120.0 Joint 3

60.0

170.53

λ · (1.0/2)

60.0

60.0

λ · 1.0 300.0

60.0 120.0

20.0

420.0

60.0

60.0 120.0

60.0 480.0

60.0

37.40

60.0

Joint 1

60.0

Collapse

Figure 6.25 Consecutive loading stages and sectional forces until total frame collapse

The process of determination of the collapse load is substantially the same as that followed for the cases of fixed-ended and fixed simply supported beams in Sections 5.2.1 and 5.2.2, respectively. More specifically, with a gradual increase in the parameter starting from zero, the first plastic hinge appears in a section. Next, the thus modified structure is considered to have a hinge at the corresponding position, which receives at both its edges an externally applied self-equilibrating pair of moments equal to Mpl, or at only one end 258

Frames

if this is a fixed support. Further loading increase — on the modified structure — leads to the formation of a new plastic hinge at another location in the structure. The newly modified structure is considered to have two hinges at the corresponding positions, loaded externally by the self-equilibrating moments Mpl. This procedure is continued until the developed plastic hinges make the structure either statically determined or a first-degree mechanism. In the first case, it is understood that the next plastic hinge formation following the continuously considered increase in the parameter will lead to a first-degree mechanism, and consequently to collapse. In the second case, it is considered that the structure has already collapsed. This last value of will directly provide the collapse load as well as the prevailing moment diagram at the instant of collapse in all cases (Duddeck, 1984). This diagram does not differ substantially in its qualitative characteristics from that corresponding to an elastic analysis. The whole process can be followed in the example given in Figure 6.25. g At this point it is useful to follow the lateral deformation behaviour of a simple fixed (i.e. statically indeterminate) frame, as shown in Figure 6.26, subjected to a constant uniform load of 30 kN/m, under an ever-growing horizontal concentrated force P, by following the above incremental procedure. As is shown in Figure 6.26, until the moment of creation of the first plastic hinge, the behaviour is linearly elastic. Then, the implementation of the first plastic hinge changes the system, which, under the constant external action of the plastic moment at node 3, may be further loaded, until the next plastic hinge is formed at node 4. Again, the new system affords some resistance to the ever-growing lateral load with ever-decreasing stiffness, with an additional plastic hinge at node 1, until, by further increasing the horizontal load, the final plastic hinge is created near the midspan of the girder in the already statically determinate system, which implies the collapse of the structure. It is clear that the respective displacements at the characteristic loading stages may be readily calculated on the basis of the corresponding bending moment diagram and by using the principle of virtual work appropriately. The diagram relating the acting lateral load to the developing lateral displacement describes a typical non-linear behaviour of the so-called push-over response of the frame under examination. Of course, it should be pointed out that, in order for the above procedure to be valid, the ductility requirements must be met, which means, in practice, that the relative rotation of the first developed plastic hinge (e.g. at nodes 4 and 3 in Figures 6.25 and 6.26, respectively) should not exceed the corresponding values, given in Sections 5.2.1 and 5.3, for steel and concrete (0.1 rad and 0.02—0.05 rad, respectively). Should this maximum allowable value develop at an earlier phase, the corresponding load and moment diagram should be considered to be the ultimate diagram. g If, with the progressive increase in loading, a plastic hinge is formed in an axially-only stressed tension bar with a cross-sectional area A, developing a yield force Ny ¼ fy A, then this bar, not being able to develop a higher axial force, may be removed and substituted by the corresponding forces Ny applied at both its ends. 259


P 2

EI

Mpl

EI Mpl

4.0 m

Mpl (HE-B 220) Mpl = 206.0 kN m

1

4.0 m 30.0 kN/m 3

1

4

112 kN 2

30.0 kN/m

Joint 3 194

157 kN 2

206 kN m

1

4

(c) 206

Joint 4

Joint 1

206

206

206

P: kN 178 157

206 kN m

(d)

206 kN m (c)

112 104

(d)

94

206 kN m

1

206

80

30.0 kN/m

1

206 kN m

(b) 206 44

2

4

4.0 m

(a) 50

178 kN

EI

30.0 kN/m

104 kN 2

66

3

(b)

206

94

(a) 206 kN m

206

206 Collapse

29 32

63

142 δ: mm

Figure 6.26 Push-over behaviour of a portal frame

Similarly to the above, the relative displacement of the ends of a possible tension bar is assumed to not exceed, roughly, 15% of its length during the whole gradual increment of the loading.

6.6.2

Design

The process described in the previous section considers the problem of finding the collapse load provided that the design of the structure has already been undertaken. In practice, however, the problem of plastic design, as was stated at the very beginning of the discussion, is dimensioning a structure so that the service load magnified by the factor S is smaller than or equal to the corresponding collapse load. This is ensured, according to the so-called static theorem of plastic analysis, provided that any moment 260

Frames

diagram being in equilibrium with the considered loading satisfies at any point the condition (M, N) (MR, NR)l and that buckling of the structure is avoided. Of course, provided the structure is statically redundant, an unlimited number of moment diagrams exist that are in equilibrium with the given loading. The ‘elastic solution’ also belongs naturally to these diagrams. The ‘static theorem’ can also be formulated more widely as follows. If a structure is due to a model that can be considered to be included in the structure, and which is designed on the basis of failure of some of its members through an adopted sectional force distribution satisfying everywhere the equilibrium requirements and not necessarily the compatibility conditions, then the collapse load for the model is less than (or at most equal to) the collapse load of the structure. This formulation of the static theorem is of particular value since it allows the ‘substitution’ of continuous media with framed structures, as, for example, in Section 4.2.1.2 (see also Section 6.7). According to the above, the design procedure is clearly a search procedure. An ideal (economic) solution of the plastic design problem of structures lies in finding that design selection which, according to the above process, leads to collapse loads identical to the service loads factored by S. The full achievement of this objective is feasible but not strictly necessary. In practice, a bending moment diagram is sought that is in equilibrium with the magnified loads on the one hand, and satisfies at all sections the requirements of the static theorem on the other. A diagram like this reflects the so-called moment redistribution with respect to that of the elastic analysis. Such diagrams are acquired through inserting, at first, so many hinges at the nodes and the fixed supports that the structure becomes statically determined, and then imposing the ‘selected’ plastic moments externally at their edges, so that the statically determined system can be directly solved under the given (factored) loads. If the design offers at any position resistances that are nowhere exceeded by the resulting corresponding diagram, then, according to the ‘static theorem’, the collapse load is definitely higher than or equal to the imposed external load. It is obvious that this is also still valid if — and this would indeed be desirable — the diagram exhausts in only one further location the corresponding moment resistance, thus implying an additional hinge. Of course, the most economical solution possible is sought. Moreover, it should be understood at this point that superposing the elastic solution (Mel, Nel) with a self-equilibrating stress state (M1, N1) without additional loads leads to a moment diagram M that is also in equilibrium with the existing external loads. Thus, the diagrams M ¼ S Mel þ M1 and N ¼ S Nel þ N1 can be used for application of the ‘static theorem’, and so it may be stated that superposition of the self-equilibrating stress state (M1, N1) does not affect the collapse load. An example of such a moment diagram M1 is the statically redundant moments MSP due to prestressing in a statically redundant system, as previously examined in Section 5.4. Another example is the internal forces developed in a statically redundant system due to support settlements. 261


Because of the above, it can be seen that the redistribution of moments due to creep in a statically redundant frame with different creep properties of its members (see Section 5.5.1) does not affect the collapse load of the frame. It is therefore obvious that, for design, it is neither necessary nor useful to follow the ‘elastic’ moment diagram strictly but instead an alternative, modified, diagram that is in equilibrium with the (factored) external loads. The only restriction, as previously pointed out, concerns the final relative rotation of the first created plastic hinge. g Since the rotation capacity (ductility) of the plastic hinge created depends directly on the capacity of the member to develop as large a curvature as possible, it is necessary to pay particular attention to the compressed concrete members, as is explained next. From the expression for curvature 1 "c þ "s ¼ r h and given that the presence of a compressive axial force limits the maximum ‘allowable’ value "c to the order of 2% (instead of 3.5% for pure bending), it becomes clear that the ability to develop a sufficient curvature for the requirement for plastic behaviour of a simultaneously bent and axially compressed reinforced concrete element is perceptibly limited. In order to deal with this unfavourable characteristic, a dense arrangement of stirrups (optionally set up in spiral form) may be provided. This results, as can be shown experimentally, in an increase in the compressive resistance of the concrete member and, accordingly, in an increase in the ultimate strain "c of up to three times, still depending on the cross-sectional area of the arranged stirrups. This provision is known as concrete ‘confinement’, and is usually applied to the extreme regions of each stressed element, by combined compression and bending (Paulay and Priestley, 1992). g In conclusion, any design decision leads, following the procedure in Section 6.6.1 for a specific load arrangement, to a specific collapse load, which then corresponds to a particular moment diagram that creates as many plastic hinges as required to make the structure a first-degree mechanism. This resultant collapse load is either lower or higher, or possibly equal to, the factored loads. In the first case, the selected design is unsafe. In the second case, the selected design is definitely safe, and economic as long as the deviation from the factored loads is smaller. The last case corresponds to the ideal solution. It is again pointed out that the development of sectional force diagrams due to temperature differences, support settlement or moment redistribution due to creep, although affecting the serviceability state, leaves the collapse load unaffected, since they are self-equilibrating diagrams.

6.6.3

Example

For a practical understanding of the above issues, the metal frame in Figure 6.27 is examined as an illustrative example. The loading shown in the figure is assumed to depict the factored loads for which the frame should be designed. The members of the frame are selected from the profile (HE-B) series. Obviously, the moment resistance Mpl is the side on which the fibres are under tension. 262

Frames 60 kN

163.4

60 kN 51.3

20 kN 2

HE-B 220

3

44.5

6.0 m

100.64 HE-B 140

1 3.0

6.0

HE-B 220

[M] based on constant EI

4

3.0

163.4

23.3

3.0

136.5 HE-B 140: Mpl = 61.1 kN m HE-B 220: Mpl = 205.9 kN m

166.47 24.52

0.12

166.47

0.40

0.80 0.80

0.58

49.01 120.0 [M] based on selected cross-sections 4.69

0.61

127.26 205.9

68.63

343.17

1.0

1/3

205.9

205.9

1.0

4.0

205.9

[Mr]

2.0 [M1]

(· λ)

205.9 205.9

9.98

892.23 By superposing the two diagrams the ‘static theorem’ is everywhere satisfied λu = 70.52 > 60.0

Figure 6.27 Design check on the basis of the static theorem

6.6.3.1 Elastic consideration Assuming constant rigidity (EI), the bending moment diagram is determined according to the statically redundant system. Considering the highest moments of each member as the required Mpl, the profile sections HE-B 140 for the inclined column and HE-B 220 for the remaining two members are selected. Since the diagram with the selected sections in Figure 6.27 satisfies the requirements of the static theorem everywhere, it can be concluded that the collapse loads of the frame are higher than the given loads. In order to confirm this last conclusion, the collapse load of the designed frame is sought, according to the procedure described above. The considered loading consists of the loads 0.333 (horizontal), 1 (vertical) and 1 (vertical), corresponding to the factored loads. In order to expedite the procedure described in Section 6.6.2, a thought experiment will now be undertaken. First, the elastic moment diagram, corresponding to the specific sectional properties of the frame members, is determined (see Figure 6.27). Obviously, the differences in this diagram to the one corresponding to constant EI are not significant. Next, the existing 263


ratio M/Mpl at each characteristic position in the frame is recorded. As the value of the parameter increases, it is logical to expect that the plastic hinge formation follows the sequence of the corresponding ratios M/Mpl, starting from the one with the highest value and going to the lowest one. Now, so many plastic hinges (r in number) are considered that the next one to be developed makes the structure a first-degree mechanism. In a well-designed frame, the number r of these initial plastic hinges is equal to the number of the redundant degrees of freedom of the structure — in this case equal to 3 — and converts the structure to a statically determinate one while remaining stable. In any case, however, a modified structure with r hinges may be considered, with the corresponding moments Mpl applied externally at their edges, thus leading to a bending moment diagram [Mr]. If [M1] is now the bending moment diagram of the same modified structure, due to external loads corresponding to ¼ 1 only — in this case 0.333, 1 and 1 — the value u of the parameter can be then determined so that the diagram [Mr] þ u [M1] causes an additional plastic hinge, thus leading the so-formed first-degree mechanism to collapse. In Figure 6.27, the ‘priority indices’ M/Mpl for each characteristic section of the frame, the diagram [Mr] and the diagram [M1] are shown. The next plastic hinge formation (which causes the frame to collapse) will occur at the top of the inclined column. The ultimate load factor u will result from the requirement 343.17 4.00 u ¼ 61.1, that is, u ¼ 70.52 > 60.0 (18% higher), thus justifying the static theorem prediction. According to the above, one should also check the relative rotation at the final stage of the first occurring hinge formation, i.e. that with the highest index M/Mpl, in this case 3. This relative rotation ’ can easily be obtained through the principle of virtual work, according to the relation ð Mreal 1 ’ ¼ Mvir EI where Mvir is the moment diagram of the modified structure with r hinges, due to the application of the external virtual moment 1 at the edges of hinge 3, and Mreal is the finally developed moment diagram [Mr] þ u [M1], which in this case is [Mr] þ 70.52 [M1] (Mvir and Mreal are not shown in Figure 6.27). This results in ’ ¼ 0.094 rad, which is an acceptable value according to Section 5.2.1. It is noted again that in the case where the value of the relative rotation of the first plastic hinge is not acceptable, the plastic hinge formation procedure should stop at an appropriate stage, followed by a corresponding reduction in the ultimate load.

6.6.3.2 Moment redistribution Now, a modification (redistribution) of the moments in the elastic diagram is attempted in order to obtain a more economical solution. First, hinges are placed at the same points as described in the previous section, in order to convert the system into a statically determinate one, on which the factored loads are applied. On applying an additional arbitrary self-equilibrating pair of moments to the hinges, a particular moment diagram results, which is in equilibrium with the external loads (Figure 6.28). 264

Frames 60 kN

60 kN 140.0

20 kN

6.67

140.0

HE-B 200

140.0

73.3 140.0 HE-B 100

HE-B 200

140.0

140.0 HE-B 100: Mpl = 26.0 kN m HE-B 200: Mpl = 160.1 kN m

6.67

140.0

164.32 14.15

0.02

164.32

0.54

53.21

1.03 0.80

0.33

128.3 116.86 4.58 160.1

53.36

160.1

1.0

1/3

160.1 266.83

1.0

4.0

[Mr]

2.0 [M1]

(· λ)

160.1 160.1

9.98

693.76 By superposing the two diagrams the ‘static theorem’ is everywhere satisfied λu = 60.20 > 60.00 (ideal)

Figure 6.28 Plastic design based on a ‘preselected’ moment diagram

The goal is to consider moment values for the above external pairs such that, by determining Mpl for each member according to the moment developed, a more economical solution compared with the ‘elastic’ solution is obtained. A value of 140.0 kN m is obtained after trials (the use of classic analysis software is essential) for all the external moments. The corresponding moment diagram is given in Figure 6.28. According to this diagram, the section HE-B 100 is selected for the inclined column and HE-B 200 for the remaining two members. The weight difference between the two solutions is obvious and, since the requirements of the static theorem are satisfied, the designed structure will need higher loads than the factored ones in order to collapse, and hence it is safe. For confirmation of the latter conclusion, the same ‘shortened’ procedure as described previously is followed. The ‘priority indices’ (M/Mpl) corresponding to each characteristic cross-section of the frame, the diagram [Mr] and the diagram [M1] are given in Figure 6.28. The next plastic hinge will occur again at the top of the inclined 265


column. The factor u of the collapse load will result from the requirement that 266.83 4.00 u ¼ 26.0 that is u ¼ 60.2 > 60 (Figure 6.28) It becomes clear that the economy of the solution is ascertained by this imperceptible (essentially nil) deviation of u from the ideal value of 60.0. It should be pointed out once more that the collapse loads in question are feasible only under the condition of sufficient deformation capacity (plasticity) of the cross-sections, particularly of the cross-section that is first plastified. This can always be checked by calculating the developed relative rotation of the first plastic hinge based on the principle of virtual work. Note that in the example the influence of the axial forces on the plastic moments was not taken into account, as this is negligible in practice.

6.7

Design and check of joints

The load-bearing action of a frame is based exclusively on the ability of its joints to transfer the sectional forces from one member to another. This process is not as obvious as perhaps the static analysis of a framed structure implies. The joint has to receive reliably the compressive and tensile forces as well as the corresponding shear forces of the sections of the adjacent members, remaining essentially undeformed. Taking up these forces causes a particular stress state in the joint, as the developed principal stresses in its interior — which constitute the only invariable criterion of the elastic behaviour (see Section 4.1.1) — do not result directly from considering the frame as such. A complex state that is described by two-dimensional (or even threedimensional) elasticity and not by one-dimensional elasticity (which is valid for framed structures) prevails in the joint. In steel frames, which are mainly formed by I-shaped cross-sections, where the compressive and tensile forces in the cross-section are essentially taken up by the flanges, continuing the ‘internal’ flanges inside the joint is sufficient, as shown in Figure 6.29. Thus, the joint formation as a closed, very stiff frame ensures its undeformability under the acting internal forces of the jointed members. However, due to the compressive and tensile forces acting on the joint, a ‘third’ force acting at both the external and the internal junctions of the joint needs to be provided in order to ensure local equilibrium. By limiting the examination in an external joint of a column and a girder, and with regard to moments causing tension at the outside fibres — as always happens for gravity loads — the system needs a compression force acting in the diagonal sense of the joint configuration, as shown in Figure 6.29. This compressive force cannot possibly be offered by the joint panel corresponding to the web of the I-section, because of the risk of buckling and, for this reason, a profiled segment (stiffener) is placed in the diagonal sense, in order to carry more reliably the arising compression force. In contrast, moments causing tension at the internal fibres, as can occur under 266

Frames

Compression

Tension

Layout of stiffening member

Compression

Tension

Figure 6.29 Load transfer in a steel frame joint

some forms of horizontal loading (e.g. earthquake), require the presence of a tensile diagonal force that can be provided by the node panel without any problem. g Turning now to concrete frames, it is useful to examine the possibility of approaching the two-dimensional stress state in the joint through an appropriate truss action. This idea is based on the fact that a two-dimensional stress state is composed of principal compressive and tensile stresses, as was mentioned in Section 4.1.1, and thus the internal forces acting in the structure can generally be represented by the members of a statically determinate truss formation. As has also been pointed out in Section 4.2.1.2, the compression members of this configuration correspond to the compressed uncracked concrete regions, whereas the tension members correspond to the reinforcement bars that should be placed in order to substitute for the cracked regions due to tension. The choice of a statically determinate configuration instead of a statically redundant one is preferable, in order to have a clear view of the restored equilibrium, regardless of the compatibility requirements. Indeed, ensuring equilibrium through a logically composed 267

Structural systems: behaviour and design Stirrup reinforcement

Compression

Tension

Stirrup reinforcement

Compression

Tension

Figure 6.30 Truss action in a concrete frame joint

model leads, according to Section 6.6.2, to a limit load smaller than the actual one (static theorem of plastic analysis). This logic can also be applied in the design and check of joints (Figure 6.30). Again, by limiting the examination to an external joint of a column and a girder, where the moments of connected members cause tension at the outside fibres, it is obvious from Figure 6.30 that the balance of forces in the interior of the joint is ensured without problem. This results because the diagonal region of the joint can offer the compression bar needed for equilibrium at both the exterior and interior edges. However, in the case where the internal fibres of the jointed members are in tension, a diagonal tension member is required, as shown in Figure 6.30, and this, of course, means that the stirrups of both the horizontal and the vertical members have to continue to the interior of the node, in order to receive the required diagonal tensile force (see Figure 4.18). Similar truss configurations can also be formed in other cases of joints (see Figure 6.30). 268

Frames

References Duddeck H. (1984) Traglasttheorie der Stabwerke — Beton Kalender, Teil II. Berlin: Ernst. Franz G., Scha¨fer K. (1988) Konstruktionslehre des Stahlbetons, Band II, B. Berlin: Springer-Verlag. Paulay T., Priestley M.J.N. (1992) Seismic Design of Reinforced Concrete and MASONRY Buildings. New York: John Wiley. Rosman R. (1983) Erdbeben-widerstandsfa¨higes Bauen. Berlin: Ernst. Salvadori M. and Levy M. (1967) Structural Design in Architecture. Englewood Cliffs, NJ: Prentice-Hall.

269

7 The influence of deformations on the state of stress — elastic stability 7.1

Overview

This chapter deals with problems that arise from the presence of high axial compressive forces, due mainly to gravity loads, in beams and frames. These problems link to the fact that the equilibrium of all structures is realised physically, without exception, at their deformed state. Thus, under certain conditions, it can be the case that the analysis based on the geometry of the undeformed structure, according to the usual practice ( first-order analysis), leads to erroneous and unsafe sectional forces. It is understandable that in such cases the design must necessarily take into account the influence of deformations on the state of stress of the structure; this is known as second-order analysis. The aim of this chapter is the practical understanding and the quantitative assessment, using first-order analysis, of such a state of stress of beams and frames.

7.2.

Buckling of bars

To understand the influence of the deflection on attaining the equilibrium in a compressed member, it is absolutely essential to first examine the centrally compressed bar, even though such a case is clearly a virtual one that cannot be encountered in reality. It is assumed that the centrally compressed straight bar is subjected to a pair of equal and opposite forces P applied to its free ends (Figure 7.1). It is intuitively immediately understandable that, for a ‘small’ value of the force P, a light horizontal transverse action will cause a transverse deformation, expressed through the deflection f, which will disappear once this action is removed. The understanding of this phenomenon is crucial. It is clear that if a beam is obliged to curve due to a transverse action, then it will develop an ‘internal’ bending resistance Mi ¼ EI/r, which should be in equilibrium with the bending moment Me induced by the external loads (r is obviously a function of f ). Of course Me ¼ P f þ M1, where M1 denotes the moment due to the transverse action. Thus, if Mi ¼ Me, and given that Mi > P f, once the transverse action is removed, the internal resistance takes precedence over the external moment and the bar reverts to its initial straight configuration. Assuming now that the force P is gradually increased, it would at some point reach a value Pcr at which a minimal transverse force Structural systems: behaviour and design 978-0-7277-4105-9



P

Mi = EI/r

Transverse action EI

(M1)

Pcr

f

Me = P · f + M1 Equilibrium: Mi = Me

P

P

Pcr No transverse action is practically needed in order to cause a curvature

Figure 7.1 Buckling as the limit state of equilibrium

will suffice (and therefore a minimal moment M1 too) to produce a deflection f in the bar. Therefore, for any imposed curvature 1/r on the bar, Mi ¼ Pcr f i.e. the bar will equilibrate at any arbitrary position, without offering resistance to any change in its curvature (see Figure 7.1). This automatically means that the bar will fail, given that under the action of the pair of sectional forces (N ¼ Pcr and M ¼ EI/ r ¼ Pcr f ) the strength of the cross-section will be exhausted as soon as the curvature (i.e. the deflection f ) reaches the appropriate value (see Section 6.5). The load Pcr is called the buckling load and, although it is an absolutely theoretical concept — given that in reality a perfectly central application of a compressive force cannot occur and the bar cannot be perfectly rectilinear — it plays an important role in dealing with practical problems concerning the assessment of the influence of deformations on the response, as will be explained below. The above relationship allows a rough approximation of Pcr by considering that 1/r 8 f/L2. However, the resulting value of 8 EI/L2 is about 20% less than the exact one. The reason for this deviation will be explained later. g For the precise determination of the critical load Pcr a more general reconsideration of the situation than the one that has just been described is appropriate. The same beam as above, subjected to equal and opposite compressive axial loads P, is considered, this time with a transverse load p(x) as well (Figure 7.2). In order to be deflected by w(x), the beam opposes a transverse resistance equal to qi ¼ EI (dw4/dx4) (see Section 2.3.6), and the fact that the beam is in equilibrium means that the external load p(x) is counterbalanced by the resistance qi, i.e. p(x) ¼ qi. The presence of axial forces P throughout the curved beam signals the development of transverse deviation forces pd, which add to the load p(x). According to Section 2.2.7, 272

Influence of deformations on the state of stress (elastic stability) p(x) P

P x w

EI EI(d4w/dx4): transverse resistance qi

Equilibrium: p(x) + P/r = qi (loading resistance)

P P P/r

p(x) = 0

Equation of buckling

Figure 7.2 Equilibrium of a compressed beam in a deformed state due to bending

this force is given by pd ¼ P/r ¼ P (dw2/dx2) which is obtained by adopting for the curvature the approximate expression (dw2/dx2) (a positive magnitude), instead of the strict one 1 w00 ¼ r ð1 þ w0 Þ3=2 The equilibrium must therefore be pd þ p(x) ¼ qi, which can be written as dw4 dw2 þ P ¼ pðxÞ dx2 dx4 This equation reflects the equilibrium of the compressed beam, in the presence of the transverse load p(x). However, in the previously examined case the transverse load p(x) is absent, so the equation is written as EI

dw4 dw2 þ P ¼0 dx2 dx4 This is the classical differential equation for buckling, which expresses, for the deformed state of the bar, the equilibrium between the deviation forces P (dw2/dx2) and the transverse bending resistance of the beam EI (dw4/dx4), both of which are mobilised through the imposed deformation w(x). The solution of this equation has the form: EI

w ¼ C1 sin ðkxÞ þ C2 cos ðkxÞ þ C3 x þ C4 where k¼

pffiffiffiffiffiffiffiffiffiffi P=EI 273


The treatment of this solution with respect to the coefficients C is carried out by taking into account the existing boundary conditions of the bar, which means the way in which the bar is supported, with the aim, of course, of determining the critical load Pcr. (The importance of the boundary conditions may be understood through the above rough calculation of the critical load. The deviation of the value by 20% is due to the fact that at the ends of the bar the curvature was considered to be 8 f/L2 (and not zero as it ought to be, because of the null bending response prevailing there). This treatment results in the following expression for the buckling load: Pcr ¼

p2 EI s2k

The magnitude sk, which is known as the effective length, depends on the way in which the bar is supported, and it may be considered as the distance of those points of the deformation line of the bar which exhibit zero curvature, i.e. the inflection points. At these points the bending moment of the bar is zero. Thus, for a bar hinged at both ends sk ¼ L, for a cantilever sk ¼ 2 L, for a fixed simply supported bar sk ¼ 0.70 L and for a fixed-ended bar sk ¼ 0.50 L (Figure 7.3). In all cases should be ensured that the point of application of the compressive force P can be axially displaced, as shown in Figure 7.3. In the case of a fixed-ended bar with one end free to move transversally, sk ¼ L. As is apparent from the last equation, the critical load increases (i.e. the risk of buckling reduces) as EI increases and sk decreases. It is also interesting to consider the critical axial stress cr, which is developed when the buckling load is reached: cr ¼

Pcr p2 E ¼ 2 A P

P

P

P

sk = 0.70 · L

sk = L

P

sk = 0.50 · L

sk = 2 · L

sk = L

Slenderness λ = sk √A/I Pcr = E · A · π2/λ2 = EI · π2/s2k

Figure 7.3 The influence of the support type on the value of the buckling load

274

Influence of deformations on the state of stress (elastic stability)

where is a dimensionless quantity, the slenderness, which is the ratio of the effective length sk to the radius of inertia i of the cross-section: pffiffiffiffiffiffiffiffi i ¼ I=A ¼ sk/i The slenderness is the measure of dangerousness of the axial compressive response of a bar — the greater the value of , the lower the compressive stresses that the bar can carry, thus leaving the possibility of the cross-section taking on higher stresses unexploited. g In the above expression for the buckling load Pcr a constant rigidity EI is assumed over the entire bar length, which may not correspond to reality. For this reason, a more general method for determining the critical load Pcr that is of particular importance practically is examined below for the determination of the critical load Pcr. This method may be applied in any case, and is known as Vianello’s method (Menn, 1990). It is assumed that the bar, which is axially loaded with the load P, has an initial eccentricity equal to w0(x) (Figure 7.4). Under the bending moments M0 ¼ P w0(x) the resulting deflection w1 at a certain point of the bar can be calculated directly as a function of P on the basis of the principle of virtual work (see Section 2.3.3). The moments M1 ¼ P w1(x), in turn, also produce an additional deflection w2. The procedure, provided the load P is less than the critical one, converges, and the final deformation w is w ¼ w0 þ w1 þ w2 þ . . . If the function w0(x), and consequently also w1(x), is affine to the buckling shape of the bar, then it can be proved that w1 ¼

P w Pcr 0

ðaÞ

Moreover, w ¼ w0

1 1 P=Pcr

ðbÞ

Due either to initial deformation or to transverse load P

w0 w1

P

P

w0

First-order deflection due to compressive load

W Final deflection curve

If w0 = w1 then P = Pcr

ξ = P/Pcr, w = w0/(1 – ξ)

P

Figure 7.4 The criterion for determining the critical compressive load

275


Equation (a) allows the direct determination of the critical load Pcr, as for P ¼ Pcr it must be that w1 ¼ w0. More precisely, for a selected position x ¼ x0, e.g. where the maximum w occurs, the deflection w1(x0) is calculated as a function of the load P and the corresponding value w0(x0). Then, by equating the two deformations w1(x0) and w0(x0), the critical load Pcr is obtained directly (see Figure 7.4). Equation (b) allows the assessment of the influence of the deformations on the response, as will be explained later. It should be noted that the concept of w0 is not necessarily linked to an initially deflected beam exhibiting some sort of imperfection. It can also be simply considered as the first-order deformation produced by an arbitrary transverse load acting on the beam. The elastic deformation w0(x) thus developed necessarily satisfies the appropriate boundary conditions, and this ensures (as discussed previously) that the final result will approximate the ‘accurate’ one as closely as possible. Thus, as considered above, the buckling load Pcr is the load which, for an initial deformation w0(x) respecting the boundary conditions of the beam, produces through first-order analysis the same deformation as the initial one at some selected point (usually the point having the maximum w). g The following example (Figure 7.5) illustrates the procedure more precisely. In order to assess the buckling load of the fixed-ended beam, with one end freely movable as in Figure 7.5, the developed deflection w0(x) under a transverse load q is first determined. This load is calculated, on the basis of the corresponding bending diagram [M0], as the bending diagram of a simply supported beam loaded with [M0]/EI, according to Mohr’s theorem (see Section 2.3.6). The maximum deflection at midspan (x ¼ L/2) is w0 ¼ 0.0026 q L4/EI. This deflection w0(x) is considered as the initial one, and the axial load P, after developing the corresponding bending moment diagram [M], produces at the same q

[M0]

0.0003

0.0011

0.0018

0.0024

0.0026

q · L /12

0.0024

q · L /12

0.0018

2

0.0011

0.0003

( · P · q · L4/EI ) 2

P

[M] L First-order moment diagram due to the compressive load

0.125 · L EI w0 The initial deformation due to the transverse load satisfies the boundary conditions

0.125 · L 1

[M1]

0.125 · L Virtual loading

The critical load causes a first-order deflection equal to the initial one

Figure 7.5 Estimation of the critical compressive load

276


point (x ¼ L/2) the deflection w1. This deflection is calculated, according to the principle of virtual work, according to Figure 7.5, and is given by ðL M M1 L2 dx ¼ 0:025 P w0 w1 ¼ EI EI 0 The critical buckling load results from the requirement w1 ¼ w0 at the specific point considered. It is obtained as Pcr ¼ 40:0

EI L2

This value is just 1% higher than the ‘accurate’ one.

7.3

The influence of deformation on the response of beams (second-order theory)

As is clear from the foregoing, the presence of axial forces in a beam de facto modifies the bending moment diagram and, consequently, the deformations too, which are anyway caused by the bending response. The beam, which carries a transverse loading and is subjected to compressive axial loads P (Figure 7.6), develops a deflection line that obeys the differential equation examined in Section 7.2. It is clear that both the deformation developed and the resulting bending response of the beam are greater than those determined on the basis of first-order theory. Repeating once again the result of Section 7.2, if ¼ P/Pcr and W1 represents the deformation of the beam according to a first-order result, then the final deformation

P

P

W1 w

P

P W1

Final deformation [second order]

Final deflection curve

Initial deformation

First-order deflection curve ξ = P/Pcr

w = W1/(1 – ξ) Increased moments relative to first-order values

w0 P

P

Under constant axial load the superposition is valid

P

w1

P

First-order deflection curve First-order deformation: W1 = w0 + w1

Figure 7.6 Estimation of the final deformation based on the value of the critical load

277


w of the beam will be w ¼ W1

1 1

The presence of a compressive load imposes on the initially (to first order) developed deformation W1 the magnifying factor 1/(1 ), which increases steadily as the compressive load approaches the value Pcr. Thus, for example, for a compressive force equal to 50% of Pcr the value of W1 is doubled, while for 80% of Pcr the value of W1 is quintupled. It is also found that, as the compressive force remains constant, the deformation due to a transverse load is proportional to it, i.e. the principle of superposition is valid, but only under the above condition. If M1 represents the first-order bending moments, the actually developed bending moments M in the beam are M ¼ M1 þ P w

g

In order to reinforce the understanding of the concept of W1 in the above expression for the final deformation w, a compressed beam subjected to both an initial deformation w0 and a transverse load (see Figure 7.6) is considered. Assuming that a deformation w1 is produced to first-order by the load, then, by superposing the individual actions (which is legitimate, given that they are both referred to the same compressive force P), according to Section 7.2, w ¼ w0

1 1 1 1 þ w1 ¼ ðw0 þ w1 Þ ¼ W1 1 1 1 1

i.e. W1 is the sum of the first-order products w0 and w1. g The above analysis is now applied to three illustrative examples. Example 1. A column with a fixed base and height L, in addition to a transverse load q, is subjected at its top to a horizontal force H, a moment M0 and a compressive load P (Figure 7.7). The horizontal deflection wA that is developed at the top of the column is calculated on the basis of the corresponding first-order deflection W1A : W1A ¼

q L4 H L3 M0 L2 þ 8 EI 3 EI 2 EI

Based on the buckling load Pcr ¼

p2 EI 4 L2

it is obtained that 1

wA ¼ W1A 1 278

4 P L2 p2 EI

Influence of deformations on the state of stress (elastic stability) P

M0

H A

Final deformation

First-order deformation L

q

EI

W1A

F

wA

Figure 7.7 Assessment of the final deformation

It is clear that the presence of the compressive force increases the deformability of the column. The bending moment at the base of the column F is given by ! q L2 F F A M ¼ M1 þ P w ¼ þ H L M0 þ P wA 2 Example 2. The simply supported structure shown in Figure 7.8 is subjected at its ends to the loads P, and develops in the midspan a deflection wm, resulting from the corresponding first-order deflection W1m , by applying the magnifying factor 1/(1 ). W1m is given by W1m ¼

P e L2 8 EI

and given that Pcr ¼

p2 EI L2

P

P e

m

W1m First-order deformation

Wm Final deformation

Figure 7.8 Assessment of the final deformation

279


A

H=1

Final deformation L

EI

wA

kR = 1/w A

F The stiffness decreases as the compressive load approaches the critical value

Figure 7.9 Assessment of the transverse stiffness of the cantilever

it results that P e L2 w ¼ 8 EI

1

m

1

P L2 p2 EI

The bending moment in the middle of the beam is m m Mm ¼ Mm 1 þPw ¼PeþPw

Example 3. The horizontal stiffness kR at the top of the column shown in Figure 7.9 is obviously influenced by the applied compressive load P. According to first-order theory, this stiffness is equal to kH ¼ 3 EI/L3 (see Figure 3.24). The horizontal force H ¼ 1 produces, according to Example 1, a displacement at the top equal to wA ¼

L3 3 EI

1

4 P L2 1 2 p EI and consequently the corresponding stiffness kR, i.e. the opposing resistance of the cantilever for an imposed unit displacement on its end, is 1/wA, which is written as 4 P L2 P kR ¼ kH 1 2 ¼ kH 1 Pcr p EI It can thus be seen how the presence of a compressive force leads to a reduction in the beam’s stiffness. Indeed, by approaching the critical buckling value, this stiffness tends to disappear. The resulting bending moment at the base F due to a horizontal force H at the top is: MF ¼ H L þ P wA 280

g


Thus it is clear that for a compressed beam any eccentricity in the compressive load distinguishes it from a centrally compressed beam, in that it develops a specific deformation and bending response, which should be taken over by the corresponding cross-section. In a purely axially compressed bar no bending deformation occurs until the compressive load reaches its critical value, and thus the smallest disturbance is fatal for the structure. Therefore, the check of the strength capacity of a beam subjected to a compressive service load P, having first-order deformations and bending moments W1 and M1, respectively, for a global safety factor applied to service loads, will be carried through on the basis of an axial force N¼P and a bending moment M ¼ M1 þ P W1

1 P 1 Pcr

It should be checked, of course, that the pair (M, N) is ‘covered’ by the capacity of the section, as described in Section 6.5. g Having examined the compressed beam, it is useful now to consider a transversely loaded beam under an additional tensile axial load (Figure 7.10). The presence of tensile forces P throughout the curved length of the beam means — analogously to Section 7.2 (see Figure 7.2) — the development of transverse deviation forces pd ¼ P/r which, having the same sense as the bending resistance of the beam, oppose the load p(x). The equilibrium requires now that p(x) ¼ qi þ pd, and the corresponding equation takes the form EI

dw4 dw2 P ¼ pðxÞ dx2 dx4

P

P w W1 Final deflection curve First-order deflection curve ξ = P/Pcr w = W1/(1 + ξ) Reduced moments relative to the first-order ones Valid superposition under a constant axial load

Figure 7.10 Assessment of the final deformation under a tensile axial load

281


In this case it is understandable that both the developing deformation and the resulting bending response of the beam will be less than those determined on the basis of the first-order theory. The final deformation w is derived through an approach analogous to the one described previously, and is given by the relation w ¼ W1

1 1þ

ð ¼ P=Pcr Þ

It is clear that the reduction coefficient 1/(1 þ ) as applied to the first-order deformation W1 due to the transverse load, may take, depending on the acting tensile load, some small value. It is found here also that, under a constant tensile force, the deformation w due to a transverse load is proportional to it, i.e. the principle of superposition is still valid. The developed bending moments M of the beam are M ¼ M1 P w where M1 are the corresponding first-order moments. The model of the tensioned beam has a direct application both in the analysis of suspension bridges and in calculating the torsional response of thin-walled beams due to prevented warping, as will be examined in Chapters 9 and 13, respectively.

7.4

The influence of deformation on the response of frames

The columns of frames are compressed members par excellence and, when their compressive response increases significantly with respect to their slenderness, their behaviour as frame members leads generally to a reduction in the horizontal stiffness of the frame (Figure 7.11). Of course, when this stiffness disappears, the applied vertical loads have reached their critical values, which leads directly to collapse. The purpose of this section is to show how one can obtain a very good approximation of the frame response under the influence of occurring deformations, using only firstorder analysis.

Vertical loads reduce the lateral resistance offered to horizontal loads

Figure 7.11 The influence of vertical loads on the behaviour in response to lateral loads

282


7.4.1

One-storey frames

The behaviour of a frame with significant axial forces on its columns against lateral displacements may illustrated using with the model shown in Figure 7.12 (Franz et al., 1991). The frame, which is loaded only by the vertical loads P1 and P2 at its joints, may be considered as equivalently consisting of two interconnected, geometrically identical systems. The first system, having its members connected through hinges at their ends, has all the vertical loads applied at its nodes, being itself connected through a hinged bar to the unloaded actual frame (second system) at the level of the girder. First, the critical sum of vertical loads has to be determined that renders the system elastically unstable, i.e. ready to collapse. An imposed small horizontal displacement causes an oblique position of the hinged columns, as shown in Figure 7.12. From the existing equilibrium of the nodes, involving the vertical loads and the inclined compression forces of the columns, arise horizontal forces Ha that act on the monolithic frame. The frame obviously opposes a certain resistance Hp against the imposition of the displacement . From the figure it can be seen that the action Ha consists of the sum of the horizontal components of the inclined compressive forces N in the columns. Given the small inclination angles of the columns, X X X X P tan ¼ P tanð=hÞ P ð=hÞ ¼ ðP=hÞ Ha ¼ where h is the column height. P1

P2

h

Total deviation force P1

Resistance of the frame

P2

δ Ha P1 h

ψ2

P2

Hp

δ

ψ1 Ha < Hp

kH: horizontal stiffness

At the critical load the deviation force equals the resistance of the frame Critical load = kH · h (columns of equal height)

Figure 7.12 Fictitious load-carrying mechanism for the critical vertical load

283


If kH represents the lateral stiffness of the frame, then the resistance Hp will be Hp ¼ kH It should be noted that the resistance of the frame Hp does not depend on the vertical loads, whereas the action Ha does depend both on the vertical loads and on the angle of inclination of the columns. For a relatively low value of vertical loads Hp > Ha and, after removing the displacement , the system returns to its initial undeformed configuration. However, if the loads take values that for any imposed are Hp ¼ Ha, the critical load will be reached and, consequently, collapse will occur. In this case, from the previous equations it can be deduced that the critical loads must satisfy the relationship P (P/h)cr ¼ kH and, if the columns have the same height h, P ( P)cr ¼ kH h g Having considered how the above model works, it is useful to confirm the above results by examining the actual frame more directly (Figure 7.13). A horizontal displacement is again considered, which is imposed on the frame in Figure 7.13, which has a total P vertical load ( P) acting on its nodes. The oblique action of P the compressive force N of the columns in their inclined position under the loads ( P) is equivalent to P the action of a horizontal force Ha ¼ P (/h) on the undeformed configuration of the frame. If this force suffices to P exhaust the resistance of the frame Hp ¼ kH , i.e. if Ha ¼ Hp, then the critical load ( P)cr has been reached. If Ha < Hp, then obviously the frame comes back to its initial configuration. P1

P2

P3

Resistance of the frame

Total deviation force

Hp

δ

δ

δ

Ha

h

kH: horizontal stiffness At the critical load the deviation force equals the resistance of the frame Critical load = kH · h (columns of equal height) P1

P2

P3 Total deviation force

δ

δ P1

δ P2

Ha P3

The deviation force results from the horizontal components of the oblique column forces

Figure 7.13 Determination of the critical vertical load of the frame

284

Influence of deformations on the state of stress (elastic stability) Initial displacement δ0 ∆δ Additional displacement P1

The additional displacement is due to the fictitious horizontal loading

P2 Ha

Figure 7.14 Increment in the initial displacement due to the influence of vertical loads

P Thus, it becomes apparent that the vertical loads acquire their critical value ( P)cr when the activated horizontal force Ha, due to an arbitrary deviation , causes the very same , following a first-order analysis. g It should now be understood that an initial (unintentional) constructional eccentricity 0 in the frame, with respect to the undeformed geometrical configuration, will in the presence of (significant) vertical loads P inevitably increase by (Figure 7.14). The final equilibrium configuration of the frame with a total displacement ¼ 0 þ , may be conceived as a consequence of a fictitious external horizontal force Ha acting at the level of the beam. According to the above, this is equal to X þ P 0 Ha ¼ h This force imposes on the frame the additional displacement and is given by Ha ¼ kH According to the previous expression for ( obtained that

P

P)cr for columns of equal height, it is

1 ¼ 0 P ð PÞcr P 1 P Thus, the final displacement ¼ 0 þ is ¼ 0

1 P P 1 P ð PÞcr

It is found that the initial deviation is magnified, because of the presence of compressive loads, by the same factor that is applied in the case of beams (see Section 7.3). The initial presence of a horizontal force H0 automatically means the existence of an initial deviation 0 and, according to the above, the development of a further deformation leading to an additional response (Figure 7.15). The total response may be conceived as being due to a fictitious horizontal force H which, caused by the 285

Structural systems: behaviour and design Total displacement δ Initial displacement δ0 ∆δ Additional displacement P1

P2

P1

P2 H

H0

Fictitious loading H = H0 + Ha Ha The column shear forces balance the horizontal load H

H

The column shear forces balance the fictitious load

The fictitious loading causes the total displacement

... but they are reduced by the horizontal components of the inclined column forces

Figure 7.15 A fictitious horizontal force as an equivalent action on the frame

oblique position of the columns, is given by H ¼ H0 þ Ha ¼ H0 þ

X P h

This force obviously produces the total displacement (see Figure 7.15), which means that H ¼ kH and, consequently, ¼

H0 P P kH h

This result may be easily verified as being identical to the previous one for columns of the same height. If this is not the case, then ¼

286

H0 PP kH h


The above expression for H takes the following form after substituting from the equation above for columns of the same height: H ¼ H0

1 P P 1 P ð PÞcr

i.e. the final bending response is derived from the first-order one due to the forces H0 by applying the standard magnifying factor: 1 P P 1 P ð PÞcr From the above results the activated additional force Ha can be written as 1 Ha ¼ H0 P ð PÞcr P 1 P It should be made clear that this horizontal force Ha is fictitious, and its purpose is simply to be able to represent the equilibrium in the deformed state (second-order theory) by using first-order analysis, based on the (fictitious) horizontal load H ¼ H0 þ Ha applied on the undeformed frame. The column shear forces, of course, obviously equilibrate the force H0 and not the force H. However, this is apparently in conflict with the fact that the increased bending response of the columns due to the forces H leads to increased shear forces, which should equilibrate H rather than H0. The ‘paradox’ is removed if the inclined position of columns is taken into account, whereby the horizontal components of the compressive axial forces do actually reduce the ‘increased’ shear forces due to the force H. Thus, the actual shear forces do indeed equilibrate the horizontal force H0 (see Figure 7.15). g Clearly, the resistance (stiffness) offered byP a frame when a unit horizontal displacement ( ¼ 1) is imposed in the presence of loads P is equal to the force R that produces such a displacement. According to the last results, for columns having the same height P P R ¼ kH h while for columns of unequal height R ¼ kH

XP h

Thus, it has clearly been shown how the horizontal stiffness of the frame is reduced due to the vertical loads (see Figure 7.11). This stiffness vanishes, of course, when the critical load is reached. 287


7.4.2

Multi-storey frames

7.4.2.1 The influence of deformation The behaviour of multi-storey frames is examined in a similar manner to that of onestorey frames. Clearly, an initially oblique position of the columns under the presence of considerable axial forces, even with a small deviation from the vertical due either to a constructional imperfection or to the first-order effect of the application of horizontal forces, raises some additional moments, which increase this deviation even more. The final deviations from the vertical cause the activation of ‘fictitious’ horizontal forces at the joints of the frame, which may be considered as being applied to the undeformed frame, producing (by first-order means) the very same deformations that have mobilised them. The problem in multi-storey frames lies simply in the fact that at each storey level there is an unknown final horizontal drift, which necessarily has to be determined. The procedure is illustrated using the example of a three-storey frame P 7.16) P (Figure , P2 and (Franz et al., 1991). At each storey level the total vertical loads P 1 P P3 are acting, while the initial clockwise deviation angles of the columns from the vertical are 10, 20 and 30, respectively. It is clear that these initial angles will increase and reach their final values 1, 2 and 3. Consequently, the initial storey drifts 10, 20 and 30 will finally attain the final values 1, 2 and 3, respectively. It is understandable that in the final equilibrium state the inclined compressive forces of the columns, according to previous considerations (see Section 7.4.1), provide, through their horizontal components, deviation forces of the same sense, because those resulting on the lower side of the corresponding girder are always greater than those on the upper side, as can be seen in Figure 7.16. P3 (H0,3)

P3 Ha,3 δ3

δ30 θ30

ΣP3

P2 (H0,2)

Ha,2 δ2

δ20 θ20

Σ(P2 + P3) h2

θ2 Σ(P2 + P3)

P1 (H0,1)

ΣP3

θ3

h3

θ10

Ha,1

δ1

δ10 h1

θ1 Σ(P1 + P2+ P3)

The vertical loads cause an increase in the initial deviations (possibly due to horizontal forces)

Figure 7.16 The activation of ‘fictitious’ horizontal forces

288

Additional fictitious forces for the first-order calculation


Thus the following deviation forces act at the girder level of each storey: P Ha,3 ¼ [3 P3] P P P Ha,2 ¼ [2 ( P3 þ P2) 3 P3] P P P P P Ha,1 ¼ [1 ( P3 þ P2 þ P1) 2 ( P3 þ P2)] The final deformation of the frame will consist of the initial deformation plus the one produced by the forces Ha,1, Ha,2 and Ha,3, following of course a first-order procedure. Introducing the flexibility coefficient ik as that displacement of level i which is produced by a unit horizontal force acting at level k, as shown in Figure 7.17, it can be written that (see Figure 7.16) 1 ¼ 10 þ Ha;1 11 þ Ha;2 12 þ Ha;3 13 ¼ 1 h1 2 ¼ 20 þ Ha;1 21 þ Ha;2 22 þ Ha;3 23 ¼ 1 h1 þ 2 h2 3 ¼ 30 þ Ha;1 31 þ Ha;2 32 þ Ha;3 33 ¼ 1 h1 þ 2 h2 þ 3 h3 Substituting the above expressions for Ha,1, Ha,2 and Ha,3, a linear system of three equations with respect to the unknown deviations 1, 2 and 3 is obtained. Thus, the effective actions Ha,1, Ha,2 and Ha,3 can be readily determined and the state of stress of the frame may be deduced according to classical first-order procedures. In this way, the final deformation and state of stress of the frame may be found as the result of the coexistence of the initial deformations 10, 20 and 30 and the vertical loads. It is clear that if 10, 20 and 30 are due to the action of horizontal loads H0,1, H0,2 and H0,3 (see Figure 7.16) under first-order conditions, then the fictitious forces Ha,1, Ha,2 and Ha,3 resulting from the above system have to be added to the initial ones. Thus the final response of the frame will, under first-order consideration, correspond to the horizontal forces H1 ¼ H0,1 þ Ha,1 H2 ¼ H0,2 þ Ha,2 H3 ¼ H0,3 þ Ha,3 1

Level 3 δ33

δ32

δ31

δ23

δ22

δ21

δ13

δ12

1

Level 2

1

Level 1 δ11

Flexibility coefficients

Figure 7.17 Description of the horizontal flexibility of the frame

289


7.4.2.2 Critical load A factor is now sought which, by magnifying the vertical loads of the frame, makes them critical for the elastic stability of the frame. Clearly, in this case the vertical loads should have such values that the activated forces Ha,1, Ha,2 and Ha,3 due to an arbitrary deviation configuration 1, 2 and 3 can produce these very same displacements, following a first-order procedure. By omitting the initial deviations, the equations given above (see Section 7.4.2.1) may be rewritten as 1 ¼ Ha;1 11 þ Ha;2 12 þ Ha;3 13 ¼ 1 h1 2 ¼ Ha;1 21 þ Ha;2 22 þ Ha;3 23 ¼ 1 h1 þ 2 h2 3 ¼ Ha;1 31 þ Ha;2 32 þ Ha;3 33 ¼ 1 h1 þ 2 h2 þ 3 h3 where Ha,3 ¼ [3

P

P3] P P P Ha,2 ¼ [2 ( P3 þ P2) 3 P3] P P P P P Ha,1 ¼ [1 ( P3 þ P2 þ P1) 2 ( P3 þ P2)]

It is clear that the above relationships lead to a homogeneous system with respect to 1, 2 and 3, i.e. a linear system in which all the constant terms are equal to zero. As it is well known, such a system has a solution different from zero only if the determinant of the coefficients of the unknowns 1, 2 and 3 is equal to zero. This requirement leads to a third-degree algebraic equation with respect to , and it is obvious that the minimum real value of satisfying this equation will represent the sought factor.

7.5

Lateral buckling of beams

The resultant force of the compressive stresses on a beam subjected to gravity loads represents a danger for the ‘stability’ of the upper part of the beam section, and consequently for the whole beam, given the reduced moment of inertia of this part (and of the beam) about the vertical axis of symmetry. The situation is best visualised using the example of a truss in which all the compression members have a buckling length sK (see Section 7.2) approximately equal to their actual length, whereas the whole upper (compressed) chord has as ‘buckling length’ equal to the whole span length for a possible deformation out of the truss plane. Under certain conditions the upper part (flange) of a symmetrical beam may bend out of the plane of symmetry, even if the beam loads act in this plane (Figure 7.18). It is understandable that this deformation is inhibited by the torsional stiffness of the beam, which is obviously mobilised; while, on the other hand, the presence of the longitudinal internal tension force in the lower part of the beam plays a relieving role. This situation can become crucial for the stability of the beam if the bending moment induced by the beam loads reaches a certain critical value. According to a classical result referring to a simply supported rectangular beam, with ends restrained against torsion, and loaded in its vertical 290


Truss visualisation

Very high slenderness in the lateral direction

Through the lateral deformation the torsional resistance is mobilised

Applying the load at the bottom side relieves the response of the beam

Figure 7.18 Visualisation of the lateral buckling

symmetry plane, this value Mcr depends on the moment of inertia Iy of the beam section about its vertical axis, the torsional moment of inertia IT and the beam length L (Gaylord and Gaylord, 1957): qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Mcr ¼ G IT EIy L where the dimensionless coefficient depends on the form of the bending diagram M and takes the following values: . . . .

for constant M, ¼ 1.0 for parabolic M (uniform load), ¼ 1.13 for triangular M (concentrated load at midspan), ¼ 1.35 for a moment M applied at one end, ¼ 1.77.

The above expression is approximately valid also for I-beams having a section of double symmetry. It should be noted that the side on which the beam load is applied influences the examined response, as may be seen in Figure 7.18. Applying the load on the top side leads to a more adverse response than applying it on the bottom side, as in the last case a ‘relieving’ torsional moment is present throughout the beam length. g The case is now examined where a simply supported beam of depth h with a single plane of symmetry (the vertical one) has an initial deformation which is characterised by the horizontal deviations e0 and eu of the upper and lower part of the section (flange), respectively. The upper and lower part actually provide the bending resistance to the lateral response through their corresponding moments of inertia Io and Iu about the 291

Structural systems: behaviour and design M D

P

w

Final deformation

e0 I0 h

Z

IT Iu

e0

eu Midspan section

w

Initial deformation

eu L D = Z = P = M/h

D M

P

h

h Z

Analogy with the initially deformed fictitious beam P EI* e* w Final deformation according to second-order theory L

Figure 7.19 Towards the assessment of the final lateral deformation

vertical axis of symmetry (Figure 7.19). Under the action of the compressive force P ¼ M/h arising from an externally applied moment M, the deviation e0 will increase and reach the value w. The beam will withstand this deviation and not lose its stability, and the value of w determines the response of the upper section under the compressive force P and the bending moment M ¼ P w acting in the corresponding horizontal plane. The whole problem may be described by analogy to a fictitious compressed and transversely loaded simply supported beam, such as the one examined in Section 7.2. This fictitious beam has an appropriate moment of inertia I together with an appropriate initial deformation e and is subjected to the same compressive force P as that of the upper part of the actual beam (Mann, 1976). At first, the critical compressive force PK of the upper part (flange), which is fatal for the stability of the considered beam (PK ¼ Mcr/h), may be calculated with satisfactory accuracy through the familiar expression for the buckling load: PK ¼

p2 EI L2

where I ¼ Io ðc1 þ c2 Þ 292


and sffiffiffiffiffiffiffiffiffiffiffiffi 1 I c1 ¼ 1 u 2 Io sffiffiffiffiffiffiffiffiffiffiffiffi I c2 ¼ 1 þ u Io sffiffiffiffiffiffiffiffiffiffiffiffi L G IT ¼ h E Io Now, after expressing the fictitious initial deformation e as P 1 2 2 eu c2 þ 1 c2 e ¼ e0 1 þ P þ PK c 2 eo the final deformation w is found to be w ¼ e

1 1

P PK

It is remarkable that this result is identical to equation (b) in Section 7.2. pffiffiffi It may be observed that in the case when Io ¼ Iu, and where c1 ¼ 0 and c2 ¼ 2, the last expression for PK coincides with the one given previously for the critical bending moment of a beam having a rectangular section (or one with double symmetry). It should be noted that, in the case of a reinforced concrete beam, the practically negligible horizontal rigidity of its lower part, due to the existing cracked state, may be taken into account by considering Iu ¼ 0 in the above expressions. g The problem of lateral stability is usually resolved by inserting a horizontal bracing connecting the compressed longitudinal upper parts of two adjacent beams, or the upper chords of two adjacent trusses (Figure 7.20), in order to make a combined element, which exhibits a strong lateral rigidity in its compressed part. This combined element may serve as ‘lateral support’ for the remaining beams or trusses (see Figure 7.20).

7.6

Plastic analysis

7.6.1

General remarks

When a frame is to be dimensioned for horizontal forces accompanied by strong compressive column forces on the basis of the ultimate limit state design (see Section 6.6), it is necessary to take the influence of the deformations into account. In particular, the final displaced configuration of the frame must be assessed, in order to establish those fictitious horizontal forces that will enable the analysis of the frame according to first-order considerations (see Section 7.4). 293


Bracing

Beam or truss

Beam or truss

Figure 7.20 Lateral bracing of beams or trusses

For a practical treatment of the problem, an appropriate factor is actually sought which, applying to all loads on the frame, will lead the frame to collapse on the basis of a corresponding bending moment diagram, possibly also taking into account the interaction of the internal forces M and N (see Section 6.5). Through this moment diagram, the determination of the storey drift is also possible, using the principle of virtual work. The limit value of the horizontal force, which acts externally at each girder level, therefore includes the fictitious horizontal force corresponding to the deformed (inclined) position of the columns (this force obviously not acting in reality). Thus the removal of this force from the limit horizontal force at each girder level allows the determination of the actual ultimate horizontal forces, in the context of the deformed configuration, as well as of the load magnifying factor leading to the collapse of the structure (Duddeck, 1984). The following example illustrates the basic points described above.

7.6.2

Numerical example

The steel frame shown in Figure 7.21 is made up of the cross-section HE-B 200 and is acted on by the vertical load P, as well as by the horizontal force 0.1 P. The ultimate value of the load P leading to failure is to be determined. The limit moment diagram (a) which leads the frame to collapse may be easily deduced from the ‘elastic’ one (see Section 6.6). In particular, on the statically determinate frame having a hinge in place of a fixed support, the plastic moment of the cross-section is applied externally. Then, the horizontal load is sought which will cause a plastic hinge also at the top of the column. This should balance the column shear force, as may be deduced from the horizontal equilibrium of the girder: 0.1 P ¼ 2 160.0/7.50 ¼ 42.66 kN Thus the corresponding vertical load has the value P ¼ 426.6 kN. 294

Influence of deformations on the state of stress (elastic stability) P 1

42.70

3.37

160.0

0.1 · P

HE-B 200 I = 5700 cm4 Mpl = 160 kN m

7.50 m

δ

1

7.50

7.50

(a)

160.0

2.50 m

4.13 Elastic diagram

160.0 Ultimate moment diagram (first-order)

Virtual loading

0.1 · P Equilibrium in the ultimate moment diagram 2 · 160.0/7.50 = 42.70 δ

42.70

31.0 kN

11.70 kN Fictitious horizontal load (Omission of the interaction (M, N))

First-order ultimate horizontal force

Second-order ultimate horizontal force

Figure 7.21 Determination of the ultimate value of the horizontal load

However, as it has been explained, this ultimate horizontal force will necessarily include the fictitious horizontal force 426.6 (/7.50), where is the horizontal displacement of the frame. The latter is readily determined, according to the principle of virtual work, on the basis of the bending diagram shown in Figure 7.21(a), considering as the virtual loading a unit horizontal load acting on the statically determinate frame (see Figure 7.21). It is obtained that (see Section 2.3.3) 1 ¼ (160.07.50/2320.07.50/6)7.5/(2.15700)þ(160.07.50/3)2.5/(2.15700) ¼ 0.205 m Thus the fictitious horizontal force is equal to 426.6 (0.205/7.50) ¼ 11.66 kN, and consequently the actual ultimate force is 42.66 11.66 ¼ 31.0 kN and the ultimate vertical load is 31.0 10 ¼ 310 kN. It is also be interesting to check the ‘distance’ of this load from the critical buckling load of the frame. As the (first-order) lateral stiffness of the frame is kH ¼ 261.8 kN/m, the critical load will be (see Section 7.4) Pcr ¼ 261.8 7.5 ¼ 1963.5 kN 310.0 kN. g 295


However, it should be noted that the influence of the axial load on the ultimate bending strength has been omitted in the above. For the cross-section type used, HE-B, the following relationship may be used for the interaction between M and N (see Section 6.5.1): N M ¼ Mpl 1:1 1 Npl Given that, in this case, Npl ¼ 1601 kN, the previous equation for the determination of P becomes 0.1 P ¼ 2 160 1.1 (1 P/1601)/7.50 from which it is obtained that P ¼ 364.0 kN, instead of the previously found value of 426.6 kN. Now, the diagram shown in Figure 7.21(a), which will be used for the determination of , instead of the plastic moment 160 kN m, has the moment 160 1.1 (1 364.0/ 1601) ¼ 136.0 kN m, and on this basis it is obtained that ¼ 0.205 136.0/160.0 ¼ 0.174 m Consequently, the ultimate horizontal force is actually equal to 36.4 364.0 (0.174/ 7.50) ¼ 27.95 kN and the ultimate vertical load is 27.95 10 ¼ 279.5 kN, which is certainly reduced from the value of 310 kN.

References Duddeck H. (1984) Traglasttheorie der Stabwerke — Beton Kalender, Teil II. Berlin: Ernst. Franz G., Hampe E., Scha¨fer K. (1991) Konstruktionslehre des Stahlbetons, Band II, B. Berlin: SpringerVerlag. Gaylord E.H., Gaylord C.N. (1957) Design of Steel Structures. New York: McGraw-Hill. Mann W. (1976) Kippnachweis und Kippaussteifung von schlanken Stahlbeton- und Spannbetontra¨gern. Beton- und Stahlbetonbau 2, 37—42. Menn C. (1990) Prestressed Concrete Bridges. Basel: Birkhaüser Verlag.

