Federico Guarracino, Alastair C. Walker-Energy methods in structural mechanics _ a comprehensive introduction to matrix and finite element methods of analysis-Thomas Telford Ltd (1999).pdf

Energy methods in structural mechanics A comprehensive introduction to matrix and ®nite element methods of analysis

F. Guarracino and A. Walker, FREng

Thomas Telford

Published by Thomas Telford Publishing, Thomas Telford Limited, 1 Heron Quay, London E14 4JD. URL: http://www.t-telford.co.uk Distributors for Thomas Telford books are USA: ASCE Press, 1801 Alexander Bell Drive, Reston, VA 20191-4400 Japan: Maruzen Co. Ltd, Book Department, 3±10 Nihonbashi 2-chome, Chuo-ku, Tokyo 103 Australia: DA Books and Journals, 648 Whitehorse Road, Mitcham 3132, Victoria First published 1999

A catalogue record for this book is available from the British Library ISBN: 0 7277 2757 5 # F. Guarracino and A. Walker, and Thomas Telford Limited, 1999 All rights, including translation, reserved. Except for fair copying, 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 Books Publisher, Thomas Telford Publishing, Thomas Telford Limited, 1 Heron Quay, London E14 4JD. This book is published on the understanding that the authors are 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 re¯ect the views of the publishers. Typeset by Academic ‡ Technical Typesetting, Bristol Printed and bound in Great Britain by Bookcraft (Bath) Limited

Preface The last two centuries have witnessed an impressive development in the size and complexity of engineering structures. During the early part of the 19th century much of the design analysis was based on empiricism and experience. The latter half of the 1800s saw the inception of methods giving insight to the forces generated in frame structures due to external e€ects, such as gravity, wind and thermal radiation. Supreme examples of the complexity of structures that could be constructed on the basis of these hand calculation methods are the Ei€el Tower (Fig. 1) and the Forth Bridge (Fig. 2). These structures are evidently robust and their continued use is a tribute to the skill of the engineers to apply the analysis methods available to them and to ensure that the structures incorporated adequate levels of safety to compensate for the engineers' ignorance of the detailed forces and deformations in the elements of their structures. The methods that were available to structural and mechanical engineers from 1860 to 1950 were based essentially on energy principles and on the requirements of equilibrium of forces and compatibility of deformations. These factors, basically, still underlie all the current methods of structural analysis. The major change in the intervening years since the last quarter of the 19th century has been the possibility of applying the methods to even more complex forms of structures and the possibility of extending the linear analysis into the non-linear regime to examine the failure modes of the structures. The advent of structures such as aircraft and airships, and their development during the period 1900±1930 emphasised the need for the design of minimum weight structures which implied that the very large levels of conservatism incorporated in bridge and building design had to be reduced. The methods available to airship designers were adequate to allow them to calculate the forces in the quite complex shapes of wires and beams that formed the frames of the ships. An example of this is given in Fig. 3, which shows an airship called the R27 on the right. The photograph shows the bulkhead between

iv

Preface

Fig. 1. Ei€el Tower

gas bags. The bulkhead was formed by an outer ring frame supported from an inner hub by tensioned wires, much like the construction of a bicycle wheel. The novelist Nevil Shute1 was an engineer working on a later airship called the R100, a sister ship of the ill-fated R101, which had a similar bulkhead, and he describes how he and a colleague worked together for many weeks calculating day after day the tensions in the wires until eventually their analysis converged to satisfy force equilibrium and deformation compatibility requirements. This higher degree of analysis complexity was required for airships to maintain their low weights but, in general, for bridges and framed 1

Nevil Shute, Slide Rule, Heinemann, London, 1954.

Preface

v

Fig. 2. Forth Rail Bridge (courtesy of ICE Library)

buildings the level of analysis was much less sophisticated which meant that the estimation of the real forces in the members of structures could be quite inaccurate. However, reliance on experience and factors of safety enabled a satisfactory approach to the design and construction

Fig. 3. Structural frame in R27 (courtesy of Imperial War Museum)

vi

Preface

of bridges during the ®rst half of the 20th century. The analysis of complex structures such as aircraft and ships was based on formulations derived from experience, testing and the results of research and specialist analysis performed on elements such as plates and curved shells. It was widely appreciated that the availability of analysis capabilities that would supply many more accurate details on the deformations and loads in structural elements in complex structures, and that could be applied in a commercial environment, would provide engineers with the facility to design structures with speci®ed levels of safety and with greater economy of materials. More particularly, however, the growth of the aerospace industry, with supersonic military and passenger aircraft ¯ying at very great altitudes, and the need to design huge o€shore structures and ships with high levels of structural eciency and the ability to withstand extreme environmental loads, provided a signi®cant spur to develop more e€ective structural analysis capabilities. The major innovation that has enabled success to be achieved came not from the mechanical or structural engineers but from electronic engineers, mathematicians and physicists. They have made available extremely powerful computing capabilities that enable tasks, such as those tackled by Nevil Shute, to be completed in a few seconds instead of weeks and at very much greater structural detail. The present generation of structural and mechanical engineers has access to powerful computers and programs that enable designers routinely to evaluate the strains and displacements in very complex structures and thus to re®ne the form of their structure to obtain greater economic and technical eciency. Figs 4±6 are a few examples of structures that have been designed during the last 20 years using computer-based analysis methods. This book presents a uni®ed development of the analysis principles and methods that are currently embodied in computer analysis programs. The development is founded on fundamental energy principles, particularly the law of conservation of energy, and proceeds through a uniform yet essentially rigorous approach by gradually introducing the reader to variational principles and thus deriving the underlying approaches in frame and ®nite element analysis. It is the authors' experience in teaching structural analysis methods, and applying them in practice, that in text books the presentations of the basis of energy methods and frame and ®nite element analysis can too often be confusing and lead engineers to misunderstand the results they obtain from the application of these methods. In fact, in many cases, because engineers do not have a sucient understanding of the principles underlying ®nite elements, they treat computer analysis as a `black

Preface

vii

Fig. 4. Tsing Ma suspension bridge (courtesy of Mott MacDonald)

box' and take an uncritical approach to the results from the analyses. This, we feel, is dangerous since a computer analysis will nearly always achieve some result from the calculations, but this does not necessarily imply that the result is appropriate to the structure being designed. In preparing this book we have followed a logical structure that will enable the reader to follow the development of the ®nite element and frame analysis methods from fundamental energy principles through to the type of formulations that are embedded in commercial ®nite element programs. The emphasis in the development is on strain, compatibility of deformations and energy. Stress is considered as a less important, secondary concept calculated from the strains in a structure, so that it practically plays no part in the theoretical basis of the analysis methods. Our intention is to provide engineers with an understanding that allows them to interpret results from ®nite element analyses and to determine what the results mean in a physical sense. The presentation in the text encompasses methods of structural analysis that can be used to provide simpli®ed formulations in particular cases and enable the investigation of parametric variation without recourse to lengthy computation. Our experience is that these methods, such as those proposed by Rayleigh, Cotterill, Castigliano and others, are very valuable during preliminary design where a `feel' for the structure's capability to resist the applied loads is being investigated. The text shows that these simplifying methods are one step towards the computer-based methods used in the design oce. By basing the development in this text ®rmly on

viii

Preface

Fig. 5. O€shore structure

energy principles it is clear how the principles of Cotterill±Castigliano and the trial function methods proposed by Rayleigh and Ritz lead directly to practical computer methods without any recourse to additional and somewhat misleading aspects such as the concept, sometimes introduced in other treatments, of cutting a structure up into totally independent elements and then imposing equilibrium at discrete points, i.e. the node joining elements. This book is essentially an introductory text and therefore deals with a linear elastic treatment of structural analysis. However, we feel it is important that the reader is made aware of at least one of the limitations of this form of analysis. As structural forms become evermore ecient in their use of materials the possibility of failure by buckling becomes an important factor. This aspect is introduced in the last

Preface

ix

Fig. 6. Space shuttle craft

chapter of this book to show how geometric non-linearity can be introduced into ®nite element and frame analysis. Of course, this type of non-linearity leading to failure in practical structures is only one facet of the limit state behaviour of structures. The other facets such as material non-linearity, local buckling and the e€ects of the variations in structural geometry are reserved for consideration in a future text. F. Guarracino A. Walker

Acknowledgement This book has been a long time in gestation. During that period, our wives have shown considerable forbearance over the time we have spent discussing the contents of the book and developing the presentation. We are truly grateful to Angela and Barbara, without whose patience and encouragement this book would never have seen the light of day.

Contents 1

Work and energy Introduction 1.1 Work and energy 1.2 The principle of conservation of energy 1.3 Equilibrium of mechanical systems

2

Kinematics and equilibrium of systems of rigid bodies Introduction 2.1 Equilibrium of rigid bodies 2.2 A general in®nitesimal displacement kinematics for systems of rigid bodies 2.3 Reactions of constraints 2.4 Internal reactions for a straight beam 2.5 A three-dimensional example Suggested exercise problems

11 11 11

Deformation of bodies and material properties Introduction 3.1 Deformation properties in one dimension 3.2 One-dimensional thermal strain 3.3 Three-dimensional strain 3.4 Strain±displacement relationships 3.5 Simpli®cations possible when deformations are very small 3.6 Deformation in the neighbourhood of an arbitrary point 3.7 Transformation of the components of strain with the change of reference frame 3.8 Principal directions of deformation. Maximum and minimum extensions 3.9 Plane strain 3.10 Strain±displacement compatibility conditions

49 49 50 58 60 63

3

1 1 1 5 9

20 33 35 43 46

67 70 73 75 78 79

xii

Contents

3.11 Strain energy 3.12 Yield criteria Suggested exercise problems 4

5

6

7

81 87 91

Theory of elastic deformation of beams Introduction 4.1 Deformation of axially loaded bars 4.2 Deformation of beams 4.3 The theory of very small displacements 4.4 Euler's method for the analysis of beam deformations 4.5 The boundary conditions associated with Euler's equation of beam deformation 4.6 Example of application of Euler's method of beam deformation for a beam with a variable depth 4.7 Approximate solution of Euler's equation of beam deformation using the method of ®nite di€erences 4.8 Applied displacements and constraint reactions 4.9 E€ects of temperature changes Suggested exercise problems

93 93 95 98 102 105

General principles in the analysis of linear elastic structures Introduction 5.1 The principle of superposition of the e€ects in the linear theory of elasticity (existence and uniqueness of the solution) 5.2 Reciprocal theorems in the linear theory of elasticity

141 141

Total potential energy Introduction 6.1 The principle of the stationary value of total potential energy 6.2 Two important remarks regarding the principle of stationary value of total potential energy 6.3 Cotterill±Castigliano's ®rst theorem 6.4 Cotterill±Castigliano's second theorem in the linear theory of elasticity 6.5 Numerical example: frame analysis by Cotterill±Castigliano's theorem Suggested exercise problems

151 151

154 155

The method of trial functions Introduction 7.1 The basis of the method of trial functions

169 169 170

111 116 121 126 130 137

142 147

151

158 160 166

Contents

7.2 The method of Rayleigh±Ritz 7.3 The quality of the trial function 7.4 Numerical example: the method of trial functions applied to a cantilever with linearly varying depth 7.5 The systematic search for a trial function 7.6 Localised Rayleigh functions 7.7 Numerical example: application of Rayleigh localised trial functions to the analysis of a frame Suggested exercise problems 8

9

10

xiii

173 178 183 187 189 199 203

Matrix analysis of pin-jointed trussed structures Introduction 8.1 Plane pin-jointed structures as systems of axially loaded bars 8.2 Matrix formulation of the elastic equilibrium equations for plane pin-jointed structures 8.3 Numerical example: analysis of a truss 8.4 Matrix analysis of pin-jointed space structures Suggested exercise problems

205 205

Matrix analysis of rigid-jointed framed structures Introduction 9.1 Plane rigid-jointed frame structures as systems of beams 9.2 Matrix formulation of the elastic equilibrium equations for plane rigid-jointed frames 9.3 Numerical example: analysis of a rigid jointed frame 9.4 Matrix analysis of space frames 9.5 Equivalent joint loads 9.6 Restraint displacements 9.7 Some ®nal remarks on the matrix analysis of structures Suggested exercise problems

233 233

Analysis of thin plates Introduction 10.1 Kirchho€'s model of plate deformation 10.2 Equilibrium of stretched plates 10.3 Equilibrium of bent plates. 10.4 An example of the method of trial function for the solution of bending plates. 10.5 More examples of application of simple trial functions to plate bending

267 267 269 270 274

206 212 221 226 230

233 240 245 249 254 259 261 263

286 290

xiv

Contents

10.6 Localised Rayleigh±Ritz method applied to plate bending 11

12

The theory of ®nite elements Introduction 11.1 Sub-division of a structure into ®nite elements 11.2 Displacement functions and shape functions 11.3 Element mapping and intrinsic coordinates Ð isoparametric elements 11.4 Convergence criteria for the displacement functions Ð the patch test 11.5 Some general families of shape functions 11.6 Strain energy, work of applied loads and equilibrium equations 11.7 Evaluation of sti€ness matrices and equivalent nodal loads Ð numerical integration 11.8 Numerical example: analysis of a compressed foundation

295 303 303 305 307 314 318 323 330 335 341

Stability of equilibrium and non-linear deformations of beam±columns Introduction 12.1 The concept of stability. Initial buckling of columns 12.2 Energy criterion of stability 12.3 Some illustrative examples of stability conditions. Branching points and limit points. 12.4 Euler's equation of beam deformation in the presence of axial loads 12.5 Analytical treatment of some examples of columns 12.6 Method of trial functions for the initial buckling of columns 12.7 Localised trial functions applied to the initial buckling of struts 12.8 Method of trial functions for the initial buckling of frames 12.9 Analysis of a frame loaded by vertical and lateral loads 12.10 Localised trial function analysis of frame buckling Suggested exercise problems

404 408 417

Index

421

353 353 353 357 358 365 372 384 391 399

1.

Work and energy

Introduction

This chapter introduces the concepts of work and energy that are the basic building blocks of the methods of analysis that are developed later in the textbook. We take as self-evident the concept of force and follow Newton's1 approach that a force is an in¯uence that causes changes in displacements and velocities within a de®ned mechanical system. A force may also cause deformation of a body and is represented by a vector that impinges on a body along a speci®ed line of action. This chapter proposes a de®nition for equilibrium and establishes a test for the equilibrium of a body in a given frame of reference.

1.1

Work and energy

The concept of work W is basically related to the idea of a force moving its point of application along a certain path. For example, given a force P (assumed constant) and a displacement s ˆ …A2 ÿ A1 † of its point of application (Fig. 1.1), we will de®ne the work done by the force as the scalar product W ˆ P  s ˆ P  …A2 ÿ A1 †

…1:1†

which can also be expressed in terms of Cartesian components W ˆ Px sx ‡ Fy sy

…1:2†

Similarly, the elementary work of a force moving along a certain path s at a certain time t can be de®ned by dW ˆ Ps …t† ds

…1:3†

The coordinate ds is tangential to the path s at the time t and Ps …t† is the component of the force P at the time t in the same direction (Fig. 1.2). 1

Newton, Isaac (Woolsthorpe, Lincholnshire, 1642±London, 1727), English physicist and mathematician, one of the most prominent ®gures in the history of modern science.

2

Energy methods in structural mechanics Sx

A2

Sy A1

y

P Py x

Px

Fig. 1.1

The work done between the positions a0 and a1 is consequently written as a…1

Wˆ

Ps ds

…1:4†

a0

From a dimensional point of view the work W is the product of a force multiplied by a length (that is ‰F  LŠ) and this concept encompasses moment times rotation, and pressure times change of volume: these all being products characterised by the same physical dimensions. Work is performed when we apply a moment to the stopper in order to open a bottle, or when we in¯ate a balloon by increasing its internal pressure (Fig. 1.3). It appears thus natural to introduce the concept of energy as the capability of a system to perform work.

a1

a0 Ps

Fig. 1.2

P

Work and energy

3

m

P

Fig. 1.3

The variation of energy
E is de®ned as the rate of work performed by, or on the system
E ˆ W

…1:5†

The energy E constitutes the measure of the capability of performing work stored in the system itself. It is worth noting that, according to this de®nition, we are concerned only with changes of energy. In fact, as we will see later, the absolute value of energy of a system is not relevant to our purposes. Di€erent forms of energy can be recognised in nature. A body with a mass m moving at a certain speed v can move another resting body by hitting it; the ®rst body is said to possess a form of energy relative to the stationary body that is called kinetic energy (Fig. 1.4). The moving body can deform an elastic spring which, in turn, can push the impinging body away by exerting a force on it (Fig. 1.5). The spring is thus able to possess what is de®ned as elastic energy. Heat is a form of energy: this can be illustrated by heating a pot of water and arranging the pressure of the vapour to lift a weight, which v*2

m2 m2 m1 v*1

v1 m1

Fig. 1.4

4

Energy methods in structural mechanics

v

m

v

m

l0

m

l < l0

Fig. 1.5

was of course the primary source of power of the Industrial Revolution. This is illustrated diagrammatically in Fig. 1.6. Other examples of energy are found in chemical and nuclear reactions. Appropriate mathematical expressions can be derived to quantify the energy gained or lost by a system. We say, for example, that within a certain coordinate frame a mass m moving at a velocity v possesses the kinetic energy described by the formula K ˆ 12 mv2

…1:6†

Notice that the velocity must be measured relative to some frame of reference which for many engineering calculations can conveniently be taken as ®xed to the Earth's surface. It is convenient to use the convention that when the body is stationary in that particular reference frame it has zero kinetic energy. However, relative to another frame, say one ®xed on the Sun, the body has considerable velocity and is seen to possess kinetic energy. In a nuclear reaction an absolute energy can be attributed, according to the theory of relativity, to a mass m by E ˆ mc2

…1:7†

W

H2O

Fig. 1.6

Work and energy

5

where c is the velocity of light and is an absolute value irrespective of the frame of reference. To sum up, it is very important to bear in mind that energy is a general property attributed to all systems and bodies in nature. The universal nature of energy provides the foundation for the treatment of structural mechanics which is developed here.

1.2

The principle of conservation of energy

The principle of conservation of energy, also known as the First Law of Thermodynamics, is a general axiom founded on the presumption that, in the absence of any interaction with the environment, the energy of a given system remains constant in the passage from one state to another. In other words, energy is not created or destroyed in a system isolated from its environment and, consequently, the only possible variation is given by the exact amount of exchange of energy or, what is equivalent, work with other systems. In the following text such a system, which experiences only internal changes of state without interaction with external agents, is called a closed system. Strictly speaking, no perfectly closed systems are found in nature, since there is always some interaction between all material bodies of the universe across ®nite systems. Nevertheless, the concept of a closed system is a very useful approximation, both from a theoretical and a practical point of view, and is employed extensively in the development of our analytical methods. Everyone has experienced the exchange of heat by conduction that takes place between two bodies at di€erent temperatures or the phenomenon of heat generation by friction. The principle of conservation of energy requires that in a closed system in the ®rst case the sum of the heat of the two bodies remains constant, as well as in the second case the sum of the work performed against friction and the heat produced is a constant value. This principle can be summarised in the following mathematical statement, valid for any closed system E ˆ const:

…1:8†

As we are going to deal with simple thermodynamic systems, attention will now be restricted to two basic ways of exchanging energy: namely, mechanical work W and heating Q. Henceforth, we can write the principle of conservation of energy in the following form
E ˆ W ‡ Q

…1:9†

6

Energy methods in structural mechanics P

P

l l'

Fig. 1.7

where
E is the change of the level of energy in the system, W is the work done on or by the system and Q is the amount of heat exchanged. We will assume that the positive sign is for the work done on the system and the heat given to it; the negative sign holds for work done by the system and heat taken from it. Equation (1.9) is sucient to link the variations of energy of a given system to its changes of state, that is the changes of temperature, position or shape of its component parts. In fact, the state of a mechanical system can be described by a number of independent geometrical variables together with one thermodynamic variable, say the temperature T. In mechanics these geometrical variables are called generalised coordinates and the number of such coordinates n required to specify completely the state of the system is equal to the number of degrees of freedom less, of course, the geometrical constraints imposed at the system boundary. This is exempli®ed later in the text. For the sake of clarity, let us now consider the simple structural element shown in Fig. 1.7, i.e. a bar subject to forces P at both ends. It is natural to assume as the parameters de®ning the state of the system to be its actual length l 0 and its temperature T. The initial temperature of the element is T0 and its corresponding length is l. A zero value of energy is attributed to this initial state which is acceptable since we are interested in measuring, by de®nition, the variation of energy. When the system undergoes an in®nitesimal change of state, from the principle of conservation of energy (equation (1.9)) we have dE ˆ dW ‡ dQ

…1:10†

where, by de®nition, dW ˆ P dl

…1:11†

and dl is the variation in length of the member. If the material is linearly elastic, the following relationship can be stated between the deformation and both the applied force and variation of temperature
l ˆ l 0 ÿ l ˆ

P ‡ l…T ÿ T0 † k

…1:12†

Work and energy

7

where k, the sti€ness, and , the coecient of linear thermal expansion, are constants for the material and the element, at least for small changes in temperature. Further, the following relationship can be assumed between the heat exchanged by the system and its temperature: dQ ˆ C dT

…1:13†

provided the coecient of heat capacity C is not a function of temperature or sti€ness; which experimentally is accurate for a limited temperature range and small deformations. The requirement in equation (1.13) is adequate for the type of mechanics considered here. By means of equations (1.11), (1.12) and (1.13), equation (1.10) becomes dE ˆ C dT ‡ k…
l ÿ l …T ÿ T0 †† d
l

…1:14†

and by a simple integration we get E…T; l 0 † ˆ C …T ÿ T0 † ‡ 12 k
l 2 ÿ kl
l …T ÿ T0 † ‡ A

…1:15†

As we have attributed the zero value of energy to the state de®ned by T ˆ T0 and l 0 ˆ l, we have in this case E…T0 ; l† ˆ C…T0 ÿ T0 † ‡ 12 k …l ÿ l†2 ÿ kl …l ÿ l† …T0 ÿ T0 † ‡ A ˆ0

…1:16†

The value of the constant A is, therefore Aˆ0

…1:17†

E…T; l 0 † ˆ C…T ÿ T0 † ‡ 12 k
l 2 ÿ kl
l …T ÿ T0 †

…1:18†

and This expression constitutes the de®nition of the internal energy E of the structural member in term of its variables of state, i.e. its actual length l 0 and its temperature T. In the same manner, from equation (1.9), we can derive the expression of the energy possessed by a body of mass m de®ned as a point moving along a straight path at a certain velocity v, as illustrated in Fig. 1.8. Let us assume that the body is motionless at the beginning of the path and attribute zero energy to this initial state. Subsequent states of the system will be de®ned by a single variable; the most convenient variable is the position x along the straight path. According to the second of Newton's laws, a force must be applied to the body in order to move it along its direction.

8

Energy methods in structural mechanics v=0 m

x

m

v

Fig. 1.8

This force will produce a variation of velocity with the time t given by the formula dx_ …1:19† dt where the force F…t† is considered as a function of time and the dot represents the derivative with respect to the time, i.e. F…t† ˆ m

dx …1:20† dt Equation (1.9) states, in absence of any exchange of heat energy, that v  x_ 

dE ˆ dW

…1:21†

The work done, dW, is therefore given by the formula dx_ dx dt Thus, using equation (1.20) dW ˆ m

…1:22†

dE ˆ mx_ dx_

…1:23†

This equation can be integrated between the velocity limits x_ ˆ 0 and x_ ˆ v, and yields E ˆ 12 mv2

…1:24†

This is the expression that was stated earlier, see equation (1.6), and represents the energy possessed by the body due to its velocity v, i.e. the rate of change of its state variable x with respect to time.

Work and energy

9

On account of this relationship we can de®ne this form of energy as kinetic energy, and the derivation of equation (1.24) provides a basis for the de®nition of equation (1.6).

1.3

Equilibrium of mechanical systems

A mechanical system is said to be in static equilibrium when it is motionless and experiences no change of state within a speci®ed frame of reference, i.e. all the geometrical variables required to specify completely the position of the system in the space do not change for all time. This is a common sense de®nition that provides engineers with a valuable operational approach to the analysis and design of real structures. In the case of simple thermodynamic systems this intuitive concept can be extended to all the state variables of the system; that is to say, the geometrical variables plus the temperature T. We want now to establish a condition of equilibrium for mechanical systems in terms of energy. This will form the basis of our ability to predict the state of equilibrium of bodies subjected to any speci®ed system of forces. Later, the condition of equilibrium is further extended to enable the analysis of complex structures to evaluate their global and local deformations. Without loss of generality to our purposes we shall restrict our attention to the case of absence of heat exchange and assume dQ ˆ 0. Evidently, if a motionless system has no rate of change in its geometrical variables xi with time, this means dxi ˆ0 dt

…1:25†

The kinetic energy of the system is thus constant and can be set equal to zero. That is X 1 _2 Kˆ …1:26† 2 mi xi ˆ 0 i

where mi are the generalised masses of the system which, by de®nition, are ®nite and positive quantities. In this case a generalised mass can be thought of as asssociated with a certain geometrical variable xi . Vice versa, if equation (1.26) holds then equation (1.25) must be true, since the mass must always be positive, and the two statements are therefore equivalent. This means that we can state the constant zero value of kinetic energy as the condition of equilibrium for a mechanical system. As we assumed dQ ˆ 0, the energy of a mechanical system in equilibrium reduces itself to the strain energy U. This kind of energy is

10

Energy methods in structural mechanics

related to the deformation of the system: in equation (1.18), for example, it is represented by the term 1=2k
l 2 ÿ kl0
l …T ÿ T0 †. This aspect of the energy is considered in greater detail in chapter 3. For the moment, we can state that for a mechanical system in equilibrium we have, from the principle of conservation of energy (equation (1.10)), dE ˆ dU ˆ dW

…1:27†

This equation means that any in®nitesimal increment of work done on the system tends to produce only deformation of the system itself and does not produce any kind of motion. Importantly, equation (1.27) can be used to obtain a practical test of equilibrium for mechanical systems. Let us consider a mechanical system in a state of equilibrium and suppose that on account of the geometrical constraints it could actually be able to move. We can then apply to it a trial displacement ®eld; this is de®ned to be a ®eld of displacements that have in®nitesimal magnitude and are compatible with the constraints on the system, but are imaginary and do not actually occur. Such a trial ®eld will be denoted with the symbol , and the same symbol will be used for all the variations in any quantity a€ected by the same set of displacements. These quantities will be called virtual to underline the fact that they could take place but in actual fact are imaginary. As the system has been de®ned to be in equilibrium, equation (1.27) holds for any trial set of displacements and we have U ˆ W

…1:28†

This relationship is at the same time a condition and a test of equilibrium. Thus, if the condition in equation (1.28) is satis®ed for any ®eld of displacements then equation (1.27) is true and we have dK ˆ 0: this means that the kinetic energy of the system is constant and equal to zero and the system is in equilibrium. Alternatively, we can simply specify any possible set of displacements of the system and verify that equation (1.28) is satis®ed: this operation constitutes a practical test of equilibrium. This very important approach, which is central to engineering mechanics, is explored in detail throughout this book. In the following chapter our study starts with the equilibrium of simple systems of rigid bodies.

2.

Kinematics and equilibrium of systems of rigid bodies

Introduction

Although strictly speaking all bodies and engineering structures are deformable, common sense tells us that some systems can be treated as being composed of components that are essentially rigid, that is undeformable. An example of such a system would be a collection of billiard balls that move about the table and travel quite large distances compared to their diameters. When the balls collide there will be some deformation of the diameter, but this is so small compared to the dimension of the diameter and to the distances moved that the deformations can be completely ignored in the calculation of the movements of the balls. This approach is very important in engineering mechanics since it is not uncommon for a structure to experience both global and local displacements. Another example of this division is the movement of an aircraft that as part of its function may travel many thousands of kilometres in ¯ight and have extensive vertical movement in response to air currents and turbulence. These same aerodynamic e€ects will also cause ¯exing of the wings. The deformations in the ¯exing are very small compared to the global body movements so that the global movements are usually calculated on the basis that the aircraft is a rigid body. The local deformations of the wing are calculated in separate analyses with the aircraft modelled as being stationary in a de®ned frame of reference. This chapter presents the practice of using energy principles to evaluate the equilibrium of systems of rigid bodies with respect to applied loading.

2.1

Equilibrium of rigid bodies

We will de®ne as rigid a body whose distance between any pair of its points remains unchanged for any transformation and for all time.

12

Energy methods in structural mechanics

This idealisation is very useful in the analysis of many systems which can be approximated to be rigid on account of their very limited deformations. More generally speaking, we can say that any displacements within the body are negligible compared to the gross displacement experienced by the body itself. According to this de®nition, for a rigid body, or system of bodies, dl ˆ 0 for any pair of its points and thus Uˆ0

and

dU ˆ 0

…2:1†

No strain energy can therefore be stored in an undeformable body. The condition of equilibrium (equation (1.28)) becomes W ˆ 0

…2:2†

which requires the work done by the external forces for any ®eld of virtual displacements to be zero. This equation is generally known as the principle of stationary work. In the case of rigid bodies it is possible to perform this test of equilibrium in a very simple manner as the components of any in®nitesimal displacement ®eld can be represented in terms of linear diagrams. Let us start from the analysis of a rigid body free from any constraints on its movements in the two-dimensional Cartesian space (i.e. in our case a plane whose points are referred to coordinates relative to two mutually orthogonal axis; essentially this idea was introduced by Descartes1 ). Its positions will be completely de®ned by the coordinates of two of its points or, alternatively, by the coordinates of one point and an angle of rotation in the plane. We say that the body possesses three degrees of freedom and we call these parameters (i.e. the two coordinates of a ®xed point and the angle of rotation) Lagrangian coordinates (after Lagrange2 ). The numbers of degrees of freedom can be determined, at least for simple collections of bodies, by counting the number of coordinates that are needed to draw the ®xed shapes of the bodies in some arbitrary position. To ®x our ideas consider a simple rigid bar (Fig. 2.1) subject to two forces P…A† and P…B† at the ends A and B. As an e€ect of its de®nition, a rigid bar is one that cannot deform by bending or stretching and can only experience rigid body movements. To describe the position of the bar in the x±y space we choose as Lagrangian coordinates, for example, the coordinates of the extremity 1

Descartes, Rene (La Haye, Touraine, 1596 ± Stockholm, 1650), French mathematician and philosopher. 2 Lagrange, Joseph-Louis (Turin, 1736 ± Paris, 1813), Italian mathematician of French origins.

Kinematics and equilibrium of systems of rigid bodies

13

B C P (A)

P (B)

A

y d (C)

l

φ

x

Fig. 2.1

A and the angle of rotation  along axis orthogonal to the x±y plane. The angle of rotation is assumed positive when anticlockwise. This description of the system means that the coordinates of any point, say C, along the axis of the bar can be expressed with respect to the coordinates of the point A, x…A† and y…A†, to the angle  and to the distance d…C† from the end point A x…C† ˆ x…A† ‡ d…C† cos…†

…2:3†

y…C† ˆ y…A† ‡ d…C† sin…†

…2:4†

Now consider a possible displacement and refer to Fig. 2.2. The new coordinates x0 …C† and y0 …C† of a point at distance d…C† from the extremity A are given by the relationships x0 …C† ˆ x0 …A† ‡ d…C† cos…0 † 0

0

…2:5†

0

y …C† ˆ y …A† ‡ d…C† sin… †

…2:6†

where x0 …A†, y0 …A† and 0 are the coordinates of the displaced point A and the angle assumed by the displaced bar, respectively. B C

δu (C)

A

δu (A) y

φ' A

φ

Fig. 2.2

B C

x

14

Energy methods in structural mechanics

If instead of an actual ®nite displacement we consider a virtual displacement ®eld, the virtual movement can be de®ned by 0 ˆ  ‡  Therefore, equations (2.5) and (2.6) become

…2:7† 3

x0 …C† ˆ x0 …A† ‡ d…C†…cos…† cos…† ÿ sin…† sin…††

…2:8†

y0 …C† ˆ y0 …A† ‡ d…C†…sin…† cos…† ‡ cos…† sin…††

…2:9†

Now, since the virtual displacement ®eld is required to be in®nitesimal, for small displacements expand the terms sin…† and cos…† in the following manner cos…0† sin…0† 2  ÿ  ‡


…2:10† 1! 2! sin…0† cos…0† 2  ‡  ‡


…2:11† cos…† ˆ cos…0† ÿ 1! 2! and take into consideration only the ®rst order terms in the above series; that is, we can assume 2 , 4 ; . . .  1 and 3 , 5 ; . . .  , and hence sin…† ˆ sin…0† ‡

sin…†  

…2:12†

cos…†  1

…2:13†

Thus equations (2.8) and (2.9) become x0 …C† ˆ x0 …A† ‡ d…C† cos…† ÿ d…C† sin…†  0

0

y …C† ˆ y …A† ‡ d…C† sin…† ‡ d…C† cos…† 

…2:14† …2:15†

or, more conveniently, x0 …C† ˆ x0 …A† ‡ dx …C† ÿ dy …C† 

…2:16†

y…C† ˆ y…A† ‡ dy …C† ‡ dx …C† 

…2:17†

where dx …C† ˆ d…C† cos…† and dy …C† ˆ d…C† sin…† are, respectively, the components of the distance d…C† along the axes x and y. Let us denote with ux ˆ x0 ÿ x and uy ˆ y0 ÿ y the components of the virtual displacement of a chosen point. By subtracting equation (2.3) from equation (2.16) and equation (2.4) from equation (2.17) we obtain

3

ux …C† ˆ ux …A† ÿ dy …C† 

…2:18†

uy …C† ˆ uy …A† ‡ dx …C† 

…2:19†

Note: cos… ‡ † ˆ cos  cos ÿ sin  sin and sin… ‡ † ˆ sin  cos ‡ cos  sin .

Kinematics and equilibrium of systems of rigid bodies

15 δux

B

δφ A

δuy δφ

Fig. 2.3

It is evident that if the components of displacement are drawn for all the points of the element we get two straight lines that constitute the diagrams of the displacement components, as shown in Fig. 2.3. Let us now write down the condition of equilibrium equation (2.2) following on from the relationships in equations (2.18) and (2.19). De®ning Px and Py , respectively, as the components of the generic force P along the axes x and y, we have W ˆ Px …A† ux …A† ‡ Py …A† uy …A† ‡ Px …B†‰ux …A† ÿ ly Š ‡ Py …B†‰uy …A† ‡ lx Š ˆ 0

…2:20†

and grouping the terms, we have W ˆ ‰Px …A† ‡ Px …B†Š ux …A† ‡ ‰Py …A† ‡ Py …B†Š uy …A† ‡ ‰ÿPx …B†ly ‡ Py …B†lx Š  ˆ 0

…2:21†

By de®nition the virtual variation of the Lagrangian coordinates ux …A†, uy …A† and  are totally arbitrary and independent. Therefore, in order to satisfy equation (2.21), each of the terms separately must equal zero, and equation (2.21) can be replaced by the following three equations Px …A† ‡ Px …B† ˆ 0

…2:22†

Py …A† ‡ Py …B† ˆ 0

…2:23†

ÿPx …B†ly ‡ Py …B†lx ˆ 0

…2:24†

16

Energy methods in structural mechanics

Notice that three independent coordinates have been identi®ed as being sucient to describe the actual position of the body and this, by the requirement of zero virtual work, will automatically give three independent equations of equilibrium. The equilibrium conditions may also be expressed as Px …A† ˆ ÿPx …B†

…2:25†

Py …A† ˆ ÿPy …B†

…2:26†

ly Py …B† ˆ lx Px …B†

…2:27†

which means that the forces P…A† and P…B† must have the same direction of the axis of the element and be opposite in direction for the system to be in equilibrium, as shown in Fig. 2.4. The application of zero virtual work shown above provides exactly the same result as we would have arrived at using Newton's Laws. In fact, equations (2.22) to (2.24) are equivalent to . . .

Summation P of vertical components of the forces being equal to zero, i.e. Py ˆ 0. Summation P of horizontal components of the forces being equal to zero, i.e. Px ˆ 0 Summation of the P moments about some point in the plan being equal to zero, i.e. M ˆ 0.

However, the approach leading to equations (2.25) to (2.27) is somewhat more advantageous than the Newtonian method. A merit of the energy approach is that in the case of complex systems it may be relatively simpler for engineers to decide on the numbers of degrees of freedom than to decide on how many equilibrium equations are valid. In fact, the energy method automatically determines the correct number and format of the equilibrium equations. So far we have dealt with a rigid but unconstrained body. As a further step and as an example of a constrained body let us now consider the same rigid bar hinged at one end (Fig. 2.5). B P(B)

ly

A P(A)

Fig. 2.4

lx

Kinematics and equilibrium of systems of rigid bodies

A

17

B P(B)

y

x

Fig. 2.5

It is evident that a vector displacement u…A† of the extremity A is not possible any more, so ux …A† and uy …A† are both zero. We are thus left with a single degree of freedom, determined by the Lagrangian coordinate . Any virtual displacement ®eld will be de®ned by the variation . Applying a force P…B† at the point B, the work requirement in equation (2.2) can be written W ˆ ÿPx …B†ly  ‡ Py …B†lx  ˆ 0

…2:28†

yielding the condition of equilibrium ly Py …B† ˆ lx Px …B†

…2:29†

This requires that the direction of the applied force is along the axis of the bar. A further and particular merit of the energy approach is that it automatically rules out the presence of constraint reactions in absence of friction. The situation becomes slightly more complicated dealing with a system of rigid bodies mutually constrained. As usual, it is ®rst of all necessary to identify the number of Lagrangian parameters required to completely de®ne the position of the system in the Cartesian space. We have already seen that in the Cartesian two-dimensional space a rigid unconstrained body possesses three degrees of freedom. With reference to the hinged rigid bar in Fig. 2.5 this number has reduced to one because a point of the bar is ®xed in the space and no displacements are allowed at that point. From observation, we can intuitively state a formula that provides the number of degrees of freedom of a rigid body in two-dimensional

18

Energy methods in structural mechanics C A

B

y

x

Fig. 2.6

space, i.e. f ˆ3ÿc

…2:30†

where f is the total number of degrees of freedom and c is the number of simple constraints applied. In the case of the hinged bar (Fig. 2.5), c ˆ 2 because two components of translation are eliminated at the point A. If we are dealing with a system of rigid bodies mutually constrained, equation (2.30) may be re-written as f ˆ 3b ÿ c

…2:31†

where b is the number of rigid bodies forming the system. It must be emphasised, however, that in this case we can also have relative constraints, i.e. constraints limiting the mutual positions of the bodies. As an example of this, consider the system shown in Fig. 2.6. This ®gure shows two rigid bars mutually hinged. One bar is also attached to the foundation by another hinge and the other is constrained to slide parallel to the foundation at the point C. The number of simple constraints is thus given by the sum of two, due to the hinge, plus one, due to the sliding support, plus two, due to the internal hinge (once the position of one of the two members is de®ned the internal hinge suppresses two degrees of freedom for the other). We have therefore f ˆ 3b ÿ c ˆ 3  2 ÿ …2 ‡ 1 ‡ 2† ˆ 6 ÿ 5 ˆ 1

…2:32†

This means that all the con®gurations of the system can be determined by a single coordinate, which can be the angle of rotation of the ®rst rigid bar member  or the y-coordinate of the central hinge, y…B†. This can be con®rmed by sketching the bodies at some arbitrary position and noting that only one angle has to be de®ned to enable the sketch to be completed, as is shown in Fig. 2.7.

Kinematics and equilibrium of systems of rigid bodies B

19

C

δφ

A

δuy (B) y

x

Fig. 2.7

Once the diagrams of the displacement components have been drawn, see Fig. 2.8, the equation of equilibrium (2.2) can be very straightforwardly written for any system of forces acting upon the system. It is worth noting that we have zero displacements for points A and B, which are therefore ®xed centres of rotation, and consequently we see zero values in both diagrams. Moreover, we read the same value for the components of displacement in correspondence to the internal hinge, which is a relative centre of rotation. This concept is more fully clari®ed in the following section, where kinematics of systems of rigid bodies for in®nitesimal displacement ®elds will be systematically developed. δux

δuy

δφ

Fig. 2.8

20

Energy methods in structural mechanics

2.2

A general in®nitesimal displacement kinematics for systems of rigid bodies

So far we have assumed that a rigid body in two-dimensional Cartesian space, i.e. a plane, possesses three degrees of freedom and that for an in®nitesimal rotation we may assume sin…†  

…2:33†

cos…†  1

…2:34†

We have subsequently discussed some very simple examples of systems of rigid bodies. Now a more general theory for the kinematics of rigid bodies will be developed which will allow us to deal with any system of rigid bodies mutually connected in the two-dimensional space. Once these concepts are fully stated, the extension to the three-dimensional Cartesian space will be brie¯y discussed. We begin our discourse by introducing the following theorem, proposed by Euler4 : Every displacement of a two-dimensional rigid body in a plane can be reduced to a translation or a rotation about an axis normal to the plane.

This is true in general and holds both for in®nitesimal and ®nite displacement ®elds. We are not going to demonstrate this theorem, but the reader will ®nd it easy to verify the generality of the theorem by trial sketches. In fact, once two points A and B have been chosen on a rigid body, by de®nition the line A±B will retain the same length for any displacement of the body. De®ning A0 ±B 0 as the position of the line after the displacement has occurred, it is straightforward to recognise that in every instance we can bring A±B on A0 ±B 0 (and therefore the whole body) by means of a simple translation or a rotation. Conversely, it is worth noting that, dealing with in®nitesimal displacement ®elds, we can take the normal to the radius for the arc, as shown in Fig. 2.9; actually, this is the geometrical meaning of formulae (2.12) and (2.13), i.e. taking the ®rst-order terms only in the series of equations (2.10) and (2.11). In the present section we will assume that all the displacements are in®nitesimal and their components will be simply indicated as ux and uy . Rotations from the reference con®guration, which also are assumed to be in®nitesimal, will be simply indicated by . In this case the kinematics of a free body is extremely simple; once we have de®ned the centre of rotation (which is a proper one in the 4

Euler, Leonhard (Basel, 1707 ± Petersburg, 1783), Swiss mathematician, author of several fundamental contributions to the mechanics of structures.

Kinematics and equilibrium of systems of rigid bodies

21

A A'

d

φ C

Fig. 2.9

case of a rotation and an improper one in the case of a simple translation), we can project this point on the axes where we are drawing the components of displacement. When the centre is a proper one, its displacement is zero, otherwise the centre is said to be at in®nity and this means that all the points have the same components of displacement. In fact, a translation is nothing more than an act of rotation having its centre at an in®nite distance, being  ˆ 0. The displacement of any of the points of the body, shown in Fig. 2.9, can be given with reference to the displacement of the centre of rotation, according to equations (2.18) and (2.19) ux …A† ˆ ux …C† ÿ dy …C†

…2:35†

uy …A† ˆ uy …C† ‡ dx …C†

…2:36†

where dx …A† and dy …A† are, respectively, the components of the distance d from the centre of rotation C to the point A along the axes x and y. As stated, in the case of a simple translation the centre is an improper point at in®nity,  ˆ 0 and we have ux …A† ˆ ux …B†

…2:37†

uy …A† ˆ uy …B†

…2:38†

where A and B are any pair of points on the body. Note that the diagrams of the displacement components form two straight lines as was pointed out in the previous section and is illustrated in Fig. 2.10. With regard to the constraints a distinction is made between external and internal constraints. External constraints have the e€ect of limiting

22

Energy methods in structural mechanics ux

A

dy (A)

φ

dx (A)

C uy

φ

Fig. 2.10

the displacement of the body with reference to the points of the plane, which are considered ®xed. Internal constraints have the e€ect of limiting the displacement of the body with reference to the points of another body of the system. Therefore the internal constraints a€ect relative displacements. External and internal constraints both can be classi®ed according to the number of degrees of freedom they a€ect. For example the ®xed end of the rigid rod in Fig. 2.11 prevents the displacement

A

B

y

x

Fig. 2.11

ux (A) = uy (A) = φ = 0

Kinematics and equilibrium of systems of rigid bodies A

23

B ux (A) = uy (A) = 0

φ≠0

φ

y

x

Fig. 2.12

A

y

ux (A) ≠ 0 uy (A) = 0

φ≠0 x

Fig. 2.13

of the point A and any rotation, and therefore suppresses all the three degrees of freedom of the rod. A hinge acts to prevent any displacement of the point to which it is applied but leaves open the possibility of rotation (Fig. 2.12). A roller prevents any displacements of the point P along its normal axis (Fig. 2.13). A two-pinned bar prevents, for in®nitesimal displacement ®elds, any displacements along its axis and therefore acts like a roller (Fig. 2.14).

A

y

ux (A) ≠ 0 uy (A) = 0

φ≠0 x

Fig. 2.14

24

Energy methods in structural mechanics

Whatever their shapes, it is evident that the constraints described above can be classi®ed according to the degrees of freedom they suppress in the plane and can be therefore categorised as simple, double or triple. Having referred the displacements of all the points of a rigid body to the position of its centre of rotation, we can now examine the kinematics of a constrained body through the conditions imposed by the constraints on its centre of rotation. In Fig. 2.15 the most common Name

Type

Notes

Roller

Simple

ux 6ˆ 0 uy ˆ 0  6ˆ 0

The centre of rotation must belong to the normal to the roller's direction

Simple

ux 6ˆ 0 uy ˆ 0  6ˆ 0

The centre of rotation must belong to the axis of the bar

Double

ux ˆ 0 uy ˆ 0  6ˆ 0

The centre of rotation is the pinned point

Double

ux 6ˆ 0 uy ˆ 0 ˆ0

The centre of rotation is improper and lays at the in®nite on the direction of the bars

Triple

ux ˆ 0 uy ˆ 0 ˆ0

No centres of rotation are allowed

y x Two-pinned bar

y x Hinge

y x Pair of parallel two-pinned bars

y x Fixed end

y x

Fig. 2.15

Kinematics and equilibrium of systems of rigid bodies

25

Fig. 2.16

types of constraints have been collected, together with their e€ects on the centres of rotation of a rigid body. Of course, it may happen that the constraints imposed on the body forbid completely the existence of a centre of rotation and in this case the body is ®xed in the de®ned space. This is the case, for example, of the body constrained by three two-pinned bars, as shown in Fig. 2.16. This happens because the centre of rotation should belong to the axis of each of the two-pinned bars but they do not concur in any point. Therefore the centre of rotation cannot exist and the body is prevented from any movement. By de®nition, a simple constraint imposes a single condition on the centre of rotation, while a double constraint imposes two conditions and a triple constraint imposes three. As the position of a proper or improper centre of rotation is de®ned by two conditions, it is clear that at least three simple constraints are needed (or, what is equivalent, a double plus a simple one or a triple one) to ®x a body in the space. In fact, a double constraint is fully equivalent to two simple constraints and a triple is fully equivalent to three simple ones. This is in accordance with the formula (2.30) stated earlier f ˆ3ÿc provided, however, that the constraints are all e€ective. In other words, if we consider the example shown in Fig. 2.17 we will discover that one two-pinned bar is ine€ective as it requires the centre of rotation to be on the axes of the other two two-pinned bars. The centre of rotation is therefore placed in C and the beam still possesses a degree of freedom. It must thus be borne in mind that the term c in equation (2.30) (as well as in equation (2.31)) is the number of the e€ective simple constraints applied. We have also stated that the internal constraints restrict the displacements of a rigid body with reference to another

26

Energy methods in structural mechanics

C

Fig. 2.17

one and therefore a€ect relative displacements instead of absolute ones. Nevertheless the internal constraints can still be calculated as shown in the equation (2.31) f ˆ 3b ÿ c since for any part of the system they suppress the same number of degrees of freedom as if they were external. For example, consider the system shown in Fig. 2.18. It is evident that the internal hinge still suppresses two degrees of freedom for the element BC whose position is de®ned by the position of the element AB and by the angle of rotation 2 . Given the presence of internal constraints, discovering if a constraint is e€ective or not becomes a slightly more complicated matter for a general system of rigid bodies. In fact, we must recognise along with the presence of absolute centres of rotation the presence of relative centres of rotation. Nevertheless it is quite simple to state two theorems regarding the position of the absolute and relative centres of rotation for any in®nitesimal displacement ®eld. Let us consider two rigid bodies A and B and their in®nitesimal ®elds of displacement in the plane. To ®x our ideas we can make reference to Fig. 2.19. A

B

φ2

C

Fig. 2.18

Kinematics and equilibrium of systems of rigid bodies

27 ux

cB

B

cAB

cA

A

uy

Fig. 2.19

According to what we have previously said we can imagine that the respective displacement ®elds can be reduced to rotations with regard to the centres CA and CB . Now, consider the relative displacement ®elds, i.e. the displacements of the points of B with respect to A and the displacements of the points of A with respect to B. In order to draw the relative displacements of the points of B with respect to A we can imagine A to be ®xed in space and express the relative displacement ®eld with regards to a relative centre of rotation CAB . Moreover, we can imagine B to be ®xed in space and express the relative displacements of the points of A with respect to B with regard to the same centre, being the two relative ®elds equal in magnitude but opposite in direction. As a direct consequence of this fact we have CAB  CBA

…2:39†

28

Energy methods in structural mechanics

This is because that in two-dimensional space the relative centre of rotation is a point which has the same absolute displacement either considered belonging to A or to B. Since we know that the displacement of a point P in the ®rst case is orthogonal to the direction CA ±P and in the second to the directions CB ±P we are led to the conclusion that the relative centre CAB must lie on the line CA ±CB . We can therefore state the ®rst of the proposed theorems If each of two rigid bodies A and B undergoes an in®nitesimal displacement ®eld in twodimensional space, the absolute centres of rotation CA and CB and the relative centre CAB must all lie on the same straight line.

We call this theorem the ®rst rule of concatenated displacements for rigid bodies. The second rule of concatenated displacements for rigid bodies, i.e. the second of the proposed theorems, follows in a straightforward manner from the ®rst one. If we consider three rigid bodies A, B and C and their in®nitesimal displacement ®elds in the plane we can recognise the existence of three absolute centres of rotation CA , CB and CC and of three relative centres of rotation CAB , CAC and CBC . We already know that relative centres must be aligned with the corresponding absolute ones. Let us consider one of the bodies, say A, to represent the ®xed space. The relative centre CAB becomes the absolute centre CB and the relative centre CAC becomes the absolute centre CC . Therefore CBC must be aligned with CB and CC and so also must be the relative centres CAB , CAC and CBC . Thus the second rule of concatenated displacements for rigid bodies can be written as: For in®nitesimal displacement ®elds, the relative centres of rotation of any triplet of rigid bodies must lie on a straight line.

The ®rst and the second rule of concatenated displacements for rigid bodies allow us to deal quite simply with the kinematics of any system of rigid bodies. For example, consider the system of rigid rods in Fig. 2.20. This is a system of three rods constrained to a foundation by two hinges and mutually connected by two more hinges. The number of double constraints is four and we get from formula (2.31) f ˆ 3b ÿ c ˆ 3  3 ÿ 2  4 ˆ 1

…2:40†

This means that we have at least one degree of freedom for the system, which therefore can actually move. In order to discover if the constraints are all e€ective, we can perform the analysis of its kinematics by means of the two rules stated. First of all locate the absolute centres of rotation C1 and C3 ,

Kinematics and equilibrium of systems of rigid bodies B

29

C 2

1

3

A

D

y

x

Fig. 2.20

which must coincide with the pinned ends A and D, and thereafter the relative centres C12 and C23 , which have to be located at the hinges B and C. This is shown in Fig. 2.21. The ®rst rule of concatenated displacements now enables us to de®ne the position of the absolute centre C2 . This centre must lie at the intersection of the lines C1 ±C12 and C23 ±C3 , as it is shown in Fig. 2.22. No constraints can be classi®ed as ine€ective and the knowledge of these centres is sucient to draw the diagrams of the components of displacement of the system. These diagrams depend on the parameter chosen B ≡ C12

C23 ≡ C 2

1

A ≡ C1

C3 ≡ D

y

x

Fig. 2.21

3

30

Energy methods in structural mechanics ux

C2

C12

C23

φ C3

C1

uy

φ

Fig. 2.22

to represent the single degree of freedom available, e.g. the angle of rotation of the rod 1. Now, consider a ®nal aspect of concatenated displacement. So far we can assert that for a system of n rigid bodies undergoing a certain ®eld of in®nitesimal displacements we have n absolute centres of rotation. We can also arrange all the centres of rotation in a symmetrical matrix 0 1 C11


C1n .. .. A @ ... …2:41† . . Cn1


Cnn where the centres of rotation o€-diagonal in the matrix are the relative ones. Therefore, the number of relative centres is n…n ÿ 1† 2 and the total number of centres is n…n ‡ 1† 2

…2:42†

…2:43†

Kinematics and equilibrium of systems of rigid bodies A

B

C 1

D

31

E 2

y

x

Fig. 2.23

We can now re¯ect on the fact that we usually take into consideration systems whose ®elds of displacement are dependent on a single parameter; that is, one degree of freedom activated at a time. In order to de®ne any diagram of displacement components for such systems we must draw n straight lines and we need to de®ne 2n ÿ 1 points. We are thus led to the conclusion that in order to draw these diagrams we do not need to locate all the centres of rotation but we just need to know the position of 2n ÿ 1 centres. Finally, it is necessary to underline that a careful study of the centres of rotation of the system must be always carried out, as this is the only way to investigate the kinematics of rigid bodies in a consistent and rigorous manner. In fact, as we have already pointed out, we can often have an odd layout of the constraints themselves. For example, let us consider the structure in Fig. 2.23. In this system we have two rigid rods constrained by two double constraints and three simple ones. If equation (2.31) is applied, regardless of the e€ectiveness of the constraints, we have f ˆ 3b ÿ c ˆ 3  2 ÿ …2 ‡ 2 ‡ 1 ‡ 1 ‡ 1† ˆ ÿ1

…2:44†

The fact that the number of degrees of freedom is negative might suggest that the system is not only allowed to move, but it has one redundant constraint, too. However, if we carry on the study of the centres of rotation, as illustrated in Fig. 2.24 we discover that rod 1 cannot move because its absolute centre of rotation must belong to the axes of the two-pinned bars and at the same time it must be an improper centre on the direction of the pair of parallel two-pinned bars. As the twopinned bars are not aligned with the pair of parallel ones this centre cannot exist. Moreover, each of these two-pinned bars is sucient to forbid the existence of the centre C1 and therefore one is redundant. The hinge in D can therefore be considered as an absolute centre of rotation for rod 2 and this is the case, as the roller in E requires this

32

Energy methods in structural mechanics

C1 ≡ ∞ B

C

D

E

1

2

A y

x

Fig. 2.24

centre to be aligned with its axis. It is evident that the roller constitutes an ine€ective constraint. Therefore, element 2 can actually move and its ®eld of displacements is de®ned by a single degree of freedom, i.e. the angle of rotation , as is shown in Fig. 2.25. Equation (2.31) gives the correct result if we drop from the calculation the redundant constraint and the ine€ective one f ˆ 3b ÿ c ˆ 3  2 ÿ …2 ‡ 2 ‡ 1† ˆ 1

…2:45†

To summarise with respect to the kinematics of systems of rigid bodies, we can have two types of constraints (a) (b)

e€ective constraints, ones that are able to suppress some degrees of freedom of the body ine€ective or redundant constraints, ones that a€ect degrees of freedom already suppressed by other constraints. A

B

1

C

2

D

φ

y

x

Fig. 2.25

E

Kinematics and equilibrium of systems of rigid bodies

33

The only way to discover the e€ectiveness of a constraint is to conduct a careful study of the kinematics of the system.

2.3

Reactions of constraints

So far we have assumed that the condition of equilibrium for the systems of rigid bodies is given by the principle of stationary work (equation (2.2)), that is, W ˆ 0 which requires the work done by the external forces for any ®eld of virtual displacements to be zero. If we remember the de®nition of virtual displacement ®eld, i.e. a ®eld of displacements that is in®nitesimal and compatible with the constraints of the system, but is imaginary and does not actually occur, we are immediately led to the conclusion that the principle of stationary work can only be applied to those systems that possess at least one degree of freedom. This is a requirement because a system that does not possess any degrees of freedom must be considered in equilibrium irrespective of the particular set of applied forces, since we will always have ui …A† ˆ 0

…2:46†

for any of its points and therefore equation (1.26), which has been proposed as the de®nition of equilibrium, X 1 _2 Kˆ 2 mi xi ˆ 0 i

always holds true. However, the principle of stationary work can also be applied to evaluate the forces that the constraints must apply to the system to keep it in equilibrium. This approach can be de®ned as the principle of substitution of constraints and establishes an equivalence between a geometrical entity, i.e. a constraint, and a static entity, i.e. a force. As an example, consider the three-pinned arch shown in Fig. 2.26, that is subjected to the vertical force P at the hinge B. Equation (2.31) gives f ˆ 3b ÿ c ˆ 3  2 ÿ …2 ‡ 2 ‡ 2† ˆ 0

…2:47†

as there are no ine€ective constraints, since the centres C1 , C2 and C12 are not aligned. Therefore the system is in equilibrium regardless of the particular set of applied forces. As the hinge at A can be imagined as being composed of two orthogonal rollers, as pointed out in Fig. 2.27, we can conceptually suppress the hinge at A and leave the vertical roller only (Fig. 2.28).

34

Energy methods in structural mechanics

C12 ≡ B

l

P

C1 ≡ A

C2 ≡ C

y

x

Fig. 2.26

In this situation the system can move to take up a displaced shape, and we can evaluate, using the principle of stationary work (equation (2.2)), the vertical force exerted on A in order to maintain the system in equilibrium. This force can be thought of as being equivalent to the e€ect of the horizontal roller. With reference to Fig. 2.28, equation (2.2) can be written as
‡ R
ˆ 0 …2:48† 2 In fact, once we have assigned an arbitrary vertical displacement ÿ
to the point A, the vertical displacement of the point B is seen to be ÿ
=2 and from equation (2.48) we have W ˆ P

P …2:49† 2 The negative sign means that the force R has the opposite direction to the force P. We call R the reaction of the horizontal roller. The procedure is therefore very simple Rˆÿ

A A

Fig. 2.27

Kinematics and equilibrium of systems of rigid bodies

35

C12

P

C1 ≡ C2

R uy

δuy (A) = –Δ

δuy (B) = – Δ 2

Fig. 2.28 In order to obtain the reaction of a certain constraint we can suppress that constraint and draw the in®nitesimal displacement ®eld thus allowed (if any). The reaction is represented by the force that, acting instead of the constraint, renders the total work of all the applied actions to be zero. The reaction is considered to act in the same direction of the suppressed constraint.

Of course the procedure is meaningful when there are no redundant constraints. If the suppression of a constraint does not enable the system to have a displacement ®eld, we are not able to identify any work in equation (2.2) and cannot evaluate the reaction. Actually, it should be noticed that in the presence of redundant constraints a system of rigid bodies is generally in equilibrium and the fact that we are not able to calculate the reaction of some constraints is only due to the idealisation of rigidity. In the following text we will learn that dropping this simpli®cation allows the determination of the reaction of all of the constraints. For the moment, we say that systems whose constraint reactions can be evaluated according to the scheme of rigid bodies are called statically determinate, otherwise they are called statically indeterminate.

2.4

Internal reactions for a straight beam

In the previous section it was assumed that external constraints are able to transmit certain forces to the bodies in order to keep them in equilibrium. In the same manner we can assume that a certain part

36

Energy methods in structural mechanics

P z

A

B

y

l

P

–

P 2

–

P 2

Fig. 2.29

of a body transmits a system of forces to the remaining part of it in order to keep that part of the body in equilibrium. The material in this section is illustrated, for clarity, by the simply supported beam shown in Fig. 2.29. Using the principle of stationary work (equation (2.2)) we can simply determine the reactions transmitted by the supports at A and B, i.e. P …2:50† 2 We can also contemplate an imaginary vertical cut through the beam at the section marked Y±Y, at a distance z from the support A. In this situation we can consider the portion of beam between the support A and the section Y±Y as a free body subjected to the force ÿP=2 at A, as shown in Fig. 2.30. Notice that a free body is de®ned to be RA ˆ RB ˆ ÿ

a portion of a structure, or system, that is abstracted from the structure and the continuity of the structure between the free body and the rest of the structure is represented by forces applied at the boundary of the free body.

Thus the portion of beam shown in Fig. 2.30 will be subjected to the resultant of the forces transmitted by the portion of beam on its right side, which must be able to maintain it in equilibrium. Looking at the case of a beam contained in the plane z±y, let us decompose this resultant force in terms of a horizontal component N

Kinematics and equilibrium of systems of rigid bodies

37

z Y

A

P

Y

–P

Y'

2

B

Y'

–P 2

M N S

Fig. 2.30

and a vertical component S both applied to the centroid of the cross section Y±Y, together with a moment M, as is customary for any system of forces in the plane. We will call these components axial force, shear force and bending moment, respectively, in the beam at the section Y±Y. Their values are calculated using the principle of stationary work, equation (2.2), and are P Pz and Mˆ …2:51† 2 2 If the principle of zero work is applied to the rigid beam on the right hand of the cut, we ®nd that the resultant of the forces transmitted to the portion of beam on the right side of the cut must have the same magnitude but opposite direction of the resultant of the forces applied to the left side of the cut; which is in accordance with the third of Newton's Laws. Therefore, we have N ˆ 0;

Sˆ

P Pz and MR ˆ ÿ …2:52† 2 2 where the superscript R denotes that the forces are applied to the right side of the cut. Notice that no superscripts were applied to equation (2.51). This is because it is conventional to assume that the axial force, the shear force and the bending moment, i.e. the so called characteristic of internal action, are the components of the force acting on the left side of the cut. They are normally referred to a Cartesian reference frame in which the z-axis coincides with the axis of the beam, aligned with the centroids of the normal cross section (which for simplicity here is thought to be uniform along the beam length), and the x and y-axes coincide N R ˆ 0;

SR ˆ ÿ

38

Energy methods in structural mechanics

x

z

y

Fig. 2.31

with the principal axes of inertia of the cross sections. The convention adopted in this book is for the z axis to be positive when measured from the left-hand end of the beam and the y axis is ordinarily arranged to be vertical with downwards positive, as shown in Fig. 2.31. Of course, this is purely a convenient convention and others can be equivalently adopted, provided they enable consistent communication. By sequentially applying the equilibrium equation (2.2) to any chosen portion of the beam under consideration, it is easy to recognise that (a)

(b)

(c)

the axial force at any section is equal to the opposite of the sum of all the horizontal components of the external forces acting to the left side of that section or to the sum of all the horizontal components of the external forces acting to the right side of that section the shear force at any section is equal to the opposite of the sum of all the vertical components of the external forces acting to left side of that section or to the sum of all the vertical components of the external forces acting to right side of that section the bending moment at any section is equal to the opposite of the sum of all the moments yielded by the external forces which act to the left side of that section with reference to the centroid of the section itself or to the sum of all the moments yielded by the external forces which act to the right side of that section.

It is therefore quite an easy procedure to determine the values of the characteristics of the internal action at any section. At the same time it can be valuable to obtain a graphical representation of the distribution of the various forces acting along the beam by plotting their distribution along the length of the beam. These plots are

Kinematics and equilibrium of systems of rigid bodies

39

w (z) z

y

l

Fig. 2.32

commonly referred to as axial force, shear force and bending moment diagrams. To simplify the preparation of the diagrams it is very useful to establish some general relationships between distributed forces, shear force and bending moment. Distributed force is nothing more than a system of forces that act over at least part of the length of the beam. This type of force is generally measured per unit length and is represented by a function q…z† of the position z along the axis of the beam. For example, a special case of a uniformly distributed force is given by the self-weight w…z† of the simple supported beam of Fig. 2.32, while a non-uniformly distributed force is exempli®ed by the cantilever beam shown in Fig. 2.33 (for the sake of simplicity both these forces are considered to act normally to the beam axis). Given a displacement ®eld uy …z†, the work of a distributed force q…z† between the positions z0 and z1 is obviously provided by the integral z…1

Wˆ

q…z†uy …z† dz

…2:53†

zo

as shown in Fig. 2.34.

q (t) z

y

l

Fig. 2.33

40

Energy methods in structural mechanics q (z) z0

z1

z uy (z)

y

Fig. 2.34

Let us now take into consideration an in®nitesimal element of beam obtained by means of a conceptual cut at two very close cross sections (Fig. 2.35). The element under consideration will be subject to an axial force N, a shear force S and a bending moment M at its left hand end and to an axial force N ‡ dN, a shear force S ‡ dS and a bending moment M ‡ dM at its right end. Since the length dz is very small, we can neglect the amount by which the distributed vertical load q…z† varies over the element. The equilibrium of the in®nitesimal part of beam can be once more investigated by means of the principle of stationary work, equation (2.2). There are three degrees of freedom in the z±y plane, so  W ˆ …ÿN ‡ N ‡ dN†  uz ‡ …ÿS ‡ S ‡ dS ‡ q…z† dz†  uy ‡ …ÿM ‡ M ‡ dM ÿ S dz†  ˆ 0

…2:54†

In this equation the second order in®nitesimal quantities, namely the products of the type dS dz, are neglected. q (z)

M

M + dM N + dN

N

S

S + dS dz

Fig. 2.35

Kinematics and equilibrium of systems of rigid bodies

41

Equation (2.54) simply leads to the following three equations dN ˆ 0

…2:55†

dS ˆ ÿq…z† …2:56† dz dM ˆS …2:57† dz which must hold for the equilibrium of any portion of the beam. In particular note that equation (2.56) implies that the slope of the shear force distribution is equal in magnitude but opposite in sign to the intensity of the vertically distributed lateral load at any position along the beam. Similarly, note that equation (2.57) implies that the slope of the bending moment distribution is equal to the value of the shear force at any position along the beam. However, it is obvious that the presence of an applied vertical point force P (or couple M) at a certain section A implies that for A0 and A00 , the sections immediately on either side of the conceptual cut, we have S…A00 † ˆ S…A0 † ‡ P…A† 00

0

M…A † ˆ M…A † ‡ M…A†

…2:58† …2:59†

Therefore, according to equations (2.56) and (2.57), an applied vertical point force P leads to an abrupt step change in both the shear force distribution and the slope of the bending moment distribution, while an applied couple M causes an abrupt step change in the bending moment distribution but leaves the slope unchanged. At this point a simple step by step procedure can be outlined for plotting the shear force and bending moment diagrams in cases of statically determined systems of beams (a) (b) (c)

®nd the reactions of the constraints at the ends of each beam ®nd the values of the shear force at key points along the beam (e.g. points of application of point forces) plot the shear force diagram according to equations (2.56) and (2.58).

Notice that equation (2.56) immediately leads to the following relationship z…2 Sy …z2 † ÿ Sy …z1 † ˆ ÿq…z† dz …2:60† z1

for any couple of positions which do not include an applied vertical point force.

42

Energy methods in structural mechanics

(d )

locate the sections where the shear force is zero. These sections, according to equation (2.57) are positions of maximum or minimum bending moment (e) ®nd the values of the bending moment at key points along the beam ( f ) plot the bending moment diagram according to equations (2.57) and (2.59).

Notice that equation (2.57) also leads to an useful integral relationship: z…2

M …z2 † ÿ M …z1 † ˆ

S …z† dz

…2:61†

z1

for any two sections of the beam which do not include an applied couple. An example of shear force and bending moment diagrams for a system of beams is shown in Fig. 2.36. The presence of axially directed distributed or point loads and the consequent procedure for drawing the diagram of the axial force can be treated by means of trivial modi®cations of the previous equations and is left to the reader. q

l

P

l/2

l/2

+P

+ P + ql

2

2

–

– Pl – ql 2

P 2

S

2

2

+ Pl 4

Fig. 2.36

M

Kinematics and equilibrium of systems of rigid bodies

2.5

43

A three-dimensional example

Up to now examples in this chapter have been restricted to those in two-dimensional Cartesian space, i.e. to rigid bodies which can be analysed with reference to a plane of symmetry that contains also the applied forces and constraints. This may seem a severe restriction, but actually it is not, as a very large number of structures can be analysed as formed of substructures which can be treated as a plane. However, the concepts which enabled the treatment of equilibrium in the preceding sections remain exactly the same in the case of rigid bodies in three-dimensional Cartesian space. The condition of equilibrium is obviously the principle of stationary work (equation (2.2)), that is W ˆ 0 for any ®eld of virtual displacements. Of course the virtual displacements take place in the three-dimensional space, but apart from this fact all the simpli®cations deriving from their being in®nitesimal hold true in the same manner. In order to illustrate this fact with a practical example, consider the beam shown in Fig. 2.37.

1

A

3

P

x B

2 C

6

y z l 5

Fig. 2.37

44

Energy methods in structural mechanics b

P 3

6

h

2

Section A

4

Section B

5

Section C

Fig. 2.38

This is a simple steel I beam constrained by six two-pinned short bars and subject to a force P at the middle of the span l. The exact position of the constraints and of the applied force is shown in Fig. 2.38 with reference to the cross sections at A, B and C. It is easy to infer that a rigid body in the three-dimensional space possesses six degrees of freedom. For example, once we have assumed the three orthogonal axes x, y and z to be immovable on the beam, it is clear that the position of the beam in the space will be determined by the position of these axes. As the position of the axes is uniquely identi®ed by, say, the coordinates of their origin O0 and by the three angles of rotation that the axes themselves form with reference to another reference frame ®xed in the space, it follows that six parameters are sucient to de®ne the position of the beam in the space. The situation is shown in Fig. 2.39. Thus, six simple constraints may be sucient to suppress any degree of freedom, provided they are all e€ective. This fact, as has been said before, can be investigated only by analysing the kinematic of the body. With reference to section A of Fig. 2.38, it appears clear that this section can only rotate about the centre given by the intersection of the directions of the pinned bars 2 and 3. The same happens to section C, but section B is allowed to rotate about a centre which must lie along the axis of the pinned bar 4 and therefore cannot be aligned with those of sections A and C. We can therefore conclude that the beam cannot undergo any rotation about any axis parallel to the axis z. As any translation of the sections A and C is forbidden, too, we see that the origin O0 of the axes x, y and z cannot move in the plane x±y as well. Three degrees of freedom consequently must be eliminated. On account of the presence of pinned bars 2 and 5 any rotation about the x axis must be excluded and the same happens to any rotation about the y axis on account of the presence of pinned bars

Kinematics and equilibrium of systems of rigid bodies

z

45

y

ζ

α3 α2 O'

α1

x O

η

ξ

Fig. 2.39

3 and 6. Finally, no translations along the z axis are allowed by the pinned bar 1 and therefore all the degrees of freedom must be considered as eliminated. The constraints are all e€ective. In order to ®nd the reaction of the constraint represented by the twopinned bar 4, we can suppress it and draw the in®nitesimal displacement ®eld thus allowed. This turns to be a rotation about an axis identi®ed by the centres de®ned by the intersection of the directions of the pinned bars 2 and 3 and by the intersection of the directions of the pinned bars 5 and 6, respectively. Thus, all sections can rotate about the trace of this axis in their respective planes and the situation for section B is shown in Fig. 2.40. The equilibrium equation (2.2) can be therefore written as  W ˆ Pb  ‡ Rb  ˆ 0

…2:62†

and we have R ˆ ÿP

…2:63†

where the negative sign indicates that the reaction R is opposite in direction to the force P. The reader will ®nd very easily, by the same way, that the reactions of all the other constraints are null.

46

Energy methods in structural mechanics

P

δφ

C

b·δφ δφ

b·δφ b

R

Fig. 2.40

W l P

θ

W

l/2

W

Fig. 2.41

Suggested exercise problems 1.

2.

A simple pin-jointed frame composed of rigid bars is shown in Fig. 2.41. The bars AB and CD are at an angle  to the vertical. Use the principle of stationary work to derive a relationship between the weight W of the bars and the force P required to maintain equilibrium of the frame. [Ans: P ˆ 4W tan ] Two weightless rigid bars are shown in Fig. 2.42. One is pivoted at A and the other at B. The bars contact without friction at C. Use the principle of stationary work to derive a relationship between the forces P and W and the angle  to maintain equilibrium. [Ans:  ˆ cosÿ1 …W=2P†]

Kinematics and equilibrium of systems of rigid bodies

47

A

l

θ l/2 B l/2

C l/2

P W

Fig. 2.42

h W l/2 A

W

B

l

Fig. 2.43

3.

4.

A bar AB of weight W is suspended from a rigid foundation by two extensionless but ¯exible wires, each of length l. A second bar has pins attached at a distance l=2, as shown in Fig. 2.43. Use the principle of stationary work to determine the distance h that maintains equilibrium of the two bars if it is assumed that no friction occurs between the pins and the wire. [Ans: h ˆ 0:711l ] Two solid rigid cylinders of the same density lie inside the bore of a rigid cylinder, as shown in Fig. 2.44. Use the principle of stationary work to derive the value of the angle  that corresponds to equilibrium of the system. [Ans:  ˆ 18:38]

48

Energy methods in structural mechanics

θ

4D

2D D

Fig. 2.44

375 mm 200 m

m

50 mm 75 mm

P 2 kN

m

400 m m 200 m

Fig. 2.45

5.

A clamping ®xture is shown diagrammatically in Fig. 2.45. The elements of the ®xture can be treated as being rigid bars and the pin connections as being completely frictionless. Use the principle of stationary work to calculate the force P that is required to generate the clamping force of 2 kN. [Ans: P ˆ 0:445 kN]

3.

Deformation of bodies and material properties

Introduction

The previous chapter has presented the application of energy principles to the analysis of systems of rigid bodies. These are systems in which the deformations of the bodies are insigni®cant when compared to the movements of the component bodies in the system. At that stage we illustrated this with the example of the collision of two billiard balls in which the deformations at the contact area are negligibly small compared to the distances moved by the balls. The rigid body approach allows us to predict the macroscopic, or global, movements of the components parts of the systems. However, no body is ever completely rigid and we can consider the analysis of most systems at two scales; one is the macroscopic that, as described above, is concerned with the movements of the bodies as if they never change their shape. The other scale is rather microscopic in which we are concerned with the detailed changes of shape of the bodies comprising the system. In this scale the colliding billiard balls are `frozen' in macroscopic space and only the local deformations in the balls as they impact are considered. In reality, of course, systems operate simultaneously in the macroscopic and microscopic scales but it simpli®es analysis, usually without signi®cant loss of accuracy, to separate the analytical treatment of the two scales. It must be noted that there are also many systems like, for example, rubber bands, which experience deformations on a large scale and whose analysis cannot be performed according to the earlier outlined procedures. These systems require a special treatment and there are several theories in structural mechanics which allow us to tackle such analyses. However, on account of their complexity, these methods of analysis are de®nitely out of the scope of the present book and the interested reader can ®nd excellent introductory textbooks to this fascinating area of mechanics.

50

Energy methods in structural mechanics

Before moving on to formulate the energy involved in the deformation of a body we need to examine how materials deform and to characterise the deformations, and the forces that cause the deformations. The approach followed is to subject a sample of the material to a simple one-dimensional test and to generalise, using mathematical constructs, the results from the test to obtain a description of the properties of the material for deformations in three dimensions.

3.1

Deformation properties in one dimension

The approach followed for most materials is exempli®ed here by reference to testing of a typical structural carbon steel. Figure 3.1 shows a specimen machined to a suitable shape for the uniaxial test. The shape of the specimens, Figs 3.1(a) and (b), rectangular and round section, respectively, have evolved during a century of material testing. The objective of the shape is to ensure that the gauge length, lg , has uniform conditions resulting from the axial force, P, applied to the specimen in the test machine and is not in¯uenced by the complex conditions of forces in the parts of the specimen gripped by the testing machine, as shown hatched in Fig. 3.1. The test section shown in Fig. 3.1(a) has thickness t and breadth b giving an area A. During a typical test to evaluate the quasi-static properties of the material the axial tensile load is applied at a slow rate that eliminates any dynamic e€ects and causes the gauge length gradually to extend from lg to lg ‡
lg . The change of length is measured by an instrument t P

P b

lg (a)

P

P d

lg (b)

Fig. 3.1

Deformation of bodies and material properties

51

Δ1 Δ2

P

Δ3

Δlg

Fig. 3.2

P

that is called an extensometer and is attached to the surface of the test specimen. Several types of extensometers have been developed during the past century to measure the very small change of length, typically
lg  0:05 to 0:5 mm. The instrument may be based on mechanical magni®cation of the changes of the gauge length or it may use the change of electric resistance of a length of wire that is constrained to change its length proportionally with the change of the gauge length. Figure 3.2 shows typical results from tests on specimens with di€erent values of cross-sectional area, say by increasing the thickness. It is evident that the initial part of the curves is fairly linear and the load at which the curve reaches a maximum is dependent on the area. Figure 3.3 shows comparable results from tensile tests specimens in which the area is maintained constant and the gauge length is changed. Moreover the load at which the curves reach a maximum is unaffected by the gauge length but the extension is a€ected by the changes.

lg1

lg2

lg3 lg3 > lg2 > lg1

Δlg

Fig. 3.3

Energy methods in structural mechanics

P/A

52

Δlg /lg

Fig. 3.4

It is also evident from tests that if the ordinates of the graphs of the results are normalised, as shown in Fig. 3.4, by dividing the applied load by the area of the specimen at the gauge length and by dividing the extension by the gauge length, then all the curves coincide. The normalisation de®nes two parameters that are used widely in structural analysis, namely, strain; " 
lg =lg

and

stress;   P=A:

If in the above de®nition of strain the initial gauge length is used, then what is normally de®ned as the nominal strain is obtained. A di€erent measure of strain, called the true strain, can be obtained by dividing the elongation of the bar by the actual gauge length, which increases as the tensile load is applied. In the development of this book we will always make reference to the nominal strain. Stress is a parameter that can be derived simply in the analysis of the results of a tensile test by measuring the applied load and dividing it by the corresponding initial value of the area of the cross-section of the gauge length, which gives the nominal stress (a more exact value of the axial stress, known as the true stress, can be calculated by using the actual area of the bar, which can become signi®cantly less than the initial one). The de®nition of stress depends on the assumption that the load is uniformly spread across the cross section of the gauge length. In general, the state of stress is complex in practical structures and since it cannot be measured directly it must be inferred from measured values of strains and deformations using assumed or measured values of material properties. Alternatively, strains can be measured directly on the surface of a structure by arranging the readings from extensometers to be normalised by dividing the measured

Deformation of bodies and material properties

53

Fig. 3.5

elongation of the gauge length by the gauge length of the instrument. Modern electrical resistance strain gauges can have very short gauge lengths, currently as small as lg  2 mm, and the electronic monitoring of the changes in resistance of the gauges directly provide strain values, usually up to a maximum value of strain of 0:05, i.e. 5%. An example of the measurement of strains on a tensile test specimen using microgauges is shown in Fig. 3.5. Other methods, such as the use of laser interferometry, also enable the direct measurement of strains in engineering structures. Figure 3.6 illustrates the full range of results that are obtained from a tensile test on a structural grade carbon steel. The linear part of the graph extends up to a strain value of about 0:002, that is 0:2%, and thereafter the stress±strain graph becomes non-linear; this is called yielding. The ®gure shows the stress increasing gradually with increasing strain up to a strain level of about 0:04, i.e.

54

Energy methods in structural mechanics 600

Stress, σ (N/mm2)

500 400 300 200 100 0

0

1·0000

2·0000 Strain, ε (%)

3·0000

4·0000

(a) 600

Stress, σ (N/mm2)

500 400 300 200 100 0

0

2·0000 4·0000 6·0000 8·0000 10·0000 12·0000 14·0000 16·0000 18·0000 20·0000 Strain, ε (%) (b)

Fig. 3.6

4%, this is called strain hardening. In some structural steels the stress does not increase with increasing strain following the onset of yielding, up to a strain of about 2%. This is called the yield plateau, or LuÈder's plateau after the famous German physicist. The maximum load is reached at a strain of about 8±10%, as is shown in Fig. 3.6(b), at which stage the cross-sectional area of the tensile specimen reduces rapidly and the specimen fractures, this is sometimes called necking. The ultimate load, at which necking begins, is usually about 10± 20% greater than the load at which yielding is initiated. Fracture

Deformation of bodies and material properties

Stress, σ

E

C

σ0

55

B

σP

loading (ii)

unloading (i)

εP

A

D

D

Strain, ε (a)

C

Stress, σ

σ0

ε0

B

A

Strain, ε (b)

Fig. 3.7

occurs usually at about a strain of 18±25% and is a measure of the ductility of the steel. Close to the yield strain the test specimen completely recovers its original geometry when the load is removed. In other words the material properties are reversible; this portion of the stress±strain properties are classi®ed as elastic. Thus for the elastic condition, as shown in Fig. 3.7(a), provided that the load is kept to less than the value at which linearity is lost (that is the proportional limit of strain, "p ), the variation of load, increasing and decreasing, is directly related to the strain. In many structural steels the proportional limit and the loss of elasticity due to the onset of yielding are virtually coincident and thus the

56

Energy methods in structural mechanics

limiting elastic condition for design assessment is taken to be the yield strain, or stress whichever is appropriate. An approximation to the stress±strain curve is made that considerably simpli®es analysis and is particularly relevant to the methods presented in this text. This is that stress±strain curve up the yield strain is modelled as being linear, and this gives rise to the linear elastic conditions that underlie the later developments in this text. If the load in Fig. 3.7(a) is reduced during the test, at a strain greater than the yield strain, "0 , the graph follows the line (ii) in Fig. 3.7(a) until the load is completely removed at point D. In other words, after the yield strain is reached the material no longer has a unique stress±strain relationship and the loading and unloading paths are not reversible back to the original material condition of zero stress and strain. If the test specimen is reloaded from D it will follow the slope of the original elastic line, A±B, until there is a loss of proportionality and the stress±strain graph intersects with the yield line at E. Thus, provided the change of strain due to external loading, starting from zero load value, is less than the yield strain for the material, the material may be regarded as elastic no matter what its previous history of loading may have been. Tests show that for many elastic materials the stress±strain properties are the same in compression as in tension. Figure 3.7(b) shows the stress±strain curve for a material, typi®ed by an aluminium alloy, that has a much more marked non-linear stress± strain curve: from points A to B the relationship is fairly linear, but from B to C there is a distinct curve. The limit of proportionality at B does not necessarily indicate the limit of elastic behaviour of the material. The material may load and unload elastically along the line A±C±A. A pseudo elastic property is often constructed for such a non-linear elastic material, in that the stress±strain relationship is taken to be linear up to the load corresponding to point C, as shown by AD in Fig. 3.7(b). The analysis methods developed in this book are concerned only with the linear elastic part of the stress±strain curve of the material and with predicting the limiting levels of loading at which the material no longer remains elastic. The methods therefore apply to the great majority of engineering analyses that are performed as a basis for design. The fact that the material does not fracture at the yield strain and that the engineering structure can carry increased load before material fracture actually occurs is used by structural engineers as a measure of in-built safety for their structure. The elastic material model that is used as the basis for the analysis methods developed here is shown in Fig. 3.8.

Deformation of bodies and material properties

57

σ

σ0

α

E = tanα

ε0

ε

Fig. 3.8

The strain is directly proportional to the load applied in the tensile test, the constant of proportionality is called the material modulus, or Young's modulus.1 The material modulus is given the symbol E. The limiting allowable level of strain for the elastic analysis is the yield strain, "0 . Thus the strain is related to the applied load in a simple uniaxial loading test by P  ˆ …3:1† AE E This equation is often referred to as Hooke's law, after Hooke,2 who ®rst proposed this load±de¯ection proportionality. The values of E for a typical carbon steel range from 195 000 to 210 000 N/mm2 and commonly a value of 205 000 N/mm2 is used in design analyses. The yield strains for steels widely used in structural applications vary from 0:12% to 0:25% depending of the grade of material. Special heat-treated steels can have yield strains of up to 0:5%. Stress conventionally plays a central role in structural analysis. Many text books in strength of material and in structural mechanics make great use of stress in the development of the theory of structures and in the presentation of the limiting strength of materials. Mathematical descriptions of the equilibrium of components or complete structures "ˆ

1 Young, Thomas (Milverton, Somerset, 1773 ± London, 1829), English natural philosopher. 2 Hooke, Robert (Freshwater, Isle of Wight, 1635 ± London, 1703), English physicist.

58

Energy methods in structural mechanics

are often based on the components of stress in the structure, etc. A structural analysis then has the objective of evaluating the magnitudes of these stresses throughout the structure and comparing the greatest magnitude with the maximum allowable stress for the particular material from which the structure is composed. This comparison then determines the maximum allowable value of the loading that may be applied to the structure. In this text, which is primarily concerned with linear elastic structural analysis using energy methods, the basis of the analysis methods is not stress but the deformation, and therefore strain ®elds, that exist in structure and their components. The arbiter of the allowable loading is not the level of the maximum stress, but the level of the maximum mechanical strain in comparison with the allowable maximum strain for the material. This text therefore does not use the concept of stress widely, not for any reason other than that from this standpoint in most cases there is no real need to make resort to it.

3.2

One-dimensional thermal strain

All structural materials tend to expand when heated, as illustrated in Fig. 3.9. The rollers in the ®gure represent that the specimen is not restricted from axial movement. The thermal strain "t is de®ned as the ratio of the extension of a gauge length to the original gauge length, that is "t ˆ


lg lg

…3:2†

The ratio of the change of length to the original length is found experimentally to be linearly related to the change of temperature, for temperature ranges up to about 100 8C. Thus, "t ˆ 
T

i:e:
lg ˆ lg 
T

…3:3†

where  is the coecient of linear expansion of the material.
T is the change of temperature. The coecient of linear expansion varies slightly with temperature, but is taken to be constant in the analysis methods developed in this

lg

Δlg

ΔT

Fig. 3.9

Deformation of bodies and material properties

59

Coefficient of thermal expansion α, 10–6 °C –1

16

14

12

0

100

200

300

Temperature °C

Fig. 3.10

book. The value of the coecient for a structural steel in the range 0 < T < 100 8C is usually taken as  ˆ 1:25  10ÿ5 mm=mm=8C

…3:4†

The actual variation of the coecient of expansion with the temperature is shown in Fig. 3.10 for a high strength carbon steel. Notice that the thermal strain equals the yield strain at a temperature change of "
T ˆ 0 …3:5†  that is, for a medium grade structural steel 1:5  10ÿ3
T ˆ : ˆ 120 8C …3:6† 1 25  10ÿ5 The foregoing remarks relate to changes of length that result from changes of temperature where there are no constraints or applied forces. The situation illustrated in Fig. 3.11 is one where the change of temperature occurs simultaneously with the application of an external force P. The strain " induced in the bar is the sum of the mechanical strain due to the force and the thermal strain due to the change of temperature. ΔT P

Fig. 3.11

P

60

Energy methods in structural mechanics

Fig. 3.12

Thus P ‡ 
T …3:7† EA The constraint can take the form of displacement restrictions on the boundaries of the body. An extreme example is illustrated in Fig. 3.12 in which the ends of a bar are prevented from movement, i.e.
l ˆ 0 and therefore the total strain in the bar is zero. "ˆ

P ‡ 
T ˆ 0 and P ˆ ÿEA
T …3:8† EA The negative sign implies that the force is compressive to maintain zero total strain for the increase in temperature.

3.3

Three-dimensional strain

Section 3.1 has presented the observations of the change of length of the gauge length of a tensile test specimen when viewed in the direction of the applied load. However, test observations show that the crosssection of the gauge length also changes its dimensions. That is, as illustrated in Fig. 3.13, the application of the load P causes changes in the dimensions normal to the direction of the load. The thickness t changes to t ‡
t and the breadth changes from b to b ‡
b. If we de®ne a set of orthogonal coordinates, x, y and z as

t

t + Δt

P

P

lg

y

x z

Fig. 3.13

lg + Δlg

b

b + Δb

Deformation of bodies and material properties

61

shown in Fig. 3.13, the strains in the three directions are "x ˆ


lg ; lg

"y ˆ


b ; b

"z ˆ


t t

…3:9†

Observations show that that
t and
b are negative for positive values of
l. The test results show that the strain normal to the direction of the applied load is related to the strain in the direction of the load by a constant, called Poisson's ratio, , after S. D. Poisson.3 Thus, ÿ"y ÿ"z ˆ …3:10† ˆ "x "x The values of Poisson's ratio measured in tests for the elastic part of the stress±strain curve of a typical carbon steel are about 0:27±0:31 and 0:3 is commonly used in design analyses for steel structures. Poisson's ratio increases to about 0:5 for strains much greater than the yield strain. During the application of the load P the volume of the gauge length will change due to the elastic strains. The change of volume
V is
V ˆ l…1 ‡ "x †b…1 ‡ "y †t…1 ‡ "z † ÿ lbt ˆ lbt…"x ‡ "y ‡ "z † ‡ f …"2 ; "3 ; . . .†

…3:11†

Since the strain is very small for elastic deformations, the higher order terms in the equation can be ignored as being insigni®cantly small and the volumetric strain "v that is the ratio of the change of volume to the original volume, is
V  "x ‡ "y ‡ "z ˆ "x …1 ÿ 2† …3:12† V Uniaxial tests are commonly carried out to establish relevant material properties for use in design analyses; typical results have been discussed above. It is possible to generalise these results to describe the possible e€ects of applying simultaneously load in the three orthogonal directions. The practical diculties of ensuring that the loads are applied is such a manner that the test specimen has uniform conditions throughout its gauge volume have meant that there is only a limited range of experimental con®rmation of the theoretical development that now follows. However, the application of the theory to the analysis of real structures and the measurement of strains "v 

3

Poisson, SimeÂon-Denis (Pithiviers, Loiret, 1781 ± Paris, 1840), French mathematician, physicist and astronomer.

62

Energy methods in structural mechanics Py Pz

Px Ay

y x Az

Ax Px

Pz

z

Py

Fig. 3.14

on the structures have provided a high degree of con®dence that in fact the generalisation of uniaxial tests to triaxial loading is valid. Figure 3.14 shows a test specimen subjected to three loads, Px , Py and Pz applied in the x, y and z directions such that the conditions are uniform throughout the gauge volume. A uniform change of temperature,
T, is also applied concurrently with the change of loading. The strain, "x , in the x direction is "x ˆ

Py Pz Px ÿ ÿ ‡ 
T EAx EAy EAz

…3:13†

Similarly the strains in the y and z directions are4

4

"y ˆ

Py Px Pz ÿ ÿ ‡ 
T EAy EAx EAz

…3:14†

"z ˆ

Pz Px Py ÿ ÿ ‡ 
T EAz EAx EAy

…3:15†

In most books equations (3.13)±(3.15) are stated in terms of the nominal stresses xx ˆ Px =Ax , yy ˆ Py =Ay and zz ˆ Pz =Az , i.e. yy zz  ÿ ‡ 
T "x ˆ xx ÿ E E E yy xx zz "y ˆ ÿ ÿ ‡ 
T E E E yy   "z ˆ zz ÿ xx ÿ ‡ 
T E E E

Deformation of bodies and material properties

These relationships can be written in matrix format as 3 2 Px 3 2 3 2 36 EAx 7 2 7 
T "x 1 ÿ ÿ 6 6 P 7 7 6 7 6 76 y 7 6 7 ‡ 4 
T 5 4 "y 5 ˆ 4 ÿ 1 ÿ 56 6 EAy 7 7 
T "z ÿ ÿ 1 6 4 Pz 5

63

…3:16†

EAz

In arriving at the relationship above, it has been assumed that the material modulus is constant in all directions: this de®nes an isotropic material and is a reasonable description for a vast class of materials used in structural engineering. There are other engineering materials, such as ®bre composites, for which the material modulus varies from one direction in the body to another, this is classed as anisotropic material. However, this introductory book is concerned only with the development of analysis methods relevant to linearly elastic isotropic materials.

3.4

Strain±displacement relationships

During the change of shape of a body as a result of imposed loading, strains are incurred. These strains vary throughout the volume of the body and can be related to the displacements of the points of the body. The relationship is illustrated using the planar body shown in Fig. 3.15. Figure 3.15(a) shows the original shape of the planar body. An element is identi®ed on the body of length dl between two points A and B. For some reason, which for the present purpose does not need to be speci®ed, the body changes shape as shown in Fig. 3.15(b). The points A and B move to A0 and B 0 , respectively, and the length of

A'

l

A

(a)

Fig. 3.15

B

A

B'

l'

B

(b)

64

Energy methods in structural mechanics y B' ux

A'

uy + duy x

uy

ux + dux B A x

Fig. 3.16

the element changes to l 0 . The change of length, and therefore the strain, incurred by the element can be related to the local displacements. A rectangular coordinate system (x; y) is identi®ed with the element, as shown in Fig. 3.16, to simplify the presentation. The point A initially has coordinates …x; 0† and moves to A0 …x ‡ ux ; uy †; the point B moves from …x ‡ dx; 0† to B 0 …x ‡ dx ‡ ux ‡ dux ; uy ‡ duy †. The original length of the element is l ˆ dx

…3:17†

and the deformed length is l 0 ˆ ‰…dx ‡ dux †2 ‡ du2y Š1=2    2  2 quy qux qux 1=2 ˆ 1‡ ‡ ‡2 dx qx qx qx Thus the strain induced in the element is   2  2  quy l0 ÿ l qux qux 1=2 ˆ 1‡ ‡ ‡2 ÿ1 EAB ˆ l qx qx qx

…3:18†

…3:19†

Notice that in equation (3.19) the strain which occurred to the element AB laying in the x direction was indicated as EAB and not as "x as was done for equation (3.9). This is due to two facts that will be fully clari®ed in the following presentation: ®rstly, the reference coordinate system x, y, z, for the specimen in Fig. 3.13 was oriented along the principal directions of deformation, which in that particular case were the same throughout the whole body; secondly, the notation "ij refers to the components of strain in the small displacement theory

Deformation of bodies and material properties

65

of elasticity, where, as is true for a very large number of engineering applications, substantial simpli®cations can be assumed on account of the fact that deformations are extremely small. Notice also that in equation (3.19) the relationships are concerned only with relative local displacements in the body. The global, i.e. rigid body, displacements do not incur strains. The generalised strain±displacement relationships can be derived by considering a general displacement of an element in a three-dimensional body subjected to a general form of loading. Consider a straight segment between the points A and B of length ds at some position in the undeformed three-dimensional body. The point A initially has coordinates (x; y; z) and the point B initially has coordinates …x ‡ dx; y ‡ dy; z ‡ dz†. When the body is subjected to a deformation the position, orientation and length of the element changes and its length becomes ds 0 such that we can write EAB ˆ

jA0 B 0 j ÿ jABj ds0 ÿ ds ˆ jABj ds

…3:20†

After deformation has occurred, the new coordinates of the generic point of original coordinates x, y and z, are, evidently, x0 ˆ x ‡ ux …x; y; z† y0 ˆ y ‡ uy …x; y; z†

…3:21†

z0 ˆ z ‡ uz …x; y; z† where, once again, ux , uy and uz are the components of the displacement of the point along the Cartesian axes x, y and z. Without any loss of generality, we shall consider the displacement a continuous vectorial function, together with the partial derivatives of its components. Applying equation (3.21) to the points A and B and expanding the right-hand side by Taylor's series and retaining only ®rst-order in®nitesimal terms, we have   qux qu qu 0 dx ˆ 1 ‡ dx ‡ x dy ‡ x dz qx qy qz   quy quy quy …3:22† dy0 ˆ dx ‡ 1 ‡ dy ‡ dz qx qy qz   qu qu qu dz0 ˆ z dx ‡ z dy ‡ 1 ‡ z dz qx qy qz

66

Energy methods in structural mechanics

The square of the length of the element AB results jABj2 ˆ ds2 ˆ dx2 ‡ dy2 ‡ dz2

…3:23†

and that of the deformed element A0 B 0 is jA0 B 0 j2 ˆ ds02 ˆ dx02 ‡ dy02 ‡ dz02

…3:24†

so that, from equation (3.22) we have jA0 B 0 j2 ÿ jABj2 ˆ ds02 ÿ ds2 ˆ 2…exx dx2 ‡ eyy dy2 ‡ ezz dz2 ‡ 2exy dx dy ‡ 2exz dx dz ‡ 2eyz dy dz†

…3:25†

where 

  2  2  quy qux 2 quz ‡ ‡ exx qx qx qx    2  2  quy 1 qux 2 quy quz ‡ eyy ˆ ‡ ‡ qy 2 qy qy qy  2  2  2  quy qu 1 qux quz ezz ˆ z ‡ ‡ ‡ qz 2 qz qz qz   1 quy qux qux qux quy quy quz quz ‡ ‡ ‡ ‡ exy ˆ eyx ˆ 2 qx qy qx qy qx qy qx qy   1 quz qux qux qux quy quy quz quz ‡ ‡ ‡ ‡ exz ˆ ezx ˆ 2 qx qz qx qz qx qz qx qz   1 quz quy qux qux quy quy quz quz ‡ ‡ ‡ ‡ eyz ˆ ezy ˆ 2 qy qz qy qz qy qz qy qz qu 1 ˆ x‡ qx 2

…3:26† …3:27† …3:28† …3:29† …3:30† …3:31†

The terms eij , de®ned by equations (3.26)±(3.31) with i; j 2 ‰x; y; zŠ, are components of strain or Green's strains (after Green5 ) and equations (3.26)±(3.31) themselves represent generalised strain±displacement relationships.