296

8 Arches 8.1

Basic characteristics of structural behaviour

Arches represent that structural type which transfers gravity loads to two support points mainly through axial compressive forces, thus limiting the bending as much as possible and offering aesthetically a particularly satisfactory result. This type of structure is used either for roofing relatively large areas or as a bridge over large obstacles (rivers or valleys with abrupt bends) without intervening supports. In this last case, the live load is applied to a horizontal beam lying above or below the arch, transferring the vertical loads directly to the arch through intermediate elements (usually vertical), acting as supports for the beam (Figure 8.1). The transfer of gravity loads without bending to the supports is achieved if the arch axis precisely follows the pressure line corresponding to these loads. It is useful to design the arch to always follow the pressure line corresponding to the permanent loads. It should be recalled that by the ‘pressure line’ between the support points of the arch, we mean that line along which only compressive axial forces are transmitted that balance the externally applied gravity loads at any point, so that there is no need for the structure to develop bending moments (and shear forces) in order to take up these loads (see Sections 2.2.7 and 6.2.1). In this sense, for a specific vertical load g there exists an infinite number of such ‘pressure lines’, all of them governed by the equation y(x) ¼ k M0g (x) which means that each diagram affine to the moment diagram M0g (x) of the corresponding simply supported beam represents a pressure line. The corresponding developing horizontal component H of the axial force at the ends of the pressure line is equal to 1/k. The higher the arch (i.e. the higher the value of k) is, the smaller the horizontal thrust H (Figure 8.2). However, it should be noted, as in Section 6.1.1, that the compressive axial force N along the pressure line increases towards the supports, although its horizontal component remains constant and equal to H everywhere. Then (see Figure 8.2), N ¼ H/cos . It is clear, in general, that the pressure line for a specific loading is specified only if a third point is also found on the arch which the pressure line passes through (see Figure 8.2). For example, an arch having at its midspan the rise f (i.e. y(L/2) ¼ f ) exhibits a value of k which, according to the last equation, is determined Structural systems: behaviour and design 978-0-7277-4105-9



Figure 8.1 The formation of typical arches

as follows: k ¼ f/M0g (L/2) The rise f of the arches is usually selected between one-fifth and one-tenth of the arch span. An arch with immovable supports designed in the above way essentially develops the same axial compression forces, regardless of its statical form (three-hinged, two-hinged or fixed arch). Naturally, the above horizontal thrusts, as applied to the ground with an outward direction, may cause a relative horizontal shift of the supports (Figure 8.3). The arch is particularly sensitive to this relative shift, and is stressed in bending, depending on the degree of static redundancy. Thus, while the three-hinged arch remains unaffected, the two-hinged arch and, in particular, the fixed arch develop an increasing response. A reduction in f/L causes a notable increase in the sensitivity of the relative displacement (shift) of the supports, by developing bending moments that cause tension at the bottom fibres. However, in g

g

N N H = 1/k

α

H = 1/k

Mg0(x)

Mg0

x

g Through the supports passes an infinite number of pressure lines

(f /Mg0) · Mg0(x) Mg0/f

f

y

The pressure line is determined once a third point on the arch is specified

Figure 8.2 Arch design following the pressure line

298

Mg0/f

Arches

Initial thrust acting on the ground

H

H

The deformability of the soil causes a relative shift of the supports and consequently a bending response due to the thrust decrease

Prestressing with the initial thrust force ensures the absence of bending

Figure 8.3 Effects due to the horizontal yielding of the supports, and the prevention of such yielding

order for the outwards applied horizontal thrust to be taken up ‘internally’, it is possible to insert a cable tie in the arch, prestressed with a force equal to the horizontal thrust for immovable supports (see Figure 8.3). In that way, not only is the development of bending prevented but also, in the case of a concrete arch, a redistribution of internal forces due to creep is excluded for permanent loads, as explained previously in Section 6.2.1 (see Figure 6.7). It is obvious, then, that the reactions of the arch at the supports consist only of two vertical forces, which is particularly favourable for the foundation. g It can now be seen that if an arch designed to coincide with the pressure line for the permanent loads g is loaded with an additional live load p applied at an arbitrary position, then the arch axis will no longer follow the pressure line corresponding to the new loading (g þ p), so that bending will be developed (Figure 8.4). This ‘new’ pressure line may be determined so that it passes through a specified intermediate point of the arch with zero bending moment. The magnitude of the bending moment results as the moment of the compression force of the new pressure line with respect to the point of the arch being considered (see Figure 6.4). The fibres which will be tensioned by the bending are obviously determined according to the relative position of the pressure line with respect to the point of the arch being considered (see Figure 8.4). g

g

Initial loading

g+p g Pressure line for the new loading

New loading Pressure line for the initial loading

The bending causes tension at the top fibres

Figure 8.4 Effects due to changes in the pressure line

299


At this point it should be noted that, particularly for materials with a limited tensile strength (unreinforced concrete, masonry), bending is not prohibited in principle, provided that the compression force resultant is acting in the core of the crosssection. Thus, if the ‘new pressure line’ due to a live load lies within the core of all sections of the arch, then no tensile stresses are developed. This is examined more specifically in the following example. A two-hinged arch with span L will bear the permanent uniform load g as well as the uniform live load p (Figure 8.5). Before the height of the arch is determined, it is clear that any arch configuration that follows the moment diagram of the simply supported beam, that is, any second-degree parabolic curve passing from the support points, essentially develops only axial compressive stresses. Once the height of the arch f is determined, however, then the Resultant compressive force

g

M (f/Mg0)

·

f

H

(M/H)

~H

Mg0(x)

~H

H = Mg0/f h

Mg0 = g · L2/8 L The arch does not develop bending p/2 p

p/2 g p/2

Pressure line for (g + p) Passing through the core of the section means bending without tensile stresses

p/2 M=0

H=0 Tension on the bottom fibres p+g

Tension on the top fibres L/2 g p/2 p/2

Mg + p

p · L2/64 Bending diagram valid also for the actual live load p

Figure 8.5 Load-bearing function and stressing of an arch under an additional live load

300

Arches

equation of the pressure line, and hence the arch axis, will take the form y(x) ¼ (8 f/q L2) M0g (x) while the horizontal component H of the reaction will be H ¼ q L2/(8 f ). It is clear that the axial force at the top of the arch also has the same value. The arch, of course, should also be examined for the live load p. It is obvious that considering the live load over the whole length of the arch will proportionally increase the axial stressing without causing bending, because the form of the arch will still be coincident with the pressure line that corresponds to the loading g þ p. However, as has been already mentioned for single-storey frames with inclined legs (see Section 6.2.3), the most unfavourable layout for the live load corresponds to the loading of only the half of the structure with the load p (see Figure 8.5). This loading comprises a symmetric part and an antisymmetric part, both being equal to p/2. The symmetric loading increases the axial compression without causing bending moments. The antisymmetric loading causes zero axial stresses at the top of the arch, making, in consequence, the horizontal components of the support reactions, as well as the bending moment at the crown, null (see Section 2.4.2). Moreover, according to the antisymmetric loading, the resulting vertical displacement at the crown is also zero. The pressure line corresponding to the total load g þ p will pass, as stated previously, through the midpoint of the arch, where the bending moment is zero. Thus, the factored moment diagram of the simply supported beam can be appropriately determined so that the pressure line passes through this point (see Figure 8.5). The equation of this ‘new’ curve is y ðxÞ ¼ ½ f=M0g þ p ðL=2Þ M0g þ p ðxÞ Consideration of the pressure line for the loading g þ p, despite the fact that the arch was formed for a pressure line corresponding to the load g, is important for an understanding of arch stressing. As also shown in Figure 8.5, the deviation of the arch axis from the ‘new’ pressure line is clearly hints at the development of bending moments that cause tension at the bottom fibres in the left half of the arch and at the top fibres in the right half. The absence of horizontal forces at the crown and at both ends of the arch under the antisymmetric loading p/2 allows one to conclude that the maximum moment developing in each half of the arch corresponds to MF ¼ ( p/2) (L/2)2/8 ¼ p L2/64 Of course, for a masonry arch, the already determined ‘new’ pressure line should not pass outside the core of each section (see Figure 8.5). Considering approximately (but nevertheless conservatively) that at a quarter of the arch span the compression force due to g þ p/2 is equal to H ¼ (g þ p/2) L2/8 f, it is concluded that, for a cross-sectional depth equal to h, h M p L2 8f pf ¼ ¼ 2 6 H 8 ðg þ p=2Þ 64 ðg þ p=2Þ L 301

Structural systems: behaviour and design p

p/2 p/2 Parabola of second degree p/2

p/2 p/2

M=0

M ≈ 0.970 · M0

H=0 M L/2

M0 = p · L2/64

Bending diagram valid also for the actual live load p

Figure 8.6 Fixed arch behaviour due to antisymmetric loading

From the above condition and by introducing the non-dimensional ratio ¼ h/f, the following limit live load value can be obtained: 4g 32 Arch bending is favoured by the ‘free prestressing’ offered by the axial forces, so that the eventual reinforcement needed is more favourable than in a beam or even a frame girder, because of the presence of compression force. Considering now fixed arches, the antisymmetric loading p/2 induces bending moments at the supports, which cause tension at the corresponding top and bottom fibres, as shown in Figure 8.6. This bending intensity is very little affected by a reduction or an increase in the ratio f/L, being near the mean value of 0.970 ( p/2) (L/2)2/8 ¼ 0.015 pL2 (see Figure 8.6). g p¼

It should be noted that, due to the existing slope of the arch axis, the distribution of selfweight over a horizontal projection is not uniform, being increased towards both ends of the arch (Figure 8.7). This fact, also depicted in Figure 8.2, leads to a pressure line that is different to the second-degree parabolic one, and hence the arch with a parabolic axis develops some bending that is significantly weaker as the ratio f/L decreases. g An arch with a large span develops high compressive forces under permanent loads, which cause elastic shortening of its axis, resulting in deviation from the pressure line, and hence bending, so that, strictly speaking, purely axial stressing of the arch is g

Realistic distribution of self-weight

Figure 8.7 Non-uniform distribution of the self-weight

302

Arches g

H

H HN

(Reactions)

Decrease in horizontal thrust

The pressure line coincides with the arch axis g

The pressure line for (H – HN)

The pressure line for H

Deformed axis

H – HN

H – HN

H – HN

Tension at the bottom fibres

Upward shifting of pressure line due to thrust decrease

Figure 8.8 Cause of bending response due to shortening of the arch axis

only theoretically feasible, even under permanent loads (Figure 8.8). In the case of a two-hinged arch, shortening of its length will require self-induced horizontal reactions directed outwards, in order to restore immovable supports in the statically determined system, i.e. in the same sense as the horizontal reactions which are developed due to a horizontal outward yielding of the supports, as previously considered. This will naturally result in a reduction in the horizontal thrust of the arch under the permanent loading, and consequently the corresponding pressure line must be considered to be shifted upwards from its initial position — i.e. the arch axis — in order to conform to the decreased value of H. This upward deviation of the pressure line from the arch axis means, as can be seen in Figure 8.8, that the development of bending moments causes tension at the internal fibres of the arch. The crown deflection wc of an arch developing a horizontal thrust H under a uniform load g, due to the elastic shortening of the arch axis, may be roughly estimated from the expression (Menn, 1990) wc ¼

H L ½1 þ 3ð f=LÞ2 EA 4 ð f=LÞ

where A represents the cross-sectional area of the arch. In the case of a two-hinged arch, the bending moment Mc induced at the crown by the above deflection (Figure 8.9) may be assessed from the expression 8 EI wc L2 where I represents the moment of inertia of the arch section. MðhÞ c ¼

303

Structural systems: behaviour and design Ms(f) Two-hinged arch

Mc(h)

Fixed arch

Mc(f)

Figure 8.9 Assessment of the bending response due to shortening of the arch axis

The corresponding bending response in the case of a fixed arch Mc and Ms at the crown and the fixed supports, respectively, may be approximately assessed as (Menn, 1990) 16 EI wc L2 32 EI MðfÞ wc s ¼ L2 At this point, it should be emphasised that in large concrete arches the presence of creep ‘following’ the intense compressive stressing of the concrete always creates additional shortening of the arch axis, which means deviations from the ideal pressure line and, hence, bending. This is also an appropriate point to mention that awareness of the creep phenomenon itself happened about 100 years ago, when a particularly shallow three-hinged concrete arch of 75 m span with a rise-to-span ratio f/L equal to 15, designed by the then very young French engineer Eugene Freyssinet, showed some months after the completion of the bridge, and despite a successful decentering procedure, an unexpected sinking of 13 cm at the crown. This fact gave rise to the idea of an ‘unacceptable’ reduction in the modulus of elasticity of concrete with time, which allowed Freyssinet to conceive of and study the phenomenon of creep (see Section 1.2.2.1). MðfÞ c ¼

8.2

Elastic stability — second-order theory

The bearing capacity of an arch is based on the development of high compression forces, which means that particular attention must be paid to ensure its elastic stability (buckling), on the one hand, and to the fact that its response may be modified by the development of transverse deformations — according to second-order considerations — on the other, as for example in the case of the antisymmetric loading p/2 studied in the previous section. Arch buckling leads, generally, to an antisymmetrical deformed configuration (Figure 8.10). A conservative estimate of the critical compressive force Hcr (thrust) of the arch under a uniform load may result from the following expression (O’Connor, 1971): Hcr ¼ C 304

EI L2

Arches qcr

Buckling deformation

Hcr

Hcr

Arch axis

p g

Pressure line for (g + p)

e

w

w

e

Deformed arch axis

Undeformed arch axis

Tension on the bottom fibres

Tension on the top fibres

Figure 8.10 Deformed state due to the application of a live load

where the factor C depends directly on the ratio f/L as well as on the arch type. A fixed arch develops roughly double the critical compressive force of the two-hinged arch, and this force increases as long as the ratio f/L decreases. The latter shows, for values of f/L greater than 0.2, practically the same critical thrust as the three-hinged arch. Indicative values of the factor C are given in Table 8.1. As mentioned previously, any loading q, the pressure line of which deviates from the arch axis, causes bending. However, the transverse deformation caused by this bending augments the initial deviation further, consequently causing an additional bending, and so on, until the final convergence. Thus, the final bending depends on the final transverse deformation of the arch (second-order theory). This final bending obviously corresponds to the deviation of the deformed arch axis from the pressure line of loading q, which can immediately be seen from an examination of the two-hinged arch example (see Figure 8.10). Table 8.1 Values of factor C f/L

0.1

0.2

0.3

0.4

0.5

Fixed-arch Two-hinged arch Three-hinged arch

75.2 35.0 27.8

63.2 28.8 24.8

47.7 20.0 20.0

35.1 13.6 13.6

24.8 9.2 9.2

305


It is obvious that, due to the total load g þ p, a downward shift w will take place at the points on the left half of the arch, due simply to the antisymmetrical part p/2 (see Figure 8.10). The pressure line, however, corresponding to the total load g þ p, passes in the same left half above the undeformed arch axis at a distance e, and the total divergence from the actual arch axis thus comprises the sum of the above two deviations, i.e. w þ e, and is not due to w alone, as is the case for the rectilinear members according to the second-order theory (see Section 7.3). Given that the pressure line g þ p passes in the examined left part above the deflected arch axis, the resulting bending will cause tension at the bottom fibres, while in the right part, where it lies under the deflected axis, the top fibres will be tensioned (see Figure 8.10). A calculation with regard to second-order conditions is generally necessary. Naturally, the more the developing compressive force H diverges from the critical value Hcr for buckling, the more the additional bending moment decreases. Considering a twohinged arch again, it is possible to assess the final deflection w (occurring at the quarter point of the arch span), on the basis of the deviation e of the pressure line from the undeformed arch axis. Thus, according to the previous section (see Figure 8.5), e¼

MF p L2 8f 1 f ¼ ¼ 2 ðg=pÞ þ 1=2 8 H 64 ðg þ p=2Þ L

The deflection w may be estimated (somewhat conservatively) through the equation (Franz et al., 1991) w¼e

H=Hcr 1 H=Hcr

According to the above, the final value M of the ‘first-order’ bending moment MF ¼ p L2/64 will be equal to M ¼ MF

e þ w p L2 1 ¼ e 64 1 H=Hcr

where an analogous ‘magnification factor’ is applied to the initial value of the bending moment MF, as also is the case for beams and frames, as discussed in Chapter 7. The compressive force H corresponds to the total load g þ p and — as previously explained — is due to the symmetric part only (g þ p/2): H¼

ðg þ p=2Þ L2 8f

The estimated values H and M may then represent a basis for the preliminary design checking of the arch. However, using a global safety factor applied to service loads, the check should be made analogously to Section 7.3, for an axial force Nd ¼ H 306

Arches

and a bending moment Md ¼

p L2 1 1 H=HK 64

g

At this point, the possibility of an out-of-plane buckling of a free-standing arch will be considered, in the sense already examined in Section 7.5, with regard to the lateral buckling of beams. For arches, the degree of end fixity has a major influence on this buckling behaviour, the fixed arch having in this respect a definite advantage over the two-hinged arch. Although in an out-of-plane deformation the torsional rigidity of the arch offers some help, this resistance can be neglected in the preliminary design and, instead, the moment of inertia of the section Iy about a vertical axis will be considered as the main resistance factor. Thus, for a fixed arch, taking as the ‘effective length’ the half-length of the arch axis, in analogy to the corresponding basic buckling formula according to Section 7.1, the following expression may be used in order to approximately estimate the critical compressive axial force NK for the out-of-plane buckling: NK ffi

4 2 EIy 2 2

L2 ½1 þ ð8=3Þ ð f=LÞ

¼

EIy 39:5 2 2 L2 ½1 þ ð8=3Þ ð f=LÞ

In the case of two-hinged arches, as used in bridge construction, especially when the load-bearing roadway lies under the arch (see Section 8.4), in order to ensure (or enhance) the lateral stability of the two closely arranged arch ribs, a bracing must be arranged between them, consisting usually of horizontal ribs perpendicular to the plane of the arches and, obviously, high enough in order not to interfere with the traffic (Figure 8.11). In this way, a ‘curved frame’ is formed, the buckling behaviour of which may be judged if considered to be plane, and investigated, as Figure 8.11 suggests, according to the procedure described in Section 7.4.2.

Axial compressive force N N

N

N

N

Figure 8.11 Lateral bracing of arches

307


8.3

The girder-stiffened arch system

As mentioned at the beginning of this chapter, the transfer of traffic load from one point to another through an arch imposes the requirement for the construction of a horizontal, or at least slightly inclined, deck (or girder). The deck transfers its loads to the arch usually through vertical elements that provide, primarily, a direct support to the deck (see Figure 8.1). Initially, the structural function of the horizontal girder was limited simply to bearing and transferring the live loads onto the arch via the vertical supporting members. At some point, however, it became clear that the whole structural action would be much more effective if the combined load bearing of the horizontal girder and the arch could be considered. This is explained below (Figure 8.12). The unfavourable arrangement of the live load in one-half of the structure, say the left part, understandably causes a lowering of the left part of the arch and a raising of the right part. Unfavourable bending of the arch due to this deformation is considerably limited if the horizontal beam is made to follow it. However, in order for the beam to deform in the same way, it is clear that it must receive downward forces in

p g

EI G EI A

The arch deforms and the girder follows but it requires the application of forces

The arch receives the opposite forces

The final arch loading tends to become uniform

Figure 8.12 Static operation of the girder arch system under a live load

308

Arches p

M = p · L2/64

Two-hinged arch

M ~0.97 · M Fixed arch ~0.56 · M

Figure 8.13 The effect of a live load on arch bending

the left part and upward forces in the right. These forces will obviously be provided by the arch, which will itself then receive by reaction the opposite forces, i.e. additional upward and downward forces will be applied to the left and right parts of the arch, respectively. As a result, the unfavourable unilateral loading of the arch that exists without participation by the horizontal girder tends to become uniform because of the correspondence of deformations, so that the bending of the arch is clearly limited (see Figure 8.12). It can be seen that the bending moment M of the arch, which, according to Section 8.1, is due only to the antisymmetric part ( p/2) of the live load, will now be distributed proportionally between the arch (MA) and the girder (MG), according to the bending stiffness of the two elements (see Section 3.2.10), i.e., MG ¼ M

EIG EIG þ EIA

EIA M ¼M G EI þ EIA

(a)

A

In the extreme case where the girder stiffness EIG is much higher than the arch stiffness EIA, bending will be experienced almost entirely by the beam, while in the opposite case it will be experienced entirely by the arch. As neither case is aesthetically satisfactory, the solution EIG EIA is usually adopted, whereby the bending response is almost equally distributed. An estimate for the magnitude M of the distributed bending in the examined case is equal to p L2/64 or p L2/113.6 in the two-hinged or fixed arch, respectively, according to Section 8.1 (Figure 8.13). If, now, the arch is formed according to the pressure line for permanent loads, it can be seen that the girder will function essentially as a continuous beam on unyielding supports for these loads. However, as pointed out in Section 8.1, under permanent loads, especially for long spans, a transverse deformation due to axial shortening of the arch is developed. The deflection wc at the crown (Figure 8.14), assessed in Section 8.1, will also be imposed on the girder, thus causing additional bending stresses. 309

Structural systems: behaviour and design L EI G EI A

wc

Figure 8.14 Check of the deflection of the arch top

Assuming fixed-end supports for the girder and the arch too, the beam bending may be estimated for preliminary design purposes, through the expression 24 EIG wc (see Section 3.3.3) L2 causing tensile stresses in the bottom fibres of the midspan and in the top fibres of the fixed supports. It is obvious that the response of the girder as a continuous beam on unyielding supports should be superposed to the above. The corresponding bending response of the fixed arch Mc and Ms at the crown and the fixed supports, respectively, may be assessed as in Section 8.1 (Menn, 1990): MG ¼

16 EIA 32 EIA w M ¼ wc c s L2 L2 where IA represents the moment of inertia of the arch. g Mc ¼

The association between the girder and the arch increases perceptibly the buckling axial force of the arch. A conservative estimate of the critical axial force Ncr of the fixed arch is given by the expression (see Menn, 1990) Ncr ¼

2 EIc ð0:7 L=2Þ2

where EIc ¼ (EIG þ EIA)/2 may be considered, or even its half value, allowing for possible cracking. In the case of a concrete structure, the girder is usually prestressed in order to ensure uncracked conditions in the serviceability state and thus a higher available stiffness EIG, as well as being able to deal more effectively with the developing alternating bending moments, depending on the left or right placement of the live load. Attention must also be paid here to the modified bending under the live load p, due to the developing transverse deformations of the arch (second-order effect), which, of course, also affects the girder, and is shared between them according to the above equations (a). The distributed bending moment M may now be roughly calculated, analogously to Section 8.2, following the expression M ¼ MF

1 1 N=Ncr

where MF represents the ‘first-order’ maximum span moment of the arch and N is the axial force due to the permanent load and live load. 310

Arches

8.4

The tied-arch system

It is possible, and often desirable, to suspend the roadway from twin arch ribs from below through hangers which transfer the applied loads on the deck upwards to the arch, instead of supporting the girder on the arch from above. The so-formed spatial bridge structure may be considered as basically a plane arch system, such as the one depicted in Figure 8.1, where the deck is represented by the horizontal ‘tie’ element, which, being suspended from the arch, may also be treated structurally as a ‘girder’ stressed both axially and in bending. This type of bridge is used whenever the free space under the traffic deck is restricted, as is the case for low-level crossings, over rivers, etc., or even if the available foundation soil is not suitable for conventional arch foundations. As all the permanent loads are transferred to the arch through the hangers, the arch may be formed according to the pressure line of the loads, in the same way as discussed in Section 8.1. The relatively dense layout of hangers ensures that the arch receives the applied load on the ‘girder’ in a rather uniform manner, so that the parabolic form of the arch is usually imposed. Given that the arch for permanent (uniform) loads is stiff, it can be seen that the hangers may be appropriately stressed (prestressed) by the reaction forces they offer to the girder, and considered as a continuous beam over unyielding supports. Then, it is appropriate to also prestress the tie girder in order to provide the arch with the horizontal thrust required for a normal two-hinged arch. In this way, as has been explained in Section 8.1, no horizontal thrust is transferred to the foundation, which essentially receives vertical forces only. Although the deck may be made of concrete, which, if prestressed as mentioned, can withstand the high tensile force the tie has to offer, it is more suitable for the arch, for both constructional as well as for aesthetic reasons, to be made of steel. Although all the characteristics of the structural behaviour of the arch itself, the stability issues included, as described Sections 8.1 and 8.2, are obviously also valid for this type of arch system, it would nevertheless be useful to again examine the carrying mechanism for the live load. As mentioned in Section 8.1, for the most unfavourable layout of the live load p, namely the loading of only one half of the girder, given that the symmetric part ( p/2) causes no bending, it is the antisymmetric part ( p/2) that needs to be examined (Figure 8.15). Under this antisymmetric load the equivalent half-system acts approximately like two simply supported beams connected through a tension element and subjected to a load applied to the lower one. It is obvious from Figure 8.15 that the resulting bending moment M ¼ p L2/64 has to be taken up from both the girder and the arch. According to Section 3.2.10, if MA and MG represent the bending regions of the girder and the arch, respectively, then they may be roughly assessed as 1 1 MA ¼ M MG ¼ M 1þ 1 þ 1= where 48 EG IG ðL=2Þ3 f ¼ þ 48 EA IA 2 EH AH ðL=2Þ3 and the superscripts G, A and H refer to the girder, the arch and the hanger, respectively. 311

Structural systems: behaviour and design M=0

M=0

p/2

g

L/2

H H = Mg0/f

p/2

f

M = p · L2/64

g EA · IA E H · AH p/2

E G · IG

p/2 p/2

Figure 8.15 Load-bearing action of a tied arch for permanent and live loads

It is clear that both the girder and the arch have a more or less substantial bending response to undergo, the reason for this being obviously the fact that both elements take up the transferred forces only through bending. However, the situation changes if the truss-like layout of diagonal hangers is applied, as in Figure 8.16. Then, the live load on the girder can be taken up in a much more favourable way, because the whole system acts as a truss, which is structurally more

p

p/2 The hanger forces are practically taken up only through bending p/2

f

p/2 p/2

p/2 Possibility to take up the live loads through a truss action

M=0

H H = Mg0/f

g

Figure 8.16 Favourable load-bearing action through a diagonal layout

312

g

Arches

q

Mq0/f

R/2

R = q · f/tan2 ω f Mq0/f

R/2

q

Figure 8.17 Network tied-arch layout

effective than a bending element. Although the bending in both the girder and the arch is not eliminated, it is nevertheless significantly reduced, to a value which is about 15% of that corresponding to vertical hangers, thus leading to an analogously smaller deflection. Moreover, it is clear that the diagonal layout of hangers leaves the axial forces in the arch and the girder tie for the permanent loads (as well as the symmetric part of live load) essentially unaffected. A further noticeable reduction in the deflection and in the bending moments in both the girder and the arch is possible if a network of diagonal hangers is used, as shown in Figure 8.17 (Tveit, 1966). The hangers have a constant slope, and cross each other at least twice. In this case, the axial forces both in the arch crown and the tie are slightly changed. As can be seen from Figure 8.17, the existing hanger tensile forces under permanent loads at the midspan section suggest that, for horizontal equilibrium, the crown compressive force and the tie tensile force have to be increased and decreased by half of the resultant horizontal force of the hangers (see Figure 8.17). It is clear that suitable computer software has to be used for a thorough analysis of the system.

References Franz G., Hampe E., Scha¨fer K. (1991) Konstruktionslehre des Stahlbetons, Band II, B. Berlin: SpringerVerlag. Menn C. (1990) Prestressed Concrete Bridges. Basel: Birkhaüser Verlag. O’Connor C. (1971) Design of Bridge Superstructure. New York: John Wiley. Tveit P. (1966) Design of network arches. Structural Engineer 44(7), 247—259.

313

9 Cable structures 9.1

Overview

The main constituent element of cable structures is the high-strength steel cables that present particularly favourable load-bearing properties. The high tensile strength of steel cables, their main characteristic, allows them to take up large axial forces with small cross-sections, and thus they are more economical in terms of price/offered strength than structural steel. As the cable lengths used are very large with regard to their cross-section, the cables present an inherent flexibility, and thus do not offer resistance to bending moments or compressive forces. Therefore, cables can only develop tension, and this allows the complete exploitation of the cable cross-section, as the danger of buckling does not exist. The aforementioned characteristics generally make cable structures the most economic ones. Cable structures present unlimited possibilities for the creation of the most versatile forms, and are particularly useful for economically covering large areas under roof forms, and as effective basic bearing members in bridges having very large spans. In this chapter the main load-bearing characteristics of various, mainly ‘plane’, forms of cable structures are examined, including cable—beam roofing systems, suspended roofs having relatively large spans, and suspension and cable-stayed bridges. The basic structural characteristic of cables is that they automatically acquire the so-called ‘funicular form’ for every load pattern acting on them, whether the loads are concentrated or distributed. This funicular form implies the development of exclusively tensile forces at each point of the cable, which directly raises the problem of safely taking up the anchoring forces of the cable at both its ends. The solution of this problem is of great economic and aesthetic importance, and plays a decisive role in the whole structural design (Figure 9.1). Of course, it should be pointed out that for a given plane loading there are two families of funicular forms that can take up this loading exclusively by axial forces. The first form develops exclusively tensile forces and concerns the ‘cables’, while the second develops exclusively compressive forces and concerns the ‘arches’. It is obvious that the two form families are mirror images of each other. As became clear in Chapter 8, an arch is a structure having a funicular shape for some system of coplanar loads, and for a different loading the same arch will no longer have the same funicular structure, resulting in the development of bending. This does not happen in the case of a cable, wherein the new system of loads will continue to cause exclusively tensile stresses. However, there is the serious disadvantage that the cable Structural systems: behaviour and design

Copyright Thomas Telford Limited # 2010


Funicular form (tension) Problem raised in taking up the anchor forces A

A

Solution I

Solution II Equilibrium at node A

High anchor force

Less anchor force

Taking up the same loading through an arch (compression)

Cable sensitivity Change in loading means change in the geometry

Wind uplift forces may overtake the gravity loads

Figure 9.1 The cable as a funicular form for carrying vertical loads

will automatically adopt a new funicular form because of a lack of bending stiffness (contrary to what happens in an arch), which means a more or less drastic change in the form of the cable. A typical example of this problem is the application of wind uplift forces on cables that have been appropriately formed for gravity loads (see Figure 9.1). The upwardly applied wind loads may possibly exceed the existing permanent loads, thus leading to the danger of an intense change in the cable form. A first measure for tackling this problem could be to increase the dead weight merely by adding more weight along the length of the cable. However, this is an uneconomic solution (Figure 9.2). The structurally proper solution is to ‘stiffen the cable’ by covering the already suspended cable with concrete of constant cross-section and applying additional tension to the cable itself as prestressing. The vertical upward deviation forces that are thus developed will set the so-formed inverse concrete arch under pure axial compression. This structural configuration, by retaining the form of the suspension cable practically unaffected, allows the additional gravity loads to be borne as well as the upward loads. 316

Cable structures Adding dead weight removes somehow the sensitivity without representing a structurally sound solution

Covering with a concrete thin layer and afterwards further tensioning (prestressing)

Deviation forces due to prestressing

Development of initial compressive stresses Self-contained prestressing does not affect the anchorage forces

Additional loads Arch working in tension

Typical arch action (compression)

Figure 9.2 The need for stabilisation of the cable form

For the vertical gravity loads, the already prestressed arch, acting in an inverse way to its typical function (see Chapter 8), receives an additional tension, which is suppressed by the already existing compression forces, while the potential upward loads are taken up according to the typical arch action (see Figure 9.2). In both cases, the bending stiffness of the arch ensures the non-deformability of the cable, even for additional asymmetric loads. This topic is treated more thoroughly in Section 9.6.

9.2

Cable—beam structures

A purely cable plane structure that acts as a single-span beam basically has the form of a truss, which, even before accepting the gravity loads that it is intended to receive, is in equilibrium, with all its members being under tension through prestressing. In such a truss, in which compression members cannot naturally exist, the upper and lower chords will necessarily take a curved form, concave upward and downward, respectively. The reason for this is that the chords, each of which consists of a single cable, will make the funicular form due to the tensile forces of the intermediate members that are applied on them; these forces are directed downwards for the top chord and upwards for the 317


Without external loading, initial tensioning (prestressing) in all members

Compressed struts

Funicular form for the acting forces

Funicular form for the acting forces

Figure 9.3 Cable—beam formation

bottom chord (Figure 9.3). It is clear that, at each node, any existing external forces, along with the concurrent axial forces, must be in equilibrium. In a similar beam-like structure, the cables of the upper and bottom chord may alternatively be connected through vertical axially compressed struts, where the curvatures of the two cables are obviously reversed (see Figure 9.3). In any case, the structure should be subjected to such an initial tensile stress state through prestressing that the resulting compressive forces due to the imposition of external loads (gravity, wind, etc.) will be suppressed by the tensile forces introduced during prestressing. This will prevent the development of compressive forces anywhere within the structure, which might lead to slack members and, consequently, to failure of the whole system (Figure 9.4). It should be clearly understood that the design and behaviour of this type of cable structure are based on the combined influence of the introduced prestressing in all their members and the mandatory consideration of equilibrium at the deformed state, i.e. at the final positions of the nodes. Thus a cable, which in the initial state of prestressing is horizontal and symmetrically placed with respect to the vertical axis of symmetry of the structure, can, with suitable prestressing, offer in the case of antisymmetrical loading the required (for equilibrium) vertical force along the axis, through the inclined position that the cable adopts with deformation. In order to illustrate this last condition better, the symmetrical system shown in Figure 9.5 is considered. The system is composed of an upper and bottom cable and two vertical struts. It is assumed that the system is prestressed with the prestressing forces given in the figure. The reactions at the supports of the system are zero. The application of an antisymmetrical loading causes antisymmetrical reactions which, due to equilibrium at the support nodes, require the development of certain axial forces in 318

Cable structures F

Te

ns

F

ion

Compression

C

Compression

n

sio

res

p om

F Compression

Co

mp

res

sio

n

on

nsi

Te

Tension

F

F Tension

Taking up of compressive forces is impossible without prior prestressing of cables P

P

Te

Te

ns

ion

Te

ns

F

ns

Compression

ns

Te

Tension Compression

ion

n

sio

n Te

ion

Tension

ion

Equilibrium n sio

n

Te

Tension

Figure 9.4 The importance of prestressing in the formation of cable structures L a

P

Te

ns

ion

Te

ns

Compression

Te

Compression

Tension n

io ns

ion

F ion

Tension

ns

Te

Tension

ion

s en

T

P F Stressless

ion

Te

ns

F Te

ns

s res

mp

ion

Co

Te

ns

F · (a/L)

ion

ion

F · (a/L)

Tension Impossible vertical equilibrium

ion

ss

re mp

F · (a/L)

Stressless Co Antisymmetric loading

F F

n Tensio

F Te

ns

n

P

ion

io ns

Te

P

Te

ion

F · (a/L)

n Tensio

Ten

Deformed configuration

F · (a/L) F · (a/L)

Tension

T

Te

n

sio

ns

ion ens

ns

ion

Tension

Equilibrium possible thanks to the inclined axial forces

Figure 9.5 The need for prestressing in maintaining the equilibrium in the deformed configuration

319


the inclined cables. This, however, makes the maintenance of the overall equilibrium impossible if the system is examined in the undeformed configuration. Indeed, the tensile forces of the horizontal cables resulting exclusively from the initial prestressing (the loading itself induces no forces) are unable to offer the required upwards vertical force F (1 a/L) for equilibrium. It is clear that only after considering the deformed inclined position of the horizontal cables, with their appropriate prestressing forces, can the required total vertical force in question be offered (see Figure 9.5). Of course, it is understood that the prestressing force should be sufficient such that the equilibrium will be obtained with acceptable values of the nodal deformations. g Usually, when determining the permanent loads in a symmetrical structure and for preliminary design purposes, the deformed geometry may be ignored, by following the general procedure described below. However, for live loads with an arbitrary layout, one must make an estimation using special software programs that take into account the deformed geometry (position) of the nodes, when checking for equilibrium, through successive iterations. Note that, due to the large deformations and the ‘geometric non-linearity’ involved, the superposition principle is not generally valid. The preliminary design is performed first for the permanent loads, thus determining a functional and aesthetically acceptable geometrical form in combination with the adopted anchoring layout for the relevant tensile forces. It is clear that all the above require the explicit determination of the introduced prestressing, which thus constitutes an essential parameter for the preliminary design. In any case, finalisation of the geometry in the initial prestressed state under the permanent loads will mean that the equilibrium requirements are satisfied in the precisely determined nodal positions, but this can be resolved only with the use of a suitable software program. However, for preliminary design purposes, the determination of the required cable prestressing can be done as described below. As a first step, the tensile and compressive axial forces that are in equilibrium with the considered external nodal permanent loads are determined. These axial forces, which, along with the external loads, keep each node in equilibrium, may be determined by analysing the system as a plane frame with rigid joints (by means of a typical software program for framed structures) and by considering an extremely small bending stiffness of the members, so that the developing bending moments and shear forces are limited to practically vanishing values. The sought-after self-equilibrating state of prestressing should be such that, if superposed on the results obtained in the first step, it will lead all the members to a tensile stress state. This prestressing state may be determined, as a second step, by imposing a uniform temperature decrease (or increase, if needed), e.g. of 1008C in a selected group of bars, and then searching for the minimum possible factor for this adopted temperature, so that the superposition of the axial forces due to prestressing and those due to the permanent loads ensures for all members of the structure the maintenance of a predetermined minimum safe amount of tension. 320

Cable structures 100 kN

100 kN 100 kN

100 kN

+18

4.5

–1

.0

2.40

–64.2 7 +6. 73

. 61

+28.5

–1 61 .73 .7 6 + +28.5

.4 09 –1

1.20

6 +5

–1 09 .4

2.40

+18

+79.4 .0

+79.4

+5 6

4.5

Analysis as a skeleton frame

–15

.1

–153

3.1

24.0 m

+78

.0

+33.0

+54.6

3 +0 .3 9

3.

4

2.3

+

+2

+1

.54 +8 +33.0

.54

+78

+81.0

+91.3 +8

4

. 23

3

+1

+81.0

9 .3 +0

λ·

2.3

.0

Prestressing: temperature reduction of 100°C for all the diagonal members

+54

.6

(λ = 10.0)

Figure 9.6 Determination of the required prestressing of cables: example 1

The application of the above technique is now illustrated in the following two examples: (1) In this example (Figure 9.6) each concentrated load of 100 kN consists of a permanent load of 35 kN and a snow load of 65 kN. (2) In this example (Figure 9.7) each concentrated load of 60 kN consists of a permanent load of 20 kN and a snow load of 40 kN.

60

–307

.2 – 313

–29.9

–29.9

–301.2 –301.2 –303.2

60

+304

–29.9

60

–29.9

0.6

–29.9

+31

–29.9

3.7

–307

–29.9

3.2 .2 –30

–31

60

60

60

60 kN

.7

Analysis as a skeleton frame

0.6 +31 304.1

.1 +3 .2 + 00.2 +298.3 +298.3 +300

+31.8

+31.2

+30.9 +30.9

+31.2

+31.8

–3.42

+31.2

–3.15

–3.02

+30.9

–2.98

.8

+30.9

–3.02

+32

+31.2

–3.15

λ·

+

+31.8

–3.42

32.8

+31.8

+32

.8

.8

+32

Prestressing: temperature increase of 100°C for all the vertical members

(λ = 10.0)

Figure 9.7 Determination of the required prestressing of cables: example 2

321


In both cases the resulting factor is equal to 10.0, as can easily be concluded from the behaviour of the two extreme diagonals (Example 1, see Figure 9.6) and of the two intermediate members of the upper chord (Example 2, see Figure 9.7). The check for non-symmetric loads, i.e. the check for developing deformations and for possible slack members (i.e. compression), is done, as explained above, by considering the equilibrium in the deformed state through the use of a special software program that takes into account the geometric non-linearity of the system. In the two examples examined, this means the consideration of unilateral snow load. In Example 1 (see Figure 9.6) the first two loads from the left are equal to 100 kN and the other two loads are equal to 35 kN, while in Example 2 (see Figure 9.7) the first four loads from the left are equal to 60 kN and the other three loads are equal to 20 kN. It is clear that, in each case, the cables retain the prestressing that was previously determined (with ¼ 10.0). It is checked that in both cases all cables are found to be under tension.

9.3

The freely suspended cable

According to the previous section, a freely suspended cable of span L under a uniform permanent load g (self-weight, etc.) takes a form affine to the bending moment diagram M0(x) of a simply supported beam under the load g. If Hg is the horizontal component of the anchoring force at both ends of the cable and f is the sag of the cable at its midlength, then Hg ¼ g L2/8 f and, according to Section 2.2.7, the ordinates y(x) of the cable form are yðxÞ ¼ M0g =Hg i.e. a parabolic line (Figure 9.8). Hg = Mg0/f

Hg

g

y

f Funicular form g

Mg0 Mg p

g

Hp + Hg

Hg + Hp y η

η = (Mp – Hp · y)/(Hg + Hp) p

Mp

Figure 9.8 Determination of the funicular form under an additional load

322

Cable structures

For an additional load p the cable is deformed further, with ordinates , which brings an additional horizontal component Hp to the cable force, and the cable takes a new funicular form with ordinates ( y þ ) that are expressed, just as previously, through the equation yþ ¼

M0g þ M0p Hg þ Hp

Based on the above expression for y, the additional ordinates result as (see Figure 9.8) ¼

M0p Hp y Hg þ Hp

In this expression for the force Hp is unknown. To determine Hp the relationship must be found, on the basis of Hooke’s law, between the geometric elongation of the cable due to the ordinates , its elastic characteristics (i.e. the elasticity modulus Ec) and the crosssectional are Ac. The following result may then be obtained (Timoshenko and Young, 1965): ð ðL Hp g 1 L d2 Ls ¼ dx dx Hg 0 2 0 dx2 Ac Ec As, in the present context, the interest is mainly in the covering of large areas and in suspension bridges, where the ratio ¼ f/L lies between 1/8 and 1/12, the above equation may be simplified, to a good approximation, as ðL Hp g L ¼ dx ðaÞ Ac Ec s Hg 0 where the length Ls is given by the expression Ls ¼ L ð1 þ 8 2 Þ It must be noted that the length Ls used in the above ‘constitutive equation’ of the cable does not represent the actual length of the cable Lc. This can be estimated using the following equation: 8 2 Lc ¼ L 1 þ 3 It is clear that from the last expression for that the integral in equation (a) above can be determined with respect to the unknown value Hp. It is useful now to introduce, instead of Hp, the ratio Z¼

Hp Hg

as the unknown quantity, and the dimensionless parameters ¼

p g 323


"¼

Hg Ac Ec

where p is equal either to p or p/2 depending on whether the load p is extended over the entire length of the cable or is located in its extreme half, respectively. The former live load layout leads to the minimum cable force, whereas the latter one leads to the maximum deflection, at approximately a quarter of the span. The parameter " represents the strain of the cable at its lowest point due to the dead load g. Appropriate substitution in equation (a) leads to the following quadratic equation with respect to Z: 16 2 1 16 2 1 2 ¼0 Z Z þ 1þ 2 2 3 1þ8 " 3 1þ8 " After the corresponding determination of Z (i.e. of Hp) the values of may be readily obtained as shown above. It results that ðL=2Þ ¼ f

Z Zþ1

ðL=4Þ ¼ f

0:50 0:75 Z Zþ1

or

with respect to the valid layout of live load p as above. g An approximate, but more practical, estimate of the increase f in the established sag f of the cable, due to a uniform load p along its whole length, can be obtained by considering its elongation Lc according to Hooke’s law (Timoshenko and Young, 1965): Hp L 16 f 2 1þ 2 Lc ¼ 3 L Ac Ec where Hp ¼

p L2 8f

However, the change Lc may also be obtained from the above expression for the cable length Lc due to the change f in its sag (i.e. as the first derivative with respect to f ), as Lc ¼

16 f f 3 L

From these last two equations it results that p L2 3 pL f ¼ 1þ ¼ 2 kc 8 Ac Ec 16 324

Cable structures

where kc ¼

8 Ac Ec 3 L 1þ 16 2

expresses the stiffness of the cable, i.e. the uniformly distributed load ( p L) required to cause a unit vertical displacement f ¼ 1 at its midpoint.

9.4

Prestressed cable nets

As explained in Section 9.2, the geometry of the system shown in Figure 9.9 is ensured by the prestressing of its vertical members. The prestressing should be uniformly distributed in order to obtain a parabolic form for both the upper and bottom cables. It is clear that if this distributed prestressing is equal to u (kN/m), it should be valid that u¼

H1 H2 ¼ R1 R2

where H1 and H2 are the horizontal components of the axial force in each cable, and R1 and R2 are the corresponding radii of the curvatures. Nothing will change in the above-described developing forces if the following progressive reformation of the system is followed. First, cable 1 is rotated by 908 and moves vertically downwards until its midpoint coincides with the midpoint of cable 2, coming into contact underneath (Figure 9.10(a)). Next, the cables are considered to be arranged parallel to one other at the same time and at small distances a in both directions, while always maintaining the contact of cables 1 and 2. The resulting network can constitute the cover for a specific area (Figure 9.10(b)). It is understood that the distributed load u is applied downwards on cable 1 and upwards on cable 2, and the cables develop the forces H1 and H2, respectively, according to the above equation (Figure 9.10(c)). The curved surface created by cables 1 and 2 is a hyperbolic paraboloid and has specific geometrical properties that have particular implications on the bearing H1

H1 R1 1

1 u 2 u 2 H2

R2

H2

u = H1/R1 = H2/R2

Figure 9.9 The stress state of the prestressed cable system

325

Structural systems: behaviour and design H1

1 R1 u u

u 2

R2

1

H2

2

H2 u

(a)

(c)

H1

The state of prestressing in the cable net is described by the condition u = H1/R1 = H2/R2

a 2

H1

1 2

H2 a

H2 1

H1

(b)

Figure 9.10 Creation of a prestressed cable network space system

behaviour of shells, as is examined extensively in Chapter 12. It should be noted that, in the constructional practice of such a prestressed cable network, only cables 2 are prestressed, and this automatically sets cables 1 in a prestressed state as well. It is clear that cables 1 and 2 must give up their forces to the outline of the considered space. Therefore, in order for the structural members at the perimeter to be able to take up these forces in an appropriate manner, they must be formed according to the funicular form, thus developing analogously either compression or tension (Figure 9.11). In the first case the members will be composed of continuous arches concave inwardly, while in the second case they will be composed of cables concave outwardly, both having a parabolic form, given that the loads received from the cables are practically uniform. Moreover, it is understood that the end points of

Layout of compressed arch

2 1

Layout of tensioned cable

Figure 9.11 Formation of the bearing elements at the boundary

326

Cable structures (q1 · a) 1 (q2 · a) 2

Figure 9.12 Distribution of the vertical load in both directions

cables 1 will generally be higher than those of cables 2, and this certainly means an analogous formation of the peripheral bearing element. The prestressing forces H1 and H2 must have such values that the imposition of permanent and live loads always leaves both cable types under tensile stress. This is checked through the following consideration (Figure 9.12). A uniformly distributed gravity load q (kN/m2) applied to the spatial system may be considered as being distributed to a load q1 (kN/m2) along the direction of cables 1 and to a load q2 (kN/m) distributed along the direction of cables 2, causing additional tensile forces in cables 1 and additional compressive forces in cables 2. Now, the requirement for loads q1 and q2, besides offering in total the load q, is to cause the same vertical deformation in cables 1 and 2 at their midpoint, thus allowing the estimation of their value for preliminary design needs. This last requirement can be written, according to Section 9.2, as f ð1Þ ¼

ðq1 aÞ L1 ð1Þ kc

¼ f ð2Þ ¼

ðq2 aÞ L2 ð2Þ

kc

that is q1 L2 kð1Þ ¼ c q2 L1 kð2Þ c where kð1Þ c ¼

8 Að1Þ Ec c 3 L1 1 þ 16 21

and kð2Þ c ¼

8 Að2Þ Ec c 3 L2 1 þ 16 22

The last equation for q1 and q2, in combination with the requirement that q1 þ q2 ¼ q, allows their direct determination. 327


Cable 1 develops a total tensile force N1 that is equal to N1 ¼ H1 þ (q1 a) R1, while the total force N2 of cable 2 is equal to N2 ¼ H2 (q2 a) R2. Of course, it should be checked whether the force N1 can be taken by the cross-section of cable 1, as well as whether the force N2 remains tensile with a sufficient margin of safety so that cable 2 does not become slack.

9.5

Suspension bridges — the suspended girder

For bridging a span, an individual cable is, by itself, functionally useless. However, cable may be used to the suspend a deck carrying traffic loads, in the form of a suspension bridge, wherein it takes up the total permanent and live load acting on it and transfers this load to the supporting towers and anchorage (Figure 9.13). The suspension is realised through vertical tension elements (hangers) placed at small distances along the two longitudinal edges of the deck. The deck can be considered as a beam (a flexible one), given its very long length with respect to its two other transverse dimensions

Hangers

Tower f

Anchorage

Stiffening girder

Anchorage L

Figure 9.13 Structural system of a suspension bridge

328

Cable structures

(depth and width), and it is obvious that the two cables used in the two created planes of the hangers may be considered as a single one in order to examine the load-bearing function of the system under vertical loads (see Figure 9.13). Usually a suspension bridge has a typical configuration of three spans — the main span and two adjacent minor spans. This gives the optimum layout for bridging a large obstacle by using two intermediate support points (the high bridge pylons). The pylons offer a support for the suspension cables at their top, while at the deck level they allow a corresponding support for the girders from either side. Although these girders may be constructed continuously over the pylon’s supports, it is more usual for them to be considered (and constructed) as simply supported ones (see Figure 9.13). The tension Hg on the cable is always adjusted with respect to the desirable sag f so that, for the existing total permanent load g, the girder (deck) is absolutely horizontal. According to Section 9.2, Hg ¼ g L2/8 f. The bending stress on the deck girder under the permanent loads can in practical terms be considered as minimal due to the small distances between the hangers and the care taken to ensure that they are in an absolutely horizontal position. However, an additional live load p applied on the girder tends to deform the cable, and it is clear that this additional deformation of the cable will also be entirely imposed to the girder itself, given that the hangers, with their relatively small length and the relatively small additional axial force they have to carry, may be assumed to be nondeformable (Figure 9.14). The girder should be designed to limit through its rigidity the cable deformation due to the live load p, on the one side, and to resist the bending imposed on it by this very same deformation , on the other side, as previously explained. Thus the imposition of a live load p on the beam causes an additional deformation both in the cable and in the beam, which results in an increase in the horizontal component of the cable force by Hp, as well as an increase in the bending moment of the suspended beam (see Figure 9.14). To determine the magnitudes of these variables, the deformed geometry of the system should be taken into account. The aim of the following analysis is precisely the essential comprehension and practical solution of this problem (see Stavridis, 2008). The girder is in equilibrium under the following loads (Figure 9.15): (1) the selfweight g (permanent load) acting downwards, (2) a live load p applied over a certain length and acting downwards, and (3) the actions qc(x) of the tension hangers acting upwards. Loads g and p can be considered as being uniformly distributed, but the load qc(x) is non-uniform along the length of the girder. Applying a live load p on the girder makes the cable deform vertically through the hangers and, because the hangers will not be uniformly tensioned, the cable will receive non-uniform forces (see Figure 9.14), which will lead to a new funicular form that is different to the previous (parabolic) one. These same forces will obviously be applied with opposite sign on the girder, constituting precisely the actions qc(x), as shown in Figure 9.15. The distances between the hangers are so small that the above consideration of a distributed applied load qc(x) on the cable as well as on the girder is justified. Thus, if z(x) represents the ordinates of the cable position after application of the vertical 329

Structural systems: behaviour and design p

Additional live load g

x

Self-weight

1/r : due to permanent load

η(x) z(x) = y + η

y(x) Undeformable hangers

Hg: cable force due to permanent load Hp: increase of cable force due to live load

η(x) x

z(x)

Loads on the cable

qc S

qc

S

Vertical equilibrium qc · ∆x = H · ∆(dz/dx)

H = Hg + Hp: horizontal component of axial force S

Figure 9.14 Cable deformation and the acting forces

forces qc(x) (i.e. of its new funicular form) and H is the horizontal component of the total axial force of the cable, then, from the equilibrium of an elementary part of the cable along the vertical direction (see Figure 9.14), the following basic equation results (see Section 7.1): d2 z H dx2 In this equation qc(x) is the downward force, and the term dz2/dx2 represents the curvature (1/r) of the cable, which has a positive value. Thus the above equation is qc ðxÞ ¼

p

g

qc Forces acting on the girder

Figure 9.15 Actions applied to the girder

330

Cable structures

nothing more than the generally valid relationship q ¼ H (1/r) connecting the distributed load on a cable with its tensile force and its curvature, which applies everywhere a funicular form, be it in a prestressing cable (see Section 4.3.1.1), an arch (see Section 2.2.7) or even a shell (see Sections 12.2, or 12.3), etc. It is clear that, in this case (see Figure 9.14), z(x) ¼ y(x) þ (x) and H ¼ H g þ Hp According to the above, the total load q(x) applied to the beam, considered as positive downwards, is (see Figure 9.15) q(x) ¼ qc(x) þ g þ p This load is related to the girder deformation and the beam rigidity EI according to Section 2.3.6: d4 ¼ qðxÞ dx4 On the basis of the previous equations the same load may be written as EI

d2 d2 y qðxÞ ¼ 2 ðHg þ Hp Þ þ Hp 2 þ p dx dx Given that the term d2y/dx2 is equal to the negative curvature (1/r) of the cable under the permanent loads g (i.e. is equal to 8 f/L2), the above differential equation for the suspended girder takes the final form Hp d4 d2 EI 4 2 ðHg þ Hp Þ ¼ p r dx dx According to what was examined in Section 7.2, this equation represents the behaviour of a fictitious simply supported beam under a transversely applied load ( p Hp/r) and subjected to an axial tensile force (Hg þ Hp), according to the second-order theory (Figure 9.16). The deflection curve (x) may be sufficiently satisfactorily approximated, according to Section 7.2, by the equation 1 ¼ W1 1þ p Hp/r Hg + Hp

Hg + Hp Fictitious beam: the produced deflection must satisfy the constitutive equation for the cable

Figure 9.16 Actions on the fictitious simply supported beam

331


where W1 is the deflection curve of a simply supported beam under the transversely applied load ( p Hp/r) and is equal to ¼ (Hg þ Hp)/Pcr. Then, of course, Pcr ¼ 2 EI/L2. It is therefore clear that, for a given load p having a specific application zone over the suspended girder, the cable should develop such an additional force Hp that the deflection curve (x) resulting from the analysis of the fictitious tensioned beam (see Figure 9.16) satisfies the constitutive equation (a) for the cable deformation as formulated in Section 9.3: ð Hp g L L ¼ dx Ac Ec s Hg 0 Two cases of the application of the live load p are now examined according to the usual preliminary design practice: . Case 1: the live load is applied on the extreme half of the girder. . Case 2: the live load is applied over the entire length of the girder’s length. In the first case the aim is to determine the largest deformation of the cable and the girder, and, consequently, the maximum bending moment. The second case leads to the determination of the maximum cable force (Figure 9.17). In Case 1 the load p may be analysed as a symmetric load p/2 applied over the entire length of the girder and as an antisymmetric load p/2, as shown in Figure 9.17. Then, the deflection curve (x) of the fictitious tensioned beam may be considered as resulting from superposition of the deflection curves sym and ant corresponding to the symmetric load ( p/2 Hp/r) and the antisymmetric load p/2, respectively (see Figure 9.17). Note that the superposition principle is valid within the second-order theory, provided that the axial force remains constant for all superposed systems. Therefore, in this case the above constitutive equation is written as ð L ðL Hp g L ¼ sym dx þ ant dx Ac Ec s Hg 0 0 As ant is an antisymmetric function, the second integral will be zero, and the equation is simplified to ð L Hp g Ls ¼ sym dx Hg Ac Ec 0 Obviously this last equation is also valid in Case 2, wherein the live load p is applied over the entire whole length of the girder, and here the fictitious tensioned beam is loaded with ( p Hp/r). As mentioned previously, the deflection curve sym may be determined to a very good approximation through the expression sym ¼ W1 332

1 1þ

Cable structures First case: live load over the left half of the span p Hp/r Hg + Hp

Hg + Hp Fictitious beam

The antisymmetric loading is ineffective p/2

p/2 Hp/r

Hg + Hp

Final loading of the fictitious beam

Hg + Hp

Second case: live load over the whole span p Hp/r

Hg + Hp

Fictitious beam

Hg + Hp

Figure 9.17 Actions on the fictitious beam depending on the live load position

where W1 is the deflection curve of the simply supported beam under the load ( p Hp/r). The load p is equal to p/2 or to p, depending on whether one-half or the whole length of the girder is loaded. The moment diagram M01 for the simply supported beam is, of course, parabolic, with a maximum value ( p Hp/r) L2/8, and the deflection curve W1 is, according to Mohr’s theorem (see Section 2.3.6), equal to the moment diagram of the beam loaded by an external load (M01 /EI): 3 4 ð p Hp =rÞ L4 x x x 2 þ W1 ðxÞ ¼ L L L 24 EI Now the ratio Z, as the unknown quantity instead of Hp, and the dimensionless parameters , and ", as used in the previous section, will be used again. Moreover, the following dimensionless parameter G is introduced: Hg L2 G¼ EI 333


Substitution in the above constitutive equation for the cable leads, finally, to the following quadratic equation with respect to the dimensionless ratio Z: 2 p þþ1 Z ¼0 Z2 þ G where ¼

8 p2 2 15 " 1 þ 8 2

The maximum cable force (Hg þ Hp) thus occurs when the whole length of the girder is loaded with the live load, and the value of this force is obtained directly from the above equation by setting ¼ p/g. The maximum bending moment Mmax and the maximum deflection max of the suspended girder are determined by loading the half-length of the girder. These magnitudes can be evaluated for the purposes of preliminary analysis at the quarter of the span. By splitting the considered loading into a symmetric and an antisymmetric part of equal halves, as explained previously, it is noted that, because of the antisymmetric loading, no additional cable force develops. Moreover, the beam behaves like a simply supported beam with a half-span and subjected to half the live load. Obviously, the additional cable force Hp is due to the loading of the whole girder with the load p/2, and this force can be determined using the above equation by setting ¼ p/(2g). Then, superposing the relevant magnitudes at the quarter and at the middle of the span in the full- and the half-length beam, respectively, the following expressions are obtained: Mmax ¼ Msym þ Mant It is understandable that Msym ¼ M0sym ðHg þ Hp Þ sym and Mant ¼ M0ant ðHg þ Hp Þ ant Using the dimensionless parameters introduced so far, the following expressions are obtained: g L2 p2 G ðZ þ 1Þ Z 0:75 0:0742 2 Msym ¼ 2 8 p þ G ðZ þ 1Þ 2 2 gL p G ðZ þ 1Þ 0:125 0:013 Mant ¼ 8 4 p2 þ G ðZ þ 1Þ and sym 334

5 g L4 p2 Z 2 0:7125 ¼ 2 384 EI p þ G ðZ þ 1Þ

Cable structures

ant ¼

5 g L4 4 p2 0:03125 384 EI 4 p2 þ G ðZ þ 1Þ

Note that in the above results the bending moments and the deflections are expressed in terms of the corresponding quantities of the stiffening girder under the permanent load g, considered as a simply supported beam of span L. Obviously, the maximum deformation of the cable will be: max ¼ sym þ ant

g

9.6

Stiffening the suspension cable

9.6.1

Conceptual analysis

As explained in Section 9.1, the automatic adoption of the funicular form for any loading pattern acting on a cable constitutes an advantage in that it makes the cable the most efficient load-bearing element from a strength exploitation point of view. However, it constitutes a weakness too, because of the excessive deformability of the cable under the application of an additional load with a different pattern from the permanent one. It was shown in Section 9.5 how this sensitivity of the cable is overcome by the use of a horizontal girder which, for any additional (live) loading, restricts the deformability of the cable decisively, due to its appropriately chosen bending stiffness. It has already been mentioned in Section 9.1 (see Figure 9.2) how the stiffening of the freely suspended cable may also be achieved by another means, namely by embedding the hanging cable in a relatively thin band of concrete, which is prestressed with an appropriate force so that the resulting inverted arch is under compressive stress. The gravity loads to be applied later will cause tensile forces on the inverted arch. These, however, will be suppressed by the pre-existing compressive forces due to prestressing. More specifically, consider an orthogonal concrete cross-section with width b and height d, and a total cross-sectional area of the cables of Ap. The cables are supposed to have no bond with the surrounding concrete. Assuming that, with concrete ribbon casting, the load transferred to the cable having the section Ap is g (the concrete weight plus the cable self-weight), the cable must be prestressed with a force Hg ¼ g L2/(8 f ) in order to establish the predetermined sag f over the span to be covered L (Figure 9.18(a)). If, after hardening of the concrete, the cable is stressed further by an additional prestressing force HV, then the inverted arch with section (b d) will receive, all over its length, from the cable a uniform upward deviation force uV ¼ HV/r (Figure 9.18(b)), and under this load a uniform compressive stress V will develop: V ¼

u V L2 1 8f bd

this being obviously equal to HV/(b d). 335

Structural systems: behaviour and design L Hg = g · r

Hg f

(a) Actions on the cable

g HV

HV

uV = HV/r (b) Actions on the inverted arch

uV

Figure 9.18 Conceptual layout for the inverted arch

For each additionally applied gravity load p that is smaller than uV the arch ribbon will carry a total upward load equal to (uV p). That is, the arch will always be under compression without bending, while possible upward loads due to wind pressure will be carried normally by its operation as a typical arch. Of course, this assumption is simplified because the cable and the concrete band constitute a system that carries all additional loads in common, and hence the need arises to determine the stress distribution between the cable and the concrete section. Although this last assumption that the inverted arch can carry the gravity load p may be practically plausible, in the case when the live load does not extend to the full length of the arch but only up to the midspan (a loading that is mostly unfavourable in many cases, as examined previously) the combined action of the arch and the cable must be considered, in order to estimate the bending response in the arch and the stiffening effect of the above layout. It may be understood that the concrete—cable system has essentially the same characteristics as the suspended girder system considered in Section 9.5, in that the role of the girder as a stiffening element may be represented by the bending stiffness of the arch. More specifically, the fact that the common deflection of the cable and the arch is determined by both the axial rigidity of the cable and the bending rigidity of the arch itself, leads to the conclusion that this system behaves like a fictitious suspension system with a separate cable and stiffening girder. Indeed, in such a system, after an initial geometry for the cable sag under permanent loads has been established through the introduction of an appropriate cable force, a later applied live load to the stiffening beam produces identical deflections to the cable and the girder at each point, due to the presence of the vertical hangers. It may be concluded that the stress-ribbon system exhibits exactly the same characteristics as the suspension 336

Cable structures

r Hw = w · r

Hw = Hg + Hv f

w = g + uv EI L

Figure 9.19 Fictitious suspended beam

bridge and, consequently, this may be used as a fictitious model for the analysis of the system examined. Of course, in the fictitious system of a suspended beam with a cable cross-section Ap and a beam cross-section (b d), the ‘permanent’ load ‘g’ considered for the system must be one that establishes the sag f of the cable. In the present case this load is equal to the sum of the load g and the deviation load HV/r due to prestressing. Indeed, the fictitious system of the suspended beam under the permanent load w ¼ (g þ HV/r) develops in the cable an axial force (g þ HV/r) L2/ (8 f ) ¼ Hg þ HV, which obviously results in the same value as for the initial cable force in the actual system (Figure 9.19). In this way the results obtained in Section 9.5 for the stress state that develops in suspension bridges due to an unfavourable layout of the live load can be applied directly also in the present case, in order to estimate the bending moment M of the inverted arch as well as the maximum deflection in the case where the live load is extended only up to the midspan position. Of course, when checking the stress state in the orthogonal section of the arch, the simultaneous action of a compressive force N must also be taken into account. The value of this compressive force is given by ðuV p=2Þ L p L ¼ HV 8 2 8 It is obvious that, when covering areas with large spans, the ratio ¼ f/L may be even smaller than that for suspension bridges. At this point, one may consider the the so-called stress-ribbon bridges (which are mainly pedestrian bridges) in the same way as described above, where, in order to retain some maximum allowable slopes of the ‘inverted arch’ for pedestrian use, the ratio f/L must be limited to the value 1/50 (see Stavridis, 2010). N¼

9.6.2

Numerical example

In order to illustrate numerically the content of the previous section, the following purely indicative numerical example is presented. 337


A hanging cable having a span of L ¼ 60.0 m, a sag f ¼ 5.0 m and a cross-sectional area Ap ¼ 42 cm2 is to carry a permanent load g ¼ 5.0 kN/m and a uniformly distributed live load p ¼ 3.0 kN/m. The cable has a radius of curvature equal to r ¼ 60.02/ (8 5.0) ¼ 90.0 m. . According to Section 9.3, in order to find the maximum cable deflection the live load is placed over the left half of the span. The relevant parameters are calculated as: 3:0 ¼ 0:30 2 5:0

¼

Hg ¼ "¼

5:0 60:02 ¼ 450:0 kN 8 5:0

450:0 ¼ 0:00051 0:0042 2:1 108

Using the corresponding quadratic equation these lead to the value Z ¼ Hp/ Hg ¼ 0.294 and, according to the corresponding equation given in Section 9.3, it is obtained that

L 0:50 0:60 0:75 0:294 ¼ 5:0 ¼ 0:307 m 4 0:294 þ 1

. It is now assumed that the cable is stiffened by means of a steel beam having the profile HE-B 450, with a moment of inertia equal to I ¼ 79890 cm4 and a section modulus W ¼ 3550 cm3. According to Section 9.4, the relevant additional parameter G is G¼

450:0 60:02 ¼ 9:656 2:1 108 0:0007989

and, for the same position of the live load, through the corresponding quadratic equation it is found that Z ¼ Hp/Hg ¼ 0.290 (which is practically the same as before). According now to the corresponding equations given in Section 9.4, the maximum deflection is found to be noticeably reduced: max ¼ sym þ ant ¼ 0.0016 þ 0.0717 ¼ 0.073 m. With regard to the bending response of the beam, from the equations given in Section 9.4 it is found that Mmax ¼ 135.0 kN m, which leads to a moderate maximum bending stress of 135.0/0.00355 ¼ 38 028 kN/m2. . Finally, the case is examined where the cable is embedded in a concrete ribbon 20 cm thick, according to Section 9.6.1. The weight of the concrete for a width of 1.0 m, fits with the assumed permanent load, namely g ¼ 0.20 25.0 ¼ 5.0 kN/m. Applying now an additional prestressing HV ¼ 850.0 kN, the fictitious suspended beam, as previously explained, is considered, consisting of the very same 20 cm concrete thick ribbon, and subjected to the fictitious permanent load w ¼ 5.0 þ 850.0/ 90.0 ¼ 14.44 kN/m. 338

Cable structures

Following the same procedure as above, the corresponding parameters are calculated as 3:0 ¼ 0:208 14:44 450:0 þ 850:0 "¼ ¼ 0:00147 0:0042 2:1 108 and

¼

ð450:0 þ 850:0Þ 60:02 ¼ 33:427 2:1 108 0:0006667 For the same position of live load, through the corresponding quadratic equation from Section 9.4, the value Z ¼ Hp/Hg is calculated on the basis of the value /2 as being Z ¼ 0.099. The maximum deflection is now found to be 0.12 m and the maximum bending moment to be M ¼ 19.72 kN m. The corresponding compressive force is, according to Section 9.6.1,

G¼

N ¼ 850:0

3:0 60:0 ¼ 715:0 kN 2 8 ð1=12Þ

It is clear that the resultant compressive force, having an eccentricity of 19.72/ 715.0 ¼ 0.028 m ( L/10. Finally, it is noted that the deflection at the cantilever corner examined previously (see Figure 11.17) for a uniform load is twice that of the adjacent cantilevers. Where there is a possible need for prestressing, the remarks made in Sections 4.4 and 11.2.1.2 are valid.