5

Green, George (Sneinton, Nottingham, 1793 ± ibidem, 1841), English physicist and mathematician.

Deformation of bodies and material properties

67

In fact, by virtue of equations (3.20) and (3.25) we can relate the strain EAB occurred to the element AB to the components of strain eij as follows6 EAB …1 ‡ 12 EAB † ds2 ˆ exx dx2 ‡ eyy dy2 ‡ ezz dz2 ‡ 2exy dx dy ‡ 2exz dx dz ‡ 2eyz dy dz

…3:32†

By dividing any term by ds2 we can also write EAB …1 ‡ 12 EAB † ˆ exx 2x ‡ eyy 2y ‡ ezz 2z ‡ 2exy x y ‡ 2exz x z ‡ 2eyz y z

…3:33†

where x , y and z are the direction cosines of the vector AB. In case the linear element before deformation is parallel to the x axis, we have x ˆ 1 and y ˆ x ˆ 0. Therefore from equation (3.33) we get p EAB ˆ 1 ‡ 2exx ÿ 1 …3:34† which, by virtue of equation (3.26), coincides with equation (3.19). The strain±displacement relationships (3.26)±(3.31) are absolutely general. However, they form non-linear relationships of the partial derivatives of the components of displacement. This is a major complication and even if progress in various engineering ®elds has necessitated and produced detailed study of non-linear problems, a vast class of engineering problems have been treated successfully by means of the linear theory of elasticity. This is accomplished by linearising the above mentioned relationships, as will be seen in the next section.

3.5

Simpli®cations possible when deformations are very small

The vast majority of materials used in structural engineering are actually subjected only to extremely small deformations. Therefore it

6

In fact, it is jA0 B0 j2 ÿ jABj2 ˆ …jA0 B0 j ÿ jABj ‡ jABj†2 ÿ jABj2 ˆ …EAB jABj ‡ jABj†2 ÿ jABj2 ˆ …EAB ds ‡ ds†2 ÿ ds2 2 ˆ …2EAB ‡ EAB † ds2

ˆ EAB …2 ‡ EAB † ds2

68

Energy methods in structural mechanics

may be assumed that components of displacement and their derivatives are insigni®cant compared with unity and that any product involving these terms can be ignored. This means that we can rewrite the strain±displacement relationships (3.26)±(3.31) in the following form qux qx quy ˆ qy

"xx ˆ

…3:35†

"yy

…3:36†

"zz ˆ

quz qz

  1 quy qux ‡ 2 qx qy   1 quz qux ‡ "xz ˆ "zx ˆ qz 2 qx   1 quz quy "yz ˆ "zy ˆ ‡ 2 qy qz "xy ˆ "yx ˆ

…3:37† …3:38† …3:39† …3:40†

where the terms "ij , de®ned by equations (3.35)±(3.40) with i, j 2 ‰x; y; zŠ, are components of strain for small deformation and are the basis of the linear theory of elasticity. It is worth noting that in compact form equations (3.35)±(3.40) can be often written in a summarised form as   1 qui quj "ij ˆ ‡ …3:41† 2 qxj qxi with i, j 2 ‰x; y; zŠ and the convention that xx  x, xy  y and xz  z. The components "xx , "yy and "zz have a simple geometrical meaning. In fact, let us consider once more a linear element of the body which before deformation is parallel to the x axis and has length dx. After deformation, the projection of this linear element on the same axis is given by the ®rst of equations (3.22), i.e.   qux 0 dx ˆ 1 ‡ dx ˆ …1 ‡ "xx † dx …3:42† qx Therefore, we have dx0 ÿ dx …3:43† dx that is "xx represents the relative increase in the projection on the x axis of a linear element which before deformation was parallel to this axis. "xx ˆ

Deformation of bodies and material properties

69

y

Pyx

dy

Pxy

Pxy

Pyx dx x

Fig. 3.17

Moreover, the components "xy , "xz and "yz can be simply related to an additional form of strain called shear strain. This form can be demonstrated, for example, by reference to the deformation of a small element of material in which the forces are parallel to the faces of the element, see Fig. 3.17. A practical example of this form of strain occurs in the deformations of a thin-walled cylinder subjected purely to torsional loading, shown in Fig. 3.18. The perimeters A±B and C±D move angularly relative to each other and the material between closely spaced perimeters will experience the `lozenging' shown in Fig. 3.18. Considering the general deformation at a point on a body, restricted here for clarity to a plane section, the point moves from O to O0 . The ®bres dx and dy experience extension, but, in general, they also experience relative rotation, as shown in Fig. 3.19.

A T

Fig. 3.18

T

D

B

C

70

Energy methods in structural mechanics y

) dy

∂ux ∂y

dy

(1 +

∂uy ∂y

φ2 φ1

∂uy ∂x

dx

o'

v u

(1 +

∂ux ∂x

) dx

o

x

Fig. 3.19

The rotations from the horizontal and vertical are, respectively,7   quy =qx quy qu qu  1ÿ x  x 1  tan 1 ˆ qx qx qx …1 ‡ qux =qx† …3:44†   quy qux =qy qux qux  1ÿ  2  tan 2 ˆ …1 ‡ quy =qy† qy qy qy and the relative rotation …1 ‡ 2 † results in quy qux ‡ ˆ 2"xy 1 ‡ 2 ˆ qx qy

…3:45†

Sometimes the relative rotation …1 ‡ 2 † is de®ned as engineering shear strain xy , that is quy qux ‡ …3:46†

xy ˆ qx qy As the treatment of structural analysis in this book is almost entirely concerned with problems which do not justify the more complicated non-linear theory (the only partial exception being constituted by the presentation of buckling phenomena in chapter 12), in the following we will always make reference to the simpli®cations introduced in this section.

3.6

Deformation in the neighbourhood of an arbitrary point

Always with reference to a deformed body, let us consider the displacement of a point B in the neighbourhood of a generic point A. 7

We recall that for small values of x it is 1=…1 ‡ x†  1 ÿ x.

Deformation of bodies and material properties

71

We again make reference to a generic Cartesian rectangular coordinate system x; y; z. By a simple Taylor's expansion with origin at location A, if only ®rst order in®nitesimal terms are retained, we have qux qux ux …B† ˆ ux …A† ‡ ‰x…B† ÿ x…A†Š ‡ qx A qy A qux
‰y…B† ÿ y…A†Š ‡ qz ‰z…B† ÿ z…A†Š A quy quy uy …B† ˆ uy …A† ‡ ‰x…B† ÿ x…A†Š ‡ qx A qy A …3:47† quy
‰y…B† ÿ y…A†Š ‡ qz ‰z…B† ÿ z…A†Š A quz quz ‰x…B† ÿ x…A†Š ‡ uz …B† ˆ uz …A† ‡ qx A qy A quz
‰y…B† ÿ y…A†Š ‡ qz ‰z…B† ÿ z…A†Š A or also, for brevity,

qux BA qux BA qux BA x ‡ y ‡ z qx A qy A qz A quy BA quy BA quy BA uy …B† ˆ uy …A† ‡ x ‡ y ‡ z qx A qy A qz A qu qu qu uz …B† ˆ uz …A† ‡ z xBA ‡ z yBA ‡ z zBA qx A qy A qz A

ux …B† ˆ ux …A† ‡

…3:48†

The above expressions can also be written as ux …B† ˆ ux …A† ‡ "xx xBA ‡ …"xy ÿ !z †yBA ‡ …"xz ‡ !y †zBA uy …B† ˆ uy …A† ‡ …"yx ‡ !z †xBA ‡ "yy yBA ‡ …"yz ÿ !x †zBA …3:49† uz …B† ˆ uz …A† ‡ …"zx ÿ !y †xBA ‡ …"zy ‡ !x †yBA ‡ "zz zBA where

  1 quz quy ÿ !x ˆ 2 qy qz   1 qux quz !y ˆ ÿ 2 qz qx   1 quy qux !z ˆ ÿ 2 qx qy

…3:50† …3:51† …3:52†

72

Energy methods in structural mechanics

and all the partial derivatives in equations (3.35)±(3.40) and (3.50)± (3.52) are evaluated at A. This means that all coecients occurring in equation (3.49), i.e. ui …A†, "ij and !i , have to be considered constants and equal to their value at A. It follows that an in®nitesimal small region surrounding an arbitrary point A on account of deformation is in general subject to the sum of (a) (b)

(c)

a simple translation given by the displacement vector u…A†; a rotation with components !x , !y and !z . In fact in the theory of small displacements, as we have already seen in chapter 2, it is easy to realise that the product of a skew-symmetric matrix ‰!Š by the position vector, i.e. 3 32 2 r3 2 0 ÿ!z !y ux xBA 6 7 4 ury 5 ˆ 6 0 ÿ!x 7 …3:53† 4 !z 54 yBA 5 r BA ÿ!y !x 0 uz z always represents a rigid rotation displacement ®eld about the origin A; changes in dimensions and shape given by the displacement ®eld 32 BA 3 2 3 2 "xx "xy "xz ux x 76 BA 7 6 7 6" " " …3:54† 5 4 uy 5 ˆ 4 yx yy yz 54 y  BA " " " uz z zx zy zz

A' z

u (A) A

y

x

Fig. 3.20

Deformation of bodies and material properties

73

Thus, apart from a translational and rotational movement, the deformation of an in®nitely small region surrounding A is described by equation (3.54). This is a linear transformation, where straight lines remain straight, planes remain plane (and both remain parallel if they were parallel before occurrence of deformation), and in general any second-order surfaces transform into second order surfaces, that is, for example, a sphere transforms into an ellipsoid, as is shown in Fig. 3.20.

3.7

Transformation of the components of strain with the change of reference frame

The transformation of the expression of the components of strain "ij with the change of reference frame, that is, for example, in the passage from a Cartesian system of orthogonal coordinates x1 , y1 , z1 to another Cartesian system x2 , y2 , z2 with a di€erent origin and with rotated axes, can be pursued very easily with a minimum of matrix algebra. In fact, look at the situation in Fig. 3.21 and consider a vector v of components vxz , vyz , vzz with reference to the second Cartesian frame x2 , y2 , z2 . In the ®rst reference frame x1 , y1 , z1 the same vector has components vx1 , vy1 , vz1 which are related to the components vxz , vyz , vzz by the formula 2 3 2 32 3 nx2 x1 nx2 y1 nx2 z1 vx2 vx1 4 vy 5 ˆ 4 ny x ny y ny z 54 vy 5 …3:55† 1 2 1 2 1 2 1 2 nz2 x1 nz2 y1 nz2 z1 vz1 vz2 z2 v z1 x2 y2 y1

x1

Fig. 3.21

74

Energy methods in structural mechanics

where nx2 x1 ; . . . ; nz2 z1 are the direction cosines of the axes of the second system with respect to the ®rst one. Hence, in the passage from x2 ; y2 ; z2 to x1 ; y1 ; z1 the pure strain displacement ®eld given by equation (3.54) becomes 32  3 2  3 2 nx2 x1 nx2 y1 nx2 z1 ux2 ux1 7 6 u 7 6 n 6  7 …3:56† 4 y1 5 ˆ 4 y2 x1 ny2 y1 ny2 z1 54 uy2 5 uz1

nz2 x1

nz2 y1

nz2 z1

uz2

By remembering from geometry that the matrix ‰nŠ of the director cosines is orthogonal, i.e. its inverse coincides with its transpose that is ‰nŠÿ1 ˆ ‰nŠT , the inverse relationship of the above equation (3.56), results in 3 2 3 2  3 2 nx2 x1 nx2 y1 nx2 z1 T ux1 ux2 7 6  7 6 u 7 6 n …3:57† 4 y2 5 ˆ 4 y2 x1 ny2 y1 ny2 z1 5 4 uy1 5 uz2

nz2 x1

nz2 y1

that is, by virtue of equation 2  3 2 nx2 x1 nx2 y1 ux2 6  7 6 6 uy 7 ˆ6 ny x ny y 2 1 4 25 4 2 1  nz2 x1 nz2 y1 uz2 2 BA 3 x1 6 BA 7 7
6 4 y1 5

nz2 z1

uz1

(3.54) 3 2 nx2 z1 T "x1 x1 7 6 6 ny2 z1 7 5 4 "y1 x1 nz2 z1 "z1 x1

"x1 y1

"x1 z1

3

"y1 y1

7 "y1 z1 7 5

"z1 y1

"z1 z1

…3:58†

zBA 1 Thus, the following relationship holds true 2  3 2 3 2 nx2 x1 nx2 y1 nx2 z1 T "x1 x1 "x1 y1 ux2 6  7 6 7 6 6 uy 7 ˆ 6 ny x ny y ny z 7 6 "y x "y y 2 1 2 1 5 4 1 1 1 1 4 25 4 2 1  nz2 x1 nz2 y1 nz2 z1 "z1 x1 "z1 y1 uz2 2 32 BA 3 nx2 x1 nx2 y1 nx2 z1 x2 6 76 BA 7 76 7
6 4 ny2 x1 ny2 y1 ny2 z1 54 y2 5 nz2 x1 nz2 y1 nz2 z1 zBA 2

"x1 z1

3

7 "y1 z1 7 5 "z1 z1

…3:59†

and by comparison with equation (3.54) the transformation law of

Deformation of bodies and material properties

75

the components of strain with the change of coordinate system results in 2 3 2 3 nx2 x1 nx2 y1 nx2 z1 T "x2 x2 "x2 y2 "x2 z2 6 7 6 7 6 "y x "y y "y z 7 ˆ6 ny x ny y ny z 7 2 2 2 2 2 2 2 1 2 1 2 1 4 5 4 5 "z2 x2 "z2 y2 "z2 z2 nz2 x1 nz2 y1 nz2 z1 2 3 "x1 x1 "x1 y1 "x1 z1 6 7 7
6 4 "y1 x1 "y1 y1 "y1 z1 5 "z1 x1 "z1 y1 "z1 z1 2 3 nx2 x1 nx2 y1 nx2 z1 6 7 7 …3:60†
6 4 ny2 x1 ny2 y1 ny2 z1 5 nz2 x1 nz2 y1 nz2 z1 Finally, as they will be useful in the next section, we can develop these matrix products in extenso and write "x2 x2 ˆ "x1 x1 n2x2 x1 ‡ "y1 y1 n2x2 y1 ‡ "z1 z1 n2x2 z1 ‡ 2 "x1 y1 nx2 x1 nx2 y1

"x2 y2

‡ 2 "x1 z1 nx2 x1 nx2 z1 ‡ 2 "y1 z1 nx2 y1 nx2 z1 ... ... ˆ "y2 x2 ˆ 2 "x1 x1 nx2 x1 ny2 x1 ‡ 2 "y1 y1 nx2 y1 ny2 y1

…3:61†

‡ 2 "z1 z1 nx2 z1 ny2 z1 ‡ 2 "x1 y1 …nx2 x1 ny2 y1 ‡ ny2 x1 nx2 y1 † ‡ 2 "x1 z1 …nx2 x1 ny2 z1 ‡ ny2 x1 nx2 z1 † ‡ 2 "y1 z1 …nx2 y1 ny2 z1 ‡ ny2 y1 nx2 z1 † ... ...

3.8

Principal directions of deformation. Maximum and minimum extensions

In this section we are concerned with ®nding one or more directions in which the extension given by equation (3.43), i.e. "xx ˆ

dx0 ÿ dx dx

assumes an extremal value, that is, the result is a maximum or a minimum.

76

Energy methods in structural mechanics

This study can be performed very easily with the aid of the law of transformation (3.61). In fact the problem amounts to ®nding the values of n x1 , n y1 and n z1 for which the ®rst of equations (3.61) becomes a maximum or a minimum. As by de®nition of direction cosines the following relationship must hold true n2x1 ‡ n2y1 ‡ n2z1 ˆ 1

…3:62†

the problem is one of extremum with constraints and, as is known from calculus, it can be dealt with by means of Lagrange multipliers (named after the eminent mathematician Joseph-Louis Lagrange, see chapter 2). In other words, we are led to searching the stationary value of the following function L…n x1 ; n y1 ; n z1 ; "† ˆ " ÿ "…n2 x1 ‡ n2 y1 ‡ n2 z1 ÿ 1†

…3:63†

where " is a constant, i.e. the Lagrange multiplier. Therefore, if we equate to zero the partial derivatives of expression (3.63) with respect to n x1 , n y1 and n z1 , we obtain the following system of three homogeneous algebraic equations 8 > < …"x1 x1 ÿ "†n x1 ‡ "x1 y1 n y1 ‡ "x1 z1 n z1 ˆ 0 "x1 y1 n x1 ‡ …"y1 y1 ÿ "†n y1 ‡ "y1 z1 n z1 ˆ 0 …3:64† > : "x1 z1 n x1 ‡ "y1 z1 n y1 ‡ …"z1 z1 ÿ "†n z1 ˆ 0 The zero solution of this system has no interest on account of the condition (3.62). Consequently, in order to ®nd the non-zero solutions of equation (3.64) we must equate the determinant of its coecients to zero, that is "x x ÿ " "x1 y1 "x1 z1 1 1 "y1 y1 ÿ " "y1 z1 ˆ 0 …3:65† det "x1 y1 "x z "y z "z z ÿ " 1 1

1 1

1 1

or, in expanded form "3 ÿ E1 "2 ‡ E2 " ÿ E3 ˆ 0

…3:66†

where E1 ˆ "x1 x1 ‡ "y1 y1 ‡ "z1 z1

…3:67†

E2 ˆ "x1 x1 "y1 y1 ‡ "x1 x1 "z1 z1 ‡ "y1 y1 "z1 z1 ÿ "2x1 y1 ÿ "2x1 z1 ÿ "2y1 z1 …3:68† E3 ˆ "x1 x1 "y1 y1 "z1 z1 ÿ …"x1 x1 "2y1 z1 ‡ "y1 y1 "2x1 z1 ‡ "z1 z1 "2x1 y1 ÿ "x1 y1 "x1 z1 "y1 z1 †

…3:69†

Deformation of bodies and material properties

77

Because the principal extensions are independent of the choice of direction of the coordinate axes, also the coecients E1 , E2 and E3 result independent from this choice of the coordinate axes and are named strain invariants. From a mathematical standpoint, solving equation (3.65) means ®nding the eigenvalues of the problem ‰"Š ÿ "‰IŠ ˆ 0

…3:70†

Equation (3.66) is a cubic in the Lagrange multiplier " and, as it is obtained from a symmetric matrix, it can be demonstrated that it admits always three real values, which are the eigenvalues of the problem, equation (3.70). Corresponding to these values, we ®nd three sets of direction cosines (that is, for example, n x1 , n y1 , n z1 ; n x1 , n y1 , n z1 and n x1 , n y1 , n z1 ) which represent the eigenvectors of equation (3.70). By virtue of a well known property of eigenvectors, the three directions associated to these sets of direction cosines are mutually perpendicular. We are therefore led to the conclusion that for every point of the body it is always possible to ®nd three perpendicular directions in which the components of strain "ii assume stationary value, that is are a maximum or a minimum relative. The values assumed by the Lagrangian multiplier, i.e. the roots "1 , "2 , "3 of equation (3.66), represent exactly these components and are called principal components of strain. In order to illustrate this point, let us consider again the ®rst of equations (3.61). By grouping its right-hand terms, it can be written as " ˆ …"x1 x1 n x1 ‡ "x1 y1 n y1 ‡ "x1 z1 n z1 †nx1 ‡ …"x1 y1 n x1 ‡ "y1 y1 n y1 ‡ "y1 z1 n z1 †n y1 ‡ …"x1 z1 n x1 ‡ "y1 z1 n y1 ‡ "z1 z1 n z1 †n z1

…3:71†

Given that "1 and the set of direction cosines nx1 , ny1 , nz1 satisfy equations (3.64), then equation (3.71) becomes " ˆ "1 n2 x1 ‡ "1 n2 y1 ‡ "1 n2 z1 ˆ "1

…3:72†

and, by repeating the same procedure, we ®nd also that " ˆ "2 and " ˆ "3 . An important property of the principal directions of deformation is that along these directions the components " , " , " of strain, i.e. the shear strain components, are zero. To show this property the

78

Energy methods in structural mechanics

fourth of equations (3.61) can be written as " ˆ 2…"x1 x1 n x1 ‡ "x1 y1 n y1 ‡ "x1 z1 n z1 †n x1 ‡ 2…"x1 y1 n x1 ‡ "y1 y1 n y1 ‡ "y1 z1 n z1 †n y1 ‡ 2…"x1 z1 n x1 ‡ "y1 z1 n y1 ‡ "z1 z1 n z1 †n z1

…3:73†

Again, as "1 and the set of direction cosines n x1 , n y1 , n z1 satisfy equations (3.64), equation (3.73) becomes " ˆ 2"1 …n x1 n x1 ‡ n y1 n y1 ‡ n z1 n z1 † ˆ 0

…3:74†

since the axes ,  and  are orthogonal. By applying the same line of reasoning to the ®fth and the sixth of equation (3.61) we conclude that " ˆ " ˆ 0

3.9

…3:75†

Plane strain

There are structural forms that are very long, in the z direction, compared to their other dimensions. Two examples are shown in Fig. 3.22. Figure 3.22(a) represents a part of a very long dam wall that is supported at its base and at its ends. The ends are prevented from movement in the z direction and the loading is uniform along the length of the dam face. We may deduce that, on account of the length, there is symmetry of deformation at any section along the dam and that it may reasonably be assumed that there is no displacement along the z axis and therefore no variation of deformations or strains along the

dz z

z

(a)

Fig. 3.22

(b)

Deformation of bodies and material properties

79

dam wall. In fact, the deformation of the wall can be assumed to be the same at any cross section and can be de®ned completely by the nonzero components of displacement ux and uy , which are assumed to be functions of x and y only. Thus, the response of the wall to the applied loads may be analysed using a representative length dz of the wall as shown in the ®gure. Similar conditions may be assumed for a very long tunnel that has a uniform pressure along its length from the surrounding soil, see Fig. 3.22(b). These cases are normally referred to as plane strain and the assumption of zero variation of strain along the length can signi®cantly simplify the analysis. However, it must be remembered that the results pertain only to deformations far from the ends and that near the ends the deformations will actually be three-dimensional. Formally, the above assumptions can be written as ux ˆ ux …x; y†

uy ˆ uy …x; y†

uz ˆ 0

…3:76†

and the expressions (3.35)±(3.40) of the strain components can be simpli®ed as follows "xx ˆ

qux qx

…3:77†

"yy ˆ

quy qy

…3:78†

quz ˆ0 qz   1 qux quy ‡ "xy ˆ 2 qy qx   1 quy quz ‡ ˆ0 "yz ˆ 2 qz qy   1 qux quz "xz ˆ ‡ ˆ0 2 qz qx "zz ˆ

3.10

Strain±displacement compatibility conditions

…3:79† …3:80† …3:81† …3:82†

The strain±displacement relationships in equations (3.26)±(3.31) describe a complete derivation of the strains from the displacement ®eld throughout a body. However, one may wonder if an arbitrarily chosen set of six continuous functions eij ˆ fij …x; y; z† de®ned in the domain of the body always represent an actual deformation of the

80

Energy methods in structural mechanics

continuous medium. Evidently this cannot be the case as there is a disparity between the number of strain components (six) and those of the components of the displacement ®eld (three) and therefore not all the strain components can be independent. Hence, it can be shown that the functions eij ˆ fij …x; y; z† will de®ne the deformation of a body, thus satisfying all the continuity conditions, only if they ful®l a set of six di€erential equations, called strain compatibility conditions. These conditions may be derived in a straightforward manner when deformations are very small, i.e. with reference to the simpli®ed strain±displacement relationships (3.35)±(3.40). For the sake of simplicity, the derivation of the compatibility conditions is exempli®ed with regard to a case of plain strain. So, with reference to equations (3.77), (3.78) and (3.80) "xx

qu ˆ x qx

"yy

quy ˆ qy

"xy

  1 quy qux ‡ ˆ 2 qx qy

and di€erentiating "xx twice with respect to y we obtain q2 "xx q3 ux ˆ qy2 qx qy2

…3:83†

In similar manner we have q2 "yy q3 uy ˆ qx2 qy qx2

  q2 "xy 1 q3 ux q3 uy ˆ ‡ qx qy 2 qx qy2 qx2 qy

…3:84†

Using the derivatives of the displacements shown above, it is evident that the condition of compatibility of the strains to ensure continuity of the displacements is 2 q2 "xy q2 "xx q "yy ˆ0 ‡ ÿ 2 qx qy qy2 qx2

…3:85†

This condition is obviously a necessary one for the continuity of deformation, but it is possible to show that it also is sucient. However, the discussion of the conditions of solvability of the set of equations of strain compatibility conditions is somewhat more complicated and goes beyond the scope of this book. The compatibility relationship (equation (3.85)) is for a case of plane strain. Nevertheless, by applying a similar procedure we can easily get the complete set of conditions relating to a general three-dimensional

Deformation of bodies and material properties z

dz

Px dx

81

Px

dy y

x

Fig. 3.23

body, i.e. 2 q2 "xy q2 "xx q "yy ˆ0 ‡ ÿ 2 qx qy qy2 qx2

q2 "xx q2 "zz q2 "xz ˆ0 ‡ ÿ 2 qx qz qz2 qx2 q2 "yy q2 "zz q2 "yz ˆ0 ‡ ÿ 2 qy qz qz2 qy2

…3:86†

q2 "yz q2 "xx q2 "xz q2 "xy ÿ ÿ ˆ0 ‡ qy qz qx qy qx qz qx2 2 2 2 q2 "xz q "yy q "xy q "yz ÿ ÿ ˆ0 ‡ qx qz qy qz qx qy qy2

q2 "xy q2 "zz q2 "yz q2 "zx ÿ ÿ ˆ0 ‡ qx qy qx qz qy qz qz2 These conditions are generally known as Beltrami's equations.8

3.11

Strain energy

This text presents the development of methods for the analysis of structures using the concept of strain energy and work as a basis. The formulations of strain energy for a linear elastic body are presented in this section. Figure 3.23 shows an in®nitesimal element of material as some position in a continuous body. The element has dimensions dx, dy and dz. A force Px is applied to the element in the x direction. The value of the force starts from zero and gradually reaches its ®nal value Px . The 8

Beltrami, Eugenio (Cremona, 1835 ± Rome, 1900), Italian mathematician.

82

Energy methods in structural mechanics

increment change of strain energy,
U, stored by the element is equal to the work done W by the force during the loading process on account of the incremental change of length, from zero to dl ˆ "xx dx, of the element in the x direction. Thus "…xx


U ˆ W ˆ

Px d"xx dx

…3:87†

0

As for a linearly elastic material the relationship between the force and the strain is, according to the test results reviewed in section 3.3, "xx ˆ

Px E dy dz

…3:88†

The incremental change in strain energy for the elementary volume is therefore "…xx


U ˆ

Px d"xx dx 0 "…xx

E"xx d"xx dx dy dz ˆ

ˆ 0

E 2 " dV 2 xx

…3:89†

This is a noteworthy result. In fact, whenever we deal with a linear load±de¯ection relationship, as is the case in equation (3.88), which is described by a relationship of the type
l ˆ "xx l ˆ Px =k

…3:90†

the work performed by the load Px in the loading process from 0 to its ®nal value Px is
l …

Wˆ


l …

Px d
l ˆ 0

k
l d
l ˆ 0

1 1 k
l 2 ˆ Px
l 2 2

…3:91†

i.e. it amounts to half the product of the ®nal value of the force Px by the ®nal value of the displacement
l. This fact is of fundamental importance in the linear theory of elasticity and constitutes the statement of Clapeyron's theorem, named after Clapeyron9 who investigated the phenomenon in a wider context. Clapeyron, BenoõÃ t-Paul-EÂmile (Paris, 1799 ± ibidem, 1864), French engineer and physicist. 9

Deformation of bodies and material properties

83

Pz z

dz

Px Py

dx Pz

Px

dy y

x

Fig. 3.24

Given that the element under consideration is an in®nitesimal one, equation (3.89) can be employed to de®ne the strain energy per unit volume at a point, which is called strain energy density. This is simply done by letting ' ˆ U=dV and assuming the constant which de®nes the reference value of the strain energy to be zero in the unconstrained con®guration. It follows immediately that the value of strain energy for a body with a volume V, uniform strain "xx and constant value of modulus E, is … … E 2 U ˆ ' dV ˆ "xx dV …3:92† 2 V

V

We next consider the situation where also forces in the y and z directions are applied. The examination of the test results in the previous sections showed that forces applied normal to each other in¯uence the strains through a constant, called Poisson's ratio, . Figure 3.24 shows an elementary volume of material subjected to forces in the x, y and z directions. The strains related to these forces and directions are Py Px Pz ÿ ÿ E dy dz E dx dz E dx dy Py Px Pz ÿ ÿ "yy ˆ E dx dz E dy dz E dx dy Py Pz Px ÿ ÿ "zz ˆ E dx dy E dy dz E dx dz "xx ˆ

…3:93† …3:94† …3:95†

By superposing the e€ects of these three forces and integrating them over the volume of a body of volume V, if we assume an homogeneous and isotropic condition for the material constants, E and , that is,

84

Energy methods in structural mechanics

they are constant throughout the body and are constant in every direction at every point in the body, we obtain … E Uˆ ‰…1 ÿ †…"2xx ‡ "2yy ‡ "2zz † 2…1 ‡ †…1 ÿ 2† V

‡ 2 …"xx "yy ‡ "xx "zz ‡ "yy "zz †Š dV

…3:96†

It is interesting to note that if we consider the situation where the strains in the y and z directions are zero, that is "yy ˆ "zz ˆ 0, we conclude that this is possible provided that   Py  Px ˆ 1 ÿ  dy dz dx dz …3:97†   Pz  Px ˆ dx dy 1 ÿ  dy dz and the value of the strain energy is … E…1 ÿ † "2 dV Uˆ 2…1 ‡ †…1 ÿ 2† xx

…3:98†

V

The development can be extended to include the situation where there are direct and shear strains in the elementary volume, still assuming an homogeneous and isotropic condition for the material constants E and . Before doing so, consider the shear loading shown in Fig. 3.25, similar to that already considered in Fig. 3.17. In such a case it is found from testing that

xy ˆ

Pxy Pyx ˆ G dy dz G dx dz

…3:99†

y Pyx

Pxy

Pxy

dy

Pyx

dx

Fig. 3.25

x

Deformation of bodies and material properties y

85

Pyx

y Pyx

D

C PAC

Pxy

Pxy

Pxy

A Pyx

PAC

B

Pyx

x (a)

Pxy

x (b)

Fig. 3.26

where the constant G is the shear modulus of elasticity of the material. Equation (3.99) is normally referred to as Hooke's law for shearing. This law applies also when more than one loading (shear or normal) is acting, i.e. shear forces do not in¯uence each other, nor in¯uence or are in¯uenced by normal forces. The constant G can be easily related to the constants E and . In fact, let us consider that the in®nitesimal element has the shape of a square, that is dx ˆ dy. As this in®nitesimal element is in equilibrium, it is easy to conclude that we must have jPxy j ˆ jPyx j. Moreover, if we ideally make a cut along the diagonal line BD, as shown in Figure 3.26(a), and write the equilibrium equations for the triangular sub-elements ABD and BCD, we are led to the conclusion that on the surfaces of the cut pcan  only be present normal tensile forces PAC of magnitude equal to 2Pxy , see Fig. 3.26(b). By repeating the same reasoning for an ideal cut along the diagonal AC, we ®nd that along the direction pBD  and normally to the cut a compressive force of modulus PBD ˆ 2Pxy is acting, see Fig. 3.27. Thus, applying Hooke's law for axial loading to the direction AC we have "AC , the component of strain along the direction AC, Pxy P P ÿ  pBD ˆ …1 ‡ † "AC ˆ pAC E 2 dy dz E 2 dx dz E dy dz

…3:100†

The law of transformation of the components of strain with the change of reference axes, equations (3.61), applied to the case of pure shear strain at hand yields

xy "AC ˆ "xy ˆ …3:101† 2

86

Energy methods in structural mechanics Pyx

y

Pxy

PBD PBD

Pxy

Pyx

x

Fig. 3.27

and allows us to write

xy Pxy ˆ …1 ‡ † 2 E dy dz

…3:102†

At this point, a simple comparison between equation (3.99) and equation (3.102) provides the relationship between the constant G and the constants E and , that is Gˆ

E 2…1 ‡ †

…3:103†

The expression of the strain energy for the situation where all the strains, direct and shear, are present in the elementary volume is now simple. In fact, given that there is no interaction between Hooke's law for direct and shear strains, we can simply compute the work done by the shear forces on the elementary volume, that is   quy 1 qux 1 1 Wˆ Pxy dx ‡ Pyx dy ˆ Pxy xy dx ˆ Pyx xy dy 2 2 2 qx qy …3:104† given that for the element to be in equilibrium we must have Pxy dx ˆ Pyx dy and provided we made reference to Clapeyron's theorem. By virtue of Hooke's Law, equation (3.99), we can write equation (3.104) as 1 2 E W ˆ G xy "2 dx dy dz dx dy dz ˆ 2 …1 ‡ † xy

…3:105†

Finally, by superposing the e€ects of the shear forces Pxy , Pyz , Pxz , Pzx , Pyx and Pzy and by integrating over the domain of a body with volume V we obtain the general expression of the change in strain

Deformation of bodies and material properties

energy related to the shear strains … E Uˆ …"2xy ‡ "2xz ‡ "2yz † dV …1 ‡ †

87

…3:106†

V

which, added to equation (3.96), gives … E …1 ÿ †…"2xx ‡ "2yy ‡ "2zz † Uˆ 2…1 ‡ †…1 ÿ 2† V

‡ 2 …"xx "yy ‡ "xx "zz ‡ "yy "zz †  2…1 ÿ 2† 2 2 2 …"xy ‡ "xz ‡ "yz † dV ‡ …1 ‡ †

…3:107†

At the end of this section it is worth noting that the strain energy is a scalar quantity and by its own nature it is independent of the particular reference frame. Hence, the expression of the strain energy density at a point must give the same value with respect to any system of coordinate axes. In particular, when we make reference to a system of axes coincident with the principal directions of strain , ,  we can write 'ˆ

E ‰…1 ÿ †…"2 ‡ "2 ‡ "2 † 2…1 ‡ †…1 ÿ 2† ‡ 2…" " ‡ " " ‡ " " †Š

…3:108†

because the shear strain components are zero in this particular reference frame.

3.12

Yield criteria

The uniaxial tests on ductile material described in section 3.1 show that the linear load±extension property of the material is limited to certain values of strain. At the value of strain, "0 , which we can call the yield strain, the force±displacement properties become non-linear. The onset of the non-linear conditions have for many years been generally regarded as a limiting condition in structural design. However, relatively recent developments in mechanics of materials and the alliance of the non-linear material properties assessment with more advanced numerical techniques of analysis have enabled the actual limiting state of failure of structures to be investigated and, in some cases, incorporated in design analysis. These non-linear analyses can be extremely complex and involve local and overall buckling of the structure as material yielding spreads. This level of analysis is outside the scope of this text, but the following provides criteria that have been developed which can

88

Energy methods in structural mechanics

be used to identify the limit of validity of the linear elastic analysis methods. The primary requirement in the development of the yield criteria is to extend the observations from uniaxial tension tests to the complex conditions of a three-dimensional strain ®eld that exists in the interior and on the surface of a structure subjected to various forms of loading. As a matter of fact, there are no currently available tests that can validate uniquely or directly any speci®c yield criterion. The usual approach is to particularise the three-dimensional yield criterion using the yield strain derived from a uniaxial test on the material from which the structure is composed, then apply the yield criterion, with some reduction for safety, to the linear elastic analysis results to determine the allowable level of the external loads that can be sustained in practice. Experience has shown this approach to provide a fairly reasonable basis for structural practice. The strain energy formulation developed in the previous sections has been related to a general state of three-dimensional strain at some point in the interior of a structure. The strain state encompasses direct and shear strains. The relative magnitudes of these components of strain will vary with the choice of direction of the axes, x, y and z. As we have seen, it is possible to choose the axes such that there are no components of shear strain in the description of the strain state. Here we identify these strains as "1 , "2 and "3 . The strain energy formulation, in terms of the principle strains is given by equation (3.108), i.e. 'ˆ

E 2…1 ‡ †…1 ÿ 2† 2

2

2


‰…1 ÿ †…"1 ‡ "2 ‡ "3 † ‡ 2…"1 "2 ‡ "1 "3 ‡ "2 "3 †Š where ' is the strain energy density at a point in the structure and the strain energy U for the whole structure is obtained by integrating the strain energy density across the volume of the structure. … U ˆ ' dV …3:109† V

Since we are usually interested in the initial yielding at some point in the structure we will use the strain energy density as the relevant property to be considered in the development of a yield criterion. Several yield criteria have been proposed over the years and some of these will be reviewed here.

Deformation of bodies and material properties

89

Maximum principle strain

This criterion is attributed to Saint-Venant10 and essentially states that yield will occur for a three-dimensional strain state at a point in a structure when the largest of the principal strains at the point equals the yield strain measured in a uniaxial tension test on a corresponding material. Thus yielding will occur when "0 ˆ "1  "2  "3

…3:110†

The St Venant criterion has been found in practice to be particularly relevant to brittle materials in which there is very little extension in the material after the onset of yielding.

Maximum strain energy criterion

This criterion was proposed by Beltrami and postulates that yielding will be initiated in the three-dimensional state of strain at a point in a body when the strain energy density at that point equals the strain energy density at the yield strain in a uniaxial tensile test on a corresponding material. The state of principal strains at the inset of yielding of a uniaxial test specimen is "1 ˆ "0 ;

"2 ˆ "3 ˆ ÿ"0

…3:111†

and the corresponding strain energy density '0 is, according to equation (3.110), E 2 " …3:112† 2 0 Thus equating the strain energy density at some point in the structure with the strain energy density at yield for a uniaxial tensile test specimen of corresponding material, we have '0 ˆ

"eff ˆ "0 where "eff is the e€ective strain de®ned by  1 ‰…1 ÿ †…"21 ‡ "22 ‡ "23 † "eff  …1 ‡ †…1 ÿ 2† 1=2 ‡ 2…"1 "2 ‡ "1 "3 ‡ "2 "3 †Š

10

…3:113†

…3:114†

Saint-Venant, AdheÂmar-Jean-Claude Barre de (Villiers, Paris, 1797 ± Saint Ouan, Loir-et-Cher, 1886), French mathematician and engineer.

90

Energy methods in structural mechanics

and constitutes a measure of the actual strain state with respect to yielding according to Beltrami.

Von Mises criterion

This yield criterion, like that proposed by Beltrami, is based on the strain energy density at some point in the structure. However, in the von Mises11 criterion the particular component of the strain energy is that which is associated with distortion of the material. The strain energy density formulation in equation (3.108) can be reformulated as  E 1 'ˆ …" ‡ "2 ‡ "3 †2 2 3…1 ÿ 2† 1  1 2 2 2 ‰…" ÿ "2 † ‡ …"1 ÿ "3 † ‡ …"2 ÿ "3 † Š ‡ …3:115† 3…1 ‡ † 1 By remembering equation (3.12) it is easy to infer that the ®rst term in equation (3.115) can be associated with volumetric change owing to the strain state by means of the material constant K, the bulk modulus. K

E 6…1 ÿ 2†

…3:116†

The second term in equation (3.115) can be therefore considered to be associated with the strain energy of distortion of the material due to the strain state. The von Mises criterion postulates that yielding will occur when the distortional strain energy density in a three-dimensional state of strain equals the distortional strain energy density at the yield strain in a uniaxial tensile test specimen of the corresponding material. Thus the strain energy density 'd associated with distortion is 'd ˆ

E 1 ‰…" ÿ "2 †2 ‡ …"1 ÿ "3 †2 ‡ …"2 ÿ "3 †2 Š 2 3…1 ‡ † 1

…3:117†

The corresponding distortional strain energy density 'od for a tensile specimen is E 2 …1 ‡ † 2 "0 2 3 Hence, the von Mises criterion is

11

'od ˆ

…3:118†

"eff ˆ "0

…3:119†

Mises, Richard von (Lemberg, Austria, 1883 ± Boston, 1953), German mathematician and engineer.

Deformation of bodies and material properties

91

where 1 "eff ˆ p ‰…"1 ÿ "2 †2 ‡ …"1 ÿ "3 †2 ‡ …"2 ÿ "3 †2 Š1=2 …3:120† 2…1 ‡ † Experience has shown that the von Mises criterion provides results that accord reasonably with tests on biaxial tests on ductile materials, such as structural steels and aluminium alloys. It is accepted that this criterion is more appropriate for ductile materials than that proposed by Beltrami and is preferred for application in structural practice.

Suggested exercise problems 1.

A tensile test round specimen with a 15 mm diameter gauge section is shown in Fig. 3.28. The specimen was subjected to a axial load P and the change of length,
l for the gauge length, lg ˆ 50 mm, was measured using a mechanical extensometer that was accurate to 0:005 mm. The change of diameter
d was also measured. The following are the results from the test:

P (kN) 9:9 20:1 28:45 35:8 45:9 52:9 53:20 53:4 52:9 53:2
l (mm) 0:02 0:04 0:06 0:07 0:09 0:11 0:15 0:20 0:29 0:52
d (mm) 0:01 0:02 0:02 0:03 0:04 0:05 0:06 0:09 0:16 0:27

2.

Plot these results on appropriate scales and calculate the averaged values of the material modulus, the yield strain and Poisson's ratio. [Ans: "0 ˆ 2:2  10ÿ3 ,  ˆ 0:44, E ˆ 2:56  105 N/mm2 ] A section of structure is essentially operating in plane strain. The strains in orthogonal x and y directions are "xx ˆ 1:78  10ÿ3 , "yy ˆ ÿ1:28  10ÿ3 and "xy ˆ 1:66  10ÿ3 . The negative sign denotes compressive strain. Calculate the principal strains in the plane and determine, using the von Mises criterion, if the strain state in the material exceeds the elastic limit. The material has a yield strain, "0 ˆ 2:15  10ÿ3 in a tensile test. [Ans: "1 ˆ ÿ2:0  10ÿ3 , "2 ˆ 2:5  10ÿ3 , "eff ˆ 3  10ÿ3 ] 15 mm dia. P

P

lg = 50 mm

Fig. 3.28

4.

Theory of elastic deformation of beams

Introduction

Bars and beams are ubiquitous structural elements, acting as parts of frames and grid structures. Essentially a beam, or bar, is a structural component whose length is signi®cantly greater than its breadth and width. Conventionally a bar is an axially loaded element, while a beam has loads applied normal to its longitudinal axis. A beam may be supported at various positions along its length and in various manners, as is seen later in the text. Up to this stage we have examined the behaviour of elastic materials, the interpretation of tensile tests and the concept of strain in one, two and three dimensions. Chapter 1 also considered the concept of equilibrium for general systems. Essentially, the objective in structural mechanics is to determine the manner in which speci®c structures or components are deformed by external actions and to enable design calculations to evaluate the maximum levels of loading and deformations that can safely be allowed in practice. The analysis of the response of a beam to a speci®c pattern of applied loads requires that . . .

The beam as a whole and all of its parts must be in equilibrium with the applied loading. The deformations of the beam must be continuous and compatible with the restraints of the supports. The strain energy must conform with the properties relevant to the material (in our case the conditions of elasticity).

The exact solution of the deformations of a generic structural element is, in general, very dicult to achieve for solids with arbitrary shapes. The approach is simpli®ed by de®ning classes of elements, for example bars, beams, plates and shells, whose shapes have generic features which enable us to make assumptions to reduce the complexity of the analysis. Of course, the introduction of such assumptions

94

Energy methods in structural mechanics q

Fig. 4.1

implies a certain degree of approximation, but experience and testing has shown that the assumptions result in predictions for the deformations of elastic and many plastic structural components within an accuracy acceptable for engineering applications. Starting here with bars and beams, which are essentially three dimensional, with length, depth and breadth, we seek to simplify the analysis by the assumption about the way in which these elements deform in response to loading. We consider beams composed of materials with linear elastic properties and loaded such that they remain in the range of small deformations. Simple observation shows that if one were to draw a line around the surface of an unloaded beam or bar, say with rectangular cross section, the line being composed of straight parts and normal to the axis of the element, then when the beam is loaded and deforms, the line is still continuous and the originally straight lateral parts are still reasonably straight. This is shown in Fig. 4.1. The above observation leads to a generalisation that a plane section imagined as cut through the beam before loading will remain plane after loading.

This is usually called Bernoulli's model of beam bending1 and serves the very important purpose of reducing the actual three-dimensional deformations to an idealised one-dimensional model in which the variation of the deformations can be described with reference to the behaviour of the longitudinal axis of the beam. It goes without saying that the same model applies to the simpler case of axially loaded bars that are also introduced in this chapter. Notice that bars loaded axially in compression can exhibit a form of instability, or buckling, and this is considered in detail in chapter 12. An alternative but essentially identical approach to assuming plane sections remain plane, is to make a simplifying assumption with regard to the strain distribution induced in the beam during loading. For a 1

Bernoulli, Jacob (Basel, 1654 ± ibidem, 1705), Swiss mathematician.

Theory of elastic deformation of beams

95

normally loaded element the simplest and acceptably accurate assumption in the realm of linear elasticity is that the strain varies linearly through the beam in the direction of any normal to the longitudinal axis.

It will be seen that this assumption is in fact equivalent to the Bernoulli assumption of plane sections. Of course, for beams composed of materials with more complex characteristics, or elements made from a composite of di€erent materials, the assumption of linear distribution of strain has to be carefully validated by comparisons with the deformations predicted by more exact analysis and with the corresponding results from carefully conducted tests. In the following two sections we consider bars and beams of uniform cross section: this is purely for the sake of simplicity, but the same treatment can be applied to bar and beams which vary in shape along their longitudinal axes as shown in an example in section 4.6.

4.1

Deformation of axially loaded bars

We start the mathematical treatment of the deformation of one-dimensional elements by considering axial loading only, i.e. the deformation of bars. Following what has been said in the introduction to this chapter and without loss of generality, a bar is de®ned here to be a straight element with a uniform cross section and to be axially loaded. Reference is also made to a Cartesian frame with the z-axis coincident with the axis of the bar, represented by the straight line through the centroids of the cross sections. The x and y-axes are arranged to coincide with the principal axes of inertia of the cross section, as shown in Fig. 4.2. N

x

z N y

Fig. 4.2

96

Energy methods in structural mechanics N

N

dl dl' l l'

dA

(a) dN

dl (b)

Fig. 4.3

According to Bernoulli's model, we assume that, after deformation, any cross section remains plane and orthogonal to the curve representing the deformed axis of the beam.