11.2.3 Four-side supported plates Four-side supported plates are the most commonly used type of slabs in building construction. They are supported perimetrically on beams or walls, and are usually arranged orthogonally, being connected to each other in a continuous, monolithic way.

11.2.3.1 Stress state The four-side supported slabs transfer their loads in both directions, differentiating their load-carrying response depending on whether the load is uniformly distributed or concentrated. In the case of a uniform load, the shorter direction also receives the largest bending response, as has been explored for grids (see Figure 10.2). It should be clarified, however, that, with respect to a uniform load, the concept of ‘rigidity’ along the x or y sense concerns the distributed load px or py, respectively, which causes a unit deflection at the midpoint of a characteristic strip of unit width. This rigidity is expressed as k (EI)/L4, where the factor k is equal to 76.8 or 384, depending on whether the length L refers to a simply supported or to a fixed-end direction, respectively, and I ¼ d3/12. Thus, it may be considered that the load p is ‘distributed’ to the loads px and py, i.e. p ¼ px þ py, while the requirement for common deflection of the two mutually orthogonal strips requires (according also to Section 10.2) that py px ¼ ðstiffnessÞx ðstiffnessÞy 377

Structural systems: behaviour and design Table 11.1 Bending moments of orthogonal plates under a uniform distributed load Ly =Lx

1.0 1.50 1

Perimetrically simply supported plates

Fixed plates

mx =p L2x

my =mx

mx =p L2x

my =mx

mex =p L2x

mey =p L2x

0.037 0.073 0.125

1.0 0.38 0

0.018 0.034 0.042

1.0 0.30 0

0.052 0.076 0.083

0.052 0.057 0.057

that is px ðstiffnessÞx ¼ py ðstiffnessÞy Thus, under conditions of uniform support, in the case where Ly ¼ 2 Lx, px ¼ 24 py ¼ 16 py As briefly pointed out in Section 11.2, the bending stress state due to a concentrated load is independent of the dimensions of the plate and is proportional to the load. Tables 11.1 and 11.2, which refer to Figure 11.19, show the bending moments of perimetrically simply supported plates and fixed plates, for both their midpoints and their fixed supports, under a uniform and a concentrated load, for three characteristic aspect ratios. The case of an extremely oblong ground plan has already been examined in the case of two-side supported plates, but it is also included here for purposes of comparison. It can be shown that the twisting moment mxy (developed only at the free support) is, for the examined side ratios, roughly equal to 4% of the total load of the plate. With respect to a uniform load, it is observed that the moment mx at the midpoint of a square plate is barely 30% of the corresponding moment for a two-side supported plate in the case of simple supports, and 43% of the corresponding moment for the case of fixed supports. However, these percentages are increased if the beneficial influence of the twisting moments is ignored, which is equivalent to omitting the corresponding term in the differential equation for the plate (see Section 11.1). In the case of ribbed Table 11.2 Bending moments of orthogonal plates under a concentrated load Ly =Lx

1.0 1.50 1

378


Fixed plates

mx =P

my =mx

mx =P

my =mx

mex =P

mey =P

0.274 0.311 0.327

1.0 0.83 0.76

0.233 0.257 0.26

1.0 0.88 0.85

0.103 0.127 0.170

0.103 0.034

Plates mey

my

my

mex

mx

mex

mx

mey Lx

Lx

(a)

(b)

Figure 11.19 Characteristic locations of bending moments in orthogonal plates: (a) simple supports; (b) fixed supports

plates this influence is deliberately not taken into account, as will be examined later. Table 11.3 gives the corresponding data for the case of a uniform load where the twisting moments are neglected. The developing bending moments are, of course, increased, but the ratio of the moments in the two directions remains substantially the same. An estimate of the reactions on the sides of the perimeter can be obtained by evaluating the load acting on the regions created by the ‘bisectors’ of the corners with supported sides and the midline of the two opposite longer sides (Figure 11.20). In the case where a fixed side is connected to a simply supported one, the ‘bisectors’ are set at an angle of 608 to the fixed side. g In building construction, four-side supported plates are usually set in a continuous orthogonal arrangement, being supported on beams or, possibly, walls. Thus their support along the length of the beams (or walls) can be considered as a simple one, and only a vertically distributed load is transferred, while the plates themselves are mutually monolithically connected. As the plates receive a permanent load g, as well as a movable live load p, there is a need to determine the most adverse bending moments both at the centre of the plates and over their supports, i.e. in the common contact lines (Figure 11.21). Table 11.3 Bending moments, with the twisting moments neglected, under a uniform load Ly =Lx

1.0 1.50 2.0


Fixed plates

mx =p L2x

my =mx

mx =p L2x

my =mx

mex =p L2x

mey =p L2x

0.077 0.128 0.142

1.0 0.38 0.15

0.025 0.043 0.045

1.0 0.31 0.06

0.056 0.085 0.088

0.056 0.055 0.052

379


45

45

45

60

60

45

45

45

45

60 60

60

45

60

Figure 11.20 Distribution of the reactions at the sides

For this purpose, three distinct loading-layout cases may be considered, namely: (1) loading with ( g þ p/2), (2) loading with p/2, and (3) loading with ( g þ p). For each individual plate, the maximum bending moments in the centre can be estimated by superposing the load in case (1), with fixed conditions all over their perimeter, and the load in case (2), assuming simple support conditions over the same support lines. The bending moments at each internal support (tension on the top side) can be estimated from load case (3), with fixed conditions all around, and taking always the mean of the values from each side (see Figure 11.21). Span moments (bottom fibres tensioned)

g + p/2

Support moments (top fibres tensioned)

g + p/2

g+p

Fixed supports g + p/2

g+p

Fixed supports

g + p/2

g+p

g+p

Average of support moments

g, p

g, p

p/2

p/2

Internal supports

Simple support

g, p

p/2

g, p

p/2

Figure 11.21 Evaluation of the bending moments in continuous plates

380

0.75 < Lx/Ly < 1.30

Plates

11.2.3.2 Design The reinforcement is determined on the basis of the bending response. The fact that bending occurs in both directions in the centre of a slab implies that the reinforcement bars should be placed in two layers, one just over the other, in the x and y directions respectively. Obviously this results in a differentiated effective depth h (or lever arm z) for the two senses. As is the case along the length of the supports, the bending moments develop only in the transverse direction; they are obviously covered by only one reinforcement layer, placed normally on the top side. To estimate the reinforcement, the equation for two-side supported plates (see Section 11.2.1.2) may be applied for each direction. With regard now to the deformations, these can be determined through the ‘distributed loads’ px and py as has been explained previously: px ¼ p

L4x L4x þ L4y

L4y py ¼ p 4 Lx þ L4y The deformation w at the midpoint of a strip of unit width is 12 L4x;y 1 w ¼ px;y k E d3 where k is determined according to Section 11.2.3.1. However, the estimated value of w due to permanent loads must be increased by w ’ due to creep, and the final result cannot exceed 1/250th of the corresponding span length. This condition is usually met for plate thicknesses d > L/30. g In the rather rare case where, due to larger dimensions of the plate, the application of prestressing is required, the permanent load g of the plate should be taken up by the deviation forces of the cables. The required deviation forces can be ensured by a uniform parallel arrangement of cables in one direction only, this direction being the shorter one. In this way the cables offer the largest possible deviation forces for the same prestressing force, given that u ¼ 8 P f/L2 (kN/m2) (Figure 11.22). It should also be observed that, because the usually prevailing conditions do not allow a free shift of the ‘prestressing front’, uniform compressive stresses due to prestressing within the plane of slab must be excluded, as explained previously (see Section 11.2.1.2),

11.2.4 Ribbed plates If in a slab only the compression zone of the concrete is maintained, with the remainder of the material being supplemented in the form of parallel ribs arranged in both directions, a statically more favourable system is acquired. This arrangement has, of course, a higher structural depth, and hence smaller internal compressive and tensile forces, offering the possibility of using an even smaller thickness for the compressed 381

Structural systems: behaviour and design Layout A P

P

P

P

Layout B P

P

P

P

P

P

P

P

With layout A a given total prestressing force may outbalance a greater load on the slab

Figure 11.22 Layout for prestressing in a four-sided plate

plate and less tension reinforcement (Figure 11.23). Moreover, the deformations of this structural system will obviously be smaller than those of a solid plate. If the plate is two-side supported, the parallel ribs are ordered in a dense layout only in the load-bearing direction. However, one up to three transverse ribs still have to be placed in the other direction (depending on the span), for reasons of wider distribution of the eventual live loads, as explained for grids (see Chapter 10). For an equal axial distance e between the ribs, under the self-weight G of a zone of unit length and width e, and a live load p, the developing bending moment Mr on the rib considered is Mr ¼ (G þ p e) L2/8, for a two-side supported plate. This moment is, of course, taken by the beam-like rib with depth d. The distance e does not have to exceed 70 cm (see Figure 11.23). In the four-side supported plate the ribs are arranged, as already mentioned, along the x and y directions. Provided that the same distance e ( 52°

ϕ > 52°: Nθ tensile ϕ < 52°: Nθ compressive ϕ > 45°: Nθ tensile ϕ < 45°: Nθ compressive

p: ϕ > 45°

Figure 12.25 Behaviour of the ring membrane actions

from the above expressions for N, as ’ ¼ 51.508 for the self-weight and ’ ¼ 458 for the uniform load p. g In the case where the shell base is a horizontal circular plan, the meridional membrane forces N’ at its boundary also have to act on the element that supports the shell, but in the opposite direction, i.e. outwards (Figure 12.26). This element may normally be a circular beam (ring) which, under the action of the horizontal components of N’, will

Shell

H (∆H)

ME

(∆ϕ)

(∆ϕ)

(∆H)

H Ring The elongation of the ring and the shortening of the boundary creates a gap

ME The boundary forces act like redundant quantities: they restore the gap

Figure 12.26 Interaction between the shell and the ring beam

427


develop only tension without bending, since it represents the funicular form for such a radial uniform loading (see Section 2.2.7 and Figure 2.42). Consequently, its fibres tend to elongate, and do not conform to the deformation of the shell boundary as is implied by its ring forces N. If the ring beam lies over the limit of the approximately 528 angle, the shell boundary will shorten (N < 0), whereas under this limit it will elongate (N > 0). Hence, in both these cases there will be a gap in the deformation of the two systems, although in the latter the gap will be less. Even in the case of a hemispherical shell, where the horizontal component of N’ vanishes and the ring beam is not loaded within its plane, the existence of the tensile ring shell forces N leads again to a gap between the ring beam and the shell. Re-establishment of the deformation compatibility requires application of appropriate additional forces on the shell boundary, which also act on the ring beam but in the opposite sense. These forces are suggested by those which the ring beam can sustain according to its stiffness, and these, obviously, are horizontally distributed radial forces H as well as radially distributed torsional loads ME (see Figure 12.26). This means that the shell will be subjected to equal and opposite boundary loads, with the torsional loads ME of the beam clearly representing bending moments for the shell. An analytical examination reveals that, due to the double curvature of the shell, this disturbance of the membrane state does not extend significantly into the body of the shell. Rather, it is restricted to its boundary region, and thus does not have an essential influence on the preliminary design of the shell. Despite this, a thorough examination of the stress state and deformation of the boundary region is necessary in order to have a better understanding of its response (Billington, 1965).

12.4.1.1 Shell boundary Generally, the shell boundary exhibits, under its loads, a radial horizontal displacement H and a rotation of angle ’ (see Figure 12.26). Due to its own weight g and under a membrane state of stress it is R2 g 1 cos ’ sin ’ H ¼ Ed 1 þ cos ’ ’ ¼

2Rg sin ’ Ed

It is well known that H is produced by the strain " ¼ N/E d of the horizontal boundary ring, which has the radius R sin ’ and is subjected to the action of N. Consequently, H ¼ " (R sin ’) It should be noted that H is inwards when ’ 51.508. The development of ’ results in an increase of the end slope at the boundary of the shell. 428

Shells

According to the previous analysis, for a constant vertical load p uniformly distributed all over the shell, N R2 p H ¼ " R sin ’ ¼ cosð2 ’Þ sin ’ R sin ¼ 2Ed Ed Due to a uniform horizontal force H (kN/m) in the outward radial direction, H ¼

2 R2 sin2 ’ H Edc

2 R2 sin ’ H E d c2 pffiffiffiffiffiffiffiffiffi where c ¼ 0:76 R d, according to Section 12.3.1. Here, H is directed outwards, whereas the tangent slope at the edge is decreased due to ’. Due to a uniformly distributed bending moment ME applying tension to the inner boundary fibres of the shell, ’ ¼

H ¼

2 R2 sin ’ ME E d c2

4 R2 ME E d c3 where H is directed outwards, whereas the tangent slope at the edge is decreased due to ’. In the above expressions, it should be noted that the values of H and ’ are reciprocally equal, according to the Betti—Maxwell theorem (see Section 2.3.5). ’ ¼

12.4.1.2 Ring beam It is already known that a ring beam (section data: AR, IR) subjected to a distributed radial force H develops only an axial force NR ¼ H R and no bending at all, due to its funicular behaviour under that loading (see Section 2.2.7 and Figure 12.27). PLAN VIEW ∆H

H

H R NR = p · R

∆H = (R 2/EA) · H

NR

No bending response

Figure 12.27 Behaviour of a ring beam under internal pressure

429

Structural systems: behaviour and design PLAN VIEW

ME

ME

R

M = ME · R

M No torsional response

Self-equilibrating loading Sections rotate in the direction of the loading

Figure 12.28 Behaviour of a ring beam under a distributed torsional load

The strain in the ring beam is "R ¼ NR/(E AR), the elongation of its circumference is "R 2R, and its radial displacement is H ¼ "R R ¼

R2 H E AR

On the other hand, the application of a torsional load ME uniformly distributed along the entire length of the periphery simply causes an equally directed rotation ’ of its sections, ’ ¼

R2 ME E IR

and a constant bending moment M ¼ ME R, without creating any torsional moments, as shown in Figure 12.28. Under these conditions, the centroid of the beam section does not undergo a radial displacement H. In view of the above results, the evaluation of the magnitudes H and ME becomes possible by imposing compatibility of the displacements at the shell boundary and the ring beam, following a procedure identical to the force method, which was presented in Chapter 3. However, it should be noted here that the displacements H have a much stronger influence on the magnitudes H and ME than do the rotations ’. g At this point it should be emphasised that the bending response of the shell edge is essentially identical to that of a cylindrical shell of the same thickness, which circumscribes the spherical shell and is tangential to it at their boundaries (Figure 12.29). The cylindrical shell in question obviously has the same radius as the spherical one, i.e. equal to R. This result, which can be confirmed through a more detailed analysis, has clearly to do with the fact that the characteristic length c appears in the above expressions for the deformations of the spherical shell boundary. 430

Shells

d ∆H/sin ϕ

d

∆H ME

R

Tangential cylindrical shell ϕ

Estimated bending response using the cylindrical shell lies on the safe side

Figure 12.29 Identical bending behaviour at the boundary of a spherical and a cylindrical shell

This length was first introduced in the examination of the behaviour of cylindrical shells (see Section 12.3.1). In this way, and for ’ > 208, the bending moment at the spherical shell boundary can be approximately determined according to the results of Section 12.3.1, through the projection (H/sin ’) of the horizontal displacement H (due to its membrane state) on its radial direction. The value of the projection coincides with the radial displacement of the cylindrical shell that is tangentially equivalent. For a spherical shell with a fixed boundary, the moment is expressed as ME ¼ 0:29 E d2

H R sin ’

while for a hinged boundary it is Mmax ¼ 0:092 E d2

H R sin ’

H in the above equations takes the value which corresponds to the prevailing loading, which is either the self-weight or a uniform load. The resulting distributed thrust H (kN/m) acting on the shell boundary and directed outwards can be determined directly from the corresponding equations of the previous discussion. g As has already been pointed out, the resulting bending disturbance of the pure membrane state in a spherical shell is, in any case, restricted to a relatively narrow region by its boundary, having a width of the order of magnitude of the characteristic length c. Prestressing of the ring beam can restore the incompatibility of the deformations. This, additionally, solves the problem of taking on the tensile stresses in the case of a concrete beam, and also restricts the dimensions of its section (Figure 12.30). 431


Elongation of the ring and shortening of the boundary create a gap Nϕ

Nϕ R

P/R

Nϕ

Nϕ

H

H

P/R

Deviation force on the ring

Prestressing the ring with force P The ring receives an inward radial pressure (P/R – H) and the gap is eliminated

P/R

PLAN VIEW

Figure 12.30 Prestressing of the ring beam

By placing a circular tendon prestressed with a force P in a concrete ring beam, the ring will be automatically subjected to an inward radial uniform pressure P/R. If this pressure is greater than the already existing pressure H, the ring will develop an axial compressive force NR ¼ (P/R) R H R ¼ P H R, and consequently it will exhibit an inward radial displacement (see above): NR R2 P H R¼ H ¼ R E AR E AR The prestressing force P can be selected in conjunction with the other structural parameters, so that, according to the previous analysis, the resulting gap between the ring beam and the shell boundary vanishes (see Figure 12.30).

12.4.2 Shallow shells In order to confine the covered volume, as well as for functional or aesthetic reasons, a restriction of the dome height is often necessary, and thus the shell is characterised as 432

Shells N N N N d

f f

a a

Figure 12.31 Structural layout of a shallow shell

shallow. This implies a decrease in curvatures and, consequently, an increase of the membrane compression forces. In such a shell, where the height f is less than about one-fifth of the covered span a, the radius of curvature can be approximately expressed as R¼

f 2 þ ða=2Þ2 2f

or simply R ¼ (a/2)2/(2 f ), if f < a/10 (Figure 12.31). Such a shell can cover an orthogonal, usually square, area, where the resulting boundary curves are sections of the spherical surface cut by vertical planes corresponding to the four sides of the shell base. Due to the self-weight g, the resulting meridional membrane forces N’ along any ‘parallel’ are, according to the relations given in Section 12.4.1, N’ ¼

p a2 g R ¼g 2 p a sin ’ 2

So, it can be concluded that this force remains constant essentially over the entire shell surface. On the basis of this result and according to the analysis in Section 12.4.1, the ring force N, which is obviously compressive, is equal to the meridional one: N ¼ g (R/2). In the case of an orthogonal base, the above result may also be confirmed by considering that the shallow shell consists of shallow arches which transfer their load in both directions through compressive forces. These compressive forces N (per unit of width) are equal to g a2 gR ¼ N¼ 2 8f 2 according to the previous approximate expression for the radius of curvature. g 433

Structural systems: behaviour and design Table 12.1 Estimation of the bending moment (Me =p d2 ) at the fixed boundary of a shallow shell a=d f=d

100

150

200

250

0.2 0.1 0.05

4.20 12.80 33.50

5.20 16.70 43.80

5.90 19.90 53.20

6.50 22.80 62.00

In such a shell, having essentially the form of a curved orthogonal plate, the bending response should be taken into consideration even during the preliminary design phase, especially when its boundary is restrained from rotation. An estimation of the bending moments developed at the boundary can be made on the basis of an analogy to the ‘tangential’ cylindrical shell, as previously explained (see Section 12.2.1 and Figure 12.29). This bending moment is found to be Me ¼ 0:29 p R d and causes tension of the upper fibres. Note that, because of the shell shallowness, the self-weight can be considered essentially as a uniformly distributed load. This result lies, generally, on the safe side, i.e. the bending moment is overestimated. However, in order to estimate directly the influence on the bending response of the ratios f/a and a/d using more accurate analytical methods (Stavridis and Armenakas, 1988), it is useful to provide Table 12.1, which gives shallow shells with square bases and fixed boundaries. The table values refer to the dimensionless magnitude Me =ð p d2 Þ. In contrast to plates, the dependence of the bending response on the shell thickness is clear here. Poisson’s ratio has been assumed to be null. It should be noted that, in the case where the lateral transverse displacement of the boundaries is not constrained — which means that no forces N’ can develop at the boundary — the values in Table 12.1 must be doubled. g In the compressed dome, and particularly in the shallow dome, the danger of buckling deserves special attention. Of course, the existence of double curvature is a relieving factor which increases the critical load. This can be estimated using the relation (Schmidt, 1961) pD ¼ 0.15 E (d/R)2 From this relation it is seen that an increase in the radius of curvature decreases the load-carrying capacity of the shell, whereas an increase in the shell thickness increases the critical load. One way to increase the critical load is to place transverse ribs along the mutually orthogonal directions, which increases the average shell thickness, as is often seen in older structures. g 434

Shells

Vertial load g

N

N

S0

S0 Z ω

S0 H

f

A

L S0

S0 S0

H = S0 · L/4 Boundary arch Usually offered by a tie rod

Figure 12.32 Load-carrying behaviour at the boundary of a shell with a polygonal base

As previously mentioned, a spherical dome can cover a triangular, square or polygonal area. The sections of the shell surface on the corresponding planes passing through the sides of the polygonal base create arch-like openings with favourable lighting conditions (Figure 12.32). For the proper structural design of the shell boundary, it should be taken into account that forces cannot be applied in a direction which is normal to the boundary and tangential to the shell, while shear forces should be applied along the boundary, taking up the whole vertical load on the shell, if equilibrium has to be maintained through only a membrane state of stress. The fact that membrane forces N’ acting transversely to the arch-like boundary cannot exist means that the vertical load in the boundary regions will be taken only by the membrane forces N, which is parallel to the arch-like edges (see Figure 12.32). If Rb represents the radius of curvature of the corresponding arch, then according to the basic membrane relation given in Section 12.2 N ¼ g Rb Shear forces have to vanish along each axis of symmetry of the shell. So, in the arch-like boundary segment with span length L and height f, they vary from zero at the crown to a maximum of S0 at its ends. Clearly, the integral sum of their vertical components over all the boundary arches must equilibrate the total load g A of the shell, where A denotes the area of the polygonal base. Assuming a linear distribution of the shear forces along each boundary arch, the approximate maximum value is directly obtained as 1 3gA S0 ¼ k 4f 435


where k is the total number of boundary arches (see Figure 12.32) (Salvadori and Levy, 1967). These shearing forces can be directly applied to the shell by an arch of suitable section, which may be adjusted to the shell boundary. Such an arch cannot transmit forces other than those lying in its own plane; therefore, it has to take on the shearing forces of the shell boundary along its axis and transfer them to the ends of the arch, which are also the support points of the shell. These forces are obviously directed towards the ends, and consequently produce an outward horizontal thrust H. This thrust can be estimated on the basis of the adopted linear distribution of shearing forces as H ¼ S0L/4, and has to be borne either by a fixed support or through a tie connecting the ends of each boundary arch (see Figure 12.32). At each shell corner, the developed membrane shear forces S0 produce a tensile force Z acting in a direction perpendicular to the angle and equal to Z ¼ S0/tan (!/2) This tensile force has to be supported by appropriate reinforcement.

12.5

Hyperbolic paraboloid shells

12.5.1 Overview These shells belong to the translational shells category. Their midsurface is generated by a downwards curved parabola (generatrix), which ‘slides’ on another fixed parabola curved upwards (guide), keeping its plane always parallel to itself. The created surface has the form of a ‘saddle’ and, obviously, has a negative curvature, in contrast to the shells examined previously (Figure 12.33). Such shells, apart from their impressive aesthetics resulting from the alternating curvature, show a characteristic of particular constructional importance because of their negative Gaussian curvature, i.e. they may also be generated using straight lines.

Directrix Generatrix

Generatrix Directrix ϕ1 ϕ2 ∆12 ϕ1 · ϕ2 ∆12

= constant

Plane of parallel reference

Figure 12.33 Creation of hyperbolic paraboloid shells

436

Plane of parallel reference

Shells

Thus, if a straight line, regarded as a generatrix, moves in space with one of its edges, sliding over another fixed straight line, keeping itself parallel to a fixed vertical plane on the one hand and with a constant rate of variation of its slope over the travelled distance by its sliding end on the other, then this line will produce the same surface as that previously described (see Figure 12.33). This will be explained in detail immediately below. What will, however, be pointed out first is that an understanding of this ‘dual’ nature of the hyperbolic paraboloid shell is of paramount importance for the comprehension of its load-carrying potential, as well as for the design of these shells in general, as they are able to cover orthogonal (and other) layouts with an unlimited variety of morphologies. In all cases, the possibility of creating the surface by using exclusively straight-line segments is a crucial constructional advantage which gives these shells a clear economic precedence. As an understanding of the geometry of this shell is necessary for understanding its structural behaviour, it is appropriate first to examine it through the following analysis.

12.5.2 Perception of the geometry The analytic description of the creation of the surface by the two parabolas is based on Figure 12.34. In the considered orthogonal system Oxyz the ‘generatrix’ parabola with a span and rise equal to 2a and f1, respectively, has its plane parallel to the plane Oxz, while the ‘guide’ parabola curved in the opposite direction lies in the plane Oyz, having a span and rise (sag) equal to 2b and f2, respectively. The coordinate z of the created surface is determined from the relation z¼

x2 y2 2R1 2R2

where R1 and R2 are the radii of curvature of the two parabolas, assumed to be constants, and equal to R1 ¼ a2/2 f1 and R2 ¼ b2/2 f2, respectively. It is assumed that both parabolas are ‘shallow’ in the sense described in Section 12.4.2.

2a f2

R2 b a

2b

O

x

O

x

f1 R1

Generatrix

Directrix

y y

z

z 2

2

Equation of surface: z = x /2 · R1 – y /2 · R2

Figure 12.34 Creation of a shell through a parabolic generatrix

437

Structural systems: behaviour and design All vertical planes parallel to Ou and Ov intersect the curved surface along straight lines Surface equation: z = x2/2R1 – y2/2 · R2

R2 v ω

O

The two created surfaces are identical

x

ω u

R1

Surface equation: z = k · u · v Torsion of surface: k = 2/(R1 + R2)

y z

tan2 ω = R2/R1

z v

c O

ϕ1

2ω

ϕ2 a a

Torsion: k = ∆ϕ/c

u

a

Figure 12.35 Creation of a shell through a straight generatrix and the appearance of twist

It can be seen that the so-created surface covers a rectangular ground plan of dimensions (2 a) (2 b). Every section of this surface with a vertical plane is a parabola, while a section with a horizontal plane produces a hyperbola. If a vertical plane is now considered whose trace on the horizontal plane Oxy forms an angle ! with the x axis such that tan2 ! ¼ R2/R1, then this plane intersects the shell along a straight line. In addition, if the symmetric of this plane with respect to the x axis is considered, its intersection with the shell is also a straight line. In other words, the x axis bisects the angle 2 ! formed by the traces of the above two planes over the horizontal plane Oxy (Figure 12.35). If these two traces are considered as skew coordinate axes Ou and Ov, then it can be proved that the coordinates z of the surface may also be obtained from the expression z¼kuv where k ¼ 2/(R1 þ R2). It is clear that the intersection of the surface with a plane parallel to Ouz or parallel to Ovz is a straight line. The same shell can be assumed to be created if a straight line Oa, initially coinciding with axis Ou, is transposed in space while remaining parallel to the plane Ouz, with its edge always lying on the axis Ov. During this line movement, the ratio of ’, the change in the angle of its slope ’ relative to the plane Ouv, to the corresponding distance travelled by its edge O on the axis Ov should be kept constant and equal to k (see Figure 12.35). It is clear that k represents the constant twist of the surface relative to the coordinate system Ouv considered: k ¼ @ 2 z=@u @v

g

The transition from one geometry to the other may also take place. 438

Shells z

All vertical planes parallel to Ou and Ov intersect the curved surface along straight lines

O

A

u

2ω

C′

B

v

a

b f

C

Equation of surface: z = k · u · v Torsion of surface: k = f/(a · b)

Figure 12.36 Creation of a hyperbolic paraboloid surface over a skew ground plan

Assume that the shell is intended to cover a parallelogram ground plan OACB with sides a, b enclosing an angle 2 !, while the skew axes Ou and Ov coincide with the two sides of the parallelogram (Figure12.36). Corresponding to the edge C with skew coordinates u ¼ a, v ¼ b, a point C0 is defined with coordinate z ¼ f and connected through a line segment to the two other edges A and B. The hyperbolic paraboloid shell can be created if the segment OA moves parallel to the plane Ouz, inclined towards the lines OB and AC0 . The coordinates z of the so-formed surface are z ¼ k u v, where k ¼ f/a b. It is clear that any vertical plane parallel either to the axis Ou or to Ov intersects this surface along a straight line, thus making it possible for the surface to cover any parallelogram ground plan. This same surface may now be depicted through a system of mutually orthogonal parabolas with opposite curvatures in the following way (Figure 12.37). If the axis Ox is considered as the bisecting line of the angle 2 !, and the axis Oy is drawn perpendicular to it, then each of the planes Oxz and Oyz intersects the surface along parabolas. The intersection with the plane Oxz shows an upward curvature with a radius R1 ¼ 2 cos2 !=k while the intersection with the plane Oyz shows a downward curvature with a radius R2 ¼ 2 sin2 !=k Thus, the coordinates z of this same surface referred to the orthogonal system Oxy can be expressed through the radii of curvature R1 and R2, as previously shown. From the above analysis, it is clear that a hyperbolic paraboloid shell with equal radii of curvature R1 ¼ R2 ¼ R, being referred to the orthogonal axes Oxy, can also be considered as created from the straight generatrices representing its intersections with 439

Structural systems: behaviour and design All vertical planes parallel to Ox and Oy intersect the curved surface along straight lines z y

z=k·u·v k = f/(a · b)

O

R2

ω

A

u

ω R2 = 2 · sin2 ω/k R1 = 2 · cos2 ω/k

R1 C′

b Bisecting the angle 2ω

x 2ω B

v

a

C

The surface created from the two parabolas is identical with the initial one

Remark: In case of rhomboid layout (a = b) the axis Ox coincides with the line OC, but the two radii of curvature are not equal

Figure 12.37 Restoration of parabolic generatrices in a shell over a rhomboid ground plan

those vertical and mutually perpendicular planes Ouz and Ovz which bisect the right angles of the axes Ox and Oy. The twist k of the surface corresponding to the equation z ¼ k u v is, according to the above, equal to k ¼ 1/R. In order to acquire a more direct understanding of the load-carrying behaviour, shells with equal radii of curvature in both the x and y directions will now be considered.

12.5.3 Considerations of equilibrium In the orthogonal layout shown in Figure 12.38, a2/2 f1 ¼ b2/2 f2 ¼ R (see Section 12.5.2). The shell, as can be seen from the figure, extends between the four parabolas corresponding to the sides of the orthogonal base. The twist txy of the surface with reference to the system Oxy is zero, as can be deduced from the expression @ 2 z=@x @y, which implies that the tangential shear forces S in the directions x and y are also zero. The equilibrium of a curved quadrilateral element (see Figure 12.38) shows that the load p may be carried by the compressive forces, as well as by the tensile forces Nx and Ny. This is clear from the fact that the load p causes compression of downward-curved parabolas, making them act as ‘arches’, while simultaneously it applies tension to the upward-curved parabolas, making them behave as ‘cables’. From the equilibrium equation in Section 12.2, it can be deduced that the membrane forces Nx and Ny have a common value equal to p R/2. However, in order for this membrane state in the shell to exist, the above axial membrane forces should be appropriately introduced at the boundaries of the shell. If 440

Shells Vertical load p

Cable action (p · R/2)

Boundary arches x

R

b f2 Arch action (p · R/2)

f1 R a p · R/2

y

Arch action p · R/2 Cable action The membrane state with simultaneous arch and cable action applied on vertical boundary arches is problematic

Figure 12.38 Load-carrying action of a hyperbolic paraboloid shell with parabolic edges

the shell is unyieldingly supported along its parabolic boundaries, then this is feasible. However, if the shell is supported on, for example, vertical parabolic arches, these are not positioned to carry forces perpendicular to their plane, and thus the purely membrane state of stress is impossible. The very fact, of course, that these arches are able to carry vertical loads means that vertical shear forces will necessarily be introduced in the shell, thus leading unavoidably to bending. Another support possibility is the existence of unyielding supports along the ends of downward-curved (i.e. compressed) parabolic ‘arches’, while the edges of the upwardcurved ones are free (Figure 12.39). Then, the load p has to be supported entirely by the compressed parabolas, which will develop a compressive force Nx equal to p R, double the previous value (see Section 12.2). Of course, unyielding supports may also be considered only at the edges of the tensioned parabolas, the sides of the other direction being left free. Then, the load p going through the tensioned parabolic trajectories with the axial force Ny ¼ p R should be carried by prestressed tendons embedded in the concrete (see Section 9.1). g It is now of interest to examine whether the above-discussed case of a shell carrying its load p through compression paths (arches) can be realised without transferring transversal loads to the edge-supporting arches, which has been previously shown as 441

Structural systems: behaviour and design Vertical load p The vertical load is carried only through the arch action

Free boundary Unyielding boundary

Unyielding boundary

b

p·R

b

Plan view

p·R

a Free boundary

a

Arch action (p · R) Arch action

p·R

Figure 12.39 Feasible membrane equilibrium using arch action and free opposite boundaries

problematic, while ensuring that a pure membrane state develops (see Franz and Scha¨fer, 1988). This is achieved on the basis of the remark that the cuts of the shell with vertical planes mutually perpendicular and with an angle of 458 to the axes x and y are straight lines, as mentioned previously (see Section 12.5.2). First, it is again assumed that the load p is carried by the axial compressive force Nx ¼ p R in one direction only, as in Figure 12.39, keeping the two other opposite edges free from transverse loads. This state of stress will be superposed on another one, as explained in the following two cases. In the first case, where the ground plan is assumed to be square, a self-equilibrating loading is further considered, as in Figure 12.40. This loading consists of a distributed load p R acting in the same sense perpendicularly to the opposite edges of the compressed parabolas, being equilibrated by tangential shearing forces applied at the two other opposite sides. As is evident from Figure 12.40, the normal forces applied to the two boundaries may be analysed in two perpendicular directions at 458 as tensile or compressive forces, which are transferred directly to the neighbouring boundary where they are balanced by the shearing forces acting there. Thus, this additional loading is self-equilibrated within the shell itself, and therefore able to be superposed on the former state of stress. It is now obvious that one boundary of the compressed parabolic paths will be free of forces, whereas on the opposite side an external compressive force 2 p R is required through an appropriate support (see Figure 12.40). It is also clear that the existing arches, at the other two opposite sides, are able to receive — and to transmit — the longitudinal shearing forces without problem. It can also be seen that the required compressive force Nx ¼ p R is applied to the free edge, due to the simultaneous tensile forces inclined by 458 which meet on the boundary (see Figure 12.40), so that the overall compressive action of the parabolic shell arches is not disturbed. The second case, where one side in the ground plan is twice as long as the other, also allows the development of a purely membrane state by considering the application of 442

Shells Release of one shell edge from externally applied normal forces p·R

io

·p

ns

·R on si

·p

es

7 io

0.

ns

70

Te n

a

p·R

pr

p·R

om

p·R

C

Plan view

0.

a

70

7

Te

·R

n

n sio es pr

m Co

p·R

p·R (Self-equilibrated loading)

They are offered by the boundary arches p·R

Te

n

ns

sio

io

n

es

pr

m

Co on

2·p·R

om C

n

io

ns

Te

pr

es

si

p·R

p·R They are offered by the boundary arches Thanks to the shell geometry compressive forces are introduced, although the edge remains free

Figure 12.40 Feasible membrane equilibrium in a square layout with a free edge

additional self-equilibrating loading tensile forces Nx ¼ p R to the shorter sides of the shell (Figure 12.41). This load, which eliminates, of course, the initially obtained corresponding compressive forces, may be considered as coexisting with the self-equilibrated shearing forces on each oblong side of the shell. The analysis of longitudinal tensile forces at both short sides in components at an angle of 458 makes the establishment of the internal equilibrium possible. This is achieved due to the selected directions being implemented in straight lines on the shell, which allows the direct transfer of forces between the oblong boundaries, where they are balanced by the shearing forces acting there. These shearing forces can be imposed by the corresponding boundary arches while the short sides remain free, as the tensile forces balancing there provide the compressive forces Nx ¼ p R, leading to the compressive carrying action of the arches in the oblong sense (see Figure 12.41). 443

Structural systems: behaviour and design Release of two shell edges from the external action of normal forces p·R

p·R They are offered by the boundary arches Plan view p·R

p·R

Te

io

n io n

ns

ss pr e

on

om C

io n

n io

pr es s

s es pr

om

p·R

They are offered by the boundary arches

om

Introduction of compressive forces, although the edge remains free

Te

n

io

C

on

on

p·R

si

on

Te ns i

es

p·R

i ns Te

C

ns Te

p·R

pr

on

p·R

om

i ns Te

p·R

p·R

n

C

p·R p·R

io

ns

ns

io

n

Te

2·a

Te ns i

a

p·R

(Self-equilibrated loading)

Figure 12.41 Feasible membrane equilibrium in an oblong layout with two free edges

12.5.4 Hypar shells with straight edges To cover orthogonal areas, hypar shells are usually formed with straight boundaries. The shell is necessarily referred to a system of axes Oxy comprising two adjacent sides of the orthogonal base OABC (Figure 12.42). With the origin O kept fixed, the three other edges can be moved vertically at any distance, as shown in Figure 12.42. The shell is created through the constant displacement of side OA0 in parallel to the plane Oxz, inclined towards the sides OB0 and A0 C0 , so as to result in position B0 C0 .

Edge beams (f3) C′

B′

p

C

Dir

ec

A′

f2 y

f1

B z b O

tio

ny

S

x

0

S

S

S

N y = 0 on i t c Stronger slopes e ir

D Vertical equilibrium: S = p/(2 · k)

Torsion: k = (f3 – f1 – f2)/(a · b)

Figure 12.42 Load-carrying shear mechanism of a hypar shell with straight edges

444

=

x

A a

Nx

Shells

The twist k of the surface thus created is k¼

f3 f1 f2 ab

The segments ‘f ’ are considered signed quantities with respect to the axis z. According to Section 12.5.2 and given that 2 ! ¼ 908, both surface curvatures (i.e. ‘1/R’) are equal to k. These, understandably, appear along the skew directions at an angle 2 !/2 ¼ 458 relative to the axis Ox. It should be noted that, in this case, the equation of the surface is not expressed by z ¼ k x y but by z ¼ k x y þ ( f1/a) x þ ( f2/b) y which does not interfere at all with its geometry, as the twist of the surface according to the expression @ 2 z=@x @y remains equal to k. Considering now the equilibrium of an orthogonal shell element in the system Oxy under the vertical load p, it can be seen that the only membrane mechanism through which this load may be carried is the tangential shearing forces S acting along the sides of the element (see Figure 12.42). This happens because the shell element presents null curvatures about the two orthogonal axes (i.e. x and y), and, consequently, any developing internal forces Nx and Ny cannot offer a component opposing the load, because they are collinear for any corresponding pair of opposite element sides. Thus, it follows that the above forces cannot be developed, so that the basic equation for membrane equilibrium in Section 12.2 yields the following constant value for S: S¼

p 2k

Indeed, the difference in tilt at opposite sides of the element allows the forces S to offer an upward component, which equilibrates the vertical load p (see Figure 12.42). These forces S must necessarily be applied at the straight edges of the shell too, which is possible only if edge beams are arranged along each boundary. It should be noted here that the previous conclusion about zero membrane forces Nx and Ny is strictly valid only if the load p acts perpendicularly to the shell surface. Since the acting load is vertical, it is clear that for small slopes of the surface the above conclusion is essentially valid. ‘Small slopes’ is synonymous with the existence of ‘small curvatures’, which in turn means that the shell should be ‘shallow’, and this is the case if the edge beams have a slope not exceeding 188. For steeper slopes, the vertical load p has a non-negligible component in the direction of the straight generatrices, leading to the development of axial forces in their direction and, consequently, the membrane shearing forces are no longer constant. However, for most cases of roofing, the hypar shells used can be considered ‘shallow’. Of course, irrespective of the above impact of the equilibrium consideration for the axes x and y, it should not be overlooked that, due to existing curvatures of 458 in both oblique senses, axial compressive and tensile forces N that are equal in magnitude 445

Structural systems: behaviour and design Compressive arch action (p · R/2) (f3)

Edge beams

Equal radii of curvature R

p Dir

ec

f2 y

f1

x

ny

S S x

S

b

tio

Nx

a p · R/2

=

0

S S

N y = on 0 cti e r i D S = p/(2 · k) = p · R/2

O Tensile cable action (p · R/2)

(1)

p · R/2 Plan

(

R p·

/2)

·√

2

Torsion: k = (f3 – f1 – f2)/(a · b) = 1/R

The compressive, tensile and shearing forces have the same value

Figure 12.43 Load-carrying mechanism through the arch and suspension action of a shell

are developed, which, according to Sections 12.5.2 and 12.5.3, are pR p ¼ 2 2k i.e. equal to the membrane shear forces (Figure 12.43). Thus, the shell always works as a group of compressed ‘arches’ and suspended ‘cables’ both developing the same axial force. These forces reaching the boundary may be considered as being applied to the vertical sides of unit length of the orthogonal pffiffiffi triangular element, as shown in Figure 12.43, the hypotenuse of which is 2. The equilibrium of this element, given that the component of forces N perpendicular to the boundary is zero, confirms the necessity for the membrane shearing force to act along the boundary and to be equal to N. This is exactly the shearing force found above, running through the whole shell, which should, therefore, also be applied at the boundary. The edge beams which transmit these shear forces to the shell are obviously subjected to the opposite forces, and contribute their sum throughout the boundary length to each support of the shell. As shown in Figure 12.44, the forces on the edge beams accumulate to their lowest points, where they are taken up by appropriately designed columns that transfer the total load of the shell to them. The edge beams develop a progressively increasing compression towards their low supports, and in all cases care must be taken to carry their self-weight, usually by providing a continuous support. g N¼

The question arises, however, of whether the edge beams that will carry the above shearing forces can be designed free of intermediate support. In other words, is a layout of boundary beams which are supported (fixed) only at their corresponding lower ends and remain free all along their length, and, through the use of large cantilevers, provide an imposing aesthetic result, possible? 446

Shells Shearing actions (per m) on the edge beams (equal) S

S

S

S

The shearing actions are always summed up on the lowest corners

Figure 12.44 Forces acting on the edge beams of the straight boundaries

The fact is that, although according to the membrane state these cantilevers do not carry any part of the shell vertical loads but only longitudinal loads, they nevertheless have a significant self-weight, which they are not able to carry as long they are acting as cantilevers. The question, therefore, which essentially arises is whether the shell can participate in carrying the self-weight of the edge beams and, if so, with what consequences for its state of stress. The answer to this question is in principle positive, and is based on the fact that there are straight generatrices between the edge beams which can act as cables. These ‘cables’ can act against the tendency of the cantilevers to deform, by opposing tensile forces along the boundary, which relieve the cantilever from the response due to its selfweight. These forces are carried essentially unchanged from a given boundary edge to the opposite one (Schlaich, 1970). As shown by the model in Figure 12.45 — which is provided with an optional support at the high-point C to take into account a possible asymmetric loading — the system of the two adjacent ‘cantilevers’ OAB, which is subjected to the vertical loads of their self-weight, is also acted on by the forces of the tensioned cables at every discrete level. These cable forces Z counteract the tendency of the ‘cantilevers’ to tilt about the axis AB, as they cause an opposite bending moment to M0 due to their selfweight, thus considerably relieving them. For simplicity, the structure is assumed to be symmetrical about the vertical plane passing through the axis AB. An approximate estimation of the tensile cable forces can be made by assuming that the cantilevers are elastically undeformable and behave like rigid bodies rotating freely about the hinge axis of their support AB (Figure 12.46). It can be seen that the system of cables actually consists of prestressed tendons anchored at the opposite edge beams. Thus, for each level i, the forces Zi are determined on the basis of the angle ! between the cables considered in plan view, the difference zi of their ends, their projection length x on the vertical plane OC, the corresponding cable length Li, the distance ei of the corresponding resultant Ri of forces Zi at each node from the support axis AB 447

Structural systems: behaviour and design Carrying the weight of the boundary beam through the shell O

i

i+1

C i+2

Zi + 1

B Zi + 1

i i+1

Optional

Zi + 1 Zi + 1

i+2

A

The cable forces relieve the cantilevers, and in this way strongly decrease the bending action of the loads Symmetry about the vertical plane AB

Figure 12.45 Self-weight of the edge beams is taken up by ‘cable action’

and, finally, the cross-sectional area Fi of the ‘cables’, which may at first represent the tributary shell strips between them. Thus (Schlaich, 1970), X 1 Ki Zi ¼ M0 2 cosð!=2Þ i Ki ei where Ki ¼

Fi li ðx sin þ zi cos Þ L3i

and x ¼

ðABÞ 2 tanð!=2Þ

As mentioned previously, these tensile forces Zi should be taken up by prestressing (Figure 12.47). By introducing the appropriate prestressing forces along the straight generatrices, a purely membrane state in the shell may be feasible in practice, given that each compressive force applied at the shell boundary is not spread out but remains in the ‘strip’ of the corresponding generatrix. In addition, the absence of curvature in the tendons essentially eliminates the friction losses. g The synthesis of individual shells with a square ground plan in wider forms makes the covering of larger areas with a small number of supports possible. With the appropriate juxtaposition of shell parts, edge beams are needed only at the free boundaries in practice, because at the ‘interior’ straight boundaries of each shell, the required shearing forces S for the equilibrium are provided mutually by the adjacent shells. 448

Shells The cable forces acting on part AOB O

i

∆zi

C

i+1

Ziv

i+2

i

Ri

Ziu

B

Zi + 1

Gi i + 1

Ri + 1

Ri + 2 Zi + 2 i+2

Gi + 1i + 2

i+1

Gi + 2 Ziv = Ziu

i

y

Ii

A The cable forces Z are determined from the condition that the moments of the R values about the axis AB balance the moments of the G values

The cable forces have to be equal in each direction because of symmetry about the vertical plane AB v B

Ri = 2 · Zi · cos(ω/2)

i+1

i+1

i+2

Zi

i

ω/2

ω

O

Rhomboid plan

i

i+2

i

Ri

C

Zi

i+1

i+2

i+2 i ∆x

i+1

A u

Figure 12.46 Assessment of the required tensile (prestressing) cable forces

The required tensile forces are provided through prestressing O C B Optional

A

Figure 12.47 Realisation of cable action through prestressing of a shell

449

Structural systems: behaviour and design The horizontal acmes (beams) carry the self-equilibrating actions from the adjacent shells Actions on the boundaries

n

Edge beams ch

o cti

a

n

ctio

ea

l Cab

d

Ar

e

Ti

ro

Tie rod

Figure 12.48 Layout of four hypar shells to cover a square ground plan

The covering of a square ground plan using four individual shallow hypar shells is shown in Figure 12.48. Each shell develops a shearing state of stress in the sense of its generatrices, which, as explained above, is caused by the simultaneous presence of the equal tensile and compressive forces of the ‘arch’ and ‘suspension’ action, respectively. It is concluded that the two imaginary cut strips along the top horizontal edges are themselves in equilibrium under the oppositely acting shearing forces, as shown in Figure 12.48. Thus, the required shearing forces along the interior boundary of each shell are provided by the adjacent shell. However, the required shearing forces at the ‘external’ boundaries have to be provided by edge beams which, being acted on by the opposite forces, have to give up their resultants to the corner columns. These resultants have a horizontal component acting along the diagonals of the ground plan, according to the arch action indicated in Figure 12.48 (see Section 8.1). The equilibrium of the column top necessitates the provision of a horizontal diagonal action in order to avoid an overstressing of both the columns and the edge beams, and this is accomplished by providing diagonal ties that are prestressed accordingly. g Despite the fact that the preliminary design of hypar shells may be based on the assumption that the loads are carried by membrane action, it should be noted that various incompatibilities arising from this consideration may also affect the bending action of the shell. Thus, it is noted, among other things, that, for example, nonuniform loadings may cause bending in the shell, or even that the mechanism through which the self-weight of the straight edge beams is carried through membrane action may never be complete. As has been emphasised elsewhere, in order to determine the final state of stress of the shell, the use of appropriate software is necessary, so that all details of its constructional formation can be taken into account. 450

Shells

12.5.5 Elastic stability The presence of compressive forces in hypar shells constitutes a danger of buckling which should be evaluated sufficiently during their preliminary design. In the case of shells with parabolic boundaries, it is clear that the presence of tensile forces along the upward-curved parabolas clearly contributes to the stability of the embedded ‘arches’. Thus, the use, in this case, of the expression for the critical load of a cylindrical shell having the same dimensions in plan as well as the same curvature will definitely lead to safe results (see Section 12.3.2). In the case of hypar shells with straight boundaries that are rigidly supported, having ground plan dimensions a and b (a b), a height difference f and a thickness d, the critical load pK may be assessed from the relation (Beles and Soare, 1972) f 2 d2 a2 b2 If there are also edge beams having a moment of inertia equal to I, the critical load is greater. This can be assessed from the relation sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi f 2 d2 2 24 I pK ¼ 2 E 2 2 1þ 3 3 a b d a pK ¼ 0:40 E

12.6

Conoidal shells

Along with the translational shells are also classified those shells which, in order to cover a rectangular ground plan between two parabolic fronts over two opposite sides of the rectangle, are formed through the segment of the other side as a generatrix by using the two parabolic edges as ‘guides’. This generatrix moves in parallel to a vertical plane, passing through the corresponding side (usually the oblong one), inclined towards the two parabolic guides, thus offering obvious advantages regarding the construction of the shell formwork (Figure 12.49). One of the two parabolic fronts is deliberately made higher than the other, which could also simply be a straight. The thus-formed conoidal shell can be supported on all four sides, i.e. at its parabolic edges through appropriately formed arches, and at the straight sides either through free-standing beams or rigid continuous supports. The preliminary design of such a shell can be based on considering it as an arch having a height fm equal to the average height of the two front parabolas. If L is the span of the parabolic arch, the compressive force per unit metre of the shell due to the vertical load g is Nd ¼ g L2/(8 fm) In addition, the critical buckling load can be assessed on the basis of the average radius of curvature R ¼ L2/8 fm, from the corresponding relation for cylindrical shells (see Section 12.3.2). In the case when the shell is not directly supported along its straight edges of length B, it can be assumed that the above compressive ‘arch’ forces of the shell are carried by the deep beams, which are formed along the edge region with an assumed structural height of B/4 and act as simply supported beams (see Figure 12.49). 451

Structural systems: behaviour and design Boundary arch Tie rod f2 N

N

N

N

f1 Tie rod

L

B Boundary arch

Average arch height: fm = (f1 + f2)/2

Beam action of the edge strip

Figure 12.49 Load-bearing action of conoidal shells

A not unusual use of conoidal shells is that of a shelter in the form of a cantilever, fixed at its parabolic edge and having all its three other straight edges free (Figure 12.50). Its structural function may be conceived as that of a cantilever of length B having a parabolic cross-section, which presents an advantageous increase in its ‘moment of inertia’ towards the fixed end, where the maximum tensile and compressive stresses are developed.

Thickness d

Mtr Nz Ten s

ile f

Mb

Mb Mtr

f

orce

s

Compressive forces L

Figure 12.50 Structural action of a conoidal shell as a cantilever

452

B

Shells

According to the above approach, the maximum tensile force Nz which results under a vertical load g is (see Kollar, 1984) B2 g=2 B2 g Nz ¼ ¼ 1:90 0:265 f f and is developed at the top of the supported parabolic edge. Beyond this ‘membrane’ load-carrying mechanism with regard to the cantilever moment Me ¼ g B2/2, a bending moment Mb of much less intensity also has to be considered, being assessed as Mb ¼ (d/f )2 Me. Furthermore, a bending moment Mtran in the transverse direction has to be taken into account, which causes tension at the top fibres since it increases the transversal curvature of the shell. Its value at the middle of span L may be assessed as Mtran ¼ 0.13 g L2 In the case when such shells are arranged monolithically connected in order to shelter larger spaces (see Figure 12.50), the transversal moments are limited to about 25% of the above value, with alternating signs across each field, causing a tension at the bottom fibres over the top of the connected ‘arches’.

References Beles A.A., Soare M.V. (1972) Berechnung von Schalentragwerken. Wiesbaden: Bauverlag. Billington D. (1965) Thin Shell Concrete Structures. New York: McGraw-Hill. Ciesielski R., Mitzel A., Stachurrski W., Suwalski J., Zmudzinski Z. (1970) Beha¨lter, Bunker, Silos, Schornsteine, Fernsehtu¨rme und Freileitungsmaste. Berlin: Wilhelm Ernst. ¨gge W. (1960) Stresses in Shells. Berlin: Springer-Verlag. Flu Franz G., Scha¨fer K. (1988) Konstruktionslehre des Stahlbetons, Band II, A. Berlin: Springer-Verlag. Kollar L. (1984) Schalenkonstruktionen — Beton — Kalender 1984 — Teil II. Berlin: Wilhelm Ernst. ¨ger A. (1963) Zur praktischen Berechnung der axial gedru ¨ckten Kreiszylinderschale. Stahlbau 32, Pflu 161—164. ¨ger A. (1966) Zur praktischen Berechnung der axial gedru ¨ckten Kreiszylinderschale unter MantelPflu druck. Stahlbau 35, 249—252. Salvadori M. and Levy M. (1967) Structural design in Architecture. Englewood Cliffs, NJ: Prentice-Hall. ¨tzten Randtra¨gern. Beton Schlaich J. (1970) Zum Tragverhalten von Hyparschalen mit nicht unterstu u. Stahlbetonbau 3, 54—63. ¨mmten Schalenmodellen aus Schmidt H. (1961) Ergebnisse von Beulversuchen mit doppeltgekru Aluminium. IASS Symposium Delft, pp. 159—181. Seide P. (1981) Stability of cylindrical reinforced concrete shells. American Concrete Institute ACI — SP 67, 43—62. Stavridis L.T., Armenakas A.E. (1988) Analysis of shallow shells with rectangular projection: analysis. Journal of Engineering Mechanics ASCE, 114(6), 943—952. Timoshenko S. (1956) Strength of Materials, Part II. Princeton, NJ: Van Nostrand.

453

13 Thin-walled beams 13.1

General characteristics

The concept of a thin-walled beam is applied in a wide variety of structural forms. These beams are used mainly in bridges, but they find application in building construction as well. The cross-section of a thin-walled beam is formed by using thin surface elements — mainly plane elements — that are monolithically connected to one other at their common edges. They thus form beam-type oblong members, resulting in a huge saving in material compared with a corresponding solid cross-section of the same constructional height and, simultaneously, have better load-carrying properties. It is for this reason that bridge superstructures, either suspended or not, are formed almost exclusively of thin-walled beams. As in beams having a solid cross-section, in thin-walled beams the dimension of length L is clearly larger than the maximum transverse dimension B, with a length/ width ratio (L/B) of more than 5, while the thickness t of the plane elements is thin enough that t/B 0.1 (Figure 13.1). Thus, the beam is formulated as a folded structure according to Section 11.6, and it can basically be analysed using suitable computer software. However, the main goal of the present chapter is the examination of this structural type as a linear beam element so that its load-bearing action can be appraised and understood accordingly. The main difference between the thin-walled and the solid beam, and which causes the biggest difficulties in analysing the thin-walled beam as a beam element, lies in the way the torsion is taken up. Not only shear but also longitudinal normal stresses contribute, a fact that is not foreseen in the so-called ‘classical technical theory’ of solid beams. The development of torsional stresses in thin-walled beams occurs both in rectilinear bridges, due to the eccentricity of traffic loads with respect to the crosssectional axis, and in curved bridges, where only the self-weight leads to a torsional response. Thin-walled cross-sections are distinguished as ‘open’ and ‘closed’ sections, each category having its own load-carrying characteristics. The closed cross-sections are also called ‘box sections’ due to their closed form. g The comparison between a closed and an open double-T section, as they are used in bridge design with respect to the evaluation of their structural behaviour, shows clear advantages for the former, without excluding at all the effectiveness of the latter, in a number of cases. Structural systems: behaviour and design 978-0-7277-4105-9


Structural systems: behaviour and design Thin-walled sections L

Thickness t B

B

L/B > 5 t/B < 0.1

Figure 13.1 The formation of thin-walled beams

In a beam with a double-T section, the possibility of developing compressive forces in the bottom region over the supports, either in a continuous system or in a cantilever (Figure 13.2), is certainly limited in comparison with a box section with the same depth (see Figure 13.1), as the bottom plate of the box can participate fully in taking up the compressive force, something that the double-T section cannot do. This is automatically translated as a restriction of the bearable tensile force by the top plate (in the case of concrete, through suitable reinforcement, prestressed or not — see Chapters 4 and 5), resulting in a limited corresponding bending resistance of the crosssection. g As stated previously, in beams made of concrete, prestressing is necessary for spans longer than 15—20 m. Apart from the fact that the bending resistance of a closed section is clearly greater, another ‘unfavourable’ characteristic of the open section appears in its use under service conditions, due to the combination of a smaller core depth and a higher positioned centroidal axis, as explained below. Consider the midspan section of a simply supported prestressed beam. The prestressing compressive force P that is initially applied to the cable trace is finally shifted upwards to a distance a ¼ Mg/P, due to the self-weight bending moment Mg, as discussed in Section 4.3.1 (Figure 13.3). However, in an open section this position is closer to the bottom core limit than is the case in a corresponding closed section. Hence the diagram of compressive stresses shows a greater difference between the extreme fibres in the open cross-section than is the case in the corresponding closed cross-section. Now, as already mentioned in Section 4.1.1, the curvature 1/r of

Need for a bottom flange in order to take up the compressive force Double-T section: limited possibility

Figure 13.2 The need for a closed cross-section formation

456

Thin-walled beams ∆σ

h

P

ku

Centroidal axis

Final position of compressive force Mg/P

P

Initial position of compressive force (g = 0)

Core extent

1/r = ∆σ/E · h ∆σ ko

h

Centroidal axis

P

Final position of compressive force

ku

Mg/P P

Initial position of compressive force (g = 0)

The larger stress difference causes larger permanent deformations

Figure 13.3 The influence of the type of section on the variation in stress due to prestressing across the depth

the beam is expressed as |"o "u|/h, i.e. as ||/(E h). Thus the open section presents larger deformations due to prestressing than does the closed one, and this difference becomes two to three times larger under the influence of creep (see Section 1.2.2.1), which is, optically at least, annoying (see Figure 13.3). g The last and perhaps most important difference between the two types of cross-section is related to torsion. A torsional moment MT applied to the open section is taken up, according to the classic technical theory, by the three oblong orthogonal parts through the development of shear stresses, the maximum values max of which occur at the external longitudinal sides (Figure 13.4), as explained in Section 2.5.3. Thus max ¼ (3 MT,i)/(B t2), where B is the length of each oblong side and MT,i is that part of MT which is carried by each part and offered naturally by the couple of the oppositely directed stresses . Note that the distribution of MT to the parts is proportional to their torsional inertia IT ¼ B t3/3 (see Section 3.2.10). However, due to the small lever arm (roughly 2 t/3) of the above couple, the possibility of taking up a torsional moment through this mechanism is always limited (see Section 2.5.3).

v = τ · t = MT/(2 · Fk) τ

T t

τ MT

t

Smaller lever arms lead to larger stresses

MT

Larger lever arms lead to smaller stresses

τ Higher torsional stiffness

Torsional moment is shared to the section walls proportionally to their torsional rigidity

More favourable uptake of torsion than in an open section

Figure 13.4 The basic torsional behaviour of open and closed cross-sections

457


Conversely, in a closed section, shear stresses of the same magnitude with a constant flow all over the profile have the ability to take up a much larger torsional moment, due to the relatively large lever arm they can offer. Thus, in each wall of the section, uniformly distributed shear stresses 1, 2, 3 and 4 are developed, each of which has the characteristic property of presenting a constant shear flow v ¼ ( t), i.e. a constant shear force per unit length (see Figure 13.4). Thus v ¼ 1 t1 ¼ 2 t2 ¼ 3 t3 ¼ 4 t4, while, according to the Bredt’s formula, v¼

MT 2 Fk

where Fk is the area encompassed by the midline of the cross-section (see Section 2.5.3). Certainly, the torsional rigidity of a closed section is much greater than that of an open section having the same cross-sectional area, as for constant thickness t, the torsional inertia IT is equal to 4 t F2k =L , where L is the perimeter of the midline of the closed cross-section (see Section 2.5.3). This generally means that a closed section has smaller torsional deformations than a corresponding open section. g However, the possibilities of the open section to take up torsional moments are not exhausted by the classical shear stress mechanism mentioned previously, also known as the St Venant mechanism. In addition to this mechanism there is another that is not predicted by classical technical theory and which helps the St Venant mechanism in taking up the torsional moment. However, despite this, the ability of an open section to carry torsional moments remains limited compared to the ability of the closed section. In a last comparison of the two types of cross-section, the possibility of the participation of the two webs in receiving an eccentrically applied load will be examined, based on a gross estimate of the additional mechanism of torsional resistance mentioned above for open sections, a detailed examination being reserved for later. Both types of cross-section have a width b and a height h (Figure 13.5). Consider a concentrated load Q applied at the midsection of a simply supported beam, placed precisely on the left web of each section. The load is split into a symmetric and an antisymmetric load, as shown in Figure 13.5. The symmetric loading causes bending and shearing of the beam, whereas the antisymmetric load causes torsion. Examining first the beam with an open section located to the left of its midpoint, it may be concluded that for the symmetric loading the shear force Q/2 acts through the downward forces Q/4 at each web, while the plate shear flows have the directions shown in Figure 13.5 and vanish at the midpoint of the plate. In the antisymmetric loading, the torsional moment (Q b/4) is counterbalanced by a couple of opposing shear forces in the webs, which are equal to (Q b/4)/b ¼ Q/4, and the shear senses in the plate that result are as shown in Figure 13.5. (At this point, it should be noted that the directions of the shearing stresses generally correspond to those followed by an analogous hydraulic flow, regardless of whether they originate from shear or torsion.) Of course, the presence of these shear forces implies bending of the lefthand web downwards and the right-hand web upwards, and clearly this opposite 458

Thin-walled beams Q · b/4 Q/2 Q/2 Q

Q/2 Q/2

Q/2

h

Q/2

b

Q · b/4

Q/4 Q/4

Q/2

Q/4 Q/4

Q · b/4

The right web remains practically uninvolved in uptaking the load

Q/8

Q/4 Q/4

Q/2

Q/8

Q · b/4

Participation also of the right web due to better torsional behaviour

Figure 13.5 The uptake of an eccentric load by beams having an open and a closed cross-section

bending of the two webs will cause such rotations at their ends that Bernoulli’s law of plane sections is automatically ruled out. This is a rough description of the mobilised bending mechanism of an open cross-section, which generally takes up a bigger share of the torsional moment than does the St Venant mechanism. This leads to the conclusion that the left-hand web receives all the shear force Q/2, while the righthand web remains unstressed (see Figure 13.5). The beam with a closed cross-section shows a better behaviour. Again, a cross-section to the left of the beam midpoint is examined. In the symmetric loading the shear force Q/2 again causes shear forces of Q/4 in the webs, and the shear stresses in the top and bottom slab result according to the above rule of ‘hydraulic flow’, as shown in Figure 13.5. In the antisymmetric loading, however, the torsional moment (Q b/4) causes a peripheral shear flow according to Bredt, which develops in each web a shear force (see Figure 13.5) Qb Q h¼ 4 ð2 b hÞ 8 instead of Q/4, as in the open cross-section. Thus, the left-hand web receives a shear force (3 Q/8), i.e. 75% of the shear force Q/2, while the right-hand web receives a 459


shear force Q/8, i.e. the remaining 25%. It is understandable that the superior behaviour of the closed cross-section in the distribution of an eccentrically applied load is owed to a more favourable uptake (with smaller shear forces) of the torsional moment (Q b/4) (see Figure 13.5). The basic principles governing the behaviour of thin-walled cross-sections will now be examined in a more systematic way.