Consider the bar shown in Fig. 4.3(a), which is in equilibrium with the two opposite axial forces N applied at the ends. First of all it is worth noticing that we can obtain an elementary length of the bar by means of two cuts normal to the z axis at two very closely spaced cross sections dl apart. If we rely upon the smallness of the deformation of such a slice of the bar and consider each of the parts of the bar as rigid bodies, we are led to the conclusion that for the equilibrium it is necessary to have the same axial force N applied at any section. This fact follows from equation (2.55). We will also assume that the axis of the initially straight bar remains a straight line. The implications of this assumption are examined in chapter 12. Given the adopted model of deformation, we can extract an in®nitesimal sectional area from the body of the bar and derive from its behaviour a relationship which describes the behaviour of the whole bar, see Fig. 4.3(b). We are primarily going to deal with linearly elastic materials and from tensile tests we know that the relationship between the length of the in®nitesimal specimen before the deformation, dl, due to the axial force dN acting on its section dA, and the length after the

Theory of elastic deformation of beams

deformation has occurred, dl 0 , is given by the formula   1 dN dl dl 0 ˆ 1 ‡ E dA

97

…4:1†

where E is Young's modulus for the material, see equation (3.1). In the bar, every elementary length between two chosen cross sections has the same elongation which is the consequence of the assumption that plane sections remain plane. Therefore, as … N ˆ dN …4:2† A

and

… dA

Aˆ

…4:3†

A

we can sum up all the contributions in the cross section to obtain … 1 0   dN C B N 0 A C B … …4:4† dl ˆ @1 ‡ A dl ˆ 1 ‡ EA dl E dA A

provided, of course, that E is constant. This constitutes the deformation formulation for an in®nitesimal element of the bar between two very near cross sections under the axial force N. As the axial force is constant for the whole length of the bar, we can integrate over the length l and obtain the relationship between the applied axial force and the elongation of the whole bar   … … … N N l ˆ dl 0 ˆ dl ˆ 1 ‡ 1‡ dl EA EA l

 ˆ 1‡

l



l

N l EA

…4:5†

The energy stored in the bar is simply given by the work done by the axial forces N during the process of loading, i.e.
l …

UˆW ˆ

N…
l† d
l 0

where N…
l† is the axial force during the loading process.

…4:6†

98

Energy methods in structural mechanics

Equation (4.5) gives us …l 0 ÿ l† EA …4:7† l and we can write, remembering Clapeyron's theorem for linearly elastic materials, see equation (3.91), Nˆ

UˆW ˆ

4.2

1 EA 0 …l ÿ l†2 2 l

…4:8†

Deformation of beams

Having examined the behaviour of bars, we now turn our attention to the deformation of beams subjected to a variety of types of loading. Start by considering a beam to be essentially a straight structural element of uniform cross section capable of resisting shear forces and bending moments about the two principal axes of inertia in the plane of its generic cross section. The shear forces and the bending moments are considered here to be induced only by end forces and transverse loads, i.e. normal to the longitudinal axis of the bar. As before, a Cartesian reference frame is adopted with the z-axis coincident with the axis of the bar, represented by the straight line through the centroids of the cross sections along the beam. The x and y-axis are taken initially to coincide with the principal axes of inertia of the cross section, see Fig. 4.4. As we are primarily interested in establishing some relationships that can describe the bending deformations of a beam, regardless of the particular loading applied, let us consider, for the sake of generality, the beam shown in Fig. 4.5.

Mx My

x Sx y

Sy

Fig. 4.4

Theory of elastic deformation of beams

99

z x y

M

M

dz l

Fig. 4.5

This beam is subject to the couples M and ÿM about the x-axis at the left and right ends of the beam, respectively. These couples are equal in magnitude and opposite in sign and therefore in equilibrium. This beam is classed as being in pure bending. Following the same line of reasoning adopted for the axially loaded bar of Fig. 4.3, i.e. conceptually cutting the beam in two parts at any cross section, we can easily deduce that the same bending moment is applied to every cross section of the beam. Also in this case we will assume, according to Bernoulli's model, that any cross section remains plane and orthogonal to the axis after the deformation has occurred due to the bending moment.

In contrast to the case of the axially loaded bar, which remains straight, we will assume that in this case the axis of the beam becomes an arc of a circle. Moreover, we will also assume that this arc of circle retains its original length, so that we can consider that no stretching of the axis of the beam has occurred.

Now consider Fig. 4.6, where an element of the beam between two cross sections at distance dz is shown. As a consequence of the assumptions stated above, it is evident that after deformation has occurred, the right side face of the element has rotated through an angle d about the axis x relative to the left side face. In this framework, it is evident that the ®bres of the beam have increased or decreased their length proportionally to their distance from the x axis, according to the simple rule dl 0 ˆ dz ‡ y

…4:9†

100

Energy methods in structural mechanics dφ

h

z y dz

Fig. 4.6

Notice that the ®bres below the axis have increased their length, as the sign ‡ holds for their distance from the axis, while the ®bres above the axis have decreased their length, being governed by negative values of y. We can now refer to the elongation, equation (4.1), to obtain a relationship between the bending moment M and the radius of curvature r of the axis of the beam after deformation has occurred. Thus, from equation (4.1) we have …dl 0 ÿ dz† E dA …4:10† dz and this means that an axial force dN acts on the typical in®nitesimal element of the cross section extracted from the beam. The magnitude of this force is obviously dependent on the deformation. Substituting equation (4.9) in equation (4.10) we obtain y d dN ˆ E dA …4:11† dz We can now sum all the contributions of the elementary forces on the cross section to obtain the resulting axial force N and bending moment M. Thus, … … … y d d E dA ˆ E y dA N ˆ dN ˆ …4:12† dz dz A A A … … 2 … y d d E dA ˆ E y2 dA …4:13† M ˆ y dN ˆ dz dz dN ˆ

A

A

A

Since we assumed the reference system x±y to have its origin at the centroid of the cross section, then from this condition we have … y dA ˆ 0 …4:14† A

Theory of elastic deformation of beams

101

and therefore Nˆ0

…4:15†

Notice that for pure moment the neutral axis, that is the position where the longitudinal ®bres neither extend nor contract, passes through the centroid of the cross section of the beam. The integral term on the right hand part of equation (4.13) is termed the second moment of area Ix , sometimes called the moment of inertia, … …4:16† Ix  y2 dA A

In this case the second moment of area has been evaluated about the xaxis. The magnitude of this parameter will vary according to the orientation of the axis about which the second moment of area is calculated. Now, we can write d EI …4:17† dz x which yields the desired relationship between the bending moment M and the radius of curvature r. As we have Mˆ

d 1 ˆ dz r it follows

…4:18†

M 1 ˆ EIx r

…4:19†

Equations (4.17) and (4.19) constitute key relationships for beams subjected purely to a bending moment. It is a simple matter to calculate the relative rotation between the ends of the beam of Fig. 4.5. As the bending moment is constant, we can integrate equation (4.17) over the whole length l of the beam and obtain …l

…l M M M ˆ dz ˆ dz ˆ l EIx EIx EIx 0

…4:20†

0

The strain energy stored in the beam is given by the work done by the couple M during the process of loading and is … U ˆ W ˆ M…† d 0

…4:21†

102

Energy methods in structural mechanics

which we can also write, by means of equation (4.20) and Clapeyron's theorem, equation (3.91), UˆW ˆ

1 EIx 2  2 l

…4:22†

So far we have dealt with a bar subjected to a constant axial force and a beam subjected to a constant bending moment about the axis x. It is signi®cant that all the relationships obtained above are linear and therefore the behaviour of a beam subjected to both bending moment and axial force can be obtained very straightforwardly by adding the e€ects of the axial load to those of the bending moment.

In the next section we will see why this can be allowed from a geometrical point of view in the framework of the theory of very small displacements. Following from their derivation, it is also clear that equations (4.4), (4.17) and (4.19) hold true for any element of beam between two in®nitesimally near cross sections. Therefore they can also be applied to the analysis of beams subjected to axial forces N and bending moments M that vary along the axis z.

4.3

The theory of very small displacements

Following the development in the previous section we can now begin the study of the deformations of simple systems of beams subjected to a variety of types of loading and constraints. A foreword is, however, necessary to introduce the basis on which the theory presented here is founded, i.e. the theory of very small displacements, also called the theory of in®nitesmal displacements. In order to ®x our ideas, by very small we may for example intend displacements of the beam normal to its longitudinal axis that are less than about 1% of its length. In this manner the strains will remain in the elastic range for most engineering materials. This theory of very small displacements is essentially based on two hypotheses, namely .

.

the displacements of a deformable body are essentially negligible in comparison with its dimensions and therefore e€ects from separate loading states can be geometrically added with reference to the position and form of the element before loading on account of the smallness of the displacement ®eld the forces acting on the body can be always referred to the unstrained, i.e. undeformed con®guration.

These hypotheses are normally suciently accurate for the great majority of engineering applications and allow a substantial simpli®cation for analysing most cases of structural deformation. From a mathematical

Theory of elastic deformation of beams

103

dz 0

φ

z uy uy + duy ds

y

Fig. 4.7

point of view, the ®rst hypothesis corresponds to the neglection of all the second and higher order terms in the derivatives of the displacement ®eld. This point can be clari®ed, for example, with respect to the expression of the curvature of the axis of a beam. In fact, with reference to Fig. 4.7, it is   duy  ˆ arctan ÿ dz s  2 q duy dz ds ˆ dz2 ‡ du2y ˆ 1 ‡ dz …4:23† d2 uy 1 d d 1 dz2 s  ˆ ÿ s ˆ ˆ    2 3  r ds dz duy 2 duy 1‡ 1‡ dz dz where uy is the displacement of the beam centroid in the y direction and is generally a function of z, i.e. uy …z†. Remember that conventionally, anticlockwise rotations are considered to be positive. As stated above, typically in the practical engineering application of beams the de¯ection is less than 1% of the length and the slope is also negligible in comparison with unity, that is for an initially straight bar with the axis coincident with the z axis  2 duy 1 …4:24† dz

104

Energy methods in structural mechanics

Equation (4.24) leads to the linearisation of equation (4.23), for the displaced axis of the beam 2

d uy 1 ÿ 2 …4:25† r dz This approximation has also a particular geometrical meaning, as it assigns, to the same order of approximation, the identical length l to both the undeformed …uy ˆ 0† and the deformed …uy ˆ uy …z†† axis of the beam between the values z ˆ 0 and z ˆ l. In fact, for the undeformed state, uy ˆ 0, it follows that 0

…l

l ˆ dz ˆ l

…4:26†

0

while for uy ˆ uy …z† we have (Fig. 4.7) s  2 …l q …l duy 1‡ dz dz2 ‡ du2y ˆ l0 ˆ dz 0

…4:27†

0

which becomes, on account of the negligibility of the term in equation (4.24), 0

…l

l ˆ dz ˆ l

…4:28†

0

Therefore the line of equation y ˆ uy …z† has the same length of its projection on the z-axis and we can consider the horizontal displacement of the left end of the cantilever of Fig. 4.8, to be zero when subject to bending due to the applied couple M. The second of the hypotheses stated above can be clari®ed by saying that we can consider a force N which is perfectly axial before the deformation to be still axial after the deformation has occurred; or if the force N is normal to the beam before deformation it will still be normal after the deformation has occurred as it can be always referred to the unstrained con®guration of the beam. This is shown in Fig. 4.9.

M

Fig. 4.8

Theory of elastic deformation of beams

105

N

P

Fig. 4.9

It is evident from this example that in this theory the non-linear e€ects (which eventually lead to buckling) cannot be considered. Buckling is considered in chapter 12 using an appropriately modi®ed description of beam de¯ections. Other implications will be discussed as we further the study of deformable bodies.

4.4

Euler's method for the analysis of beam deformations

In the previous sections expressions have been derived for the strain energy of a beam that deforms under the in¯uence of axial forces and bending moments. Now we consider the methods with which we can extend these expressions to analyse the behaviour of actual beams subjected to speci®ed loading and constraints. From a general point of view there are two distinct approaches to this analysis process, i.e. . .

use an energy expression directly (typically to seek an approximate solution) use the di€erential equations of equilibrium which can be derived from the variational statement, equation (1.28).

The advantage of the latter approach is that when a solution is obtained to the di€erential equation describing the equilibrium of the beam the answer can be considered exact, at least for the model adopted. The disadvantage is that such solutions usually are restricted to idealised boundary conditions and only to limited variations in the geometry of the beam and its types of supports. Nevertheless, the range of exact solutions that have been developed for beams is fairly extensive and forms a valuable counterpart to the direct use of the energybased methods. One particular value of the exact solution of the di€erential equations is that such results provide a standard against which to compare and validate the accuracy and use of approximate methods, such as ®nite elements, involving direct use of the energy functional and trial functions. Thus, the approach followed in this text is similar to that in current structural engineering practice in which the deformations of beam with

106

Energy methods in structural mechanics q

A

z

y

B

Fig. 4.10

simple boundary conditions, uniform geometry and loading are evaluated using simple expressions derived from the di€erential equation of equilibrium developed in this section. The deformations of more complex conditions of beam geometry and loading are usually calculated by means of approximate methods, such as the ®nite element methods developed later in this text, but the exact solution of simpli®ed beam conditions is always kept as a validation benchmark. It is also possible to obtain approximate solutions from the di€erential equation of beam equilibrium by using the method of ®nite di€erences. This was a quite popular and successful approach in the years 1930±1950. However, the ®nite element method has essentially eclipsed the ®nite di€erence approach for structural analysis, although the latter remains the basis for computer modelling of ¯uid dynamics. For the sake of completeness, section 4.7 presents the ®nite di€erence method as applied to the analysis of the deformations of beams. From a purely theoretical standpoint, it may be noted2 that the ®nite di€erence and the ®nite element method, introduced later in this text, can both be considered to be essentially forms of assumed trial functions, with the former applied to the functions involved in the di€erential equations of equilibrium derived from the energy formulation and the latter applied directly to the energy functionals. In order to derive the di€erential equation of equilibrium for straight beams we apply the classical variational calculus to the energy functional. This route is essentially due to the eminent Swiss mathematician Euler (actually, after a suggestion by Bernoulli3 ) hence the classi®cation of this development can be stated as Euler's method of analysis of beam deformations. Consider the simple supported beam shown in Fig. 4.10, subject to a uniformly distributed vertical load per unit length q. 2

Croll, J. and Walker, A. C. 1971 Finite di€erence method as a localised Ritz process. Intnl J. of Eng. Math. 3, 155. 3 Bernoulli, Daniel (GroÈningen, 1700 ± Basel, 1782), Swiss mathematician and physicist, nephew of Jacob.

Theory of elastic deformation of beams

107

In order to study its state of equilibrium and deformation, we can make reference to the general condition, equation (1.28) U ˆ W and write down the equivalence between the variation of the strain energy and the variation of the work done by the external forces, both due to a virtual ®eld of displacements u. Writing the variation of the work done by the external forces is quite a simple matter …l W ˆ q uy dz

…4:29†

0

where uy is, as usual, the displacement of the beam axis in the y direction. However, writing the variation of the strain energy requires a little more development. Following the approach in equations (4.17) and (4.22) for a beam subject to constant bending moment, we can write the energy stored in an in®nitesimally long element of beam as 1 …d†2 Udz ˆ EIx 2 dz

…4:30†

which holds true provided the curvature, and therefore the bending moment, is evaluated point by point along the beam. In fact, in general the moment and therefore the curvature varies along the length of the beam. The energy stored in the whole simply supported beam is given by the integral …l 1 …d†2 Uˆ EIx 2 dz

…4:31†

0

or, equivalently,  2 …l 1 d EIx dz Uˆ 2 dz

…4:32†

0

which, on account of equations (4.18) and (4.25), becomes  2  2 2 …l …l d uy 1 1 1 Uˆ EIx dz ˆ EIx dz 2 r 2 dz2 0

0

…4:33†

108

Energy methods in structural mechanics

The variation of this expression with regards to the displacement ®eld uy is4 …l

U ˆ EIx u00y u00y dz

…4:34†

0

where, for the sake of enabling a succinct presentation, we have written d2 uy  u00y …4:35† 2 dz We can therefore write the condition of equilibrium (equation (1.28)) as …l

EIx u00y

u00y

0

…l dz ˆ q uy dz

…4:36†

0

This variational expression can be manipulated to obtain a perfectly equivalent di€erential equation. Thus, integrating the left hand

4 The application of the calculus of variations provides a means to minimise the energy functional and is summarised as follows. Suppose we are given a functional U in the integral form   …  …  q q ; ::: dV ‡ G ; ; ::: dS U… † ˆ F ; qx qx V

S

where F and G are functions of …x; . . .† and S is the boundary of the closed region V. Suppose also that the variations in are taken among the admissible set of functions satisfying the general boundary conditions H1 … † ˆ 0

on S1

H2 … † ˆ 0

on S2

where S1 ‡ S2 ˆ S. Then, for a small admissible variation from to ‡  we de®ne the corresponding ®rst variation in U by    … qF qF q  ‡  ‡


dV U ˆ q q…q =qx† qx V

…

‡ S

   qG qG q  ‡  ‡


dS q q…q =qx† qx

which stems from a straightforward series expansion.

Theory of elastic deformation of beams

109

member by parts, we obtain …l

EIx u00y

u00y

dz ˆ

‰EIx u00y

u0y Šl0

…l

0 ÿ EIx u000 y uy dz

0

…4:37†

0

and integrating the right hand term of equation (4.37) we have …l

EIx u00y

u00y

dz ˆ

‰EIx u00y

u0y Šl0

ÿ

‰EIx u000 y

0

uy Šl0

…l

‡ EIx u0000 y uy dz 0

(4.38)

Now, remembering that a virtual displacement ®eld is, by de®nition, in®nitesimal and compatible with the constraints on the beam, let us have a closer look at the two ®rst right hand terms. The ®rst term, i.e. ‰EIx u00y u0y Šl0

…4:39†

is identically zero, since the expression ÿEIx u00y

…4:40†

represents, on account of the equations (4.25), (4.18) and (4.17), the bending moment M…z† at the ends of the beam. In fact, for a beam with simple supports at A and B and loaded such that no couples are applied at A and B the bending moment at the ends is zero and equation (4.39) is evidently zero. The second term on the right hand side of equation (4.38), i.e. l ‰EIx u000 y uy Š0

…4:41†

is also equal to zero, as the variation must vanish on the simple supports at A and B which do not allow any vertical displacement. Thus, the condition of equilibrium (equation (4.36)) reduces to …l

EIx u0000 y

…l uy dz ÿ q uy dz ˆ 0

0

…4:42†

0

which, since the virtual displacement ®eld uy is arbitrary, is equivalent to the fourth order di€erential equation EIx u0000 y ÿq ˆ0

…4:43†

This equation is usually called Euler's equation of beam deformation, as its solution represents the deformed con®guration of the elastic beam under the speci®ed lateral load distribution q. In the present example of the beam shown in Fig. 4.10, the term EIx , which is

110

Energy methods in structural mechanics

commonly referred as the ¯exural sti€ness of the beam, is a constant, therefore the solution to equation (4.43) is trivially obtained by integrating equation (4.43) four times to give uy …z† ˆ

qz4 ‡ Az3 ‡ Bz2 ‡ Cz ‡ D 24EIx

…4:44†

where A, B, C and D are constants of integration. Now we impose on equation (4.44) the geometrical and mechanical constraints (i.e. in terms of displacements and forces, respectively) at the simple supports A and B, which allowed us to drop the terms in equations (4.39) and (4.41). This process enables the evaluation of the constants of integration. These constraints require that uy …z† ˆ 0 M…z† ˆ

for z ˆ 0

ÿEIx u00y …z†

ˆ0

and z ˆ l for

zˆ0

…4:45† and z ˆ l

…4:46†

Therefore we have a non-homogeneous linear system of four equations in the four unknown constants A, B, C and D. The system is 8 uy …0† ˆ D ˆ 0 > > > > > EIx u00y …0† ˆ EIx 2B ˆ 0 > > > < ql 4 …4:47† u …l† ˆ ‡ Al 3 ‡ Bl 2 ‡ Cl ‡ D ˆ 0 y > > 24EI x > > > 2 > > > EI u00 …l† ˆ ql ‡ 6EI Al ‡ 2EI B ˆ 0 : x y x x 2 or, in matrix form, 2 32 3 2 3 A 0 0 0 0 1 6 0 2 0 0 76 B 7 6 7 0 6 76 7 6 7 …4:48† 6 3 2 76 7 ˆ 6 7 4 4l 5 4 5 4 l l 1 C ÿql =…24EIx † 5 6l

2

0 0

D

ÿql 2 =…2EIx †

and yields Aˆÿ Cˆ

ql ; 12EIx

ql 3 ; 24EIx

Bˆ0 Dˆ0

…4:49† …4:50†

Finally, we get uy …z† ˆ

q …z4 ÿ 2lz3 ‡ ql 3 z† 24EIx

…4:51†

Theory of elastic deformation of beams

111

q

A

z B

y l

Fig. 4.11

This fourth-order parabolic expression represents the ®eld of vertical displacements, uy …z†, which the axis of the beam undergoes to assume its con®guration of equilibrium under the assigned uniformly distributed load q. As the axis was originally straight, equation (4.51) also prescribes the deformed con®guration of the beam, as shown in Fig. 4.11. The solution of the di€erential equation (4.43) is exact and therefore within Bernoulli's assumptions, this method of analysis predicts exactly the de¯ections of the initially straight beam. Up to now we have illustrated the Bernoulli model of beam bending and shown that we can obtain the con®guration of equilibrium of a simple supported beam under a speci®ed form of loading, according to the general condition equation (1.28). In the next section we will see that Euler's equation for beam deformation, equation (4.43), can be employed for the bending of any other simply constrained beam, provided appropriate boundary conditions are taken into account.

4.5

The boundary conditions associated with Euler's equation of beam deformation

For the sake of clarity we have considered so far a simply-supported beam. However, the procedure which led from the variational condition of equilibrium (equation (4.36)) to the Euler equation of equilibrium (4.43) is a general one and can be applied to any kind of variously constrained beam. The resulting di€erential equation (4.43) does not change, but the boundary conditions of the problem can assume di€erent forms. Therefore, let us now consider a generic beam, whose bending behaviour is described according to Bernoulli's model of plane sections. For the moment the treatment is not restricted to any speci®c applied constraints. Consider a beam subjected to a generic distributed vertical load per unit length q…z†, as shown in Fig. 4.12. The load q…z† can vary along the axis of the beam but is always orthogonal to it and is taken to act in the y direction.

112

Energy methods in structural mechanics

q (z)

z

y l

Fig. 4.12

Irrespective of the form of constraints applied at both ends of the beam, it is straightforward to write the general condition of equilibrium (equation (1.28)) in the form shown in equation (4.36). Integrating twice by parts the terms on the left hand side of equation (4.36) we obtain equation (4.38) where, of course, the virtual displacement ®eld has to be in®nitesimal and compatible with the constraints applied to the beam. Let us now have a closer look at the ®rst two terms on the right hand side of equation (4.38). We already know that M…z† ˆ ÿEIx u00y …z†

…4:52†

and, if we consider Fig. 4.13, we can readily deduce that in the theory of in®nitesimal displacements …z†  tg…† ˆ ÿ

duy …z† dz

…4:53†

0

z

φ duy

dz

y

Fig. 4.13

Theory of elastic deformation of beams

113

That is, the rotation of a generic cross section can be identi®ed with the negative value of the slope of the axis of the beam. Therefore we can attribute a precise physical meaning to the terms which appear in equation (4.38), as follows EIx u00y ˆ ÿM

…4:54†

u0y

…4:55†

ˆ ÿ

It remains to elucidate the meaning of the term EIx u000 y …z†

…4:56†

Remembering equation (4.52), from the relationship equation (2.57), which links the slope of the bending moment diagram to the value of the shear force, we have dM…z† ˆ ÿEIx u000 …4:57† y …z† ˆ S dz Thus it is clear that the ®rst two right hand terms of equation (4.38) are zero whatever the constraints applied at the beam's ends. This is exempli®ed by a simple cantilever, as shown in Fig. 4.14. We have at the left hand end, uy …0† ˆ 0

and …0† ˆ ÿu0y …0† ˆ 0

…4:58†

and at the right hand end M…l† ˆ ÿEIx u00y …l† ˆ 0

and S…l† ˆ ÿEIx u000 y …l† ˆ 0

…4:59†

so that both terms ‰EIx u00y u0y Šl0

l and ‰EIx u000 y uy Š0

…4:60†

are zero. Therefore, also in this case Euler's equation of beam equilibrium (4.43) is valid. Of course, while integrating the di€erential

q (z)

z

y

Fig. 4.14

114

Energy methods in structural mechanics

P

A

l/2

B

l/2

Fig. 4.15

equation of equilibrium, equation (4.43), we must take into account the appropriate boundary conditions, i.e. equations (4.58) and (4.59), to evaluate the constants of integration. The reader may ®nd it edifying to verify that the terms in equation (4.60) vanish for any beam in the y±z plane with various constraints at the ends, provided the beam is indeed ®xed in the frame of reference when considered as a rigid body.5 Another interesting example is that given in Fig. 4.15. This is a beam clamped at the left-hand end and simply supported at the right-hand end. It is subject to a point load P at the middle of the span. In order to derive an expression for its de¯ection, it is convenient to divide the whole span into two half-spans of length l=2. Let us consider the two half-spans separately and denote the vertical displacement ®eld of the ®rst half-span as u1y and that of the second one as u2y . The condition of equilibrium (1.28) can be written as follows l=2 …

EIx u001y

u001y

…l dz ‡

0

EIx u002y u002y dz ˆ Puy …l=2†

…4:61†

l=2

where, of course, the displacement at the mid point of the whole span is unique and is denoted by u1y …l=2† ˆ u2y …l=2† ˆ uy …l=2†. After integrating equation (4.61) twice by parts, as discussed above, we get l=2

l=2

000 l ‰EIx u001y u01y Š0 ‡‰EIx u002y u02y Šll=2 ÿ‰EIx u000 1y u1y Š0 ÿ‰EIx u2y u2y Šl=2 l=2 …

‡ 0 5

EIx u0000 1y

…l u1y dz ‡

EIx u0000 2y u2y dz ˆ Puy …l=2†

…4:62†

l=2

It is worth pointing out that for the cantilever in Fig. 4.8, which actually corresponds to the situation in Fig. 4.5, it is found that uy ˆ ÿ…M=2EI†z2 . This represents an arc of a circle only under the assumptions of small displacement theory developed in section 4.3.

Theory of elastic deformation of beams

115

Now, taking into account the following boundary conditions u1y …0† ˆ 0

and …0† ˆ u01y …0† ˆ 0

…4:63†

u1y …l=2† ˆ u2y …l=2†

…4:64†

1 …l=2† ˆ 2 …l=2† ˆ u01y …l=2† ˆ u02y …l=2†

…4:65†

M1 …l=2† ˆ M2 …l=2† ) ÿEIx u001y …l=2† ˆ ÿEIx u002y …l=2†

…4:66†

000 EIx u000 1y …l=2† ÿ EIx u2y …l=2† ˆ P

…4:67†

u2y …l† ˆ 0

and M2 …l† ˆ

ÿEIx u002y …l†

ˆ0

…4:68†

which impose the geometrical and mechanical aspects of the constraints at both ends of the beam and require the continuity at the mid-point of the span, we are led to the equation l=2 …

EIx u0000 1y

…l u1y dz ‡

0

EIx u0000 2y u2y dz ˆ 0

…4:69†

l=2

This expression is clearly equivalent to the following set of di€erential equations   l 0000 …4:70† EIx u1y ˆ 0 for z 2 0; 2   l 0000 …4:71† EIx u2y ˆ 0 for z 2 ; l 2 provided the conditions given by equations (4.63)±(4.68) are applied to the general solutions   l 3 2 …4:72† u1y …z† ˆ Az ‡ Bz ‡ Cz ‡ D for z 2 0; 2   l u2y …z† ˆ Ez3 ‡ Fz2 ‡ Gz ‡ H for z 2 ; l …4:73† 2 in order to de®ne the value of the eight constants A±H. In detail, we have u1y …0† ˆ 0

and …0† ˆ u01y …0† ˆ 0

…4:74†

u1y …l=2† ˆ u2y …l=2†

…4:75†

1 …l=2† ˆ 2 …l=2† ˆ u01y …l=2† ˆ u02y …l=2†

…4:76†

M1 …l=2† ˆ M2 …l=2† ) ÿ EIx u001y …l=2† ˆ ÿEIx u002y …l=2†

…4:77†

116

Energy methods in structural mechanics 000 EIx u000 1y …l=2† ÿ EIx u2y …l=2† ˆ P

u2y …l† ˆ 0 and

M2 …l† ˆ

…4:78†

ÿEIx u002y …l†

Finally, we can write 11 P 3 3 Pl 2 z ÿ z u1y …z† ˆ 96 EIx 32 EIx

ˆ0

…4:79†

 l z 2 0; 2

…4:80†



for

5 P 3 5 Pl 2 Pl 2 Pl 3 z ‡ z ÿ z‡ 96 EIx 32 EIx 8EIx 48EIx   l for z 2 ; l 2

u2y …z† ˆ ÿ

…4:81†

These two expressions completely de®ne the lateral displacement ®eld of the propped cantilever under analysis. Notice that the illustrated procedure can be extended in a straightforward manner to any system of concentrated loads.

4.6

Example of application of Euler's method of beam deformation for a beam with a variable depth

We have seen above that it is fairly straightforward to obtain an analytical solution to Euler's di€erential equation (4.43) for a beam with a uniform ¯exural sti€ness EIx . The analysis becomes more complicated when we consider beams with variable loading and geometry. As an illustration of this, the following analysis considers a cantilever with a linearly varying depth subjected to a load at its tip, as shown in Fig. 4.16. The ®gure shows that the cantilever has a length l. At the built-in end the depth of the section is d1 and at the tip the depth is d2 . The uniform

P

z

d1

dx

d2

y l

Fig. 4.16

Theory of elastic deformation of beams

117

breadth of the beam is b, so that the second moment of area of the cross section at some position z along the beam is Ix …z† ˆ

bd 3 …z† 12

…4:82†

The depth at position z is d…z† ˆ d1 ‡ …d2 ÿ d1 †

z l

…4:83†

By introducing the non-dimensional variables   d2 =d1 and   z=l, we can write d…; † ˆ d1 ‰1 ‡ … ÿ 1†Š

…4:84†

where  2 ‰0; 1Š. Also notice that for a generic function f from calculus we have df df d 1 df ˆ ˆ dz d dz l d

…4:85†

For an in®nitesimal element of beam between two cross sections at distance dz we can consider the depth as constant and we can again make reference to Bernoulli's model of beam deformation introduced in section 4.2. Retracing the steps of section 4.4, we can therefore write the di€erential equation governing the deformation of the beam as   2 1 d EIx …† d uy …† ˆ q…† …4:86† l 2 d 2 l2 d 2 As in the present case no distributed load q…† is applied (we consider the weight of the cantilever negligible in comparison to the applied force P), by integrating twice equation (4.86) we get   2 1 d EIx …† d uy …† ˆ Al …4:87† l d l2 d 2   2 EIx …† d uy …† …4:88† ˆ Al 2 ‡ Bl d 2 l2 The mechanical boundary conditions are   2 1 d EIx …† d uy …† S…1† ˆ ÿ ˆP l d l2 d 2   2 EIx …† d uy …† ˆ0 M…1† ˆ ÿ l2 d 2

…4:89† …4:90†

118

Energy methods in structural mechanics

and, consequently, we have ÿAl ˆ P

)

ÿPl ‡ Bl ˆ 0

P l

…4:91†

BˆP

…4:92†

Aˆÿ )

so that   2 EIx …† d uy …† ˆ Pl…1 ÿ † l2 d 2 By virtue of equations (4.82) and (4.84) we can write   2 1 d uy …† Pl 12Pl 1ÿ …1 ÿ † ˆ ˆ EIx …† l 2 d 2 Eb d13 ‰1 ‡ … ÿ 1†Š3

…4:93†

…4:94†

The slope of the beam can be obtained by integrating equation (4.94) as follows 1 duy …† l d   12Pl 2 2 ÿ  ‡ 2 ÿ 2 ˆÿ ‡C Eb d13 2… ÿ 1†2 …1 ‡  ÿ †2

…† ˆ ÿ

…4:95†

The constant C is evaluated using the following natural boundary condition …† ˆ ÿ

1 duy …† ˆ0 l d

at  ˆ 0

…4:96†

Hence, we have Aˆÿ

…2 ÿ † 2… ÿ 1†2

…4:97†

and equation (4.95) becomes …; † ˆ ÿ

  1 duy …† 12Pl 2 …2 ‡  ÿ 2† ˆÿ l d Eb d13 2…1 ‡  ÿ †2

…4:98†

The de¯ection function for the beam is obtained from equation (4.98) once again by integration, thus   12Pl 3 f1 …; † uy …; † ˆ ‡D …4:99† Eb d13 f2 …; †

Theory of elastic deformation of beams

119

where f1 …; † ˆ 3  2 ‡ 2  ÿ 42  2 ‡ 5 2 ÿ 3 ‡  ÿ 2 2 ‡ 2 ‡ 2…1 ‡  ÿ 2† ln…1 ‡  ÿ †

…4:100†

f2 …; † ˆ 2…1 ‡  ÿ †… ÿ 1†3 Evaluating the constant D for the boundary condition uy …0; † ˆ 0 we have Dˆÿ

 2… ÿ 1†3

…4:101† …4:102†

so that equation (4.99) is 3 2 …3 ÿ 42 ‡ 5 ÿ 2† 2 ‡ 2…1 ÿ † 7 6 12Pl 3 4 ‡2…1 ‡  ÿ † ln…1 ‡  ÿ †5 …4:103† uy …; † ˆ Eb d13 2…1 ‡  ÿ †… ÿ 1†3 The most easily recognised result from equation (4.103) is the value of the de¯ection of the tip of a cantilever with a uniform depth d1 . In fact, for  ˆ 1 and  ˆ 1 equation (4.103) gives uy …1; 1† ˆ

4Pl 3 Eb d13

…4:104†

Figures 4.17(a)±(c) show typical results for the de¯ection along the beam for values of  ˆ 1, 0.5 and 0.05, respectively. In these ®gures the value of the de¯ection has been normalised with respect to the de¯ection of the tip of a cantilever with a uniform depth, that is all the calculated values have been divided by equation (4.104). The value  ˆ 0:05 has been chosen since the value of  ˆ 0 causes the tip of the cantilever to have no depth and therefore the de¯ection becomes unrealistically large at the tip. The variation of the normalised de¯ection at the tip of the cantilever with the parameter  is given by   3  ÿ 42 ‡ 3 ‡ 2 ln…† u~y …1; † ˆ 3 …4:105† 2… ÿ 1†3 Figure 4.18 shows results from equation (4.105). It is clear that for di€erent and more complicate variations of the depth of the section along the beam length the integration of the di€erential equation can pose some problems. In such a case, if we wished we could integrate the equation of equilibrium numerically. However,

120

Energy methods in structural mechanics 1·0

α=1 0·8

u (ζ)

0·6 0·4 0·2 0 0

0·2

0·4

ζ (a)

0·6

0·8

1·0

0·6

0·8

1·0

0·6

0·8

1·0

2·0

α = 0·5

u (ζ)

1·5

1·0

0·5

0 0

0·2

0·4

ζ (b)

6

α = 0·05 5

u (ζ)

4 3 2 1 0 0

Fig. 4.17

0·2

0·4

ζ (c)

Theory of elastic deformation of beams

121

Fig. 4.18

following such an approach a simple formula for use in design calculations would not be readily available. The next section shows an alternative approach to obtaining a solution to the Euler di€erential equation of deformation.

4.7

Approximate solution of Euler's equation of beam deformation using the method of ®nite di€erences

In the next chapters much space will be dedicated to the development of the method of trial functions, including the method of ®nite elements. In fact this latter method has in recent years become the main approach to evaluating the deformations of beams, plates and shells with practical forms of geometry, boundary conditions and loading. Several exact analytical solutions, usually for simpli®ed boundary conditions and loading, of the di€erential equations of equilibrium derived from the corresponding energy expression for beams, plates and shells can provide valuable information for validating the ®nite element numerical modelling and are generally used to assess the accuracy of the results. It is possible, nevertheless, to obtain approximate solutions for the di€erential equations of equilibrium for a variety of geometries and loading conditions. This can be achieved using the ®nite di€erence method. Nowadays this method is not used widely in practical engineering design analysis, largely because of the greater versatility of the ®nite element method and the large number of available commercial computer programs that incorporate the ®nite element method. However, for the sake of completeness in this text, the present section presents the basis of the di€erence method and exempli®es its use for the determination of the deformations of beams based on the Euler equation of equilibrium.

122

Energy methods in structural mechanics f (x)

(n +1) (n –1)

(n)

h

x

Fig. 4.19

Figure 4.19 shows the plot of a function f(x) of a variable x. The slope of the curve at some position x is obtained analytically by di€erentiation of the function, i.e. df …x† …4:106† dx The slope can be approximated as follows. The range of the function is sub-divided to form a series of nodes along the curve describing the function. The nodes are usually, but not necessarily, equidistant from each other. In the ®gure the nodes are separated by a distance h. The value of the function at node n is fn and the slope at that node is approximated by f 0 …x† ˆ

1 ‰f ÿ fn ÿ 1 Š …4:107† 2h n ‡ 1 Intuitively, as the nodes become very closely spaced, i.e. h ! 0, the approximate value for the slope approaches the analytical value. The higher order derivatives can be obtained using the same approach. For example, the second derivative is written as the ratio between the increment of the ®rst derivative between the mid-points of the arcs n ÿ 1; n and n; n ‡ 1 and the distance h, namely       d df …x† fn ‡ 1 ÿ fn fn ÿ fn ÿ 1 00 f n …x† ˆ ÿ h  dx dx x ˆ xn h h f n0 …xn † 

ˆ

1 ‰ fn ‡ 1 ÿ 2fn ‡ fn ÿ 1 Š h2

…4:108†

Theory of elastic deformation of beams

123

The fourth derivative is approximated by 1 ‰ f n00‡ 1 ÿ 2f n00 ‡ f n00ÿ 1 Š h2 1 ˆ 4 ‰ fn ÿ 2 ÿ 4fn ÿ 1 ‡ 6fn ÿ 4fn ‡ 1 ‡ fn ‡ 2 Š h

f n0000 

…4:109†

The ®nite di€erence method is exempli®ed here by application to a beam with uniform cross section, simply supported at both ends and subjected to a uniformly distributed load q. The accuracy of the result is judged by comparison of the value of the mid-span de¯ection with the corresponding value from the analytical solution of the Euler deformation equation, see equation (4.51),   l 5ql 4 ql 4 uy ˆ ˆ : 2 384EIx 76 8EIx

Three internal nodes

Figure 4.20 shows the arrangement of the nodes along the beam, in this case the model includes three internal nodes, i.e. between the end boundary conditions. The boundary conditions are chosen so that displacement in the y direction is zero at z ˆ 0 and l and that the moment, and therefore the curvature, is also zero at these positions. Thus u0 ˆ u4 ˆ 0

and

uÿ1 ˆ ÿ u1 ;

u5 ˆ ÿ u3

…4:110†

The Euler equation (4.43) can be written in terms of ®nite di€erences at

l q

EI

z y –1

0

1

2

h=

Fig. 4.20

l 4

3

4

5

124

Energy methods in structural mechanics

each of the internal nodes by virtue of equation (4.109). Hence, Node 1 uÿ1 ÿ 4 u0 ‡ 6 u1 ÿ 4 u2 ‡ u3 ˆ

qh4 EIx

and by setting l hˆ ; 4

Q

ql 4 256EIx

we have, taking into account equation (4.110), 5 u1 ÿ 4 u2 ‡ u3 ˆ Q Similarly, Node 2 u2 ÿ 4 u3 ˆ Q ÿ4 u1 ‡ 6 Node 3 u2 ‡ 5 u3 ˆ Q u1 ÿ 4 The equations describing the relationships between the de¯ections at the nodes and the applied loading can be put into matrix form, as ‰AŠ‰ un Š ˆ ‰Qn Š where

2

5 ‰AŠ ˆ 4 ÿ4 1

ÿ4 6 ÿ4

…4:111† 2 3 3 2 3 u1 1 1 ÿ4 5 ‰un Š ˆ 4 u2 5 ‰Qn Š ˆ Q4 1 5 u3 1 5

…4:112†

The values of the de¯ections at the nodes may be obtained from equation (4.111) by rearranging the equation as ‰ un Š ˆ ‰AŠÿ1 ‰Qn Š which gives 2 : 3 2 3 25 u1 4 u2 5 ˆ Q4 3:5 5 2:5 u3 and the de¯ection at the mid-span is u2 ˆ

3:5 ql 4 ql 4 ˆ : 256 EIx 73 1EIx

This is within about 5% of the analytical result shown above.

Theory of elastic deformation of beams

125

Five internal nodes Ð with assumed symmetry

The accuracy of the numerical solution can be improved, if required, by increasing the number of nodes included in the model. However, this will increase the numerical e€ort and to maintain a minimum e€ort it is worthwhile to use any form of geometric symmetry that will occur in the beam. This is illustrated in this example by increasing the number of internal nodes to ®ve and by incorporating symmetry of deformations at the mid-span. The arrangement of the ®nite di€erence model is shown in Fig. 4.21. The symmetry is readily incorporated by specifying u4 ˆ u2 ;

u5 ˆ u1

Now l hˆ ; 6

Qˆ

ql 4 1296EIx

Thus, dealing with each of the nodes in turn, a matrix equation, similar to equation (4.111), can be obtained. In this case 2 3 5 ÿ4 1 6 7 ‰AŠ ˆ 4 ÿ4 7 ÿ4 5 2 ÿ8 6 and the solution is 2 : 3 8 75 6 : 7 ‰ un Š ˆ Q4 15 0 5 17:25 The de¯ection at mid-span is u3 ˆ

17:25 ql 4 ql 4 ˆ 1296 EIx 75:1EIx

Symmetry

–1

Fig. 4.21

0

1

2

3

4

5

6

7

126

Energy methods in structural mechanics

which is within 2% of the exact analytical value given above. Again, if required, an even greater degree of accuracy can be obtained by increasing the number of nodes in the model.

4.8

Applied displacements and constraint reactions

Another class of actions capable of producing deformations on elastic structures is now introduced. However, these actions cannot be classi®ed as forces or thermal loads but they can be classi®ed as applied displacements. In order to ®x our ideas, let us consider the beam with built-in ends shown in Fig. 4.22. We assume that in its initial con®guration the beam is undeformed and we attribute the zero value to the energy of this status. Let us now suppose that as part of the loading procedure the built-in end B is forced to undergo a vertical displacement u…B† and a rotation  …B†, as shown in Fig. 4.23. Even in the absence of applied loads, the beam is thus forced to assume a deformed con®guration. The deformations can be examined, as described above, by means of Euler's equation of beam deformation (4.43) in its homogeneous form, that is EIx u0000 y ˆ 0

…4:113†

provided the following boundary conditions are imposed uy …0† ˆ 0

and u0y …l† ˆ 0

at A

…4:114†

uy …l† ˆ u and ÿu0y …l† ˆ  at B

…4:115†

B

A

Fig. 4.22

B

uy

A

φ

Fig. 4.23

Theory of elastic deformation of beams

127

After some simple integration and evaluation of the constants, we get      2  3 u 3 u 2 ‡ uy …z† ˆ ÿ 2 ‡ 3 z ‡ z …4:116† l l2 l l On account of this deformation the beam will possess a certain level of strain energy U. This can be calculated using the expression in equation (4.33) …l 1 Uˆ EIx u00y 2 dz 2 0

which, by means of equation (4.116) yields  2  u 12 1 4 12 u2 ‡ 2 ‡ 3 U ˆ EIx 2 l l l

…4:117†

However, one could object that the expression of the internal energy (equation (4.117)) should actually be zero, as the principle of conservation of energy requires that
U ˆ U ˆ W

…4:118†

and in our case there are no externally applied forces, and so it would seem W ˆ0

…4:119†

But obviously this is not really the actual situation. In fact, a displacement applied to its constraints requires the structure to assume a certain con®guration. Remembering what has been presented in chapter 2, section 2.3, we can have two cases, namely (a) (b)

the system is statically determinate, i.e. the constraints are strictly only sucient to prevent any rigid body displacements (there are no redundant constraints) the system is statically indeterminate, i.e. there are more constraints than degrees of freedom (some constraints are redundant).

In the ®rst case the structure can move bodily without deforming, in accordance to the kinematics of rigid bodies studied in chapter 2. This is the case shown in Fig. 4.24. Of course no strains are induced, as the reader can easily verify by integrating Euler's equation of beam deformation with the following

128

Energy methods in structural mechanics

uy

Fig. 4.24

boundary conditions uy …0† ˆ 0

and EIx u00y ˆ 0

uy …l† ˆ u and

EIx u00y …l†

ˆ0

at A at B

…4:120† …4:121†

We know as well that, in this case, in the absence of applied forces the reactions of constraints are zero (section 2.3) and therefore
U is e€ectively zero. The case of the structure in Figs 4.22 and 4.23 is, on the contrary, that of a statically indeterminate system. In this case some of the constraints are redundant and a rigid body displacement ®eld cannot take place. An imposed displacement and/or rotation of the built-in end B can only occur by means of a deformation of the structure. Moreover, we discover that at B we have non-zero values of the bending moment and of the shear force, which represent the reactions of the constraint. Thus, we have   6 u 4  M…B† ˆ M…l† ˆ ÿEIx u00y …l† ˆ EIx 2 ‡ 6ˆ 0 …4:122† l l   12 u 6  R…B† ˆ S…l† ˆ ÿEIx u000 6ˆ 0 …4:123† …l† ˆ EI ‡ x y l3 l2 Notice that also at A, i.e. for z ˆ 0, we also have non-zero values for the reactions at the ®xed ends. These are   6 u 2 00  M…A† ˆ ÿM…0† ˆ EIx uy …0† ˆ EIx 2 ‡ 6ˆ 0 …4:124† l l   12 u 6 000  R…A† ˆ ÿS…0† ˆ EIx uy …0† ˆ EIx ÿ 3 ÿ 2 6ˆ 0 …4:125† l l We will now see that the de¯ection of the beam can be related to the reactions arising at the constraints. First of all it must be pointed out that the deformed con®guration of the beam is the same as would have obtained by considering a cantilever ®xed at A and subject to a force R  at the end B, where the values of the moment and force and a couple M are given by equations (4.122) and (4.123). This would have clearly meant imposing the following two mechanical conditions at B instead

Theory of elastic deformation of beams

129

R

M

Fig. 4.25

of the two geometrical conditions (equation (4.115))   6 u 4 00  ÿEIx uy …l† ˆ M…B† ˆ EIx 2 ‡ l l   12 u 6  ÿEIx u000 …l† ˆ R…B† ˆ EI ‡ x y l3 l2

…4:126† …4:127†

and the result would have been seen to be identical. Because of this result, let us evaluate the strain energy for the beam shown in Fig. 4.25, which is therefore perfectly equivalent to the beam in Fig. 4.23. We have, by Clapeyron's theorem 1 1  R u‡ M  2 2     1 12 u 6 1 6 u 4  ˆ EIx  ‡ 2 u ‡ EIx 2 ‡ 2 2 l l3 l l

Wˆ

and after a little algebra  2  u 12 1 4 12 u2 U ˆ EIx ‡ 2 ‡ 3 2 l l l

…4:128†

…4:129†

This expression turns out to be identical to that in equation (4.117). It is now clear that the strain energy stored in the statically indeterminate beam can be related to the work done by the reactions of the constraints along the deformation process.