13.2

The basic assumption of a non-deformable cross-section

A thin-walled beam with an open cross-section that is subjected to a torsional moment presents a perceptible distortion in each cross-sectional plane (Figure 13.6). More specifically, the longitudinal fibres of the beam are deformed along their length so that no cross-section remains planar but undergoes warping. The reason for this will be explained later. However, the new outline of the cross-section, mapped over its initially single plane, gives a form that is rotated with respect to its initial position, and does not present any deformation relative to its initial form (see Figure 13.6). This is the socalled assumption of a ‘non-deformable cross-section’. Despite the fact that this assumption can be satisfied implicitly, the insertion of transverse diaphragms at certain reasonable distances over the length of the beam may be additionally required, which, given the negligible rigidity transversely to their plane, does not affect the deformability of the longitudinal fibres of the beam, i.e. its warping. The reason why a distortion of the cross-section occurs is that a torsional moment applied to an open section is taken up not only by the local shear flow in each of its constituent thin-walled elements (as mentioned in Sections 2.5.3 and 13.1) but also, and mainly, by their bending within their plane, as shown for example in Figure 13.6. This involves, of course, a bending rotation of the sections of the elements within their plane, and the ‘synthesis’ of these bending rotations of the elements leads to warping of the cross-section. It is important to note that the presence of warping should not be confused with the requirement of non-deformability of the cross-section. For beams having a closed cross-section, the magnitude of warping is very limited. It should be noted, however, that the assumption of the non-deformability of closed crosssections that is adopted throughout this chapter may potentially be abandoned for

Plan view

Undeformable section

Deviation from the plane section (warping)

Figure 13.6 Warping and non-deformability of the cross-section

460

Thin-walled beams

reasons of constructional practice reasons, with corresponding consequences on the structural behaviour, as will be examined in Chapter 14.

13.3

Shear centre

The concept of a shear centre is of fundamental importance in describing the behaviour of thin-walled cross-sections in torsion. In the example shown in Figure 13.7, the thinwalled beam with a channel section is fixed at its left-hand end and is subjected to a vertical distributed load, positioned over the web. At the fixed end, besides the bending moment, this load causes a shear flow over the cross-section of the beam. In this case, the determination of the senses of shear forces in the flanges requires more than the analogy with a hydraulic flow (see above). To establish the longitudinal

p

p Twist

A V

V

A

A L

B

B MT

B Development of torsion

p

p·a

p

MT = p · a · L The torsion modifies the wall shearing stresses

Shear centre S a a = MT/V

S

V

V

No torsion developed

Figure 13.7 The shear centre in a thin-walled beam

461


equilibrium of a cut-out part of the top and bottom flange that is acted on by different longitudinal normal forces requires a shear force on its internal face. The sense of the shear stresses on the plane of the cross-section is obtained, based on the above, according to Cauchy’s theorem (see Section 4.1.1). If, now, the beam is considered as a free body under the distributed uniform load, the shear flow forces and the normal stresses on the fixed-end section, then the equilibrium is not satisfied because the shear forces of the two flanges produce an unbalanced moment. It is clear that to establish the equilibrium a torsional moment must be applied from the fixed end of the beam, with a vector directed towards the fixed end (see Figure 13.7). The twisting angle of each cross-section will be denoted by a vector with opposite sign. In order to exclude the above torsion, the plane of the distributed load must be shifted to a suitable distance a on the outside, thus passing from the point S as it appears in Figure 13.7. With this loading arrangement, the three applied shear forces on the walls at the fixed cross-section will be in equilibrium with the loads, both in the vertical direction as well as with the developing moments with respect to point S, so that no torsional moment need be imposed by the fixed end to maintain equilibrium. Thus, it is obvious that the beam does not twist. The point S is called the shear centre of the cross-section, and it is the point to which torsion is always referred (see Figure 13.7). Any load that passes through the shear centre does not cause torsion but only a bending deformation, whereas the action of each load, acting at a distance e from it, is equivalent to that of the same load shifted to the shear centre plus a torsional moment. This moment, when the load p is a distributed one, constitutes a distributed torsional moment mD ¼ p e along the beam axis (see Figure 13.7), whereas when the force P is concentrated it constitutes a concentrated torsional moment P e. It is obvious that the development of a torsional reaction on the beam will cause an additional shear flow in the walls of the cross-section. Clearly, if a thin-walled section has one axis of symmetry, then its shear centre must lie on this axis. It should also be pointed out that the shear centre is the point about which the cross-section rotates. In other words, it is the point of the cross-section that under pure torsion remains immovable. The above means of ‘qualitatively’ searching for the shear centre (i.e. through the development of shear flows) can, in principle, be applied to any cross-section. Of course, the analytical determination of the shear centre (Zbirohowski-Koscia, 1967) is cumbersome and must be performed using suitable computer software. Software programs of similar nature may also be employed to determine the shear flow distribution in the walls of a closed cross-section.

13.4

Warping of thin-walled beams and the stress state due to its prevention

The uptake of a torsional moment by a thin-walled beam causes warping of its cross-section, as mentioned in Section 13.2. However, the different mechanisms of 462

Thin-walled beams

taking up a torsional moment in open and closed sections involve some characteristic differences in the behaviour of thin-walled beams, depending on their type, as shown below.

13.4.1 Open sections The beam shown in Figure 13.8 having an I cross-section is examined. This is considered in free body equilibrium under two equal and opposite torsional moments, as shown below. It is understandable that the constant torsional moment at each cross-section of the beam is taken up by the total of the shear stresses developed in each wall of the crosssection. These shear stresses vanish, of course, at the middle of each wall thickness, so that the midplane of each wall is free from shear stresses (see Section 2.5.3). Under the influence of the torsional moments, each cross-section of the beam, including the extreme ones, will rotate about its centre of symmetry (shear centre), which means that the section flanges will move in opposite directions. As shown in Figure 13.8, the assumption that the extreme cross-sections will retain their initial plane leads to a certain shear deformation of the beam flanges, which naturally is not Acute angle

Right angle Middle plane of section Acute angle

Retaining of the plane section implies changing of the right angle, but this is impossible Right angle Retaining of the right angle implies deviation from the plane section, i.e. warping

Right angle

The section develops exclusively shearing stresses

Figure 13.8 Compulsory warping of a free beam under torsion

463

Structural systems: behaviour and design Plan view

A B

MT Top flange

Bottom flange

τω

σω A

A

σω B Self-equilibrating σω

MT

τS

τω B

τω τω

Offering of torsional moment M T,S (St Venant)

MT

Offering of torsional moment M T,ω (warping prevention)

Figure 13.9 The development of the longitudinal stress state due to restrained warping

possible, given the stressless state of the midplane of the section. Hence it is concluded that the occurrence of the above shear stresses in the section walls for the uptake of torsion involves the warping of the cross-section. It is clear that no other stress needs to develop in the element examined. g The above stress state changes drastically if one end of the beam, say the left-hand one, is fixed while at the free end the previous torsional loading is maintained. It is understood that at the fixed end the cross-section cannot warp, and obviously remains planar (Figure 13.9). For the stress state prevailing at the extreme free cross-section, all that has been mentioned previously is valid. The applied torsional moment MT causes a rotating shear flow in each wall. The total developed torsional contribution of the walls according to St Venant is MT,S. Obviously, at the extreme cross-section MT ¼ MT,S, but this is not the case for any other point in the beam. Because of the developing rotation of the cross-sections by , which naturally varies along the length of the beam, the two section flanges are bent within their 464

Thin-walled beams

plane in opposite directions. For the particular sense of MT the top flange is bent to the left and the bottom flange to the right. This bending of the flanges, i.e. the developing curvature, is naturally accompanied by longitudinal stresses ! that vary along the length of beam, as the curvature also varies, being maximum at the fixed end and zero at the free end, where obviously normal stresses cannot be developed. It is clear that this variation in longitudinal normal stresses ! leads, for equilibrium reasons, to the development of shear stresses ! in both the top and the bottom flange (see Figure 13.9). Thus, at each position of the top or the bottom flange, a bending moment is developed corresponding to a linearly varying diagram of longitudinal normal stresses, as well as a transverse shear force in each flange for each position. It can be seen from Figure 13.9 that, while the developing longitudinal normal stresses ! in the two flanges are selfequilibrating, i.e. they do not create an axial force in the cross-section, the shear forces in the two flanges offer a torsional moment MT,!. This moment, together with the moment MT,S, constitutes the imposed torsional moment MT: MT ¼ MT,S þ MT,! It is obvious that the contribution of MT,S to the total moment MT of the cross-section decreases as one moves away from the free end, where of course it constitutes 100%. It becomes clear that both the stresses ! and the developing ! along the length of the beam, which finally give rise to the contribution of MT,! in taking up the torsional moment MT, owe their existence to the prevention of the cross-sectional warping at the fixed end. In general, any prevention of warping of the cross-section in places where torsion develops causes the above stresses. It is clear that the higher this distortion, the bigger the stresses, and vice versa. A detailed examination of warping is, therefore, absolutely necessary, because the consequences of its prevention constitute the only reason why the technical theory of torsion (St Venant) needs to be revised. As a matter of fact, the latter predicts the sectional distortion, but not the stresses ! and ! that appear due to its prevention.

13.4.2 Closed sections Before examining warping in detail, it is important to point out that the above contribution of shear stresses ! to the total torsional moment MT applies only to open sections. The reason for this is that in closed sections the developing ! constitute a self-equilibrating system of stresses (in the same manner as the stresses !) and, consequently, do not contribute to the uptake of the torsional moment of the corresponding section. Thus, in closed sections the developed shear flow, according to Bredt, is actually the only mechanism of taking up the applied torsional moment. This is now shown (Figure 13.10) for the same cantilever beam examined previously. The applied torsional moment is considered to be introduced through a transverse diaphragm that is assumed to be very stiff in its own plane. However, the diaphragm raises no resistance to transverse displacements, such as those that accompany the warping of section. The diaphragm transmits (according to Bredt) a peripheral shear 465


A

B

B τS A

D τS (Bredt)

C

B M

MT

MT σω τω

σω τω

Due to Bredt τω A

σω

B

τω

σω τω

M

τω

Due to additionally imposed rotation of section

τω

A Reduced warping

τω

τω Self-equilibrating τω

(Compression) B σω

The torsional moment is taken up only through the Bredt stresses

Lower warping stresses (self-equilibrating) A (Tension)

Figure 13.10 Restrained warping and the longitudinal stress state in a closed cross-section

flow to the walls of the closed section (see Section 13.1). Thus the cross-section rotates about the shear centre due to the applied torsional moment by an angle that reduces to zero at the fixed support. To elucidate the deformation of the beam, the top plate is first considered cut out from the webs, with the midpoint M of its left side simply pinned against the horizontal displacement (see Figure 13.10). The plate is in equilibrium under the shear actions (according to Bredt) at both ends and the shear forces (according to Cauchy) on its longitudinal sides. Under the above forces the plate is deformed in shear and the two end sides acquire a slope with respect to their initial (undeformed) 466

Thin-walled beams

positions, thus causing a deviation from the cross-sectional plane and, therefore, warping. Certainly, this deviation (warping) is reduced by the imposed movement of the right-hand side of plate ‘downwards’, with point M being fixed, due to the anticlockwise rotation of the cross-section of the loaded end (see Figure 13.10). Thus it is clear that the developed warping in a closed cross-section is definitely smaller than that in an open one. It is clear that the existence of fixity does not allow the left-hand side to develop this slope (i.e. to warp), and therefore in order for the plane section to maintain planar, longitudinal stresses ! must develop (see Figure 13.10) over the entire perimeter of the section, and the same applies for each individual wall of the cross-section. It is thus clear that each wall is subjected to bending. The developing longitudinal stresses are smaller than those of an analogous open section, because the warping of the closed section is reduced. Here also the stresses ! constitute a self-equilibrating system, and the fact that they obviously vanish at the free end means that they keep reducing all along the length of the beam, thus giving rise to the development of shear stresses !. Indeed, by cutting the top and bottom plates, as well as the two webs at their midwidth along some length of the beam, the need to develop longitudinal shear stresses is apparent from the longitudinal equilibrium of the free part, which in turn (according to Cauchy’s theorem) produces corresponding shear stresses ! on the cross-section itself. However, as can be seen from Figure 13.10, the total of ! on the cross-section is self-equilibrating, and thus it does not contribute to the uptake of the torsional moment of the corresponding section, although it modifies to a small degree the existing shear stresses according to Bredt. Hence, in closed sections, MT ¼ MT,S As to the length over which the stresses ! and ! extend, this is of the order of the cross-sectional width, given the self-equilibrating character of ! and the nondeformablity of the cross-section, according to the corresponding conclusion of elasticity theory (the St Venant theorem). As the additional stresses ! and ! do not particularly modify the stress state obtained by means of the technical theory, they may generally be omitted in the preliminary design stage, provided that the ‘non-deformability’ of the cross-section is ascertained.

13.4.3 Analysis of warping If each point of the cross-section outline is characterised by its distance s on it, measured from a certain constant point, then the longitudinal displacement of a point s of the section that is found on the abscissa x of the beam may be represented by the function u(x, s), while (x) is the twisting angle at the specific position (x) of the beam (Figure 13.11). As is already clear from the above, the warping of a cross-section at position x results from the new positions u(x, s) of its points (x, s), after the longitudinal deformation of the 467

Structural systems: behaviour and design Beam axis

x

s θ(x)

u(x, s) ω(s)

Location of section [x = x1]

u(x1, s) = (dθ/dx)[x = x1] · ω(s) The diagram ω(s) represents the section warping due to a unit torsional moment Typical ω(s) diagrams

∫ ω(s) · t · ds = 0

Limited values of ‘ω’ in closed sections

Figure 13.11 Warping behaviour of cross-sections

beam fibres. It is found that (Zbirohowski-Koscia, 1967) d !ðsÞ dx where the ‘warping function’ !(s) is referenced at each point of cross-section outline and is determined on the basis of the position of its shear centre. Thus each thinwalled cross-section has its own characteristic diagram of the warping function !(s). This diagram shows for each point of the cross-section how much this point will be shifted along the beam axis due to a unit change d/dx in the twisting angle of the beam. The diagram !(s) (dimensions m2) always satisfies the equation ð !ðsÞ t ds ¼ 0 uðx; sÞ ¼

The determination of the warping diagram !(s) for a cross-section is a cumbersome process and should be done using a specialised software program. The qualitative warping diagrams of some characteristic cross-sections are shown in Figure 13.11. g It should be noted that there are some thin-walled cross-sections that do not develop any warping under the torsional response, and consequently they are never subjected to the stresses ! and ! that arise due to the restraint of warping. The following cross-sections belong to this category (Figure 13.12): . cross-sections having a centre of symmetry — the shear centre coincides with this point . open cross-sections having walls passing through a common point — the shear centre coincides with this point . all the closed cross-sections that can be circumscribed to a circle having a constant wall thickness (triangular, etc.) 468

Thin-walled beams Sections that do not warp under a torsional moment S

S S t4

t1

t4

t1 S

t3

S

S t2

S: shear centre

t = constant

t2

t3

Figure 13.12 Cross-sections that do not develop warping

. more generally, if in a closed cross-section the wall thicknesses are considered as vectors and the resultants of these vectors at each corner pass through a common point, then the cross-section does not undergo warping.

13.4.4 Longitudinal stresses due to prevention of warping It is reiterated that the longitudinal normal stresses ! due to the restraint of warping are not predicted by the classical technical theory of beams, as the only normal stresses that appear therein refer either to bending moments or to axial forces. In each case the developed stresses are statically equivalent to the occurring bending moment or axial force. As the stresses ! are self-equilibrating, they do not have any static equivalence to any force or moment, i.e. they do not contribute to the global equilibrium of the structure. However, it has been shown that they may be expressed in the same way as the normal stresses in technical theory, through a sectional magnitude M! called the bimoment (Figure 13.13). In absolute analogy with the technical theory of bending, it is found that for each point of the cross-section the longitudinal warping stress can be expressed as (Vlasov, 1961) ! ¼ where I! ¼

M! !ðsÞ I! ð

!2 dA A

The magnitude I! (dimensions m6) is called the warping constant, and is obtained by integrating over each elementary part of the surface dA of the cross-section. It is understood that !(s) plays the role of ‘distances from the neutral axis’, as is the case in pure bending. The similarity with pure bending is obvious. It is observed that, for a given value of the bimoment, the distribution of ! is identical to the distribution of !(s). 469

Structural systems: behaviour and design Beam axis Mω(x) x θ(x)

σω σω

σω(s) = (Mω/Iω) · ω(s) MT σω

Self-equilibrating σω

Distribution of ‘σ’ identical with that of ‘ω’

Bimoment Mω = ∫ σω(s) · ω(s) · dA = –(EIω) · (d2θ/dx2) Iω = ∫ ω2 · dA Much less in closed than in open sections The bimoment represents quantitatively the self-equilibrating longitudinal normal stresses

Figure 13.13 Size and significance of the bimoment

The analytical determination of I! is very cumbersome and should be performed using specialised computer software. The bimoment itself (dimensions kN m2), the physical meaning of which is covered in the following two sections, is defined, analogously to the bending moment in technical theory, through the expression ð M! ¼ ! ðsÞ !ðsÞ dA A

If (x) is the function of the twisting angle along the length of the beam, then d2 dx2 As closed cross-sections exhibit a clearly smaller warping compared to open crosssections, they clearly also have a much more ‘moderate’ warping function !(s), and consequently their warping constant I! is comparatively smaller. Thus in closed crosssections the normal stresses ! are limited. M! ¼ EI!

13.5

The bimoment concept

In a thin-walled beam the bimoment concept is directly related to the self-equilibrating system of the normal stresses that arise from the prevention of warping of the crosssection under a torsional moment. It is clear that preventing a constructional warping of even a single section (e.g. the end of one beam) is sufficient to exert a restraining effect on the warping of all other sections of the beam. It should also be understood that this prevention of warping activates the bending rigidity of the section walls, and 470

Thin-walled beams Activation of the bending and torsional stiffness of the vertical strips Inducing bending in the strips MT

MT

Self-equilibrating longitudinal stresses Bimoment Restoring the identical strains and stresses of the adjacent sides

Summed up to M T,S

MT

Statically equivalent to M T,ω M T = M T,S + M T,ω

Figure 13.14 An example of the physical meaning of the bimoment

automatically causes longitudinal normal stresses that are not negligible, being in any case obligatory in the inherent self-equilibrium along the longitudinal sense, given that no external axial force is present. It may be considered that these normal stresses represent the bimoment itself. The fact that a torsional moment generally causes bending, thus giving rise to the concept of the bimoment, besides what was examined in Section 13.4.1, is illustrated in the model shown in Figure 13.14. The vertical beam with an open profile is fixed at its bottom end, while the top end is rigidly connected to a horizontal diaphragm that is non-deformable within its own plane and has no transverse bending rigidity. A torsional moment MT is applied on the diaphragm. The beam may initially be considered as being composed of mutually independent vertical plane strips that exhibit a bending rigidity only within their plane and have a torsional rigidity according to St Venant. The forced movement of the horizontal diaphragm under the torsional moment, consisting of a displacement and a rotation, causes a shear force and a torsional moment on each wall segment according to its own bending and torsional rigidity, respectively. The horizontal diaphragm is in equilibrium under the externally applied torque, as well as the reacting shear forces and torsional moments of the walls, which are acting in opposite senses. It is clear that these shear forces will amount to a total moment MT,! because, for equilibrium 471


reasons, it is impossible to offer any other force. MT,!, together with the summation of the St Venant moments MT,S, will then balance the moment MT: MT ¼ MT,! þ MT,S The development of a shear force at the top of each wall segment causes bending, leading to axial normal stresses. These stresses are obviously self-equilibrating, i.e. they have no resultant along the longitudinal direction, as no axial force is acting on the segment. This situation will not change even when a self-equilibrating system of longitudinal shear forces (see Figure 13.14) is imposed along the free sides of the wall segments in order to restore the incompatible deformations occurring there. This set of self-equilibrating axial stresses is actually equivalent to the existence of a bimoment. Returning now to the last equation, it is again pointed out that a torsional moment MT is composed of a St Venant moment MT,S and a warping moment MT,!. It is understandable that the relationship between these two contributions depends not only on the cross-section itself but also on the length of the beam. For example, as the length increases, the ratio of the bending to the torsional rigidity for each wall segment decreases, and consequently the percentage contribution of MT,! with respect to MT,S in taking up the torsional moment MT decreases too.

13.6

Two theorems of the bimoment

Theorem 1 A force P parallel to the beam axis and applied to a point on the cross-section having a warping value ! causes a bimoment equal to M! ¼ P ! (Figure 13.15). (The compressive loads are considered to be negative.) In the beam with an I cross-section, where three self-equilibrating forces are applied (N ¼ 0), a bimoment equal to M! ¼ 2 (P/2) !1 is developed. This bimoment obviously causes axial stresses according to the basic equation given in Section 13.4.4, contrary to what would be predicted using the technical theory. The equal and opposite compressive forces P on the beam in Figure 13.15, which can be considered as prestressing forces, cause a positive bimoment M! ¼ P !1, (P < 0, !1 < 0) and, consequently, tensile normal stresses at the ends of the cross-section (! > 0), contrary to what would be concluded according to the technical theory. However, a uniform compressive stress p0 applied on the entire the cross-section does not cause a bimoment, because (see Section 13.4.3) ð ð M! ¼ p0 dA ! ¼ p0 ! dA ¼ 0 and, therefore, the whole section is acted on by the above compressive stress. Theorem 2 A moment M applied on a plane parallel to the beam axis at a distance e from the shear centre causes a bimoment M! ¼ M e. (The bimoment is considered positive if the vector of the applied moment M is directed towards the shear centre.) 472

Thin-walled beams P/2

P

P P/2

P Mω = 2 · (P/2) · ω1

ω1 [ω]

[ω]

Mω = P · ω1

ω1

ω1

The presence of bimoment means development of self-equilibrating stresses M M

Direction of moment vector showing away from S

S h/2

Mω = M · h

h/2

Figure 13.15 The development of a bimoment in the absence of a torsional load

According to the example shown in Figure 13.15, even though the two externally applied moments M on the beam are in equilibrium, they cause a bimoment M! ¼ M (h/2) M (h/2) ¼ M h and accordingly they give rise to a longitudinal stress response.

13.7

Warping shear stresses

As explained previously, the warping shear stresses ! are required in order to ensure the equilibrium, given the variation in the normal stresses ! along the length of the beam. As stated, their distribution over the cross-section is not self-equilibrating, as is the case for !, but leads to the warping torsional moment MT,!. It should be noted, however, that this particular effect occurs only in open sections. As shown in Section 13.4.2, the stresses ! in closed sections constitute a self-equilibrating system, which affects the distribution of shear stresses according to Bredt. However, it does not contribute to the uptake of the torsional moment MT, which is taken up exclusively by the mechanism of Bredt’s peripheral shear flow. 473


τω

τω

MT

τω

MT

τω

MT

MT

Indicative distribution of warping shearing stresses

Figure 13.16 Shear stresses due to the prevention of distortion

It should be mentioned here that a warping moment MT,! is also present in closed cross-sections, and this leads to self-equilibrating shear stresses !, as explained previously. The analytical determination of the distribution of stresses ! in the cross-section (open or closed) based on the moment MT,! is generally cumbersome, and for this reason specialised computer software must be used. The distribution of the warping shear stresses ! in some typical cross-sections is shown in Figure 13.16.

13.8

The governing equation for torsion and its practical treatment

Consider a suitably supported rectilinear thin-walled beam that is subjected to a distributed torsional load mD, and, possibly, to a concentrated torque T as well. The beam is referenced to axes with an abscissa x, as shown in Figure 13.13. The characteristic deformation magnitude on the basis of which all the torsional magnitudes of the beam are expressed is the rotation (x). Thus it is found that the warping torsional moment MT,! is expressed as MT;! ¼ EI!

d3 dx3

while the torsional moment according to St Venant MT,S is MT;S ¼ G ID

d dx

Hence the equation established in Section 13.4 (MT ¼ MT,! þ MT,S) may be written as MT ¼ G ID

d d3 EI! 3 dx dx

and, given that the equilibrium relation is, in effect, dMT ¼ mD dx 474

Thin-walled beams

the last equation takes the form (Vlasov, 1961) d4 d2 G I ¼ mD D dx2 dx4 It should be pointed out that this equation is valid for both open and closed crosssections, despite the fact that in closed cross-sections the size of MT,! leads, as previously mentioned, to self-equilibrating shear stresses !. It is, however, possible that the last result may be also obtained from the equation EI!

¼ T,! þ T,S which is valid for both open and closed cross-sections. Moreover, as the bimoment M! in Section 13.4 is expressed as M! ¼ EI! (d2/dx2) the following relationship between the bimoment and the warping torsional moment is obtained directly from the above: dM! dx This above equation is the fundamental relationship that governs the torsional behaviour of thin-walled cross-sections and, of course, in order to determine all the above torsional magnitudes in terms of the function (x), the satisfaction of the pertinent respective boundary conditions is required. Thus, the following boundary conditions are distinguished: MT;! ¼

(1) In the case of a fully fixed cross-section at position x ¼0 d/dx ¼ 0 (2) In the case where twisting of the cross-section, but not its warping, is prevented, which means that in the considered cross-section no longitudinal warping stresses ! can be developed, then ¼ 0 and M! ¼ 0, and on the basis of the previous equations ¼0 d2/dx2 ¼ 0 (3) In the case of a completely free cross-section M! ¼ 0 and MT ¼ 0, and therefore on the basis of previous equations d2/dx2 ¼ 0 d d3 EI! 3 ¼ 0 dx dx The boundary conditions for other cases can be worked out based on the above. Thus, for example, when in the place of the internal support of a continuous beam instead G ID

475

Structural systems: behaviour and design Table 13.1 Analogy between the magnitudes in the bending and the torsional response !

Beam in bending Displacement w Tensile force H Moment of inertia I Uniform load p Concentrated transverse load P Vertical component of the tensile force Hv Bending moment M Shear force (vertical) Q

Beam in torsion ! ! ! ! ! ! ! !

Rotation angle Torsional rigidity Warping constant Uniform torsional moment Concentrated torque Torsional moment according to St Venant Bimoment Torsional moment

G IT I! mD T MT,S M! MT

! ! ! MT;! ¼ V Hv (see Figure 13.17).

two beams are linked, the arranged bearings, with or without a diaphragm placed above the support, prohibit the end rotation, and thus (1) ¼ (2) ¼ 0, while due to the continuity of normal stresses ! the equality of bimoments, M!(1) ¼ M!(2) must also be required at the same position. g Analogy Torsional response

Bending response P=T

mD ID

Iω

H

T

H = G · ID I = Iω

θ

w=θ

M = Mω

V H HV = M T,S

H

HV

M T,ω H HV

V

V

M T,ω

V = HV + M T,ω Boundary conditions

Prevention of rotation (θ = 0) Free warping (Mω = 0)

Imposition of rotation θ Free warping (Mω = 0)

Prevention of rotation (θ = 0) Prevention of warping (Mω ≠ 0)

Imposition of rotation θ Prevention of warping (Mω ≠ 0)

w=θ

w=θ

Figure 13.17 The analogy between the torsional and the bending stress states of a tensioned beam

476

Thin-walled beams

Although the solution of the above basic equation governing the torsional stress state of a beam through the classical analytical treatment of differential equations is, in principle, feasible, it is very awkward and impractical, especially for more complex cases. However, it should be noted that the form of the equation in question is precisely the same as the equation for a tension beam under transverse loading according to the second-order theory, as already examined in Section 7.2 (Roik, 1983). This equation is dw4 dw2 H ¼ pðxÞ dx2 dx4 This observation is of crucial importance because it establishes a complete equivalence between the two problems, and this allows the torsional problem to be confronted using the conceptually much simpler problem of beam bending, for which many accessible software programs exist. This equivalence, and the corresponding dimensions, is illustrated in Table 13.1 and Figure 13.17. It is clear that all the boundary conditions of the problem that are expressed through the specific requirements for the rotation angle , the bimoment M! or the torsional moment MT, may be attributed to the corresponding terms for the beam in bending (see Figure 13.17). This result provides an overall and naturally more direct illustration of the torsional problem. EI

13.9

Examples

Example 1 The aim is to determine the torsional stiffness of a cantilever of length 10.0 m, i.e. to determine the torsional moment required at its free end that will cause a rotation ¼ 1 rad. The behaviour of both an open and a closed cross-section, having the same external dimensions and the same cross-sectional area, is examined, considering in each case the possibility of preventing or not the warping of the free end. It is assumed that the profile of the cross-section remains undeformable. Figure 13.18 shows the details of approaching the torsional behaviour through the established analogy with the beam in bending based on the second-order theory (see Section 13.8). The torsional stiffness is determined as the shear force that develops due to an imposed support settlement on the bending model of the beam, after the assumed twisting angle at the free end. It can be seen that in the open section the prevention of warping at the free end increases considerably the torsional rigidity, which is significantly higher than the value predicted according to the classic expression after St Venant (G ID/L). In the closed cross-section these differences are much smaller and thus have a much lesser practical importance. Example 2 Consider the stress state of a simply supported concrete bridge, with a span of 40 m and an open cross-section. The bridge is loaded at the midspan with an eccentric load of 1000 kN applied directly on the top of one web. Assume that the profile of the cross-section remains undeformable due to a suitable insertion of transverse diaphragms (Figure 13.19). 477

Structural systems: behaviour and design t = 0.15 E = 3.0 · 107 kN/m2

G = 1.25 · 107 kN/m2

t = 0.30

2.50 m

Iω = 0.1318 m6 ID = 0.026 m4

St Venant torsional stiffness G · ID/L = 32 375 kN m/rad

1.50 m A

N = G · ID = 323 750 kN

B ? MT

Imposition of rotation θ = 1 rad Prevention of warping

(∆T = –899.3°C) VB = 85 950 kN = MT

w=θ=1m

10.0 m A

G · ID = 323 750 kN

B ? MT

Imposition of rotation θ = 1 rad Free warping

VB = 49 594 kN = MT w=θ=1m

t = 0.15 2.50 m

E = 3.0 · 107 kN m

G = 1.25 · 107 kN m

St Venant torsional stiffness G · ID/L = 1.33 · 106 kN m/rad

Iω = 0.022 m6 ID = 1.064 m4 1.50 m A

N = G · ID = 1.33 · 107 kN

B ? MT

Imposition of rotation θ = 1 rad Prevention of warping

(∆T = –36 944°C) VB = 1.392 · 106 kN = M T w=θ=1m

10.0 m A Imposition of rotation θ = 1 rad Free warping

G · ID = 1.33 · 107 kN

B ? MT

VB = 1.360 · 106 kN = M T w=θ=1m

Figure 13.18 Determination of the torsional stiffness of a thin-walled beam

The load is analysed exactly as described in Section 13.1, i.e. by considering a symmetric and an antisymmetric load. For the sake of later comparison, it is noted that the symmetric load, which consists of the two loads of 500 kN, causes at the midspan a maximum tensile stress of 6300 kN/m2. In this respect the interest is, of course, focused on the antisymmetric loads of 500 kN (see Figure 13.19), a loading that causes a corresponding torsion. It is supposed that the layout of the bearings can offer to the beam the required reactions for the development of the torsional moment of 1425 kN m at its ends. The analysis is done according to the analogy established in Section 13.8, and is shown in Figure 13.19. The resulting bending moment of 18 946 kN m at the midspan represents the bimoment, and this is used as the basis for determining the longitudinal warping stresses, as detailed in Section 13.4.4. Based on the corresponding diagram [!], the greatest tensile stress is 4121 kN/m2, i.e. 65% of the tensile stress caused by bending. The contribution of MT,S, 478

Thin-walled beams 500 kN

500 kN ID = 0.253 m4 Iω = 24.18 m6

t = 0.25

2.50

t = 0.50 m 5.70

E = 3.0 · 107 kN/m2 G = 0.4 · E

10.50 m Plan view

10.50 1000 kN

40.0 m [M T]

1425

1425 500 · 5.70 = 2850 kN m 2850 kN G · ID = 3036 · 103 kN Midspan: Mω = 18 946 kN/m2 M T,S = 0 M T,ω = 1425 kN m

18 946 kN m Rotation at the end: ϕ = 0.23 · 10–3 rad

[M] Rotation at midspan: ϕ=0

1425

1425 End: M T,S = ϕ(GID) = 698 kN m M T,ω = 1425 – 698 = 727 kN m

[V]

2.77 [ω] 5.26

[σω] 18 946 · 5.26/24.18 = 4121 kN/m2

Figure 13.19 Determination of the stress state in a thin-walled girder due to an eccentric load

i.e. of the St Venant shear flow, in taking up the torsional moment MT is determined according to Section 13.8. As shown in Figure 13.19, this contribution is zero at the midspan, while at the ends it amounts to 49%. The corresponding percentages for the warping torsional moment MT,! are 100% and 51%, respectively.

References Roik K. (1983) Vorlesungen u¨ber Stahlbau. Grundlagen. Berlin: Wilhelm Ernst. Vlasov V.Z. (1961) Thin Walled Elastic Beams. London: Oldbourne Press. Zbirohowski-Koscia K. (1967) Thin Walled Beams — From Theory to Practice. London: Crosby Lockwood.

479

14 Box girders 14.1

General

Box girders, used mainly for bridges, are thin-walled beams having a closed section, and, as such, they obey everything in the previous chapter, on one condition: in order for the basic equation as well as for all the related concepts (bimoment, etc.) concerning torsional response to be valid, the undeformability of the box section must be ensured. This condition, for beams with relatively small-section dimensions, can be guaranteed by a certain wall thickness, but for larger dimensions, stiff diaphragms are required over each support, as well as at two or three places along the span. If these diaphragms are not desirable for constructional reasons, then the uptake of torsional loads induces deformation of the section, and, consequently, an additional bending response in the longitudinal as well as in the transverse sense is developed. In this chapter, rectilinear beams are first examined, followed by curved beams.

14.2

Rectilinear girders

14.2.1 General loading case In the typical box section shown in Figure 14.1, an eccentric layout of the traffic load is considered. This is specified in each case by the load standards followed. It is appropriate to consider those external actions applied to the nodes A and B of the upper plate, which eliminate both their displacement and rotation. These ‘fixing actions’ (see Section 3.3.1) consist of longitudinally distributed vertical loads qA and qB, as well as longitudinally distributed moments mA and mB, which are determined directly from the fixed-end actions on the assumed clamped nodes A and B. Thus, following Section 3.3.1, it can be considered that the state of stress of the girder results from the superposition of the fixed state I containing the loads with the appropriate nodal actions and of the state II containing only these nodal actions qA and mA, together with qB and mB, acting, however, in the opposite sense. The fixed state I, as a trivial state, is directly determinable, so that state II is predominantly considered. This state — provided the section is symmetric — can always be split into symmetric and antisymmetric parts. The symmetric part consists of the symmetrically distributed loads qR/2 and moments mR/2 applied to the edges A and B, respectively. The antisymmetric part consists of the mutually opposite loads qS/2, as well as the moments in the same sense, mS/2 (see Figure 14.1). Structural systems: behaviour and design



mA A

mB

B Fixed state

qA

qB

b0 qA qR = qA + qB mR = mA + mB

mA

qS = qA – qB mS = mA – mB

qR/2

mB Nodal loading

qR/2

mR/2

qB

qS/2

mR/2

qS/2

mS/2

Symmetric loading

mS/2 Antisymmetic loading

Figure 14.1 Derivation of nodal actions on a beam profile from the acting eccentric loading

The symmetric loading activates the longitudinal bending stiffness of the girder, and results in the well-known sectional force diagrams for bending moments and shear. Moreover, it causes a transverse response, if the equilibrium of a segmental part of unit length is considered. This girder strip is acted on by the resultant of the shear flows applied to each of its two faces (Figure 14.2). Given that the shear flow v is expressed as v ¼ V S/I, then dv/dx ¼ (S/I) (dV/dx) ¼ (S/I) qR so that the above resultant is exactly the differential shear flow dv acting on the strip, equal to dv ¼ (S/I) qR. It should be noted that S represents the static moment of inertia of the cut-out part of the section being considered,with respect to its centroidal axis. Thus, the girder strip is in equilibrium under the symmetric loads qR/2, as well as under mR/2, and the differential shear flow dv. The loads qR/2 and dv do not induce qR

qR/2

qR/2

mR/2

V

dv

V + dV

mR/2 dv Differential shear flow (= resultant from the two faces)

dx ∆x = 1

Flow directions follow the hydraulic analogy

Figure 14.2 Transverse response under symmetric loading

482

dv = (S/I) · qR

Box girders

any bending at all, but only axial stresses. The loads mR/2 cause both bending and axial stresses (see Figure 14.2). This state of stress may be determined using classic computer software for plane frames. g When speaking of the symmetric loading of a box girder, two points should be made regarding haunched bridge girders either in a cantilevered system or, analogously, in a continuous one over its internal supports — in addition to what has already been mentioned in Sections 4.4 and 5.1 — mainly with respect to the free cantilevered method of construction. The first remark concerns the variation in the tensile axial forces Z in the top slab of a cantilever, which is associated with the longitudinal shearing flow v, as shown in Figure 14.3. As explained above, the inclined compressive force of the bottom slab helps with its vertical component, in offering to the cross-section the required resistance to the induced vertical shear force, thus achieving, for a parabolic curving of the bottom slab, an approximately constant shear flow throughout the cantilever length. By cutting out from the web an elementary segment of the top slab subjected to the axial tensile forces, as well as to the acting shear flow, and considering its horizontal equilibrium,

∆x

τ · bw

Z + ∆Z

M

z

Z τ · bw

V D ≈ M/z Constant shear flow τ · bw

3

2

1

1

2

Accumulated prestressing force

τ · bw = ∆Z/∆x

3

Tensile force Z 3 2 1

3 2

3

1st

2nd

3rd

etc.

segment pu = D/r D ≈ M/Z Transverse response

pu

Figure 14.3 Structural behaviour of a cantilevered haunched girder

483


it is concluded that v ¼ bw ¼

Z x

where

1 M z b ¼ V (see Section 4.4) z z x w

The above relations are quite generally valid. In the case of a constant depth (z/x ¼ 0), the tensile force Z exhibits a parabolic variation along the girder (z/x ¼ V/z). However, in the case of a variable depth leading to a constant flow v, the tensile force Z is linearly varied, which, with regard to the balanced cantilevered construction for continuous bridges, mentioned in Section 5.1 and suggested in Figure 14.3, means that an essentially constant amount of prestressing ‘dosage’ needs to be added at each consecutive construction stage, thus representing a constructional advantage. It should be pointed out that the prestressing reinforcement for the free cantilevered process (see also Section 5.5.2) is placed principally on the top slab of the girder section. The second remark regards the fact that the ever-changing inclination of the compression force of the bottom slab, due to bending, creates an inward deviation pressure pu ¼ D/r (see Figure 14.3), which stresses the bottom slab in the transverse sense, obviously counteracting its self-weight, but necessarily affecting the whole closed transverse segmental strip in bending, as shown in Figure 14.3. g As the symmetric loading does not present any other peculiarities, the whole problem is shifted to the antisymmetric loading part. This acts as a distributed torsional load, mD ¼ (qS/2) b0 þ mS (see Figure 14.1), resulting in a particular torsional moment diagram (Figure 14.4). At each position of the beam, the torsional moment developed is taken up by the corresponding Bredt shear flow. Thus, for each wall of the section considered in its longitudinal sense, a shear force diagram can be assigned. If, now, the non-deformability of sections can be ensured through the layout of transverse diaphragms, as mentioned previously, an additional longitudinal response consisting of axial and shearing stresses will take place, according to Chapter 13. The fact that the section is closed leads to a limited response arising from the restraining of the limited warping of the section.

[M T]

[M T]

The Bredt shear flow induces shear forces in each section wall

Figure 14.4 Influence of the support layout on the distribution of the torsional response

484

Box girders

However, the presence of the transverse diaphragms is generally not desirable for constructional reasons. Thus, the girder section must be considered as deformable, the development of torsional response is no longer governed by the basic equation of Section 13.8, and, consequently, the structure has to be considered, accordingly, as a folded system. In this case, as will be examined in the following section, the abovementioned additional response to the shear flow, according to Bredt, consists of a longitudinal bending for each wall, as well as of a transverse response for the section profile.

14.2.2 Response due to the deformability of the profile section The imposed torsional loading (mD ¼ (qS/2) b0 þ mS), due to the antisymmetric couple, causes deformation of the closed section, causing longitudinal bending of the section walls (Figure 14.5), which is coupled with the resulting transverse bending of the section profile itself (Schlaich and Scheef, 1982). It is again pointed out that this response is in addition to the initial Bredt shear flow in the section walls. In order to now investigate the influence of the deformability of the section on the response, the equilibrium of a cut-out girder strip of unit length is first considered with the antisymmetric loads qS/2 and mS/2 acting at the section edges A and B (see Figure 14.5). It is clear that the segment is in equilibrium under the above external

qS/2

qS/2

mS/2

mS/2 A

Antisymmetic loading

Torsional moment diagram

B

Equilibrium of section stripe

Bredt shear flow induces shear forces in the section walls qS/2 b0

Self-equilibrating system

v = MT/(2 · Fk) [kN/m] qS/2

mS/2 v

∆v v

∆v

mS/2

Differential shear flow (≈ resultant from the two faces) ∆v = (qS · b0 + 2 · mS)/(4 · Fk)

∆x = 1

Figure 14.5 Analysis of antisymmetric loading

485


forces and the differential shear flow v, which is obtained as the resultant of the Bredt shear flows on the two faces of the strip considered. Given that v¼

MT 2 Fk

then v MT 1 m ¼ D ¼ x x 2 Fk 2 Fk and, consequently, v ¼

qS b0 þ 2 mS 4 Fk

According to the section dimensions given in Figure 14.5, Fk ¼ d (b0 þ bu)/2. Obviously, the examined strip tends to deform under the forces qS/2, mS/2 and v. This deformation consists essentially of a change in length of its diagonals, and induces an additional state of stress for each section wall. So, it is considered appropriate — following Section 3.3 — to insert a hinged strut of infinite axial stiffness in the sense of the augmenting diagonal, i.e. a non-deformable one (Figure 14.6). With the insertion of this element throughout the length of the girder, the deformability of the section is eliminated, whereas the section itself develops a bending and axial state of stress, the former being induced by the acting moments mS/2. It is clear that this diagonal element develops a tensile axial force R (kN/m) distributed along the length of the

Cut out strip in equilibrium qS/2

qS/2

∆v

mS/2

mS/2 Differential shear flow

qS/2

qS/2

mS/2

mS/2 A=

∞

The fictitious member is tensioned with a force R

R R Opposite actions cause a longitudinal and transverse response

Additional response to the Bredt shearing one

Figure 14.6 Additional girder response due to antisymmetric loading

486

Box girders

R0 Rw

R

R Rw

Ru Applied self-equilibrating shear forces for restoring the incompatibility of stresses

Ro

Rw an

am

Rw

sp

Be

Ru

Figure 14.7 Taking up the diagonal loads through the longitudinal wall beams

girder. Applying now the opposite of the acting forces R on the corresponding longitudinal edges of the girder in absence of the diagonal strut, it is obvious that the superposition of the thus resulting state of stress with that of the blocked strip will give the final response (see Figure 14.6). g For the examination of this last ‘diagonal loading’ it is first considered that the forces R act on a girder having hinged connections at the section edges instead of monolithic ones (Figure 14.7). The forces R may be resolved equivalently at each edge into the two concurring walls; therefore, each of them can be considered as a longitudinal beam loaded by the corresponding component, developing bending moments M0 and corresponding normal stresses , according to the classical theory of bending. However, the resulting strains /E at the common edges of the walls are not equal — as they should be — and for this reason some distributed longitudinal forces have to be additionally introduced along the edges of each wall, in order to establish the strain compatibility at each edge (see Figure 14.7). It is clear that in this way the initially determined normal stresses will be changed. By following the above analysis, the determination of the longitudinal axial stresses is possible, e.g. for the left web of the ‘hinged’ box section in Figure 14.8 they can be determined through the classical bending formula on the basis of the aforementioned moment M0. In this case, the moment of inertia I used is slightly larger than the normal value Iw for the web by a factor kw, and the new ‘neutral axis’ lies at a distance y0 from the top fibre, which is somewhat less than its ‘normal value’, i.e. half of the web height (Schlaich and Scheef, 1982). The moment M0 results from the loading of the left web with the corresponding component Rw of the ‘diagonal force’ R (see Figure 14.7). It is found that Rw ¼ R bw/s. 487

Structural systems: behaviour and design b b0 Rw bw

R

Rw

R

s

t0 tw

Iw tu αo = (t0 · b3)/(tw · bw · b02) αu = (tu · bu)/(tw · bw)

bu σ0

β = (bu/b0)

y0 bw

σu

Inducing the bending moment M0 σ0 = (M0/I*) · y0 σu = (M0/I*) · (bw – y0)

Rw

‘Neutral axis’ an

am

sp

Effective moment of inertia I* = Iw · kw

Be

Figure 14.8 Bending response of the longitudinal section walls in a ‘hinged’ system

Thus, 0 ¼ u ¼

M0 y0 I

M0 ðbw y0 Þ I

where I ¼ Iw kw Furthermore (see Figure 14.8), 2 ½ð0 þ 2Þ ðu þ 2Þ 1 ð1 þ Þ ð3 þ 3 þ 0 þ u Þ 1 þ 2 þ u b y0 ¼ 3 þ 3 þ 0 þ u w kw ¼

Thus, the longitudinal web beam obeys the following typical differential equation (see Section 2.3.6): EI 488

d4 w ¼ Rw dx4

Box girders

w δ R

r

r

R rw

w

The frame opposes a resistance r to the increase in its diagonal

Deflections w induce an increase δ in the diagonal δ = w · D

r=δ·C

rw = r · (bw/s) = K · w

The web acts as if resting on an elastic base Rw

(I* = Iw · kw)

rw = K · w bw

w End diaphragm

Rw

End diaphragm E, I*

w

K

The deflection w refers only to the profile deformation The presence of a diaphragm means a corresponding support (w = 0)

Figure 14.9 Structural action of a section wall in the monolithic system

where Rw and I represent the initial distributed web loading and the equivalent moment of inertia, respectively, as explained above. It is noted that w represents the in-plane deflection of the web, and it is due merely to the assumed deformability of the section (see Figure 14.8). Only now may the monolithic connection of the section walls be taken into account. The bending deformation of the hinged section profile, under the action of diagonal forces R, has as a consequence an increase of its diagonal length by (Figure 14.9). However, this change cannot be realised without resistance, given that the transverse stiffness C of the closed monolithic section profile is automatically mobilised. This stiffness is expressed through the relation r ¼ C, as that diagonal pair of forces r required to produce the deformation ¼ 1. The stiffness C clearly depends on the geometric and elastic data of the section, and may be determined using typical computer software for plane frames. Thus, the tendency of the hinged section profile to be deformed by is counteracted by the resistance r of the monolithic closed frame, and in this way the web is subjected to, in addition to its ‘initial’ loading Rw, the loading of the component rw of the force r, obviously with opposite sense. Thus, rw ¼ r (bw/s) (see Figure 14.9). Taking into account that the bending deformation w of the web can be expressed through the diagonal deformation by the relation ¼ w D (see Figure 14.9), where D¼

2 bw ð1 þ Þ ð2 þ 2 þ 2 2 þ 0 þ u 2 Þ 3 þ 3 þ 0 þ u s 489


it is found that the component rw is also expressed through w by the relation rw ¼ w D C ðbw =sÞ ¼ w K where K ¼ D C ðbw =sÞ The differential equation for the web beam can now be adjusted to the restored rigid section profile, and can be written as d4 w ¼ Rw rw ¼ Rw w K dx4 and, finally, takes the form EI

d4 w þ K w ¼ Rw dx4 This equation is recognised as the typical equation of a beam on an elastic foundation with a modulus of subgrade reaction K (see Section 17.3.3.1). Indeed, the web is carried by the elastic support offered by the profile resistance when undergoing diagonal deformation (see Figure 14.9). g EI

It is, of course, clear that the presence of a single concentrated load on the web will lead to an applied concentrated moment and a corresponding shearing flow (Bredt) and, finally, to a concentrated load applied to the model of the elastically supported beam (Figure 14.10). If this eccentric load is the only one acting on the girder, then, in the above equation, the distributed load Rw on the right-hand side does not exist. In this case, it is clear that the diagonal actions ‘R’ on the girder consist of a single pair of equal and opposite forces. These forces, if a transverse diaphragm exists at this location, are taken up directly by it, without causing any ‘secondary’ stresses in the girder, as would happen in an ‘undeformable section’. If such a transverse diaphragm does not exist, then the diagonal forces are taken up by the deformable section of the girder itself, as has been previously examined in detail (see Figure 14.10) (Menn, 1990). Recapitulating, the deformation of the web under the component Rw of the diagonal loading R induces the transverse stiffness of the closed profile. This stiffness arises as a load rw acting on the web in the opposite sense, being linearly related to its deflection, i.e. as occurs in a beam resting on an elastic foundation. It is clear that the thus-resulting longitudinal bending response of the web will lead, through the developed axial stresses according to the above relations, to the edge stresses of the remaining section walls, as at each edge the strains " and, consequently, the stresses " E are common. g Regarding the boundary conditions that have to be fulfilled by the deflection w, it should be pointed out that w specifically concerns only that deformation which arises from the deformability of the section profile and not the overall deflection of the web beam, to which the torsional twist angle of the section also contributes as a rigid body. Thus, at each position, where the profile remains undeformed because of the presence of a transverse diaphragm, the condition w ¼ 0 has to be satisfied. This means that, at the 490

Box girders P/2 P/2

P

P/2 P/2 [MT]

Due to antisymmetric loading P/2 R

P/2

Equilibrium of girder strip Rw

R

Rw

E, I*

Figure 14.10 Behaviour of a box girder under an eccentric concentrated load

corresponding point of the elastically supported beam model, an unyielding support must be inserted (see Figure 14.10). Moreover, at a fixed position of the girder, the following condition is imposed: dw ¼0 dx Finally, the treatment of a girder support with no transverse diaphragm under the developed support reactions arising from torsional response is examined (Figure 14.11). It should be noted that, in this case, no support needs to be provided for the model of the elastically supported beam but, instead, the developed vertical reactions should be considered as antisymmetric loads. These exert a concentrated torsional moment and, according to the above, they lead to a concentrated force on the elastically supported web model. g 491

Structural systems: behaviour and design Development of torsional reaction

Interior support Q Q

Q

[MT]

Rw

Rw R R

Q Q

Equilibrium of girder strip

Figure 14.11 Behaviour at a support without a transverse diaphragm

Of course, the simultaneous transverse response of the section profile should not be overlooked. Based on the deformation w of the web, the diagonal loading r of the closed rigid profile, as shown in Figure 14.12, can be readily determined, thus allowing the evaluation of the transverse response of the section. According to the above, s r¼wK bw

14.2.3 Numerical example A simply supported orthogonal box girder with a 40.0 m span, shown in Figure 14.13, is examined. At the midspan and over the left web a concentrated load of 1000 kN is r

w δ r w Deflections w induce an increase δ in the diagonal δ = w · D

Imposing an increase δ in the frame diagonal requires action of the forces r r=δ·C Transverse response

Figure 14.12 Transverse response due to profile deformation

492

Box girders

500 kN

1 kN/m

0.25

2.50 m

Diagonal increase 4.65 · 10–5 m2/kN C = 1/4.65 · 10–5 = 21 524 kN/m2

500 kN

0.50 0.16

1 kN/m

6.20 m

Equilibrium of girder stripe

500 kN

100 · 2.50 kN

500 kN

620 kN

Differential flow 100 kN/m

250 kN 250 kN

100 · 6.20 kN

620 kN 250 kN

End diaphragm

E, I*

End diaphragm

w K = 12 025 kN/m2 –3

w = 1.26 · 10

m

514 kN m

[M]

Longitudinal web stresses σo = 1201 kN/m2 σu = 1344 kN/m2 σo (Compression)

σo (Tension)

40.5 kN/m

Transverse response σu (Tension)

σu (Compression)

r = w · D · C (kN/m)

Figure 14.13 Antisymmetric loading of a girder caused by an eccentric load

considered, which, according to Figure 14.10, induces torsion through the antisymmetric couple of 500 kN. On the basis of the expressions in Section 14.2.2, the following cross-section data are determined: ¼ 1, 0 ¼ 1.24, u ¼ 0.79, kw ¼ 1.002, y0 ¼ 1.18, D ¼ 1.495, s ¼ 6.69 m as well as Iw ¼ 0.504 m4 Thus obtaining I ¼ 0.504 1.002 ¼ 0.505 m4 493


Moreover, the stiffness C (kN/m2) of the unit length of the girder strip with respect to a unit increase (1.0 m) of the diagonal distance of its edges is determined as C ¼ 21 524.0 kN/m2. From the equilibrium of the girder strip at midspan, the web is acted on by a concentrated force of 250.0 kN. More specifically, the strip is in equilibrium under the antisymmetric couple of 500 kN as well as the peripheral differential shear flow v, which, according to Section 14.2.2, is v¼

500 2 6:20 ¼ 100 kN=m 4 2:50 6:20

The force Rw acting on the left web is Rw ¼ 500.0 v 2.50 ¼ 250.0 kN (see Figure 14.13) The web, as explained previously, acts as a beam on an elastic foundation with a modulus of subgrade reaction equal to K ¼ 1.495 21 524.0 2.50/6.69 ¼ 12 025 kN/m2. The corresponding bending moment diagram is obtained, as can be seen in Figure 14.13, exhibiting a maximum value at midspan equal to 514 kN m. Moreover, the corresponding deflection is 1.26 10 —3 m. Thus, the normal stresses at the top and bottom fibres are obtained, from the relations in the previous section, as 0 ¼ 514.0 1.18/0.505 ¼ 1201 kN/m2 (compression) u ¼ 514.0 (2.50 1.18)/0.505 ¼ 1344 kN/m2

(tension)

The above longitudinal stresses are solely due to the torsional response of the girder, by taking into account the deformability of its cross-section. The complement of the stress image for the whole section is shown in Figure 14.13. The development of the deflection w causes an increase in the diagonal length of the profile, which requires the action of the corresponding diagonal forces r. These are, according to the above, r ¼ 1.26 10 —3 1.495 21 524.0 ¼ 40.5 kN/m On the basis of these forces, the corresponding bending, axial and shearing responses of the closed girder section can be readily deduced (see Figure 14.13).

14.2.4 Summary According to the analysis above, the examination of the behaviour of rectilinear box girders under an eccentric loading consists of the following: . The eccentric loading is reduced through the ‘trivial fixed state’, represented by the transverse bending state [m0], into a non-symmetric nodal loading. . The non-symmetric nodal loading is split into a symmetric and an antisymmetric part. . The symmetric part causes a longitudinal bending and shearing response determined by the corresponding sectional force diagrams [M] and [V], as well as a transverse 494

Box girders

response concerning a transverse girder strip of unit length arising from a selfequilibrated loading consisting of the acting loads and the differential shear flow. The transverse response is represented by the corresponding bending moment, shear force and axial force diagrams [msym], [vsym] and [nsym], respectively, plotted along the section profile. . The antisymmetric part causes a torsional response along the girder, represented by a torsional diagram [MT], which enables the determination of the Bredt shear flow at each position. This shear flow causes, in turn, on the one hand, shear of the section walls, depicted by a the diagram [VT] for each wall, and, on the other hand, a longitudinal ‘diagonal loading’ {R}, arising from the resulting differential shear flow of a strip. To this last response, the transverse loading arising from fixation of the section edges has to be added, represented by the diagrams [mf], [vf] and [nf]. . The ‘diagonal loading’ {R} causes a longitudinal bending and shear response, represented by the sectional force diagrams [MR] and [VR], respectively, arising from the consideration of a characteristic section wall (usually the web), as a beam resting on an elastic foundation. This beam is supported only at places where the deformation of the section profile is prevented. . The above response is always accompanied by the transverse response of the section profile, determined on the basis of its mobilised ‘diagonal stiffness’, and is represented by the corresponding sectional diagrams [mR], [vR] and [nR] along the profile perimeter. Thus, for the response of each wall, the following diagrams must be taken into account: . . . . .

Longitudinal bending: [M] þ [MR]. Longitudinal shear: [V] þ [VT] þ [VR]. Transverse bending: [m0] þ [msym] þ [mf] þ [mR]. Transverse axial response: [nsym] þ [nf] þ [nR]. Transverse shear: [vsym] þ [vf] þ [vR].

The transverse shear is usually of secondary importance.

14.3

Curved girders

14.3.1 General The box girder is particularly suitable for bridges curved in plan, as these are subjected to a permanent torsional response caused even by non-eccentric loads, e.g. the self-weight of the girder. The skeletal model used for a curved girder is of the grid type (see Chapter 10), i.e. a plane structure loaded transversely to its plane, and, as such, it develops three sectional forces at each position, namely a vertical shear force V, a bending moment MB and a torsional moment MT (Figure 14.14). As will be shown immediately below, an equilibrium interrelation exists between bending and torsional moments, and thus any redistribution of the bending response 495

Structural systems: behaviour and design MT q V

MB

Figure 14.14 Curved beam model with its developed sectional forces

of the girder, for reasons associated with its plastic analysis (see Section 6.6.2), has, necessarily, to be accompanied by an adjustment of the corresponding torsional moments.

14.3.2 Determination of the bending and torsional response 14.3.2.1 Evaluation of the equilibrium equations The girder is assumed to have a constant radius of curvature equal to R. The loading consists of a vertical distributed load q passing through the shear centre of the crosssection, as well as of a distributed torsional moment mD (Figure 14.15). It should be recalled that, exactly as in the case for grids, any part of the beam must satisfy three

The change in the bending moment vector means an additional torsional load mT = MB/R

M T + dM T q

MB + dMB

dϕ MB + dMB V MB

MD ds V + dV

MB

M T · ds = MB · dϕ

PLAN VIEW Bending equilibrium Usually negligible

MT

d2MB /ds2 = –q + (1/R) · (dM T/ds) Analogy with determination of M: d2M/ds2 = –q

Torsional equilibrium dM T/ds = –mD – MB/R Analogy with determination of V: dV/ds = –q

Figure 14.15 Evaluation of equations of equilibrium

496

Box girders

conditions of equilibrium, i.e. those with respect to vertical forces, and those with respect to the projections of moment vectors on two arbitrary horizontal axes. The equilibrium equations of an elementary segment of length ds forming an angle d’ (1/R ¼ d’/ds), as shown in Figure 14.15, can be written in the form: dV ¼ q (equilibrium of vertical forces) ds dMT M ¼ mD B (equilibrium of moment vectors, on a tangential axis) ds R dMB M ¼ V þ T (equilibrium of moment vectors, perpendicular to the axis) ds R It should be understood here, according to Figure 14.15, that the vectorial change in a bending moment between two adjacent sections is equivalent to the action of a distributed torsional load. This can also be recognised from the presence of the term MB/R on the right-hand side of the second equation above. In other words, the presence of a bending moment at the two edges of a curved beam segment means that, apart from mD, an additional distributed torsional load MB/R is acting on the beam. In an analogous way, the vectorial change in the torsional moment MT is related to bending development, as can be seen from the third equation (Menn, 1990). The elimination of the shear force V from the above equations provides the two final equations, which clearly describe the interdependence of the bending and torsional moments of the girder: d2 MB 1 dMT ¼ q ðaÞ R ds ds2 dMT MB ðbÞ ¼ þ mD ds R If the arc span length L is much smaller than the radius of curvature R (i.e. L/R < 0.3) and, moreover, the distributed torsional load mD is much smaller than q R, then, by expressing MB as q L2/c — where c is of the order of about 10 — from equation (b) it may be concluded that 1 dMT 2 R ds ¼ L þ mD q R2 c R q which means that the term (1/R) (dMT/ds) on the right-hand side of equation (a) is much smaller than q. Thus, the first equation may be more simply written as d2 MB ¼ q ds2 i.e. resembling the equilibrium relation of a rectilinear beam between the bending moment and the load (see Section 2.2.3). This means that, under the above conditions, 497


the bending moments of the curved girder may be approximated by the bending moments MB of a straight beam of span L equal to the arch length of the girder. Once the moments MB are determined in this way, then the right-hand side of equation (b) takes a concrete value, and the equation itself refers directly to the equilibrium relation of a beam, dQ/ds ¼ p, between shear forces and the distributed load. This means that the torsional moments MT may result as the shear forces of a straight beam of length L carrying the load MB/R þ mD (see Figure 14.15). If the above conditions regarding the ratios L/R and mD/q R are not met and a better approximation is needed, then, on the basis of the above initial values of MB, one can return to equation (a) and evaluate the term 1 dMT 1 MB ¼q þ mD q R ds R R as a new loading of the beam, in order to obtain ‘improved’ values of the bending moments MB, and afterwards, through equation (b), to determine anew the torsional moments MT as shear forces for the loading MB þ mD R The repetition of the above procedure converges rapidly.