In other words, the strain energy stored in the beam results from the work that we must perform against the sti€ness of the beam in order to apply a given displacement at the built-end B. Therefore U is not zero, and the principle of conservation of energy is not violated as W is also not zero. It should be noted that the magnitudes of the reactions, equations (4.122) and (4.123), are linear functions of the bending sti€ness EIx . As a consequence, very rigid structures induce very high values of the bending moments and shear forces, even for quite small displacements of the constraints. At the end of this section we shall derive

130

Energy methods in structural mechanics B

uy

A

φ

Fig. 4.26

the magnitudes of the restraint reactions when a vertical displacement  are applied at the end A of the beam shown in u…A† and a rotation …A† Fig. 4.26. This is a situation really complementary to that shown in Fig. 4.23 and has been dealt with in the preceding presentation. However trivial its solution actually is, it is of considerable importance for the matrix treatment of frames which we will develop in chapter 9 and is it therefore valuable to that development to present here the above mentioned results in some detail. We have from the above analysis   6u 4  M…A† ˆ ÿM…0† ˆ EIx ÿ 2 ‡ …4:130† l l   12u 6  R…A† ˆ ÿS…0† ˆ EIx ÿ 2 …4:131† l3 l   6u 2  …4:132† M…B† ˆ M…l† ˆ EIx ÿ 2 ‡ l l   12u 6  R…B† ˆ S…l† ˆ EIx ÿ 3 ‡ 2 l l

4.9

E€ects of temperature changes

…4:133†

In this, the ®nal section of this chapter, we wish to deal brie¯y with the e€ects of temperature changes on bars and beams. In section 1.2 we presented the expression of the energy of a bar which we assume is in equilibrium under the actions of equal and opposite axial forces N and ÿN and a temperature increase
T ˆ T ÿ T0 . This is, see equation

Theory of elastic deformation of beams

131

(1.13), E…T; l 0 † ˆ C…T ÿ T0 † ‡ 12 k
l 2 ÿ kl
l…T ÿ T0 † We attribute the zero value of energy to the state de®ned by T ˆ T0 and l 0 ˆ l. The relationship between the elongation
l ˆ l 0 ÿ l and the applied forces and variation of temperature is given by equation (1.12)
l ˆ

F ‡ l…T ÿ T0 † k

where k is the axial sti€ness of the bar and  is the coecient of linear thermal expansion, which for small changes in temperature can be considered constants for the material and the element. We can now give a more precise meaning to the axial sti€ness k of the bar and, remembering equation (4.5), write 1 EA 2 EA
l ÿ l
l…T ÿ T0 † …4:134† 2 l l Elucidating equation (4.134) it is noted that the ®rst term on the right hand side of this equation, that is E…T; l 0 † ˆ C…T ÿ T0 † ‡

C…T ÿ T0 †

…4:135†

represents the heat stored in the bar and is directly related to the variation of temperature
T ˆ T ÿ T0 . The second and third terms, i.e. 1 EA 2 EA
l ÿ l
l…T ÿ T0 † …4:136† 2 l l represent the di€erence between the equation (4.8) for the elastic energy U which would have been stored in the bar if the elongation
l ˆ l 0 ÿ l had been a consequence of the application of only the axial forces N and ÿN and a correction term. This term can be rearranged as EA l
l…T ÿ T0 † ˆ Nl…T ÿ T0 † …4:137† l and is seen to represent the work done by the axial load N as an e€ect of that part of the elongation due to the thermal expansion only. This work is not actually stored as elastic energy of the bar and is subtracted from equation (4.8). The system is de®ned to be in equilibrium and therefore the kinetic energy K has zero value. Therefore, from a purely kinematic point of view the e€ect of the temperature increase
T ˆ T ÿ T0 can be reduced to the elongation l…T ÿ T0 †

…4:138†

132

Energy methods in structural mechanics ΔT = T – T0

l

Fig. 4.27

As seen in section 3.2, an elongation which is related to equation (4.138) constitutes a strain ®eld which does not depend on the applied loads. Such strains are generally called inelastic strains. In the case of thermal strains we can easily characterise them using the assumption that the material is isotropic and has the same thermal expansion properties in all directions. Thus we can say that the meaning of the coecient of linear thermal expansion  is that for a unit rise in temperature the normal strain produced in all directions is equal to , that is "xx ˆ "yy ˆ "zz ˆ 
T

…4:139†

If the body is free to expand this state of strain is not associated with elastic energy and the only energy stored in it is represented by the heat expressed in equation (4.135). The unit change in volume "v , is given by the expression "v ˆ "xx ‡ "yy ‡ "zz ˆ 3
T

…4:140†

However, we may now also consider that the body is not free to expand. For example, let us refer to the bar shown in Fig. 4.27. This bar is clamped at each end and is subject to a temperature increase
T ˆ T ÿ T0 . The end constraints prevent any changes in length and thus react to produce a state of strain which, superposed on the thermal strain, makes the total strain equal to zero. From a mathematical viewpoint we have N lˆ0 …4:141† EA where N is the axial force induced in the bar due to the constraint reactions, as shown in Fig. 4.28. The value of the axial force is seen to be
l ˆ l
T ‡

N ˆ ÿ
T EA

…4:142†

that is a compressive force which produces an axial deformation of the same value but opposite in sign to that induced by the thermal expansion. The energy of the bar E…T† is E…T† ˆ C…T ÿ T0 † ‡

1 EA ‰…T ÿ T0 †l Š2 2 l

…4:143†

Theory of elastic deformation of beams N

133

N

l l + αl ΔT

Fig. 4.28

and is evidently a function only of the temperature T (the length l is ®xed). The magnitude of the reactive forces that are activated by temperature changes may be very high; this is exempli®ed by a steel bar 0.5 m long, whose sectional area is 2500 mm2 . If we de®ne this bar to be rigidly ®xed between two walls, the force N given by equation (4.142) is equal to 244 kN (for steel bodies between 0 8C and 100 8C the coecient  can be taken equal to 11:6  10ÿ6 8Cÿ1 and the modulus can taken as 2:1  105 N/mm2 ). For this reason the forces induced by thermal strains must always be carefully evaluated in all those structures that are exposed to direct sunlight, like bridges and tall buildings. As we stated in chapter 1, our study is limited purely to the mechanical behaviour of bodies. Therefore we do not concern ourselves with the heat exchanged between the environment and the body under analysis, provided all the transformations take place very slowly and the body assumes at any stage the same temperature as the environment. This condition is satis®ed for practically all the structures in normal engineering practice and we can limit ourselves to the kinematic e€ects of temperature changes, i.e. the expansion and thermal strains introduced above. In this manner we can simply determine the deformed state of a simple supported beam which is exposed to direct sun rays, such as is shown in Fig. 4.29. For the sake of simplicity, consider a rectangular cross section and suppose that the temperature of the upper face of the beam, which is directly exposed to the sun, is T2 , while the temperature of the lower T2 T1 A

B l

Fig. 4.29

134

Energy methods in structural mechanics T2

x H

T1

y

Fig. 4.30

face is T1 , such that T2 > T1 > T0

…4:144†

T0 is, as usual, the reference temperature. We suppose also that the temperature gradient is constant through the depth of the cross section, that is the temperature varies linearly between the ®bres at y ˆ ÿH=2 and the ®bres at y ˆ H=2, as shown in Fig. 4.30. It is convenient to think of such a state of thermal strain as being equivalent to the sum of a uniform temperature variation, that is T ‡ T2 T ˆ 1 …4:145† 2  i.e. and a linear temperature variation between the values T and ÿT, …T ‡ T2 † T2 ÿ T1 ˆ T ˆ T2 ÿ 1 …4:146† 2 2 …T ‡ T2 † T1 ÿ T2 ÿT ˆ T1 ÿ 1 …4:147† ˆ 2 2 at ÿH=2 and H=2, respectively. The situation is pictured in Fig. 4.31. T2

T

=

T1

Fig. 4.31

+T

+

–T

T=

T1 + T2 2

T=

T2 – T1 2

–T =

T1 – T2 2

Theory of elastic deformation of beams

135

T2

T1 A

B

Fig. 4.32

 we ®nd also With reference to the uniform temperature variation T, that the resulting thermal strains are uniform through the depth of the beam and we have "z ˆ T …4:148† On the other hand, with reference to the linear temperature variation  we ®nd that the thermal strains also vary linearly between T and ÿT, through the depth of the beam and we have 2y "z ˆ ÿT …4:149† H It is clear that this expression corresponds to a curvature of the longitudinal axis of the beam given by 1 1 ˆ ÿT …4:150† r H which is constant along the longitudinal axis of the beam. Therefore the simply supported beam being considered here deforms in response to the temperature variation as shown in Fig. 4.32, and the displacements at A and B are …l

 uz …B† ˆ T dz ˆ Tl

uz …A† ˆ 0

…4:151†

0 l=2 …

…A† ˆ 0 l=2 …

…B† ˆ 0

1 l dz ˆ T H 2H

…4:152†

1 l ÿT dz ˆ ÿT H 2H

…4:153†

T

It is worth noting that in the case where the same variation of temperature is applied to a ®xed-ended beam (Fig. 4.33) the displacements described in equations (4.151)±(4.153) are not allowed any more.

136

Energy methods in structural mechanics T2 A

B T1 l

Fig. 4.33

In order to evaluate the reactions that arise at the ends, we can once again imagine the thermal strains as given by the sum of the contributions in equations (4.149) and (4.150). With reference to the uniform axial strain ®eld, equation (4.149), we have already found that the constraint reactions are represented by the forces Ry …A† ˆ 
T EA Ry …B† ˆ ÿ
T EA …4:154† With reference to the linear strain ®eld, equation (4.149), by recalling equation (4.33) we can write the elastic energy of the beam as   …l 1 1 1 2  ‡ T Uˆ EIx dz 2 r H 0

  …l 1 1 2 ˆ EIx ÿu00y ‡ T dz 2 H

…4:155†

0

where the term between brackets is the elastic curvature, i.e. the algebraic di€erence between the total curvature and the thermal one. The condition of equilibrium is   1 00  u00y dz ˆ 0 U ˆ EIx ÿuy ‡ T H …l

…4:156†

0

and, following the same procedure as presented in section 4.4, we are led to the conclusion that this condition is equivalent to the fourth order di€erential equation   1 00 EIx ÿu00y ‡ T ˆ0 …4:157† H The general solution of this equation is in the form   1 2 z ‡ Cz ‡ D uy …z† ˆ Az3 ‡ B ÿ T H

…4:158†

Theory of elastic deformation of beams

137

and the ful®lment of the boundary conditions u…z† ˆ 0

for

zˆ0

…z† ˆ ÿu0y …z† ˆ 0

and z ˆ l

for

zˆ0

…4:159†

and z ˆ l

…4:160†

produces uy …z† ˆ 0

…4:161†

This means that apparently the beam does not undergo any deformation. In reality the elastic strains compensate exactly the thermal ones at any point of the beam. The bending moment resulting from equation (4.157) is seen to be constant and has the same value of the reactive moments at the constraints A and B, that is 1 M…A† ˆ ÿM…0† ˆ EIx u00y …0† ˆ EIx T H 1 M…B† ˆ M…l† ˆ ÿEIx u00y …0† ˆ ÿEIx T H

…4:162† …4:163†

Suggested exercise problems 1.

A section through the constant thickness wall of a dam which is composed primarily of reinforced concrete is shown in Fig. 4.34. The wall is subjected to the pressure of the water. Ignoring the e€ects of the end conditions of the dam we can make an initial estimate of the de¯ection of the top of the dam wall using a unit length of the wall. Calculate this de¯ection
assuming the equivalent uniform modulus of the reinforced concrete as Ec ˆ 40 kN/mm2 . [Ans:
ˆ 11:4 mm]

3m Δ

40 m

Fig. 4.34

35 m

138

Energy methods in structural mechanics 400 mm

q 20 mm C

A

20 mm 1·5 m

B

D

20 mm t

D 10 m

7·5 m

30 m 15 m (a)

(b)

Fig. 4.35

2.

3.

A bridge consists of two longitudinal girders joined by a transverse beam carrying the load. The loading on the girders can be represented by the uniform distributed load shown in Fig. 4.35(a). The girders have the cross-section geometry shown in Fig. 4.35(b) and are made from steel with a modulus of 205 kN/ mm2 . The bottom ¯anges of the girders are reinforced by increasing the thickness from 20 mm to …20 ‡ t† mm so that the longitudinal strain in the ¯ange at the mid-length, position C in Fig. 4.35(a), is equal to the strain at the quarter length, position D in Fig. 4.35(a). Calculate the required thickness t of the reinforcement and also the reduction in the de¯ection that will result from adding the reinforcing plate. [Ans: t ˆ 23:9 mm, the mid-span de¯ection of the reinforced beam is 76% of the beam without reinforcement] The main beam of a bridge is supported at mid-span by a column sitting on an elastic foundation, as shown in Fig. 4.36(a). This is equivalent to the idealisation shown in Fig. 4.36(b) where the column is represented by a spring with a sti€ness of k [N/mm]. q

q

EI

EI

(a)

Fig. 4.36

k

(b)

Theory of elastic deformation of beams

139

Direction of fluid flow 30 m

30 m

30 m

A

B

Fig. 4.37

4.

5.

Develop an expression for the vertical de¯ection of the mid-span of the bridge deck, relating the de¯ection
, the ¯exural sti€ness EI, the length l, and the uniform distributed load q. [Ans:
ˆ 5ql 4 =…384EI ‡ 8kl 3 †] A pipe carrying ¯uid with a unit weight, f ˆ 11 000 N/m3 , has an outside diameter of 1 m and a wall thickness of 15 mm. The pipe is supported at 30 m intervals as shown in Fig. 4.37. The ¯uid ¯ows through the pipe such that it half ®lls the pipe. The ¯uid has a temperature of 80 8C and it may be assumed that the conduction of the heat from the ¯uid to the atmosphere through the pipe is such that the temperature varies linearly from 80 8C at the bottom of the pipe to ambient temperature of 20 8C at the top, at end A. Calculate the de¯ection of the pipe at mid-span when it is ®rst installed empty, the unit weight of steel is s ˆ 77 000 N/m3 . Calculate also the change of de¯ection as the ¯uid ¯ows through the pipe and the temperature reached at the equilibrium conditions speci®ed above. [Ans:
1 ˆ 6:3 mm,
2 ˆ 13 mm] A mechanical component comprises two straight bars joined rigidly at B. The bar BC is subjected to a uniformly distributed load, q ˆ 2000 N/m, as shown in Fig. 4.38. Both bars are made A

50 mm diameter 2m q B

C 3m

Fig. 4.38

140

Energy methods in structural mechanics

from 50 mm diameter round steel bar and are connected to rigid foundations at A, B and C. The modulus for steel is 205 kN/ mm2 . Develop a solution to enable the evaluation of the rotation, c of the joint at C. Evaluate also the maximum strain "m experienced by the elements in the structure. [Ans: c ˆ 0:029 rad, "m ˆ 1:91  10ÿ3 ]

5.

General principles in the analysis of linear elastic structures

Introduction

This text is primarily concerned with the presentation of methods of analysis of structures and structural components, such as beams, that have a linear relationship between the levels of the applied loads and the e€ects of these loads with regards to the strains and deformations of the structure and all its components. It is common experience to expect that if a beam is loaded by a force normal to its axis the deformation of the beam will be in the direction of the load. If we double the magnitude of the force we would expect that the level of the deformation would increase by a factor of two. Essentially the common experience forms the basis for linear behaviour of structures. We should be aware, however, that structures do not always obey this linear correspondence between load and e€ect. In chapter 12 we consider the situation of beam and frame buckling where the application of a load aligned with the axis of the beam can result in deformations normal to the axis. The level of the deformations is in such a case nonlinearly related to the level of the load. In other words, if we increase the level of the load the magnitude of the deformations are increased disproportionally. Another source of non-linear relation between loading and consequent deformation would result from non-linear material properties, either a non-linear stress±strain path or yielding of the material at a constant stress. This source of non-linearity is not included in the presentations of this text. The linear behaviour of structures and its analysis constitutes a central feature of structural engineering and this text considers a wide range of types of structures and components that exhibit this form of behaviour. There are some general principles that form the fundamental basis of this linear behaviour and of the analysis methods developed to explain and predict the response of structure to applied loading. This chapter presents these principles of superposition and

142

Energy methods in structural mechanics

reciprocity, in a general manner. Although the principles are illustrated with regard to simple beams, it should be remembered that the principles are completely general and apply to all forms of structures, such as trusses, frames and plates.

5.1

The principle of superposition of the e€ects in the linear theory of elasticity (existence and uniqueness of the solution)

So far we have introduced the ®rst examples of elastic structures, i.e. bars and beams, and dealt with their treatment using energy principles. It is now timely to give some attention to the framework in which we are operating, i.e. the linear theory of elasticity. In order to make completely clear the foundations of this theory, let us recall what we have already determined in the previous chapter, namely .

.

equations (4.7) and (4.19) describing the elastic behaviour of Bernoulli's model of beam deformations are both linear. This means that e€ects resulting from axial forces or bending moments are proportional to the intensity of these actions on the element operating in the theory of in®nitesimal (very small) displacements allows the displacement and strain ®elds corresponding to di€erent load conditions simply to be added.

These are the two pillars on which the linear theory of elasticity is founded, as indicated in Fig. 5.1. The ®rst of the points above, which involves the linearity of the constitutive relationships of the model, is clearly a physical assumption,

Fig. 5.1

Theory of infinitesimal displacements

Linearised constitutive laws

Linear theory of elasticity

General principles in the analysis of linear elastic structures

143

P q A B

l/2

l/2

Fig. 5.2

while the second, related to the linearisation of the displacement and strain ®elds, is essentially a mathematical axiom. As may be inferred from its description, in this theory all the equations are linear, with comprehensible and acceptable simpli®cations in the treatment of a vast class of structural problems. First of all it is clear that in this framework the principle of superposition of e€ects holds true. In other words we are allowed to add and subtract the e€ects due to di€erent separate loading conditions. In its simplest form this means that if we, say, were to double the magnitude of the external forces, i.e. by adding to a certain set of loading a second set of loading equal to the ®rst one, we would expect the consequent deformations also to increase by a factor of two. As a further example, let us consider the beam shown in Fig. 5.2, loaded by a uniformly distributed load q and by a concentrated load P at the middle of the span. With respect to any e€ect, we can analyse the beam model shown in Fig. 5.3 and add the resulting solution to the one already obtained for the beam shown in Fig. 4.15 and analysed in chapter 4. The ®eld of lateral displacements, i.e. normal to the longitudinal axis of the beam, for the beam in Fig. 5.3, subject to the uniformly distributed load q is given by the solution of Euler's equation of beam deformation, (4.43), under the following boundary conditions uy …0† ˆ 0

and …0† ˆ 0 at A

uy …l† ˆ 0

and M…l† ˆ 0

…5:1†

at B q

A

Fig. 5.3

B

144

Energy methods in structural mechanics

From equation (4.44) we have uy …z† ˆ

qz2 …2z2 ÿ 5 l z ‡ 3l 2 † 48EIx

…5:2†

The solution for the beam in Fig. 5.2, subject to both the evenly distributed load q and the concentrated load P at the mid-point of the span, is thus given by the simple addition of equation (5.2) to equations (4.80) and (4.81), i.e. u1y …z† ˆ

z2 ‰ÿ9Pl ‡ 6ql 2 ‡ …11P ÿ 10ql†z ‡ 4qz2 Š 96EIx

…5:3†

for z 2 ‰0; l=2Š and u2y …z† ˆ

…l ÿ z† ‰2Pl 2 ÿ 10Plz ‡ …5P ‡ 6ql†z2 ÿ 4qz3 Š 96EIx

…5:4†

for z 2 ‰l=2; lŠ. Of course we would have obtained the same expressions by writing the condition of equilibrium for the beam subjected to both these loads simultaneously and developing all the calculations accordingly. However, as will become clearer in the following, the principle of superposition of the e€ects is a very powerful tool not only from a practical, but also from a theoretical point of view. For example, let us ®x our attention on two apparently very simple questions . .

Does the general condition of equilibrium (1.28) always lead to a solution of the problem? Is this hypothetical solution unique?

In the linear theory of elasticity the answer to both these questions is yes. With regards to the ®rst question, the demonstration of such a statement is quite a complicated matter from a general point of view and it has been fully established only in relatively recent times. Nevertheless, it is very easy to reach the correct conclusion in many speci®c problems, given the linearity of the equations that describe the models. This has been the case, for example, of Euler's equation of beam deformation, (4.43). With regards to the second question, the answer was provided by Kirchho€ 1 about the middle of the 19th century and its demonstration 1

Kirchho€, Gustav Robert (KoÈnigsberg, 1824 ± Berlin, 1887), German mathematician and physicist.

General principles in the analysis of linear elastic structures

145

Pi

V u=0

Fig. 5.4

constitutes an instructive application of the principle of superposition of e€ects. In order to demonstrate the assertion of uniqueness, let us start our reasoning from the most general possible standpoint. Let V be a generic linearly elastic body with certain geometrical constraints on the boundary and subject to a generic system of forces P i , as shown in Fig. 5.4. Let us initially assume the existence of two distinct solutions of this problem, say A and B. Also we assume that these solutions are de®ned by the ®elds of displacements uA …x; y; z† and uB …x; y; z†. As we know, the variation of the strain energy of the body is dependent on its state of deformation only, that is …
U ˆ f …"ij † dV …5:5† V

Moreover, in chapter 3 we noted that in the linear theory of elasticity this strain energy is a quadratic function of the strain components "ij , see equation (3.107), …  E …1 ÿ †…"2xx ‡ "2yy ‡ "2zz † Uˆ 2…1 ‡ †…1 ÿ 2† V

‡ 2…"xx "yy ‡ "xx "zz ‡ "yy "zz †  2…1 ÿ 2† 2 2 2 ‡ …"xy ‡ "xz ‡ "yz † dV …1 ‡ †

…5:6†

Since these components are functions of the displacement ®eld u…x; y; z†, we can derive a di€erent strain ®eld for each of the two

146

Energy methods in structural mechanics

hypothetical solutions of the elastic problem at hand quA x qx


 A  1 quy quA ‡ x ˆ 2 qx qy


quB ˆ x qx


 B  1 quy quBx ‡ ˆ 2 qx qy




"A xx ˆ "A xy "Bxx "Bxy

…5:7†

…5:8†

Finally, as the principle of superposition of e€ects holds true, we can write the principle of conservation of energy with reference to the difference between the solutions A and B of the problem. This di€erence is characterised by the strain ®eld B "ij ˆ "A ij ÿ "ij

…5:9†

and by the total zero sum of external actions, that is Pi ÿ Pi ˆ 0

…5:10†

The principle of conservation of energy, equation (1.9), can be therefore written as
E ˆ
U ˆ W ˆ 0

…5:11†

and requires
U ˆ 0, because of the zero value of the external work, W ˆ 0. But, on account of equation (5.6), we have … E B 2 A B 2 Uˆ ‰…1 ÿ †‰…"A xx ÿ "xx † ‡ …"yy ÿ "yy † 2…1 ‡ †…1 ÿ 2† V

‡

…"A zz

ÿ

"Bzz †2 Š

B A B ‡ 2  ‰…"A xx ÿ "xx † …"yy ÿ "yy †

B A B A B A B ‡ …"A xx ÿ "xx †…"zz ÿ "zz † ‡ …"yy ÿ "yy † …"zz ÿ "zz †Š

‡

 2…1 ÿ 2† A B 2 A B 2 ‰…"xy ÿ "Bxy †2 ‡ …"A ÿ " † ‡ …" ÿ " † Š dV xz xz yz yz …1 ‡ † …5:12†

General principles in the analysis of linear elastic structures

147

B where
U is zero only if "A ij ˆ "ij . With reference to the relationships in equations (5.7) and (5.8), this means that the displacement ®elds uA …x; y; z† and uB …x; y; z†, which represent two di€erent solutions of the problem of the elastic equilibrium, must di€er at most by a constant, corresponding to a rigid displacement of the whole body. Therefore, if, as is the usual condition, the geometrical constraints are such as to prevent any rigid displacements, the two solutions A and B must coincide and the uniqueness of the solution is demonstrated.

5.2

Reciprocal theorems in the linear theory of elasticity

Apart from the self-consistency of the theory, from a technical point of view the success of the mathematical theory of linear elasticity lies undoubtedly on the extensive availability of powerful and practical tools for the analysis of engineering structures. Among these tools a very important place is occupied by the so-called reciprocal theorems. In order to introduce these theorems, let us consider the simple supported beam shown in Fig. 5.5, which we consider to be loaded in sequence ®rst by the group of forces P1 and then by the group of forces P2 . The work done by all these forces is  ˆ W11 ‡ W22 ‡ W12 W …5:13† where W11 is the component work from the group P1 , W22 is the component work from the group P2 (remember that for both these components of work Clapeyron's theorem holds true) and W12 is the so-called displacement work of P1 , i.e. the work of P1 due to the displacement P1

A

B

P1

A

Fig. 5.5

P2

B

148

Energy methods in structural mechanics P2

A

B

P2

P1

A

B

Fig. 5.6

®eld produced by P2 . Sometimes the component work, i.e. the work done by a force along its own deformation, is also called eigenwork (the word eigen is a German word and means `own'). If we reverse the order of loading by ®rst applying P2 and then P1 , we have, as shown in Fig. 5.6,  ˆW ‡W ‡W W …5:14† 22

11

21

where W21 is the displacement work of P2 . Of course, since the principle of superposition of the e€ects holds true, the ®nal con®guration of the beam will be identical in both cases and we have from the principle of conservation of energy   ˆW
U ˆ W …5:15† By comparison of equations (5.13) and (5.14) we are led to the conclusion that W12 ˆ W21

…5:16†

This is called Betti's theorem, which can be stated as follows: Given two forces or two groups of forces acting on a linearly elastic structure, the work done by the ®rst force, or group of forces, on account of the displacements due to the second force is equal to the work done by the second force, or group of forces, on account of the displacements due to the ®rst force.

A second theorem, due to Maxwell, can be straightforwardly derived by Betti's theorem. It states that for the same load applied at di€erent points A and B we must have the condition uAB ˆ uBA

…5:17†

General principles in the analysis of linear elastic structures

149

where uAB is the generalised displacement 2 at A due to the force at B and uBA is the generalised displacement at B due to the same force at A. In fact, if P is a generic load applied both at A and B, we have from equation (5.16) WAB ˆ WBA

)

P uAB ˆ P uBA

…5:18†

and we immediately get the relationship (equation (5.17)).

2

Here as generalised displacements we intend the components of displacement enabling the applied forces to perform work. In this sense we can introduce generalised forces as well. Thus, generalised forces and corresponding generalised displacements (or vice versa) are quantities characterised by the fact that the sum of their products always represents a true work.

6.

Total potential energy

Introduction

Up to this point we have concentrated on the development of methods that enable the solution for systems of rigid bodies and for beam deformations in a closed form. The latter is based on Euler's method to derive the di€erential equation of equilibrium. In some circumstances of restricted forms of loading and boundary conditions the di€erential equation may admit a solution in a closed form. In general, however, it has been found dicult to obtain an exact solution for several of the practical ranges of beam geometry and loading types. For this reason we often have to turn to approximate methods of analysis to provide information for practical designs in structural engineering. The foundation of the energy principles that underlie many of the approximate methods developed in later chapters is introduced here.

6.1

The principle of the stationary value of total potential energy

At this point in the presentation we can introduce what is usually called the principle of stationary value of total potential energy. From a theoretical point of view, the reader will ®nd this principle to be nothing more than a direct and formal derivation of what has previously been presented in chapter 1. Nonetheless, this principle has a broad ®eld of applications in mechanics and is central to methods to obtain solutions to several problems in structural mechanics. Particularly, the principle provides the basis for establishing approximate solutions by means of trial functions, that, in turn, lead directly to the ®nite element method of analysis. With reference to a closed mechanical system in equilibrium and in the absence of heat exchange, we have from the principle of conservation of energy, equation (1.21), dE ˆ dW

152

Energy methods in structural mechanics

and for a virtual displacement ®eld, i.e. a ®eld of displacements which are in®nitesimal and compatible with the constraints on the system, but are imaginary and do not actually occur E ˆ W

…6:1†

From this equation we derived, see equation (1.28) U ˆ W

,

K ˆ 0

as a condition as well as a test of equilibrium, where, of course, K is the kinetic energy of the system. We are now going to show that it is straightforward to build a special functional whose variation within the ®eld of kinematically admissible functions coincides with equation (1.28) and is therefore stationary when the system is in equilibrium. In other words, if we write ˆUÿW

…6:2†

we have  ˆ 0

,

U ÿ W ˆ 0

…6:3†

Therefore, if we call the functional  the total potential energy of the system, we can enunciate the principle of stationary value of total potential energy Given a closed mechanical system, isolated from its environment such that no exchange of heat energy occurs across the system boundary, the stationary values of the total potential energy represent con®gurations of equilibrium for the system itself.

This holds true from a mathematical standpoint. From a physical point of view the principle of stationary value of total potential energy can be clari®ed with reference to a very simple pictorial representation. Thus, let us consider Fig. 6.1, where a landscape of hills and valleys is represented. Our mechanical system is simply simulated by a ball subjected to a gravitational ®eld. Everywhere in the landscape the ball is therefore subjected to a vertical force P having the following constant value P ˆ mg

…6:4†

where m is the mass of the ball and g is the gravitational acceleration. The force P is directed downwards into the plane. If we suppose the landscape surface to be perfectly smooth and frictionless, we can assume that the reaction of the surface can only act normally to the surface itself. As no strain energy U can be stored in the system, the condition of equilibrium, equation (1.28), becomes W ˆ 0

…6:5†

Total potential energy

153

Φ (s) E D

A B C

h (s)

C

s

Fig. 6.1

where the work W is performed only by the force acting on the system, i.e. the gravitational force P. The reaction R of the supporting surface cannot perform work, as by its de®nition its direction is orthogonal to any virtual displacement, which, by de®nition, must be tangential to the landscape surface. Since the gravitational force P is directed downwards, its virtual work has value only in the case where a virtual displacement possesses a vertical component. We are consequently led to the self-evident conclusion that the ball in the landscape represented in Fig. 6.1 cannot be in equilibrium except at the top of a hill, the bottom of a valley or at a point of in¯ection. In order to express the condition of equilibrium (equation (1.28)) with reference to the total potential energy of the system, we can write …s† ˆ ÿPh…s†

…6:6†

where h(s) is the height of the landscape path, described by the parameter s. The condition of stationary value of the total potential energy is …s† ˆ ÿP h…s† ˆ ÿP  u…s† ˆ ÿW ˆ 0

…6:7†

and coincides with zero value for the virtual work performed by the gravity force P, as shown in Fig. 6.2. It is worth noting that the height function h(s) can be measured with reference to some base level C. We already know that the value C of the constant cannot a€ect the equilibrium, as we are interested only in variations. From a physical point of view this means that only relative heights are relevant and it does not matter if the whole landscape is raised of a constant quantity everywhere.

154

Energy methods in structural mechanics δu δu

δh (s)

P P Equilibrium

No equilibrium

Fig. 6.2

In equation (6.6) the total potential energy of the system coincides with the load potential ÿPh…s†, which represents the capability of the gravitational force P to perform work on account of the height h…s† of the ball. If no work tends to be performed by the gravitational force, the principle of conservation of energy assures that no kinetic energy (which by de®nition of this scheme is the only form of energy that the ball can possess on account of its mass) can be gained by the system and it is in equilibrium. The question as to whether the equilibrium state is stable or not is considered in chapter 12, it suces here to note that the conditions and examples relative to structural systems presented in this book are actually stable if not otherwise stated.

6.2

Two important remarks regarding the principle of stationary value of total potential energy

Immediately after the introduction of the principle of stationary value of total potential energy, it is valuable to review two important aspects of the principle (a)

(b)

The principle of the stationary value of total potential energy is a formal representation of a state equation (i.e. the condition of equilibrium, equation (1.28)) by means of extremum conditions on a speci®cally designed functional. It is a mathematical device and, despite the fact that we can often give a pictorial representation of the terms of this functional, there is actually no need for it to have a precise physical meaning. As a generality, not all systems can have a total potential energy functional. Such a functional is a property of socalled conservative systems, i.e. those system in which the work, equation (1.4) performed by the forces acting on the system depends only on the end points and not upon the path joining them. However, in the theory developed in this book, all the structural systems taken into consideration are conservative by de®nition. Unless otherwise explicitly stated, we consider

Total potential energy

155

only closed mechanical system in which there are no sources or sinks of energy and the processes of deformation of bodies are totally reversible without transformation of mechanical work into heat and vice versa. Under these circumstances the existence of relevant potential functionals is always assured. Of course, once again the reader has be aware that no such structures actually exist in nature; but in engineering we deal with models of real situations. The rules of behaviour for these models are set by ourselves, and from an engineering point of view we are satis®ed if they produce, as they do, results adequately accurate for our needs of design and safety in practical applications. The satisfactory nature of the models has to be obtained by comparing the predictions of structural behaviour from the models with corresponding results from tests.

6.3

Cotterill±Castigliano's ®rst theorem

This section introduces one of the most straightforward developments from the principle of stationary value of total potential energy, i.e. Cotterill±Castigliano's ®rst theorem1;2 on the derivatives of the strain energy function U. By means of this theorem it is possible to evaluate the loads acting on an elastic system in equilibrium by means of a function of its strain energy. This is the ®rst of many practically valuable methods derived from the principle of stationary value of total potential enegy and others are introduced in following sections. The development is illustrated by taking into consideration the cantilever shown in Fig. 6.3. The total potential energy of this system is given by …uy † ˆ U…uy † ÿ P


…6:8†

and is a functional of the ®eld of vertical displacements uy …z†, according to Bernoulli's model of beam. As we have seen in chapter 4, the strain energy functional U…uy † can be expressed as …l 1 EIx u002 U…uy † ˆ y dz 2

…6:9†

0

Also, the displacement ®eld uy …z† is a function of the de¯ection at the end
, and the expression for this function can be easily obtained by requiring satisfaction of the following geometrical and mechanical

1 2

Cotterill, James Henry (Norfolk, 1836 ± Parkstone, Dorset, 1922), English engineer. Castigliano, Carlo Alberto (Asti, 1847 ± Milan, 1884), Italian engineer.

156

Energy methods in structural mechanics

P z A

B Δ

φ (B)

y

l

Fig. 6.3

constraints to the general integral of Euler's equation of beam deformation uy …0† ˆ 0

and …0† ˆ ÿu0y …0† ˆ 0

M…l† ˆ ÿEIx u y00…l† ˆ 0

and uy …l† ˆ


…6:10† …6:11†

Remember that the general integral of Euler's equation of beam deformation in the absence of point loads along the beam is the third-order expression uy …z† ˆ Az3 ‡ Bz2 ‡ Cz ‡ D

…6:12†

as can be easily deduced from equation (4.44). The required expression for the displacement ®eld uy …z† as a function of the tip de¯ection
of the cantilever is therefore
3
uy …z† ˆ ÿ 3 z3 ‡ 2 z2 …6:13† 2l 2l By making use of equation (6.13), the functional for the total potential energy, equation (6.8), can be rewritten as a simple function of the de¯ection
…
† ˆ U…
† ÿ P


…6:14†

The condition of extremum condition for this function is, of course, d dU ˆ ÿPˆ0 …6:15† d
d
and allows us to state mathematically Cotterill±Castigliano's ®rst theorem: dU ˆP d


…6:16†

Total potential energy

157

Therefore, in order to evaluate the force P that produces a speci®c de¯ection
at the end of the cantilever, according to equation (6.16) all we need to do is to substitute the second derivative of equation (6.13) with respect to z in equation (6.9) and successively take the ®rst derivative of the resulting function with respect to
. Thus, we have   …l 1 3
3
2 U…
† ˆ EIx ÿ 3 z ‡ 2 dz 2 l l

…6:17†

0

and, after a little of algebra, we obtain dU 3EIx
ˆ d
l3

…6:18†

which provides the appropriate value of the force P. Of course Cotterill±Castigliano's ®rst theorem has a much broader ®eld of application than a simple cantilever subjected to one point load. We can illustrate this by considering an elastic structure loaded by n generalised forces Pk …k ˆ 1; . . . ; n†. From equation (6.2) we have ˆUÿ

n X

Pk
k

…6:19†

kˆ1

where
k are the generalised displacements corresponding to the generalised forces Pk . Provided we can express the strain energy U as a function of these generalised displacements, we have n X …
1 ; . . . ;
n † ˆ U…
1 ; . . . ;
n † ÿ Pk
k …6:20† kˆ1

and the n conditions for the total potential energy  to be stationary are qU ˆ Pk q
k

…6:21†

Cotterill±Castigliano's ®rst theorem can be consequently stated in complete generality as follows The partial derivative of the strain energy of an elastic system with respect to a generalised displacement at a certain point gives the value of the corresponding generalised force at the same point.

It is worth noticing that the Cotterill±Castigliano ®rst theorem, as well as the principle of stationary value of total potential energy, does not

158

Energy methods in structural mechanics

require any hypothesis with regard to in®nitesimal displacements or the linearity of the elastic force±displacement relationships.

6.4

Cotterill±Castigliano's second theorem in the linear theory of elasticity

Even if they have no fundamental basis in energy principles, there are several theorems in structural mechanics that can be stated in the linear theory of elasticity by means of purely mathematical relationships. This is the case of Cotterill±Castigliano's second theorem, which is formally a dual relationship of Cotterill±Castigliano's ®rst theorem. This second theorem requires that the problem at hand is studied within the framework of the theory of in®nitesimal displacements and restricted to linearly elastic force±displacement relationships. In fact, only in such circumstances can we make reference to the principle of superposition of e€ects. If we consider an elastic structure loaded by n generalised forces Pk (k ˆ 1; . . . ; n), we can divide the loading into two groups. The ®rst group is composed only of the force Pi , and the second group contains all the remaining forces, Pj … j ˆ 1; . . . ; i ÿ 1; i ‡ 1; . . . ; n†. Then the strain energy U of the structure is U ˆ WI ‡ WI;II ‡ WII ˆ 12 Pi
i;I ‡ Pi
i;II ‡ WII

…6:22†

where WI is the component of work of the force Pi , i.e. the work done by the force Pi along the line of the generalised displacement
i;I and due to the application of Pi only; WI;II is the displacement work done by the force Pi along the generalised displacement
i;II due to the second group of loading; and WII is the component of work of the second group of loading, Pj , on account of the generalised displacements due to the second group of loading itself. As all the force±displacement relationships are linear, we can write P
i;I ˆ i …6:23† K where K is a sti€ness parameter. Therefore we can express the work WI as 1 1 P WI ˆ Pi
i;I ˆ Pi i 2 2 K and we get

…6:24†

1 P U ˆ Pi i ‡ Pi
i;II ‡ WII …6:25† 2 K At this point, provided we can express the strain energy U as a function of the n generalised forces Pk …k ˆ 1; . . . ; n†, we can write down the

Total potential energy

159

partial derivative of equation (6.25) with respect to Pi as qU Pi ˆ ‡
i;II qPi K

…6:26†

Finally, taking into account equation (6.23) once again, we have qU ˆ
i;I ‡
i;II ˆ
i qPi

…6:27†

This relationship constitutes the formal expression of Cotterill± Castigliano's second theorem. This theorem can be stated as follows The partial derivative of the strain energy of an elastic system with respect to a generalised force at a certain point gives the value of the corresponding generalised displacement at the same point.

By the manner in which we have obtained this result, it is clear that it holds true only in the framework of the linear theory of elasticity and from a physical point of view it does not rely upon a real energy principle. In conclusion it is instructive to apply Cotterill±Castigliano's second theorem to the cantilever of Fig. 6.3 in order to ®nd the de¯ection
due to the load P at B. In order to do so we have to express the strain energy U of the cantilever as a function of the force P. As we know that ÿEIx u00y ˆ M…z†

…6:28†

we have from equation (6.9) 1 Uˆ 2

…l 0

M 2 …z† dz EIx

…6:29†

Making reference to Fig. 6.4, it is clear that M…z† ˆ P…l ÿ z†

…6:30†

and we can write 1 U…P† ˆ 2

…l 0

P2 …l ÿ z†2 dz EIx

…6:31†

performing the required mathematical operations, we have U…P† ˆ

P 2l 3 6EIx

…6:32†

160

Energy methods in structural mechanics

M (z) = P (l – z) l–t

P

A

B z

y l

Fig. 6.4

and from Cotterill±Castigliano's second theorem, equation (6.27)
ˆ

6.5

dU Pl 3 ˆ dP 3EIx

…6:33†

Numerical example: frame analysis by Cotterill±Castigliano's theorem

A very simple frame is shown in Fig. 6.5 to which a vertical load P is applied at mid-span of the horizontal member. We consider the elements to be axially undeformable. This is called a portal frame and the horizontal and vertical members have cross sections shown in Fig. 6.5(b), respectively. Given that the members are made from a steel with a yield strain of "0 ˆ 0:12% and a modulus of E ˆ 2:1  105 N=mm2 , we wish to calculate the maximum value of the load P that can be applied to the frame such that the maximum strain does not exceed 0:7"0 . The ®rst step in this example is to calculate the distribution of moments along each of the members for a general portal frame, shown in Fig. 6.6. The free-body diagram is shown in Fig. 6.7. The symmetry conditions for the forces at C results in Vc ˆ 0 Equilibrium of the forces at sections C and A gives P and HC ˆ ÿHA VA ˆ ÿ 2

…6:34† …6:35†

Total potential energy P

2·5 m

161

2·5 m

1 B 1

2

5m

2

C

A (a)

10 mm X

X 125 mm

10 mm 100 mm

8 mm

12 mm

X

X

150 mm

200 mm

Section 2 – 2

Section 1 – 1 (b)

Fig. 6.5

The variation of bending moments along the horizontal and vertical members, i.e. zones 1 and 2, respectively, are Zone 1 M1 …z1 † ˆ MC ‡

P z 2 1

…6:36†

162

Energy methods in structural mechanics P

l/2

l/2

EI1

l

EI2

Fig. 6.6

l/2 P/2

B

MC

C

Zone 1

HC

z1

z2

VC

Zone 2

l

A

HA MA VA

Fig. 6.7

Total potential energy

163

Zone 2 M2 …z2 † ˆ MC ‡

Pl ÿ HC z2 4

…6:36b†

where, of course, at the present stage MC and HC are unknowns. The strain energy U is 1 Uˆ 2EI1

l=2 …

1 6 4 2EI1

…l 1 ‡ M22 …z2 † dz2 2EI2 0

0

2 ˆ

M12 …z1 † dz1

l=2 …

…l

3

7 M12 …z1 †dz1 ‡ I M22 …z2 †dz2 5

0

…6:37†

0

where I  I1 =I2 . Substituting equation (6.36) into equation (6.37) and after integration the strain energy is obtained. This is  1 l l2 l3 Uˆ MC2 ‡ MC P ‡ P2 2EI1 2 8 96  3 3  l2 l3 2 2 l 2 2 l ÿ MC HC l ÿ PHC ‡ HC ‡ I MC l ‡ MC P ‡ P 2 16 4 3 The deformation conditions for the frame are, because of symmetry Zero rotation at the position C, i.e. C ˆ 0 Zero horizontal displacement at position C, i.e. wC ˆ 0 Thus from the second of Cotterill±Castigliano's theorems, C ˆ

qU ˆ0 qMC

Hence, from the strain energy    qU 1 1 2 1 Pl ‡ MC l ‡ I 2MC l ‡ Pl 2 ÿ HC l 2 ˆ ˆ0 qMC 2EI1 8 2 …6:38† For the horizontal de¯ection at C, it is wC ˆ

qU ˆ0 qHC

164

Energy methods in structural mechanics

Therefore,

   qU 1 2 1 3 3 2 ˆ I ˆ0 H l ÿ MC l ÿ Pl qHC 2EI1 3 C 4

…6:39†

From equations (6.38) and (6.39) HC ˆ

3P 8…2 ‡ I †

MC ˆ ÿ

…I ‡ 1† Pl 4…2 ‡ I †

…6:40†

The moments along the members can be calculated using equations (6.40) and (6.36), thus 1 1 Pl MB ˆ Pl MA ˆ ÿ 8…2 ‡ I† 4…2 ‡ I † The variation of the moment is linear between these values as shown in Fig. 6.8. The maximum strain due to the applied load P can occur at position B in member AB or at position C in member BC and ®rst it is necessary to evaluate the moments at these positions. The second moments of area of members AB and BC about their XX axes, i.e. I2 and I1 , respectively, are I1 ˆ

1 12

1  125  2243 ÿ 12  115  2003 ˆ 4:04  107 mm4

I2 ˆ

1 12

1  100  1703 ÿ 12  92  1503 ˆ 1:51  107 mm4 1

MB = 4 (I + 2) Pl B 1 MB = 4 (I + 2)

Pl B

l/3

A M = – 1 Pl A 8 (2 + I)

Fig. 6.8

C (I + 1)

MC = – 4 (I + 2) Pl

Total potential energy

165

hence Iˆ

I1 ˆ 2:68 I2

Taking ®rst the condition at position B for member AB, the moment is MB ˆ 0:0534Pl ˆ 267P the strain " is "ˆ

My EI

where y is the distance from the XX axis to the outer surface of the ¯ange. Thus, 267  P  85 "ˆ :  0:7  1:2  10ÿ3 2 1  105  1:51  107 The maximum allowable value of the load, P, if the strain in member AB at position B is not to exceed the speci®ed level is P ˆ 11:7  104 N Considering now the strain at position C on the member BC, 983  P  112  0:7  1:2  10ÿ3 "ˆ : 5 7 : 2 1  10  4 04  10 hence P ˆ 6:47  104 N It is evident that the lesser of the loads calculated above is that which governs the permissible loading condition. The vertical de¯ection of the member BC at the point of application of the load P, i.e. vc , can be calculated from the strain energy using again Cotterill±Castigliano's second theorem vC ˆ

qU q…P=2†

   qU 1 1 1 3 1 3 1 1 2 2 3 ˆ M l ‡ Pl ‡ I Pl ‡ MC l ÿ HC l q…P=2† 2EI1 8 C 48 8 2 4 Hence, the vertical de¯ection at C is vC ˆ

1 ‡ 2I Pl 3 2 ‡ I 96EI1

166

Energy methods in structural mechanics

Thus vC ˆ

…1 ‡ 2  2:68† 6:47  104  50003 ˆ 13:49 mm …2 ‡ 2:68† 96  2:1  105  4:04  107

The de¯ection is generally assessed on the basis of a proportion of the length of the structural member, thus vC 1 ˆ 370 l This result can be considered acceptable as codi®ed guidelines give admissible levels of de¯ection as vC 1  180 l

Suggested exercise problems 1.

An arch is shown in Fig. 6.9; it has a mean radius R, is pinned at A and C. A horizontal load P is applied at C. It is free to move horizontally at C. The cross section of the arch is rectangular as shown in Fig. 6.9. Develop expressions for the maximum moment, Mmax the vertical de¯ection at B,
vB and the horizontal de¯ection at C,
hC . [Ans: Mmax ˆ 0:311PR,
vB ˆ 1:73PR3 =Ebd 3 ,
hC ˆ 8:66PR3 =Ebd 3 ] B d

X

R X

C

A

d b Section X – X

Fig. 6.9

P

Total potential energy P

167

P

A B

θ

θ

R

C

D 2P cos θ

Fig. 6.10

2.

3.

A section of pipe is subjected to continuous line loads. The section through the pipe is shown in Fig. 6.10, where the loads P are the value of the line load per unit length of the pipe. Treating the pipe as a plane strain problem, develop an expression for the maximum moment Mmax in the ring for  ˆ 608, and the position of the maximum moment. [Ans: Mmax ˆ 0:091PR at 1208 from A] A rigid jointed frame is connected to a rigid foundation, as shown in Fig. 6.11. The members of the frame are made from tube with outside diameter D, and wall thickness t. Develop an expression for the allowable load qa such that the maximum strain in the

q D

l

t l

60°

Fig. 6.11

168

Energy methods in structural mechanics 1·0 m

1·0 m B

10 kN C

1m

A

0·5 m

Fig. 6.12

4.

frame is not to exceed the material yield strain "0 of the tube material. [Ans: qa ˆ ED2 t"0 =16:8l 2 ] A diving board consists of a beam attached to a frame, as shown in Fig. 6.12. The board is to be proof tested by applying a 10 kN load at the end C. Analysing the frame as being pin-jointed, calculate the vertical and horizontal de¯ections at C,
hC ,
vC , that would be expected to occur during the proof test. The beam has a second moment of area of 5  104 mm4 and the bars have a cross-sectional area 200 mm2 . E ˆ 200 kN/mm2 . [Ans:
hC ˆ 0:13 mm,
vC ˆ 0:87 mm]

7.

The method of trial functions

Introduction

It is evident from the development of energy principles in structural analysis and the examples presented in chapter 6 that these methods can be very powerful in developing solutions to practical situations in structural mechanics. The applicability of the methods, however, depends on having available information, such as the variation in the de¯ection or of the moment of the beam or the frame being analysed. In many practical circumstances the precise variation of these parameters is not available, but it would be very valuable if the energy methods could still be applied. In this chapter we shall show how this can be achieved. It is appreciated that the solutions obtained by the use of approximate methods are themselves approximations to the exact situation of deformation or forces in the structural model; but in any case in engineering the prediction of structural behaviour will always be approximate to some extent since knowledge on material properties, loading and construction dimensions is never exact and, consequently, precisely represented in the mathematical model. Therefore, provided we have some exact solutions available, usually related to idealised structural forms and loading, with which to calibrate and validate the approximate methods, the use of energy methods with approximate variations in de¯ections and internal forces is a very powerful approach that underlies current structural engineering practice. This practice makes considerable use of approximate energy methods, usually in the form of the ®nite element method, that are available in commercially produced computer programs. The present chapter outlines the method of trial functions, outlines how suitable trial functions may be identi®ed and will enable the reader to understand how the ®nite element method is e€ectively an automated way of incorporating special trial functions in structural analysis.

170

7.1

Energy methods in structural mechanics

The basis of the method of trial functions

From the outset we must be aware that the method of trial functions is an approximate approach that enables engineers to obtain predictions for the response of complex structures to practical loading where no exact solution to the behaviour of the structural model can be obtained. The degree of approximation and how to estimate this is considered in the following text. So far we have dealt with basic ideas in the theory of elastic structures and in particular with the treatment of the Bernoulli model for a beam. This is a particularly simple model of the behaviour of a one-dimensional structure, with an approximation to the strain distribution that is quite satisfactory from an engineering point of view. As we have seen in chapter 4, the general expression for the displacement ®eld of a generic beam with constant bending sti€ness EIx is yielded by the fourth-order di€erential equation (4.43). However, even in the simple case in which we need to take into consideration a straight beam whose section is variable along the axis and subjected to a distributed load q…z†, the value of which is dependent on z, we have seen in section 4.6 that we are led to a di€erent and more complicated equation, equation (4.86), that can be also written as   d2 uy …z† d2 EIx …z† ˆ q…z† …7:1† dz2 dz2 The solution of this equation cannot now be expressed as a fourthorder polynomial as was the case for equation (4.44), and the possibility of obtaining a closed form is strictly dependent on the particular expressions of the functions EIx …z†and q(z) for the problem at hand. This in the general case is not possible. Thus even this very simple case clearly illustrates the necessity for establishing an approximate method of solution. As was pointed out in chapter 6, the ability to provide approximate solutions is one of the most important applications of the principle of stationary value of total potential energy, i.e. determining approximate solutions to the problem of elastic equilibrium. This important application of the principle of stationary potential energy is made by assuming equations (6.2) and (6.3) as starting points and making use of trial functions. This application is described below. The trial function is essentially a possible representation of the unknown quantity that has been used to describe the total potential energy functional; in general this quantity is the displacement ®eld of the structure or its elements. Under some circumstances, the quantity is also called the shape or coordinate function.

The method of trial functions

171

In order to clarify our ideas, let us consider the simply supported beam already studied in chapter 4 and depicted in Fig. 4.10. We found that the exact solution to this problem is given by equation (4.51) q uy …z† ˆ …z4 ÿ 2lz3 ‡ l 3 z† 24EIx and is a fourth-order polynomial expression. The con®guration of equilibrium of this beam can also be described in terms of the total potential energy functional, i.e. …uy † ˆ 0

…7:2†

where the potential energy …uy † according to equation (6.2), is given by …l …l 1 002 …uy † ˆ EIx uy dz ÿ quy dz 2 0

…7:3†

0

This expression is clearly a functional (i.e. a function of a function) of the lateral displacement uy …z† of the axis of the beam. If treated with the usual methods of the calculus of variations, the condition in equation (7.2) would lead us to the Euler equation of beam deformation, see equation (4.43), together with the boundary conditions uy ˆ 0

for

z ˆ 0 and

M…z† ˆ ÿEIx u00y ˆ 0

for

zˆl

…7:4†

zˆ0

and z ˆ l

…7:5†

However, let us now imagine that there is actually no simple solution to this structural model posed by equations (4.43), (7.4) and (7.5). In this circumstance we could use our engineering experience of the displacement of beams subjected to lateral loading and guess a suitable displacement shape that could act as a trial displacement ®eld. The analysis process is initiated by substituting the trial displacement ®eld uy …z† in equation (7.3). In order to be admissible, the trial displacement ®eld must satisfy the geometrical boundary conditions in equation (7.4). We may choose, i.e. guess, a simple displacement ®eld which is de®ned in shape but not in amplitude, for example z uy ˆ a sin …7:6† l as is shown in Fig. 7.1. Substituting the trial function, equation (7.6), into the potential energy function, equation (7.3), we have …l …l  z002 z 1  EIx a sin dz …7:7† …uy † ˆ dz ÿ qa sin 2 l l 0

0

172

Energy methods in structural mechanics q

A

uy* = a sin

πz l

B

a

l

Fig. 7.1

This process achieves a very important simpli®cation in the analysis we have reduced the functional of the total potential energy to a simple function of the parameter a, which regulates the amplitude of the shape function, equation (7.6).

This means that the condition of equilibrium (equation (7.2)) can be satis®ed by minimising a simple function rather than a functional, i.e. the integral of the function of the parameter a. This is a situation very much more amenable to algebraic manipulation than is the integral equation for equilibrium, equation (7.2). Thus in the present case we have …l …l  z002 z 1 …a† ˆ dz EIx a sin dz ÿ qa sin 2 l l 0

0

…l …l 4   z 1 2 2 z ˆÿ dz ÿ qa sin dz EIx a 4 sin 2 l l l 0

ˆ

0

4 ql EIx a2 ÿ 2 a 3  4l

…7:8†

and the condition of equilibrium, equation (7.2), yields d…a† ˆ0 da

)

aˆ4

ql 4 EIx 5

…7:9†

This relationship de®nes the value of the parameter a in terms of the applied load and the geometric and elastic material parameters of the beam and is the amplitude of the trial displacement ®eld uy .

The method of trial functions

173

We have thus obtained the solution of our problem without making reference to the calculus of variation or solving di€erential equations. Of course, we must bear in mind that this is only an approximate solution for the structural model of the beam; nevertheless, it may furnish surprisingly good results. For example, if we compare the vertical de¯ection at the middle of the span obtained by the procedure of trial function to the exact solution, given by equation (4.51) for z ˆ l=2, we have ql 4 uy …l=2† ˆ 0:0130208 EIx

(exact solution)

ql 4 uy …l=2† ˆ 0:0130711 EIx

(trial function)

…7:10†

That is the guessed trial function predicts a de¯ection at the mid-span of the beam with an error of ‡0.39% compared to the exact value. With regard to the bending moment we have, M…l=2† ˆ ÿEIx u00y …l=2† ˆ 0:025000ql 2 M  …l=2† ˆ ÿEIx uy 00 …l=2† ˆ 0:025801ql 2

…7:10b†

That is, the approximate method overestimates the maximum moments with an error of ‡3.2%. Of course this is a very simple example and a fortunate case, due to the fact that the shape function, equation (7.6), happens to ®t the exact displacement ®eld very well and it is therefore a good choice. Generally, engineers through their observations have a good facility to describe the probable shape of the deformations that will occur in their structure when the loading is applied. This facility is an excellent basis for `guessing' an appropriate trial function for the energy method described above. However, in general, we may not always be so fortunate in our choice of trial function as was the case for the example analysis described, and we should pay a good deal of attention to the procedure that is explained in the following section.

7.2

The method of Rayleigh±Ritz

In the previous section we have taken a ®rst approach to the method of trial function for the solution of the problem of elastic equilibrium in the case of a simply supported beam. We now generalise the method by presenting a more comprehensive procedure that can be developed from the simple idea introduced above: this is called the method of

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Energy methods in structural mechanics

Rayleigh±Ritz.1;2 Section 7.1 has shown that the goal of the method of trial functions consists of commuting the minimisation of a functional, which requires a variational treatment and leads to a di€erential formulation, to the minimisation of a function, which is achieved by means of elementary derivatives and leads to an algebraic problem. Algebraic formulations are very much easier to solve than ones involving functionals. The Rayleigh±Ritz procedure consists of the following steps (a) (b)

(c) (d)

Write the functional  of the total potential energy of the system. Choose a trial function for the displacement ®eld; the function must satisfy the geometrical boundary conditions (in the case of a beam those relative to de¯ection and slope) but it may or may not satisfy the mechanical boundary conditions (i.e. regarding moment and shear). This shape function must be dependent on a ®nite number of parameters a1 ; . . . ; an . Insert the trial function in the potential energy functional  and integrate over the dimension of the structure. Minimise the obtained function …a1 ; . . . ; an † with respect to the parameters a1 ; . . . ; an . We now have a linear system of n equations in the n unknowns a1 ; . . . ; an given by q q ˆ


ˆ ˆ0 qa1 qan

(e)

The solution of this system of equations de®nes the actual value of the shape function and constitutes the approximate solution of the problem of elastic equilibrium at hand.

Actually, the outlined procedure retraces the steps of the example already given in the previous section, with the di€erence that we can now have a trial function depending on several parameters or degrees of freedom, also termed generalised coordinates. The procedure outlined above is exempli®ed now by the case of a beam encastre at one end and simply supported at the other, as shown in Fig. 7.2. We consider this beam subjected to a distributed load q of constant value, which will enable us to compare the results given by the Rayleigh±Ritz method with the exact solution of the model.

1 Lord Rayleigh (Strutt, John William) (Langford Grove, Essex, 1842 ± Witham, Essex, 1919), English physicist. 2 Ritz, Walther (Sion, Valais, 1878 ± GoÈttingen, 1909), Swiss physicist.

The method of trial functions

175

q

A

B l

Fig. 7.2

For the sake of clarity of presentation, we will proceed step by step following the sequence presented above. (a) The functional of the total potential energy of the system is …l …l 1 002 …uy † ˆ EIx uy dz ÿ qu dz 2 0

…7:11†

0

(b) Select a shape function with several degrees of freedom, for example we choose here ®ve variables of the type     z z  ‡ b1 cos uy ˆ a0 ‡ a1 sin l l     2z 2z ‡ b2 cos …7:12† ‡ a2 sin l l where the variables are a0 , a1 , b1 , a2 , b2 . This shape function must satisfy the geometrical boundary conditions at A and B in Fig. 7.2, i.e. uy ˆ 0 and ÿu0y ˆ 0 uy …z† ˆ 0

for

zˆ0

for z ˆ l

…7:13† …7:14†

As we have

     z  z ÿ b1 sin uy 0 ˆ a1 cos l l l l     2 2z 2 2z ÿ b2 sin ‡ a2 cos l l l l

…7:15†

the geometrical conditions (7.13) and (7.14) lead to the system 8  > < uy …0† ˆ a0 ‡ b1 ‡ b2 ˆ 0 uy …l† ˆ a0 ÿ b1 ‡ b2 ˆ 0 …7:16† > : 0 ÿuy …0† ˆ ÿa1 ÿ 2a2 ˆ 0

176

Energy methods in structural mechanics

This system allows us, for example, to express a0 , a1 and b1 as functions of a2 and b2 and we are left with a shape function that is de®ned by two independent degrees of freedom only. That is ao ˆ ÿb2 a1 ˆ ÿ2a2

…7:17†

b1 ˆ 0 and consequently the suitable trial function is          2z 2z z  ÿ 1 ‡ a2 sin ÿ 2 sin uy ˆ b2 cos l l l

…7:18†

(c) We substitute the trial function, equation (7.18), into the energy functional, equation (7.11) and integrate over the length l of the beam …uy †

…l …l 1  00 2 ˆ EIx uy dz ÿ qu dz 2 0

0

    2 …l 1 4 2z ˆ EIx b2 2 cos 2 l l 0

   2 42 2z 22 z ÿa2 2 sin ‡ 2 sin dz l l l l 

   …l   2z ÿ1 ÿ q b2 cos l 0



‡ a2 sin



2z l



 ÿ 2 sin

z l

 dz

…7:19†

obtaining, once all the mathematical operations have been performed, …a2 ; b2 † ˆ

5EIx 4 2 4EIx 4 2 16EIx 3 4ql a ‡ qlb2  a2 ‡ 3  b2 ‡  a2 b2 ‡  2 l5 l 3l 3 …7:20†

We have therefore accomplished the replacement of the functional, equation (7.19) by the approximating function of the two parameters a2 and b2 , shown in equation (7.20).

The method of trial functions

177

(d ) We minimise the obtained function …a2 ; b2 † with respect to the parameters a2 and b2 , getting the linear system q 10EIx 4 16EIx 3 4ql ˆ0 ˆ 3  a2 ‡  b2 ‡ 3 qa2  l 3l

…7:21†

q 16EIx 3 8EI ˆ  a2 ‡ 3 x 4 b2 ‡ ql ˆ 0 3 qb2 3l l

This system of two simultaneous algebraic equations can also be stated in matrix form, as      10 16=3 a2 4= ql 4 ˆ 4  EIx 1 16=3 8 b2 (e) The solution of the system of equations (7.21) yields the values of the parameters a2 and b2 1 15ql 4 EIx 3 16 ÿ 452   ql 4 12 45 b2 ˆ ÿ ‡ EIx 2 162 ÿ 454 128 ÿ 3602 a2 ˆ

…7:22† …7:23†

Back substituting the results in equations (7.22) and (7.23) in the trial function, equation (7.18) we have the expression for the approximate solution of the problem of the elastic equilibrium for the beam shown in Fig. 7.2. It is instructive to compare the values of the de¯ection, of the slope and of the bending moment at the middle of the span furnished by the two term trial function, equation (7.18), with those derived from the exact solution, see equation (5.2). With regard to the de¯ection at the mid-span we have ql 4 uy …l=2† ˆ 0:0052083 EIx

(exact solution)

uy …l=2†

(trial function)

ql 4 ˆ 0:0043468 EIx

…7:24†

That is the trial function approximates the de¯ection with an error of ÿ16.5%. Regarding the slope we have 3

ql …l=2† ˆ ÿu0y …l=2† ˆ ÿ0:0052083 EIx 0

 …l=2† ˆ

ÿuy 0 …l=2†

ql 3 ˆ ÿ0:0070998 EIx

(exact solution) …7:25† (trial function)

178

Energy methods in structural mechanics

That is, the trial function over-estimates the slope at the mid-span with an absolute error of ‡36.3%. Finally, with regard to the bending moment we have M…l=2† ˆ ÿEIx u00y …l=2† ˆ 0:0625000ql 2

(exact solution)

M  …l=2† ˆ ÿEIx uy 00 …l=2† ˆ 0:0634988ql 2

(trial function)

…7:26†

showing that the trial function provides a good estimate for the moment at mid-span and has an error of ‡1.6%. It is worth noting that, despite the fact that we have employed a trial function with ®ve degrees of freedom, we have obtained an approximate de¯ection that is 16.6% larger that the exact one, while in the previous section we got an error of 0.39% using a trial function with only one degree of freedom. Moreover, in the example shown above the approximation for the slope is much worse than that for the de¯ection and the approximation for the bending moment is far better than those for the de¯ection and the slope. This variability of the degree of accuracy of the approximate results is due to the capability of the particular chosen trial function to ®t the exact solution and its derivatives. This aspect of the method underlies much of the problems associated with approximating techniques in structural mechanics and will be discussed to more detail in the next section.

7.3

The quality of the trial function

So far we have presented the results obtained for the beam shown in Fig. 7.2 by means of the trial function described in equation (7.12) in comparison to the exact solution given by the expression (equation (5.2)). In order to make some simple observations about the quality of the results yielded by the Rayleigh±Ritz method, let us take into account another trial function, once again of the type       z z nz  uy ˆ a0 ‡ a1 sin ‡ b1 cos ‡


‡ an sin l l l   nz …7:27† ‡ bn cos l It is natural to expect that the quality of the approximation tends to increase along with the number of degrees of freedom since the additional terms may be thought of as giving greater ¯exibility to the functions to meet the requirements for the minimum energy conditions. To check this hypothesis we assume a seven degrees of freedom

The method of trial functions

179

Table 7.1 N

uy …l=2†

 …l=2†

My …l=2†

5 7 exact

0:004347ql 4 =EIx 0:005097ql 4 =EIx 0:005208ql 4 =EIx

ÿ0:007100ql 3 =EIx ÿ0:004711ql 3 =EIx ÿ0:005208ql 3 =EIx

0:063499ql 2 0:059914ql 2 0:062500ql 2

trial function, i.e. uy

     z z 2z ‡ b1 cos ‡ a2 sin ˆ a0 ‡ a1 sin l l l       2z 3z 3z ‡ a3 sin ‡ b3 cos ‡ b2 cos l l l 

…7:28†

Following exactly the same procedure of the previous section we are able to show in Table 7.1, the values of the de¯ection, slope and bending moment at the middle of the span of the beam in Fig. 7.2. These parameters are written as functions of the overall number N …N ˆ 2n ‡ 1† of degrees of freedom that comprises the trial function, where the last row gives the exact analytical values, uy …l=2†, y …l=2† and My …l=2†. For the sake of simplicity Table 7.1 can be also expressed in terms of percentage di€erences between the approximate solutions and their exact counterpart, as shown in Table 7.2. Table 7.2 allows us to make some interesting considerations about the quality of the approximations corresponding to the di€erent values of N. First of all, let us ®x our attention on the de¯ection. It appears that taking into account two more degrees of freedom leads to a better approximation to the exact value. However, as we have already noticed before, in both these cases the approximation is worse than in the case of the simple supported beam of the ®rst section, where we employed only a single degree of freedom trial function! We can therefore conclude that in the Rayleigh±Ritz method a major role is played by the shape of the trial function chosen. If this shape is very near to the exact solution, the quality of the obtainable approximation tends to be satisfactory even with a small number of degrees of freedom. Table 7.2 N


…uy ÿ uy †%


…y ÿ y †%


…My ÿ My †%

5 7

ÿ16.54% ÿ2.13%

‡36.32% ÿ9.54%

‡1.60% ÿ4.14%

180

Energy methods in structural mechanics da (y0) dz

a y0

b

db (y0) dz

y z

Fig. 7.3

Secondly, let us consider the percentage di€erences relating to the bending moment. Here the situation is quite di€erent because increasing the number of degrees of freedom does not lead to a better approximation. By means of a trial function with ®ve degrees of freedom we have an error of 1.60% only, while by means of a trial function with seven degrees of freedom the error changes sign and increases in absolute value to 4.14%. This behaviour can be easily explained as follows. The fact that the trial function satis®es the geometrical boundary conditions and is generally near in shape to the exact solution does not necessarily imply everywhere a good approximation to the mechanical parameters (i.e. those concerning the internal forces), such as the bending moment, which depends on the derivatives of the de¯ected shape. In other words, as shown in Fig. 7.3, we can have trial functions which do not di€er very much from the exact curve on average, but whose gradients can di€er signi®cantly from the exact one at many points. As a consequence of this discourse it must be pointed out that another source of error is introduced by the mechanical boundary conditions, that is the conditions relating to the forces at the boundaries. It was emphasised earlier that the method of Rayleigh±Ritz does not require the respect of the mechanical boundary conditions. In the case at hand, i.e. the Bernoulli model of the beam, these conditions involve the derivatives of the displacement ®eld of the beam axis. For example, it can be seen that every term in the chosen shape function, equation (7.27) violates the mechanical condition M…l† ˆ 0 at B. Actually, we have M  …z† ˆ ÿEIx uy 00 …z† and the second derivative of equation (7.27) is not zero at z ˆ l for a generic set of values of the parameters a0 , a1 , b1 , . . . , an , bn . However, we could try to satisfy the

The method of trial functions

181

mechanical conditions together with the geometrical ones. Intuitively, this should result in a better approximation but at the expense of greater algebraic work. However, as we are going to see, this is not necessarily the case. In order to investigate this point, let us go back to equation (7.12) and impose the requirement of satisfaction of the mechanical condition M  ˆ ÿEIx uy 00 ˆ 0

for

zˆl

…7:29†

as well as the geometric requirements in equations (7.13) and (7.14). Now we have the following system of four equations in the ®ve parameters a0 , a1 , b1 , a2 , b2 8  u …0† ˆ a0 ‡ b1 ‡ b2 ˆ 0 > > < y uy …l† ˆ a0 ÿ b1 ‡ b2 ˆ 0 …7:30† 0 …0† ˆ ÿa1 ÿ 2a2 ˆ 0 ÿu > > : y  00 ÿuy …l† ˆ b1 ÿ 4b2 ˆ 0 which can be solved in terms of a0 , a1 , b1 , b2 . Thus, we get a0 ˆ 0 a1 ˆ ÿ2a2 b1 ˆ 0

…7:31†

b2 ˆ 0 As a consequence, the trial function becomes      2z z  ÿ 2 sin uy ˆ a2 sin l l

…7:32†

If we substitute equation (7.32) in potential energy functional, equation (7.11) and integrate, as before, over the length l of the beam, we obtain 5EIx 4 2 4ql a a2 ‡ …7:33†  2 l3 It should be noted that by imposing an additional condition, i.e. the moment requirement, another degree of freedom has been suppressed and so the substitution of the approximating function, equation (7.32), into the energy functional, equation (7.11) results in a function only of the parameter, a2 . We minimise the function …a2 † with respect to the parameter a2 , and get …a2 † ˆ

d 10EIx 4 4ql ˆ0 ˆ a2 ‡ da2  l3

…7:34†

182

Energy methods in structural mechanics

which gives the value of the parameter a2 as a2 ˆ ÿ

2ql 4 5EIx 5

…7:35†

Finally, back substituting the result in equation (7.35) in the trial function, equation (7.32) we have the solution corresponding to this case in which all the geometrical and mechanical boundary conditions are satis®ed. However, if we take into account for comparison the values of the de¯ection, slope and bending moment at the middle of the span, we have to recognise that the approximations obtained, i.e. 4

ql uy …l=2† ˆ 0:0026142 EIx 3

ql  …l=2† ˆ 0:0082128 EIx

…7:36† …7:37†

M  …l=2† ˆ ÿEIx uy 00 …l=2† ˆ 0:0258012ql 2 …7:38† result in errors of ÿ49.8%, ‡57.7% and ÿ58.7%, respectively. It is clear that these results are far worse than those in the case in which, with the same trial function, only geometrical boundary conditions have been ful®lled. Even if this is a limiting case (i.e. we have assumed too few degrees of freedom), nevertheless it must be underlined that the error due to the shape of the trial function can outweigh that due to non-compliance with the boundary conditions. Indeed it is not inconceivable that a trial function satisfying also the mechanical boundary conditions can give inferior results compared with a function that satis®es the geometrical boundary conditions only but is generally nearer in shape to the exact solution. The conclusion from this assessment of the value of trial functions is that one must choose the functions with great care and using as guidance the maximum information on the probable deformations of the structure. Even then it is possible that the results will not correspond closely with the exact solution. Nevertheless, even with one degree of freedom trial function the method provides a quick initial solution to structural problems that can be used as the basis for later, more re®ned methods. However, it is evident that it would be valuable if a systematic approach could be obtained to guide the choice of trial functions. This is the subject of sections 7.5 and 7.6, in which an automated approach is introduced by means of the concept of localised functions.

The method of trial functions

7.4

183

Numerical example: the method of trial functions applied to a cantilever with linearly varying depth

Up to now all applications have been concerned with beams having uniform cross sections and material properties that do not vary along the beam. We now go back to an example of the type of equation (7.1) and expand on this to illustrate the capability of the trial function method to provide a formulation also for the de¯ection of a cantilever beam in which the depth of the section varies linearly along its length. The analytical solution for this beam was presented in section 4.6, here we use the method of trial functions. This method is such that a simple function will always result from the algebra. However, as we have seen, the accuracy of the prediction for the deformations may not always be acceptable and the results must be analysed critically. Therefore, let us now present in detail the analysis of the beam with the tapered depth using the method of trial function. First, a one degree of freedom function is used. We recall that the natural boundary conditions are given by the equations (4.96) and (4.101), i.e. uy ˆ 0

and

ˆÿ

duy ˆ0 dz

at z ˆ 0

In terms of the non-dimensional variable   z=l, a one degree of freedom trial function that satis®es these conditions is      …7:39† uy …† ˆ a1 1 ÿ cos 2 where a1 ˆ uy …1† is the de¯ection of the tip of the cantilever beam. The ¯exural strain energy of the beam is  2   2  …1 EIx …z† d uy …z† 2 EIx …† d uy …† 2 Uˆ dz ˆ d 2 dz2 d 2 2l 3 …l 0

…7:40†

0

and note again that the ¯exural sti€ness EIx …† must be contained within the integral sign since for this beam the sti€ness varies along the beam. From equations (4.82) and (4.83) we can write the second moment of area for the beam as  3 bd1 Ix ˆ ‰1 ‡ … ÿ 1† Š3 ˆ Ix ‰1 ‡ … ÿ 1† Š3 …7:41† 12 where Ix is the second moment of area corresponding to the crosssection geometry of the cantilever at the built-in end and   d2 =d1 .

184

Energy methods in structural mechanics

The strain energy of ¯exure can now be written as  2  …1 d uy E Ix U ˆ 3 ‰1 ‡ … ÿ 1† Š3 d 2l d 2

…7:42†

0

Substituting the trial function in the strain energy gives  E Ix 2 2 … ÿ 12†…1 ‡  ÿ 2 ‡ 3 † Uˆ 3 128 2l

 3 2 3 ‡ ‰ …1 ÿ † ÿ 2…1 ÿ  † ‡ 6…1 ÿ †Š a21 …7:43† 16

The total potential energy  of the deformation of the cantilever comprises the strain energy due to the ¯exure of the beam and the loss of potential of the load P applied at the tip of the cantilever. Thus  ˆ U ÿ Puy …l† ˆ U ÿ Pa1

…7:44†

The equilibrium condition is obtained from the stationary value of the potential energy with respect to the degree of freedom a1 , that is d ˆ0 da1

…7:45†

Thus, the condition of stationarity of the potential function gives  E Ix 2 2 … ÿ 12†…1 ‡  ÿ 2 ‡ 3 † 128 l3  3 2 3 ‡ ‰ …1 ÿ † ÿ 2…1 ÿ  † ‡ 6…1 ÿ †Š a1 ÿ P ˆ 0 16 (7.46) Substituting the relevant value for  for a particular beam geometry in this equilibrium equation yields the value for the de¯ection at the tip of the cantilever and the shape of the de¯ection along the length of the beam. Example values are ˆ1

)

Pl 3 uy …1† ˆ : 3 02EI1

 ˆ 0:5

)

Pl 3 uy …1† ˆ : 1 95EI1

ˆ0

)

Pl 3 uy …1† ˆ : 1 31EI1

…7:47†

The method of trial functions

185

Fig. 7.4

It may be seen that in comparison with the exact values, obtained from the solution of the di€erential equation, the de¯ection at the tip of a cantilever with uniform depth … ˆ 1† obtained by means of the trial function provides a very good result, but that as the value of  is reduced the results from the approximate method diverge from the exact value. This is shown graphically in Fig. 7.4 where the value for the tip de¯ection from the one degree of freedom trial function is compared to the exact result for corresponding values of . The approximation is reasonable, i.e. to within 5%, up to a value of  ˆ 0:5. Again, we can try to improve the accuracy of the result from the approximate method by increasing the number of degrees of freedom in the trial function. This allows greater ¯exibility to the trial function to ®t the exact de¯ected shape of the cantilever. The following illustrates this by using a two degree of freedom trial function, i.e.        3  uy …† ˆ a1 1 ÿ cos ‡ a2 1 ÿ cos …7:48† 2 2 where a1 and a2 are the degrees of freedom, or generalised coordinates of the trial function. The shapes of the functions corresponding to these coordinates are shown in Fig. 7.5. This function is substituted in the total potential energy functional  and the equilibrium equations are obtained, as before, from the

186

Energy methods in structural mechanics 2·0 [1 – cos πζ/2] [1 – cos 3πζ/2] 1·5

1·0

0·5

0 0

0·2

0·4

ζ

0·6

0·8

1·0

Fig. 7.5

stationary condition of the potential energy. In this case two equations are obtained from the conditions q ˆ 0; qa1

q ˆ0 qa2

…7:49†

The equilibrium equations can be described in matrix notation as 

f1 …†

f2 …†

f3 …†

f4 …†



a1 a2



  Pl 3 1 ˆ0 ÿ 2E Ix 1

…7:50†

where f1 …†; . . . ; f4 …† are functions only of the variable  describing the degree of taper in the beam. These functions are f1 …† ˆ

3 2 16  …1

ÿ  ‡ 2 ÿ 3 † ÿ 34 …1 ÿ 3 ‡ 32 ÿ 3 †

1 4  …1 ‡  ‡ 2 ‡ 3 † ‡ 64

f2 …† ˆ

135 2 64  …1

2 3 ÿ  ‡ 23 2 ÿ 23 3 † ÿ 27 4 …1 ÿ 3 ‡ 3 ÿ  †

f3 …† ˆ

135 2 64  …1

2 3 ÿ  ‡ 23 2 ÿ 23 3 † ÿ 27 4 …1 ÿ 3 ‡ 3 ÿ  †

f4 …† ˆ

27 2 16  …1

ÿ  ‡ 2 ÿ 3 † ÿ 94 …3 ÿ  ‡ 2 ÿ 33 †

4 2 3 ‡ 81 64  …1 ‡  ‡  ‡  †

…7:51†

The method of trial functions

187

2·5 Exact solution 2·0

Two degrees of freedom One degree of freedom

uy (1)

1·5

1·0

0·5

0 0

0·2

0·4

α

0·6

0·8

1·0

Fig. 7.6

The values for the generalised coordinates in the trial function are simply calculated using   Pl 3 1 ÿ f2 …† a1 ˆ f1 …† 2EI1 …7:52†   Pl 3 f3 …† ÿ f1 …† a2 ˆ 2EI1 f2 …†f3 …† ÿ f4 …†f1 …† and the de¯ection at the tip of the cantilever is obtained from uy …1† ˆ a1 ‡ a2

…7:53†

Figure 7.6 shows the e€ect of including two degrees of freedom in the trial function; the agreement to within 5% of the exact value for the de¯ection at the tip of the cantilever now extends in the range 0:2 >  > 1. If greater accuracy was required across the full range of the parameter  it could be achieved by increasing the numbers of degrees of freedom of the trial function. Of course, we underline again that this would be at the expense of greatly increased algebraic complexity in the analysis.

7.5

The systematic search for a trial function

On the basis of what we have noticed in the previous sections in this chapter, it should be clear that in general terms to ®nd a trial function is, in the ®rst instance, a matter of good judgement, experience and an

188

Energy methods in structural mechanics

appreciation of the physical meaning of the problem. Generally, engineers develop with practice a good `feel' for the de¯ected shapes of structures. However, one may wonder if there is a systematic way of ®nding such a function and, more importantly, if this function tends to converge to the exact solution along with the increase of the number of degrees of freedom taken into consideration. The answer is yes, but unfortunately this is a quite complicated matter and cannot be reasoned on the basis of simple mathematics, to which this book is restricted. We will simply state that in order to establish some criteria of convergence for the trial function, in the ®rst instance we must deal with a special set of approximation functions, also named coordinate functions. These coordinate functions fi are characterised by the following properties . . .

each fi satis®es the geometrical boundary conditions for any N, the elements f1 ; f2 ; . . . ; fN are linearly independent for any solution g of the problem, we can always ®nd an approximation function a1 f1 ‡ a2 f2 ‡


‡ aN fN which di€ers, according to a certain measure, in®nitely little from g.

Under certain assumptions, it is possible to demonstrate that all the choices of a set of coordinate functions ensure the convergence of the Rayleigh±Ritz approximation to the exact solution in the limit. However, it is clear that for a ®nite number of coordinate functions fi , one choice of functions might give better accuracy over the another. Moreover, from a technical point of view, it must again be underlined that we can obtain satisfactory results (with care!) even from a choice of approximation functions that are not coordinate functions. Going back to the examples presented above, we could have demonstrated that for the beam shown in Fig. 7.1 we indeed have the convergence to the exact solution of the approximate function   N X nz  uy …z† ˆ where n is odd …7:54† an sin l nˆ1 since

   nz an sin l

…7:55†

is a set of coordinate functions. Each of the terms in the series, equation (7.54) ful®ls the geometrical boundary conditions speci®ed in equation (7.4). On the other hand, with regard to the beam shown in Fig. 7.2 the choice of the approximate function described in equation

The method of trial functions

189

(7.27), i.e. uy

      z z nz ‡ b1 cos ‡


‡ an sin ˆ a0 ‡ a1 sin l l l   nz ‡ bn cos …7:56† l

is not constructed from a set of coordinate functions. In fact, not one of the terms independently satis®es the geometrical boundary conditions speci®ed in equations (7.13) and (7.14).

7.6

Localised Rayleigh functions

It is fairly evident from the foregoing examples and analyses that the Rayleigh±Ritz technique cannot be simply and directly automated for general use in practical structural analysis. Nevertheless, the power of the method lends value to the search for an automated application. This section presents the basis whereby the method of trial functions, as applied to beams, may incorporate simple functions that, through the stationary conditions of the potential energy of the structure, can combine to describe the de¯ected shape of beams with a variety of boundary and loading conditions to a high degree of accuracy. The approach is introduced here as the localised Rayleigh±Ritz technique, but it may be seen essentially as the progenitor of the ®nite element method. Suppose we sub-divide a beam into a number of regions. The end points of these regions are called nodes and are numbered sequentially along the beam, as shown in Fig. 7.7. Now, we consider a trial function centred on one of these nodes, i, as shown in Fig. 7.8. The trial function has the following attributes . . .

unit de¯ection at node i zero de¯ection at all other nodes zero slope at all nodes.

l 1 1

2 2 region 2

Fig. 7.7

3 3

4 4

5 5

6 6

node 5

7 7

8 8

9 9

10 10

11

190

Energy methods in structural mechanics li–1

i– 2

li

i– 1

i

fi

i +1

i+ 2 z

1

zi–1

zi

y

Fig. 7.8

The analysis is simpli®ed if we identify a local coordinate system for the region on either side of node i as shown in Fig. 7.8, the coordinate system is 0  z i ÿ 1  li ÿ 1

0  z i  li

…7:57†

The presentation is further simpli®ed if we again non-dimensionalise the coordinate system by de®ning z z i ÿ 1  i ÿ 1 and i  i …7:58† li ÿ 1 li Thus, the coordinate system is (as shown in Fig. 7.9) 0  i ÿ 1  1

and 0  i  1

…7:59†

The trial function between nodes i ÿ 1 and i is obtained very easily. The boundary conditions that ensure the trial function has a degree Element i– 1

i– 1

i

ui– 1

0

ζi– 1

Fig. 7.9

z

ui

0

y

i+ 1

ζi

The method of trial functions

191

of freedom at node i and nowhere else, are fi ÿ 1 …i ÿ 1 † ˆ 0

at i ÿ 1 ˆ 0

fi ÿ 1 …i ÿ 1 † ˆ 1

at i ÿ 1 ˆ 1

dfi ÿ 1 …i ÿ 1 † ˆ0 di ÿ 1

…7:60†

at i ÿ 1 ˆ 0 and 1

A simple polynomial function satisfying these requirements is fi ÿ 1 …i ÿ 1 † ˆ …3i2ÿ 1 ÿ 2i3ÿ 1 †

…7:61†

The trial function fi between nodes i and i+1 is obtained in the same way, and the corresponding polynomial function is fi …i † ˆ 1 ÿ 3i2 ‡ 2i3

…7:62†

that satis®es the requirements fi …i † ˆ 1 at

i ˆ 0

fi …i † ˆ 0 at

i ˆ 1

dfi …i † ˆ 0 at di

…7:63†

i ˆ 0 and 1

With this trial function the de¯ection of the beam is obtained by multiplying the trial function by the de¯ection, ui , at node i as shown in Fig. 7.9. It is clear in the ®gure that the de¯ection of the beam in the region between nodes i ÿ 1 and i is governed only by the values of the de¯ections at these nodes. This region between the nodes is called an element and the de¯ection function for element i ÿ 1 can be obtained by combining equations (7.61) and (7.62), i.e. uiyÿ 1 …1 † ˆ ui ÿ 1 …1 ÿ 3i2ÿ 1 ‡ 2i3ÿ 1 † ‡ ui …3i2ÿ 1 ÿ 2i3ÿ 1 †

…7:64†

The ¯exural strain energy in this element of the beam is Ui ÿ 1

EI ˆ x 2

li ÿ 1

… 

EI ˆ 3x 2li ÿ 1

0

 d2 uiyÿ 1 …zi ÿ 1 † 2 dzi ÿ 1 dz2i ÿ 1

 …1  2 i ÿ 1 d uy …i ÿ 1 † 2 di ÿ 1 di2ÿ 1 0

EI ˆ 3 x 12…ÿ ui ÿ 1 ‡ ui †2 2li ÿ 1

…7:65†

192

Energy methods in structural mechanics P 1

1

2

2

3

3

Elements

4

4

5

Nodes

l

Fig. 7.10

The potential for automating this approach is now evident since each element has the same form of ¯exural strain energy with respect to the length of the element and the de¯ections at the nodes bounding the element. We can exemplify the application of this approach by using the beam shown in Fig. 7.10, where for convenience the element lengths are all equal. The strain energy of the whole beam is Uˆ

EIx 12‰…ÿ u1 ‡ u2 †2 ‡ …ÿ u2 ‡ u3 †2 3 2…l=4† ‡ …ÿ u3 ‡ u4 †2 ‡ …ÿ u4 ‡ u5 †2 Š

…7:66†

and the potential energy  on account of the concentrated load P at mid-span is …7:67†

 ˆ U ÿ P u3

Now, the boundary conditions for the simply supported beam are u1 ˆ 0

and

u5 ˆ 0

…7:68†

Hence the potential energy of the beam in these terms is ˆ

384EIx 2 ‰ u2 ‡ …ÿ u2 ‡ u3 †2 ‡ …ÿ u3 ‡ u4 †2 ‡ …ÿ u4 †2 Š ÿ P u3 l3 …7:69†

The equilibrium conditions result from the stationary value of the total potential energy, that is q ˆ 0; q u2

q ˆ 0; q u3

q ˆ0 q u4

…7:70†

The method of trial functions

193

Hence the equilibrium conditions may be put in matrix form as 32 3 2 3 2 u2 0 2 ÿ1 0 3 Pl 6 7 76 7 6 …7:71† 2 ÿ1 54 u3 5 ÿ 415 ˆ 0 4 ÿ1 768EIx 0 0 ÿ1 2 u4 From this we obtain the solution u3 ˆ

Pl 3 768EIx

u2 ˆ u4 ˆ

and

u3 2

…7:72†

It is evident that this value for the de¯ection at the mid-length of the beam is quite di€erent from the exact analytical solution of Euler's equation of beam deformation, i.e. uy …l=2† ˆ

Pl 3 48EI

…7:73†

At this point if we wish to improve the accuracy of the predicted value of the de¯ection we have a choice, we can either . .

increase the complexity of the combined trial functions through increasing the degrees of freedom by augmenting the number of elements, or increase the complexity of the trial functions and the number of degrees of freedom within each element.

Let us take the ®rst approach. It is evident that if we double the number of elements in the beam, as shown in Fig. 7.11, the equilibrium conditions are 32 3 2 2 3 u2 2 ÿ1 0 0 0 0 0 0 76 7 6 6 7 6 ÿ1 607 6 7 2 ÿ1 0 0 0 07 76 u3 7 6 6 7 76 7 6 6 7 7 7 6 0 ÿ1 607 6 2 ÿ1 0 0 0 76 u4 7 6 6 7 3 76 7 6 6 7 Pl 76 u5 7 ÿ 6 0 6 7 0 ÿ1 2 ÿ1 0 0 76 7 1536EI 6 1 7 ˆ 0 6 x6 7 76 7 6 6 0 607 6 7 0 0 ÿ1 2 ÿ1 07 76 u6 7 6 6 7 76 7 6 6 7 7 7 6 0 607 6 0 0 0 ÿ1 2 ÿ1 54 u7 5 4 4 5 0

0

0

0

0

ÿ1

2

u8

0

(7.74)

194

Energy methods in structural mechanics P 1

2

1

2

3

3

4

4

5

5

6

6

7

7

Elements

8

8

9

Nodes

l

Fig. 7.11

The result is that the de¯ection at mid-span is u5 ˆ

Pl 3 768EIx

u u2 ˆ u8 ˆ 5 ; 4

and u u3 ˆ u7 ˆ 5 ; 2

3 u u4 ˆ u6 ˆ 5 4

…7:75†

which does not di€er from equation (7.72). It is therefore clear that this approach is not very fruitful. The reason lies in the fact that the choice of the localised trial function was such as not to allow any rotations at the nodes, as a result of the third of the boundary conditions, equations (7.60) and (7.63). In other words, the representation of the de¯ection makes the beam unrealistically over-sti€ened. As mentioned before, the alternative approach to improve the accuracy of the localised method is to increase the complexity of the trial function within an element. Thus, we now extend the trial function for an element of the length of the beam to include two degrees of freedom at each node. These degrees of freedom relate to the de¯ection at the node i, ui , and the slope at that node, i . Following the approach shown above for the derivation of the trial function for the de¯ection, the shape for the trial function for the slope at node i is shown in Fig. 7.12. The corresponding boundary conditions are gi ÿ 1 …i ÿ 1 † ˆ 0

ÿ

at i ÿ 1 ˆ 0; 1

dgi ÿ 1 …i ÿ 1 † ˆ0 di ÿ 1

at i ÿ 1 ˆ 0

d gi ÿ 1 …i ÿ 1 † ˆ1 di ÿ 1

at i ÿ 1 ˆ 1

…7:76†

The method of trial functions g (ζi– 1) i– 2

i– 1

g (ζi) i

gi

i+ 1

i+ 2 z

1

ζi– 1

195

ζi

y

Fig. 7.12

gi …i † ˆ 0 ÿ

at i ˆ 0; 1

dgi …i † ˆ1 di

at i ˆ 0

dgi …i † ˆ0 di

at i ˆ 1

…7:76† cont:

The simplest polynomial expressions that ®t these boundary conditions are gi ÿ 1 …i ÿ 1 † ˆ …i2ÿ 1 ÿ i3ÿ 1 † gi …i † ˆ …ÿi ‡ 2i2 ÿ i3 †

…7:77†

The de¯ected shape along a particular element is now a function of the de¯ections and slopes at the nodes bounding the element. Thus, for element i ÿ 1, bounded by nodes i ÿ 1 and i, we have, by combining equations (7.61) and (7.77) (notice that as  ˆ ÿ…duy =d†
…1=ln †, for simplicity in the presentation we de®ne n  n ln ) uiyÿ 1 …i ÿ 1 † ˆ ui ÿ 1 …1 ÿ 3i2ÿ 1 ‡ 2i3ÿ 1 † ‡ i ÿ 1 …ÿi ÿ 1 ‡ 2i2ÿ 1 ÿ i3ÿ 1 † ‡ ui …3i2ÿ 1 ÿ 2i3ÿ 1 † ‡ i …i2ÿ 1 ÿ i3ÿ 1 †

…7:78†

196

Energy methods in structural mechanics

As before, the ¯exural strain energy Ui ÿ 1 for this element is Ui ÿ 1

EI ˆ 3x 2li ÿ 1 ˆ

…1  0

d2 uiyÿ 1 …i ÿ 1 † di2ÿ 1

2

di ÿ 1

EIx ‰12 u2i ÿ 1 ÿ 12 ui ÿ 1 i ÿ 1 ÿ 12 ui ÿ 1 i ‡ 42i ÿ 1 2li3ÿ 1 ‡ 12 ui i ÿ 1 ‡ 4i ÿ 1 i ‡ 12 u2i ‡ 12 ui i ‡ 42i Š

It is worth noticing that the results from evaluating the partial derivatives of the strain energy with respect to the degrees of freedom ui ÿ 1 , i ÿ 1 , ui and i , in matrix notation are 2 3 qUi ÿ 1 6 q 7 6 ui ÿ 1 7 32 2 3 6 qUi ÿ 1 7 ui ÿ 1 24 ÿ12 ÿ24 ÿ12 6 7 7 6 6 q 7 8 12 47 EIx 6 76 i ÿ 1 7 6 ÿ12 6 iÿ1 7 …7:79† 76 7 6 7ˆ 3 6 6 qUi ÿ 1 7 2li ÿ 1 4 ÿ24 12 24 12 54 ui 5 6 7 6 q 7 ÿ12 4 12 8 i 6 ui 7 4 qU 5 iÿ1 qi Essentially this is the basis for the sti€ness matrix for a typical element. In the analysis of the simply supported beam, given the increased complexity of the trial function, we may restrict ourselves with a subdivision of the beam into only two regions, as shown in Fig. 7.13. The total potential energy is  ˆ U ÿ P u2

…7:80†

and again, once the boundary conditions for the simply supported beam, i.e. u1 ˆ 0 and u3 ˆ 0, have been imposed, the equilibrium P 1

1

2

l

Fig. 7.13

Elements

2

3

Nodes

The method of trial functions

197

conditions result from the stationary value of the total potential energy, that is q ˆ 0; q1

q ˆ 0; q u2

q ˆ 0; q2

q ˆ0 q3

…7:81†

In matrix form the reader will ®nd it easy to con®rm that we have 32  3 2 3 2 0 8 12 4 0 1 3 6 7 7 6 12 7 6 48 0 ÿ12 76 u2 7 Pl 6 1 7 6 …7:82† 76  7 ÿ 6 7ˆ0 6 4 4 0 16 4 54 2 5 4EIx 4 0 5 0 0 ÿ12 4 8 3 The solution of this matrix equation is 2 : 3 2 3 ÿ0 125 1 7 6 u 7 Pl 3 6 6 0:08333 7 6 27 6 7 6 7ˆ 5 4 2 5 4EIx 4 0 0:125 3 and the de¯ection at mid-span is u2 ˆ

Pl 3 48EIx

…7:83†

This corresponds exactly with the result from the analytical solution of Euler's beam deformation equation, primarily because that solution shows the beam deformation can be described by a cubic function of z and the localised trial functions used here include cubic terms. In conclusion, it is worth noting that a further step to introduce the automation of the procedure can consist in applying the natural boundary conditions after the stationarity of the total potential energy has been required. In the case at hand it would have meant ®rst writing the results of equation (7.81) as 2 32 3 2 3 u1 24 ÿ12 24 ÿ12 0 0 0 6 ÿ12 7 6 7 6 7 8 12 4 0 0 76 1 7 6 607 6 76 7 7 3 6 6 ÿ24 617 12 48 0 ÿ24 ÿ12 76 u2 7 6 76 7 ÿ Pl 6 7 ˆ 0 6 ÿ12 6 7 6 7 4 0 16 12 47 6 76 2 7 4EIx 6 0 7 6 76 7 6 7 4 0 405 0 ÿ24 12 24 12 54 u3 5  0 0 ÿ12 4 12 8 0 3 …7:84†

198

Energy methods in structural mechanics P 1

1

Elements

2

2

3

Nodes

l

Fig. 7.14

and then setting u1 ˆ 0 and u3 ˆ 0 by eliminating the rows and columns relating to u1 and u3 , which gives the matrix shown in equation (7.82). In this manner the formulation can very easily be adjusted to enable the analysis of beam with other boundary conditions than simply supported. This can be exempli®ed with regard to a cantilever beam, encastre at node 1 and having a concentrated load P at node 3, see Fig. 7.14. The natural boundary conditions are u1 ˆ 0 and 1 ˆ 0 at node 1 …7:85† Thus, we can simply eliminate the ®rst two rows and columns from the matrix (7.84) and set the contribution of the load P at node 3. The equilibrium equations are 32 3 2 2 3 u2 48 0 ÿ24 ÿ12 0 3 6 7 7 6 0 16 7 6  12 4 76 2 7 Pl 6 0 7 6 …7:86† 76 7 ÿ 6 6 7ˆ0 4 ÿ24 12 24 12 54 u3 5 4EIx 4 1 5 ÿ12 4 12 8 0 3 and the result from this matrix equation is u3 ˆ

Pl 3 3EIx

…7:87†

To sum up, we see that with an accurate choice of the shape functions, even with only two elements this approach of localised functions provides the exact answer for the de¯ection of the cantilever and the central de¯ection of a simply supported beam. However, perhaps even more impressive is the ease with which the resulting matrix formulation can be particularised to incorporate the relevant boundary conditions for the structure. It is this capability of the localised Rayleigh±Ritz

The method of trial functions

199

method, that has enabled it to be programmed for computers and used in such an easy manner by engineers for a great variety of types of structures. Chapter 11 in this book introduces formally the ®nite element method which directly derives from the localised trial function approach and formulations presented here.