14.3.2.2 Summary . First, the ‘stretched’ girder is considered. . On the basis of vertical loads, the bending moment diagram MB is considered. . The ‘straight’ girder is loaded with the torsional load MB/R, according to the sense dictated by the composition of the vectors MB, as applied to the corresponding elementary curved girder segment considered in plan view. The distribution of this torsional loading is obviously identical to the corresponding bending diagram [MB]. Any existing load mD must be added. . The torsional loads MB/R are considered as distributed vertical downward loads, as long as the bending moments MB cause tension at the bottom fibres; otherwise, they are directed upwards. The possibly present mD may also be considered as a distributed vertical load with the same direction as MB/R, if its sense coincides with that of the torsional load MB/R; otherwise, it has the opposite direction. . The torsional moment diagram [MT] is identical to the diagram of shear forces for the above fictitious loading, but not with respect to its sign. g It is clear that the above equations concern only the equilibrium, and, as they do not take into account the compatibility of deformations in the case of redundant supports, they do not lead to a strictly accurate solution, unless the ratio ¼ EI/G IT is considered equal to zero. Nevertheless, their use for purposes of preliminary design is quite appropriate. It is noted that for a simply supported curved girder and for a ratio L/R less than 0.3, which means a central angle not greater than 208, the 498

Box girders

discrepancies from the ‘exact’ solution are less than 1%. Moreover, for a fixed-ended girder, and thus for the approximately equal internal spans of a continuous girder with L/R 0.3 and 5, the discrepancies are less than 1.5%. Furthermore, as mentioned initially, the above equations may have a direct utility when, for the purposes of plastic design, the bending moments obtained from a computer solution have to be altered, in which case the redistribution of torsional moments must necessarily satisfy the equilibrium criterion.

14.3.2.3 Application examples The above procedure is applied to the following examples. In the first example (Figure 14.16) a simply supported girder allowing the take up of torsional moments at its supports and a fixed-end girder are considered. Figure 14.16 shows that the fixed-end girder develops a clearly lower torsional response, while its torsional reactions are zero. This may be understood on the basis of the remark made in Section 2.3.6 that, as the beam is loaded by its bending moment diagram, its reactions PLAN VIEW L

PLAN VIEW

q (kN/m)

q (kN/m)

[M B] [M B]

q · L2/12

L

q · L2/8 q · L2/24 MB MB

MB

MB

MB

MB MB/R

MB

MB

MB/R

Distribution of torsional load Torsional reaction

MB/R

MB/R

MB/R

Analogy between torsional and transverse load MB/R

Reaction = torsional reaction

[M T]

Reaction = 0 = torsional reaction (see Section 2.3.6)

[M T]

(Shear force diagrams due to transverse load)

Figure 14.16 Determination of the bending and torsional response of a curved girder

499

Structural systems: behaviour and design PLAN VIEW 30.0 m

PLAN VIEW 40.0 m

10.0 m

g = 160 kN/m

g = 160 kN/m 8000 kN m

128 000 kN m [MB]

[MB] 14 000 kN m

MB/R

MB MB

MB

MB MB MB/R

MB/R

MB MB

MB Distribution of torsional load

(1 713 000/R) kN m

MB/R

(213 000/R) kN m Analogy between torsional and transverse load MB/R

MB/R

(213 000/R) kN

(1 713 000/R) kN

Torsional moment diagram identical to the shear force one

Figure 14.17 Evaluation of bending and torsional response with cantilever action

(specifically the torsional ones) must correspond to the developed end rotation angles, which are obviously zero. In both cases examined, attention must be focused on the correct determination of the loading sense of the torsional loading MB/R, as shown in Figure 14.16. Regarding the internal span of a continuous beam, the consideration of a fixed-end beam instead does not essentially alter the torsional response. In the second example (Figure 14.17), a beam is examined which can develop a torsional reaction only at its left support, being in the first case supported on an intermediate point, whereas in the second case it acts as a cantilever. Despite the fact that the intermediate support cannot develop a torsional reaction, it relieves the beam significantly with respect to the torsional response, since the torsional reaction in the case of a cantilever is about eight times greater. Again, due attention is paid to the direction of the torsional loading MB/R, according to Figure 14.17. g 500

Box girders

Suspension forces

Figure 14.18 Layout of a semicircular cable-stayed bridge

It is now appropriate to examine how a girder behaves when subjected to a distributed torsional moment mD along its whole length. First, it is obvious that a straight girder, where the ratio L/R is equal to zero, develops only torsional response and no bending at all. However, as the arc length L over a certain chord, and consequently the ratio L/R, increase, the torsional response is progressively reduced, being accompanied by everincreasing bending. This can be confirmed using equations (a) and (b) in Section 14.3.2, which are, of course, valid without making any assumption regarding the geometry of the curved girder. However, if a girder in the form of a ring, i.e. exhibiting the maximum possible value of the ratio L/R, namely equal to 2 p, is subjected to a uniformly distributed torsional self-equilibrating loading moment mD, it develops no torsion at all but only constant bending, as was ascertained in Section 12.4.1 (see Figure 12.28). Such a ring develops a constant twisting angle all along its perimeter. The above property of a ring makes possible the bridging of a span length 2 R through a girder in the form of semicircle of radius R, suspended along its internal edge in order to exclusively develop a bending response without any torsion. This goal, however, is achieved under the condition that the girder is being subjected to an appropriate bending moment at both its supported ends, and this is in turn directly accomplished if the girder is connected to two rectilinear beams of appropriate length, as shown in Figure 14.18 (Schlaich and Seidel, 1988). More specifically, the eccentrically suspended semicircular girder, under its selfweight g, is subjected to a constant torsional load mD ¼ g e, where e is the distance of the centroid to the suspension line (Figure 14.19). According to Section 12.4.1 (see Figure 12.28), the equilibrium condition of the curved girder leads to the exclusive development of the bending response, if a bending moment MB ¼ mD R can be provided at both its ends. This specific bending moment, which will be acting throughout the curved girder, can be provided in practice by the fixed-support moment gS L2s /8 of a fixed simply supported beam. This is rigidly connected at the 501

Structural systems: behaviour and design Centroid

PLAN VIEW

mD = g · e

g R Suspension points M = mD · R gs ·

Ls2/8

M = mD · R gs · Ls2/8

Ls gs

Ls

gs

Figure 14.19 Structural action of the system shown in Figure 14.18

two girder ends, having an appropriate self-weight gS and length Ls, both determined through the obvious requirement that mD R ¼ gS L2s /8 (see Figure 14.19).

14.3.3 The response of cross-section walls The torsional response referring to the skeletal model of the girder as examined in Section 14.3.2 implies, beyond the Bredt peripheral flow, an additional response for the box section walls. This response results from the way that gravity loads are introduced to the girder. First, it should be noted that the acting compressive forces D and tensile forces Z in the curved top and bottom flanges, respectively, cause distributed deviation forces q, according to the well-known relation (distributed deviation force) ¼ (axial force)/(radius of curvature) It is clear from Figure 14.20 that the transversely distributed equal and opposite forces qD and qZ, which the top and bottom slabs are obliged to carry, respectively, create a torsional load per unit of curved length, which is simply that resulting from the vectorial variation of bending moments, as examined in the previous section (Menn, 1990): qD ¼

D M ¼ B R hR

qZ ¼

Z M ¼ B R hR

502

Box girders D dϕ

dϕ qD · ds

ds

σ(MB)

D

Z

qZ · ds

h R = ds/dϕ

b

Z qD = D/R = MB/(h · R) qZ = Z/R

Girder strip of unit length

MB/(h · R) h

dv · h

Deviation forces

MB/(h · R) The deviation forces create torsional loading MB/(2 · h · R)

dv · b

Differential shear flow

1

MB/(2 · h · R)

MB /(2 · b · R)

1

The torsional loading is balanced by the differential shear flow

Charging of the interior web S

MB/(2 · b · R)

MB/(2 · h · R)

MB/(2 · h · R)

s

S = MB · s/(2 · b · h · R)

Additional response to the Bredt shear stress state

Figure 14.20 Introduction of a torsional response in a box girder walls

and, given that D ¼ Z, MB R Thus, it can be seen that, even without the action of an externally applied torsional moment mD, merely the existence of bending along a curved axis implies — as previously found — the imposition of a distributed torsional load MB/R, which causes, of course, a torsional response, according to equation (b) in Section 14.3.2: qD h ¼

dMT M ¼ B ds R Of course, how the loads qD and qZ are introduced to the girder has to be examined, and, in particular, how and with what consequences a transverse sectional strip is 503


self-equilibrated under the above loads. The cut-out strip of unit length, receiving the forces qD and qZ at its top and bottom sides, respectively, is in equilibrium with the developed Bredt shear flow at both its faces (see Figure 14.20). The resultant of these two flows is the so-called differential shear dv referred to the unit of transverse length (see Section 14.2.1). According to Bredt’s formula (see Section 2.5.2), dv dMT 1 MB ¼ ¼ ds ds 2 b h 2bhR Thus, the strip, being in equilibrium as a plane structure, under the loads qD, qZ and dv, gives rise to the self-equilibrated diagonal loading of the profile rffiffiffiffiffiffiffiffiffiffiffiffiffiffi MB 1 1 S¼ þ 2R b2 h2 as shown in Figure 14.20, which causes longitudinal bending of the walls as well as transverse bending of the section profile, as examined in detail for the rectilinear girder. Although the analogy is not quite accurate, for preliminary design needs and for limited curvatures, it can be considered that the left web wall takes the downward uniform load Sw ¼ MB/2 R b, acting, as in the case of a rectilinear beam, like a beam resting on an elastic foundation with a ‘subgrade modulus’ K, on the basis of the stiffness of the section profile against a diagonal length change. Of course, if the resulting response is high, transverse diaphragms should be provided. Then, the deviation forces of the top and bottom slabs that trigger the whole response will be transferred through corresponding horizontal bending to each diaphragm. This, in turn, will transfer a concentrated torque to the section, by excluding at the same time its deformation that would otherwise take place. The above diagonal loading of the girder cross-section, when transverse diaphragms are not present, makes it clear that — as shown in Figure 14.20 for the longitudinal response — the ‘internal’ web is burdened whereas the ‘external’ one is relieved. However, it is interesting to note that in the case of an open section, as shown in Figure 14.21, quite the opposite happens, as explained below. More specifically, the consideration of an elementary curved segment, as depicted in Figure 14.21, shows that the action of bending moments MB is equivalent — as previously explained — to the application of a distributed torsional load mD ¼ MB/R. It can be seen that this torsional load can be taken up by the ‘opposite bending’ of the two webs, through the resolution in the continuously distributed vertical loads q ¼ MB/R b which obviously give an additional load to the external web while decreasing the load on the internal one. If it is assumed that the existing vertical load q is equally distributed between the two webs, then their bending response will be obtained as MB,ext ¼ (1/2) MB(q) þ MB(q ) MB,int ¼ (1/2) MB(q) MB(q ) for the external and the internal web, respectively. 504

Box girders MB MB

mD = MB/R

Transfer of the torsional load to the webs q* = MB/(R · b)

D MB

dϕ qD · ds

q*

Zex

MB Change in the exterior web D Zin

ds Deviation force in slab b

Zex

qD = D/R

Deviation forces in web

qZ,in = Zin/R

qZ,ex = Zex/R

1

Figure 14.21 Uptake of a torsional load by the webs of an open section

The compressive forces in the slab due to the bending moments MB compel it to take up ‘outward’ deviation forces, whereas the existence of compressive and — especially — tensile stresses in the webs create transverse distributed deviation forces which equilibrate those of the slab, causing at the same time transverse bending of the open frame of the section, as shown in Figure 14.21 (Menn, 1990).

14.3.4 The influence of prestressing in curved girders As was made clear in Chapter 4 on beams, the layout of prestressing tendons aims to provide deviation forces that counteract the gravity loads for which the structure has to be designed. In the case of curved girders, gravity loads indirectly create torsional loads through the existing curvature, eventually being increased by their eccentric layout on the bridge deck. Thus, it is useful to examine how the presence of prestressing in a curved box girder influences its torsional response and, moreover, to consider the possibility of torsional relief through an appropriate tendon layout. Given that the tendons are arranged within the thickness of each web by using all of its available depth, it is first pointed out that a cable, inclined with respect to the girder axis, implies torsion (Figure 14.22). This is explained by the fact that the inclined compressive force on the section has a component in its own plane, which causes a torsional moment equal to the product of this component by its distance to the shear centre of the section. The fact that, in the usual prestressing layouts, the prestressing does not induce torsion arises because the different torsional contributions of the 505


P

P

P

Application of torsional load through deviation forces

P

P S

The component of the compressive force causes torsion

P If the compressive force acts perpendicularly to the section, no torsion is developed

Figure 14.22 Torsional response due to tendon inclination

cables cancel each other. However, an appropriate change in the cable inclinations may induce a torsional response that could relieve that arising from the loads. Of course, a cable lying in the top or bottom slab with an inclined position to the girder axis may also cause a torsional moment with respect to the shear centre through the component of its compressive action on the section (see Figure 14.22). Moreover, in a curved beam, where the prestressing cable runs parallel to the shear centre axis and is anchored at its two ends, no torsional loads are imposed, as the prestressing force does not offer anywhere a component on the corresponding section plane (see Figure 14.22). On the basis of the foregoing, it can be seen that in a curved statically determinate girder an arbitrary tendon layout with a given prestressing force causes exactly the same response (i.e. diagrams of sectional forces MB, Q and MT) as that developed in the corresponding rectilinear (i.e. ‘stretched out’) girder. Why this happens is now explained. A statically determinate structure containing a prestressed tendon appropriately anchored, if subjected only to the prestressing forces (anchor and deviation forces), 506

Box girders Prestressing of a statically determinate beam

P Curving the beam induces torsion under the applied loads

No torsion developed

P Curving the beam does not cause torsion

Figure 14.23 External loads and prestressing in a curved statically determinate beam

does not develop any reaction at the points of support. Hence, any bending or torsional moment, or even shear force of a section, results from the prestressing force acting on it. The point of application and the direction of this force always, as explained in Section 4.3.1.1, correspond to the appropriate trace and local tangent of the tendon. If the component of this action on the section plane does not pass from the shear centre, then a torsional moment is caused, as previously mentioned. Otherwise, no torsion is imposed. This situation remains unchanged if the initially considered straight girder becomes curved (Figure 14.23). Thus, while a rectilinear statically determinate girder, under, for example, its selfweight, will develop torsion if it becomes curved, a prestressing tendon not producing torsion in the rectilinear girder will continue to do so, even if the girder becomes curved. Accordingly, it may be concluded that in a curved statically determinate girder, if the prestressing deviation forces completely counterbalance the external loads (e.g. self-weight), bending will be eliminated but torsion will not. It should, however, be noted at this point that the ‘bending moment’ MP due to prestressing should not be perceived as creating (see Section 14.3.2) a torsional load MP/R, just because the bending moment MP is not ‘genuine’, i.e. it is not directly required for the equilibrium, as recalled in Figure 14.24. Of course, all the above is changed once the girder becomes statically indeterminate. A curved continuous girder, for example, with a prestressing tendon that in the assumed ‘stretched’ structure does not induce torsional moments develops a torsional response, as explained below, and depicted in Figure 14.25. The continuous girder considered in Figure 14.25 has simple supports at both its ends that are capable of developing torsional reactions, whereas it is simply supported in the middle. The primary — statically determinate — structure obtained through removal of the intermediate support under the existing deviation and anchor forces does not develop any torsion, as already explained. However, the redundant vertical force that must be applied in order to eliminate the deflection occurring at the midpoint of the 507

Structural systems: behaviour and design Equilibrium requires the action of a moment g

No moment is required for equilibrium

P u P

Application of loads

Application of prestressing

Figure 14.24 Prestressing in a statically determinate beam implies an internal moment

primary structure does induce a torsional response. This externally imposed redundant force, which differs very little from the corresponding reaction of the rectilinear continuous beam, provided that the values L/R as well as are small enough (see Section 14.3.2.1), will induce a triangular bending moment diagram in the primary

PLAN VIEW In the ‘stretched’ girder no torsion is developed

R Statically determinate structure In the curved girder no torsion is developed

Deformation due to deviation forces An applied force R at the midspan is required

R

R

PLAN VIEW

[MSP]

The force R on the curved girder causes torsion

Bending diagram due to R Statically indeterminate prestressing moments

MSP/R

MSP/R

Torsional loading due to R

Substitute transverse loading

‘Shear force’ diagram [MT] Torsional moment diagram for a prestressed curved girder due exclusively to redundant prestressing moments [MSP]

Figure 14.25 Influence of prestressing on a curved redundant beam

508

Box girders

structure. This represents the statically indeterminate prestressing moments [MSP], as explained in Section 5.4.1. The bending diagram [MSP] applied to the curved statically determinate girder will automatically produce the torsional loading [MSP/R]. The implied torsional response will then be obtained as the shear force diagram of the corresponding rectilinear beam, due to the transverse triangular loading [MSP/R] (see Figure 14.25). It is clear, then, that a prestressing tendon in a curved continuous girder causes a torsional response solely due to the statically indeterminate prestressing moments MSP of the corresponding rectilinear beam. Thus, according to the above and referring to Figure 14.26, the torsional moment diagram [MT] of a curved statically indeterminate prestressed girder can be obtained

P

e f

u P Deviation forces L u

P

[MB] P·f MSP = MB – P · e

R Deformation due to deformation forces Required force R = P · (f – e )/L

[MSP]

R

P · (f – e ) Bending diagram due to R (Statically indeterminate prestressing moments) MSP/R

Torsional loading due to R MSP/R R The force R induces torsion

Substitute transverse loading

‘Shear force’ diagram [M T]

Torsional moment diagram for a prestressed curved girder due exclusively to [MSP]

Figure 14.26 Assessment of the influence of prestressing on a statically indeterminate curved girder

509

Structural systems: behaviour and design mD,P = u · hs u hs P

P

u

Pv S

b

Constant torsional moment: +Pv · b/2

mD,P = u · b/2 Constant torsional load due to the corrresponding deviation forces

Figure 14.27 Creating torsional loading through separate prestressing cables

on the basis of the bending moment diagram [MP] due to deviation and anchor forces (see Section 5.4.1), as follows. After the statically indeterminate moments (MSP ¼ MP P e) have been determined, where P and e are the prestressing force and the tendon eccentricity at each position, respectively, the torsional load [MSP/R] is applied as a distributed vertical load on the statically determinate rectilinear girder. The resulting shear force diagram represents, then, the torsional moment diagram [MT] of the statically indeterminate curved girder.

14.3.5 Reducing the torsional response through prestressing The distributed torsional action mD, or, more generally, the development of a torsional moment diagram [MT], in a box girder, may be reduced to a lesser degree through prestressing (Menn, 1990). As has been stated previously, the presence of an inclination at some point on a prestressing cable in a girder web principally implies a torsional action. If the cable is straight, the torsional response developed is constant. If the cable is parabolic, the constant deviation forces u induce a constant distributed torsional load: mD,P ¼ u b/2 (Figure 14.27). The arrangement of a straight cable in the top or bottom slab of the girder, in a different direction to the girder axis, will also cause a constant torsional moment, while a parabolic cable, through its deviation forces, will impose a constant torsional load mD,P ¼ u hs, where hs is the distance from the shear centre (see Figure 14.27). It is clear, however, that arranging a parabolic prestressing cable in one web only implies the imposition not only of a distributed torsional load but also of a distributed vertical load, which may be undesirable if only a torsional load mD,P is to be imposed. For this reason, another parabolic cable with opposite curvature should be arranged in the other web (Figure 14.28). Analogously, in a horizontal cable layout, the girder slabs should be provided with cables of opposite curvature. The prestressing tendons can be specified separately in the first stage of design, to counteract the bending and the torsional response of the girder. In a second stage, 510

Box girders P P u h P

P u P

u

u

P

P P

mD,P = u · b

b

mD,P = u · h

Figure 14.28 Basic layouts of prestressing cables for inducing torsional action

the synthesis of the two separate tracings into a single, final, one can be attempted, by retaining, if possible, the number of cables (i.e. the total prestressing force) required for bending. In this way, a differentiated tendon curvature can be obtained for both webs, which must ensure the required upward deviation forces, on the one hand, and offer the required torsional counter-load, on the other. The above is illustrated by the example of the simply supported curved girder in Figure 14.29. It becomes clear that the arrangement of a convex cable in the ‘outer’ 36.0 m PLAN VIEW

2

MB/R

2460 kN m

1 g = 168 kN/m

544 kN m/m

R = 50 m [M T]

Prestressing the tendon does not induce any torsion Differentiating the deviation forces Aimed for torsional loading uin 1

(uex – uin)/2

uex 2

Counteracting of torsion is achieved through a differentiated tendon profile [M T] Tendon profile in web 2

Tendon profile in web 1

Significant decrease in torsion is accompanied by an additional transverse response

Figure 14.29 Relieving influence of the tendon layout on the torsional response

511

Structural systems: behaviour and design Differential shear flow

Introduction of torsion

Transverse response

SELF-WEIGHT 1

2

PRESTRESSING

(uex – uin)/2 1

2

(Increase in response)

Figure 14.30 Influence of the prestressing layout on the transverse response of a girder

26.0 m

26.0 m

2

PLAN VIEW 1

R = 50.0 m g = 155 kN/m 14 450 kN m

[M B] 7173 kN m MB/R 894 kN m

894 kN m 928 kN m

[M T] Shearing of walls (Bredt) [M T]

894 kN m Introduction of torsion

Differential shear flow (Bredt)

Transverse response

Span region 2

1

1

2

1

2

1 [M T]

Support region 2

1 1

Figure 14.31 Longitudinal and transverse response due to the self-weight of a girder

512

Box girders

web and a concave cable in the ‘inner’ web will produce a torsional load counteracting the torsional effect of the self-weight. Consequently, the final tendon profiles present a stronger curvature for the outer web than for the inner one, both of them being, of course, convex. If in the above case the deviation forces of the outer and inner web are uex and uin, respectively (uex > uin), it can be seen that, while the symmetric total deviation force (uex þ uin)/2 counteracts the gravity loads, the resulting torsional load is due to the antisymmetric loading of each web with (uex uin)/2, thus being equal to mD,P ¼ (uex uin) b/2. However, this torsional load will cause a corresponding peripheral shear flow, leading to an additional longitudinal and transverse response of the girder, according to the examination in Section 14.3.3 (Figure 14.30).

14.3.6 Summary examples It is appropriate at this point to recapitulate the complete torsional response of a singlespan prestressed curved girder, according to the following four states of stress:

2

1

PLAN VIEW

Aimed for torsional loading achieved through a differentiated tendon profile Web 2

Web 2

Web 1

Web 1 Deviation loads must conform in total to the initial tendon profile [M T]

Shearing of walls (Bredt) [M T]

Transverse response 2 [M T]

uin

uex 1

1 1


2


Transverse response

(uex – uin)/2 2

1 1

1

2 (Surcharging of section)

Figure 14.32 Influence of the differentiated tendon profile on the torsional response

513


(a) The shear response of the section walls according to Bredt for gravity loads, due to the existing curvature of the girder. (b) The additional response due to the deformation of the section profile, on the basis of the differential shear flow in (a) and the deviation forces on the top and bottom slabs. (c) The shear response of section walls according to Bredt, due to the torsional loads initiated by the eventually differentiated tendon profile in the webs. (d) The additional response due to the deformation of the section profile, on the basis of the differentiated shear flow in (c) and the antisymmetric loading of the two webs by the deviation forces of the tendons. The application of the above in the case of the single-span girder examined, as in Figures 14.29 and 14.30, shows that the transverse response of the section due to the differentiated tendon profile in the webs increases that arising from the self-weight, resulting in greater longitudinal and transverse bending. It can be seen that in the case of a statically indeterminate girder the following states of stress must, furthermore, be taken into account:

2

1

PLAN VIEW The deviation forces must conform to those resulting from the differentiated tendon profile MSP/R

[M SP] Statically indeterminate prestressing moments Shearing of walls (Bredt) [M T]

[M T] Torsional moment diagram due to prestress

2

1 1


[M T]

1

2


Transverse response

2

1 1

Figure 14.33 Influence of prestressing on the torsional response due to the static redundancy

514

Box girders 2

1 PLAN VIEW

Aimed for torsional loading Tendon layout in top slab

Tendon layout in bottom slab


Differential shear flow

Transverse response

SELF-WEIGHT 1

2

1

2

PRESTRESSING

(Relieving response)

Figure 14.34 Consequences of the arrangement of additional tendons in the top and bottom slabs

PLAN VIEW

Torsional loading due to self-weight

Aimed for torsional relief Tendon profile of top slab

u Tendon profile of bottom slab

u

Figure 14.35 Layout of additional tendons in the top and bottom slabs of a continuous girder

515


(e) The shearing response of the section walls (Bredt), due to the torsional response on the basis of the statically indeterminate prestressing moments MSP, according to Section 14.3.4. (f ) The additional response due to the deformation of the section profile on the basis of the differential shear flow and the deviation forces of the top and bottom slabs occurring in (e), according to Section 14.3.3, in the same way as in (b). Figures 14.31—14.33 illustrate the treatment of a prestressed continuous curved girder, in order to reduce the torsional response due to the action of a permanent load as the girder self-weight, following the above sequence: . steps (a) and (b) are illustrated in Figure 14.31 . steps (c) and (d) are illustrated in Figure 14.32 . steps (e) and (f ) are illustrated in Figure 14.33. g As previously explained, the torsional response due to permanent loads may clearly be reduced through the arrangement of additional tendons in the top and bottom slabs, which, by means of their deviation forces u, offer a relieving torsional moment mD,P ¼ u h, as shown in Figure 14.34. The introduction of these horizontal deviation forces, together with the differential shear flow they cause, lead, according to the foregoing, to a ‘diagonal’ state of stress, which drastically reduces that due to the permanent load (see Figure 14.34). It is clear that the above layout is much more effective for torsional reduction in a continuous girder than the previously considered differentiation of the profiles of the tendons within the webs (Figure 14.35). However, this can be achieved only through the extra cost of additional tendons. The task of restricting the torsional response through prestressing must always be undertaken for each case with regard to its technological—economic consequences, and taking account of the cost of the difficulties that will be encountered in the realisation of the prestressing layout.

References Menn C. (1990) Prestressed Concrete Bridges. Basel: Birkhaüser Verlag. Schlaich J., Scheef H. (1982) Concrete Box-girder Bridges. Zurich: IABSE. Schlaich J., Seidel J. (1988) Die Fussga¨ngerbru¨cke in Kelheim. Bauingenieur 63.

516

15 Lateral response of multi-storey systems 15.1

Introduction

In this chapter the behaviour of a three-dimensional multi-storey building system under lateral forces is examined. These forces may result either through an imposed seismic ground motion, or as an incoming wind action, or even through a temperature variation which, as will be explained later, activates the lateral stiffness of the structural system. The reason for making the consideration of temperature change an essential design issue is not only because of the obvious environmental temperature influence on the structure but also because of the fact that the shrinkage that the concrete slabs at each storey of the load-bearing system are subjected to, after the concrete has hardened, may be expressed through a temperature fall, thus causing a response of the whole system.

15.2

Formation of the system

Multi-storey systems are considered here as horizontal diaphragms (slabs) with very high rigidity within their own plane, and arranged vertically at a regular distance of a storey height and connected with each other as well as with the foundation ground through vertical elements of rigidity, according to a particular layout plan. These vertical elements consist either of frames formed by the columns and the building girders, or of single columns or shear walls, or possibly of thin-walled elements having an open or closed section (Figure 15.1). It should be noted that the diaphragm concept used below is not necessarily restricted to a solid concrete slab but can apply to any structural formation possessing a high stiffness in its own plane. The same naturally applies for all the vertical elements, which do not necessarily have to be made of concrete but can be formed of steel elements. Each of the horizontal diaphragms under the action of horizontal forces moves as a rigid body within its own plane. This movement consists of a horizontal displacement as well as a rotation, and is obviously imposed unchanged on all vertical elements, causing a corresponding response (Figure 15.2). In the present examination, which, as previously pointed out, facilitates, in principle, structural design rather than ‘accurate’ analysis, it may be assumed that all vertical elements, as the supporting skeleton of the diaphragms, constitute a set of elements of purely plane stiffness (Stavridis, 1986). Structural systems: behaviour and design


Structural systems: behaviour and design Shear wall

Vertical frame

Vertical frame

Rigid diaphragm

Vertical frame

Shear wall

Rigid diaphragm

Thin-walled element Vertical frame

Figure 15.1 Layout of horizontal diaphragms and vertical stiffness elements

By the term element of purely plane stiffness or simply plane element, it is meant that the element is stressed only by that component of the diaphragm displacement which corresponds to its plane. Thus, for example, while the same displacement and rotation vector of the diaphragm is applied to each girder of all frames in Figure 15.3, each frame is

P3 3

Diaphragm displacement

P2 2

P1 1

Figure 15.2 Horizontal movement of the diaphragms

518

Lateral response of multi-storey systems Effective component for the plane element Displacement vector of the diaphragm Rigid diaphragm

Element of plane stiffness

Rigid diaphragm

Effective component for the plane element Displacement vector of the diaphragm

Element of plane stiffness

Figure 15.3 Activation of stiffness of the plane elements through the diaphragm

assumed to be stressed only by that component of the displacement vector corresponding to the specific frame plane. Although the bending stiffness of a frame with respect both to its own plane and transversely to it can be described by two separate plane vertical elements, the clear preponderance of the stiffness in its own plane allows the omission of the corresponding transverse plane element. However, in a single column, two plane elements may be considered representative of its stiffness with respect to its two principal stiffness senses (Figure 15.4). Moreover, it may be understood that a vertical thin-walled element of either an open or closed section may be represented by the plane elements constituting the whole section. These plane elements are also able to offer the torsional stiffness of the element, as any applied torsional load may be taken up by the plane state of stress of the constituent plane elements themselves (Figure 15.4).

15.3

Lateral response

15.3.1 Treatment of load-bearing action The basic problem may be stated in the following way. It is assumed that the diaphragm of the nth level is acted upon on its centroid by a horizontal force P (Figure 15.5). The diaphragm, with reference to the orthogonal 519

Structural systems: behaviour and design nt 1

eme

el lane

P

ss

ffne e sti

Pl

ctiv

Effe

an

e

el

em

en

t2

The transverse stiffness is ignored Four plane elements

Three plane elements

The two parallel walls can take up the moment

The moment is taken up by the stiffness of the four plane elements

Figure 15.4 Expressing the stiffness of vertical members through the elements of the plane stress

4

3 y O 2

x ∆y

1

Figure 15.5 Horizontal loading of a single diaphragm

520

P

∆ω ∆x

Lateral response of multi-storey systems

Stiffness S1

Stiffness S2

causes δ=1m

causes δ=1m

Stiffness S1 causes δ=1m

Stiffness S2 causes δ=1m

The lateral stiffness S is practically unaffected by the upper-lying levels

Figure 15.6 Lateral stiffness of a plane element

axes Oxy having their origin O at the centroid of the diaphragm (i.e. its centre of mass), undergoes, through the action of the force P, horizontal movement as a rigid body, represented by the components of displacement x and y as well as by the rotation !, taken as positive if anticlockwise. This displacement is obviously resisted by the lateral stiffness of plane elements, as explained above. The lateral stiffness of each plane element is represented by the required horizontal force S acting on a specific level n, in order to induce a corresponding unit displacement (1 m) at that level. This force remains essentially the same, even if all the upper storey levels are omitted, as shown in Figure 15.6. This fact, apart from its simplifying effect, has also a further convenient consequence, as will become clear later. The resistance of the plane elements to the diaphragm movement implies for each element the action of a certain force at the corresponding level, resulting from the imposed displacements on them, while the same forces are returned in an opposite sense to the diaphragm itself. This must be in equilibrium under the above forces as well as under the directly acting force P (Figure 15.7). Given that these ‘resistance’ forces may be expressed — as will be shown later — in terms of the displacement components x, y and !, these components can be determined through the three equations of equilibrium of the diaphragm. The above components thus allow the evaluation of the forces acting on each plane element at the specific level. However, it must be ensured that the diaphragm in question can safely take up the self-equilibrating coplanar forces that it is subjected to. The simultaneous loading of some or all of the diaphragms subjected to corresponding forces P may be treated for each diaphragm separately, by neglecting all the above-lying levels of the plane elements involved, as has been previously indicated. Of course, the final response of each vertical element will be deduced from the total action of the corresponding forces determined at each level (Figure 15.8). These forces are used for the design of the plane elements and the assessment of their displacements at the various levels, as well as for the response of the corresponding diaphragm. 521


Acting forces on the plane elements (depending on ∆x, ∆y, ∆ω)

Equal and opposite forces on the diaphragm

P

∆y

∆x

∆ω

EQUILIBRIUM OF DIAPHRAGM The diaphragm is stressed

The acting forces remain the same even if the upper-lying levels are ignored

Figure 15.7 Equilibrium of the loaded horizontal diaphragm

4

P4

P3 3

P2 2

P1 1

The forces at each level are due to the individual action of ‘P ’ on the corresponding diaphragm

Figure 15.8 Treatment of the loading of all diaphragms

522


Di y

y

Element i

Element i

∆ω

xi Characteristic point (arbitrary) ei

yi

βi

∆y ∆x (negative)

O

x

O

x

Figure 15.9 Geometric and kinematic characteristics of the plane element

15.3.2 Response of a plane element Each plane element i is characterised by the angle i, described in an anticlockwise sense by the axis Ox with an arbitrarily determined positive direction on the trace of the element in the ground plan. On this trace a characteristic point with coordinates xi and yi is selected (Figure 15.9). By considering a specific level i of an element where the displacement components x, y and ! of the corresponding diagram are imposed, the corresponding displacement Di for the element may be determined, by assuming that the diaphragm, as already pointed out, remains undeformable during its movement as a rigid body: Di ¼ x cos i þ y sin i þ ! ei where ei is the (signed) distance of the trace of the element from the origin of the coordinate axes O, being equal to ei ¼ xi sin i yi cos i In the level under consideration, the element i may be structurally represented by the corresponding force S that must be applied in order to produce a unit displacement (1 m), known also as the lateral stiffness. This magnitude can be determined by calculating the displacement (m) due to a unit horizontal force (1 kN) applied to that level (Figure 15.10). Then, S ¼ 1/

(kN/m)

As has already been pointed out in Section 15.3.1, neglect of the overlying levels does not affect the above displacement in practice, or, in consequence, the corresponding stiffness (see Figure 15.10). Thus, in order for the plane element i to contribute to the imposed diaphragm displacement at the level under consideration, a horizontal force F ¼ Si Di must be 523


1 kN

δ: m

1 kN

Stiffness S = 1/δ

δ: m

F

Required force

D Induced displacement

F=S·D Practically both displacements are equal so that the corresponding level stiffness remains unaffected

Figure 15.10 Determination and evaluation of the translational stiffness of a plane element

imposed on it, i.e. F ¼ Si (x cos i þ y sin i þ ! ei)

15.3.3 General layout If the diaphragm of a specific level is subjected to a horizontal force P applied to its centroid and represented by the components Px and Py, then it may be found from its equilibrium, as previously considered, that the components x, y and ! of its displacement must satisfy the following three-equation system: K1 x þ K2 y þ K3 ! ¼ Px K2 x þ K4 y þ K5 ! ¼ Py K3 x þ K5 y þ K6 ! ¼ 0 where P Si sin i cos i, K3 ¼ Si ei cos i P P P K4 ¼ Si sin2 i, K5 ¼ Si ei sin i, K6 ¼ Si e2i

K1 ¼

P

Si cos2 i, K2 ¼

P

Once the components x, y and ! are known, the horizontal forces acting at the corresponding levels for all plane elements can be readily determined, as described in Section 15.3.2 (Stavridis, 1986). The stiffness Si at any level may be thought of as a horizontal spring in the direction of the corresponding element, which may be connected to the diaphragm at the position of the corresponding characteristic point of the element (Figure 15.11). In this way, the diaphragm acted upon by the force P maintains its equilibrium because of the developed spring forces acting on it, while the same forces are also acting on the corresponding plane elements. 524


Characteristic point of plane element

S3 y S1

O Centroid S2

x

S4

P

∆y

∆ω ∆x

Plane elements (spring) may ‘slip’ on their trace

Figure 15.11 Visualisation of the support of the loaded diaphragm through the corresponding springs

Alongside the above considerations, it is also useful to clarify the structural significance of the coefficients K1, K2, K3, K4, K5 and K6 in the above system of equations. This group of coefficients characterises the resistance offered by the system of plane elements against the horizontal movement of the diaphragm. More specifically, the terms K1 x þ K2 y þ K3 !, K2 x þ K4 y þ K5 ! and K3 x þ K5 y þ K6 ! represent, respectively, the forces Px and Py, and the torsional moment M! that must be applied at the origin O in order to cause a diaphragm displacement having the components x, y and !. In other words, the above terms represent the resistance offered by the diaphragm at point O if this is ‘obliged’ to be displaced by x, y and !. It is, of course, understood that specifying a ‘point of application’ for the torsional moment M! does not make much sense, as the moment may be applied anywhere in the diaphragm. However, the null value that is assigned to this moment in the above equation system is due to the fact that the displacement components x, y and ! are realised simply through the application of a single force P at point O, i.e. without imposing any moment M!. If for any reason the diaphragm is acted upon by an external moment M!, then this value — positive for an anticlockwise direction — must be used instead of zero. From the above equations it can be deduced that the plane elements may be ‘shifted’ anywhere along their own trace, as only their corresponding angle i and their distance ei from the origin of the coordinate axes matter. This may also be understood through the concept of spring supports introduced above, which, owing to the absolute rigidity of the diaphragm, may be connected at any point along their line of action.

15.3.4 Orthogonal layout The vertical elements in building structures are usually arranged in mutually orthogonal directions. In this case, the axes Ox and Oy are placed parallel to these two directions (Figure 15.12). Then, all the elements in the x sense have i ¼ 0 and ei ¼ yi, whereas all the elements in the y sense have i ¼ 908 and ei ¼ xi. Thus (see Section 15.3.3), P P K1 ¼ Sx, K2 ¼ 0, K3 ¼ (Sx yi) P P P P K4 ¼ Sy, K5 ¼ (Sy xi), K6 ¼ (Sx y2i ) þ (Sy x2i ) 525

Structural systems: behaviour and design PLAN VIEW S5

S1

y P O S3

S2

Centroid

x S4

S6 S7

Figure 15.12 Orthogonal layout of the plane elements

while x, y and ! are determined from the following expressions: K3 K5 þ P x K3 K1 K4 x ¼ ; K1 K1 K32 K52 þ K6 K1 K4

K3 K5 þ P y K5 K1 K4 y ¼ K4 K4 K32 K52 þ K6 K1 K4

K3 K Px þ 5 Py K K4 ! ¼ 1 2 2 K3 K5 þ K6 K1 K4 For the forces acting on the elements in the x direction at the level under consideration (see Section 15.3.2), Fx ¼ Sx (x ! yi) whereas for the elements in the y direction the forces acting are Fy ¼ Sy (y þ ! xi)

P P The magnitudes of K3 and K5, i.e. (Sx yi) and (Sy xi), respectively, play an important role. If both are zero, the diaphragm does not rotate at all (! ¼ 0), and x and y are simply P P x ¼ Px/ Sx, y ¼ Py/ Sy This result may also be interpreted and evaluated according to the spring model of the rigid diaphragm (Figure 15.13). A force Px acting on the diaphragm at point O in the x direction, such that it causes no rotation but only a displacement , must P lie on the line of action Pof the resultant of (Sx )P y ¼ 0, i.e. (SxP y) ¼ 0. the spring forces (Sx ), which means that Moreover, for equilibrium of the diaphragm, Px ¼ (Sx ) ¼ ( Sx), as has been exactly obtained before. For this examination of the behaviour in the x sense, a rigid beam may substitute for the diaphragm, as shown in Figure 15.13. Analogous results are also obtained for the y sense. 526

Lateral response of multi-storey systems PLAN VIEW S5 S5 · ∆

S5 S1 P

Centre of stiffness O Resultant

S3 y

S6 · ∆

S4

S6

S2

S6

S7 S7

x The horizontal force P acts on the diaphragm’s centroid Rigid beam

S1

S2

S3

S3 · ∆ (S1 + S2) · ∆

S4 Resultant

∆

S7 · ∆

∆

If the force P does not pass through the centre of stiffness a diaphragm rotation is developed

S4 · ∆

Figure 15.13 Simulating the diaphragm behaviour through a spring-supported rigid beam

In this way, for an arbitrary displacement imposed in the x direction, the specific line of action of the resultant of the spring forces (Sx ) may be readily determined, while for another imposed arbitrary displacement in the y direction, the spring forces (Sy ) allow the determination of the line of action of their own resultant (see Figure 15.13). It is obvious that the position of the above resultants is independent of the value , and hence for any force passing through the intersection of the above action lines, the diaphragm will not rotate but only shift in the direction of the acting force. This point, which is known as the centre of stiffness or the centre of resistance, may be paralleled to the shear centre of thin-walled sections (see Section P13.3). and K , i.e. of (Sx yi) and It is now clear that a zero value for the magnitudes K 3 5 P (Sy xi), means that the point O — representing in practice the centre of gravity of the vertical element compressive forces — coincides with the centre of stiffness. It should be made clear that all the foregoing equations are valid, independently of whether the point of application of the load P on the diaphragm — which may be its centroid — coincides or not with the centre of stiffness. However, it is clear that a possible application of a load P at a point different from the centre of stiffness means the action of a certain moment on the diaphragm, which must be balanced by the spring forces, i.e. the resistance forces offered by the plane elements. Moreover, the spring (plane element) layout must ensure the uptake of an arbitrary load P acting in any direction. The above remarks are illustrated in Figure 15.14. 527

Structural systems: behaviour and design Inapplicable layout S1

S4

Inapplicable layout S1

P Centre of gravity S3 S2

(a)

Applicable layout S1

P

P

Centre of gravity S2

Centre of gravity S3 S3

Centre of stiffness (b)

S2 Centre of stiffness (c)

Figure 15.14 Criteria for the right or wrong layout of the plane stiffness elements

In case (a) the x component of the load P cannot be balanced by the spring forces, and consequently the layout is unsafe. In case (b), although springs, i.e. plane elements, are available to the diaphragm for the uptake of any component of the load P, these are unable to provide a couple of forces, in order to balance the moment of the load with respect to the centre of stiffness, and therefore the system may easily collapse. However, in case (c) the developed spring forces can balance any component of the acting load, as well as the moment of the load P with respect to the centre of stiffness through the offered forces S1 and S3. It is now clear that, while the point of application of the load P is generally determined by the diaphragm itself, the centre of stiffness is dependent on the selected layout of the springs, i.e. the plane stiffness elements. It must be pointed out that selecting a layout of vertical elements —Pframes, shearPwalls, etc. — that exhibits small or zero values of the magnitudes (Sx y) and (Sy x) has a definite advantage over a layout where the point of application of load P, i.e. the centre of gravity of the diaphragm, is some distance from the centre of stiffness, thus causing a rotation ! of the diaphragm. This rotation is obviously the same as that occurring in each of the rigid beam models in Figure 15.13. As can be seen from the above expressions, the rotation ! causes a non-uniform distribution in the response of the plane elements, depending on their position in the ground plan (for a generalised treatment of the problem see Section 16.3.1.4).

15.4

Temperature effect

15.4.1 Treatment of load-bearing action Any temperature variation in a diaphragm always causes a certain stress state on the vertical elements. The reason for this is that a dilatation or a shrinkage of a diaphragm always imposes displacements at the corresponding level of the plane elements, which are consequently stressed (Figure 15.15). More specifically, assuming that a diaphragm is subjected to a temperature increase of T by retaining its centre of mass O absolutely fixed, it can be seen that some radial displacement will be imposed on all the vertical plane elements, which will lead to a certain shift along their trace (Figure 15.16). 528

Lateral response of multi-storey systems PLAN VIEW

δ? Temperature rise of the diaphragm δ?

Figure 15.15 Displacements of the plane elements due to a temperature variation in the diaphragm

Due to this imposed displacement, the plane elements will offer resistance at their corresponding level that will be acted on by the diaphragm, being obviously equal and opposite to that acting on the plane element itself. It is clear that the diaphragm under the above forces will generally not be in equilibrium as a free body and, therefore, at the point of fixation O, a specific force and a moment have to be applied. The action of these forces ensures the fixation of point O under the temperature variation of the diaphragm (Figure 15.16). However, given that, in reality, no external actions are applied at point O, equal and opposite actions to those above must additionally be applied at this point, in order to restore the real situation. Of course, the last actions on the ‘free’ system will cause displacement of the diaphragm with the components x, y and !, which in turn will cause an additional stress state on the plane elements, through the corresponding horizontal forces acting on their corresponding levels (Figure 15.17). The final response of the vertical elements will result as a superposition of the two above stress states (Stavridis, 1994), namely:

FIRST STAGE Plan view A

Fixation actions (–PωT)

B

O

O

D Fixed centroid

(–PyT)

C (Rise of temperature)

A

∆T

δ (–PxT)

B

A

δ

δ

B

C δ

D δ

Figure 15.16 Temperature variation under fixation of the centroid of the diaphragm

529

Structural systems: behaviour and design SECOND STAGE Imposition of effective actions on the free diaphragm Developed: ∆x, ∆y, ∆ω Plan view

Equal and opposite to the fixation actions

(PωT) (PxT) O (PyT)

The induced element actions at the corresponding level are superposed on those of the first stage

Figure 15.17 Imposition of the opposite fixation actions to the free system

. First stage. The state of stress resulting from the free dilatation of the diaphragm with its centroid O fixed through externally applied actions at this point (see Figure 15.16). This stress state is due to the imposed displacements on the specific level of the plane elements because of the radial dilatation of the diaphragm on the one hand and the temperature variation itself imposed on the corresponding girder of the plane frame on the other hand. . Second stage. The state of stress resulting in the — now — free system, through the imposition at the point O of the equal and opposite actions of the previous stage, which leads to the final displacement of the diaphragm, described by the components x, y and ! (see Figure 15.17).

15.4.2 General layout According to the foregoing examination, and on the basis of Sections 15.2 and 15.3, it may be deduced that, for the specific level considered, the horizontal displacement Di of the ith plane element referred to a characteristic point (xi, yi) selected at the middle of its trace may be expressed as follows, by conveying the influence of the two superposed states: Di ¼ [ i T T] þ [x cos i þ y sin i þ ! ei] where i ¼ xi cos i þ yi sin i, and the temperature coefficient T ¼ 10 —5/8C. The temperature variation T is considered positive if the temperature of the diaphragm increases, and vice versa. Thus, the force F acting on each plane element at the level examined is F ¼ Si Di ¼ Si T i T þ Si (x cos i þ y sin i þ ! ei) 530


The determination of x, y and ! is by the solution of an analogously structured system of equations as in the previous case of lateral loading: K1 x þ K2 y þ K3 ! ¼ PTx K2 x þ K4 y þ K5 ! ¼ PTy K3 x þ K5 y þ K6 ! ¼ PT! In the present case of temperature influence, however, the right-hand variables represent — with opposite signs — the required external actions at the point O, to comply with the aforementioned first stage: P PTx ¼ T T (cos i Si i) P PTy ¼ T T (sin i Si i) P PT! ¼ T T (Si i ei)

15.4.3 Orthogonal layout In the usual case where the plane elements are arranged in an orthogonal layout, the above expressions are simplified as follows. For the plane elements in the x sense, i ¼ xi and ei ¼ yi, whereas for the elements in the y sense, i ¼ yi and ei ¼ xi. Then, P PTx ¼ T T (Sx xi) P PTy ¼ T T (Sy yi) and PT! ¼ T T

P

(Sy xi yi Sx xi yi)

while x, y and ! are determined as K3 K5 þ ðSx xi Þ K3 K K 2 1 2 4 x ¼ T T K1 K1 K3 K5 þ K6 K1 K4 K3 K5 P þ ðSy yi Þ K5 K1 K4 y ¼ T T 2 K4 K4 K3 K52 þ K6 K1 K4 X K3 X K X ðSx xi Þ þ 5 ðSy yi Þ þ Sx xi yi Sy xi yi K K4 ! ¼ T T 1 K32 K52 þ K6 K1 K4 P

531


According to the expression for Di in Section 15.4.2, the elements in the x and y senses are subjected at the examined level to the following forces, respectively: Fx ¼ Sx Di ¼ Sx (xi T T þ x ! yi) Fy ¼ Sy Di ¼ Sy (yi T T þ y þ ! xi) It is clear, however, that the consideration of an applied temperature difference at more than one diaphragm simultaneously may be treated by superposing the respective results for each separate level, exactly as in the case of lateral loading examined in Section 15.3.1. Thus, each plane element will be subjected to a corresponding force at each specific level where the temperature variation is applied. From the above equations, it may be concluded that, contrary to what happens in the case of horizontal loading, the specific position of each plane element in the ground plan plays its own role in the development of the response, so that the element may not ‘slip’ along its axis. The symmetry in the ground plan with respect to either the x or y axis simplifies the above expressions, as the coefficients K3 and K5 have null values. The existence of a double symmetry for the axes Ox and Oy for the stiffnesses Sx and Sy, respectively, means that K3 ¼ K5 ¼ 0, and leads to an important simplification: x ¼ y ¼ ! ¼ 0. Hence, the plane elements in the x and y senses are acted upon by the forces Fx ¼ Sx xi T T and Fy ¼ Sy yi T T, respectively. The forces Fx and Fy must be taken into account for all cases during the design of the arrangement of the vertical elements — frames, shear walls — of a building structure. They meet precisely the criterion for the maximum building dimensions with regard to a monolithic construction, in order to withstand not only possible climatic variation associated with temperature change but also the ever-present concrete shrinkage of the diaphragms (slabs), being equivalent to a temperature fall of about 258C. The typically considered ‘maximum length’ of 30 m, where the ‘dilatation joints’ are normally provided, must clearly always be adjusted for, depending in each case on the plane stiffness elements themselves, as well as on their specific layout in the ground plan. It is thus clear that the design of the vertical members of a multi-storey building structure, apart from ensuring the safe uptake of any horizontal action either as seismic or wind loading, must also cover the ‘temperature variation’ of the diaphragms.

References Stavridis L. (1986) Static and dynamic analysis of multistorey systems. Technika Chronika Scientific Journal of the Technical Chamber of Greece 6(2), 187—219. Stavridis L. (1994) Beanspruchung von mehrsto¨ckigen Bauten infolge Temperatura¨nderung. Bauingenieur 69(3), 117—122.

532

16 Dynamic behaviour of discrete mass systems 16.1

Introduction

The static loads, which all load-bearing structures have to take up, may be considered as completely independent of the structure itself, as far as its deformation is concerned. This uptake of static loads is realised through the stiffness of the structure, depending on its geometric and material characteristics. The closed process (load action) ! (uptake through stiffness) is accomplished through the process of the realisation of equilibrium, i.e. of the immobility of the structure. In this static process the mass (m) of the structure obviously remains inactive. It is activated, however, as a factor of generating loads according to Newton’s second law of motion (F ¼ m ), from the moment that the structure is somehow set in motion. There are many factors that can set the structure — which is fixed to the ground — in motion, such as wind forces, earthquake ground movements, ‘sudden’ load applications, and moving loads (e.g. vehicles or pedestrians on roads or footbridges, respectively), etc. The problem created through the motion of the mass of the structure is that it implies accelerations () leading, according to d’Alembert’s principle, to inertial forces equal to m . The structure is thus stressed under these loads, which may clearly be treated as static ones in the usual manner. However, the problem lies in the fact that the above accelerations, which vary in time, are unknown from their onset. In fact, they depend on the masses and the stiffness of the system, and to determine their variation in every localised mass of the structure and at every instant of time constitutes a dynamic problem that is complicated from the onset of the motion. Thus, the whole dynamic problem may be described as follows. The load-bearing system consists of specific masses that are connected to each other as well as to the ground through a massless structural web, which, apart from its own geometric layout, exhibits specific stiffness characteristics (Figure 16.1). When the structure is set in motion, the inertia forces of the masses ( m ) on the one hand and the elastic forces of the structural web on the other are activated, depending on its stiffness characteristics and acting on the same masses. The solution to the problem of the determination of the accelerations ‘’ of the masses results from the requirement of equilibrium for each mass under the action of the inertia as well as the elastic forces. However, the difficulty in determining the motion of the masses arises from the fact that, while the elastic forces are generally proportional to the displacements of the Structural systems: behaviour and design


Structural systems: behaviour and design Mass (developing inertia forces)

Structural web (developing elastic forces)

m2

m1 · γ1

γ1

m1

γ2 m2 · γ2

Base motion (earthquake)

Figure 16.1 Masses in motion and the structural web

masses, the inertia forces are proportional to the second derivative of these displacements with respect to time. The structures constitute a priori continuous systems in the sense that their mass is continuously distributed and not concentrated at discrete locations. Nevertheless, lumping the mass on the basis of behavioural criteria is always attempted for selected locations, as the analytical treatment of the continuous mass distribution is mathematically much more cumbersome. g The possible motions of the discrete (lumped) masses generally determine the inertia forces they are subjected to, and from this point of view the latter are distinguished between translational and rotational inertia forces. In the present examination, which is, however, oriented to preliminary design needs, the inertia forces due the rotation of the masses will be neglected, except for the horizontal diaphragms of multi-storey buildings undergoing a rotation !, as explained in Section 15.2. Thus, in the plane frame shown in Figure 16.2, in order for a lateral excitation to apply, its masses are assumed — approximately — to be concentrated at each storey, only their horizontal displacement is recognised, and their rotations about an axis perpendicular to the plane of the frame, as well as their vertical displacements, are neglected as a source of inertia forces. Similarly, in the continuous beam shown in Figure 16.2, where a transverse excitation is encountered, its mass is considered concentrated at specific points, and only the vertical displacement of these points is recognised, their rotations about their own axis being neglected as a source of inertia forces. Finally, in the two-storey spatial system shown in Figure 16.2, where, a horizontal seismic excitation occurs, the mass is considered equally distributed in each horizontal diaphragm, and only the horizontal displacement components (x, y), producing inertia forces in the x and y directions, respectively, as well as the rotation !, producing an inertia moment at each storey, are recognised. 534

Dynamic behaviour of discrete mass systems

Structural web Structural web

Mass Mass

Effective motion possibilities (degrees of freedom)

Mass

Structural web

Structural web

Mass

Figure 16.2 Possible effective mass displacements (degrees of freedom)

In this way, each system is characterised on the basis of its effective displacements, which are usually referred to as degrees of freedom. Thus, in the above examples, the frame, the continuous beam and the two-storey system have, respectively, three, four and six degrees of freedom. Despite the fact that load-bearing structures usually exhibit more than one degree of freedom, the study of the single-degree-of-freedom system is absolutely crucial in order to understand all the essential characteristics and dynamic behaviour of multi-degree systems. It is necessary, then, to begin an elementary study of the dynamic behaviour of a single mass with only one ‘active’ displacement, given also that this model applies directly to the practical treatment of composite systems, particularly those with a continuous mass distribution, as shown in the last two sections of this chapter.

16.2

Single-degree-of-freedom systems

16.2.1 Dynamic equilibrium A mass m connected to the free end of a cantilevered bar of length L will be used as an example of a single-degree-of-freedom system. The bending stiffness of the bar is much less than its axial stiffness, and therefore the only possibility for displacement of the mass is transversely to the axis of the bar. Consequently, the stiffness k (kN/m) of the bar is 535


1

Static equilibrium of mass u(t1) m · ü(t1)

m k = 3 · EI/L

3

Stiffness of structural web

k · u(t1) c · u˙ (t1)

m

F(t1) c · u˙ (t1)

k · u(t1)

Figure 16.3 Dynamic equilibrium of a single mass in a cantilevered structural web

considered with respect to the displacement of its free end (Figure 16.3), and is equal to k ¼ 3 EI/L3. It is assumed that the mass is acted upon by a horizontal force function F(t) and that at a specific moment t1 it undergoes a displacement u(t1) in the same direction. Moreover, it is assumed that the mass receives an additional resistance to its motion due to what can be considered the ‘internal friction’ of the material of the bar, which is at first taken as proportional to the velocity of the mass u_ ðtÞ, and thus equal to c u_ ðtÞ. The coefficient c is called the damping coefficient, as by its nature it tends to dampen the motion of the mass. Thus, at the time t1 the mass m, according to d’Alembert’s principle, may be considered in equilibrium under the following forces: . . . .

the the the the

external force F(t1) in the same direction as u(t1) elastic force k u(t1) having a sense opposite to that of u(t1) d’Alembert’s inertia force m €uðt1 Þ with a sense opposite to that of u(t1) damping force c u_ ðt1 Þ with a sense opposite to that of u(t1).

The static equilibrium of the mass requires (see Figure 16.3) m €uðt1 Þ þ k uðt1 Þ þ c u_ ðt1 Þ ¼ Fðt1 Þ It should be pointed out that the elastic force k u(t1) coincides with that needed to shift the mass m by u(t1), according to the stiffness of its structural web.

16.2.2 Free vibration 16.2.2.1 Undamped vibration As the dynamic characteristics of the system are really independent of the external loading, their investigation is made easier if the force F(t1) is removed, and thus the free vibration of the mass m is considered after an initial arbitrary displacement is imposed at the rest position. By temporarily neglecting the ‘damping’ factor, it may be written for any time t1 that m €uðt1 Þ þ k uðt1 Þ ¼ 0 536

Dynamic behaviour of discrete mass systems Equilibrium of mass in characteristic positions A

A

A

m · ümax

m · ümax

k·A

k·A

0

0.20

0.40

A

m · ümax

m · ümax

k·A

0.60

k·A

0.80

1.00

1.20 s

T = 0.80 s ümax

u umax = A

Deceleration

Acceleration

ümax 2π · f = √k/m

f = 1/T = 1.25 Hz A 0.20 0

0.40

u˙ max A

Deceleration

0.60

0.80

u˙ max = A · (2π · f ) (u˙ = 0)

Acceleration ümax = A · (2π · f )2 = A · (k/m)

1.00

1.20 t: s A

ümax

(u˙ = 0) An increase in the stiffness of the structural web or of the mass implies an increase or decrease in the eigenfrequency, respectively

Figure 16.4 Free vibration of a single mass

For the determination of the displacement function u, the kinematic conditions of mass m at the start time are needed, i.e. its displacement u0 ¼ u(t ¼ 0) and its velocity u_ 0 ¼ u_ (t ¼ 0). The solution of the above equation shows that the mass oscillates continuously in a periodic manner about the ‘rest’ position with a deviation u(t) that varies sinusoidally with respect to time (Figure 16.4). The time needed for a full oscillation of the mass, after which the motion is repeated in the exact same way, is called the natural period T (s). The inverse magnitude f of the period is called the frequency, and f ¼ 1/T (1/s). The frequency f represents the number of full oscillations that take place within a second, the unit of this quantity being the Hertz. It is common in vibration studies to also use the term circular frequency instead of ‘frequency f ’, denoting the magnitude !, on the basis of the relation: f ¼ !/2 p The concept of frequency, or the natural frequency, of a system constitutes a basic characteristic of the system, directly influencing its dynamic behaviour towards the 537


imposition of external forces. It is found that rffiffiffiffi k (rad/s) !¼ m and, consequently, rffiffiffiffi 1 k (1/s ¼ hertz) f¼ 2p m It should be noted that the mass unit used for structures is generally the tonne (t), which is equal to 1000 kg: 1 t ¼ 1 kN s2/m, and, given a weight W in kN, its mass is m ¼ W/9.81 t As seen from the last equation, the larger the mass, the fewer the vibrations that occur in 1 s, whereas the stiffer the connection of the mass with the ground — i.e. the structural web — the higher the frequency of the system. However, the presence of the square root attenuates the effect of the above parameters. Of course, the frequency f of the system is independent of both the initial deviation u0 and the initial velocity u_ 0 of mass m. The mass m executes a steadily varying, oscillating motion (see Figure 16.4). Its displacement u(t) is expressed as uðtÞ ¼

u_ 0 sin ! t þ u0 cos ! t !

Its maximum deviation A on both sides, also called the amplitude, is found to be sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ffi u_ A ¼ u20 þ 0 ! At these positions, u ¼ A, the mass has zero velocity and maximum acceleration (or deceleration), €umax ¼ A !2 , whereas at the ‘rest position’ (u ¼ 0) it has the maximum velocity, u_ max ¼ A !. Moreover, it should be pointed out that the elastic forces k u(t1) have the opposite sense to the corresponding d’Alembert forces m €uðt1 Þ, as is implied by the basic equation itself. More specifically, as the inertia force m €uðt1 Þ is opposite to the acceleration vector €uðt1 Þ, just before the maximum deviation A, the inertia force is opposite to the maximum deceleration developed €umax , whereas just after this position the inertia force is opposite to the equal acceleration €umax . This means that in both cases the inertia force is directed to the right, i.e. in the opposite direction to the acting elastic force (k A) (see Figure 16.4) and, finally, from the equilibrium equation m €umax ¼ k umax ¼ k A, it can be checked that €umax ¼ A !2 . g It should be made clear that the selection of the cantilever-type single-mass system is only indicative of a system with mass m and stiffness of structural web k, subjected to an external force F(t), with respect to its active stiffness k. A system consisting of, for 538


Static equilibrium of mass

F(t1)

m k

m · ü(t1) u(t1)

1

Stiffness of structural web

m

k · u(t1)

c · u˙ (t1)

k · u(t1)

c · u˙ (t1)

Figure 16.5 Dynamic equilibrium of a mass carried by a spring

example, a mass connected to a spring, as shown in Figure 16.5, would be equally representative, provided, of course, that the self-weight of the mass m is ignored. However, it should be recognised that while the cantilevered mass is offered as a more illustrative model for the examination of systems consisting of a number of masses arranged in height, as in, for example, buildings or bridges, the mass—spring model is more representative of the examination of vertical dynamic actions on existing structures such as beams or plates, as in, for example, a machine permanently fixed in place. The behaviour of the mass—spring model should thus be recognised as analogous to the cantilever system, as shown in Figure 16.6.