7.7

Numerical example: application of Rayleigh localised trial functions to the analysis of a frame

This example illustrates the application of the method of localised trial functions to the analysis of the frame that previously has been analysed using the Cotterill±Castigliano theorem, see section 6.5. The frame is divided into a number of regions and, similar to the analysis in section 7.6, localised functions are proposed as trial functions. The portal frame to be analysed here is shown in Fig. 7.15. Each element is subdivided into two regions, or elements, each of equal length, i.e. ln ˆ l=2. The symmetry at position C is taken into account in this analysis to reduce the number of degrees of freedom. Again, we consider the elements to be axially undeformable. We recall from equation (7.79) that for a typical element the partial derivatives relative to the strain energy, which lead to the de®nition of

l/2 Symmetry P Node 3 B

l/2

Region 3

Region 2

I2

C

I1

Node 4

I2

Node 2 l=5m l/2

I=

Region 1

Node 1 A

Fig. 7.15

I1 I2

= 2·68

200

Energy methods in structural mechanics

the sti€ness matrix, can be written as 2 32 3 ui ÿ 1 24 ÿ12 ÿ24 ÿ12 6 7 8 12 47 EI 6 6 ÿ12 76 i ÿ 1 7 ‰K n Š ˆ 3x 6 76 7 2ln 4 ÿ24 12 24 12 54 ui 5 ÿ12 4 12 8 i where ui is the de¯ection at the node i normal to the axis of the element and i is the change of slope at that node. The junction of the regions 1 and 2 at node 2 is such that the elements have common values of de¯ection and change of slope. The junction of the vertical and horizontal members at B is such that the de¯ection of region 2 at node 3 is zero and the change of slope for regions 2 and 3 is common at node 3. Thus combining the sti€ness matrices of the three regions to provide the stationary condition for the strain energy of the portal frame 32 3 2 24 ÿ12 ÿ24 ÿ12 0 0 0 u1 76 7 6 6 ÿ12 6 7 8 12 4 0 0 0 7 76 1 7 6 76 7 6 6 ÿ24 6 7 12 48 0 ÿ12 0 0 7 76 u2 7 6 76 7 6 4EI 2 6 ÿ12 6 7  ˆ 4 0 16 4 0 0 7 ‰KŠ 76 2 7 l3 6 76 7 6 6 0 6 7 0 ÿ12 4 8…1 ‡ I† 12I 4I 7 76 3 7 6 76 7 6 6 0 6 7 0 0 0 12I 24I 12I 7 54 u4 5 4 0 0 0 0 4I 12I 8I 4 where I  I1 =I2 . The boundary conditions are u1 ˆ 1 ˆ 0 at A;

u3 ˆ 0 at B

4 ˆ 0 at C

and

Applying these boundary conditions and taking into account the term relative to the potential of the vertical load P the condition of equilibrium of one half of the symmetric frame is 2 3 2 32 3 48 0 ÿ12 0 u2 0 6 6 7 6 7  4 0 76 2 7 P 6 0 7 4EI2 6 0 16 7 6 76  7 ÿ 6 7 ˆ 0 3 4 4 5 4 5 2 ÿ12 4 8…1 ‡ I† 12I l 05 3 0

0

12I

24I

u4

1

The matrix equation can be solved to give the de¯ected shape for the portal frame corresponding to various values of I.

The method of trial functions

201

As an example, the value of I ˆ 0:0001 is analysed and the matrix equation gives the solution u4 ˆ

1 Pl 3 191:97 EI1

This value of I implies that the ¯exural sti€ness of the horizontal member of the frame is very much less than that of the vertical members. Thus the horizontal member should react to the application of the load as if it were a beam fully encastre at both ends. Note that a value of I ˆ 0 would render the matrix equation singular. The analytical formulation for the de¯ection at mid-span for such a beam is u4 ˆ

1 Pl 3 192 EI1

Thus the analysis compares nearly exactly with the analytical formulation. This is because the analytical solution which describes exactly the deformations of the members of the frame is a cubic polynomial. The trial function actually is also a cubic polynomial and therefore completely satis®es the requirement. Consider now the analysis of the portal frame shown in Fig. 6.5, which has been shown in section 6.5 to have I ˆ 2:68. We have the result from the matrix analysis as 3 2 : 2 3 u2 ÿ0 0134 7 6  7 Pl 3 6 6 0:0134 7 6 27 7 6 : 6 7ˆ 4 3 5 8EI2 4 ÿ0 0534 5 0:0423 u4 Hence the de¯ection u4 at mid-span of the horizontal member, for a load, P ˆ 64:7 kN, is Pl 3 u4 ˆ 0:0423 8EI2 ˆ

0:0423  6:47  104  50003 ˆ 13:49 mm 8  2:1  105  1:51  107

This compares exactly with corresponding value calculated using the Cotterill±Castigliano theorem. The strain at the mid-span can be calculated using the curvature of the member BC at that point. In non-dimensional coordinates, the

202

Energy methods in structural mechanics

curvature of an element is obtained from ÿ

d 2 uy d2 ˆ ÿ 2 ‰ ui ÿ 1 …1 ÿ 3 2 ‡ 2 3 † ‡ i ÿ 1 …ÿ ‡ 2 2 ÿ  3 † 2 d d ‡ ui …3 2 ÿ 2 3 † ‡ i … 2 ÿ  3 †Š ˆ ÿ‰ ui ÿ 1 …ÿ6 ‡ 12† ‡ i ÿ 1 …4 ÿ 6† ‡ ui …6 ÿ 12† ‡ i …2 ÿ 6†Š

The curvature of member BC is ÿ

2 d 2 uy 1 d uy ˆ ÿ dz2 ln2 d 2

4 ‰ u3 …ÿ6 ‡ 12† ‡ 3 …4 ÿ 6† l2 ‡ u4 …6 ÿ 12† ‡ 4 …2 ÿ 6†Š

ˆÿ

where  ˆ 0 at node 3,  ˆ 1 at node 4 and with u3 ˆ 4 ˆ 0: The result from the analysis for I ˆ 2:68 gives the curvature at C as ÿ

d 2 uy Pl Pl ˆÿ ‰2…0:0534† ÿ 6…0:0423†Š ˆ : 2 2EI2 13 605EI2 dz

The strain at the outer face of the member BC at point C is "c ˆ

6:47  104  5000  112 ˆ 0:84  10ÿ3 5 7 : : : 13 605  2 1  10  1 51  10

This compares exactly with the corresponding value from the Cotterill±Castigliano analysis, see section 6.5. It is evident that the method of trial functions can be easily applied to the analysis of frames. A more speci®c form of matrix-based analysis for frames is presented in chapter 9. This is based on the actual solution of Euler's equation for beam deformation and is the method that is usually used in structural engineering practice. However, there may be circumstances, such as the components of the frame having variable geometry that would complicate the usual frame analysis. In such a conditions the method of localised trial functions, i.e. the ®nite element method, can be used to obtained accurate solutions for the strains and deformations in the frame.

The method of trial functions

Suggested exercise problems 1.

2.

3.

203

Figure 7.16 shows a beam with a uniform ¯exural sti€ness of EI and built-in at both ends. The beam is subjected to a uniformly distributed load of q/unit length. Use the method of trial function to derive an expression for the de¯ection shape and the value of the de¯ection wc at mid-length. Compare the values obtained with the exact ones from the analytical solution of Euler's equation for beam deformation. [Ans: Trial function wc ˆ ql 4 =390EI, Analytical value wc ˆ ql 4 =384EI] The beam shown in Fig. 7.17 has a ¯exural sti€ness of EI for half its length and 2EI for the other half. The beam is subjected to a uniformly distributed load q. Use the method of trial functions to develop an expression for the de¯ection at mid-length. Apply the method of localised trial functions to provide a comparable value for the de¯ection. [Ans: One term trial function wc ˆ ql 4 =16:1EI, localised trial function with two elements along the beam, wc ˆ ql 4 =26:3EI, localised trial function with four elements along the beam, wc ˆ ql 4 =16:0EI] The beam shown in Fig. 7.18 is subjected to a distributed load, varying from Q/unit length at the end A which is simply supported, to 0 at the end B which is built-in. The beam has a uniform q A

B EI

l

Fig. 7.16

q

EI

2EI l

Fig. 7.17

l

204

Energy methods in structural mechanics Q

B

A EI

l

Fig. 7.18

¯exural sti€ness EI. Use the method of trial functions to calculate the de¯ection of the beam at mid-span. Check the answer against a two element localised trial function model. [Ans: one term trial function, wc ˆ Ql 4 =399:0EI, two element localised trial function, wc ˆ Ql 4 =347:8EI]

8.

Matrix analysis of pin-jointed trussed structures

Introduction

Foregoing chapters have been concerned with developing the basis of the analysis of bars and beams, both from the exact solution of the Euler equation of deformation and from an approximate approach. The present chapter and the next one extends the analytical solution of Euler's approach to encompass collections of bars and beams that constitute truss and frame structures. These types of structure are ubiquitous in engineering practice, including the skeletons of buildings, lattice bridges and a great variety of cranes and lifting equipment. We begin this chapter with the analysis of pin-jointed structures. These structures are collections of bars essentially subjected to axial forces which therefore constitute the primary loading in the element. However, as the joints in practical frame structures are fabricated by welding or by using groups of bolts, some bending of the individual members will be induced because of the rigidity of the joints, as well as because of the bars' self weight. This bending is generally seen as a secondary e€ect. Nevertheless, if the bars are arranged so that their axes meet in one point at each joint, it is found that the secondary bending can be considered negligible with comparison to the axial forces. Indeed, in some bridges built in the 19th century the joints actually were e€ected using pins so that the forces in the bridge members were in accord with the analysis. Essentially the function of such pin-jointed structures, or trusses, as they are commonly known, is that of carrying loads across a wide span. It is common to introduce the analysis of pin-jointed frames through consideration of the force equilibrium at the pin-joints. That method requires that there is assurance that the truss has a geometry, i.e. an arrangement of members, that allows that method of analysis to be valid. The validity rests on the establishment that the truss is statically determinate which means that the forces in the members can be

206

Energy methods in structural mechanics

calculated purely on the basis of static force equilibrium. As we know from chapter 2, if we remove a member from a statically determinate truss the structure becomes a mechanism and the equilibrium is not assured under any loading. Conversely, if additional members are added to a statically determinate truss geometry the truss is classi®ed as having statically redundant members and the forces in the members cannot be calculated solely from the force equilibrium of the pin-joints. Thus the traditional approach for introducing the analysis of pinjointed frames is applicable only to a limited range of truss geometries and, moreover, is not capable of extension to the analysis of rigidjointed frames. The method presented in this chapter is completely general and is in fact the basis of practical design analysis of frames. In the method presented here there is no requirement to investigate if the truss is statically determinate or structurally redundant, and the method is such that it can encompass all forms of truss geometry and will enable the evaluation of member forces and the deformed geometry of the truss.

8.1

Plane pin-jointed structures as systems of axially loaded bars

So far we have examined the behaviour of one-dimensional deformable bodies, like bars and beams and introduced the principle of stationary value of total potential energy as a basis for approximate energy methods, like the method of trial functions. In the present case the analysis methods are extended to the systems of several bars and beams and we start by looking at the example of a plane pin-jointed structure, also called a truss, shown in Fig. 8.1. This structure is typical of the arrangement of bars to support a pulley for lifting heavy pieces of equipment and is composed of three straight bars AB, AC and AD, pin-jointed together at the point A and attached to a vertical rigid wall by pin-joints at the ends B, C and D, respectively. We assume that the pin-joints do not o€er any resistance to rotation of the bars, either with respect to the wall or with respect to the relative rotations of the members connected at a joint. This assumes, of course, that we neglect any frictional forces acting at the contact surfaces between the pins and the members that could occur in a practical pinjoint, but such forces could be made small so that we can be assured that the above approximation is reasonable and thus no moments can arise at the ends of any bar. The only force applied to the structure is the force P at the point A, representing the load suspended from that joint. For the sake of generality we suppose at this stage that the loading is inclined to the vertical.

Matrix analysis of pin-jointed trussed structures

207

D

C

45° 30°

B

A P

y

x

Fig. 8.1

In order to deal with a structure composed of several members, it is ®rst necessary to consider its individual members, whose behaviour we are able to analyse in a relatively simple way. In the present case all the members are simple straight bars, such that for the equilibrium of each of them, taken separately, we can immediately deduce that they can only bear compression or tension, as shown in Fig. 8.2. The deformation, i.e. the kinematic behaviour, of a typical member of the pin-jointed frame is therefore quite simple and can be described by the relationship (equation (4.5))   N 0 l ˆ 1‡ l EA that represents the relationship between the initial and the ®nal lengths of the bar, l and l 0 , respectively, and the applied axial loading, N. The elastic energy stored in the individual typical member of the truss on account of its deformation is related to its change of length by equation (4.8), Uˆ

1 EA 0 …l ÿ l†2 2 l

208

Energy methods in structural mechanics N

N

Fig. 8.2

Given the behaviour of its component members, the study of the equilibrium of the whole structure can be carried out as usual with reference to the general condition of equilibrium (equation (1.28)) U ˆ W or, as we have seen in chapter 6, with reference to the principle of stationary value of the total potential energy, i.e. ˆUÿW where obviously the elastic energy of the whole structure is given by the sum of the energy stored in each of its members, i.e., X i Uˆ U …8:1† i

where i indicates the number of members in the structure. The work done by the external forces is given by W ˆ P  u…A†

…8:2†

where u…A† is the vectorial displacement of the point A. The work done by the external forces can also be expressed in terms of components of the vectorial displacement u…A† and we have W ˆ Px ux …A† ‡ Py uy …A†

…8:3†

It is evident, because we are dealing with a pin-jointed truss entirely contained in one plane, that the structure can have only two degrees of

Matrix analysis of pin-jointed trussed structures

l' i

209

ux (K ) uy (K ) K

ux (J)

uy (J) J

θi

li

y

x

Fig. 8.3

freedom, that is, the degrees of freedom are the two components of displacement of the point A, which in our example, see Fig. 8.1, is the only one free to move. Once we have de®ned its position under the applied load P, then we can immediately determine for each member how much it is stretched and the value of the axial force it is carrying (see equation (4.7)). In order to achieve this, let us relate the initial and ®nal lengths of each bar to the coordinates of its ends and to the displacement of the point A. With reference to Fig. 8.3, we can write for a generic element, whose end points are J and K q l i ˆ ‰x…K† ÿ x…J†Š2 ‡ ‰y…K† ÿ y…J†Š2 …8:4† s f‰x…K† ‡ ux …K†Š ÿ ‰x…J† ‡ ux …J†Šg2 l 0i ˆ …8:5† ‡f‰y…K† ‡ uy …K†Š ÿ ‰y…J† ‡ uy …J†Šg2 The following relationships hold true between the angle i and the coordinates of the end points J and K cos i ˆ

1 ‰x…K† ÿ x…J†Š li

…8:6†

sin i ˆ

1 ‰y…K† ÿ y…J†Š li

…8:7†

210

Energy methods in structural mechanics

thus, equation (8.5) becomes v u u 1 ‡ 2 f‰u …K† ÿ u …J†Š cos i ‡ ‰u …K† ÿ u …J†Š sin i g u x y y li x u l 0i ˆ l i u t 1 ‡ i2 f‰ux …K† ÿ ux …J†Š2 ‡ ‰uy …K† ÿ uy …J†Š2 g l …8:8† We are operating in the framework of in®nitesimal displacements and we can therefore simplify the expression in equation (8.8) by taking into account only theplinear  terms of a simple Taylor's series expansion (that is, of the type 1 ‡   1 ‡ =2). Thus, we obtain l 0i ˆ l i ‡ ‰ux …K† ÿ ux …J†Š cos i ‡ ‰uy …K† ÿ uy …J†Š sin i

…8:9†

With regard to the bars composing our structure we can write Element B±A i

sin i

cos i

ux …J†

uy …J†

ux …K†

uy (K)

0

0

1

0

0

ux …A†

uy …A†

l 0BA ˆ l BA ‡ ux …A† U BA ˆ

…8:10†

1 EABA ‰ux …A†Š2 2 l BA

…8:11†

Element C±A i

sin i

ÿ308

ÿ 12

cos i p ÿ 23

ux …J†

uy …J†

ux …K†

uy …K†

0

0

ux …A†

uy …A†

p 3 1 ux …A† ÿ uy …A† ˆl ‡ l 2 2 p 2 CA  1 EA 1 3 CA u …A† ÿ uy …A† U ˆ 2 l CA 2 2 x 0CA

CA

…8:12† …8:13†

Element D±A i ÿ458

sin i p ÿ 22

cos i

p 2 2

ux …J†

uy …J†

ux …K†

uy …K†

0

0

ux …A†

uy …A†

Matrix analysis of pin-jointed trussed structures

p p 2 2 u …A† ÿ u …A† l ˆl ‡ 2 x 2 y p   p  2 1 EADA 2 2 DA u …A† ÿ u …A† U ˆ 2 l DA 2 x 2 y 0DA

DA

211

…8:14† …8:15†

The total potential energy for our example structure is therefore  ˆ U ÿ W ˆ U BA ‡ U CA ‡ U DA ÿ W p 2 3 1 EABA 1 EACA 1 2 ‰ux …A†Š ‡ u …A† ÿ uy …A† ˆ 2 l BA 2 l CA 2 2 x p   p    2 1 EADA 2 2 u u ‡ …A† ÿ …A† 2 l DA 2 x 2 y ÿ ‰Px ux …A† ‡ Py uy …A†Š

…8:16†

It is evidently a function only of ux …A† and uy …A†. The magnitudes of the variables ux …A† and uy …A† that render the total potential energy stationary are yielded by the ful®lment of the following stationary conditions p   3 q EABA EACA 3 u …A† ÿ ˆ BA ux …A† ‡ CA u …A† qux …A† 4 x 4 y l l   EADA 1 1 ux …A† ÿ uy …A† ÿ Px ˆ 0 ‡ DA …8:17† 2 2 l p   3 q EACA 1 u …A† ÿ ˆ CA u …A† quy …A† 4 y 4 x l   EADA 1 1 u …A† ÿ ux …A† ÿ Py ˆ 0 …8:18† ‡ DA 2 y 2 l These conditions of course represent the equilibrium equations of the structure under consideration. Once we have obtained the values of ux …A† and uy …A† we know the displacements of the ends of each member, its elongation, the corresponding axial strain it has undergone and, from equation (4.7), the axial loading it is actually carrying. We can also evaluate the degree of safety of the structure. In other words we know everything of the plane truss from a mechanical point of view. It is worth noticing that the key point of the example analysis has been the choice of the number of variables that are able to describe completely the kinematics of the elastic structure. We call these variables the degrees of freedom of the structure because they allow

212

Energy methods in structural mechanics

us to determine the exact position in the plane of the structure after the deformation has taken place. In addition, quite distinct from what happens for the systems of rigid bodies examined in chapter 2 that can experience a displacement ®eld only and only if the constraints are not sucient to ®x their position in the space, elastic structures always tend to deform under loading, even if rigid body displacements are prevented by the restraints. Strictly speaking, any elastic body can be thus thought to be characterised by an in®nite number of degrees of freedom, as each of the degrees is necessary to de®ne the displacement of any of its points. However, if we can describe by means of appropriate functions the behaviour of an individual elastic element, as in the case of the elongation of axially loaded bars or the solution to Euler's equation of beam deformation, we can always analyse very complex structures by dividing them into their members and studying the overall systems with respect to a ®nite number of variables, which we call once again degrees of freedom of the system. As we have just seen, in the case of the pin-jointed structures these variables can be assumed to be the displacements of the joints, or nodal points, that link the structural components together.

8.2

Matrix formulation of the elastic equilibrium equations for plane pin-jointed structures

In the present section we will show how the equilibrium equations that we have derived in section 8.1 can be formulated in a much more compact form, one that is particularly suited to automatic processing by means of computers. Many engineering structures consist of a collection of bars and beams assembled together to provide an ecient way for resisting applied forces and carrying required loads. From a theoretical point of view these structures can always be treated as has been shown in the previous section, but following those procedures step-by-step would generally involve incredibly lengthy and timeconsuming hand calculations even, as we have seen, in the case of a fairly small number of members. However, the use of computers o€ers the possibility of analysing very complex structures in a very short time and requires minimal e€ort on the part of the engineers, provided the problems are formulated in a suitable manner. The use of linear algebra, i.e. matrix notation, stems naturally when performing calculations on a computer, as it allows very large group of numbers to be manipulated in a systematic and well-established manner. At the same time a matrix approach to the analysis of structures makes possible a comprehensive method that applies to a wide variety of types of frame structures and requires only some e€ort to

Matrix analysis of pin-jointed trussed structures

213

input the data de®ning the frame geometry, the material properties and the loading, thus minimising the possibility of errors. Many standard commercially developed programs are now available for use on even quite small computers. These programs provide a capability for calculating the displacements and strains for frames ranging from the simplest, with say three or four members, to very complex frames with hundreds of members. The intention here is to explain the mechanics that underlies the computation in the programs. As an aid to our presentation of how it is possible to formulate the equilibrium equations in a matrix form, let us once again make reference to the pin-jointed structure of Fig. 8.1. In section 8.1 the key point of our presentation was the analysis of the truss in terms of its individual members, whose behaviour has been related to the displacements of the nodal joints. Following exactly the same line of reasoning, let us try to express the same steps in a matrix format. First of all, we can recast equation (8.9) in matrix notation, which provides the relationship between the ®nal and the initial lengths of each bar as function of the displacements of its end points and of the angle  of the bar in the plane. Thus, we have 2 3 u …J† " # x 6 u …J† 7 0 0 cos i sin i 6 y 7 l i ˆ l0i ‡ ‰ÿ1 1Š 7 i i 6 4 5 u …K† 0 0 cos  sin  x uy …K† …8:19† This notation enables us to identify a matrix containing terms that provide information about the position of the i-th member in the frame, i.e. " # 0 0 cos i sin i i ‰R Š ˆ …8:20† 0 0 cos i sin i A central component in equation (8.19) is the vector containing the displacement of the member's end nodes in the general reference frame x±y, i.e. 3 2 ux …J† 6 u …J† 7 7 6 y …8:21† ‰Di Š ˆ 6 7 4 ux …K† 5 uy …K† Notice that the product between ‰Ri Š and ‰Di Š results in the vector ‰d i Š

214

Energy methods in structural mechanics u ζ (K ) u (K ) u (K ) u (J) J

uζ (J)

uη (J)

ζ

η

Fig. 8.4

containing the axial components of the nodal displacements in the local reference frame  ÿ  shown in Fig. 8.4. Therefore the relationship   u …J† ˆ ‰d i Š ˆ ‰Ri Š‰Di Š …8:22† u …K† may be properly regarded as a compatibility equation that links the displacements of the end nodes of the i-th bar in the general reference frame x±y to the axial components of displacement of the same nodes in its local reference frame  ÿ . It is worth noticing that ‰Ri ŠT ‰Ri Š ˆ ‰IŠ. It is now clear that the elongation of the bar can be expressed in matrix form as
l i ˆ l 0i ÿ l i ˆ ‰ ÿ1

1 Š‰d i Š ˆ ‰ ÿ1

1 Š‰Ri Š‰Di Š

…8:23†

which is equivalent to equation (8.19). The elastic energy stored in each bar can also be expressed in matrix notation. In fact we obtained the expression for the elastic energy, equation (4.8), by means of Clapeyron's theorem, namely by taking half of the work done by the axial loading N corresponding to the length variation
l. As the expression for N, equation (4.7), is readily stated in matrix notation, i.e. Ni ˆ

EAi i EAi
l ˆ i ‰ ÿ1 li l

1 Š‰Ri Š‰Di Š

…8:24†

we can write the strain energy in matrix format as 1 1 EAi ‰ ÿ1 U i ˆ N i
l i ˆ 2 2 li

1 Š‰Ri Š‰Di Š‰ ÿ1

1 Š‰Ri Š‰Di Š

…8:25†

In order to reach a much more meaningful expression to our purposes, we notice that, as the transpose of a scalar (in our case a real number) is the scalar itself, we can apply the transpose chain rule to

Matrix analysis of pin-jointed trussed structures

215

the ®rst three matrices from the left in equation (8.25) (whose product is a scalar quantity) and write   1 i i 1 EAi i T i T ÿ1 i ‰ ÿ1 1 Š‰Ri Š‰Di Š ‰D Š ‰R Š …8:26† U ˆ N
l ˆ 2 2 li 1 By expanding the middle product between vectors and rearranging the terms, we obtain   1 ÿ1 1 i i 1 i T i T EAi i U ˆ N
l ˆ ‰D Š ‰R Š ‰Ri Š‰Di Š …8:27† 2 2 l i ÿ1 1 The term

  1 ÿ1 EAi ‰K Š ˆ i l ÿ1 1 i

…8:28†

is commonly referred to as the sti€ness matrix of the bar. It links the axial displacements of the end points of the bar, as yielded by equation (8.22), to the axial loads applied at the same ends, namely     N…J† 1 ÿ1 EAi ˆ i ‰d i Š …8:29† l N…K† ÿ1 1 Inspection of the sti€ness matrix reveals that it is symmetric, that is the ®rst and the second rows are identical, respectively, to the ®rst and second columns. This is a general property of any linearly elastic system and can be derived from the reciprocal theorems presented in section 5.2. In fact, this property of symmetry is equivalent to stating that a given displacement applied at point K of an elastic body causes a force at point J equal to that which would have been caused at point K were the same displacement to be applied at coordinate J rather than K.

Alternatively, this result can be viewed as a direct consequence of the principle of conservation of energy. In fact, in order to be an exact di€erential, the elastic energy function U i of the generic bar must ful®l the following condition qU i qU i ˆ qu …J† qu …K†

…8:30†

which by virtue of equations (8.22) and (8.27) implies that the sti€ness matrix ‰K i Š must be symmetric. It is worth noting also that the term   1 ÿ1 EAi ‰K i Š ˆ ‰Ri ŠT i ‰Ri Š …8:31† l ÿ1 1

216

Energy methods in structural mechanics

can also be regarded as a sti€ness matrix, provided we recognise that it relates the vector of the nodal displacements ‰Di Š to the vector of the nodal forces ‰F i Š in the general reference frame. In fact, the vector ‰F i Š is expressed as 3 2 3 2 Fx …J† cos i 0     6 F …J† 7 6 sin i 0 7 7 6 7 N…J† 6 y i i T N…J† ˆ ‰R Š ‰F Š ˆ 6 7ˆ6 7 4 Fx …K† 5 4 0 N…K† cos i 5 N…K† i Fy …K† 0 sin  …8:32† By substituting equation (8.29) in equation (8.32) we obtain  i 1 ÿ1 i i T EA ‰d i Š ‰F Š ˆ ‰R Š l i ÿ1 1

…8:33†

and remembering equ (8.22), we have ‰F i Š ˆ ‰K i Š‰Di Š

…8:34†

At this point the total potential energy of the whole pin-jointed structure, which we have already expressed in equation (8.16), can easily be re-cast in matrix notation. In order to keep the procedure as general as possible (as is desirable for a procedure suitable for computer implementation), we align all the nodal displacements in the vector ‰DŠ, namely ‰DŠT ˆ ‰ux …A† uy …A† ux …B† uy …B† ux …C† uy …C† ux …D† uy …D†Š …8:35† The required matrix form of the total potential energy is then accomplished by assembling a joint sti€ness matrix that is obtained by summing contributions corresponding to each node from the member sti€ness matrices ‰K i Š. This is obviously equivalent to summing all the expressions of the elastic energy relating to each bar, being the associative law of multiplication valid for the matrix product. In fact, let us write the expression of the generic member sti€ness matrix ‰K i Š in complete detail 2

‰K i Š ˆ

cos2 i

i i EAi 6 6 sin  cos  6 l i 4 ÿ cos2 i

ÿ sin i cos i

sin i cos i

ÿ cos2 i

ÿ sin i cos i

sin2 i ÿ sin i cos i

ÿ sin i cos i cos2 i

ÿ sin2 i sin i cos i

ÿ sin2 i

sin i cos i

sin2 i

3 7 7 7 5

…8:36†

Matrix analysis of pin-jointed trussed structures

217

This generic expression is written in the general reference frame and it is evident that in order to obtain the sti€ness matrices for the bars B±A, C±A and D±A in a form that can be related to the joint displacement vector ‰DŠ, all we have to do is identify the joints J and K in the general numbering system on an individual basis and set the relative terms in the appropriate position. Therefore, we can write 2

cos2 BA 6 sin BA cos BA 6 6 6 ÿ cos2 BA 6 BA 6 BA BA EA 6 ÿ sin  cos  AB ‰K Š ˆ BA 6 6 l 0 6 6 6 6 4 0

sin BA cos BA

ÿ cos2 BA

ÿ sin BA cos BA

0

0

0

0

0

0

0

0

0

0

3

ÿ sin BA cos BA

cos2 BA

sin BA cos BA

ÿ sin2 BA

sin BA cos BA

sin2 BA

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

07 7 7 07 7 07 7 7 07 7 07 7 7 05

0

0

0

0

0

0

0

2

sin 

0

BA

ÿ sin 

BA

cos 

BA

2

BA

ÿ sin 

(8.37) 2

cos2 CA 6 sin CA cos CA 6 6 6 0 6 CA 6 EA 0 6 CA ‰K Š ˆ CA 6 6 l ÿ cos2 CA 6 6 ÿ sin CA cos CA 6 6 4 0 0

sin CA cos CA 2

sin 

CA

0

ÿ cos2 CA

0

ÿ sin 

CA

ÿ sin CA cos CA

cos 

CA

2

ÿ sin 

0

CA

0

0

0

0

0

0

0

0

0

0

0

0

0

0

0

ÿ sin CA cos CA

0

0

cos2 CA

sin CA cos CA

0

ÿ sin2 CA

0

0

sin CA cos CA

sin2 CA

0

0 0

0 0

0 0

0 0

0 0

0 0

0

3

07 7 7 07 7 07 7 7 07 7 07 7 7 05 0

(8.38) 2

cos2 DA 6 sin DA cos CA 6 6 6 0 6 DA 6 EA 0 6 DA ‰K Š ˆ DA 6 6 l 0 6 6 0 6 6 4 ÿ cos2 DA ÿ sin DA cos DA

sin DA cos DA 2

sin  0

DA

0

0

0 0

ÿ cos2 DA DA

ÿ sin 

ÿ sin DA cos DA DA

cos 

2

DA

0 0

0 0

0 0 0 0

0 0

0 0

0 0

0 0 0 0

0

0

0

0 0

0

0

ÿ sin DA cos DA ÿ sin2 DA

0 0

0 0

0 0 0 0

cos2 DA sin DA cos DA

sin DA cos DA sin2 DA

0 0 0

ÿ sin  0 0 0

3 7 7 7 7 7 7 7 7 7 7 7 7 7 5

(8.39) Thus, the following expressions are valid: U BA ˆ 12 ‰DŠT ‰K BA Š‰DŠ

…8:40†

U CA ˆ 12 ‰DŠT ‰K CA Š‰DŠ

…8:41†

U DA ˆ 12 ‰DŠT ‰K DA Š‰DŠ

…8:42†

and it follows immediately that the elastic energy stored in the frame is

218

Energy methods in structural mechanics

given by U ˆ U BA ‡ U CA ‡ U DA ˆ 12 ‰DŠT ‰K BA Š‰DŠ ‡ 12 ‰DŠT ‰K CA Š‰DŠ ‡ 12 ‰DŠT ‰K DA Š‰DŠ  ˆ 12 ‰DŠT ‰KŠ‰DŠ  is the combined sti€ness matrix, that is where ‰KŠ  ˆ ‰K BA Š ‡ ‰K CA Š ‡ ‰K DA Š ‰KŠ

…8:43† …8:44†

This is true on account of the above-mentioned associative property of the product between matrices. The total potential energy of the frame in matrix notation can now be obtained by formulating the components of the applied force as a load vector, i.e. ‰PŠT ˆ ‰ Px …A† Py …A†

0 0

0

0 0

0Š

…8:45†

which allows us to write the work of the applied loads as W ˆ ‰DŠT ‰PŠ

…8:46†

and, consequently   ˆ U ÿ W ˆ 12 ‰DŠT ‰KŠ‰DŠ ÿ ‰DŠT ‰PŠ

…8:47†

At this point it is worth noting that if the reader takes the trouble to develop the expression in equation (8.47) to its full extent by means of the values given in the tables in section 8.1, he will ®nd a somewhat different result from that in equation (8.16). This is because that up to this point we have considered as active all the possible displacements of the nodes A, B, C and D. Even if this fact may not seem to be reasonable, since the displacements of the joints B, C and D are actually prevented by the attachment to a rigid wall, the reason is that we are developing here a method suitable for an automatic procedure and our steps must conveniently correspond to several distinct phases of operation of a computer program. Similarly to that shown in section 7.6 in the case of the method of localised Rayleigh functions, there is no necessity to identify the active degrees of freedom at an early stage, as this would impose a particularisation on the sti€ness matrices of the elements and prevent the possibility of implementing a totally automatic manipulation of the data. Moreover, to enhance further the generality of the approach we can extend the applied load vector to include the reactions at the constraints. Thus the loading vector in this example becomes ‰PŠT ˆ ‰Px …A† Py …A† Rx …B† Ry …B† Rx …C† Rx …D† Ry …D†Š …8:47b†

Matrix analysis of pin-jointed trussed structures

219

In fact, once one has constructed equation (8.47), it is always possible to obtain equation (8.16) by simply setting the displacements ux and uy of the joints B, C and D to zero. However, as we are going to see, it is better to perform this operation at the next stage. Now, returning to the elastic equilibrium problem, it is clear that the stationary value of the total potential energy expression, equation (8.47), must be required with respect to the degrees of freedom set in the displacement vector ‰DŠ. In e€ect we have to impose that q ˆ0 qDj

…8:48†

where Dj is the generic component of the displacement vector ‰DŠ. It is immediately evident that equation (8.48) is a linear system of n equations in n unknowns, namely the components of the vector ‰DŠ. It can thus be written in matrix notation as  ‰KŠ‰DŠ ÿ ‰PŠ ˆ 0

…8:49†

as a direct consequence of equation (8.47). However, the system in equation (8.49) does not admit a solution since the combined sti€ness  is singular and therefore it cannot be inverted. From a matrix ‰KŠ physical point of view, this is because the movement of the structure as a rigid body is not prevented and, consequently, the solution of the problem of equilibrium cannot be obtained. It is therefore necessary to impose on the system the requirements of the geometric constraints at B, C and D. In order to do so we can write the system, equation (8.49), in the following form       ‰KI Š ‰KI;II Š ‰PI Š ‰DI Š ˆ …8:50† ‰DII Š ‰PII Š ‰KI;II ŠT ‰KII Š where we rearranged the equations in order to distinguish two subarrays in the nodal displacement vector ‰DŠ, namely ‰DI Š and ‰DII Š, corresponding, respectively, to the degrees of freedom that are active and to the degrees of freedom which are actually suppressed by the restraints. At the same time we distinguish two sub-arrays in the load vector ‰PŠ, ‰PI Š and ‰PII Š; the ®rst corresponds to the loads applied to the free joints, the second to the reactions which arise at the restrained ones. It goes without saying that the set ‰PI Š is assigned, while the set ‰PII Š is unknown, as the displacements ‰DI Š are unknown and the displacements ‰DII Š are set to zero. This happens because we can obviously assign kinematic or mechanical conditions at any degree of freedom, but never both at the same time.

220

Energy methods in structural mechanics

In the present example, the sub-array ‰DII Š corresponds to the inactive degrees of freedom ux …B†, uy …B†, ux …C†, uy …C†, ux …D† and uy …D†and we have ‰DII Š ˆ ‰0Š

…8:51†

We can sub-divide the system, equation (8.50) into two expressions …8:52† ‰KI Š‰DI Š ˆ ‰PI Š ÿ ‰KI;II Š‰0Š ‰PII Š ˆ ‰KI;II ŠT ‰DI Š ‡ ‰KII Š‰0Š

…8:53†

E€ectively, the system equation (8.52) is disjoint from equation (8.53) and the matrix format in equation (8.49) is now reduced to a set of two linear equations in the two unknowns active degrees of freedom, ux …A† and uy …A†, that is …8:54† ‰KI Š‰DI Š ˆ ‰PI Š which the reader will ®nd to be identical to equations (8.17) and (8.18), and obviously admits a unique solution. Once we have solved equation (8.54) and obtained the magnitudes of the displacements ‰DI Š, we can straightforwardly obtain the restraint reactions ‰PII Š from equation (8.53). It is worth noting that the 2  2 joint sti€ness matrix ‰KI Š is positive de®nite (and therefore non-singular), which follows immediately from the condition U ˆ 12 ‰DI ŠT ‰KI Š‰DI Š > 0

…8:55†

once any rigid body displacement has been suppressed. From a physical point of view the set of equations (8.54) requires that the sum of the elastic reactions from the end joints of each element balances the applied forces at any joint of the structure. In the present example this is composed only of the force P at A. At this stage one is allowed to wonder why we have gone through all these lengthy and apparently cumbersome calculations in order to obtain equation (8.54). The reason lies in the fact that once we have established that the conditions of equilibrium (equation (8.48)) can be expressed in the form shown in equations (8.49)±(8.54), we can directly and automatically build a similar system for any plane pin-jointed structure without bothering to repeat all the underlying reasoning in terms of energy.

As a matter of fact, the only data required to produce the equilibrium equations (8.49) are the geometry of the structure, the de®nition of the load and the form of the generic element sti€ness matrix, equation (8.36). The imposition of the restraint conditions leads subsequently and without e€ort to equation (8.54).

Matrix analysis of pin-jointed trussed structures

221

This constitutes the basis of the so-called direct sti€ness method that in the next chapter we will apply to rigid-jointed frames.

8.3

Numerical example: analysis of a truss

The example concerns the analysis of a simple truss, shown in Fig. 8.5. The structure is pinned to a rigid foundation at A and D and has a horizontal load, P1 , applied at the joint B and a vertical load, P2 , applied at C. The loads are applied proportionally such that P1 ˆ 2P2 The frame is composed of circular hollow sections, with a mean diameter of dm ˆ 250 mm and a wall thickness of t ˆ 7:5 mm. The cross-section area of the members is  A ˆ …257:52 ÿ 242:52 † ˆ 5:890  103 mm2 4 The Young modulus is E ˆ 2:1  105 N=mm2 The axial sti€nesses of members AB, BC and CD are equal and is EA 2:1  105  5:890  103 ˆ ˆ 4:12  105 N=mm k1 ˆ l 3000 P2 P1

B

C

3m

y

+ve θ A

D

3m

Fig. 8.5

x

222

Energy methods in structural mechanics

The sti€ness of the member AC, k2 , is 1 k2 ˆ p k1 2 This example can be tackled using the method presented in section 8.1 and the reader is encouraged to apply that method and thus become familiar with it. However, the main purpose of the example is to illustrate the more powerful and automated matrix approach described in section 8.2. The objective of the analysis is to evaluate the maximum allowable level of the load P1 such that the maximum strain does not exceed 30% of the material yield strain, "0 , where "0 ˆ 1:5  10ÿ3 . The direct sti€ness method is essentially concerned with assembling the sti€ness matrix of the complete structure from the component sti€ness matrices. Thus from equation (8.36) we have Member AB i ˆ AB ˆ

 2 2

0

0

60 6 ‰K AB Š ˆ k1 6 40

1 0

0

ÿ1

0

0

0

1

3

0 ÿ1 7 7 7 0 05

this sti€ness matrix is associated with the displacement vector 2

3 ux …A† 6 u …A† 7 6 y 7 ‰DAB Š ˆ 6 7 4 ux …B† 5 uy …B† Member BC i ˆ BC ˆ 0 2

1

6 0 6 ‰K BC Š ˆ k1 6 4 ÿ1 0

0

ÿ1 0

3

0

0 07 7 7 1 05

0

0 0

0

Matrix analysis of pin-jointed trussed structures

this sti€ness matrix is associated with the displacement vector 2

ux …B†

3

6 u …B† 7 6 y 7 ‰DBC Š ˆ 6 7 4 ux …C† 5 uy …C† Member CD i ˆ CD ˆ ÿ 2

 2

0

0

1 0

0 0

ÿ1 7 7 7 05

0 ÿ1

0

1

60 6 ‰K CD Š ˆ k1 6 40

0

3

0

this sti€ness matrix is associated with the displacement vector 2

ux …C†

3

6 u …C† 7 7 6 y ‰DCD Š ˆ 6 7 4 ux …D† 5 uy …D† Member AC i ˆ AC ˆ

‰K AC Š ˆ

 4 2

1

1 ÿ1

ÿ1

3

k2 6 6 1 6 2 4 ÿ1

ÿ1

1

ÿ1 7 7 7 15

ÿ1

ÿ1

1

1

1 ÿ1

this sti€ness matrix is associated with the displacement vector 2

3 ux …A† 6 u …A† 7 6 y 7 ‰DAC Š ˆ 6 7 4 ux …C† 5 uy …C†

223

224

Energy methods in structural mechanics

Following equation (8.44), the assembled sti€ness matrix, taking account of the displacement vectors, is 3 2 a a 0 0 ÿa ÿa 0 0 6 a …1 ‡ a† 0 ÿ1 ÿa ÿa 0 07 7 6 7 6 6 0 0 1 0 ÿ1 0 0 07 7 6 6 0 ÿ1 0 1 0 0 0 07 7 6  ‰KŠ ˆ k1 6 7 7 6 ÿa ÿa ÿ1 0 …1 ‡ a† a 0 0 7 6 6 ÿa ÿa 0 0 a …1 ‡ a† 0 ÿ1 7 7 6 7 6 4 0 0 0 0 0 0 0 05 0 0 0 0 0 ÿ1 0 1 p where a  1=…2 2†. The restraint boundary conditions now are required to particularise the sti€ness matrix to the truss in Fig. 8.5. These conditions are ux …A† ˆ uy …A† ˆ ux …D† ˆ uy …D† ˆ 0 Thus, eliminating the appropriate rows and columns of the sti€ness matrix we have 3 2 1 0 ÿ1 0 6 0 1 0 0 7 7  ˆ k1 6 ‰KŠ 7 6 4 ÿ1 0 …1 ‡ a† a 5 0

0

a

…1 ‡ a†

The work done by the applied loads, see equation (8.3), is W ˆ ÿP1 ux …B† ÿ P2 uy …C† Thus the matrix equation for the equilibrium state of the truss is, see equation (8.49),  ‰KŠ‰DŠ ˆ P1 ‰PŠ where 2

ux …B†

3

6 u …B† 7 6 y 7 ‰DŠ ˆ 6 7 and 4 ux …C† 5 uy …C†

2

3 ÿ1 6 0 7 6 7 ‰PŠ ˆ 6 7 4 0 5 ÿ0:5

Matrix analysis of pin-jointed trussed structures

225

The solution of the matrix is obtained as 3 2 ÿ4:328 7 6 0 7  ÿ1 P1 ‰PŠ ˆ P1 6 ‰DŠ ˆ ‰KŠ 7 6 k1 4 ÿ3:328 5 0:5 i

The strain levels, " , in each member can be obtained using equation (8.9), "i ˆ

l i0 ÿ l i li

thus, Member AB "AB ˆ

P1 u …B† ˆ 0 kl y

Member BC "BC ˆ

P1 P …ux …C† ÿ ux …B†† ˆ 1 kl EA

Member CD "CD ˆ

P1 P …ÿuy …C†† ˆ ÿ0:5 1 kl EA

Member AC P 1 P "AC ˆ p1 p …ux …C† ‡ uy …C†† ˆ ÿ1:414 1 EA k 2l 2 It is evident from the above that the maximum strain occurs in member BC and the allowable maximum load will be governed by the maximum strain in that member. Therefore, P 1:414 1 ˆ 0:3"0 EA and 0:3  1:5  10ÿ3  2:1  105  5:89  103 ˆ 394 kN 1:414 The forces in the various members can be evaluated from the strains P1 ˆ

as

N i ˆ "i EA

226

Energy methods in structural mechanics

hence N AB ˆ 0 kN;

N BC ˆ 394 kN …tens:†

N CD ˆ 197 kN …tens:†

N AC ˆ 557 kN …comp:†

Now, taking the calculated value of P1 , the displacement of the nodes can be evaluated from the matrix solution. Thus, uy …B† ˆ ÿ4:14 mm; ux …C† ˆ ÿ3:18 mm uy …C† ˆ 0:48 mm

8.4

Matrix analysis of pin-jointed space structures

The presentation in sections 8.1 and 8.2 has been concerned with the analysis of plane pin-jointed trusses, i.e. structures which were envisaged to be entirely contained in a plane together with the applied loads. This is the case when the axes of the elements are actually contained in a plane and any geometric or mechanical feature of the elements themselves is symmetric with respect to the same plane. Now we intend to illustrate the extension of the direct sti€ness method to deal with structures that exist in three-dimensional spacetrusses. The approach is based directly on the direct sti€ness method because all the energy concepts that underlie this method remain exactly the same as in the case of plane structures. A space pin-jointed structure is de®ned to be a collection of pin-jointed bars in the threedimensional Euclidean space, as exempli®ed in Fig. 8.6. The geometry and loading of the space-truss are described by reference to a general three-dimensional system of Cartesian axes x±y±z. The local axes of the system are denoted by ±±, as shown in Fig. 8.7. Similar to the method for the two-dimensional truss, we relate here the behaviour of the individual members of the space±truss, which are subjected to tension or compression loading, to the displacement of the nodal joints. The vector containing the displacements of the end nodes P

P

z

y x

Fig. 8.6

Matrix analysis of pin-jointed trussed structures K

227

ζ

J

ξ η

Fig. 8.7

of the generic i-th element in the general reference frame x±y±z is 3 2 ux …J† 6 uy …J† 7 7 6 6 u …J† 7 7 6 z i …8:56† ‰D Š ˆ 6 7 6 ux …K† 7 7 6 4 uy …K† 5 uz …K† while in the local reference frame ±± we write 3 2 u …J† 6 u …J† 7 7 6 6 u …J† 7 7 6  i ‰d Š ˆ 6 7 6 u …K† 7 7 6 4 u …K† 5 u …K†

…8:57†

As in the previous presentation, the relationship between ‰d i Š and ‰Di Š can be expressed as ‰d i Š ˆ ‰Ri Š‰Di Š

…8:58† i

where the transformation matrix ‰R Š is given by 3 2 i 0 0 0 x iy iz 7 6 i 6 x iy iz 0 0 0 7 7 6 i i 7 6 i   0 0 0 x y z 7 6 ‰Ri Š ˆ 6 i i i 7 6 0 0 0 x y z 7 7 6 7 6 0 0 ix iy iz 5 4 0 0 0 0 ix iy iz

…8:59†

228

Energy methods in structural mechanics

ik are direction cosines of angles between the local axes , ,  and the general axes x, y, z. It easy to verify that, similar to the plane truss, in the three-dimensional space the transformation matrix (8.59) is orthogonal, i.e. ‰Ri ŠT ‰Ri Š ˆ ‰IŠ

…8:60†

Since we are dealing with structures composed of bars, the only internal forces that occur on account of the deformation of the elements are the axial forces N…J† and N…K†. However, it may be useful to collect the nodal forces in the local reference frame in a six-elements array, f i , that is 3 3 2 2 0 0 7 6 0 7 6 0 7 7 6 6 7 7 6 6 6 F …J† 7 6 N…J† 7 i 7 6 7 6 …8:61† ‰f Šˆ6 7ˆ6 0 7 0 7 7 6 6 7 7 6 6 5 4 0 5 4 0 ÿF …K† N…K† so that in the local reference frame the relationship between the nodal forces ‰ f i Š and the nodal displacements ‰d i Š can be written in the following form ‰ f i Š ˆ ‰K i Š‰d i Š with

2

0

60 6 i6 60 EA ‰K i Š ˆ i 6 l 6 60 6 40 0

…8:62† 0

0

3

0

0

0

0 0

0 1

0 0

0

0

0

0

0

0

0 07 7 7 0 ÿ1 7 7 0 07 7 7 0 05

0 ÿ1

0

0

…8:63†

1

The expression in equation (8.63) represents the sti€ness matrix of a generic bar in the local reference frame and it corresponds precisely to the expression of the sti€ness matrix, equation (8.28), obtained in for plane pin-jointed structures. Inspection of matrix equation (8.63) reveals that it is symmetric, as we would expect. The vector ‰F i Š which collects the nodal forces of the generic bar in the general reference frame is expressed as ‰F i Š ˆ ‰Ri ŠT ‰ f i Š

…8:64†

Matrix analysis of pin-jointed trussed structures

229

so that from equations (8.58) and (8.62) we have ‰F i Š ˆ ‰Ri ŠT ‰K i Š‰Ri Š‰Di Š ˆ ‰K i Š‰Di Š

…8:65†

i

where K is the sti€ness matrix of the generic bar in the general reference frame, which again corresponds to the expression in equation (8.36) obtained in the earlier two-dimensional treatment. In detail we can write 2

2

ix

6 6 i i 6 x y 6 i6 6 ix iz EA i ‰K Š ˆ i 6 2 l 6 6 ÿix 6 6 6 ÿix iy 4 ÿix iz

ix iy

2

ix iz

ÿix

ÿix iy

iy

iy iz

ÿix iy

ÿiy

iy iz

iz

ÿix iz

ÿiy iz

ÿix iy

ÿix iz

ix

ix iy

ÿiy

ÿiy iz

ix iy

iy

ÿiy iz

ÿiz

ix iz

iy iz

2

2

2

2

2

2

2

ÿix iz

3

7 7 ÿiy iz 7 7 2 7 ÿiz 7 7 7 ix iz 7 7 7 iy iz 7 5 i2 z

(8.66) The solution of the space±truss problems proceeds then exactly as in the case of plane±truss problems shown in section 8.2: the overall joint  is obtained by assembling the contributions from sti€ness matrix ‰KŠ all the bars and the elastic equilibrium equation takes the now-familiar form  ‰KŠ‰DŠ ÿ ‰PŠ ˆ 0 …8:67† ‰DŠ is a 3n dimension array which contains all the degrees of freedom of the space truss under consideration, n being the overall number of  is obviously joints. The dimension of the joint sti€ness matrix ‰KŠ 3n  3n and ‰PŠ is the load vector, i.e. a 3n dimension array containing all the components of the forces applied to the joints which, for the sake of generality, can be arranged to include the reactions of the  is symmetric for the energy constraints. The joint sti€ness matrix ‰KŠ reasons explained in section 8.2. At this stage in the analysis the constraint conditions can be applied exactly as in the earlier work , i.e. by distinguishing two sub-arrays in the nodal displacement vector ‰DŠ, ‰DI Š and ‰DII Š. These sub-arrays correspond to the degrees of freedom that are active and to the degrees of freedom which are prevented by the restraints, respectively. The remainder of the procedure consists in condensing the system of equations (8.67) accordingly (see section 8.2). Once the reduced system has been solved, we can in a straightforward manner calculate the restraint reactions, the strains and the actually carried forces in any member of the space±truss.