16.2.2.2 Damped vibration Consideration of the damping coefficient c in the equation m €uðtÞ þ k uðtÞ þ c u_ ðtÞ ¼ 0 does not alter the periodicity of the oscillation of the mass; however, it does result in the continuous decrease of the amplitude in each vibration cycle. It should be noted that the system vibrates with a somewhat increased period Td, which remains constant (Figure 16.7). It is found that the correspondingly decreased circular frequency !d is equal to sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2ffi c !d ¼ !2 2m At this point it is useful to clarify some facts concerning the damping, in order to better understand its treatment. Damping is the loss of energy in each vibration cycle. This energy is absorbed by the internal friction in the structure carrying the masses, i.e. the structural web, which gives stiffness to the system. Thus, the damping force c u_ ðtÞ is acting on the mass through the structural web, as a restraining force in the 539

Structural systems: behaviour and design Equilibrium of mass in characteristic positions m · ümax

m · ümax

m · ümax

m · ümax

A A

k·A

k·A k·A

k·A

0

0.20

0.40

0.80

1.00

1.20 s

ümax

u umax = A

0.60

Deceleration

Acceleration

ümax

A 0.20 0

0.40

u˙ max A

Deceleration

0.60 u˙ max (ü = 0)

Acceleration

0.80

1.00

1.20 t: s A

ümax

ümax (u˙ = 0)

Figure 16.6 Free vibration of a mass—spring system

same direction as the acting elastic force. Of course, the direct result of damping is, as previously mentioned, a continuous decrease in the vibration’s amplitude. The description of damping as a force proportional to the velocity of the mass is not satisfactory from a physical point of view. Nevertheless, it permits — as will be shown later — a practical and effective quantitative description of the phenomenon, because of its convenient mathematical treatment. Thus, it is found that if the damping coefficient c achieves the critical value ccr ¼ 2 m !, then no vibration can be developed, and the mass moves from its position of maximum deviation directly to its resting point. Although such a situation is impossible in a structure, its consideration is nevertheless useful because it represents a reference level for the applied damping coefficient in each case. First, the so-called damping ratio is defined, as the ratio of the damping coefficient to its critical value: ¼ c/ccr. On the basis of this equation, ¼ c/(2 m !) and, consequently, qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi !d ¼ ! ð1 2 Þ 540

Dynamic behaviour of discrete mass systems The ratio λ of two consecutive amplitudes is constant: (A1/A3) = (A2/A4) = λ Damping coefficient ζ = ln(λ)/2π

λ = e2πζ

The presence of damping ζ < 10% affects the oscillation period very little (Practically: f = √k/m /2π) u

Td

Td

Td

A · e–2π · f · t A A2

A4

0

t: s A3 A1

A · e–2π · f · t

Figure 16.7 Characteristics of damped vibration

The ratio , despite the fact that it refers to the unrealistic concept of the coefficient c, has an essential physical meaning, something that may not only be deduced from the last equation but also from the fact that between two consecutive amplitudes of the decaying vibration, with a ‘time distance’ Td, ¼ ln

uðtÞ c 2p 2p ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ u ðt þ Td Þ 2 m ! 1 2

The above equation is of particular practical importance, because, from the measurement of the consecutive amplitudes of a damped vibration of a system exhibiting a constant ratio, the value of may be directly deduced. As this value, for structural systems encountered in practice, does not usually exceed 10%, it may be considered with a good approximation to be !d ¼ ! (hence Td ¼ T) on one side and ¼2p on the other. In real situations, however, for practical reasons as well as for accuracy, the amplitudes of two vibration cycles are measured as having a time distance rather equal to n T, in 541

Structural systems: behaviour and design u Td

Td

A1

Td

Td

3*Td

(n = 3)

A2

A3

Td

A4

0

t: s

(A1/A2) = (A2/A3) = (A3/A4) = ... = λ ζ = ln(A1/A4)/(2π · 3)

Figure 16.8 Practical determination of the damping ratio

which case (Figure 16.8) n ¼ ln

uðtÞ ¼n ¼2pn uðt þ n TÞ

Hence, n 2pn Finally, the expression of the displacement u(t) of a mass m, which, after being displaced by A, is left to vibrate freely with zero initial velocity, under a damping ratio (!d !), is ¼

uðtÞ ¼

A !t

e

ð sin ! t þ cos ! tÞ

16.2.3 Forced vibration The aim of the above analysis of free vibration was to depict the basic dynamic characteristics of a single-degree-of-freedom system. An understanding of these characteristics is necessary to understand the dynamic response of the system if the mass is acted on by a force varying in time, i.e. exactly as was initially considered, according to the equation of ¨ller and Keintzel, 1984): motion (Mu m €uðtÞ þ k uðtÞ þ c u_ ðtÞ ¼ FðtÞ where, after the introduction of the damping ratio , c¼2 m! It is useful to express the force F(t) in the form F(t) ¼ F1 f(t), where f(t) describes the variation of the force during the whole period of the time examined, while F1 represents the maximum value attained by this force. For the design of a structure, the response of 542


the structural web connecting the mass with the ground is of primary interest. This response depends only on the displacement u. It is understandable that the maximum response corresponds to the maximum displacement umax, which is clearly larger than that corresponding to the ‘static’ loading of the system with the force F1, this being equal to ustat ¼ F1/k Thus, if in a forced vibration the ratio DMF ¼ u/ustat is defined as the dynamic magnification factor, then its maximum value (DMF)max ¼ umax/ustat is of definite design importance. Of course, the response of the system under the loading F1 f(t) is fully described by the exact solution of the above differential equation. This solution underlines the fact that the final response is the result of a superposition of the free vibration and the effect of the force imposed on the system, being expressed by the so called Duhamel integral, as shown below. Thus, the general solution of this equation is (Biggs, 1964) u_ 0 þ ! u0 ! t u0 cos !d t þ sin !d t uðtÞ ¼ e !d ðt F1 þ fðÞ e ! ðt Þ sin !d ðt Þ d m !d 0 Obviously, direct use of this solution is not practical. However, the accessibility of special software dealing also with systems of multiple degrees of freedom allows the desired results to be obtained through a direct numerical approach rather than through the above analytic approach. Nevertheless, the above expression is very useful, because it permits direct analysis of some typical cases and allows particularly important conclusions to be obtained for the response of single-degree and, additionally, multidegree systems. A few examples of such typical cases are listed below (see Figure 16.9): . The ‘abrupt’ application of force F1, followed either by its equally ‘abrupt’ removal after some period td, or by an immediate linearly varied decrease during the same period td. . The progressive application of the force up to the value F1 and, consecutively, its decrease down to zero over a total time period td, first in a linear fashion and then in a parabolic manner. . The progressive linearly increasing application of force F1, over a time period td, followed by maintenance of this acquired value for an unlimited time. In all the above cases, the damping effect is ignored for simplicity. In the curves shown in Figure 16.9 the behaviour of the maximum value of the dynamic magnification factor (DMF)max is depicted for each of the above cases, as a function of the ratio of the characteristic time duration td to the natural period T of the vibrating system. From curves (a), it can be seen that in the case of a constant applied force, when the time reaches the value T/2, the factor (DMF)max takes its maximum value of 2, which is not exceeded in any other case. This value is also valid for any td greater than T/2. 543

Structural systems: behaviour and design 2.4 F1 2.0

td t (a)

F1

(DMF)max

1.6

td t F1

(b) 1.2

td t

(c)

1.0

F1

0.8

td t

F1 0.4 td

t

0 0

0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 td /T

Figure 16.9 Behaviour of a single-mass system under a dynamically applied load

Otherwise, for a duration td less than about T/5, the factor (DMF)max is less than unity, i.e. the maximum deviation of the mass is even less than ustat ¼ F1/k. In contrast, the ‘immediately occurring’ linear decrease of the applied force has more moderate results. Only after an unloading duration td of about 10 times the natural period T does the factor (DMF)max begin to approach the value of 2.0, while for a duration less than T/3 the maximum deviation is less than the ‘static’ one ustat. From curves (b), it can be seen that the maximum value of (DMF)max, i.e. the maximum mass deviation, occurs when the total loading—unloading duration approaches 90% of the natural period T. This value equals 1.8 or 1.5 for a parabolic or linear force variation, respectively. For lower or greater durations td, the maximum value of (DMF)max is always lower. Curve (c) represents the typical force application scheme that is encountered in many practical cases. The more the duration td falls below T, the more the value of the factor increases, never exceeding, though, the absolute maximum value of 2.0, no matter how ‘abrupt’ the loading process is. However, for any duration td greater than T, the maximum value of factor DMF does not exceed the value 1.2. The above results confirm the definition of the static loading, as given in the introduction to Chapter 1.

16.2.4 Periodic sinusoidal acting force Of particular importance for the design of structures is the case when the acting force F(t) is acting on the system in a periodic manner, according to the expression F(t) ¼ F0 sin t. The examination of the corresponding response permits one to draw useful conclusions about any kind of periodic excitation — machinery effects, human rhythmic activities, seismic action, etc. — as well as about measures required for the attenuation of their effects (dynamic isolation). 544


By applying the general solution of the previous section in the present case and ignoring the effect of damping, the following expression is obtained: uðtÞ ¼ U sinð! t þ ’Þ þ

F0 =k sin t 1 ð=!Þ2

The first term on the right-hand side is essentially identical to the corresponding expression for the general solution, where u U ¼ A and tan ’ ¼ 0 u_ 0 =! This term represents, as already mentioned, that part of the response corresponding to the free vibration of the system, whereas the second term is exclusively due to the externally applied force. Now, as consideration of even the least possible damping makes the contribution of the free vibration (i.e. the first term) soon vanish, the conclusion may directly be drawn that the maximum value of the dynamic magnification factor (DMF)max will be u 1 ðDMFÞmax ¼ ¼ ustat max 1 ð=!Þ2 It is clear that the ratio r ¼ =! plays a basic role. For values less than 0.20 and greater than 1.50, the dynamic influence is not important. More specifically, while for values of r tending to zero (static loading) the factor (DMF)max tends to 1, for a ‘high-frequency’ excitation (r > 2.50) this factor tends towards 0. These correlations are clearly shown in Figure 16.10. However, as can be deduced from the equation above, when ¼ !, the factor (DMF)max is infinite. What is actually happening in this case is that in each cycle the factor DMF is steadily increasing in time, according to the equation (DMF) ¼ ! ¼ (sin ! t ! t cos ! t) On the basis of this equation: ðDMFÞmax; ¼ ! ¼

umax 1 ¼ np F0 =k 2

where n is the ever-increasing number of cycles. The situation that occurs in the case when ¼ ! is called resonance, and is something that should, of course, be avoided. However, as an infinite value of the factor DMF corresponds to a system without any damping effect, it should pointed out that this does not happen in reality, as even a small amount of damping action reduces the response significantly. More precisely, for an existing damping factor equal to (Biggs, 1964), 1 ðDMFÞmax ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ½1 ð=!Þ2 2 þ 4 2 ð=!Þ2 where the restrictive contribution of to the response is clear (see Figure 16.10). 545

Structural systems: behaviour and design 5.00

4.50 ζ=0 4.00

3.50 ζ = 0.10

(DMF )max

3.00

2.50 ζ = 0.20

2.00

1.50

1.00 ζ = 0.50 0.50 ζ = 1.00 0

0

0.50

1.00 1.50 2.00 2.50 r = Ω/ω Ω: Imposed frequency ω: System eigenfrequency In the case of resonance (Ω/ω): (DMF )max = 1/2ζ

Figure 16.10 Behaviour of a system towards a periodic load action

Thus, in the case of resonance ( ¼ !), (DMF)max ; ¼ ! ¼ 1/(2), and, consequently, 1 F0 umax ¼ 2 k However, until the system develops the above maximum deviation of mass, a large number of cycles is needed. g Of particular interest is the assessment of the transmitted forces to the base, i.e. to the support of the structural web. From the examination of the dynamic equilibrium of a system — see Section 16.2.1 and also Figure 16.11 — the transmitted force Fg at the base of the structural web, consists of its elastic force (k u) plus the damping force (c u_ ) acting in the same sense. The maximum force transmitted to the base max Fg can be obtained according to the equation 1=2 1 þ 4 2 r2 max Fg ¼ F0 ð1 r2 Þ2 þ 4 2 r2 546

Dynamic behaviour of discrete mass systems F0 · sin Ωt m u

m·ü u

m

m

m

m · ü(t1) k · u(t1)

k·u

c · u˙

k·u

c · u˙

Fg

c · u˙ (t1)

c · u˙ (t1)

Transmission factor T = max Fg/F0

F0 · sin Ωt

k · u(t1)

Fg

Figure 16.11 Transmission of dynamic actions at the base of a structural web

The ratio T ¼ max Fg/F0 is called the transmission factor, and its dependence on the ratio r ¼ =!, as well as on the damping factor, is depicted in Figure 16.12. According to Figure 16.12, given the frequency of an imposed load, due to, for example, an installed machine, by providing to the system — through appropriate means — a natural circular frequency !, which is less than about /2.50, the transmitted force to the base is drastically limited.

16.2.5 Seismic excitation Seismic excitation consists of an imposed oscillating motion of the support of a structure. The cause of this motion is the abrupt (brittle) rupture in some region of an already 3.00

Transmission factor T

ζ=0

ζ=0

2.00

ζ = 0.2

1.00

ζ = 0.25

0 0

1.00

Ω: Imposed frequency ω: System eigenfrequency

√2 Ω/ω

2.00

3.00

Effective isolation with soft springs and low damping (Increased displacements)

Figure 16.12 Transmissibility of dynamic actions at the base of a system

547

Structural systems: behaviour and design u

x x

m · ü(t)

m

m · x¨

x m · üs(t )

(m · a) · fa(t) m

k·x

k·x

k·x

k·x

x = u – us

us

üs(t) = a · fa(t ) Accelerogram (a: Maximum ground acceleration)

Fixed base

Imposing a ground motion means essentially imposing an accelerogram, and this is turn means imposing an inertia force on a rigidly supported system

Figure 16.13 Imposition of seismic ground motion and dynamic equilibrium

existing fault in the earth’s crust, as a result of accumulated stresses. This rupture is instantaneously spread out along the fault, causing propagation of a vibration wave within the earth’s crust lasting for a period of about 15—30 s, and then coming to rest. First, it is assumed that the imposed horizontal ground motion us(t) on the support of the structure may be expressed as us(t) ¼ us,0 f(t), where us,0 represents the maximum value of the ground displacement in the region of the support. In the present case of a single-mass system, the direction of us corresponds to the sense for that of the stiffness k of the system (Figure 16.13). It is clear that the elastic force acting on the mass m will be due to the deformation of the whole system, i.e. to the relative displacement of the mass to the ground, expressed as u us. Given that the mass m is not acted on by any external load, but is subjected to the inertia force only, obviously with reference to its absolute displacement u, as well as to the elastic force, it can be written, if the damping force is ignored, that m €uðtÞ þ k ½uðtÞ us ðtÞ ¼ 0 The absolute displacement u(t) of the mass m cannot describe the deformation of the structure depending, as pointed out, exclusively on the relative displacement: x(t) ¼ u(t) us(t) Then, substituting the last equation, m ð€x þ €us Þ þ k x ¼ 0 It is clear that the response k x of the structure depends solely on the ground acceleration €us ðtÞ (see Figure 16.13). Thus, the characteristic of an earthquake, which is relevant for the design of a structure, is only the so-called accelerogram, which is recorded by special equipment each time. For a specific earthquake and within a major region, different accelerograms may be recorded for different locations, depending on the distance from the respective 548


source of the event (hypocentre) as well as on the existing soil properties along the path followed by the seismic waves to the place recorded. The accelerogram €us ðtÞ may be expressed as €us ðtÞ ¼ a fs ðtÞ where a represents the maximum occurring ground acceleration, while the function fa(t) describes its evolution in time. Now, the last equation may be written as m €x þ k x ¼ ðm aÞ fa ðtÞ The form of this equation is exactly the same as that for the forced vibration of a singledegree-system on a fixed basis (see Figure 16.13), and, as the existing damping obviously affects the relative velocity x_ of the mass m with respect to the ground, the equation may be accordingly modified as follows: m €x þ k x þ c x_ ¼ ðm aÞ fa ðtÞ It should be noted that the right-hand term refers to a ‘fictitious force’ — representing along with the force m €x the total inertia force that the mass m is subjected to. By assuming initial conditions corresponding to ‘rest’, i.e. x ¼ 0 and x_ ¼ 0, the solution ¨ller and of the last equation, analogously to Section 16.2.3, may be written as (Mu Keintzel, 1984) ð t a ! ðt Þ fa ðÞ e sin !d ðt Þ d xðtÞ ¼ !d 0 As previously explained (see Section 16.2.2.2), for values of up to 10%, the fundamental circular frequency !d does not differ substantially from the natural frequency of system ! without damping. In any case, however, for the design of a structure, it is the maximum value Sd of the relative deviation that is of paramount interest, as it obviously leads to the maximum response: ð t a fa ðÞ e ! ðt Þ sin ! ðt Þ d Sd ¼ xmax ¼ ! 0 max It can be seen that at the time that the maximum deviation xmax occurs, the acceleration (or deceleration) €x also takes its maximum value €xmax, whereas the velocity x_ — and hence the damping force — vanishes. Thus, at this moment, the elastic force k xmax acting on the mass m balances its inertia force, depending, of course, on its absolute acceleration, and is equal to m (€xmax þ €us ). Assuming that at this same moment the maximum ground acceleration is essentially attained, it can be written (Figure 16.14) that m ð€x þ €us Þmax ¼ k xmax

or

ð€x þ €us Þmax ¼ !2 xmax

Thus, the maximum value Sa of the absolute acceleration of the mass is ð t ! ðt Þ Sa ¼ ð€x þ €us Þmax ¼ a ! fa ðÞ e sin ! ðt Þ d 0

max

549

Structural systems: behaviour and design Strict consideration

Acceptable approximation Maximum acceleration

xmax m · x¨max

m(x¨ max + a) = m · Sa

m · üs(t)

Inertia force k · xmax

k · xmax

k · xmax

k · xmax Maximum response

xmax üs(t) = a · fa(t) Accelerogram

Approximation: the maximum deviation coincides with the maximum ground acceleration

Figure 16.14 Position of maximum response in a single-degree-of-freedom system

This expression is strictly valid provided that the damping coefficient does not exceed 10%. g It is thus clear that the response of a single-degree system to a specific seismic excitation — i.e. with known characteristics regarding the time development fa(t) of the ground acceleration, as well as its maximum value a — depends only on the natural frequency, or natural period, of the system and the corresponding damping factor. On the basis of the last equation and for a given accelerogram, the maximum absolute acceleration Sa can therefore be plotted as a function of the natural period T, for different values of the damping coefficient . Such a diagram represents the acceleration spectrum, and it permits, for any oscillator with a known natural period T, the direct determination of its maximum absolute acceleration. It may be easily seen that Sa ¼ Sd !2. It is useful to point out here that, according to the above expression, Sa — as well as the spectral displacement Sd — is proportional to the maximum ground acceleration a accompanying the specific accelerogram, so that if divided by a, it represents a spectrum corresponding to an accelerogram exhibiting the same evolution in time but with respect to unit maximum acceleration. Figure 16.15 shows an accelerogram recorded from the El Centro earthquake (California, USA, 1940), together with the acceleration spectrum (Sa/a) corresponding to unit ground acceleration, for two different damping coefficients. It is can now be seen that for the practical evaluation of a specific earthquake being described by its accelerogram, the acceleration spectrum Sa — or the unit spectrum Sa/a together with the ground acceleration a — represents the examined excitation quite sufficiently. Thus, for a specific system with a known natural period T, on the basis of the corresponding spectral value Sa, the maximum inertia force can be directly obtained as m Sa. This force represents the maximum shear force Va developed at 550


Ground acceleration: g

0.4

0.2

0.0

–0.2

–0.4 0

2

4

6

8

10 T: s

12

14

16

18

20

m m m m m

4.80

Static consideration m · (Sa/a) · a

Sa/a

3.60

ζ = 0.02

2.40 ζ = 0.10 1.20 m·a Base shear 0

0

0.5

1.0 T: s

1.5

2.0

Figure 16.15 Accelerogram and acceleration spectrum (El Centro, California)

the base of the structure, called the base shear: Va ¼ m S a It is clear that the above inertia force (m Sa) is identical to the maximum elastic force (k Sd) acting on the mass m (see Figure 16.15). Thus, (m Sa) ¼ k Sd ¼ k Sa/!2 The use of the acceleration spectrum therefore allows the determination of the maximum response in the structural web through a static loading, which is, of course, of particular practical importance. 551


What can be deduced from the spectrum of a specific earthquake is that there is always a range of periods which is more sensitive towards the development of the maximum inertia force, or — in other words — of the maximum shear base. As can be seen in Figure 16.15, structures with a lower stiffness being reflected in values of a natural period T greater than, say, 1.0 sec, develop a clearly limited response. The dynamic magnification factor (DMF)max, which as defined in Section 16.2.3 is the ratio of the maximum relative displacement Sd to the ‘static’ displacement (m a)/k, is now examined: Sd !2 Sd Sa ¼ ¼ ðm a=kÞ a a ð t ¼! fa ðÞ e ! ðt Þ sin ! ðt Þ d

ðDMFÞmax ¼

0

max

Written differently: Sd ¼ (a/!2) (DMF)max Sa ¼ Sd !2 ¼ a (DMF)max It should be noted that the dynamic factor (DMF)max, as can be seen from the above expression, is identical to the acceleration spectral value (Sa/a). g Assuming now, in an approximate approach, that the variation of the acceleration a in time is sinusoidal, i.e. fa(t) ¼ sin t, then, given that the two last equations are generally valid, the previously found expression for the dynamic factor (see Section 16.2.4) may be applied directly: 1 ðDMFÞ;max ¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ½1 ð=!Þ2 2 þ 4 2 ð=!Þ2 Thus, Sd; ¼ ða=!2 Þ ðDMFÞ;max and Sa; ¼ a ðDMFÞ;max The diagram in Figure 16.10 may therefore also be interpreted as an acceleration spectrum — with respect to unit ground acceleration — for any value of the ratio of the frequency (taken as constant) to the natural frequency ! of the single-degree system. The magnitude of the assumed frequency of the maximum ground acceleration depends on the geotechnical characteristics of the foundation soil. For stiff soils and rocks, ranges from 20 to 60 rad/s, for medium stiff soils, from 12 to 50 rad/s, and for soft or very soft soils, between 10 and 3 rad/s. g Considering the influence of the foundation soil, it can be seen from Figure 16.16 that its deformability influences, in addition to the maximum developed ground acceleration, 552

Dynamic behaviour of discrete mass systems 1

1

m

m k = 3 · EI/L3 Stiffness of structural web

k = (3 · EI/L3)/p Stiffness of structural web

Smaller stiffness hence greater period

ρ = 1 + 3 · EI/(kϕ · L)

Natural period T

kϕ

kϕ Resilient ground

Imposing the same accelerogram on both systems causes in the case of resilient ground smaller or greater accelerations (inertia forces) depending on the period T itself 4.80

Sa/a

3.60

2.40

1.20 (Rigid support)

0 0

Increment of inertia force

0.5

1.0 T: s

1.5

2.0

Decrement of inertia force

Figure 16.16 Influence of the deformability of the soil on the seismic response

the response of the structure itself, due to the variation of the effective inertia force of the mass. This force increases or decreases according to the region of the corresponding acceleration spectrum that the natural period of the structure belongs to, by assuming a fixed base (see Figure 16.16). In actual design practice, the so-called design spectrum, which contains the effects of a large number of relevant accelerograms for the region examined, is used as an acceleration spectrum, also making a distinction between the various soil types, given the influence of the foundation soil deformability, as explained above. Of course, the elaboration of the design spectral curves is a major issue for the regulations in a particular country. Examples of these curves are shown in Figure 16.17. The influence of the soil type on the maximum effective acceleration Sa is obvious.

16.2.6 Influence of plastic behaviour on the seismic response According to the previous section, the basic equation of motion of a single-degree-offreedom system under a seismic excitation, with respect to a fixed base, can be 553

Structural systems: behaviour and design 0.60 Rock Stiff soil Soft soil

Sa/g

0.40

0.20

0

0

0.5

1.0 T: s

1.5

2.0

Figure 16.17 Design spectrum

written as m €x þ k x þ c x_ ¼ ðm aÞ fa ðtÞ This equation expresses the static equilibrium between the externally acting inertia force (m €x þ m a fa(t)) together with the damping force on one side and the elastic force developed by the structural web due to its deformation on the other. This last force (k x) increases for as long as the displacement x of the mass increases, i.e. as for as long as the deformation of the structural web increases, according to the existing stiffness factor k. However, it is clear that a possible plastic hinge at the base of the cantilever, where the maximum bending response is developed, places an upper limit on the increase of the above-mentioned force, which remains constant afterwards, despite further increase in the displacement x (Figure 16.18). More specifically, if L is the cantilever height and Mpl represents the plastic bending strength of the structural web, then the limit F up to which the force can be transmitted to the mass, i.e. the ‘elastic force’, may be increased, F ¼ Mpl/L. This, of course, is under the condition that the relevant

Maximum acting force m · (x¨ max + a) = m · Sa

Maximum acting force (reduced) m · (Sa/q) Mpl/L

(Inertia force)

k · xmax k · xmax

Mpl/L

L

Mpl Accelerogram üs(t) = a · fa(t) xmax Elastic behaviour

The adoption of a plastic hinge limits the response of a structural web until a predetermined value

Figure 16.18 Limitation of the inertia force due to plastification

554


section of the structural web affords the required ductility (rotational deformability), so that the augmented displacement x can be realised. It is clear that the limitation of the internal — elastic — force expressed by the term k x in the above equation restricts, accordingly, the level of the effective inertia force too. This means that, although the mass displacement x may increase, the above equilibrium equation can be satisfied with less internal force and, correspondingly, less inertia force, than required by a purely elastic behaviour of the structural web. This fact may be taken into account through a corresponding reduction in the spectral acceleration value Sa, through the factor q, known as the behaviour factor. Thus, in the corresponding equations in the previous section, the magnitude Sa/q is used instead of Sa. The selected value of the factor q is essentially the designer’s decision, depending on the estimated — or even desired — ductility of the structural web. A value of q equal to 1.0 would indicate the acceptance of elastic behaviour throughout the web without any reduction in the inertia seismic forces. A steel structure may allow consideration of a behaviour factor up to 4.0. In reinforced concrete structures, with an appropriate layout of reinforcement in order to ascertain good deformability (ductility) of the ‘critical’ sections, a value of q of even 3.0 may be adopted, leading to a corresponding decrease in the acting seismic forces. However, it should be pointed out that in the design process the consideration (or not) of q greater than unity — i.e. the adoption (or not) of plastic hinges in the structural web to account for a given maximum ground acceleration — relies exclusively on the designer’s engineering judgement.

16.3

Multi-degree systems

As multi-degree systems imply, they are systems that contain more than one discrete mass. These masses are ‘carried’ by a massless structural web exhibiting a clearly determinable behaviour regarding its deformability and stiffness. Within this frame of behaviour the masses show displacements or rotations involving either negligible or non-negligible inertia forces. These last displacements are considered as ‘effective’ for the corresponding masses, and their number represents the so-called effective degrees of freedom of the system. Examples of such multi-degree systems have been mentioned in the introductory section of this chapter (see Figure 16.2).

16.3.1 The stiffness matrix 16.3.1.1 The concept of the stiffness matrix Of decisive importance for the description of the behaviour of a multi-degree system is the so-called stiffness matrix, i.e. a set of magnitudes, describing the stiffness of its structural web, which are determined as shown in the example of a three-storey plane model in Figure 16.19. Here, it is assumed that the masses m1, m2 and m3 develop in a horizontal excitation (e.g. earthquake) inertia forces with respect only to their horizontal displacements. Thus, the effective displacements of the system are the lateral storey shifts 1, 2 and 3. 555

Structural systems: behaviour and design m1

∆1 = +1 k11

k12

k13

∆1

∆2 = +1 ∆2

m2

k21

k22

k23

Structural web ∆3 = +1 ∆3

m3

k31

k=

k32

k11

k12

k13

k21

k22

k23

k31

k32

k33

k33

The stiffness matrix represents the stiffness of the structural web with respect to the effective mass displacements

Figure 16.19 The elements of a stiffness matrix

The stiffness matrix consists in the horizontal juxtaposition of three groups of forces that are determined as follows (see Figure 16.19). The first group consists of those three — horizontal — forces (k11, k21, k31) that are required to be applied correspondingly to the effective displacements in order to produce a unit shift in level 1 and zero shift in the two others (i.e. 1 ¼ 1, 2 ¼ 0, 3 ¼ 0). Analogously, the second group consists of those forces (k12, k22, k32) needed to be acted on in the sense of the effective displacements, in order to produce a unit shift in level 2 and zero shift in the two others (i.e. 1 ¼ 0, 2 ¼ 1, 3 ¼ 0). Finally, the third group contains those forces (k13, k23, k33), which, acting along the effective displacements, produce 1 ¼ 0, 2 ¼ 0, 3 ¼ 1. Each one of the above groups comprises the reactions developed if, at first, fictitious simple supports are inserted into the system, in order to restrain all its effective displacements, and then the corresponding unit-support settlement (1 m) is imposed, as shown in Figure 16.19. In order to achieve this, the use of a computer is clearly necessary. Defining the positive senses of these forces as coinciding with those of the displacements (1, 2, 3), i.e. to the right, it is clear that the forces of each group will exhibit alternating signs. 556


The structural web which carries the three masses may be represented by the stiffness matrix k by juxtaposing horizontally the above three force groups as previously mentioned, according to the following formulation: 2 3 k11 k12 k13 6 7 k ¼ 4 k21 k22 k23 5 k31 k32 k33 It is useful to become familiar with matrix operations. However, it must be pointed out that the use of matrices contributes very little, if at all, to the understanding of the static or dynamic behaviour of any system — at least at this introductory level. Nevertheless, matrices are a necessary tool for the compact writing of equilibrium equations and the relevant equations, thus facilitating their methodical numerical treatment through computer programs.

16.3.1.2 Matrix operations In the present context, of particular interest are either the so-called square matrices — having an equal number of rows and columns — or the so-called column matrices — consisting of the vertical juxtaposition of numerical elements. A square matrix k of the order N N, has N rows and N columns (see Figure 16.19). The elements of the square matrix along the diagonal starting from the upper left element constitute the so-called diagonal elements. A column matrix a of the order N 1 has N rows and 1 column. Interchange of the rows and columns of a matrix — or vice versa — leads to the transpose matrix, denoted by the letter T. Thus, while the matrix kT still has N rows and N columns (i.e. N N), matrix aT has 1 row and N columns (i.e. 1 N). The basic physical concept of a matrix is directly related to the column matrix. This is the recording of quantities in correspondence to specific magnitudes, e.g. the storey shifts of the structure described previously, or the forces acting at each storey level. The appropriate horizontal juxtaposition of such column matrices leads to square matrices, like the stiffness matrix which has been previously defined. The sum of two column matrices is a matrix of the same order, and comprises the sums of their corresponding terms. Although the addition also applies to two square matrices of the same order, the physical meaning of this operation is clearer and better understood in this context with respect to the column matrices. The multiplication of two matrices is of basic importance, and it is accomplished as shown in Figure 16.20. The multiplication of two matrices is feasible only when the number of columns in the left matrix is equal to the number of rows in the right matrix. This may also be understood from the rule that the final order of the product matrix is found by eliminating the ‘interior’ dimensions of the matrices — as shown below. The product k a is written as (N N) (N 1), and is a matrix of order N 1, whereas product a k, which is written as (N 1) (N N), does not make sense. Of 557

Structural systems: behaviour and design P=k·∆

a1

Column matrix Multiplication cT · a

a1

Columns

a3

c1

c2

a2

Rows

c3

Multiplication k·a

a2 a3

b11 b12 b13 Multiplication k·b

k11 k12 k13

k11 k12 k13

k21 k22 k23

k21 k22 k23

k31 k32 k33

k31 k32 k33

b21 b22 b23 b31 b32 b33

= c1 · a1 + c2 · a2 + c3 · a3 Number of columns cT = number of rows a

Number of columns k = number of rows a

Number of columns k = number of rows b

Figure 16.20 Layout for the multiplication of matrices

course, product aT k does have a meaning, because it is written as (1 N) (N N), which leads to a matrix of order 1 N, whereas it can be seen that product aT a, written as (1 N) (N 1), leads to a matrix of order 1 1, i.e. a single number. It is thus clear, also taking into account the schematic layout of Figure 16.20, that the multiplication of two matrices consists of the consecutive formation of matrix products of the last type, (1 N) (N 1), which, of course, represent a single number. The square matrix, the diagonal elements of which consist of unity while all other elements are null, is called the identity matrix, and is denoted by I. Consideration of the identity matrix allows reference to the basic concept of the so-called inverse matrix of a square matrix A of the order N N, which is denoted by A1. The inverse matrix is one that satisfies the equation A A1 ¼ I or (identically) A1 A ¼ I Obviously, both matrix A1 and matrix I are of the order N N. The concept of the inverse matrix is directly connected to the solution of a linear system of N equations with N unknowns. Such a system may be represented through the following matrix notation: AX¼C If matrix A consists of the coefficients of the unknowns, then column matrices X and C contain the unknowns and the ‘known terms’ of the right-hand side of the equations, respectively. Multiplying both sides by the inverse matrix A1 yields A1 A X ¼ A1 C that is, I X ¼ A1 C 558


and, X ¼ A1 C Of course, the inversion of matrix A requires a numerical procedure that is equally cumbersome as the solution of the linear system, so that the advantage in the use of the matrix notation lies only in the compact formulation and not in the computing effort required, which in any case will be undertaken using a computer.

16.3.1.3 Correlation of loading and displacements The usefulness of the matrix notation in describing in a compact manner the behaviour of a system is illustrated using the stiffness matrix k (3 3) of the structural model shown in Figure 16.21. From the physical definition of k in Section 16.3.1.1, it is easily seen that the three forces (P1, P2, P3) required to be applied to the three storey levels in order produce the corresponding displacements (1, 2, 3) can be obtained from the equation P ¼ k ðaÞ where matrix P (3 1) represents the forces (P1, P2, P3), and matrix (3 1) represents the displacements (1, 2, 3). According to the introduced ‘matrix language’, it can be seen that if the loads P are known and the displacements caused by them are sought, these can be obtained by multiplying both sides of the above equation by the inverse k1 of the stiffness matrix k, i.e. k1 P ¼ k1 k so that ¼ k1 P Clearly, this result represents nothing more than the solution of the linear system (a), without this compact matrix formulation helping the solution itself. P1

m1 ∆1

P2

P3

m2 ∆2

m3

P=k·∆

∆3 Structural web

Figure 16.21 Correlation of loading and deformation in a multi-degree system

559


16.3.1.4 Multi-storey systems At this point it is useful — for later use too — to describe through the compact matrix methodology the overall response of a multi-storey spatial system under horizontal loads, according to the previous examination in Chapter 15 (Figure 16.22). From Chapter 15, we know that the structural web of a multi-storey spatial system consists of elements of plane stiffness. Each of the plane elements is represented by its own stiffness matrix Si — as previously defined in Section 16.3.1.1 and shown in Figure 16.23. The overall stiffness matrix K of the order 9 9, referring to the whole system, relates the loads Px and Py (order 3 1) to the developed displacements x, y and ! at each diaphragm. Matrix K is constructed as follows: 3 2 K1 K2 K3 7 6 K ¼ 4 K2 K4 K5 5 K3 K5 K6 where K1 ¼ K2 ¼ K3 ¼ K4 ¼ K5 ¼ K6 ¼

P P P P P P

Si cos2 i Si sin i cos i Si ei cos i Si sin2 i Si ei sin i Si e2i

all matrices being of the order 3 3 (Stavridis, 1986). Structural web P1, ∆1

m1

y x m2

P2, ∆2

y x

Structural web

P3, ∆3

y x

Figure 16.22 Layout of a multi-storey system

560

Structural web

m3

Dynamic behaviour of discrete mass systems ∆1 = 1 S11

S12

S13

∆2 = 1 S21

S22

S23

∆3 = 1 S31

S32

S33

The lateral stiffness matrix is assembled from the reactions developed at the lateral (imaginable) supports of the plane element

Figure 16.23 Determination of the matrix of lateral stiffness for a plane element

The correspondence with the relevant expressions in Section 15.3.3 — at least with reference to a single storey — is absolute. Thus, the equation between loading and deformation of the system is written as follows: 9 8 9 2 3 8 K1 K2 K3 > > < x > = < Px > = 7 6 Py ¼ 4 K2 K4 K5 5 y > > > : ; : > ; 0 K3 K5 K6 ! It should be noted that the matrices Px and x consist of the components in the x sense of the magnitudes (P1, P2, P3) and (1, 2, 3), respectively (see Figure 16.22). This is analogous for matrices Py and y too. The matrix ! consists of the diaphragm rotations (see Section 15.3.1). It is, of course, clear that, given a multi-storey system with N diaphragms, the above equation allows the determination of the storey displacements, through inversion of matrix K of the order (3N 3N), as examined above, through use of suitable computer software. However, in the context of the examination of dynamic behaviour from a preliminary design point of view, of prime interest is the orthogonal layout of plane elements about the axes x and y. Then, the constituent matrices of the global stiffness matrix K are expressed as follows (see Section 15.3.4): P K1 ¼ Sx K2 ¼ 0

P K3 ¼ (Sx y) P K4 ¼ Sy 561


K5 ¼ K6 ¼

P P

(Sy xi) (Sx y2i ) þ

P

(Sy x2i )

Thus, the above condensed equation may be split up as follows: Px ¼ K1 x þ K3 ! Py ¼ K4 y þ K5 ! 0 ¼ K3 x þ K5 y þ K6 ! If the plane elements layout leads to nullification — even approximately — of the matrices K3 and K5, which should be the basic intention according to Section 15.3.4, then, from the last equation, ! ¼ 0, i.e. the diaphragms do not rotate, thus not subjecting plane elements to an additional response. Recall that the horizontal loads are acting on the centroid of each diaphragm, to which the coordinates x and y are also referred. By taking into account the above, P Px ¼ K1 x ¼ ( Sx) x and P Py ¼ K4 y ¼ ( Sy) y The above equations mean that, in the examined case, the displacements in both the x and y senses are independent of each other. They are obtained by applying the loads Px and Py on the two separate systems, respectively, which are formed through coupling of all the respective plane elements in each sense, carrying at each level the corresponding mass (Figure 16.24). This coupling is accomplished at each level through hinged undeformable bars, as shown in the figure. The above conclusion is of particular practical importance for both the assessment of the dynamic characteristics of the multi-storey

P1

m1

P1

m1

∆1 P2

m2

∆1 P2

m2

∆2 P3

m3

∆2

P3

m3

∆3

P = (∑ sx) · ∆ Sense x–x

∆3

P = (∑ sy) · ∆ Sense y–y

In the case where the diaphragm rotation may be ignored, the multi-storey system acts in each sense like a plane structural web

Figure 16.24 Equivalent plane system in both directions

562

Dynamic behaviour of discrete mass systems P1

m1

P1

m1 ∆1

∆1

P2

m2

P2

m2

∆2

P3

∆2

m3

P3 ∆3

m3 ∆3

Structural web P=k·∆ The ‘rectilinear’ model is used merely to represent a structural web having a stiffness matrix [k] The two structural systems exhibit a totally different behaviour under lateral loads

Figure 16.25 Simulation of a multi-degree frame with a rectilinear model

mass system and the determination of the equivalent static loading because of seismic excitation, as will be examined later in this chapter.

16.3.2 Free vibration 16.3.2.1 Plane systems The study of multi-degree systems will be continued on an illustrative basis for the three-degree model, without thereby limiting the general validity of any conclusions for systems with more degrees of freedom. At this point, it is noted that the ‘rectilinear’ structural web in Figure 16.25 should not necessarily be considered merely as a straight member carrying at some points the masses m1, m2 and m3 but as a structural system described by its stiffness matrix k. Consequently, the model may also be considered to represent a plane frame, such as that shown in the figure, and having the same stiffness matrix k (see Figure 16.23). This clarification is necessary because the lateral shift behaviour of a vertical cantilever is quite different from that of the frame, as has already been pointed out in Section 6.4.3 (see Figure 6.16). First, it is assumed that the mass mi, acted on by an external force Fi(t) in the sense of its own degree of freedom, i.e. horizontally, undergoes a displacement xi(t). The positive sense for all the forces and displacements is considered common (Figure 16.26). Both the set of forces Fi(t) and the set of displacements xi(t) may be represented by the column matrices F(t) and x, respectively. For ease of matrix manipulation only, the system of masses mi is represented by the square diagonal matrix m of order N N. In this matrix, the masses mi constitute the diagonal elements, with all the other elements being zero. It can be written as m¼Im 563

Structural systems: behaviour and design m1 Equilibrium of mass m2 S21 m2 · x¨ 2(t)

F2(t) m2

c2 · x˙ 2(t) S23

S2 = S21 + S23

Structural web

S2 balances the static force required to displace the mass by x(t) The whole of the static forces is presented by k · x(t)

m3

x(t)

Figure 16.26 Dynamic equilibrium in a multi-degree system under external loading

Just as in a single-degree system, a specific mass mi may be considered as being always in equilibrium under the following forces (see Figure 16.26): (1) the external force Fi(t) (2) the total of the elastic forces Si transmitted from the structural web to the mass, as a consequence of its deformational configuration x(t) (3) the inertia force mi €xi ðtÞ (4) the damping force ci x_ i ðtÞ. As forces (2), (3) and (4) have the opposite sense to the force Fi(t), for the equilibrium of each mass mi (see Figure 16.26), mi €xi ðtÞ þ ci x_ i ðtÞ þ Si ¼ Fi ðtÞ It is clear that the deformational configuration x(t) for the specific time instant t may be produced by a unique set of external forces depending on the stiffness of the structural web. According to the foregoing, this set of forces can be given by the expression k x(t) (column matrix N 1), and it can be seen that each of these forces equals the respective sum Si of the elastic forces acting directly on the mass mi. The simultaneous consideration of the equilibrium of all masses can now be expressed by the following matrix formulation: m x€ þ c x_ þ k x ¼ FðtÞ The damping coefficients ci are represented by the square diagonal matrix c. g As in the single-degree systems, it is expedient here to also seek the basic dynamic characteristics of the multi-degree system, through the examination of its undamped free vibration. Free vibration is evoked by imposing an arbitrary displacement at a certain mass, according to its corresponding degree of freedom (e.g. a horizontal displacement 3 ¼ ) and then leaving the system free to vibrate. 564


In this case, the global equilibrium condition is written as m x€ þ k x ¼ 0 By trying out the harmonic solution x ¼ ’ sin ! t, where the magnitude ! is, of course, unknown and, moreover, the column matrix ’ also consists of unknown quantities, and given that x€ ¼ !2 ’ sin ! t, the above equation can be written as (Tedesco et al., 1999) ½m1 k !2 I ’ ¼ 0 This equation is known as the eigenvalue equation or frequency equation of the system, and represents a linear system of N equations with respect to the N unknowns ’. According to the laws of linear algebra, in order for a non-zero solution to exist in a linear system having all of its right-hand side members equal to zero, the determinant of the coefficients of the unknowns must be equal to zero. This determinant, as can be concluded from the last equation, consists of a polynomial of Nth degree with respect to !2, for which a number of N values (roots) can be determined. It is, of course, obvious that for each eigenvalue !2 the corresponding solution of the above linear system does not simply consist of a well-defined set of values ’ but rather of an infinite group of values ’, where is an arbitrary number. It is clear that because of the computational effort required for the determination of the above magnitudes, these can only be obtained through the use of appropriate computer software. From a physical point of view, the above signifies that the N-degree system has N different eigenfrequencies !, and each one of them corresponds to a certain displacement configuration ’ for each degree of freedom, which is multiplied by an arbitrary factor, either positive or negative. This displacement configuration is called the eigenform or natural mode (Figure 16.27). The physical meaning of any of the eigenforms ’ is that m1

m1

m1

m2

m2

m2 M1

m3

1st eigenform [ω1]

K1 = M1 · ω12

M2

m3

K2 = M2 · ω22

2nd eigenform [ω2]

M3

m3

K3 = M3 · ω32

3rd eigenform [ω3]

In a structural web deformed according to some eigenform and then left free, all the masses are vibrating in phase under the corresponding eigenfrequency The whole system acts then as an ideal mass with an ideal stiffness

Figure 16.27 Eigenforms and dynamic characteristics of a multi-degree system

565


if it is somehow imposed on the system (i.e. through an appropriate application of horizontal forces) and the system is left free, then every mass of the system will vibrate like a single-mass oscillator under the same natural frequency !, which corresponds to the selected eigenform. An additional analytical property of the eigenforms ’ should be noted. For any pair of different eigenforms ’r and ’s the following is valid: ’Tr m ’s ¼ 0 ’Tr k ’s ¼ 0 which is known as the orthogonality condition of the eigenforms ’. The fact that the whole system may vibrate at any specific frequency of the N realisable eigenfrequencies !r exhibited for each mass amplitude (represented by the eigenform ’r) means that all masses are found at the same time either at the rest position or at their maximum deviation, and suggests the idea of considering an ideal (fictitious) mass Mr connected to an ideal (fictitious) stiffness Kr (see Figure 16.27). It is indeed found that for the corresponding eigenform ’r the above magnitudes may be determined as Mr ¼ ’Tr m ’r ¼

N X

mr ’2r

1

Kr ¼ ’Tr k ’r According, of course, to the characteristics of a typical single-degree system (Biggs, 1964): !2r ¼ Kr =Mr In this way, for a given multi-degree system, N single-mass oscillators may be considered, each with specific characteristics of mass, stiffness and frequency. The lowest frequency ! is called the fundamental frequency, and has more physical importance than the others, as will be shown later. More specifically, if an arbitrary horizontal displacement is imposed on any mass in any of the two systems shown in Figure 16.28, and the structure is then left free, each mass in the system will vibrate at a frequency that essentially coincides with the fundamental frequency. This fact, which may apply to any multi-degree system, leads to an important result that allows the determination of the fundamental frequency of the system, as will now be shown. If to every mass in a multi-degree system an arbitrary static force Fr of the same sense is applied, then the masses will be displaced by r (Figure 16.29), and, if the system is then left free, a harmonic motion for every mass with a natural frequency ! will be immediately established, whereby all masses take both their rest position and their maximum deviation r at the same time instant. According to the characteristics of the single-degree systems, the maximum velocity of every mass at its rest position be ! r, and, consequently, the kinetic energy of Pwill 2 1 m the masses as a whole will be r (! r) . 2 566

Dynamic behaviour of discrete mass systems m1 m1

F

F

m2

m2

m3 m3

Removal of the static force causes vibration under the fundamental natural period

Figure 16.28 The fundamental frequency

Clearly,Pthis kinetic energy should be equal to the total work done by the forces Fr, 1 which is 2 Fr r. Thus, the value of the fundamental frequency ! may be readily obtained as (Biggs, 1964) P F r 2 ! ¼P r mr 2r The computation is facilitated if the forces Fr are selected as equal (see Figure 16.29). g The damping action is logically expressed through the damping coefficient with regard to each eigenform. Usually, the value of applied to the first eigenform corresponding to

F1

m1 ∆1 ∆ = k–1 · F

F2

m2 ∆2

A = F1 · ∆1 + F2 · ∆2 + F3 · ∆3 B = m1 · ∆12 + m2 · ∆22 + m3 · ∆3 Fundamental natural frequency

F3

m3 ∆3

ω12 = A/B

Structural web It is more appropriate to select equal forces

Figure 16.29 Practical evaluation of the fundamental eigenfrequency

567


the fundamental frequency ! is that directly relating to the nature of the structural web material, in accordance with Section 16.2.2.2. For each additional eigenform, the value of may be considered to increase by 1%. For the experimental determination of the damping coefficient, as well as of the eigenfrequency of each eigenform, the same procedure may be followed as in the case of a single-degree system, i.e. by creating through appropriate forces the corresponding displacement configuration ’r and following the corresponding free vibration of the system (see Section 16.2.2.2). The measured eigenfrequencies for each eigenform should not differ essentially from those obtained theoretically, as explained above, provided that the damping coefficients do not exceed 10%.

16.3.2.2 Multi-storey spatial systems In the case of a multi-storey building with horizontal diaphragms arranged vertically over each other, and which are carried by the structural web consisting of vertical plane elements, as examined in Chapter 15, it becomes clear that the dynamic equilibrium of the diaphragm masses is related to the displacements x and y of each diaphragm, as well as to its rotation !. While displacements x and y activate the inertia forces m €x and m €y of the mass of each diaphragm, respectively, the rotation ! activates €) on the basis of the rotational moment of inertia the rotational inertia moment ( ! of the mass of the diaphragm with respect to its centroid. For the case of an equally distributed mass in the diaphragm, ¼ m (Ix þ Iy) where Ix and Iy are the moments of inertia of the diaphragm area with respect to the axes x and y, respectively. By taking into account the ‘expanded’ mass matrix (3N 3N), 3 2 m 0 0 7 ¼6 m 40 m 05 0

0

the eigenfrequency equation is analogous to that obtained previously (Stavridis, 1986): 1 K !2 I ’ ¼ 0 ½m The eigenfrequencies ! can be determined only through the use of a computer. For the case where the diaphragm rotations are negligible, even approximately, according to Section 15.3.4 the fundamental eigenfrequency of the whole system may be assessed with satisfactory accuracy on the basis of the expression for !2 given in Section 16.3.2.1, in both the x and y directions, according to the following procedure (Figure 16.30): . A plane formation is created separately for the x and y senses, consisting of the corresponding plane elements in each direction, as shown in Section 16.3.1.4. 568

Dynamic behaviour of discrete mass systems m1

F

m1

F

∆1

∆1 m2

F

m2

F ∆2

m3

F

∆2 m3

F ∆3

∆3

(F = 1)

∆ = F · [∑ (S)x]–1 Sense x–x

∆ = F · [∑ (S)y]–1 Sense y–y

A = F · (∆1 + ∆2 + ∆3) B = m1 · ∆12 + m2 · ∆22 + m3 · ∆32 ω2 = A/B

Figure 16.30 Practical determination of the fundamental eigenfrequency of a multi-storey system

. Unit loads are applied to the different levels of the structural web, for which the respective displacements are calculated. The fundamental eigenfrequency for the considered sense (x or y) is obtained by direct application of the last expression for !2 in Section 16.3.2.1. Numerical example The ground plan shown in Figure 16.31 applies for three consecutive storeys, each having a height of 3.50 m and a slab thickness equal to 25 cm. All of the frame girders have a width and depth equal to 30 cm and 80 cm, respectively, while all the columns have a section of 50/30 cm. The wall elements have a thickness of 20 cm. Considering as the effective mass m only that corresponding to the self-weight of the slab: slab self-weight ¼ 15.40 8.0 0.25 25.0 ¼ 770 kN m ¼ 770/9.81 ¼ 78.5 t

6.00 1.00

2.50

3.00

2.50

1.00

The distinction between the two plane systems is made as shown in Figure 16.30.

2.00 1.70

8.00

3.70

Figure 16.31 Plan view layout of vertical elements of stiffness

569


By applying horizontal unit loads at the corresponding levels of each plane system (F ¼ 1 kN), the following displacements result: . Plane system in the x direction: 1 ¼ 0.0477 mm, 2 ¼ 0.0298 mm, 3 ¼ 0.0117 mm . Plane system in the y direction: 1 ¼ 0.0773 mm, 2 ¼ 0.0455 mm, 3 ¼ 0.0161 mm According to the expression for !2 (see Section 16.3.2.1) . For the x—x sense: !2 ¼

1 0:0477 þ 1 0:0298 þ 1 0:0117 ¼ 344:3 s2 78:5 ð0:04772 þ 0:02982 þ 0:01172 Þ

! ¼ 18.50/s T ¼ 2 p/! ¼ 0.465 s . For the y—y sense: !2 ¼

1 0:0773 þ 1 0:0455 þ 1 0:0161 ¼ 213:1 s2 2 2 2 78:5 ð0:0773 þ 0:0455 þ 0:0161 Þ

! ¼ 14.60/s T ¼ 2 p/! ¼ 0.43 s

16.3.3 Forced vibration The response of a multi-degree system subjected to external forces is governed, as analysed in Section 16.3.2.1, by means of a system of differential equations, which ¨ller and Keintzel, 1984) has been expressed in a condensed matrix form as (Mu m x€ þ c x_ þ k x ¼ FðtÞ As in the case of free vibration, it is also found here that the multi-degree system behaves as a system of N discrete and independent single-mass systems, having a fictitious mass M, as well as a fictitious stiffness K, according to the expressions in Section 16.3.2.1 (Figure 16.32). Each such system is governed by the equation Mr €qr þ 2 r Mr !r q_ r þ Kr qr ¼ ’Tr FðtÞ which, for the usually encountered case where F(t) ¼ P f(t), i.e. when all loads exhibit the same variation in time, may be rewritten as X N ðrÞ T ’i Pi fðtÞ Mr €qr þ 2 r Mr !r q_ r þ Kr qr ¼ ’r P fðtÞ ¼ 1

Although this equation can be treated analytically, as shown in Section 16.2.3, the numeric solution obtained using suitable computer software is more practical, provided 570

Dynamic behaviour of discrete mass systems x(t) P1 · f(t)

x(t)

m1

m1

P2 · f(t)

x(t)

m1 m1

m2

m2

m2 m2 M1

P3 · f(t)

M2

φ1T · P · f(t)

M3

φ1T · P · f(t)

φ1T · P · f(t)

m3 q2

q1

m3 K1

1st eigenform x = q1 · φ1

m3

m3

K2

2nd eigenform x = q2 · φ2

q3 K3

3rd eigenform x = q3 · φ3

The sum of {x} yields the final displacements and the response of the structural web through the stiffness matrix [k]

Figure 16.32 Dynamic response of a multi-degree system under external loads

that the time steps used in the solution are smaller than 1/10th of the corresponding eigenperiod Tr ¼ 2p/!r. It should be noted that the determination of the displacements qr(t) of the N fictitious single-degree systems allows the calculation of the displacements x(t) of the masses in the multi-degree system, according to the equation x(t) ¼ q(t), where is a square matrix (N N), obtained from the juxtaposition of the N column matrices ’r (see Figure 16.32). Moreover, the fact that each eigenform may be considered to be multiplied by an arbitrary factor does not affect the resulting displacements x(t) at all, as this factor, according to the above equations, is eliminated.

16.3.4 Seismic excitation 16.3.4.1 Dynamic analysis The governing factor for the response of a system to seismic excitation is, just as in the single-degree case (see Section 16.2.5), the accelerogram €us ðtÞ, expressed as €us ðtÞ ¼ a fa ðtÞ where a represents the maximum-occurring ground acceleration (Figure 16.33). It is obvious that the inertia mass forces are caused by their absolute acceleration, whereas the elastic as well as the damping forces acting on the masses depend only on their relative displacements to the ground. Given that no external forces are acting on the system, the condensed equation for dynamic equilibrium according to 571

Structural systems: behaviour and design m1

m1

m2

m2

m3

m3

m1 · a · fa(t)

m2 · a · fa(t)

m3 · a · fa(t)

Fixed support üs(t) = a · fa(t ) Accelerogram (a: Maximum ground acceleration)

Figure 16.33 Seismic loading of a multi-degree system

the previous section is written as m x€ þ I a fa ðtÞ þ c x_ þ k x ¼ 0 This equation can also be written as m x€ þ c x_ þ k x ¼ m I a fa ðtÞ where I represents a column matrix containing N unit elements. Thus, the problem of seismic excitation is posed as a problem of forced vibration according to the previous section, the multi-degree system being supported on fixed ground and subjected to the external loads expressed by the right-hand side of the last equation (see Figure 16.33). According to the foregoing, the multi-degree system is ‘split’ into N single-mass oscillators, each one corresponding to a respective eigenfrequency ! with the accompanying eigenform ’. On this basis, the corresponding fictitious mass Mr may be determined as well as the fictitious stiffness Kr. The equation of motion for the rth ¨ller and Keintzel, 1984) eigenfrequency is written as (Mu Mr €qr þ 2 r Mr !r q_ r þ Kr qr ¼ ’Tr m I a fa ðtÞ and, dividing by Mr, €qr þ 2 !r q_ r þ !2r qr ¼ ’Tr m I

1 a fa ðtÞ Mr

This equation is identical to the equation of motion of the single-mass system in Section 16.2.5, which is repeated here: €x þ ð2 !Þ x_ þ !2 x ¼ a fa ðtÞ 572

Dynamic behaviour of discrete mass systems m1 · a · fa(t)

m1 Γi =

m2 · a · fa(t)

m2 M1

m3 · a · fa(t)

[∑ mi · ϕi] [∑ mi · ϕi2]

m3

Γ1 · M1 · a · fa(t)

qmax = Γ1 · [Sa(1)/ω12] K1

M2

Γ2 · M2 · a · fa(t )

qmax = Γ2 · [Sa(2)/ω22] K2

M3

Γ3 · M3 · a · fa(t)

qmax = Γ3 · [Sa(3)/ω32] K3

1st eigenform xm(1a)x = qmax · φ1

2nd eigenform xm(2a)x = qmax · φ2

3rd eigenform xm(3a)x = qmax · φ3

P (1) = k · xm(1a)x

P (2) = k · xm(2a)x

P (3) = k · xm(3a)x

Superposition is not valid

Figure 16.34 Seismic analysis of a multi-degree system

It is observed that the maximum ground acceleration a is now multiplied by the factor P ðrÞ m ’i 1 T ¼ P i ðrÞ

r ¼ ’r m I Mr mi ½’i 2 which is a pure number and expresses the degree to which the rth eigenform is participating in causing the response of the system (Figure 16.34), and is thus called the participation factor. Its value is less than unity, while the sum of all the N participation factors r is logically equal to 1. Usually, the first (fundamental) eigenform exhibits the greatest participation factor, which gradually decreases for the higher eigenfrequencies. According to the results for the single-mass oscillator, it is possible to determine the corresponding (qr)max through the acceleration spectrum for an earthquake, on the basis of the corresponding value SðrÞ a for the rth eigenfrequency, considering as maximum ground acceleration the value ( r a) instead of a. Hence, according to Section 16.2.5, 1 ð€qr þ €us Þmax !2 ð t 1 ! ðt Þ ¼ 2 r a ! fa ðÞ e sin ! ðt Þ d ! 0 max

ðqr Þmax ¼

that is, ðrÞ

ðqr Þmax ¼

Sa r !2r 573


Now, for each oscillator r, the mass displacements (xmax)r of the multi-degree system may be determined on the basis of the computed (qr)max (see Figure 16.34): ðxmax Þr ¼ ’r ðqr Þmax ¼

r SðrÞ a ’r !2r

Thus, a maximum deviation is assigned to each mass, corresponding to each of the N fictitious oscillators. Of course, the actual displacement of a specific mass cannot be deduced as the sum of the above ‘component’ maximum deviations (xmax)r on the basis of the N oscillators, because the corresponding (qr)max do not refer to the same time instant. Nevertheless, according to an acceptable probabilistic consideration, this mass deviation may be obtained as the square root of the sum of the squares of the above ‘components’. Considering now the determination of the maximum response in the structural web, it can be seen that for the realisation of the above displacements (xmax)r of the N masses, the action of those forces Pr is required, which are suggested by the stiffness of the structural web itself according to the equation Pr ¼ k ðxmax Þr ¼

r SðrÞ a k ’r !2r

The determination of the stiffness matrix k follows Section 16.3.1.1 It is thus possible to determine the N systems of external forces P for the maximum (real) deviations xmax corresponding to the N (fictitious) oscillators. However, as these forces do not correspond to simultaneously developed displacements, their superposition is not meaningful, and one has to use the previous treatment (the square root of the sum of their squares) in order to obtain an acceptable approximation for loading of the structural web.

16.3.4.2 Equivalent static loads The bulk of the calculations in Section 16.3.4.1 that take into account the contribution of all the eigenforms of the system in its seismic excitation are not especially manageable, particularly for preliminary design purposes. This inconvenience led to the search for a static loading acting on the masses of the system such that the resulting response approximates the ‘exact’ dynamic analysis. A practical suggestion to this end is the fact that, for a regular distribution of mass and stiffness in a structural web, it is the first eigenform that plays the governing role in the dynamic response of the web ¨ller and Keintzel, 1984). (Mu Thus, in seeking these equivalent static loads, it is assumed, as an approximation, that these may be represented by the inertia forces H to which the masses in the system are subjected, if only the first eigenform (r ¼ 1) is taken into account. According to Section 16.3.4.1 and Figure 16.35 (see also Section 16.2.5), 2 H ¼ m ½ð€ xÞmax þ I a ¼ !2 m ðxÞð1Þ max ¼ ! ðq1 Þmax m ’

574

Dynamic behaviour of discrete mass systems Equivalent static loading 1st eigenform xm(1a)x = qm(1a)x · φ1 m1 m · ω2 · x (1) 1 m ax

m1 A · m · ϕ 1 1

m2

m2 · ω2 · xm(1a)x

m2

A · m2 · ϕ2

m3

m3 · ω2 · xm(1a)x

m3

A · m3 · ϕ3

m1 A · m · h h 1 1

h1

h2

m2

m3

Ah · m2 · h2

Ah · m3 · h3

h3

Γ1 · [∑ m i · ϕ i ] · Sa

V = [∑ m i ] · Sa Base shear ∑ mi A= ·Sa ∑ mi · ϕi

V = [∑ m i ] · Sa Base shear ∑ mi Ah = ·Sa ∑ mi · hi

Figure 16.35 Determination of equivalent static loading in a multi-storey system

On the basis of the expression for (q1)max in Section 16.3.4.1, P m ’ H ¼ m ’ P i 2i Sa mi ’i

P Looking now for a more convenient expression for these forces, their sum ( H) over the whole height of the structural web is considered. It is clear that this force constitutes the sum of the actions to which the masses are subjected, transmitted directly to the ground. It is thus found that P X ð mi ’i Þ2 Sa H¼ P mi ’2i This expression reveals that the factor P X ð mi ’i Þ2 P ¼

ðmi ’i Þ mi ’2i according to Section 16.2.5, represents a mass m which, given that the value of for the first eigenform is near unity, may be considered as approximately equal to the sum M of the masses in the system, i.e. X X mi ¼ M m ¼ ðmi ’i Þ ffi Thus, X

H ¼ V ¼ M Sa 575


This V force is called the base shear, and is used in the exact same sense as in Section 16.2.5 (see Figure 16.35). Now, the expression of the equivalent static loads H is written on the basis of the above as P 1 ð mi ’i Þ2 V P Sa ¼ m ’ P H¼m’P mi ’ i mi ’ i mi ’2i and, for each mass separately, m ’ Hi ¼ V P i i mi ’i This simpler expression, though less accurate than the previous one, has nevertheless been adopted in most of the regulations (see Figure 6.35). However, in the search for an even simpler expression for the equivalent static loads for preliminary design purposes, a further simplification is undertaken, namely the consideration of the first eigenform as a straight line. Thus, by using the level heights hi instead of the eigenform coefficients ’, the following final expression is obtained: m h Hi ¼ V P i i mi hi

(see Figure 16.35)

More specifically, for a multi-storey building — as considered in Section 16.3.1.4 and from a dynamic viewpoint in Section 16.3.2.2 — which is subjected to a given ground acceleration, in order to determine the equivalent horizontal static load for each level, the following procedure should be followed: (1) On the basis of the fundamental eigenfrequency for both the x and y senses, calculated according to Section 16.3.2.2, as well as of the maximum adopted ground acceleration, first the value Sð1Þ a and then the base shear V are determined from the acceleration spectrum used separately for both senses in the two discrete plane systems. (2) The corresponding equivalent static loads are obtained through direct application of one of the two last-obtained expressions. In the case where the eigenform values ’i are used, these may be considered to be the already determined displacements for the equivalent plane system under horizontal unit loads, applied in order to assess its fundamental frequency, as described in Section 16.3.2.2. It is clear that for a uniform distribution of masses, these static loads increase continuously for higher storeys. (3) On the basis of the acting forces on each diaphragm of a storey, the response of the vertical — as well as the horizontal — elements of the multi-storey system may be readily assessed, according to Sections 15.3.1 and 15.3.2. Numerical example Continuing the example in Section 16.3.2.2, on the basis of the design spectrum in Eurocode EC8 and for soil class B, both for the sense x—x with a fundamental period T ¼ 0.34 s and for the sense y—y with a fundamental period T ¼ 0.43 s, a maximum spectral acceleration Sa ¼ 2.50 is determined. Assuming a maximum ground acceleration 576


equal to 0.16g, the following total seismic force (base shear) is obtained: V ¼ 3 78.5 0.16 9.81 2.50 ¼ 924 kN Considering an earthquake in the x direction, the equivalent static loads acting on the centroid of each of the three diaphragms (slabs) are estimated on the basis of deviations ’. According to Section 16.3.2.2: X mi ’i ¼ 78:5 ð0:0000477 þ 0:0000298 þ 0:0000117Þ ¼ 0:007 t Thus, the loads acting in the x direction are (see Figure 16.35) P1 ¼ 924 78.5 0.0000477/0.007 ¼ 494 kN P2 ¼ 924 78.5 0.0000298/0.007 ¼ 309 kN P3 ¼ 924 78.5 0.0000117/0.007 ¼ 121 kN Considering an earthquake in the y direction, the equivalent static loads are analogously estimated. According to Section 16.3.2.2: X mi ’i ¼ 78:5 ð0:0000773 þ 0:0000455 þ 0:0000161Þ ¼ 0:011 t The corresponding forces acting in the y direction are (see Figure 16.35) P1 ¼ 924 78.5 0.0000773/0.011 ¼ 512 kN P2 ¼ 924 78.5 0.0000455/0.011 ¼ 302 kN P3 ¼ 924 78.5 0.0000161/0.011 ¼ 110 kN i.e. essentially with the same distribution as the x sense. The assessment of the response of a multi-storey system under the above loads, for both the x and y senses, may be deduced by following the procedure described in Section 15.3.1.

16.4

Approximate treatment of continuous systems

As pointed out at the beginning of this chapter, the load-bearing structures have de facto a continuous mass distribution. However, the corresponding inertia forces that have to be considered in order to express the equilibrium of each element leads to unmanageable differential equations, which excludes their use in structural design. However, the discrete systems examined so far, although not corresponding absolutely to physical reality, are nevertheless able to express the dynamic behaviour of any structure, by considering its mass lumped in as many places of the structural web as desired and by following the previously described procedures. The total (continuous) mass m of a beam, for example, may be partitioned into five lumped equidistant masses (m/5) distributed over the structural web, thus creating a system with five degrees of freedom, if only those inertia forces, which correspond to transverse displacements, are counted (Figure 16.36). 577

Structural systems: behaviour and design P · f(t) P · f(t)

P · f(t)

P · f(t )

P · f(t )

P · f(t) m

P · f(t) m m

m

m

m

m

m

m

m

Structural web

Structural web keff: the total load inducing δ = 1 keff: the total load inducing δ = 1

δ=1

δ=1 EQUIVALENT OSCILLATOR Fe = λP · ∑ P

λM = (∑ ϕ2)/N λP = (∑ ϕ)/N

[ · f(t)]

Fe = λP · ∑ P [ · f(t )]

Me = λM · ∑ m

Me = λM · ∑ m

Ke = λP · keff

Ke = λP · keff

Figure 16.36 Equivalent substitution of a continuous structure through a single-mass system

In the same sense, the mass m of a plate may be divided in 25 equidistant masses (m/25), corresponding to a system with 25 degrees of freedom based only on the transverse deflections (Figure 16.36). However, the analysis of the above discrete systems requires the use of appropriate computer programs. This does not directly meet the needs of a preliminary design with respect to basic design parameters, such as, for instance, the fundamental frequency of the system, or its maximum deformation, or even its maximum acceleration under specific dynamic loads. Thus, it is appropriate — wherever possible and at least for the usual cases — to pursue the equivalent dynamic behaviour of the continuous systems through single-mass systems exhibiting the same frequency as the fundamental frequency of the examined system. This idea is based on the fact that, in most practical cases, the fundamental eigenfrequency of a system plays the governing role in its dynamic behaviour, as pointed out in Section 16.3.4.2. Figure 16.36 shows the ‘substitution’ of a beam and plate by a corresponding ‘equivalent’ single-mass system, for which the mass, stiffness and equivalent dynamic loading are to be determined (Biggs, 1964). The determination of the above parameters is based on the fact that, as has been discussed in Sections 16.3.2 and 16.3.3, the dynamic analysis of an N-degree of freedom system goes back to the analysis of N independent single-mass oscillators, 578


and, as explained above, it is the oscillator with the lowest frequency that can most successfully represent the whole system. As described in Section 16.2.2, the mass Me of this oscillator can be expressed as Me ¼ ’T1 m ’1 ¼

N X

ms ’2s ¼ M

s¼1

N X

ms

s¼1

where ’s are the deviation values of the first eigenform of the N-degree system and, in the case of a uniformly distributed total mass M in the discrete model, the mass factor M is PN ’2 M ¼ s ¼ 1 s (dimensionless number) N Thus, the mass Me of the equivalent oscillator is Me ¼ M M, expressed as a part (M < 1) of the total mass M of the system (see Figure 16.36). Assume, now, that the discrete system is subjected to the dynamic loading F(t) ¼ P f(t) where matrix P consists of the specific loads acting on the N masses of the system. According to Section 16.3.3, the loading term of the single-mass ‘component’ oscillator with the lowest frequency (i.e. the ‘fundamental oscillator’) of the N-degree system is ! N X T Ps ’s fðtÞ F1 ¼ ’1 P fðtÞ ¼ s¼1

Considering that load Fe f(t) will act on the equivalent oscillator, then Fe ¼

N X s¼1

Ps ’s ¼ P

N X

Ps

(see Figure 16.36)

s¼1

and, in the case of an equally distributed total load to the N masses of the discrete system, the load factor P will be PN ’ P ¼ s ¼ 1 s (dimensionless number) N It isPclear that in the case of a uniform distribution of load p (kN/m) on a beam of length L, Ps ¼ ( p L), whereas forPa uniformly distributed load q (kN/m2) on an orthogonal slab with dimensions (a, b), Ps ¼ (q a b). It should be noted that the basic aim of seeking an equivalent single-mass oscillator is to determine the kinematic behaviour (displacement, velocity, acceleration) of a specific critical point of the initial system, and this, for preliminary design purposes in the case of a single span beam or a plate, is obviously the central point of the structure. To this end, and extending the initially adopted notion of stiffness in Section 2.3.8, the concept of the effective stiffness keff is introduced as the total load which, under a given and prescribed distribution, causes unit displacement at the reference point of the structure 579


— i.e. at its midpoint (see Figure P 16.36). It is, then, clear that the deflection at that point, due to the total load Ps, is, for the considered distribution (see Figure 16.37), P Ps ¼ keff For a uniform load distribution, in the case of a beam with length L, keff ¼ 76.8 EI/L3, whereas in the case of a simply supported square plate with side L, keff ¼ 271 EI/L2. Thus, the sought-after stiffness Ke of the equivalent oscillator results from the requirement that the transverse displacement at the considered reference point of the continuous system, due to the static loads Ps, should P be identical to the displacement of the equivalent oscillator under the load Fe ¼ P Ps: P P Ps P Ps ¼ ¼ keff Ke and, hence, Ke ¼ P keff

(see Figure 16.36)

The fundamental frequency f1 of the continuous system, being equal to that of the equivalent oscillator, is thus obtained as sffiffiffiffiffiffi sffiffiffiffiffiffi rffiffiffiffiffiffi 1 Ke 1 P keff ¼ f1 ¼ 2p Me 2p M M where the factors P and M refer to a uniform distribution of load and mass, respectively. The two tables in Figure 16.37 give the values of the mass and load factors P and M for various cases of beams with distributed or concentrated loads, as well as for perimetrically supported plates (Biggs, 1964).

16.5

Design for avoiding annoying vibrations

16.5.1 Human activities Apart from the seismic disturbance of multi-storey systems, which has been examined in Section 16.3.4, checking for transverse vibrations of horizontal load-carrying elements — such as beams or plates in normal use by humans — constitutes a major design issue in several cases (Bachmann and Ammann, 1987). A pedestrian bridge, for example, or an office floor, or even a floor designed for sport or dance activities, should in no case undergo vibrations which are felt by the people in a disturbing way. Such vibrations occur when the frequency with which the foot strikes the floor lies near the fundamental frequency of the load-carrying structure. Within the framework of the preliminary design goals set out in this book, and without the need in the present context to dynamically analyse the human activity in itself, footbridges should be designed with a fundamental frequency outside the range 1.6—2.4 Hz and, if possible, not lower than 4.50 Hz. For floors intended for sport activities, this frequency should not be less than 6.8 Hz, whereas for dance floors the 580

Dynamic behaviour of discrete mass systems Type of loading and beam Length L

Mass factor λM

Load factor λP

Effective stiffness keff ( · EI/L3)

0.5

0.637

76.80

0.5

1.0

48

0.479

0.595

185

0.479

1.0

107

0.396

0.523

384

0.396

1.0

192

(a)

Uniform load on plate

a/b

Mass factor λM

Load factor λP

Effective stiffness keff ( · Eh3/12 · a2)

Plate perimetrically simply supported

1.0

0.31

0.45

271

0.9

0.33

0.47

248

0.8

0.35

0.49

228

0.7

0.37

0.51

216

a Plate thickness: h

0.6

0.39

0.53

212

0.5

0.41

0.55

216

Plate perimetrically fixed

1.0

0.21

0.33

870

0.9

0.23

0.34

798

0.8

0.25

0.36

757

0.7

0.27

0.38

744

0.6

0.29

0.41

778

0.5

0.31

0.43

866

b

b a Plate thickness: h (b)

Figure 16.37 Equivalent substitute parameters of single-mass systems for beams and plates

minimum value is 6.0 Hz. It is worth mentioning that, for the above uses, the maximum tolerable acceleration of the human body is about 5% of acceleration due to gravity g. Finally, in the case of office floors, where lower acceleration limits are required, a fundamental frequency of at least 7.50 Hz should be targeted (Bachmann and Ammann, 1995).

16.5.2 Vibrations induced by machines The running of machinery generally causes vibrations in the part of the structure in which it is installed, which may be felt by and annoy people standing near it. The installed machines produce continuous periodical sinusoidal loading, characterised by the maximum value of force F0 under a frequency f0, according to the data supplied by the vendor. As a measure of the acceptability level of accelerations felt by a person, the so-called sensitivity factor K can be considered, which is calculated using the empirical formula (Bachmann and Ammann, 1987) 0:80 f 2 K ¼ d pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 0:032 f 2 where d is the maximum amplitude of vibration (mm) and f is the fundamental eigenfrequency (Hz) of the structural part supporting the machine. Machines are supported on a floor structure either directly or through appropriate springs. The design goal is usually to limit the deflection max and/or the velocity vmax of the supporting structural part at values tolerable by the machine, as well as to limit 581

Structural systems: behaviour and design F m F0 t

M Direct bearing of machine T0

ω0 = 2π/T0

Dynamic loading imposed by the machine EQUIVALENT OSCILLATOR max Fe = λP · F0 To be checked: umax = Fe/(2 · ζ · Ke) u˙ max = umax · ω0 ümax = umax · ω02 Sensitivity factor K

Me = λM · M + m

Ke = λP · keff

Ke = ω02 · Me

Figure 16.38 Direct support of a machine in a continuous system

the sensitivity factor K to under 2.0, so that the floor vibration is tolerable by people standing nearby. g In the case where the machine is directly supported on the floor, the whole system may be described by a fictitious oscillator with a determinable mass Me and spring stiffness Ke, under an equivalent dynamic loading Fe (Bachmann and Ammann, 1987). This oscillator exhibits a damping coefficient identical to that of the supporting structure of the floor (Figure 16.38). The mass Me of the equivalent oscillator is determined as the sum of the equivalent vibrating mass M M of the supporting structural part (usually a slab) having the mass M and the mass m of the machine itself. So, Me ¼ M M þ m The spring stiffness Ke is expressed according to Section 16.4 through the effective stiffness keff of the supporting part, multiplied by the factor P (see Figure 16.38). If the fundamental eigenfrequency f1 of the supporting structure being calculated, according to Section 16.4, sffiffiffiffiffiffi rffiffiffiffiffiffi 1 P keff f1 ¼ 2p M M is lower than the operation frequency f0 of the machine, instead of calculating the stiffness Ke of the oscillator as P keff, it is more useful to consider it equal to Ke ¼ !20 Me ¼ ð2 p f0 Þ2 Me 582


in order to take account of higher frequency values being applied to the supporting structural part, which can lead to resonance. Then, on the basis of the adopted damping coefficient for the structure, the determination of its maximum deflection umax — and consequently of the machine itself — is possible, according to Section 16.2.4, as umax ¼

1 Fe 1 P F0 ¼ 2 Ke 2 Ke

which may be directly compared to the maximum max supplied by the vendor. Moreover, the maximum velocity and the maximum acceleration developed can be checked, according to Section 16.2.2.1, as u_ max ¼ umax ! ¼ ð2 p f0 Þ umax and €umax ¼ umax !2 ¼ ð2 p f0 Þ2 umax respectively (see Figure 16.38). However, of decisive importance for the acceptance (or not) of the dynamic behaviour of the system is the value of the sensitivity factor K, according to the foregoing formula: 0:80 f02 ffi K ¼ umax qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 1 þ 0:032 f02 This factor should not exceed a value of 2.0. g If the above solution does not lead to acceptable results, then, in order to limit the response of the supporting structure, the machine should be placed on springs with appropriately selected stiffness. The two masses, of the machine and of the structure, are separated as shown in Figure 16.39, thus constituting a two-degree system consisting of an ‘upper’ mass mM supported through a spring having stiffness kM on the ‘lower’ mass mS, being in turn supported through a spring of stiffness kS on a solid base. The quantities mS and kS for the structure can be considered to be known, but the characteristics of the ‘upper’ system have to be determined, namely the machine mass mM and the spring stiffness kM. The ‘machine mass’ is not necessarily the mass m only but can also include any appropriately attached mass mB — e.g. a concrete pedestal — required to fulfil the dynamic requirements (see Figure 16.39). The eigenfrequencies f1 and f2 of the two-degree system may be directly determined through the expression (Bachmann and Ammann, 1987) 2 0 131=2 s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 1 k k þ k k k þ k 4 k k S M S M S A5 @ M þ M þ M f1=2 ¼ 4 mM mS mM mS mM mS 8 p2 It is generally ensured whenever possible, through appropriately selected values, that the two frequencies are clearly separate, e.g. by making one a multiple of the other. This 583

Structural systems: behaviour and design m Dynamic loading imposed by the machine (M)

F

kM (S)

F0

mS To be determined kS

t

m

(M)

mB

T0

kM

Possibly required additional attached mass mB

(S) To be determined

mS kS

mM = m + mB

Fg

max F0

mM Unknown (to be determined)

kM

mM

mS

To be checked: umax Factor K

kS mS Known

kM

kS Fg

Two-degree system

If mass and spring stiffness lead to frequencies approximately equal to those of the two-degree system then the decoupling of the two oscillators is possible

Figure 16.39 Support of a machine in a continuous system through springs

allows the two systems to be decoupled, so that each mass can be considered as connected to a fixed base through its spring. Then, the frequencies f1 and f2 obtained from the above expression coincide — at least approximately — with the frequencies fM and fS of the individual oscillators, respectively. Thus, if the two requirements sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffiffi 1 kM 1 kM kM þ kS kM kM þ kS 2 4 kM kS 1=2 fM ¼ þ þ ¼ 2p mM mS mM mS mM mS 8 p2 mM and 1 fS ¼ 2p

sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sffiffiffiffiffiffi kS 1 kM kM þ kS kM kM þ kS 2 4 kM kS 1=2 þ þ þ ¼ mS mS mM mS mM mS 8 p 2 mM

lead — approximately — to technically plausible values for the sought-after quantities mM and kM, then the dynamic uncoupling of the two masses is feasible. Then, in the system [mM, kM], the maximum machine displacement as well as the maximum transmitted force on the base of the machine are first determined, according to Section 16.2.4, and, on the basis of this force, the system [mS, kS] is afterwards 584


analysed with respect to its relevant dynamic characteristics of interest, which concern the supporting structure itself. More specifically, by considering for the machine component — according to the above — its total mass (mM ¼ m þ mB), as well as the stiffness kM of its corresponding spring (see Figure 16.39), the corresponding eigenfrequency is obtained as sffiffiffiffiffiffiffi 1 kM fM ¼ 2p mM The mass of the equivalent oscillator with a damping coefficient will be equal to mS ¼ M M The spring stiffness kS is determined through the effective stiffness keff, according to Section 16.4, as kS ¼ P keff The eigenfrequency fS, representing the fundamental eigenfrequency of the supporting structure, is sffiffiffiffiffiffi 1 kS fS ¼ 2p mS If, now, the two systems are considered to be uncoupled, as previously explained, then, for the mass mM (machine), the maximum displacement umax as well as the maximum transmitted force Fg to the supporting base can be readily determined, according to Section 16.2.4, through the following equations ( ¼ 0): 1 F 1 umax ¼ ustat ¼ 0 (must be K1 K2 · (γ · h1 + γ · h2)

(a)

Pressures on the retaining front

h1

γ, ϕ

h2

γ′ = γ – γW

K1 · γ · h1 γW

The presence of water does not affect the angle of internal friction

K · (γ · h1 + γ′ · h2) · γW · h2

(b)

Figure 17.3 Earth pressures in a layered soil in the presence of a water table

17.2.2 Cohesive soils Cohesive soils (clay) constitute a consistent mass even in dry condition. They consist of particles of oblong or plate form with diameters ranging from 0.0002 mm up to 0.06 mm, having many more common contact points than non-cohesive soils. Given that the voids between the particles are much larger than the dimensions of the grains themselves, any water present cannot escape as easily as in non-cohesive soils under an applied external pressure, because of the resistance offered by the many contact points of the grains. Thus, settlement develops very slowly and may continue for greater periods of time (months or even years). A basic characteristic of cohesive soils is, apart from certain friction forces between grains, the development of cohesion forces as a result of mutually acting electrical forces between the particles. It should be pointed out that friction forces are decreased because of the existance of the water pressure u in the voids between the soil grains, called the pore pressure. The pore pressure, in the presence of a water table, is identical to the hydrostatic pressure at the point considered. Thus, the shearing strength basically consists of cohesion c (kN/m2), as well as of the friction resistance 0 tan ’. The stress 0 represents the actual normal pressure between the solid grains, which, of course, results from the externally imposed normal pressure reduced by the existing pore pressure u. The angle of internal friction ’ generally has more moderate values than in non-cohesive soils. The cohesion c in water-saturated soils has a value of the order of 10—40 kN/m2. Thus, ¼ c þ ( u) tan ’ 590

Supporting the structure on the ground Retaining front

Retaining front γW

Internal friction (angle ϕ) H Cohesion c

Possible decrease in angle ϕ Cohesion unaffected

2 · c · √K Relieving effect

2 · c · √K Relieving effect

K·γ·H Loading effect

γW · H K · (γ – γW) · H Loading effect

tan2 (45 – ϕ/2) ø K ø 1 – sin ϕ

Figure 17.4 Earth pressure in cohesive soils

The stress u is called the effective stress. It is clear that the shear strength of the cohesive soils depends, to a significant degree, on the presence of water in the pores, as well as on the cohesion c, which plays an important role, sometimes being more decisive than the internal friction. It should be pointed out that the presence of water may cause a decrease in the angle of internal friction ’, in contrast to what happens in non-cohesive soils. Moreover, it is noted that in a cohesive saturated soil, the additionally imposed pressure on the soil mass will not imply a direct increase in the contact pressure between the grains, as the extra pressure will be completely taken up by the water mass, without thereby improving the shear strength of the soil at all. The influx of water in cohesive soils generally affects their strength adversely, while a possible drying may give rise to pronounced deformations, accompanied by intense cracking, which will facilitate the further influx of water in the soil mass. In any case, however, under the same loading pressure the settlement of cohesive soils is greater than in the non-cohesive soils. Regarding the pressures on vertical retaining walls, the presence of the cohesion c definitely has a relieving role (Figure 17.4). Thus, the total horizontal force on a vertical front of height H consists of the force acting under the existing internal friction and the relieving force due to cohesion: pffiffiffiffi Eh ¼ H2 K=2 2 c H K It can be seen that the contribution of self-weight to the above force can be depicted by a triangular diagram, whereas the influence of cohesion is represented by a constant value over the height. The treatment of coefficients K is the same as in the case of noncohesive soils. Moreover, in the above expression for Eh, the possible presence of a water table within the soil mass must be taken into account by the appropriate reduction of the specific weight due to buoyancy, this effect being equivalent to the fact that the pore pressure u is equal to the exerted hydrostatic pressure. 591

Structural systems: behaviour and design Unsupported depth due to the existing cohesion Cohesion c Internal friction

H0

Cohesion c 4·c γ · √K

Internal friction

In case of the absence of internal friction, K = 1 In unsupported fronts the active pressure is suppressed by the cohesion

Figure 17.5 Effect of cohesion on the unbraced height of an open excavation

The above relation allows the estimation of the height H0 of an unsupported vertical cut, where Eh ¼ 0 must hold (Figure 17.5): pffiffiffiffi H0 ¼ 4 c= K In the absence of internal friction (’ ¼ 0), where, of course, K ¼ 1, the minimum height of an unbraced vertical cut of a cohesive soil is H0 ¼ 4 c/

17.3

Shallow foundations

It is reasonable, in order to provide foundation to a structure, to always seek its safe support at the shallowest level possible, since the cost of a foundation drastically increases with its depth. This, of course, presumes that the bearing capacity of the soil at the selected foundation level is adequate to permit the uptake of all the anticipated loads on the structure. Such a foundation is called a shallow foundation, and comprises the majority of foundations. However, if such a foundation is not feasible, a special structural layout must be provided so that the imposed loads are transmitted to lower soil layers with satisfactorily safe bearing capacity. The foundation is then characterised as a deep foundation. The cost of such a foundation constitutes a significant percentage of the total cost of the structure. What is presented below mainly concerns shallow foundations. However, some basic structural characteristics of deep foundations will be examined in the final sections of this chapter.

17.3.1 The deformational behaviour of elastic soil under vertical loads Soil may be considered as an elastic medium — possibly consisting of intermediate layers — being acted on by vertical loads on its free surface. Under these loads, soil behaves elastically. This assumption is supported by the results of geotechnical tests performed at the proposed site. Thus, each soil layer is characterised by its compression modulus Es, as well as by the corresponding Poisson ratio , determined by laboratory tests performed during the geotechnical exploration of the site. The compression modulus Es is the ‘elasticity modulus’ of the soil material, and represents a quantity that 592

Supporting the structure on the ground PLAN VIEW

α = b/a ξ = x/(a/2)

a

b

p0

Value of λ

x

ξ

x p0

w

w

α

0.0

2.0

6.0

10.0

14.0

1.0

7.04

2.06

1.00

0.39

0.28

0.50

9.60

3.69

1.31

0.78

0.55

0.20

13.19

6.72

3.00

1.88

1.34

2

Elastic subspace (ES, v)

w=

1–v · p0 · a · λ ES · 2π

Figure 17.6 Settlement of the elastic subspace due to a uniformly loaded orthogonal area

cannot be as precisely determined as in the case of, for example, steel. Its value corresponds to the geotechnical nature of the soil examined. Thus, non-cohesive soils exhibit values from 30 000 up to 300 000 kN/m2, depending on the grain size and the degree of consolidation (sand, 30 000—100 000 kN/m2; gravel, 70 000—300 000 kN/m2), whereas in cohesive clay soils the values of Es, clearly lower, are between 10 000 and 40 000 kN/m2. The Poisson ratio has much narrower variation margins and a significantly smaller influence on the deformation. Its order of magnitude for sand is about 0.3, while for clay it is about 0.4. Assuming that the examined soil mass, of theoretically infinite depth, has a uniform compression modulus Es throughout and its free surface is loaded by a uniformly distributed load p0 over a rectangular area (a b), then the settlement w developed at a distance x from its central point and in a parallel to the direction of side a, is (Figure 17.6) w¼

1 2 p a Es 2p 0

where qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1 Þ2 þ 2 þ ð1 þ Þ2 þ 2 þ ¼ ð1 Þ ln qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi þ ð1 þ Þ ln qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1 Þ2 þ 2 ð1 þ Þ2 þ 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1 Þ2 þ 2 þ ð1 Þ þ 2 ln qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ð1 þ Þ2 þ 2 ð1 þ Þ and ¼ b=a, ¼ 2x/a (Stavridis, 1997). g On the basis of the above result and in view of a later examination of soil—structure interaction, a strip of length L and constant width b is considered, being divided into 593

Structural systems: behaviour and design a = L/6

PLAN VIEW P=1

1

2

f62

3

4

5

b

6

L 1/(a · b)

f62

Elastic subspace

Figure 17.7 Influence of a loaded area on the settlement of a point lying on the soil surface

a number of n equal segments of length (L/n), as shown in Figure 17.7, whereby the mutual deformational influence of each of the above segments on each other is sought. Applying a unit vertical load P ¼ 1 consecutively on the midpoint of each segment and assuming it to be uniformly distributed over that segment with an intensity p0, p0 ¼

1 ðL=nÞ b

The settlements at the midpoint of all the other segments can be readily determined on the basis of the last relation. If the segments are numbered in ascending order from 1 to n, a number of n2 settlements fsr may be determined. fsr represents the settlement that develops in segment number s, when the segment number r is loaded with P ¼ 1. Thus, with a total number of segments equal to n ¼ 3 (Figure 17.8), loading, for example, PLAN VIEW (P1 = 1) P1

P2

P3

f11

f21

f31

* P1 1

2

3

(P2 = 1) f12

f22

f32

* P2

(P3 = 1) w1

w2

w3

f13

f23

f33

* P3 Figure 17.8 Determination of settlement due to specific concentrated loads

594

Supporting the structure on the ground

segment number 1, on the basis of the above relation, the quantities f11, f21 and f31 may be directly determined. Similarly, by loading segment number 2, the quantities f12, f22 and f32 may be calculated, and by loading segment number 3, settlements f13, f23, f33 are analogously obtained. The above deformations allow the determination of the settlement at all midpoints, if each segment is acted on by a certain load Pr. Thus, in this example (see Figure 17.8), the settlement w of each segment is, respectively, w1 ¼ f11 P1 þ f12 P2 þ f13 P3 w2 ¼ f21 P1 þ f22 P2 þ f23 P3 w3 ¼ f31 P1 þ f32 P2 þ f33 P3 The extension to a greater number of segments is obvious: w1 ¼ f11 P1 þ f12 P2 þ . . . þ f1n Pn w2 ¼ f21 P1 þ f22 P2 þ . . . þ f2n Pn ... wn ¼ fn1 P1 þ fn2 P2 þ . . . þ fnn Pn By assembling settlements ‘f ’ in a square matrix [F] (n n), which may be called the flexibility soil matrix, and further assembling settlements ‘wi’ and loads ‘Pi’ in the column matrices (n 1) {W} and {P}, respectively, the above expressions may be written in the compact matrix form (see Section 16.3.1): W¼FP The assumption of a constant compression modulus Es over the whole depth of a certain soil mass is not always realistic. The case of consecutive vertically arranged horizontal layers, each with a compression modulus corresponding to the thickness of the layer, is not unusual. On the other hand, even in a ‘homogeneous’ soil mass, the compression modulus increases approximately linearly with depth. Although for preliminary design purposes the consideration of an estimated ‘average’ compression modulus may be sufficient, a more ‘correct’ determination of the settlement w is presented below for the case of a layered soil profile (Stavridis, 2002). As a basic tool, the following result is used, regarding the settlement w developed at any corner of an orthogonal area (a b) under a uniform load p0, lying over a soil layer of thickness z, and a compression modulus Es. This layer is assumed to rest on an absolutely rigid base (Figure 17.9): wða; bÞ ¼ uðzÞ=Y where

a p0

!þ

1 !þ1 þ ln þ ln uðzÞ ¼ arctan þ !

þ1 !1 4 595

Structural systems: behaviour and design PLAN VIEW a w

Settlement at position S

w

w

p0

S

p0

b

S

b

w xS p0

xS + a/2 wS

w

Elastic subspace (ES, v)

S

b/2 S

w

p0

a

z

p0

S

xS – a/2

Figure 17.9 Settlement of an orthogonal area over a soil layer of finite thickness

and Es 1 2

¼ b=a

Y¼

¼ 2 z=a qffiffiffiffiffiffiffiffiffiffiffiffiffi ! ¼ 1 þ 2 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ 1 þ 2 þ 2 The settlement of a point S on the soil surface at a distance xs from the central point of the area (a b), in a parallel direction to side a, can be expressed as ws ¼ 2 [w(xs þ a/2, b/2) w(xs a/2, b/2)] where the quantities w(. . . , . . .) inside the brackets refer to the corner settlements of the orthogonal areas, [(xs þ a/2) (b/2)] and [(xs a/2) (b/2)], respectively, according to the above expression for w(a, b) (see Figure 17.9). By superposing the contribution of the compressibility of each soil layer on the development of the final settlement of point S (Figure 17.10), the following result is obtained: ws ¼ uðz1 Þ ð1=Y1 1=Y2 Þ þ uðz2 Þ ð1=Y2 1=Y3 Þ þ . . . þ uðzk 1 Þ ð1=Yk 1 1=Yk Þ þ uðzk Þ 1=Yk It is clear that the formulation of the foregoing mutual interaction of the adjacent strip segments, according to Figure 17.7, remains the same in the case of a stratified soil. 596

Supporting the structure on the ground S

1

E1, v1

2

E2, v2

3

E3, v3

Y = ES/1(1 – v 2) Contribution of layer 1 wS(z1, Y1)

z1

Contribution of layer 2 wS(z2, Y2) – w(z1, Y2)

z2

Contribution of layer 3

z3

wS(z3, Y3) – w(z2, Y3)

The settlement of point S results from superposition of each layer contribution

Figure 17.10 Contribution of the existing layers to the soil surface settlement

17.3.2 Rectangular spread footings In order to provide a foundation for the base of a column element, where a specific vertical force V1, a horizontal force H and a moment M1 must be transmitted, an orthogonal concrete footing with a specific area and depth is usually constructed, normally having its larger base dimension parallel to the plane of action of the moment M1 (Figure 17.11). The reason why such a footing is necessary is obvious: the direct support by the soil of a column element, developing at its base the sectional forces (V1, H, M1), is generally impossible because the soil cannot transmit to the element those stresses required for the ‘realisation’ of the above sectional forces. Thus, for the proper design of a footing the following points have to be taken into consideration: (A) The maximum developed contact stress with the soil, which should not exceed a safe value determined by the geotechnical study of the site, characterised as the allowable bearing pressure. V1

V1 M1

M1

H

H

V1 + G Ground

M1 + H · d

d

Self-weight G

H σ2

σ1

σ1

σ2 Induced soil stresses H < (V1 + G) · µ · 1.40 Safety against slip

Actions on the footing (equilibrium)

Figure 17.11 Footing equilibrium and soil pressures

597


(B) The absolute magnitude of the maximum soil settlement developed, as well as the angle of rotation of the footing. (C) The state of stress of the footing, which equilibrates as a free body under the acting sectional forces V1, H and M1, its self-weight G, as well as the acting contact pressures from the soil. The concrete footing must be reinforced and checked appropriately so that it can take up all the above forces. Of course, on the contact area between the footing and soil, a vertical force V ¼ V1 þ G, a horizontal force H as well as a moment M ¼ M1 þ H d are transmitted. It is under these forces that the above points (A) and (B) should be checked. The horizontal force H must be ‘covered’ by the ‘friction capacity’ (V ) between the soil and foundation. A safe value for the friction coefficient is 0.50 for cohesive soils and 0.60 for non-cohesive soils. However, by adopting a safety factor — usually taken as 1.40 — the following relation must be satisfied: H < (V ) (see Figure 17.11). Normally, the horizontal force H is not further involved in the design. 17.3.2.1 Soil pressures and settlements In a first attempt to estimate the developed soil pressures, the contact area of the footing is considered as a normal orthogonal section under the forces V and M. It is assumed that force V also includes the weight of the overlying soil mass. Force V is assumed to act on the centroid of the footing area (Figure 17.12). It is, of course, clear that moment M causes a shift e of the force V along its plane of action, being equal to e ¼ M/V. Thus, the pair of forces (V, M) is statically equivalent to force V alone, shifted, however, by distance e. As only compression contact pressures make sense over the foundation area, it is appropriate to check whether or not the shifted force V lies within the core of the corresponding section, i.e. whether it is e a/6. In that case, the maximum and the minimum contact (compression) stresses are, according to the bending formula (see Section 2.2.1), respectively, V M ða=2Þ V 6M þ þ 2 ¼ A If ab a b V 6M 2 min ¼ ab a b Of course, if the moment M is not acting, a uniform pressure ¼ V/(a b) is developed. In the case where the shifted force V exceeds the core limits — something that should generally be avoided — the above relations do not apply, since min becomes tensile. Then, the force V must be equilibrated by a triangular pressure diagram, having its maximum value on the nearest side to the force V, while its zero value to which it is extending will obviously lie within the foundation area (see Figure 17.12). As the distance c of the force V to the near side is c ¼ (a/2 e), the triangular diagram of the contact pressures extends over a length of 3 c, and the above maximum pressure results from the equilibrium requirement: max ¼

1 2

598

max ð3 cÞ b ¼ V

g

Supporting the structure on the ground a

M b

V

e

V

V

e = M/V

M

σmax The distribution is valid only if V falls within the core (e < a/6)

σmin The resultant of soil pressures is identical to the force V e

V

V

e = M/V

M

c 3·c

σmax V lies outside the core (e > a/6)

The resultant of the soil pressures is identical to the force V

Figure 17.12 Influence of the location of the resultant force on the resulting soil pressures

Although it is clear that all the above pressure distributions in each case satisfy the corresponding equilibrium conditions for the magnitudes V and M, it is easily seen that they lead to deformations which do not correspond to reality. More specifically, as the footing can be considered as essentially undeformable, the soil settlements, being identical with those of the foundation, should lie on a single plane exhibiting a certain rotation angle, because of the presence of the moment M. But even in the ‘simple’ case of the central application of the force V, M ¼ 0, it is clear that the resulting uniform pressure V/(a b) on the soil over a rectangular base does not lead — according to Section 17.3.1 — to a uniform settlement of all the points of the soil surface. Thus, for the case of the central action of the force V, after dividing the foundation surface into a certain number n of transverse strips of width a/n, the loads Pi for each strip have to be determined, which sum to the value V, on the one hand, and lead to the same settlement (w) for all strips, on the other hand (Figure 17.13). On the basis of the result of Section 17.3.1, the last requirement is expressed through the following equations: w ¼ f11 P1 þ f12 P2 þ . . . þ f1n Pn 599


b

P1

P2

P3

P4

P5

Those P values are sought which, having V as the resultant, induce at each strip the same settlement w

P6

V

a

a/6

Ground w

The bottom face of the foundation

Figure 17.13 More accurate determination of soil pressures and of uniform settlement

w ¼ f21 P1 þ f22 P2 þ . . . þ f2n Pn ... w ¼ fn1 P1 þ fn2 P2 þ . . . þ fnn Pn In addition, the following equilibrium condition must be valid: V ¼ P1 þ P2 þ P3 þ . . . þ Pn The n þ 1 unknown quantities P1, P2, P3, . . . , Pn and w, can be determined through the above linear system of equations. By selecting a relatively small number of strips, e.g. n ¼ 6, it can be ascertained that in an orthogonal footing under a central compression force V, forces P, instead of being equal, clearly increase towards the edges of the footing (Figure 17.14).

P1

P2

P3

P4

P5

P6

2.00 m V = 800 kN

2.00 m

2.0/6 m

193.9 kN

105.9

100.0

100.0

105.9

133.3 kN Assumption of uniform distribution

193.9 kN

Rigid footing

Elastically deformable soil

Figure 17.14 Determination of soil pressures under a uniform settlement

600


M P1

P2

P3

P4

P5

P6

V

Those P values are sought which, having as the resultants V and M, cause settlements on a single plane

a/6

Ground w ϕ The bottom face of the footing

Figure 17.15 More strict determination of soil pressures and deformation for an eccentric loading

The presence of the moment M introduces one more unknown, namely the inclination ’ of the resulting deformation plane of the soil surface (Figure 17.15). Thus, the above equations are modified as follows: w þ 0 ’ (a/n) ¼ f11 P1 þ f12 P2 þ . . . þ f1n Pn w þ 1 ’ (a/n) ¼ f21 P1 þ f22 P2 þ . . . þ f2n Pn ... w þ (n 1) ’ (a/n) ¼ fn1 P1 þ fn2 P2 þ . . . þ fnn Pn Moreover, the static equivalence of magnitudes V and M with the forces P1, P2, P3, . . . , Pn is required. Using the location of the force P1 as a reference point (see Figure 17.15), V ¼ P1 þ P2 þ P3 þ . . . þ Pn V (a/2 a/2n þ M/V) ¼ P1 (1 1) (a/n) þ P2 (2 1) (a/n) þ P3 (3 1) (a/n) þ . . . þ Pn (n 1) (a/n) It is clear that the n þ 2 unknowns P1, P2, P3, . . . , Pn, w and ’, can be determined through the above n þ 2 linear equations. Obviously, for practical setting out and solution of both the above linear systems, the use of appropriate computer software is essential. Moreover, it can be seen that the above equations are acceptable only for positive (i.e. compressive) values of the forces ‘P’. Otherwise, an iterative process must be followed. The developed soil pressures are obtained at each point by evaluating n P/a, whereas the maximum settlement is equal to w þ (n 1) ’ (a/n) (Stavridis, 2009).

17.3.2.2 Support on an elastic base A less reliable but clearly more practical alternative to describing the deformational behaviour of the soil is offered by the concept of the modulus of subgrade reaction, represented by the coefficient k. It is based on the assumption that the settlement of 601

Structural systems: behaviour and design k (kN/m2)

Neighbouring regions remain undeformed

Ground

w = 1.0 m

The loaded surface settles uniformly

The coefficient k does not constitute a soil constant as it depends on the loaded surface

Figure 17.16 Concept of subgrade modulus

any region on the soil surface under a distributed load p is uniform over its entire extent and is equal to p/k. The coefficient k (kN/m3) represents the pressure (kN/m2) required to be applied to a soil surface in order to produce a settlement of 1 m, thus expressing the stiffness of the soil surface (Figure 17.16). Although the concept of the subgrade modulus k is very convenient, being consistent with the concept of elastic support (see Section 2.3.7), it does not constitute a measurable soil property. More specifically, according to the above, the settlement of, for example, a rectangular surface under a pressure p is equal to p/k, i.e. it is independent of the surface itself. This is clearly not correct — as may easily be deduced from Figure 17.6 — since, under the same pressure p, an increase in the loaded area always leads to an increase in the corresponding settlement too (see Section 17.3.1). Thus, the determination of a subgrade modulus k has no meaning as a characteristic parameter of a given soil layer, as it depends directly on the specific loaded area. In addition, the assumption that all the points in the loaded area undergo the same settlement, while the surrounding region remains undeformed, is not realistic. However, the computational convenience, together with the opportunity for the qualitative and quantitative estimation of the structural behaviour offered by the use of the parameter k, often overweigh, in practice, its serious ‘natural disadvantages’. Certainly, a logical value of k for a specific orthogonal area may be obtained if the settlement w, at its central point, under a pressure equal to 1 kN/m2 is calculated according to Section 17.3.1, even by using the table in Figure 17.6. Of course, k ¼ 1/w. However, for a square surface with a side equal to a, it is approximately k Es/a

(kN/m3)

In this way, the soil under each single footing having an area A, may be simulated by a spring with a stiffness ks equal to ks ¼ k A, since, according to the above, a centrally 602


Ground A1

A2

A3

Subgrade modulus k

k · A1

k · A2

k · A3

Each spring stiffness is proportional to the corresponding footing area

Figure 17.17 Simulation of the soil deformability through translational springs

acting load P on the footing causes a settlement w ¼ (P/A)/k. Then, the whole structure may be considered as including all the springs corresponding to its footings, being itself naturally supported on a rigid base (Figure 17.17). g It can be seen that the induced rotation of a footing under the action of a moment M determined — as usual — by assuming a fixed column base implies a reduction in this moment, according to what has been discussed in Sections 3.2.9 and 3.3.6 regarding indeterminate structures. This footing rotation and the ensuing moment reduction needs to be assessed, since it will lead to more favourable soil stresses. Thus, whether the resilience of the footing can be described by a rotational spring with a specific elastic rotation flexibility f’, as defined in Section 2.3.7.2, i.e. as the developed footing rotation under the action of a unit moment, will be examined. Obviously, such a moment should be accompanied by a vertical load V, in order to ensure the development of contact pressures over the entire footing (Figure 17.18). In the case of a rigid footing, transmitting the sectional forces V and M over its bearing area A and, provided that only compressive pressures are developed over the entire foundation surface — meaning that the shifted resultant V falls within the core of the rectangular area — then, on the basis of a subgrade modulus k and according to Section 17.3.2.1, the settlements s1 and s2 at its two edges will be s1 ¼

max V M ða=2Þ þ ¼ Ak If k k

s2 ¼

min V M ða=2Þ ¼ Ak If k k

ðIf ¼ b a3 =12Þ

Consequently, the rotation ’ of the footing will be ’¼

s1 s2 M ¼ If k a 603

Structural systems: behaviour and design V

M

Ground

kϕ = If · k

ϕ = M/(I · k)

Soil simulation by a rotational spring is only possible if the shifted resultant V falls within the core of the footing

M b

If = b · a3/12

V

a

Figure 17.18 Conditions for the soil simulation through a rotational spring

This expression for ’ allows the determination of the coefficient of elastic rotation f’ — according to Section 2.3.7.2 — or, equivalently, of the stiffness coefficient with respect to rotation, k’ ¼ 1/f’, simulating in this way the soil through a rotational spring (see Figure 17.18): f’ ¼

1 If k

k’ ¼ If k However, it should be pointed out that, while, on the basis of the previous more exact approach to soil deformability (see Section 16.3.2.1) through the compression modulus Es, the load eccentricity may exceed the core limits of the foundation surface and yet maintain the compression contact stresses everywhere, the adoption of a subgrade modulus k does not lead to such a consequence, as can be ascertained by means of the above expressions. When the resultant force V acts outside the core, the settlement s1, according to the above, is s1 ¼

604

max ¼ k

V

M 3 a=2 bk V


and the corresponding rotation ’ of footing will be ’¼

s1 ¼ 3c

V

a M 2 b k 9 2 V

It is clear that rotation ’ is not proportional to the acting moment — depending also on the value of the vertical force V — and, therefore, in this case the soil cannot be represented by a rotational spring. g On the basis of the resistance k’ offered by the rotational spring against the imposition of a unit rotation, the reduction of the moment M previously mentioned may be estimated as follows. The rotational spring introduces the developed moment to the column, according to the rotation of their common joint, which obviously rotates together with the spring. If an external moment is considered to act on this node so that its rotation vanishes, then this moment M must clearly be equal to that corresponding to a clamped column base (Figure 17.19). Thus, in order to make this moment disappear, since it does not actually

[M] Diagram [M ] based on fixed support

Ground M

Diagram [M]

External moment

M M

Zero rotation

M Superposed on previous [M]

kϕ = If · k Stressless spring

Removal of external moment

M foot M foot

Ground

Rotation developed

Figure 17.19 Column—footing—ground interaction

605


exist, an equal and opposite moment M must be superposed, according to the process used in Section 3.3.1. The above moment M causes a rotation ’ to both the rotational spring and the column base, which may be determined from the equilibrium requirement of their common joint. Assuming the upper end of the column to be fixed (see Section 3.3.3.2): ’ 4 EI þ ’¼M f’ H Thus, the moment developed in the spring, i.e. the moment Mfoot taken up by the footing and transmitted to the ground, is Mfoot ¼

’ ¼ f’

M 4 EI f’ 1þ H

and is clearly reduced compared with M (see Figure 17.19). As can be seen, this result is identical to that based on Sections 3.2.10 and 3.3.7, namely that the moment M is distributed in proportion to the ‘adjacent’ stiffnesses: Mfoot ¼ M

1=f’ 1 4 EI þ f’ H

The moment Mfoot may be further reduced if beams are inserted that connect the lower end of the examined column with the neighbouring ones (Figure 17.20). In this case, too, the equilibrium of the joint of the rotational spring, which connects the column and the two connecting beams — with their far ends fixed — leads to the distribution of the moment M proportionally to the stiffness of the three adjacent members (see Figure 17.20). The corresponding portion Mfoot assigned to the footing surface is Mfoot ¼ M

1=f’ 1 ¼M X 4 EI 1 X 4 EI þ 1 þ f’ f’ L L

Generally it is useful to provide such ‘connecting ribs’ between the lower ends of the columns. However, these are indispensable in cases where an eccentric layout of the footing with respect to the column is constructionally imposed, e.g. along property lines where each column load V necessarily exhibits a certain eccentricity e (Figure 17.21). Of course, the isolated footings normally follow a symmetric layout with respect to the column section, in view of the possible shift of the column load V to either side, due to the possible occurrence of an alternating moment action M, e.g. during an earthquake. Returning now to the constructional eccentricity e of a footing, it can be seen that the column moment M deduced by assuming its base to be fixed must be — in the most adverse case — increased by V e and, consequently, the last expression for the 606


Connecting beams [M] Ground M

The presence of connecting beams reduces the moment on the footing

Diagram [M]

External moment

M

M

M

Zero rotation kϕ = I f · k Stressless spring

M foot Removal of external moment

M foot

Ground

Rotation developed

Figure 17.20 The structural action of the connecting ribs

moment taken up by the foundation surface becomes MþVe X 4 EI 1 þ f’ L where, in the denominator, the column member and the connecting rib (i.e. two members) have to be taken into account. It is clear that in evaluating the contact Mfoot ¼

V Connecting beam

(M + V · e) Ground e

M Aggravating diagram [M]

Mfoot Mfoot

Ground

Figure 17.21 Treatment of an eccentric footing

607


pressures according to the pair (Mfoot, V), the force V is assumed to act centrally (see Figure 17.21). g It is noted that the coefficient f’ may be determined either through its previous expression as f’ ¼ 1/(If k), which in the case of a square footing with side a, is f’ ¼ a/ (If Es), or, more ‘reliably’, through the consideration of the foundation soil as an elastic medium, as shown in Section 17.3.2.1. It should be pointed out here that the more resilient the foundation soil, the smaller is the moment Mfoot acting on the ground. This is structurally convenient, because a smaller Es generally means a lower allowable soil pressure but also a greater reduction in the ‘effective moment’ acting on the foundation surface. However, in the case of a large Es the coefficient f’ has a small value, and the reduction of the moment M þ V e is also small. On the other hand, the capacity of the soil to safely take up higher compression stresses is increased. 17.3.2.3 Dimensioning The footing is subjected as a free body to the soil pressures and to the transmitted sectional forces of the column base (equal and opposite to the corresponding reactions). Under the above forces, the footing acts like a slab loaded ‘upwards’. This loading causes bending, with the bottom face of the footing being in tension, whereas the compressed region is limited in practice by the corresponding width of the column. The tensile force, clearly equal to the compressive force, is provided by reinforcement distributed over the whole corresponding width of the footing (Figure 17.22). The dimensioning of the

B

Unreinforced footing

AS

b

>60° L

The tensile force is taken up by the concrete

M = B · (2σ1 + σ2) · L2/6

b d

M

σ2

AS

d

>45°

σ1 No shear reinforcement required

Figure 17.22 Design of a concrete footing

608


reinforcement refers to a section having as the width b the width of the column and as the depth d the depth of the footing itself (see Figure 17.22). It is clear that the maximum compressive force in the footing due to bending, as determined by the width b, must be kept below the corresponding allowable value. However, if a tensile force required by the soil pressures, for their transference to the column through the truss mechanism shown in Figure 17.22, can be provided by the concrete itself, then the footing may be constructed without reinforcement. This is the case if the inclination angle of the corresponding strut in the above truss model is at least 608 (see Figure 17.22). On the other hand, if the reinforced footing has been formed so that the resultant of the soil pressures outside the outline of the column can be directed to its edges under an angle greater than 458, then the established truss mechanism, as shown in Figure 17.22, transfers the soil pressures to the column without needing shearing reinforcement (Menn, 1990). If the above conditions are not met, then the footing must be checked not only in bending but also in punching shear, following the procedure in Section 11.5.2.

17.3.3 Foundation beams As was pointed out in Section 17.3.2, the foundation of columns on single footings should take into account not only the maximum developed soil pressures but also the settlements and rotations that are developed. It is clear that the settlement of each single footing is generally also affected by its neighbouring ones (see Section 17.3.1), except when the separation between them is more than about triple their dimensions. In this case, the footings settle independently of each other. However, it should be kept in mind that the founded structure is — almost always — a statically indeterminate one, and, as such, it is not sensitive to differential settlements of its supports. Of course, in the case of a relatively low allowable bearing pressure, the resulting footing dimensions could lead to footings lying either very near to each other or even overlapping (Figure 17.23). For all the above reasons, when designing the foundation of columns belonging to the same vertical plane as in, for example, a frame, it is frequently deemed useful to construct a beam with an appropriate width to serve as a common foundation for the

In relatively deformable soil the foundation beam layout is preferable

Figure 17.23 Incorporation of the foundation beam into the carrier system

609


group of columns under consideration. This beam, called the foundation beam, is loaded by the upward soil pressures and supported on its columns. Strictly speaking, the corresponding — multi-storey — frame containing the columns and the foundation beam is supported as a complete ‘closed’ structure on the ground (see Figure 17.23). In such a frame, the tendency of a column to settle down as a single element is prevented by the bending stiffness of the foundation beam, which shares this settlement between the other columns, clearly limiting in this way the differential settlements. Thus, a foundation beam with a depth adequate to ensure the necessary bending stiffness is clearly superior, as a foundation system, to that consisting of single spread footings. However, while the settlement behaviour of single footings is treated simply through appropriate spring supports — translational and/or rotational ones — supported on a fixed base, the treatment of a foundation beam is more complicated, as the problem consists of the determination of the soil pressures along the foundation beam itself, which implies examination of the interaction of the entire superstructure with the soil. It is clear that this interaction affects not only the response of the foundation beam but also that of the whole structure. The basic concepts and procedures which allow the investigation of the interaction between the soil and the frame with its foundation beam will be examined below.

17.3.3.1 Soil simulation according to the Winkler model The behaviour of a foundation beam resting on an elastic base may be approached through the concept of the subgrade modulus k, as defined in Section 17.3.2.2. According to this concept, assuming that a segment of a foundation beam undergoes a settlement w implies that the soil is acted on by a pressure psoil equal to psoil ¼ k w, which, representing the resistance offered by the soil to the imposition of the above settlement, is obviously acting on the beam segment itself too. Expressing this pressure through the distributed load p (kN/m) along the foundation beam of width b, psoil ¼ p/b ¼ k w and, consequently, p ¼ K w, where K ¼ b k — expressed now in kN/m2 — represents the subgrade coefficient in the longitudinal direction of the beam (Figure 17.24). Thus, the soil is simulated by a continuous distribution of springs, having the property of exhibiting for a uniform load p (kN/m) over some length a uniform settlement equal to w ¼ p/K, whereas outside this region no deformation occurs. The foundation beam having rigidity EI, under the distributed load q(x) is deflected by w(x). The resistance per unit length offered by the beam to this deformation equals EI (dw4/dx4), while that offered by the soil equals K w, and, obviously, both resistances balance the externally applied load q(x). Hence (see Figure 17.24), dw4 þ K w ¼ qðxÞ dx4 This differential equation, known as the Winkler equation, permits, through its closed solution, the determination of analytic expressions for the response of a beam segment embedded in an ‘elastic soil’ of the type examined, thus making possible the evaluation EI

610

Supporting the structure on the ground ∆x p

w = p/K

For beam width b:

K (kN/m2) K=b·k

x q(x)

Beam cross-section

Surface of elastic base Beam

b

w(x)

K·w Resistance offered by the beam: EI · (d4w/dx 4) Resistance offered by the soil: K · w

Equilibrium q(x) = EI · (d4w/dx4) + K · w

Figure 17.24 Foundation beam on a Winkler foundation

of the response of the whole frame, including, of course, the foundation beam itself. This analytic facility related to the adoption of the subgrade coefficient K makes this method of evaluating the soil—structure interaction very convenient, particularly with the use of an appropriate computer program. The bending response of the foundation beam of a three-storey frame is shown in Figure 17.25, under a uniform girder loading of 80 kN/m. The outer and inner columns have cross-sectional dimensions of 40/40 and 60/40 cm, respectively. The girders have a uniform section of 40/80 cm, whereas the section of the foundation beam is 60/120 cm. The soil is assumed to have a compression modulus Es ¼ 10 000 kN/m2, while its subgrade coefficient K, according to the foregoing, is considered equal to (see Section 17.3.2.2) K ¼ 0.60 k ¼ 0.60 (10 000/0.60) ¼ 10 000 kN/m2 The results are obtained by using appropriate computer software.

g

The following points are again emphasised. The subgrade modulus k does not constitute in any case a soil constant. Ensuring consistency with its definition requires reference to a specific surface, which may be obvious for a footing of specific dimensions, but in the case of the foundation beam is rather unclear. Moreover, loading a single foundation beam with a uniform load causes a uniform settlement over the whole length of the beam, meaning the absence of any bending along the beam, which obviously cannot be valid. However, the concept of the subgrade coefficient k, although physically not 611


80 kN/m

80 kN/m

80 kN/m

80 kN/m

80 kN/m

8.0 m

8.0 m

3.50

3.50

3.50

54

1010

1105

267

267

1105

1010

54

1.20

[M] 1517 kN m

Figure 17.25 Bending response of the foundation beam of a frame on an elastic base

consistent — as pointed out in Section 17.3.2.2 — nevertheless allows an approach to the determination of the soil pressures and to the response of the whole frame. In addition, it should also be recognised that the development of a settlement at some location on the soil surface implies soil deformation outside the considered region, because of the stiffness of the loaded beam, compensating in some way for the weakness of the soil model. The adoption of a subgrade modulus k leads to the important concept of the so-called elastic length Ls of the beam (see Sections 9.7 and 12.3.1): rffiffiffiffiffiffiffiffiffiffiffi 4 4 EI Ls ¼ bk The concept of the elastic length applies directly in the case where the foundation beam — theoretically of an infinite length — is loaded by a concentrated load F, which may represent a column load. It denotes that length over which the bending response of the beam extends from either side of the load F (Figure 17.26). Considering the soil pressure p as constant over the total double elastic length, i.e. p b ¼ F/(2 Ls), and given that the bending moment at both its ends vanishes, it is found that the bending moment under the concentrated load resulting from either of the corresponding cantilevers represents a very good approximation to the bending moment of the elastically supported long foundation beam developed in practice at this location: M ¼ ðF=2Ls Þ L2s ¼ 0:25 F Ls 612

Supporting the structure on the ground F Elastic base

EI

Coefficient K [M] ~p · Ls2/2 Ls

Ls Elastic length F

Ls = (4 · EI/b · K)1/4

M = p · Ls2/2 p = F/(2 · Ls) All along the elastic length the pressure is assumed constant

Figure 17.26 Physical meaning of the elastic length

Using the example given in Figure 17.25, the concentrated force F acting on the midpoint of the foundation beam is F 3 (5 80.0 8.0/8) 2 ¼ 2400.0 kN and its elastic length is equal to qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 4 Ls ¼ 4 2:1 107 0:086=10 000 ¼ 5:18 m Then, according to the above, the bending moment at the location of the concentrated load is M ¼ 0.25 24 00.0 5.18 ¼ 3108 kN m. The discrepancy — on the safe side — from the previously calculated value of 1517 kN m is mainly due to the fact that the foundation beam is considered here to be free, i.e. not being prevented from deforming by the stiffness of the superstructure.

17.3.3.2 Soil simulation as an elastic medium According to the soil model examined in Section 17.3.1, the structural frame is supported with its foundation beam on an infinitely extended elastic subspace, either having a constant compression modulus Es with a Poisson ratio , or consisting of a certain number of horizontal layers, with a corresponding Es for each layer. According to precedents, a foundation beam of width b immediately poses the problem of the determination of the distribution of soil pressures psoil(x) along its length. These pressures expressed as a distributed load, p(x) ¼ psoil(x) b (kN/m), must fulfil the following two conditions: (1) they must be in equilibrium with the vertical loads of the structure, the foundation beam included (2) the deformation of the foundation beam must match exactly the surface deformation of the soil due to the load p(x). 613


Working model of the examined frame

P 01 P 02 P 03

P 08 P 09

Ground 1

2

3

4

5

6

7

8

9

Plan view of the foundation beam

Sought-after imposed settlements

s1

s2 s3

s8 s9

Developed self-equilibrating reactions R1 R2 R3

R8 R9

S i = (P i0 + R i ) The induced soil settlements must coincide with those imposed on the model

Ground

Figure 17.27 Frame foundation with a foundation beam on an elastic subspace

It should be clear that both of these conditions are satisfied by the Winkler model, with the only remark that, for the reasons mentioned earlier, the description of the soil surface deformation through the insertion of springs is not absolutely reliable. Thus, in order to take into account the soil behaviour as an elastic medium in a more consistent manner, the model shown in Figure 17.27 is considered. The oblong orthogonal contact surface of a foundation beam of length L is divided into a number n of strips of equal width (L/n), at the midpoint of which the resultant of the corresponding soil pressures is acting. The whole structure is assumed to be simply supported at the above n midpoints. These rigid supports under the loads of the superstructure develop the reactions P0i , which may be assembled into the column matrix P0 (n 1). It can now be seen that, for a certain group of imposed settlements s1, s2, . . . , sn, the developed reactions of the continuously supported structure R1, R2, . . . , Rn, will constitute a self-equilibrating force system summing to zero. The fully developed reactions due to the structure loads and the settlements will be V1 ¼ R1 þ P01 ; V2 ¼ R2 þ P02 ; . . . ; Vn ¼ Rn þ P0n 614


The above forces may be considered to be acting in the opposite sense also on the ground, and, thus, the solution of the problem consists of finding the imposed settlements s1, s2, . . . , sn on the fictitiously supported frame such that the induced soil settlements under the loads V1, V2, . . . , Vn are identical to them (see Figure 17.27). This is achieved by the following procedure (Figure 17.28) (Stavridis, 2002). Imposing on the first support 1 a unit settlement s1 ¼ 1, a set of self-equilibrating reactions (k11, k21, k31, . . . , kn1) is developed. Next, a unit settlement s2 ¼ 1 is imposed on support 2, leading to another set of self-equilibrating reactions (k12, k22, k32, . . . , kn2). Proceeding consecutively in the same way up to the last support n, a unit settlement sn ¼ 1 is imposed equally, resulting in the self-equilibrating reactions k1n, k2n, k3n, . . . , knn. If the set of settlements s1, s2, . . . , sn is now imposed on the continuously supported frame, the developed reactions R1, R2, . . . , Rn can clearly be expressed through the following relations: R1 ¼ s1 k11 þ s2 k12 þ s3 k13 þ . . . þ sn k1n R2 ¼ s1 k21 þ s2 k22 þ s3 k23 þ . . . þ sn k2n R3 ¼ s1 k31 þ s2 k32 þ s3 k33 þ . . . þ sn k3n Rn ¼ s1 kn1 þ s2 kn2 þ s3 kn3 þ . . . þ sn knn Assembling the forces ‘kij’ into a square matrix Ksup (n n) and further assembling settlements ‘si’ and reactions ‘Ri’ in the (n 1) column matrices s and R, respectively, the above expressions may be written in matrix form (see Section 16.3.1): R ¼ Ksup s Also taking into account the reactions P0 due to the structure loading, as previously mentioned, then the above expressions for the totally developed reactions ‘Vi’ in the fictitiously supported frame being assembled in the (n 1) column matrix V can be expressed in matrix form as V ¼ P0 þ R ¼ P0 þ Ksup s According to the above, what remains is the requirement from the forces V1, V2, . . . , Vn, if applied in the opposite sense to the ground, to produce the soil settlements s1, s2, . . . , sn (see Figure 17.28). For this purpose, the relation describing the deformational behaviour of the soil surface is recalled from Section 17.3.1: W¼FP where, now, W ¼ s and P ¼ V. This relation, after multiplying both sides by F1 , allows the following expression for the forces V: V ¼ F1 s By substituting the above relation, the following equation is finally obtained: ðF1 þ Ksup Þ s ¼ P0 615


s1 · k11

k21

k31

kn1 (Due to s1 = 1)

k32

(Due to s2 = 1)

s2 ·

kn2 k12

k22

Total reactions:

sn · k1n

k2n

k3n

Due to: s1, s2, … , sn

Imposed settlements: knn (Due to sn = 1)

REQUIREMENT

R1 R2 R3

Rn

P 01 P 02 P 03

P 0n

V1 V2 V3

Vn

s1, s2, … , sn V1 V2 V3

Vn

Ground R1 R2 R3

Rn

s1, s2, … , sn

Figure 17.28 Practical analysis of the soil—frame interaction

By determining the settlements ‘si’ from this linear system, the forces ‘Vi’ acting on the ground are readily obtained from the penultimate relation. It is obvious that the whole procedure is feasible only by using an appropriate computer program for plane frame analysis. 616

Supporting the structure on the ground 80 kN/m

80 kN/m

80 kN/m

80 kN/m

80 kN/m

80 kN/m

3.50

3.50

2

3

4

5

6

8

9

127

8.0 m

127

752

815

5

8.0 m

7

5

1

752

1.20

815

3.50

[M] 1837 kN m

Figure 17.29 Example of frame analysis, with the frame resting on an elastic subspace

Numerical example For reasons of direct comparison, the frame examined in Section 17.3.3.1 will again be considered, under the same loading and soil characteristics (Figure 17.29). The model under consideration is supported at nine intermediate equidistant points of the foundation beam, as shown in Figure 17.29. According to the procedure described above, the following settlements are obtained: s1 ¼ 0.0571 m, s2 ¼ 0.0562 m, s3 ¼ 0.0565 m, s4 ¼ 0.0577 m, s5 ¼ 0.0587 m, s6 ¼ 0.0577 m, s7 ¼ 0.0565 m, s8 ¼ 0.0562 m, s9 ¼ 0.0571 m This means that the state of stress and the deformation of the frame examined are identical to those of the fictitiously supported model, under the existing loading and the above support settlements. The bending moment diagram of the foundation beam is shown in Figure 17.29. By comparing the results with those based on the adoption of the Winkler soil model, it is observed that in the present case where the soil is considered as an elastic half-space, the span moments have lower values, whereas the values of the bending response in the region of the central column are greater.

17.4

Pile foundations

17.4.1 Overview As mentioned in Section 17.3, deep foundations transfer the loads of the superstructure, through weak unsuitable soils, to a lower stratum with sufficient bearing strength. The 617

Structural systems: behaviour and design Every pile has a maximum allowable load depending on its dimensions as well as the soil strength Pile cap Ground

V M H

Ground

Sound stratum

Sound stratum

Uptake of load through skin friction forces

Figure 17.30 Layout and load-bearing action of piles

usual way to accomplish this transfer is through the use of piles. Obviously, piles are used when the subsurface conditions are not suitable for a shallow foundation. Piles are vertical or slightly inclined structural members made of concrete, up to 30 m long, having a circular cross-section and a diameter ranging from about 0.40 m up to 1.50 m. The piles are arranged in a group of relatively dense layout, being connected with each other at their upper end through a thick slab called a pile cap, which is further connected to the superstructure. At their lower end, the piles have a slightly enlarged base, resting on a soil layer of satisfactory bearing strength. In relatively rare cases, where this layer lies so deep that the length required from the piles would result in a prohibitive total cost, the equilibrium of the loads transmitted to the piles from the pile cap is mainly accomplished through skin friction forces — developed between the so-called friction pile — and the adjoining soil, through the whole length of the piles (Figure 17.30). Thus, the foundation system consisting of the pile cap and the piles is in equilibrium under the action of the superstructure applied to the pile cap and the forces acting on the lower end of piles from the sound stratum. It is clear that, in the case of friction piles, the role of the acting soil force at the lower pile end is undertaken by the total force of the lateral friction. Usually, there is a coexistence of friction and bearing forces in a rather uncertain proportion that does not affect the total axial force, in practice, that is required to be taken up by the pile in order to fulfil the global equilibrium conditions. The pile cap, as shown in Figure 17.30, is connected at its top to the supported element, which transmits the actions V, Mx, My (i.e. a vertical force and a moment with an arbitrary horizontal vector), as well as a horizontal force H (Figure 17.31). The piles are usually arranged in a symmetrical layout around the base of the founded element, and they are generally placed in a vertical position. With regard to the horizontal force H, this is taken up either by the vertical piles themselves, or by batter piles specially constructed for that purpose, with an angle of inclination not greater than 158. 618

Supporting the structure on the ground y

y Pile cap

x

O c.g.

x

O c.g.

A

c.g.: centre of gravity

A

N=3

N=4 Data of idealised section AS = N · A Ix = A · ∑ yi2 Iy = A · ∑ xi2

My

My Mx

Mx

V

V

Pi A

=

V AS

–

Mx · yi Ix

+

My · xi Iy

Figure 17.31 Loading of a pile system through a pile cap

17.4.2 Vertical loads The symmetrical layout of N piles around the point of application O of the load V implies the existence of the orthogonal axes of symmetry Ox and Oy. The components Mx and My are referred to these two axes (see Figure 17.31). For preliminary design purposes, it is sufficient to consider the magnitudes V, Mx, My as being applied to a rigid section, consisting of the total layout of circular section areas of the piles. Given the coordinates xi, yi of the centres of the circular pile sections, with respect to the centroid O of the section and assuming that all piles have a common section area A, the following sectional data are derived: . Total area of section: As ¼ N A. P . Moment of inertia about the Ox axis: Ix ¼ A P y2i . x2i . . Moment of inertia about the Oy axis: Iy ¼ A Thus, the consideration of biaxial bending, to which the total section is subjected, leads to the determination of the axial force Pi for each pile separately (see Figure 17.31): Pi V Mx yi My xi ¼ þ A As Ix Iy It is clear that the pile cap is in equilibrium under the loads V, Mx, My and the pile forces Pi, and must, therefore, be designed to safely take up the above actions. For preliminary design purposes, the eventual bending of the piles is ignored. Moreover, selecting a large enough thickness for the pile cap, i.e. equal to at least half of the clear distance between the piles, the bending action of the cap is excluded and the load introduced from the column is transferred to the piles through a strut-and-tie model, normally a 619

Structural systems: behaviour and design V

Concrete struts

P P

Tie reinforcement Equilibrium at the joints

P P In the presence of the moment the force V is shifted (however within the core of idealised section) and the space truss is formed accordingly

Figure 17.32 Load-bearing action of a pile cap

three-dimensional one (Figure 17.32). In this model, which depicts the load transfer through the most ‘stiff ’ — i.e. the shortest — paths, the struts represent the compressed concrete between the column and the pile heads, whereas the ties are realised through reinforcement bars, arranged within the narrow region of the pile circular section with good anchor conditions, as shown in Figure 17.32, thus ensuring the equilibrium of the formed joints. Designing in this way, that is, by taking up the design loads (i.e. service loads magnified by the corresponding safety factors) on the basis of a constructionally feasible equilibrium, ensures, according to the static theorem of the plastic analysis (see Section 6.6.2), that the collapse load is greater than that which the substitute strut-and-tie model can carry.

17.4.3 Horizontal loads The uptake of a horizontal load by a vertical pile clearly implies a bending response of the pile. The pile acts as if laterally supported by an elastic medium described by a subgrade soil coefficient kh and substituted accordingly through a continuous distribution of lateral springs, as in the case of the foundation beam (Figure 17.33). In non-cohesive soils, the coefficient kh can be considered to increase linearly to a depth t, according to the relation kh ¼ nh t/D, where D is the pile diameter and nh is a quantity representing the lateral stiffness of the soil, ranging from 2 to 18 MN/m3, depending on the layer density. In cohesive soils, the coefficient kh is considered to be constant, and is kh ¼ nh/D, where nh is expressed in terms of cohesion c as nh ¼ 160 c. Clearly, the uptake of a horizontal force, or even a moment acting on top of a pile (transmitted through the pile cap), is accomplished through an appropriate distribution of lateral soil reactions along the length of the pile, implying either tension or 620

Supporting the structure on the ground M Distribution of subgrade coefficient

H

t

kh = nh · t/D Non-cohesive soils

D

H

kh = 160 · c/D Cohesive soils

H Compressed springs

Tensioned springs

D

Pressure distribution

Bending moment distribution

Figure 17.33 Lateral response of piles

compression of the corresponding supporting springs (see Figure 17.33). In contrast to the foundation beam, where the development of tensile spring forces is not normally allowed, in the present case this is obviously meaningful, as any resulting spring force represents compressive contact pressure between the soil and the pile. It is clear that the above response may be determined using the same software as used for the analysis of foundation beams.

References Menn C. (1990) Prestressed Concrete Bridges. Basel: Birkhaüser Verlag. Stavridis L. (1997) Tragweke auf elastischem Boden. Bauimgenieur 72(12) 565—569. Stavridis L. (2002) Simplified analysis of layered soil—structure interaction. Journal of Structural Engineering (ASCE) 168(2), 224—230. Stavridis L. (2009) Rigid foundations resting on an elastic layered soil. Geotechnical and Geological Engineering 27, 407—417.

621

Bibliography Bachmann H., Ammann W. (1987) Vibrations in Structures — Induced by Man and Machines. Zurich: IABSE. Bachmann H., Ammann W. et al. (1995) Vibration Problems in Structures. Boston: Birkhaüser. Beles A.A., Soare M.V. (1972) Berechnung von Schalentragwerken. Wiesbaden: Bauverlag. Bieger K.W., Lierse J., Roth J. (1993) Stahlbeton — und Spannbetontragwerke. Berlin: Springer-Verlag. Biggs J.M. (1964) Introduction to Structural Dynamics. New York: McGraw-Hill. Billington D. (1965) Thin Shell Concrete Structures. New York: McGraw-Hill. Ciesielski R., Mitzel A., Stachurrski W., Suwalski J., Zmudzinski Z. (1970) Beha¨lter, Bunker, Silos, Schornsteine, Fernsehtu¨rme und Freileitungsmaste. Berlin: Wilhelm Ernst. Duddeck H. (1984) Traglasttheorie der Stabwerke — Beton Kalender, Teil II. Berlin: Ernst. ¨gge W. (1960) Stresses in Shells. Berlin: Springer-Verlag. Flu Franz G., Scha¨fer K. (1988) Konstruktionslehre des Stahlbetons, Band II, A. Berlin: Springer-Verlag. Franz G., Hampe E., Scha¨fer K. (1991) Konstruktionslehre des Stahlbetons, Band II, B. Berlin: Springer-Verlag. Gaylord E. H., Gaylord C. N. (1957) Design of Steel Structures. New York: McGraw-Hill. ¨sseldorf: Werner Verlag. Herzog M. (1996) Kurze baupraktische Festigkeitslehre. Du Kollar L. (1984) Schalenkonstruktionen — Beton — Kalender 1984, Teil II. Berlin: Wilhelm Ernst. Leonhardt F., Zellner W. (1980) Cable stayed bridges. IABSE Surveys, S-13/80. Mann W. (1976) Kippnachweis und Kippaussteifung von schlanken Stahlbeton — und Spannbetontra¨gernn. Beton — und Stahlbetonbau 2, 37—42. Menn C. (1990) Prestressed Concrete Bridges. Basel: Birkhaüser Verlag. ¨ller F.P., Keintzel E. (1984) Erdbebensicherung von Hochbauten. Berlin: Wilhelm Ernst. Mu Nervi P.L. (1956) Structures. New York: McGraw-Hill. O’Connor C. (1971) Design of Bridge Superstructure. New York: John Wiley. ¨ger A. (1963) Zur praktischen Berechnung der axial gedru ¨ckten Kreiszylinderschale. Stahlbau 32, Pflu 161—164. ¨ger A. (1966) Zur praktischen Berechnung der axial gedru ¨ckten Kreiszylinderschale unter Pflu Manteldruck. Stahlbau 35, 249—252. Roik K. (1983) Vorlesungen u¨ber Stahlbau. Grundlagen. Berlin: Wilhelm Ernst. Rosman R. (1983) Erdbeben—widerstandsfa¨higes Bauen. Berlin: Wilhelm Ernst. Salvadori M. and Levy M. (1967) Structural Design in Architecture. Englewood Cliffs, NJ: Prentice-Hall. Schlaich J. and Scheef H. (1982) Concrete Box-girder Bridges. Zurich: IABSE. Seide P. (1981) Stability of cylindrical reinforced concrete shells. American Concrete Institute ACI SP 67, 43—62. Stavridis L.T. (1986) Static and dynamic analysis of multistory systems. Technika Chronika Scientific Journal of the Technical Chamber of Greece 6(2), 187—219. Stavridis L.T. (1993) Statische und dynamische Analyse von orthotropen rechteckigen Platten. Stahlbau 62, 73—80. Stavridis L.T. (1994) Beanspruchung von mehrsto¨ckigen Bauten infolge Temperatura¨nderung. Bauingenieur 69(3), 117—122. Structural systems: behaviour and design


Structural systems: behaviour and design Stavridis L.T. (1997) Tragweke auf elastischem Boden. Bauimgenieur 72(12), 565—569. Stavridis L.T. (2001) Alternative layout for the prestressing of slab bridges. Journal of Bridge Engineering of the ASCE, 6(5), 324—332. Stavridis L.T. (2002) Simplified analysis of layered soil—structure interaction. Journal of Structural Engineering of the ASCE 168(2), 224—230. Stavridis L.T. (2008) A simplified analysis of the behavior of suspension bridges under live load, Structural Engineering and Mechanics 30(5), 559—576. Stavridis L.T. (2009) Rigid foundations resting on an elastic layered soil. Geotechnical and Geological Engineering 27, 407—417. Stavridis L.T. (2010) Evaluation of static response in stress-ribbon concrete pedestrian bridges, Structural Engineering and Mechanics 34(2), 213—229. Stavridis L.T., Armenakas A.E. (1988) Analysis of shallow shells with rectangular projection: applications. Journal of Engineering Mechanics of the ASCE 114(6), 943—952. Tedesko J.W., McDougal W.G., Ross C. Allen (1999) Structural Dynamics — Theory and Applications. California: Addison Wesley Longman. Timoshenko S. (1956) Strength of Materials, Part II. Princeton, NJ: Van Norstrand. Timoshenko S., Young D. (1965) Theory of Structures. New York: McGraw-Hill. Tveit P. (1966) Design of network arches. Structural Engineer 44(7), 247—259. Vlasov V.Z. (1961) Thin Walled Elastic Beams. London: Oldbourne Press. VSL (1984) VSL Stay Cables for Cable-stayed Bridges. Berne, VSL. Walther R., Miehlbradt M. (1990) Dimensionnement des Structures en Be´ton. Lausanne: Presses polytechniques et universitaires romandes. Zbirohowski—Koscia K. (1967) Thin Walled Beams — From Theory to Practice. London: Crosby Lockwood.

624

Index Page numbers in italics refer to figures separate from the corresponding text. absolute displacement, 548 acceleration spectra, 550—551, 551 accelerograms, 548—551, 551 action principles, 32 active pressure, 589 adhesion forces, 14 allowable bearing pressure, 597 allowable limits, 26 anchorage lengths, 14, 15 anchor forces, 509—510 angles of internal friction, 588 angles of skewness, 359—360 antisymmetric loads, 302, 332—335, 481, 484—487 aquatic environments, 3 arbitrary loadings, 84, 85 arches, 297—313 axis shortening, 302—304 conoidal shells, 452, 453 elastic stability, 304—307 formation, 297, 298 girders, 308—310 hyperbolic paraboloid shells, 440—442, 443, 444 membrane action in shells, 405 pressure lines, 297—298 second-order theory, 304—307 stiffness, 308—310 straight edge hypar shells, 445—446, 450 structural behaviour, 297—304 suspension cables, 335—339 tied-arch systems, 311—313 arch-like boundaries, 435 axial deformations, 172 axial flexibility, 7 axial forces, 43, 44—45, 67, 226, 253—257 box girders, 483—484, 486—487 dome shells, 432 hyperbolic paraboloid shells, 440—441 Structural systems: behaviour and design 978-0-7277-4105-9

straight edge hypar shells, 445—446 axial loads, 281 axial stiffness, 6—7, 486—487 axial stresses, 482—483, 487—489, 490 balanced cantilever methods, 206—207 bar buckling, 271—277 barrel shells, 416—423 base shear, 550—551, 575 basic material structural behaviour see material structural behaviour batter piles, 618 beams box girders, 490—491, 491, 507, 508 cable structures, 317—322, 331, 342, 343, 345—347 cantilever beams, 177—181 continuous beams, 187—221 continuous systems, 577—580, 581 cylindrical shells, 410—411, 418—421 deflection curves, 77—78 design control, 184—185 discrete mass systems, 526, 527, 534, 535, 580, 581 dome shells, 427—428, 429—432 elastic mediums, 613—616 elastic stability, 272—273, 277—282, 290—293, 294 equilibrium, 50—52, 62—64, 77—78, 90—91 external prestressing, 181—184 formation, 455, 456 frames, 223—269 grid structures, 354—357 ground structure supports, 609—617 lateral buckling, 290—293 multi-storey systems, 526, 527 non-shallow shells, 429—432 with openings, 142—144 Copyright Thomas Telford Limited # 2010 All rights reserved

Structural systems: behaviour and design beams (cont’d ) plates, 396—397, 400, 401, 402 prestressed concrete beams, 163—177 rectangular openings, 143—144 reinforced concrete, 150—163 simply supported beams, 139—185 soil simulations, 610—616 statically indeterminate structures, 130, 131 steel beams, 139—150, 188—194 straight edge hypar shells, 446, 450 suspension bridges, 331 thin-walled beams, 455—479 variable beam height, 178, 179—180, 181, 181 Winkler model, 610—613 bearing pressure (allowable), 597 bending application examples, 499—502 arches, 302—304, 309 box girders, 482—483, 487—490, 495—503, 505—510 conoidal shells, 453 composite beams, 210—211 continuous beams, 189—190, 196—198, 205—207 curved girders, 496—502 cylindrical shells, 410—413, 420—422 deflection curves, 77—78 dome shells, 429—431, 431, 434 elastic stability, 272—273 equilibrium, 43—46, 51—54, 66—67, 77—78, 89, 496—498 frames, 253, 256, 257 grid structures, 351—352, 355, 356 ground structure supports, 611—613, 612 hyperbolic paraboloid shells, 441 mixed systems, 249—251 multi-storey frames, 240—244, 246—247 one-storey multibay frames, 235—238 plates, 364—365, 370, 373—386, 391—392, 396—398 simply supported beams, 139—140, 144—153, 158—170, 173—180 single-storey, single-bay frames, 223—224, 226, 228, 234—235 statically indeterminate structures, 107—110, 112, 118, 136—138 suspension bridges, 329, 330 Betti—Maxwell theorem, 76—77, 429 bimoment, 469—473, 470 bonded tendons, 176—177 bond forces, 14 bond lengths, 14, 15

626

bond stresses, 14, 15 boundary forces, 408 box girders, 481—516 cross-section walls, 502—505 curved girders, 495—416 prestressing, 505—511, 512, 513 profile section deformability, 485—492 rectilinear girders, 481—495 torsion, 496—502, 510—511, 512, 513 buckling cable-stayed bridges, 347—348, 349 concepts, 27 cylindrical shells, 422 dome shells, 434 elastic stability, 271—277, 290—293 single-storey, single-bay frames, 233 straight edge hypar shells, 451 visualisation, 291 cable structures, 315—349 box girders, 510, 511 cable—beam structures, 317—322 cable-stayed bridges, 339—348, 349, 501—502 cylindrical shells, 415 freely suspended cables, 322—325 hyperbolic paraboloid shells, 440 plates, 388—389, 396—397, 398—399 prestressed cable nets, 325—328 simply supported beams, 179—180, 181 stiffening, 335—339 straight edge hypar shells, 446—448, 448, 449 suspension bridges, 328—335 suspension cables, 335—339 conoidal shells, 451—453 cantilevers box girders, 483—484 conoidal shells, 452 design, 376—377 discrete mass systems, 535—536, 538—539 equilibrium, 53—54 mixed systems, 249—251 multi-storey frames, 245—246 plates, 375—377 simply supported beams, 177—181 straight edge hypar shells, 446—447, 448 stresses, 375—376 thin-walled beams, 465—466 cast-in-place concrete deck plates, 207—209 Cauchy’s relation/theorem, 145—146, 418, 461—462 centroid of frame section design, 254—255 circular columns, 393

Index circular frequency, 537 circular plans, 427—428 circular plates, 365, 385, 386 clamped supports, 79—81 climatic conditions, 3—4 closed beam-like cylindrical shells, 416—418 closed cross-sections box girders, 481—516 general characteristics, 455—460 non-deformable cross-sections, 460—461 warping-based stresses, 462—463, 465—470 coefficients of elastic rotation, 79—80 coefficients of elastic supports, 79 cohesive soil types, 587, 590—592 collapse concepts, 26 continuous beams, 190—191 frames, 257—259, 261—262 column—footing—ground interactions, 605—606 column matrices, 557 columns, 223—269, 389—393, 397—398 compatibility of deformations, 96 composite beams behaviour, 217—220 bending behaviour, 210—211 construction stages, 212—214 continuous beams, 209—221 plastic analyses, 220—221 prestressing, 217—220 shear forces, 209—210, 211—212 temperature, 215—217 compound structures, 39—41 compressed beams, 214, 272—273 compressed concrete, 8, 9 compression box girders, 483—484, 502, 505—506, 507 continuous beams, 214 cylindrical shells, 414, 416—417, 419, 420, 422 dome shells, 426, 432—433 elastic stability, 275—277 frames, 226, 253—257 ground structure supports, 592—593 hyperbolic paraboloid shells, 440—443, 443, 444 membrane action in shells, 405, 409 plates, 373, 381—382, 394 simply supported beams, 165—172, 179, 180 single-storey, single-bay frames, 226 straight edge hypar shells, 445—446, 447, 450—451 concentrated forces, 236—238 concentrated loads, 2, 49—50, 51, 157, 158 box girders, 490—491

grid structures, 354 ground structure supports, 594—595 plates, 369—370, 375—378, 387 concepts, 1—28 bimoment, 469—472, 470 control/design processes, 26—28 discrete mass systems, 555—557 elastic supports, 108—110 equilibrium, 29—41, 81—84 flexibility, 81—84 loads, 1—5 material structural behaviour, 5—22 membrane action, 403—409 prestressed concrete, 18—22, 24—26 reinforced concrete, 10, 13—18, 22—23 shells, 403—409 statically indeterminate structures, 108—110 stiffness, 81—84 stiffness matrices, 555—557 stress, 5 concrete beam bending stages, 159—161 confinement, 262 continuous beams, 195—199, 207—209, 213—214, 217—219 creep, 10—12 dome shells, 432 frames, 229, 233, 240, 252—257, 262, 267—268 grid structures, 357 material structural behaviour, 8, 9, 10—18 one-storey multibay frames, 240 plates, 394 rectangular spread footings, 608—609 relaxation, 12—13 simply supported beams, 150—179 single-storey, single-bay frames, 229, 233 suspension cables, 335—337 two-side supported slabs, 371 concrete deck plates, 207—209 constant internal pressure, 409—413 constraint moments, 197—198 continuous beams, 187—221, 396—397, 534, 535 behaviour, 217—220 bending behaviour, 210—211 cable-stayed bridges, 342, 343 composite beams, 209—221 construction stages, 212—214 creep, 199—209, 218—220 design control, 198—199 fixed-ended beams, 188—191 fixed simply supported beams, 191—193 plastic analyses, 220—221

627

Structural systems: behaviour and design continuous beams (cont’d ) prestressed concrete beams, 195—199 prestressing, 217—220 shear forces, 209—210, 211—212 statically indeterminate structures, 106, 108, 109—110, 111 statically redundant systems, 200—205 steel beams, 188—194 temperature change, 215—217 tendon design/structural performance, 195—198 continuous girders, 507, 508 continuous plane structures, 361, 362 continuous plates, 380 continuous supports, 389—399, 446, 447 continuous systems, 577—580, 582—585, 584 corner regions, 367, 368, 369 coupling processes, 144—145, 249—251 cracked states, 371 cracking, 16—17, 19—21, 195, 199 creep continuous beams, 199—209, 218—220 frames, 261—262 material structural behaviour, 10—12 one-storey multibay frames, 240 single-storey, single-bay frames, 229 statically redundant systems, 200—205 structural changes, 205—209 critical external pressure, 414 critical loads, 275—279, 283—285, 290, 434, 451 critical locations of compressive forces, 169—171 critical longitudinal compressive stress, 414 critical stress limits, 422 cross-beams, 354 cross-sections bimoment, 470—472 box girders, 481—516 conoidal shells, 452 composite beams, 210 cylindrical shells, 416—418, 421—422 frames, 251—255 non-deformable cross-sections, 460—461 plates, 388—389 shear centres, 461—462 simply supported beams, 140, 141, 165—167 single-storey, single-bay frames, 233—234 straight edge hypar shells, 447—448 suspension cables, 335—337 thin-walled beams, 455—472 warping-based stresses, 462—469 cross-vaults, 422, 423 curvatures, 147, 153, 161 dome shells, 426, 432—434

628

hyperbolic paraboloid shells, 436—437, 439—440 membrane action in shells, 407—408 shells, 403—404, 426, 432—434, 436—437 straight edge hypar shells, 445—448 curved girders, 495—416 bending, 496—502 cross-section walls, 502—505 prestressing, 505—511, 512, 513 torsion, 496—502, 510—511, 512, 513 curved orthogonal plates, 433—434 curved quadrilateral elements, 440, 441 cylindrical shells, 409—423 barrel shells, 416—423 bending behaviour, 430, 431 constant internal pressure, 409—413 cylindrical tanks, 413—415 silos, 415—416 damped vibration, 539—542 deck plates, 207—209 decompression moments, 173—174 deep beams, 68, 400, 401, 402 deep foundations, 592 deflection box girders, 490—491 curves, 77—78, 331—332 plates, 376—377 deformable soil, 230—231 deformation method, 352—353 elastic supports, 130—132 fixed-end beams, 119—123 nodal action distribution, 132—133 procedure, 125—129, 130 qualitative handling, 105—107, 134—136, 107, 137, 138 simply fixed beams, 116—119 statically indeterminate structures, 114—138 unknown nodal deformations, 123—125 deformations arches, 304—306 box girders, 489—490, 492 cable—beam structures, 317—318, 319, 320 causes, 66—70 concepts, 29—41 continuous beams, 199—209 cylindrical shells, 410 deformation method, 114—138, 352—353 dome shells, 428, 431, 432 equilibrium, 29—92 force method, 94—114, 115 frames, 93—138, 256, 257, 282—290 grid structures, 87—92, 355, 356

Index ground structure supports, 592—596, 597, 601—603 internal forces handling, 41—65 lateral buckling, 290—293 material structural behaviour, 7 membrane action in shells, 408 multi-degree systems, 559 multi-storey frames, 244—249 plastic analyses, 293—296 plates, 361—362, 376—377, 381 second-order theory, 277—282 simply supported beams, 139—140, 143—144, 172 single-storey, single-bay frames, 231—233 statically indeterminate structures, 93—138 stress states, 93—138, 271—296 suspension bridges, 329—331 symmetric plane structures, 84—87 unknown nodal deformations, 123—125 degree of freedom systems, 534, 535—555 degrees of statical indeterminacy, 41 developing moment determination, 126—127, 128 deviation forces, 165—166, 167—168 box girders, 505, 509—510 plates, 388, 396—399 deviation loads, 396—397 diagonal elements, 557 diagonal forces, 486—490 diagonal loading, 503, 504 diagonal tension cracks, 153 diagonal tied-arch layouts, 312—313 discrete mass systems, 533—586 continuous systems, 577—580 design spectra, 553, 554 dynamic equilibrium, 535—536, 548, 563, 564 equilibrium, 535—536, 548 forced vibration, 542—544, 570—571 free vibration, 536—542, 563—570 human activities, 580—581 loads, 543—545, 559, 572, 574—576 machine induced vibrations, 581—585 multi-degree systems, 555—577 periodic sinusoidal acting forces, 544—547 plane elements, 560, 561 plastic behaviour, 553—555 seismic excitation, 547—555, 571—577 single-degree-of-freedom systems, 535—555 stiffness matrices, 555—563 displacement cylindrical shells, 411, 412 discrete mass systems, 534—538, 548, 550, 559 dome shells, 432, 434

elastic stability, 285 grid structures, 354—355 member kinematics, 123—124, 125 membrane action in shells, 408—409 multi-storey systems, 528—529, 529 one-storey multibay frames, 238—240 selection, 125, 126 statically indeterminate structures, 118—128, 135—138 thin-walled beams, 466 distributed loads, 381 distributed prestressing, 372 DMF see dynamic magnification factor dome shells, 423—436 double-T cross-sections, 455—456 downward acting deviation forces, 398 downward applied deviation forces, 398—399 downward applied deviation loads, 396—397 downward applied forces, 196—198, 398—399 downward-curved parabolas, 440—441, 442 downward edge loads, 400 downward shifting, 179, 180 ductility discrete mass systems, 554—555 frames, 258, 259, 260, 262 material structural behaviour, 7—8 Duhamel integral, 543 dynamic equilibrium, 535—536, 548, 563, 564 dynamic loads, 1, 2 dynamic magnification factor (DMF), 543—544, 545—546, 552 earth pressure, 2—3 cohesive soil types, 591 non-cohesive soil types, 588—589, 590 plates, 401—402 earthquakes, 3, 547—555, 571—577 eccentric footings, 606, 607—608, 607 eccentric loads box girders, 490—491 ground structure supports, 601 thin-walled beams, 458—460 edge beams cylindrical shells, 419, 420 straight edge hypar shells, 446, 447, 447, 450 edge loads, 400 edge-supporting arches, 441—442 edge zones cylindrical shells, 418—419, 422 hyperbolic paraboloid shells, 442, 443, 443 plates, 373—375, 387—389 effective degrees of freedom, 555

629

Structural systems: behaviour and design effective mass displacement, 534, 535 effective shear, 394—395, 398—399 effective stresses, 591 eigenforms, 565—566 eigenfrequencies, 565—567, 568, 569 eigenvalue equations, 565 EI value influence, 104 elastically rotating supports, 376 elastic bases, 601—608, 611—613, 612 elastic components cable-stayed bridges, 345—346 frames, 261—264 material structural behaviour, 7 steel beams, 139—147 elasticity moduli, 6, 139, 213—214, 592—593 elastic lengths, 612, 613 elastic mediums, 613—616 elastic soil, 592—596, 597 elastic stability, 451 arches, 304—307 bar buckling, 271—277 deformations, 271—296 frames, 282—290 lateral buckling, 290—293 multi-storey frames, 288—290 numerical examples, 294—296 one-storey frames, 283—287 plastic analyses, 293—296 second-order theory, 277—282 stress states, 271—296 elastic subspaces, 614—615, 617 elastic supports box girders, 490—491, 491 equilibrium, 79—81 statically indeterminate structures, 105, 106, 108—110, 130—132 element of purely plane stiffness, 518 end rotation, 120—121 equally distributed prestressed cable structures, 396—397 equilibrium, 29—92 action principles, 32 antisymmetric loadings, 84, 85, 86—87 basic assumptions, 34—35 beam equations, 77—78 bending moment diagrams, 51—54 Betti—Maxwell theorem, 76—77 box girders, 482—486, 495—498, 503—504, 507, 508 cable—beam structures, 318, 319, 320—322 compound structures, 39—41 concepts, 29—41, 81—84

630

conditions of, 30—32 continuous beams, 191, 198, 207—210, 218 cylindrical shells, 417—418, 420 deflection curves, 77—78 deformations, 29—92 discrete mass systems, 535—536, 548, 563, 564 dome shells, 426, 427 elastic stability, 271—273 elastic supports, 79—81 flexibility, 81—84 forces, 29 frames, 262 funicular structures, 57—61 grid structures, 87—92, 357—358, 358 ground structure supports, 597—598 hyperbolic paraboloid shells, 440—443, 444 internal forces handling, 41—65 loading forces, 48—50 member bending moment diagrams, 51—54 membrane action in shells, 407—408 multi-degree systems, 563, 564 multi-storey systems, 521, 522 plates, 363—364, 373—375, 392, 396—397 reaction principles, 32 sectional forces, 41—50 shell membranes, 405, 406 simply supported beams, 50—51, 160, 161, 182—183 single-degree-of-freedom systems, 535—536, 548 single-storey, single-bay frames, 229—230 statically determinate formations, 35—39 statically indeterminate structures, 93, 94, 96, 126—129 stiffness, 81—84 straight edge hypar shells, 444, 445, 448 stress, 29—92 support conditions, 32—34, 37—39 support types, 91—92 symmetric plane structures, 84—87 thin-walled beams, 461—463, 466—467, 470—474 three-hinged frames, 54—57, 58 torsion, 90—91 trusses, 61—65 virtual work, 70—76, 89—90 equivalency, 254—256 continuous systems, 577—580, 581 seismic excitation, 574—576 stiffness matrices, 562—563 thin-walled beams, 476, 476, 477 equivalent actions, 286, 287 equivalent cross-sections, 14, 15

Index Euler—Bernoulli beam theory, 139—140 external actions, 96, 97, 98, 126, 127 external deformations, 65 external fixed-ended spans, 241—242 external forces, 29, 30, 70—71, 254—256 external joints, 235—236 external loads, 196, 197, 262, 564, 571 externally-placed unbonded tendons, 176—177 external nodes, 241—242 external pressure, 414 external prestressing, 181—184 factors of safety, 26 failure states, 147—150 fictitious beams, 331—334, 336—337 ‘fictitious’ horizontal forces, 286—289 fictitious springs, 410 final end force/moment calculations, 129, 130 final stress states, 219—220 finite thickness soil layers, 595—596 first-order analyses/theories, 34, 35, 45, 271 fixation actions, 529, 530 fixed boundaries, 431, 434 fixed-ended beams, 119—123, 188—193, 216 fixed-end imposed rotation, 117 fixed-end moment, 241—242, 413 fixed-fixed beams, 120—121 fixed states, 116, 117, 119—120 fixed structures continuous beams, 191—193, 216—217 frames, 229—234, 237—238 statically indeterminate structures, 114, 115, 126, 127 fixed supports, 33, 37, 38, 91—92, 367, 378, 379 flat slab stresses plates, 389—399 prestressing, 396—399 punching shear design, 393—395 flexibility elastic stability, 289 equilibrium, 81—84 material structural behaviour, 7 soil matrices, 595 see also stiffening/stiffness flexible supports, 131—132 flexural cracks, 153 folded plates, 399—402 forced vibration, 542—544, 570—571 force method, 352—353, 359 analytical application, 98—101 EI value influence, 104 elastic support, 105, 106, 108—110

physical overview, 94—97, 98 qualitative methods, 105—107, 108, 110 result checking, 105 statically indeterminate structures, 94—115 support settlement, 102—104 temperature, 101—102 formation definition, 39 foundation beams elastic mediums, 613—616 ground structure supports, 609—617 numerical examples, 617 soil simulations, 610—616 Winkler model, 610—613 foundations, 29 flat slabs, 389 foundation beams, 609—617 ground structure supports, 592—621 horizontal loads, 620—621 lateral pile responses, 620—621 layout, 618 pile foundations, 617—621 rectangular spread footings, 597—609 vertical loads, 619—620 four-side supported plates, 377—381 design, 381 prestressing, 381, 382 stresses, 377—380 frames, 223—269 deformation method, 114—138 design, 266—268 elastic stability, 282—290 equilibrium, 54—57, 58 force method, 94—114, 115 ground structure supports, 611—617 horizontal loads, 223—231, 234—238, 242—244 inclined legs, 234—235 joint checks, 266—268 lateral stiffness, 231—234, 238—240 mixed systems, 249—251 multi-degree systems, 563 multi-storey frames, 240—251, 288—290 one-storey frames, 235—240, 283—287 plastic analyses, 257—266 plates, 390—391, 392 prestressed concrete, 256—257 reinforced concrete, 252—256 section design, 251—257 single-storey, single-bay frames, 223—235 steel sections, 251—252 stress states, 93—138 vertical loads, 223—231, 235—236, 240—242

631

Structural systems: behaviour and design free cantilever constructions, 206—207 free edge zones cylindrical shells, 418—419, 422 hyperbolic paraboloid shells, 442, 443, 443 plates, 372—373, 375, 387—389 free formations, 30, 31 freely suspended cables, 322—325 free support conditions, 367 free vibration damped vibration, 539—542 discrete mass systems, 536—542, 563—570 multi-degree systems, 563—570 multi-storey spatial systems, 568—569 numerical examples, 569—570 plane systems, 563—568 single-degree-of-freedom systems, 536—542 undamped vibration, 536—539 frequencies, 537—538, 565—568, 569 friction piles, 618 full prestressing, 372 fundamental eigenfrequencies, 566—567 fundamental frequencies, 566, 567 fundamental periods (T), 1 funicular membranes, 405, 406 funicular structures, 57—61, 315—316, 322—325, 427—428 Gaussian curvatures, 403—404, 436—437 generatrices, 436—439, 440, 451 girders arches, 308—310 box girders, 481—516 cable structures, 328—335, 345—347 cross-section walls, 502—505 curved girders, 495—416 equilibrium, 54 frames, 223—235 prestressing, 505—511, 512, 513 profile sections deformability, 485—492 rectilinear girders, 481—495 suspension bridges, 328—335 torsion, 496—502, 510—511, 512, 513 gradually imposed settlements, 204—205 granular material, 415—416 gravity loads, 1—2 multi-storey frames, 240, 241 one-storey multibay frames, 235 simply supported beams, 179, 180 grid structures, 87—92, 351—360 plates, 367, 368 skew bridges, 358—360 structural behaviour, 351—358

632

ground structure supports, 587—621 deformations, 592—596, 597 elastic soil, 592—596, 597 foundation beams, 609—617 horizontal loads, 620—621 pile foundations, 617—621 rectangular spread footings, 597—609 shallow foundations, 592—617 vertical loads, 592—596, 597, 619—620 half-beams, 144—145 half-loading, 234, 235 hardening steel, 7 hertz, 537 hinged boundaries, 431 hinges equilibrium, 33 frames, 258—262, 265—266 multi-storey frames, 244—246 one-storey multibay frames, 237—240 single-storey, single-bay frames, 226—232 hollow box sections, 91 Hooke’s law, 6 horizontal circular plans, 427—428 horizontal displacement, 238—240 horizontal flexibility, 289 horizontal forces elastic stability, 286—289 single-storey, single-bay frames, 231 statically indeterminate structures, 133 horizontal loads, 620—621 elastic stability, 295 multi-storey frames, 242—244 one-storey multibay frames, 236—238 single-storey, single-bay frames, 223—231, 234—235 horizontally distributed radial forces, 427, 428 horizontal yielding, 298—299 hypar shells, 444—448, 449, 450, 451 hyperbolic paraboloid shells, 403, 404, 436—451 creation, 436—437 equilibrium, 440—443, 444 geometry, 437—440 hypocentres, 548—549 identity matrices, 558 immovable nodes, 134—135 impact factors, 2 imposed relative displacement, 118—119, 121—123 imposed rotation, 117, 120—121, 126—127, 128 imposed settlements, 203—205

Index imposed shifts, 126—127, 128 inclined legs, 234—235 indirect loading, 61, 62 inertia forces, 534, 551, 552, 554—555 inertia moment, 399, 400, 452 instantaneous settlements, 203—204 interaction curves, 253—254 intermediate longitudinal beams, 354—357 intermediate supports, 213—214 internal deformations, 65 internal equilibrium, 443 internal forces, 351, 352, 405, 420, 421 continuous beams, 207—208, 217, 218 equilibrium, 41—65, 70—71, 84—87 frame design, 254—256 single-storey, single-bay frames, 229 internal joints, 235—236 internally developed actions, 209—210 internally-placed unbonded tendons, 176—177 internal pressure, 409—413, 415—416, 429 internal supports, 194 intersecting barrel vaults, 422—423 inverse matrices, 558—559 inverted arches, 335—336 inward pressure, 414 inward radial displacement, 432 joint check procedures, 266—268 joint equilibrium methods, 61, 62, 62 large-span beams, 198 lateral bracing, 307 lateral buckling, 290—293 lateral loads, 282 lateral pile responses, 620—621 lateral soil pressure, 414 lateral stiffness, 560, 561 multi-storey frames, 244—249 one-storey multibay frames, 238—240 single-storey, single-bay frames, 231—234 lateral transverse displacement, 434 layered soil, 589, 590 limit states, 271—272 limit tensile loads, 20—21 linear loads, 370, 371 liquefaction, 588 liquid tanks, 413 live loads, 2, 379, 380 arches, 300—302, 304—306, 308—309, 311—312 cable—beam structures, 320 cable-stayed bridges, 344 suspension bridges, 329, 330, 332—335

loads, 1—5 antisymmetric, 84—87, 302, 332—335 aquatic environments, 3 arches, 300—302, 304—306, 308—309, 311—313 box girders, 481—488, 490—491, 502—504, 505, 509—510 cable—beam structures, 320, 322 cable-stayed bridges, 344—345 conoidal shells, 451, 452, 453 climatic conditions, 3—4 continuous beams, 187—189, 191—193 cylindrical shells, 409—413, 416—418, 420, 422 discrete mass systems, 543—545, 559, 572, 574—576 dome shells, 427, 428—430, 434—435 elastic stability, 283, 295 equilibrium, 48—50, 51, 52, 84—85 frames, 257—259, 266—267 freely suspended cables, 322—323 gravity loads, 1—2 grid structures, 351, 352, 353—354, 357, 358 ground structure supports, 592—596, 597, 601, 619—621 hyperbolic paraboloid shells, 440—443, 443 membrane action in shells, 403, 405—408 multi-degree systems, 559, 564, 571, 572, 574—576 multi-storey systems, 519—525, 528—530 orthogonal plates, 370, 371 pile foundations, 619—621 plates, 361—391, 396—402 prestressed cable nets, 326—327 rectilinear girders, 481—485 simply supported beams, 164—165, 166 single-degree-of-freedom systems, 543, 544, 545, 546 single-storey, single-bay frames, 226—231 soil supports/surroundings, 2—3 special impact loads, 4 straight edge hypar shells, 444, 444, 445—446, 447, 451 supports, 2—3 suspension bridges, 329, 330, 332—335 symmetric plane structures, 84—85 temperature, 528—530 thin-walled beams, 458—460 long barrel shells, 419—420 longitudinal beams, 354—357 longitudinal bending moment, 370 longitudinal forces, 212, 417—419, 420, 421, 422 longitudinal stresses, 251—252, 414, 422, 469—470

633

Structural systems: behaviour and design machine induced vibrations, 581—585 masonry, 233, 249 mass-spring systems, 538—539 material structural behaviour concepts, 5—22 concrete, 8, 10—13 numerical examples, 22—26 prestressed concrete tension, 18—22 reinforced concrete, 13—18, 22—23 steel, 5—8, 9 maximum inertia forces, 551, 552 maximum tensile forces, 413 membrane action, 403—409 membrane forces cylindrical shells, 417—419, 422 dome shells, 424—428, 432—433, 435 hyperbolic paraboloid shells, 440—441 membrane load-carrying mechanisms, 416—418, 453 membrane states, 405—406, 421, 440—443 meridional forces, 425—428, 433 ‘midspan tendons’, 206—207 ‘modular ratios’, 210, 213—214 moduli of elasticity, 6, 139, 213—214 moduli of subgrade reaction, 601—602 moment bimoment, 470—473 box girders, 487—488, 502—503, 505, 507—510 calculations, 129, 130 conoidal shells, 452, 453 continuous beams, 191, 196—198, 202—203 cylindrical shells, 410, 411, 412, 413, 420—422 decompressive forces, 173—174 developing moment determination, 126—127, 128 discrete mass systems, 549—550 dome shells, 429, 431, 434 equilibrium, 51—54, 89 fixed-end, 241—242 frames, 236—237, 241—242, 261—262, 264—266 grid structures, 358—360 plates, 362—370, 375, 378—379, 382—388, 392, 398—400 simply supported beams, 140, 141, 149, 167 statically indeterminate structures, 126—127, 128, 134—135 suspension bridges, 333 thin-walled beams, 456—460, 469—474 zero-moment points, 236—237, 241—242 monolithic structures, 361—402 boundaries, 365—366 cantilever slabs, 375—377

634

circular plates, 385, 386 flat slabs, 389—399 folded plates, 399—402 four-side supported plates, 377—381 load-bearing action, 361—369 orthogonal plates, 361—362, 362, 365—366, 369—384, 385 plates, 361—402 prestressing, 371—375 ribbed plates, 381—384, 385 skew plates, 386—389 two-side supported slabs, 369—375 monolithic systems box girders, 487, 489—490 conoidal shells, 452, 453 continuous beams, 206—207 equilibrium, 62—63, 64 one-storey multibay frames, 235, 236 single-storey, single-bay frames, 224, 225, 234, 235 see also monolithic structures multibay frames, 235—240 multi-degree systems discrete mass systems, 555—577 dynamic equilibrium, 564 forced vibration, 570—571 free vibration, 563—570 seismic excitation, 571—577 stiffness matrices, 555—563 multi-storey frames, 240—251 elastic stability, 288—290 horizontal loads, 242—244 lateral stiffness, 244—249 mixed systems, 249—251 vertical loads, 240—242 multi-storey systems formation, 517—519 free vibration, 568—569 lateral responses, 517—532 layout, 524—525, 530—531, 560, 561 loads, 519—515, 520, 522, 525, 528—530 orthogonal layouts, 525—528, 531—532 plane elements, 521, 523—526 stiffness matrices, 560—563 temperature, 528—532 natural frequency, 537—538 natural modes, 565 natural periods, 537 negative Gaussian curvatures, 403—404, 436—437 Nervi, Pier Luigi, 367, 368 network tied-arch layouts, 313

Index nodal action distribution, 132—133 nodal action impositions, 114, 115 nodal deformations member kinematics, 124, 125 selection, 125, 126 statically indeterminate structures, 123—125 nodal load distributions, 111, 113 nodal rotation, 123 node-located displacement senses, 135—136, 138 non-cohesive soil types, 587—589, 590 non-deformable cross-sections, 460—461 non-shallow shells, 423—432 non-symmetric loads, 322 non-uniform self-weight distributions, 302 non-uniform temperature changes, 68—69 north light shells, 421, 422 oblique simply supported beams, 51, 52 oblong areas, 384, 385 oblong layouts, 443, 444 oblong plates, 399 one-bay rigid frames, 224, 225 one-storey frames, 223—240 elastic stability, 283—287 horizontal loads, 236—238 lateral stiffness, 238—240 vertical loads, 235—236 one-time statically indeterminate frames, 224 open cross-sections general characteristics, 455—460 non-deformable cross-sections, 460 warping-based stresses, 462—465, 468, 470 ‘open’ skeletal systems, 35 opposite fixation actions, 529, 530 orthogonal areas dome shells, 433—434 ground structure supports, 593, 595—596 straight edge hypar shells, 444—448, 449, 450 orthogonal full sections, 90—91 orthogonality conditions, 566 orthogonal layouts cylindrical shells, 421, 422 multi-storey systems, 525—528, 531—532 orthogonal parabolas, 440 orthogonal plates, 369—384 cantilever slabs, 375—377 four-side supported plates, 377—379 load-bearing action, 361—362, 362, 365—366 ribbed plates, 381—384, 385 two-side supported slabs, 369—375 orthogonal systems, 437—440 orthotropic plates, 382—384

paraboloid shells, 403, 404, 436—451 creation, 436—437 equilibrium, 440—443, 444 geometry, 437—440 partial prestressing, 171, 173—177, 199, 372—375 participation factors, 573 periodic sinusoidal acting forces, 544—547 peripheral prestressing, 415 peripheral shear flow, 465—466 permanent loads, 372—373, 376—382, 380, 382 arches, 311—312 cable—beam structures, 320 cable-stayed bridges, 342, 343 suspension bridges, 329, 330, 331 suspension cables, 336—337 physical slope angles, 2—3 pile caps, 618, 619—620, 619 pile foundations ground structure supports, 617—621 horizontal loads, 620—621 lateral pile responses, 620—621 layout, 618 vertical loads, 619—620 pinned supports, 231, 232 plane elements discrete mass systems, 560, 561 multi-storey systems, 518—519, 521, 523—526 stiffness matrices, 560 plane formations, 351—360 plates, 367, 368 skew bridges, 358—360 structural behaviour, 351—358 plane sections, 139—140 plane structures, 65, 84—87 plates, 361, 362 shells, 403 plane systems free vibration, 563—568 stiffness matrices, 562—563 plastic analyses, 358 continuous beams, 220—221 design, 260—262 elastic stability, 293—296 examples, 262—266, 294—296 frames, 257—266 numerical examples, 294—296 plastic behaviour, 147—150, 193, 553—555 plastic deformations, 7 plastic hinges continuous beams, 189—191, 194 frames, 258—262, 265—266 simply supported beams, 149, 150

635

Structural systems: behaviour and design plasticity, 153—154 plastic neutral axis, 148 plates, 361—402 boundaries, 365—366 cantilever slabs, 375—377 circular plates, 385, 386 continuous systems, 580, 581 equations, 361—369 flat slabs, 389—399 folded plates, 399—402 four-side supported plates, 377—381 load-bearing action, 361—369 orthogonal plates, 361—362, 362, 365—366, 369—384, 385 prestressing, 371—375 ribbed plates, 381—384, 385 skew plates, 386—389 two-side supported slabs, 369—375 Poisson ratio, 592—593 polygonal area coverage, 435 polygonal bases, 435—436 pore pressure, 590 portal frames, 226—227, 259, 260 positive Gaussian curvatures, 403—404 post-and-beam frames, 223—224 ‘preselected’ moment diagrams, 264—266 pressure cohesive soil types, 590, 591 cylindrical shells, 409—416, 422 dome shells, 429 non-cohesive soil types, 588—589, 590 rectangular spread footings, 597—598 pressure lines, 224—226, 234, 297—299 prestressed concrete frames, 256—257 loss of prestress, 21—22 material structural behaviour, 18—22 numerical examples, 24—26 simply supported beams, 179 prestressed concrete beams continuous beams, 195—199 design, 170—172, 198—199 simply supported beams, 163—177 tendon design/structural performance, 195—198 prestressing box girders, 505—511, 512, 513 cable—beam structures, 318, 319, 320—322 cable nets, 325—328 cable-stayed bridges, 342, 343, 346—347 cable structures, 388—389, 396—397, 398—399 continuous beams, 208, 214, 217—220 curved girders, 505—511, 512, 513

636

cylindrical shells, 415, 419, 420 dome shells, 431—432, 432 flat slabs, 396—399 four-side supported plates, 381, 382 material structural behaviour, 8, 9 membrane action in shells, 409 straight edge hypar shells, 447, 448, 449 two-side supported slabs, 371—375 principal curvatures, 404, 406—407 principal directions/stresses, 145—146 principal moment, 366—367, 368, 369, 385, 390 ‘priority indices’, 265—266 provisional supports, 213—214 punching shear, 393—395, 398—399 pure membrane states, 431, 442—443 push over responses, 259, 260 pylon stressing, 344—345 qualitative methods, 105—107, 108, 110, 134—138 quasi-beams, 391—392 radial directions, 385 radial displacement, 432 radial forces, 427, 428 radii of curvature, 439—440 reaction principles, 32 real loading, 70—74 rectangular ground plan coverage, 403, 404, 438 rectangular spread footings, 597—609 concrete footing design, 608—609 dimensioning, 608—609 elastic bases, 601—608 soil pressures/settlements, 598—601 rectilinear bars, 36—37 rectilinear girders, 481—495 loads, 481—485 numerical examples, 492—494 profile section deformability, 485—492 rectilinear models, 563 redundant forces, 96, 97, 98, 101, 116—118 redundant structures, 105, 106 redundant systems, 111, 113, 200—205, 507—508 reinforced concrete, 10 frame design, 252—256 material structural behaviour, 10, 13—18 numerical examples, 22—23 simply supported beams, 158, 159, 178—179 reinforced concrete beams continuous beams, 195 service conditions, 150—158 simply supported beams, 150—163 relative displacement, 118—119, 121—123, 548

Index relative rotation, 194 relaxation of concrete, 12—13 relieving action, 398—399 relieving influence, 511, 513 relocation of compressive forces, 168—169 ‘resistance’ forces, 361—365, 363 resistance moment, 375 resonance, 545 rhomboid ground plan coverage, 439, 440 ribbed plates, 381—384, 385 rib structures, 606, 607 rigid beams, 526, 527 rigidity, 109, 110, 112, 130—132, 351—352, 355—360, 377—378 grid structures, 351—352, 355—360 plates, 377—378 statically indeterminate structures, 109, 110, 112, 130—132 ring beams, 427—428, 429—432 ring forces, 416—428, 433 ring membrane actions, 426, 427 ring stresses, 415 ring tension, 409 rotation continuous beams, 191, 194 discrete mass systems, 554—555 dome shells, 427, 428 equilibrium, 79—80, 82—83 flexibility, 82—83 frames, 262 grid structures, 355—356, 357 inertia forces, 534 multi-storey frames, 241—242 one-storey multibay frames, 235 plates, 376 statically indeterminate structures, 117, 120—128 stiffness, 82—83 thin-walled beams, 464—467 rotational springs ground structure supports, 603—604 with rigidity, 131—132 ‘safety factors’, 193 second-order theory, 277—282, 304—307 sectional forces, 41—54 curved beams, 495, 496 deformations, 88—90 equilibrium, 41—54, 57—58, 62—63, 70—71, 88—90 frames, 257—259 statically indeterminate structures, 101

seismic excitation discrete mass systems, 547—555, 571—577 dynamic analyses, 571—574 equivalent static loads, 574—576 multi-degree systems, 571—577 numerical examples, 576—577 single-degree-of-freedom systems, 547—553 self-equilibrating systems, 34, 70—71 box girders, 503—504 cable—beam structures, 320—322 continuous beams, 191, 198, 207—209 hyperbolic paraboloid shells, 442—443, 443 plates, 373—375, 396—397 thin-walled beams, 470—474 self-weight arches, 302 cable-stayed bridges, 341—342 continuous beams, 208, 213 cylindrical shells, 418—419, 420 dome shells, 426—427, 433, 434 single-storey, single-bay frames, 234, 235 straight edge hypar shells, 446—447, 447, 448 semicircular cable-stayed bridges, 501—502 sensitivity factors, 581—582 service conditions material structural behaviour, 10 prestressed concrete beams, 163—173 reinforced concrete beams, 150—158 steel beams, 139—147 service states, 199, 371 settlements, 3 continuous beams, 203—205 ground structure supports, 593—599, 597, 598—601 statically indeterminate structures, 102—104 shallow foundations foundation beams, 609—617 ground structure supports, 592—617 rectangular spread footings, 597—609 shallow shells, 432—436 ‘shear-acting’ frames, 249—251 shear base, 551, 552 shear beam behaviour, 244, 246—247 shear centres, 461—462 shear flow, 465—466, 483—484, 490—491, 503—504 shear forces box girders, 509—510 continuous beams, 209—210, 211—212 cylindrical shells, 410, 411, 417—419, 420 dome shells, 435—436 equilibrium, 68, 89

637

Structural systems: behaviour and design shear forces (cont’d ) membrane action in shells, 406—407 one-storey multibay frames, 235, 237—238 plates, 393 simply supported beams, 144—145, 153—157, 162—163, 172—173, 177 single-storey, single-bay frames, 223—224 statically indeterminate structures, 126—127, 128 straight edge hypar shells, 448, 450 thin-walled beams, 457, 458, 461—462, 471—472 shearing responses, 178 shearing strain, 68 shearing strength, 588 shear mechanisms, 142—143, 444, 445, 445 shear moduli, 68 shear punching, 393—395, 398—399 shear stresses, 44, 45, 140—141 box girders, 484 thin-walled beams, 458, 461—464, 473—474 shear transfer mechanisms, 142—143 shells, 403—453 barrel shells, 416—423 canoidal shells, 451—453 constant internal pressure, 409—413 cylindrical shells, 409—423 cylindrical tanks, 413—415 design, 409 dome shells, 423—436 generation methods, 407 hypar shells, 444—448, 449, 450 hyperbolic paraboloid shells, 436—451 membrane action, 403—409 non-shallow shells, 423—432 shallow shells, 432—436 short barrel shells, 419 silos, 415—416 straight edge hypar shells, 444—448, 449, 450 surface geometry, 404—405 simple supports equilibrium, 32—33, 37—38, 79, 81, 91 plates, 378, 379 simply fixed beams, 116—119 simply supported beams, 139—185 cable structures, 331 cantilever beams, 177—181 compression, 165—172, 179, 180 continuous beams, 191—193, 215, 216, 217 design control, 184—185 equilibrium, 50—51, 52 external prestressing, 181—184 failure states, 147—150

638

partial prestressing, 173—177 prestressed concrete, 179 reinforced concrete, 150—163, 178—179 steel beams, 139—150 suspension bridges, 331 ultimate states, 158—163, 173—177 single-degree-of-freedom systems design spectra, 553, 554 discrete mass systems, 535—555 dynamic equilibrium, 535—536, 548 forced vibration, 542—544 free vibration, 536—542 periodic sinusoidal acting forces, 544—547 plastic behaviour, 553—555 seismic excitation, 547—555 single-mass systems continuous systems, 577—580, 581 free vibration, 358—359, 537 single-storey, single-bay frames, 223—235 horizontal loads, 223—231, 234—235 inclined legs, 234—235 lateral stiffness, 231—234 vertical loads, 223—231 sinusoidal acting forces, 544—547 skeletal systems, 29, 30, 35 skew bridges, 358—360, 384 skew ground plans, 439 skew layouts, 385 skew plates, 386—389 slabs see plates slenderness, 274—275 snow, 4 soils, 2—3, 230—231 cylindrical shells, 414 elastic mediums, 613—616 elastic soil deformational behaviour, 592—596, 597 foundation beams, 610—613 mechanical characteristics, 587—592 pressures, 598—601 rectangular spread footings, 597—601 seismic excitation, 552—553 settlements, 598—601 Winkler model, 610—613 space systems, 325—326 special impact loads, 4 spectral displacement, 550 spherical shells see dome shells splitting arbitrary loadings, 84, 85 springs fictitious springs, 410 ground structure supports, 602—604

Index machine induced vibrations, 583—585 multi-storey systems, 526, 527 with rigidity, 130—132 undamped vibration, 538—539 square ground plans, 422, 423 dome shells, 433, 435 straight edge hypar shells, 448, 450 square layouts, 442, 443 square matrices, 557 stabilisation of cable structures, 316—317 stabilised crack patterns, 16—17 statically determinate load-carrying action, 376 statically determinate parts, 373, 398 statically determinate structures box girders, 506—507, 508, 509—510 continuous beams, 196—197, 217—218 equilibrium, 35—40 frames, 224 statically determined frames, 256—257 statically determined problems, 407—408 statically indeterminate prestressing moment, 197—198, 509—510 statically indeterminate structures box girders, 507—509, 508, 509—510 continuous beams, 196—198 deformation method, 114—138 EI value influence, 104 elastic support, 105, 106, 108—110, 130—132 equilibrium, 37, 39, 41 fixed-end beams, 119—123 force method, 94—114, 115 frames, 224 nodal action distribution, 132—133 qualitative methods, 105—107, 108, 110, 134—138 simply fixed beams, 116—119 stress states, 93—138 support settlement, 102—104 temperature, 101—102, 116, 117, 120 unknown nodal deformations, 123—125 statically redundant forces, 93, 95 statically redundant parts, 373—375, 398 statically redundant structures, 232—233, 256—257 statically redundant systems, 200—205, 232—233 static interactions, 163—164 static loads, 1, 2, 574—576 static theorem, 153—154, 260—261, 263—264, 358 steel, 373—375, 384, 398 concepts, 5—8, 9 continuous beams, 216—217 frames, 251—252, 266—267 single-storey, single-bay frames, 233

steel beams continuous beams, 188—194, 213—214 elastic behaviour, 139—147 failure states, 147—150 fixed-ended beams, 188—191 plastic behaviour, 147—150 service conditions, 139—147 simply supported beams, 139—150, 191—193 stiffening/stiffness arches, 308—310 box girders, 486—487 cable structures, 335—339 cylindrical shells, 410, 418 discrete mass systems, 560, 561 elastic stability, 279—280 equilibrium, 81—84 frames, 256—257 grid structures, 351—352 material structural behaviour, 6—7 multi-degree systems, 569—570 multi-storey frames, 241—242, 244—249 multi-storey systems, 517—521, 523, 524 one-storey multibay frames, 238—240 plates, 364—365, 367, 368, 382—384 single-storey, single-bay frames, 228, 231—234 statically indeterminate structures, 104, 106, 109—114 suspension cables, 335—339 stiffness matrices concepts, 555—557 elements, 555—557 multi-degree systems, 555—563 multiplication, 557, 558 multi-storey systems, 560—563 operations, 557—559 stirrups, 394—395 straight beams, 410—411 straight boundary plates, 365—366 straight edge beams, 450 straight edge hypar shells, 444—448, 449, 450, 451 straight-line generators, 409 straight line segments, 436—437 stresses, 5 arches, 300—302 box girders, 482—484, 487—489, 490 cable structures, 325, 340—342, 344—345 cantilever slabs, 375—376 closed cross-sections, 462—463, 465—467 cohesive soil types, 591 concepts, 29—41 continuous beams, 200, 203—207, 215—217, 219—220

639

Structural systems: behaviour and design stresses (cont’d ) cylindrical shells, 414—415, 421, 422 deformations, 114—138, 271—296 EI value influence, 104 equilibrium, 29—92 flat slabs, 389—393, 394—395 force method, 94—114, 115 four-side supported plates, 377—380 frames, 93—138, 251—252, 282—290 grid structures, 87—92 hyperbolic paraboloid shells, 441 internal forces handling, 41—65 lateral buckling, 290—293 material structural behaviour, 11, 12 membrane action in shells, 405—406, 407—408 open cross-sections, 462—465 plastic analyses, 293—296 plates, 365 second-order theory, 277—282 simply supported beams, 165—166 stiffness, 110—114 suspension bridges, 329, 330 symmetric plane structures, 84—87 thin-walled beams, 456—458, 461—474 two-side supported slabs, 369—370, 371 warping, 461—470 stress—strain diagrams, 6, 8, 9 structural webs, 533—536, 534, 547 structure definition, 39 subgrade reaction moduli, 601—602 supports, 2—3, 32—33, 32, 33 arches, 298—299 box girders, 490—491, 491, 492 cable structures, 331 cantilever beams, 177—181 cohesive soil types, 587, 590—592 continuous beams, 191—193, 213—216, 217 deformations, 592—596, 597 design control, 184—185 elastic soil, 592—596, 597 elastic stability, 274 equilibrium, 32—34, 37—39, 50—52, 79—81, 91—92 external prestressing, 181—184 flat slabs, 389 foundation beams, 609—617 frames, 256, 257 ground structures, 587—621 horizontal loads, 620—621 hyperbolic paraboloid shells, 440—442, 443 internal supports, 194

640

membrane action in shells, 409 multi-storey systems, 524, 525, 526, 527 non-cohesive soil types, 587, 588—589, 590 pile foundations, 617—621 plates, 367, 376, 400 prestressed concrete beams, 163—177 rectangular spread footings, 597—609 reinforced concrete beams, 150—163 simply supported beams, 139—185 statically indeterminate structures, 102—105, 106, 108—110, 130—132 steel beams, 139—150 suspension bridges, 331 types, 32—33, 32, 33, 91—92 vertical loads, 592—596, 597, 619—620 surface loads, 2 suspension action, 445—446, 450 suspension bridges, 328—335 suspension cables, 335—339 symmetric loads, 84—85, 332—335, 333, 481—484 symmetric plane structures, 84—87 T see fundamental periods temperature, 4, 528—532 composite beams, 215—217 continuous beams, 215—217 equilibrium, 68—70 statically indeterminate structures, 101—102, 116, 117, 120 tendons, 195—198, 206—207, 505—507, 511, 513 tension box girders, 483—484, 486—487, 502, 505 cable-stayed bridges, 341—342 conoidal shells, 452, 453 continuous beams, 199, 216—217 cylindrical shells, 413, 415—417 elastic stability, 281 equilibrium, 45, 46 hyperbolic paraboloid shells, 442—443, 443 plates, 381—382, 384 prestressed concrete, 18—22 simply supported beams, 153 single-storey, single-bay frames, 229 straight edge hypar shells, 445—448, 449, 450—451 thin-walled beams, 455—479 bimoment, 469—473, 470 box girders, 481—516 formation, 455, 456 general characteristics, 455—460 longitudinal stresses, 469—470 non-deformable cross-sections, 460—461

Index shear centres, 461—462 shear stresses, 473—474 torsion, 457—458, 462—477 warping, 460—470 thin-walled models, 158, 159 three equilibrium conditions of free formations, 30, 31 three-hinged arch, 298, 305 three-hinged connections, 35—38 three-hinged frames, 54—58, 226—228, 230—231 tied-arch systems, 311—313 top flanges, 213—214, 217—219 torsion, 89, 90—91, 157—158, 159 application examples, 499—502 bimoment, 470—472, 473 box girders, 484—485, 495—507, 509—513 curved girders, 496—502, 510—511, 512, 513 dome shells, 427, 428, 430 equations, 474—477 equilibrium, 496—498 grid structures, 355—360, 356, 357 plates, 362—364, 367, 368, 387—388 thin-walled beams, 457—460, 462—477 warping-based stresses, 462—470 total frame collapse, 257—259 total loads, 344, 397—398, 446, 447 traffic loads, 372—373 transformed concrete sections, 159, 160 translational inertia forces, 534 translational springs, 602—603 translational stiffness, 523, 524 transpose matrices, 557 transversal loads, 441—442 transverse beams, 354—357 transverse bending moment, 370, 420—421, 422 transverse compression, 422 transverse diaphragms, 490—491, 492, 503, 504 transverse displacement, 434 transverse loads, 361, 362, 363—365 transverse stiffness, 279—280 ‘tree’ skeletal systems, 35 triangle of forces, 31, 32 triangular bending moment, 196—198, 507—508 triangular ground plans, 422, 423, 435 triangular loading, 509 triangular truss formations, 36—37 Trost’s proposal, 11, 13, 21—22 trusses elastic stability, 293, 294 equilibrium, 36—37, 61—65 frames, 267—268 simply supported beams, 153—157, 158, 178—179

twist, 438 see also torsion twisting moment, 363—366, 378—379, 384 two hinged solid bars, 124, 125 two-hinge frames, 227—232 two-side supported slabs, 369—375, 382, 399—400 design, 371 prestressing, 371—375 stresses, 369—370, 371 two-span beams, 217—218 two-storey spatial systems, 534, 535 ultimate bending moment, 149 ultimate force, 394 ultimate horizontal load values, 295 ultimate states, 158—163, 173—177, 253—254 ultimate strength, 192—193, 199 ultimate tensile loads, 18, 21 unbonded tendons, 176—177 undamped vibration, 536—539 uniform loads box girders, 504 cylindrical shells, 413 dome shells, 425, 426—428, 429, 434 equilibrium, 50—51 frames, 259, 260 ground structure supports, 593 one-storey multibay frames, 235 plates, 370, 376—379 single-storey, single-bay frames, 226—227 statically indeterminate structures, 116, 119—120 uniformly distributed forces/moment, 411, 412 uniform settlement, 599, 600, 600 uniform temperature changes, 68, 69 unknown displacement, 126—127, 128 unknown nodal deformations, 123—125, 125 unknown rotation, 126—127, 128 unyielding supports, 440—441, 442 upward applied deviation forces, 398—399 upward-curved parabolas, 451 variable-angle truss models, 157, 173 variable beam heights, 178, 179—181 vertical columns, 133 vertical loads box girders, 504, 505, 509—510 cable structures, 315—316 conoidal shells, 453 cylindrical shells, 420 dome shells, 426, 427, 429 elastic stability, 282, 283—285

641

Structural systems: behaviour and design vertical loads (cont’d ) ground structure supports, 592—596, 597, 619—620 hyperbolic paraboloid shells, 441 multi-storey frames, 240—241 one-storey multibay frames, 235—236 pile foundations, 619—620 prestressed cable nets, 326—327 single-storey, single-bay frames, 223—231 straight edge hypar shells, 444, 445 suspension bridges, 328—329 vertical shear forces, 141, 142 vertical stiffness, 517, 518, 519, 520 vertical straight beams, 410—411 vibrations annoying vibration avoidance, 580—585 damped vibration, 539—542 discrete mass systems, 547—553 forced vibration, 542—544, 570—571 free vibration, 536—542, 563—570 human activities, 580—581 machine induced, 581—585 seismic excitation, 547—553, 571—577 single-degree-of-freedom systems, 547—553 undamped vibration, 536—539 Vierendeel beams, 143—144

642

virtual work, 376—377 continuous beams, 191 equilibrium, 70—76, 89—90 multi-storey frames, 246, 247—248 simply supported beams, 153, 161 statically indeterminate structures, 104 warping, 461—467 analyses, 467—469 closed cross-sections, 462—463, 465—467 constants, 469—470 longitudinal stresses, 469—470 open cross-sections, 462—465 shear stresses, 473—474 stresses, 461—470 thin-walled beams, 460—470 water tables, 589, 590 wind forces, 3—4 Winkler model, 610—613 working stress design, 27 yielding, 298—299 yield states, 160, 161 yield stresses, 7—8, 254—255, 394—395 zero-moment points, 236—237, 241—242

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