230

Energy methods in structural mechanics

A 15 kN 1m

45°

45°

45°

45°

B

E

30 kN

C

D

Fig. 8.8

Suggested exercise problems 1.

The structure shown in Fig. 8.8 is to be analysed as a pin-jointed frame. Calculate the vertical and horizontal de¯ections at A and 1m Pb A

B

1m

C D

Fig. 8.9

Matrix analysis of pin-jointed trussed structures

2.

231

D,
Av ,
Ah ,
Ev , and
Eh due to the applied loads. All the structural members have a cross-sectional area of 200 mm2 . The modulus for steel is 205 kN=mm2 . [Ans:
Av ˆ ÿ15:5 mm,
Ah ˆ ÿ45:7 mm,
Ev ˆ ÿ28:0 mm,
Eh ˆ ÿ40:5 mm] A frame made from aluminium alloy, analysed as having pinjoints, is shown in Fig. 8.9. Calculate the forces and strains in the members BC, BD due to the application of the vertical load Pb at B. All members have the same cross-sectional area of 150 mm2 and the modulus of aluminium is 80 kN/mm2 . [Ans: "BC ˆ ÿ0:55  10ÿ3 , NBC ˆ ÿ6:62 kN (comp.), "BD ˆ ÿ1  10ÿ3 , NBD ˆ ÿ11:8 kN (comp.)]

9.

Matrix analysis of rigid-jointed framed structures

Introduction

Chapter 8 has provided an introduction to the topic of the analysis of pin-jointed frames. In that kind of structure the presence of pin-joints allows the component members of the structure to undergo relative rotation as the structure deforms under the action of the applied external loads. That possibility of relative rotation eliminates bending in the components and they experience only axial tension or compression loading. The present chapter extends the methodology established for pinjointed trusses to encompass the analysis of rigid-jointed frames. The chapter essentially follows the order of presentation in chapter 8 and ®rst introduces the mechanics and energy formulation underlying the analysis of a typical frame. The equations are then formalised in matrix notation in section 9.2, thus establishing the basis for the automated computer analysis of frames that can vary from two to several thousand components members. The procedure is exempli®ed by application to a frame with the geometry similar to that considered in section 8.3, but now with rigid joints that prevent any relative rotation of the members of the frame. The methodology for plane frames is extended to three-dimensional frames and the chapter concludes with some general aspects regarding the inclusion of distributed loading, such as member self-weight, and restrained displacements in the frame analysis method.

9.1

Plane rigid-jointed frame structures as systems of beams

In this section we shall extend the concepts introduced in the previous section and apply them to rigid-jointed frame structures, with the aid of the example two-dimensional system shown in Fig. 9.1. The geometry of this plane rigid-jointed frame is similar to that of the pin-jointed structure shown in Fig. 8.1. However, in contrast with that

234

Energy methods in structural mechanics D

C

45° 30°

B

A P

y

x

Fig. 9.1

truss, in this case we note that the members are not free to undergo relative rotation at connecting joints, thus the de®nition of rigid joint ( joints are sometimes called nodes). In particular any rotation is totally prevented at the nodes connected to the wall at B, C and D, while the connection at A preserves the relative angular spacing of the attached members AB, AC and AD at A. In other words, with rigid joints all the ends of members connected at the same joint must experience the same rotation. From a practical point of view, the nodes at B, C and D can be thought of as welded to rigid elements, while the node at A can be fabricated by welding the ends of steel elements together. In contrast to the bar members of the previous section, the elements into which we can divide the rigid-jointed structure are able to carry shear forces and moments, as well as axial forces. In fact, since the members are not free to rotate at the joints, corresponding restraint moments will occur at the ends. Notwithstanding the apparent complexity of the rigid-jointed structure, we can again analyse its behaviour in accordance with Bernoulli's model, introduced in chapter 4, by taking into account both the ¯exural and axial sti€ness of the members. This requires that the slopes at the nodal points are assumed as unknown variables together with the displacements at the nodes.

Matrix analysis of rigid-jointed framed structures

235

uy K

ux ζ J

θ

y

η x

Fig. 9.2

We should also remember that in the present framework, that is the linear theory of elasticity, the axial behaviour does not interact with the bending behaviour. The unstable e€ects that may be caused by the compressive axial forces in members will be examined in chapter 12. To analyse the bending behaviour of any individual member, modelled as a beam, (which will be superimposed on the axial deformation summarised by equations (4.5) and (4.8)), let us refer to Fig. 9.2. Here the relationship between the components of the vectorial displacement of the end point K and their contribution to the displacement u …K† that occurs orthogonal to the axis of the beam is clearly represented. From a mathematical standpoint, for the end points J and K we have u …J† ˆ ux …J† sin i ÿ uy …J† cos i i

u …K† ˆ ux …K† sin  ÿ uy …K† cos 

…9:1† i

…9:2†

where we assume that the sign convention for the transverse displacement u is according to the local reference frame ± shown in Fig. 9.2. In this local reference frame the bending de¯ection of the beam can be described as a function of the variables u …J†, u …K†, …J† and …K† by means of the same procedure that led to equation (4.116). Therefore we have   2u …J† …J† 2u …K† …K† 3 u …† ˆ ÿ ÿ ÿ  3 2 3 2 li li li li   3u …J† 2…J† 3u …K† …K† 2  ‡ ÿ i2 ‡ i ‡ ‡ i 2 l l l li ÿ …J† ‡ u …J†

…9:3†

236

Energy methods in structural mechanics

and the bending elastic energy stored in the beam on account of u …J†, u …K†, …J† and …K† is …l 1 EIu002 U ˆ  …† d 2 i

0

ˆ

6EI i li ‡ ‡

3

‰u2 …J† ‡ u2 …K†Š ÿ

12EI i li

3

‰u …J†u …K†Š

2EI i 2 ‰ …J† ‡ 2 …K† ‡ …J†…K†Š li 6EI i li

2

‰u …K†…K† ‡ u …K†…J† ÿ u …J†…J† ÿ u …J†…K†Š (9.4)

We can now superpose the axial and the ¯exural behaviour of the beam element under consideration and write the total elastic energy stored in it on account of its deformation. By adding equations (4.8) and (9.4) we have Ui ˆ

1 EAi 0i 6EI i 2 i 2 …l ÿ l † ‡ ‰u …J† ‡ u2 …K†Š 3 2 li li ÿ ‡

12EI i li

3

6EI i li

2

‰u …J†u …K†Š ‡

2EI i 2 ‰ …J† ‡ 2 …K† ‡ …J†…K†Š li

‰u …K†…K† ‡ u …K†…J†

ÿ u …J†…J† ÿ u …J†…K†Š

…9:5†

This expression for the strain energy relates to a typical member in the frame. With particular reference to the elements composing the example structure shown in Fig. 9.1 we can write Element B±A i

sin i

cos i

ux …J†

uy …J†

ux …K†

uy …K†

…J†

…K†

0

0

1

0

0

ux …A†

uy …A†

0

…A†

l 0BA ˆ l BA ‡ ux …A†

…9:6†

u …J† ˆ 0

…9:7†

u …K† ˆ ÿuy …A†

…9:8†

Matrix analysis of rigid-jointed framed structures

U BA ˆ

237

1 EABA 6EI BA 2 2 ‰u …A†Š ‡ ‰uy …A†Š x 3 2 l BA l BA ‡

2EI BA 2 6EI BA ‰ …A†Š ‡ ‰uy …A†…A†Š 2 l BA l BA

…9:9†

Element C±A i

sin i

cos i p 3 ÿ 2

ÿ308 ÿ 12

l

0CA

ˆl

ux …J†

uy …J†

ux …K† uy …K†

…J†

…K†

0

0

ux …A†

0

…A†

uy …A†

p 3 1 u …A† ÿ uy …A† ‡ 2 x 2

CA

…9:10†

u …J† ˆ 0

…9:11†

p 3 1 u …K† ˆ ÿ ux …A† ‡ …9:12† u …A† 2 2 y  p 2 1 EACA 1 3 CA u …A† ÿ uy …A† U ˆ 2 l CA 2 2 x p 2  6EI CA 1 2EI CA 3 uy …A† ‡ CA ‰2 …A†Š ‡ CA3 ÿ ux …A† ‡ 2 2 l l p      6EI CA 1 3 u …A† …A† ÿ ux …A† ‡ …9:13† ‡ CA2 2 2 y l Element D±A i

sin i p 2 ÿ 2

ÿ458

l

0DA

ˆl

cos i p 2 2 DA

ux …J†

uy …J†

ux …K† uy …K† …J†

…K†

0

0

ux …A† uy …A) 0

…A†

p p 2 2 u …A† ÿ u …A† ‡ 2 x 2 y

…9:14†

u …J† ˆ 0

…9:15†

p p 2 2 ux …A† ÿ u …A† u …K† ˆ ÿ 2 2 y

…9:16†

238

Energy methods in structural mechanics

U

DA

p 2  p 2 2 u …A† ÿ u …A† 2 x 2 y p 2  p 6EI DA 2EI DA 2 2 ‡ DA3 ux …A† ÿ uy …A† ‡ DA ‰2 …A†Š ÿ 2 2 l l p   p      6EI DA 2 2 ux …A† ÿ uy …A† …A† ÿ …9:17† ‡ DA2 2 2 l

1 EADA ˆ 2 l DA

Following the approach used in the analysis of the pin-jointed structure of the previous section, we will now write the total potential energy of the present frame which, using equations (9.9), (9.13) and (9.17) above, is evidently  ˆ U ÿ W ˆ U BA ‡ U CA ‡ U DA ÿ W ˆ

1 EABA 6EI BA 2 2EI BA 2 2 ‰u …A†Š ‡ ‰u …A†Š ‡ ‰ …A†Š x y 3 2 l BA l BA l BA p 2 3 6EI BA 1 EACA 1 ‡ BA2 ‰uy …A†…A†Š ‡ u …A† ÿ uy …A† 2 l CA 2 2 x l p     2 3 6EI CA 1 2EI CA uy …A† ‡ CA ‰2 …A†Š ‡ CA3 ÿ ux …A† ‡ 2 2 l l p      6EI CA 1 3 ‡ CA2 uy …A† …A† ÿ ux …A† ‡ 2 2 l p p 2 1 EADA 2 2 u …A† ÿ u …A† ‡ 2 l DA 2 x 2 y p  p 2 6EI DA 2EI DA 2 2 ux …A† ÿ uy …A† ‡ DA ‰2 …A†Š ‡ DA3 ÿ 2 2 l l p   p      6EI DA 2 2 u …A† ÿ u …A† …A† ÿ ‡ DA2 2 x 2 y l ÿ ‰Px ux …A† ‡ Py uy …A†Š

…9:18†

The energy is seen to be a function of the variables ux …A†, uy …A† and …A† only. As before, the values of the variables ux …A†, uy …A† and …A† that render the total potential energy stationary are yielded by the

Matrix analysis of rigid-jointed framed structures

239

ful®lment of the following extremum conditions p   3 q EABA EACA 3 ux …A†ÿ ˆ BA ux …A† ‡ CA uy …A† qux …A† 4 4 l l p       3 6EI CA 1 6EI CA 1 ux …A†ÿ uy …A† ‡ CA2 ÿ …A† ‡ CA3 2 2 2 l l   EADA 1 1 6EI DA ux …A† ÿ uy …A† ‡ DA3 ‰ux …A† ‡ uy …A†Š ‡ DA 2 2 l l p    DA  6EI 2 …A† ÿ Px ˆ 0 …9:19† ‡ DA2 ÿ 2 l q 6EI BA 6EI BA ˆ BA3 ‰2uy …A†Š ‡ BA2 ‰…A†Š quy …A† l l p   3 EACA 1 uy …A† ÿ ux …A† ‡ CA 4 4 l p    p  3 3 6EI CA 3 6EI CA uy …A† ÿ ux …A† ‡ CA2 …A† ‡ CA2 2 2 2 l l   EADA 1 1 6EI DA uy …A† ÿ ux …A† ‡ DA3 ‰uy …A† ‡ ux …A†Š ‡ DA 2 2 l l p     6EI DA 2 …A† ÿ Py ˆ 0 …9:20† ‡ DA2 ÿ 2 l q 2EI BA 6EI BA 2EI CA ˆ BA ‰2…A†Š ‡ BA2 ‰uy …A†Š ‡ CA ‰2…A†Š q…A† l l l p  CA  3 6EI 1 2EI DA ‡ CA2 ÿ ux …A† ‡ uy …A† ‡ DA ‰2…A†Š 2 2 l l p   p     6EI DA 2 2 ux …A† ÿ u …A† ˆ 0 …9:21† ‡ DA2 ÿ 2 2 y l The extremum conditions once again represent the equilibrium equations of the structure under consideration. The unique solution of this linear system of algebraic equations provides the magnitudes of the displacements of the node A along the x and y axes and its rotation. The substitution of these magnitudes in equations (4.122)±(4.125),

240

Energy methods in structural mechanics

(4.130)±(4.133), (4.7), (8.9), (9.1)±(9.3) give a complete description of the strains and the forces induced in each beam by the applied load.

9.2

Matrix formulation of the elastic equilibrium equations for plane rigid-jointed frames

Following the same line of reasoning employed in chapter 8, we now apply the direct sti€ness method of structural analysis to the rigidjointed frame shown in Fig. 9.1. The ®rst step in the development of the analysis involves identifying the overall degrees of freedom of the structure. In this case they are represented by both displacements and rotations of each joint with reference to the general frame x±y. The general displacement vector for the example frame is therefore a 12-element array, that is ‰DŠT ˆ ‰ ux …A† uy …A† …A† . . . . . . . . . ux …D† uy …D† …D†Š …9:22† while the vector containing the displacements of the end joints of the generic element is a 6-element array, i.e. 2 3 ux …J† 6 u …J† 7 6 y 7 6 7 6 7 …J† i 6 7 ‰D Š ˆ 6 …9:23† 7 u …K† 6 x 7 6 7 4 uy …K† 5 …K† As we have already seen in section 9.1, in order to take into account the elastic behaviour of the individual element it is necessary to superimpose the ¯exural sti€ness on the axial sti€ness. Therefore, in the local reference frame all the displacements of the joints are relevant to the model and we can write 3 2 u …J† 6 u …J† 7 7 6  7 6 7 6 …J† 7 …9:24† ‰d i Š ˆ 6 6 u …K† 7 7 6  7 6 4 u …K† 5 …K†

Matrix analysis of rigid-jointed framed structures

241

It can be easily shown for the generic beam that the relationship between the joint displacement vector ‰Di Š in the general reference frame and the joint displacement vector ‰d i Š in the local reference frame is governed by the following transformation matrix 2

cos i

6 6 sin i 6 6 6 0 i ‰R Š ˆ 6 6 6 0 6 6 0 4 0

sin i

0

0

0

ÿ cos i

0

0

0

0

1

0

0 i

sin i

0

0

cos 

0

0

sin i

ÿ cos i

0

0

0

0

0

3

7 07 7 7 07 7 7 07 7 07 5

…9:25†

1

so that ‰d i Š ˆ ‰Ri Š‰Di Š

…9:26†

Equation (9.26) has been derived bearing in mind equations (8.22), (9.1) and (9.2) and remembering that the slopes …J† and …K† are not a€ected in the transformation from the general reference frame x±y to the local reference frame ±. Notice that matrix (9.25) is now a 6  6 square array because it determines the transformation between two 6-element arrays, namely equations (9.23) and (9.24). (Matrix (8.20) was a 4  2 rectangular array in order to relate the 4element array in equation (8.21) to the 2-element array in equation (8.22).) A noteworthy property of matrix (9.25) is that it is orthogonal, i.e. its inverse coincides with its transpose as we have ‰Ri Šÿ1 ‰Ri Š ˆ ‰Ri ŠT ‰Ri Š ˆ ‰IŠ

…9:27†

where ‰IŠ is the unit matrix. Matrix (8.20) satis®es a similar condition but from a formal point of view we normally refer to inverse matrices only when dealing with square matrices. At this point we can relate the vector of the forces ‰ f i Š, that arise at the end points of the generic beam on account of the displacements ‰d i Š, to the displacements themselves by means of the sti€ness matrix of the beam ‰K i Š. This is accomplished by re-casting in matrix notation equations (4.122)±(4.125) and (4.130)±(4.133), and adding the axial sti€ness contributions contained in equation (8.28).

242

Energy methods in structural mechanics

We have therefore 2

2 EAi 7 6 li 7 6 7 6 7 6 7 6 0 7 6 7 6 7 6 7 6 7 6 7 6 0 7 6 7ˆ6 7 6 7 6 EAi 7 6ÿ i 7 6 l 7 6 7 6 7 6 7 6 0 7 6 7 6 7 6 5 4 0 3

6 F …J† 6 6 6 6 F …J† 6  6 6 6 6 6 M…J† 6 6 6 6 6 F …K† 6 6 6 6 6 F …K† 6  6 6 4 M…K†

0

0

12EI i l ÿ

i3

6EI i l

i2

0 ÿ

ÿ

ÿ

6EI i l

i2

0

0

EAi li

6EI i

i3

i2

ÿ

6EI i l

i2

0

4EI i li

12EI i l

EAi li

l

2EI i li

0 0

0 ÿ

12EI i li

3

6EI i li

2

0 12EI i li

3

6EI i li

2

32

3

76 u …J† 76 76 6 6EI i 7 6 ÿ i2 7 76 u …J† l 76 76 6 2EI i 7 76 76 …J† i 76 l 76 76 76 6 0 7 76 u …K† 76 76 6EI i 76 76 u …K† 2 76  li 76 76 4EI i 54 …K† li

7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 7 5

0

(9.28)

or, in compact form ‰ f i Š ˆ ‰K i Š‰d i Š

…9:29†

As the transformation matrix ‰Ri Š also governs the relationship between the nodal forces vector ‰ f i Š and ‰F i Š in the local and general reference frames, respectively, that is ‰ f i Š ˆ ‰Ri Š‰F i Š

…9:30†

Equation (9.30) can be rewritten as ‰F i Š ˆ ‰Ri ŠT ‰ f i Š ˆ ‰Ri ŠT ‰K i Š‰d i Š ˆ ‰Ri ŠT ‰K i Š‰Ri Š‰Di Š

…9:31†

Once again we can regard the term ‰K i Š ˆ ‰Ri ŠT ‰K i Š‰Ri Š

…9:32†

as the sti€ness matrix of the generic element in the general reference frame x±y. Its expression is displayed as equation (9.33). At this point the solution of our problem becomes straightforward: we can once more particularise equation (9.32) to obtain the sti€ness matrices of the elements B±A, C±A and D±A and, by identifying the position of the joints J and K in the general ordering system, we can assemble all the concurrent contributions in order to get the expression  namely of the overall joint sti€ness matrix ‰KŠ,  ˆ ‰K BA Š ‡ ‰K CA Š ‡ ‰K DA Š ‰KŠ

…9:34†

20 i 1 EA 2 i 6 B i cos  C 6B l C 6 @ 12EI i A 6 2 i 6 ‡ i3 sin  6 l 6 6 6 6 6 6 sym 6 6 6 6 6 6 6 6 sym 6 6 i  6 ‰K Š ˆ 6 6 6 6 6 sym 6 6 6 6 6 6 6 6 6 6 sym 6 6 6 6 6 6 4 sym sym

sym

sym

sym

sym

cos i

sym

2

sin i

4EI i li

li

i

2 li

6EI i

6EI

ÿ

sym

0 i 1 EA i i sin  cos  i B l C B C @ 12EI i A i i sin  cos  ÿ i3 l 0 i 1 EA 2 i sin  B li C B C @ 12EI i A 2 i ‡ i3 cos  l sin i

sym

sym

0 i 1 EA 2 i cos  B li C B C @ 12EI i 2 i A ‡ i3 sin  l

2 li

6EI i

1 EAi i i ÿ sin  cos  B li C B C @ 12EI i A i i ‡ i3 sin  cos  l

0

1 EAi 2 i ÿ i cos  B l C B C @ 12EI i 2 i A sin  ÿ 3 li

0

li

2

6EI i

cos i

sym

0 i 1 EA i i sin  cos  B li C B C @ 12EI i A i i ÿ i3 sin  cos  l 0 i 1 EA 2 i sin  B li C B C @ 12EI i A 2 i ‡ i3 cos  l

ÿ

1 EAi i i ÿ i sin  cos  B l C B C @ 12EI i A i i sin  cos  ‡ i3 l 0 1 EAi 2 i sin  ÿ B li C B C @ 12EI i A 2 i ÿ i3 cos  l

0 6EI i

2 li

7 7 sin i 7 7 7 7 7 7 7 7 7 6EI i i 7 cos  7 2 7 li 7 7 7 7 i 7 2EI 7 7 i 7 l 7 7 7 7 7 6EI i i 7 sin  7 2 7 li 7 7 7 7 7 7 i 7 6EI i7 ÿ i2 cos  7 7 l 7 7 7 7 5 4EI i i l …9:33† ÿ

3

Matrix analysis of rigid-jointed framed structures 243

244

Energy methods in structural mechanics

The load vector must be enlarged to a 12-element array to contain couples directly applied to the joints (these, however, in our particular example are absent). We have ‰PŠT ˆ ‰ Px …A† Py …A†

0 0

...

...

0 0Š

…9:35†

and from a formal point of view we have the recognisable system  ‰KŠ‰DŠ ÿ ‰PŠ ˆ 0

…9:36†

that in the present example is formed by 12 linear equations in the 12 unknown degrees of freedom ordered in the array ‰DŠ. Finally, we must take into consideration the restraints represented by the ®xed ends at B, C and D by setting to zero, as in the previous section, the terms corresponding to the inactive degrees of freedom, that is suppressing them. The reader will ®nd that the core of the system in equation (9.36) is then reduced to a system of three linear equations in the unknowns ux …A†, uy …A† and …A†, i.e. 3 2 3 2 Px …A† ux …A† 7 6 7 6 …9:37† ‰KI Š4 uy …A† 5 ˆ 4 Py …A† 5 …A†

0

which turns out to be identical to the system formed by equations (9.19), (9.20) and (9.21). On the same basis as in section 8.2, we ®nd that the matrix ‰KI Š is symmetric and positive de®nite. Once again the usefulness of the equilibrium equations in the form of equation (9.36) is evident: it can be directly applied to any number of non-aligned members making up a plane structural system (or, as we shall see later, a spatial structural system as well). The overall joint sti€ness matrix, equation (9.34) may always be obtained by referring the local sti€ness matrix of each member, equation (9.28), to the common system axes x±y and adding the corresponding contributions. By assigning appropriate boundary conditions at each joint, the system in equation (9.36) allows us to determine all the magnitudes of the unknown degrees of freedom and, consequently, strains and forces in each member. All this by means of a straightforward and practically automatic procedure. As in the case of the pin-jointed plane frame previously examined, from a physical point of view the set of equations under consideration requires that the sum of the elastic reactions from the end joints of each element balances the applied forces at any joint of the structure. In the present example, this is the force P at A.

Matrix analysis of rigid-jointed framed structures

9.3

Numerical example: analysis of a rigid jointed frame

245

The pin-jointed structure shown in Fig. 8.3 is an idealisation of the situation in which the connections of the members cannot transmit moment from one member to another. However, the majority of actual constructions would involve attaching the members to each other either by welding or bolting so that they indeed have the capacity to transfer moment. This condition is illustrated in Fig. 9.3 in which the members are welded to cover plates that ensure rigidity to the joint so that all the members meeting at that joint experience a common rotation. The structure can now be classi®ed as a plane frame, according to the de®nition in section 9.1. The objective in the present example is to exemplify the application of the direct sti€ness method and to compare the forces and deformation induced in the frame by the loads to the corresponding values for the truss. The direct sti€ness method, section 9.2, is, as for the truss, essentially concerned with assembling the sti€ness matrix of the complete structure from the component sti€ness matrices. Thus we use equations (9.25) and (9.28) to provide the component sti€ness matrices. This is exempli®ed below for member AB. Member AB i ˆ AB ˆ

 2 P2

P1

B C

A = 5·89 x 10 3 mm I = 4·61 x 10 3 mm

A

Fig. 9.3

D

246

Energy methods in structural mechanics 2

0:02 0 6 0 : 1 96 6 6 :71 6 ÿ30 0 ‰K AB Š ˆ E 6 6 ÿ0:02 0 6 6 4 0 ÿ1:96 ÿ30:71 0

ÿ30:71 0 6:14  104

ÿ0:02 0 30:71

0 ÿ1:96

0

0 30:71

1:96

30:71

3:07  104

0:02

0 0

0

3 ÿ30:71 7 0 7 7 4 3:07  10 7 7 30:71 7 7 7 5 0 4 : 6 14  10

this sti€ness matrix is associated with the displacement vector ‰DAB ŠT ˆ ‰ux …A†

uy …A† …A†

ux …B† uy …B† …B†ŠT

Preparation of similar sti€ness matrices for members BC, CD and AC can provide that basis for the sti€ness matrix for the complete frame. The equilibrium equations for the frame are then particularised by imposing the boundary constraint conditions, i.e. ux …A† ˆ uy …A† ˆ …A† ˆ 0 For the sake of brevity in the presentation of this example, these constraint conditions are taken into account in writing the component sti€ness matrices. Thus, for member AB, the sti€ness matrix, given the constraints at node A, is 2 : 3 0 02 0 30:71 6 7 ‰K AB Š ˆ E 4 0 1:96 0 5 4 : : 30 71 0 6 14  10 This component of the frame sti€ness matrix is now associated with the displacement vector ‰DAB ŠT ˆ ‰ux …B† uy …B†

…B†ŠT

Member BC i ˆ BC ˆ 0 2

1:96 6 0 6 6 6 0 BC ‰K Š ˆ E 6 6 ÿ1:96 6 6 4 0 0

Member CD i ˆ CD ˆ ÿ

 2

0 0:02 30:71 0 ÿ0:02 30:71

0 30:71 6:14  104 0 ÿ30:707 3:07  104

3 ÿ1:96 0 0 0 ÿ0:02 30:71 7 7 7 0 ÿ30:71 3:07  104 7 7 7 1:96 0 0 7 7 : 0 0:02 ÿ30 71 5 4 0 ÿ30:71 6:14  10

Matrix analysis of rigid-jointed framed structures

247

2 : 3 0 02 0 30:71 6 7 ‰K CD Š ˆ E 4 0 1:836 0 5 28:83 0 5:766  104 Member AC  4 2

i ˆ AC ˆ

3 0:70 0:69 10:86 7 6 0:70 ÿ10:86 5 ‰K AC Š ˆ E 4 0:69 10:86 ÿ10:86 4:34  104 Following equation (9.34), the sti€ness matrix for the frame, with the boundary constraints incorporated is  ˆ ‰K AB Š ‡ ‰K BC Š ‡ ‰K CD Š ‡ ‰K AC Š ‰KŠ The loading vector is similar to that for the truss taking account of the formulation of the sti€ness matrix 3 2 ÿ1 6 0 7 7 6 7 6 0 7 P1 6 7 6 ‰PŠ ˆ 7 E6 6 0 7 7 6 4 ÿ0:5 5 0 Thus the vector of the displacements of the unconstrained nodes can be obtained from  ÿ1 ‰PŠ ‰DŠ ˆ ‰KŠ The result has been calculated using the value of the load P1 evaluated for the truss, to give 3 3 2 2 ÿ3:95 mm ux …B† 6 u …B† 7 6 ÿ0:021 mm 7 7 7 6 6 y 7 7 6 6 6 …B† 7 6 0:93  10ÿ3 7 7 7ˆ6 6 6 u …C† 7 6 ÿ3:02 mm 7 7 7 6 6 x 7 7 6 6 4 uy …C† 5 4 0:45 mm 5 0:70  10ÿ4 …C†

248

Energy methods in structural mechanics

The axial loads in the members can be obtained from these displacements using equation (8.24), i.e. N AB ˆ uy …B† 

EA ÿ0:021  2:1  105  5:89  103 ˆ l0 3000

ˆ ÿ8:7 kN

EA l : : : …ÿ3 02 ‡ 3 95†  2 1  105  5:89  103 ˆ 383 kN ˆ 3000 EA N CD ˆ ÿ uy …C†  l : : ÿ0 45  2 1  105  5:89  103 ˆ ÿ185 kN ˆ 3000 1 EA N AC ˆ p …ux …C† ‡ uy …C††  p 2 2l : : : …ÿ3 021 ‡ 0 45†  2 1  105  5:89  103 ˆ ÿ530 kN ˆ 6000 In comparison with the corresponding results for the truss it may be seen that the de¯ections at the nodes and the axial forces in the members are slightly less for the frame than for the truss. This e€ect is due to the bending action in the members in the frame adding restraint to the actions of the forces to induce the de¯ections. However, for the slender members in this frame the e€ect of the bending actions are fairly small and it is evident that the analysis assuming the structure would act as a truss would be suciently accurate for engineering purposes. A more signi®cant e€ect of the structure acting as a frame is illustrated below. It is evident from the geometry of the structure modelled as acting as a truss that if the bracing member AC were to be damaged or severed the truss would collapse under the action of the loads. In e€ect the truss would deteriorate to act as a mechanism. However, the actual structure because of the bending sti€nesses of the members could withstand the e€ects of the loading, but at the penalty of increased de¯ections. Suppose for the frame the brace AC were removed, the sti€ness matrix for the structure would then be  ˆ ‰K AB Š ‡ ‰K BC Š ‡ ‰K CD Š ‰KŠ N BC ˆ …ux …C† ÿ uy …B†† 

with the boundary constraints again incorporated in the formulation  The loading vector is una€ected by this of the sti€ness matrix ‰KŠ.

Matrix analysis of rigid-jointed framed structures

249

change to the sti€ness matrix, i.e. 3 2 ÿ1 6 0 7 7 6 7 6 7 6 0 P 7 ‰PŠ ˆ 1 6 6 E6 0 7 7 7 6 4 ÿ0:5 5 0 Thus the vector of the modi®ed displacements is obtained from  ÿ1 ‰PŠ ‰DŠ ˆ ‰KŠ The result has been calculated using the value of the load P1 evaluated for the truss, to give 3 2 3 2 ux …B† ÿ74:07 mm 6 u …B† 7 6 ÿ0:54 mm 7 7 6 y 7 6 7 6 7 6 : 6 …B† 7 6 0 013 7 7 6 7ˆ6 6 u …C† 7 6 ÿ73:66 mm 7 7 6 x 7 6 7 6 7 6 4 uy …C† 5 4 0:062 mm 5 0:022 …C† Evidently, the structure is now much more ¯exible and de¯ects signi®cantly more due to the removal of the bracing member. The reader can con®rm that the loads in the members are N AB ˆ 223 kN;

N BC ˆ 169 kN

N CD ˆ 25:6 kN

Removal of the brace means that the actions of the loading is resisted primarily by the bending sti€ness of the members and this means that the axial loads induced in the members are signi®cantly reduced. Generally, for less slender structural members than considered in this example the e€ect of including the bending sti€ness will reduce the axial loads in the members compared to those calculated as if the structure is a truss.

9.4

Matrix analysis of space frames

Once one has understood the matrix treatment of space pin-jointed structures, i.e. space±trusses, dealing with the most general case of space frames becomes quite straightforward. In general a space- frame structure consists of a three-dimensional collection of rigidly-jointed beams, each of which is subjected to axial and transverse loading, as exempli®ed in Fig. 9.4.

250

Energy methods in structural mechanics q P

q P

q

Fig. 9.4

The key point of the direct sti€ness method consists, as before, in characterising the mechanical behaviour of a generic element in its local reference frame and successively in establishing the relationships that allow us to express the terms in the sti€ness matrices in the general reference frame. Once this is done, we can assemble the contributions from all the elements of the structure in the combined sti€ness matrix and solve the general system of equations with reference to the nodal loads (both directly applied and equivalent), provided the constraint conditions have been taken into account. In contrast to the case of pin-jointed structures, it is necessary to take into account the ¯exural and torsional behaviour of the generic i-th element. The vector collecting the displacements of the end nodes in the general reference frame is therefore enlarged to a 12elements array in order to encompass also the rotations about the axes x±y±z. Thus, we have ‰D i ŠT ˆ ‰ux …J† uy …J† uz …J† x …J† y …J† z …J† . . . . . . z …K†Š …9:38† In the local reference frame ±± it is ‰d i ŠT ˆ ‰u …J† u …J† u …J†  …J†  …J†  …J† . . . . . .  …K†Š …9:39† The relationship between ‰d i Šand ‰Di Š is governed by a 12  12 transformation matrix, ‰Ri Š, according to the usual product …9:40† ‰d i Š ˆ ‰Ri Š‰Di Š

Matrix analysis of rigid-jointed framed structures

The transformation matrix ‰Ri Š is given by 3 2 ‰LŠ ‰0Š ‰0Š ‰0Š 6 ‰0Š ‰LŠ ‰0Š ‰0Š 7 7 6 ‰Ri Š ˆ 6 7 4 ‰0Š ‰0Š ‰LŠ ‰0Š 5 ‰0Š

‰0Š

‰0Š

251

…9:41†

‰LŠ

where the sub-matrices ‰LŠ and ‰0Š are de®ned by 2 3 3 2 ix iy iz 0 0 0 6 i 7 7 6 i i 7 ‰LŠ ˆ 6 ‰0Š ˆ 4 0 0 0 5 4 x y z 5 0 0 0 ix iy iz

…9:42†

ik are direction cosines of angles between the local axes , ,  and the general axes x, y, z. It simple to verify that once again the transformation matrix (9.42) is orthogonal, i.e. ‰Ri ŠT ‰Ri Š ˆ ‰IŠ

…9:43†

The vector that contains the nodal forces in the local reference frame is a 12-element array. It includes the internal forces that arise at each end of the beam on account of the deformations, namely the axial forces, the shear forces along the axes  and , the bending moments about the axes  and  and the twisting moment about the axis , as shown in Fig. 9.5. Thus we have ‰ f i ŠT ˆ ‰F …J† F …J† F …J† M …J† M …J† M …J† . . . . . . M …K†Š …9:44† Mζ

Mη M ξ N Sξ

Sη Mζ

N Mξ

ξ

Fig. 9.5

Sξ

Mη

ζ η

Sη

252

Energy methods in structural mechanics

and in the local reference frame the relationship between the nodal forces ‰ f i Š and the nodal displacements ‰d i Š can be written in the following form ‰ f i Š ˆ ‰K i Š‰d i Š with

" ‰K i Š ˆ

‰ki11 Š

‰ki21 Š

…9:45† ‰ki12 Š

#

The sub-matrices kihk are de®ned by1 2 EAi 0 0 6 i 6 l 6 12EI i 6 0 6 0 3 6 li 6 6 12EI i 6 0 0 3 6 li ‰ki11 Š ˆ 6 6 6 0 0 0 6 6 6 6EI i 6 6 0 0 ÿ i2 6 l 6 4 6EI i 0 0 2 li 2 EAi 0 0 6 i 6 l 6 12EI i 6 0 6 0 3 6 li 6 6 12EI i 6 0 0 3 6 li ‰ki22 Š ˆ 6 6 6 0 0 0 6 6 6 6EI i 6 6 0 0 2 6 li 6 i 4 6EI 0 ÿ i2 0 l 1

…9:46†

‰ki22 Š 3 0

0

0

0

0 GJ i li

ÿ

6EI i li

2

0

0

4EI i li

0

0

0 7 7 6EI i 7 7 7 i2 7 l 7 7 0 7 7 7 7 0 7 7 7 7 7 0 7 7 i7 4EI 5

…9:47†

li 3

0 0 0 GJ i li

0 0 6EI i li

2

0

0

4EI i li

0

0

0

7 7 6EI 7 7 ÿ i2 7 l 7 7 7 0 7 7 7 7 0 7 7 7 7 7 0 7 7 7 4EI i 5 li i

…9:48†

Note that the relationship between the twisting moment, Mi , and the corresponding change of rotation, i , can be written as Mi ˆ …GJ=l i †i . J is the polar moment of area of the section of the structural member.

Matrix analysis of rigid-jointed framed structures

253

2

EAi 6 ÿ li 6 6 6 6 0 6 6 6 6 0 6 6 i i ‰k12 Š ˆ ‰k21 Š ˆ 6 6 6 0 6 6 6 6 0 6 6 6 4 0

3 0 ÿ

0

12EI li

i

0

3

0

ÿ

6EI i li

2

0

12EI i

0 0

0

l 0

ÿ

0

i3

ÿ

6EI i l

i2

0

GJ i li

0 0 6EI i li

2

0

0

2EI i li

0

0

0

7 7 7 6EI 7 ÿ i2 7 l 7 7 7 0 7 7 7 7 7 0 7 7 7 7 0 7 7 7 7 2EI i 5 li …9:49† i

Inspection of the member sti€ness matrix (9.46) reveals that it is symmetric, like all the sti€ness matrices encountered up to now. The subsequent procedure remains the same as in the previous cases: we can express the vector ‰F i Š that collects together the nodal forces of the generic beam in the general reference frame in the following form ‰F i Š ˆ ‰Ri ŠT ‰ f i Š

…9:50†

and from equations (9.40) and (9.45) we have …9:51† ‰F i Š ˆ ‰Ri ŠT ‰K i Š‰Ri Š‰Di Š ˆ ‰K i Š‰Di Š i  where ‰K Š is the sti€ness matrix of the generic beam in the general reference frame. At this point we can obtain the overall combined sti€ by assembling the contributions from all the beams and ness matrix ‰KŠ the elastic equilibrium equation takes the usual form, equation (9.36)  ‰KŠ‰DŠ ÿ ‰PŠ ˆ 0 Of course, as we have now included all the rotations of the joints in the degrees of freedom of the structure, ‰DŠ is a now 6n dimension array, n being the overall number of joints, and the dimension of the joint sti€ is 6n  6n. ‰PŠ is the load vector, i.e. a 6n dimension ness matrix ‰KŠ array containing both the components of the forces directly applied to the joints and the equivalent joint loads. The concept of equivalent joint load will be clari®ed in the next section.  is, It goes without saying that the combined sti€ness matrix, ‰KŠ, once more, symmetric (see section 9.2). Once the expression (equation (9.36)) is formulated, constraint conditions can be applied and the system of equations can be solved.

254

Energy methods in structural mechanics

It must be stressed that for space structures the level of computational e€ort tends to increase considerably with the number of elements and the use of a computer based automatic procedure becomes essential for the vast majority of practical cases. However, the analysis of many real cases can be successfully conducted by reducing three-dimensional problems to two-dimensional ones. This can be achieved by taking account of symmetries in the geometry and loading in the structure and treating the structure as an assemblage of plain frames.

9.5

Equivalent joint loads

At this point of the presentation we point out that in the examples of the sections 8.1±8.3 and 9.1±9.3 the self-weights of the individual members have been neglected. This is fairly common practice in the analysis and design of steel frames and trusses, where the loading and strains induced by the distributed self-weight are usually very small in comparison with the e€ects of the applied forces. However, for other types of structures these weights are not negligible and we should include them in the analysis. Moreover, not all external loads are actually applied at the joints and we may ®nd it necessary to take into account several types of lateral loading, as well as the e€ects of temperature changes and restraint displacements. As the matrix sti€ness method introduced in sections 8.2 and 9.2 uses the joint displacements as the degrees of freedom of the system, in order to reach the general matrix expression, equation (9.36),  ‰KŠ‰DŠ ÿ ‰PŠ ˆ 0 we have to replace the loads acting on the members by equivalent loads acting at the joints. These loads are called equivalent joint loads and are added to the actual joint loads to obtain the combined joint load vector in the same form of equations (8.45) or (9.35). Thereafter, the structure can be analysed using the matrix sti€ness method to evaluate the e€ect of the combined joint loads. Quite naturally, we can evaluate the vector of the equivalent joint loads making reference once again to the standard energy approach. According to the general condition of equilibrium, equation (1.28), this is easily accomplished by equating the work of the actual loads to the work of the required equivalent nodal forces for any virtual displacement of the degrees of freedom, i.e. the end nodes of the element. For example, let us consider the beam element shown in Fig. 9.6. This beam is subject to an evenly distributed load q acting in the direction  of the local reference frame ± and we want to determine the equivalent joint loads, i.e. those forces that are equivalent in energy

Matrix analysis of rigid-jointed framed structures

255

q

ζ

η

Fig. 9.6

to the actual load q with respect to the joint displacements. In other words, if ‰ f^ Š is the vector of the equivalent nodal forces acting on the beam in the local reference frame ±, we require that ^

W q ‡ W f ˆ 0

…9:52†

that is …l

q u …† d ‡ ‰d ŠT ‰ f^Š ˆ 0

…9:53†

0

As the displacement ®eld u …† at the left hand side of equation (9.53) is a function of the nodal displacement vector ‰dŠ by means of equation (9.3), and the same is obviously true if we take the variation of both terms, i.e. u …† and ‰dŠ, then equation (9.53) can be immediately solved with respect to ‰ f^Š and yields 2 3 0 6 ÿql=2 7 6 7 6 2 7 6 ql =12 7 7 ‰ f^Š ˆ 6 …9:54† 6 7 0 6 7 6 7 4 ÿql=2 5 ÿql 2 =12 Of course, equation (9.54) is the vector that contains the equivalent forces acting on the beam, and the equivalent joint loads vector ‰ p^Š is readily obtained by reversing these actions in direction. We have 3 2 0 6 ql=2 7 7 6 7 6 6 ÿql 2 =12 7 7 …9:55† ‰ p^Š ˆ 6 7 6 0 7 6 7 6 4 ql=2 5 ql 2 =12

256

Energy methods in structural mechanics

ql 2 12

ql 2 12 ql 2

–

ql 2

ql 2 12

ql 2 12 ql 2

ql 2

Fig. 9.7

With reference to Fig. 9.7, it is worth noting that the equivalent nodal forces ‰ f^Š coincide with the ®xed-end actions of the element, that is with the restraint reactions on the beam subject to the distributed load q and considered rigidly ®xed at both ends. This property applies to all types of lateral loading and it is useful to view the table in Fig. 9.8, where the actual and equivalent loading for some common types of applied loading are shown. It is also worth noting that, once we have added the equivalent joint loads to the actual joint loads in the combined joint load vector, the direct sti€ness analysis of the frame under consideration produces the same nodal displacements that we ®nd in the frame under the actual lateral loads. However, the displacements and the internal forces of the generic element must be obtained by superposing the displacements and the related internal forces that derive from this analysis to those corresponding to the ®xed-end element under the actual lateral loading. This follows from the fact that the procedure developed above for constructing the equivalent joint load vectors has been carried out by establishing the equivalence with respect to the equilibrium of the only degrees of freedom of the frame, i.e. the displacements of its joints. In order to illustrate the procedure with reference to a simple frame, consider the plane structure shown in Fig. 9.9. This frame is composed of two straight beams. The extremity A of the ®rst beam is ®xed, the node B is horizontally supported and the end C is free. Suppose that a constant distributed load qy acts on the beam A±B in the direction shown and that two vertical forces P1 and P2 act at the mid point of the span B±C and at the free end C, respectively. According to the direct sti€ness method, it is immediately recognised that overall we have nine degrees of freedom, i.e. the displacements ux

Matrix analysis of rigid-jointed framed structures

FA

FA w

MA

b

a

257

MB

FA = wa 1 – ( al )2 + 12 ( al )3 FB = wa ( al )2 – 12 ( al )3

l=a+b

2

MA = wa 12 – 16 ( al ) + 6 ( al )2 24 2

MB = wa 8 ( al ) – 6 ( al )2 24

Fb2 (3a + b)/l3

F

Fab2/l2

a

Fba2/l2

b

p

a

Fa2 (3b + a)/l3

pa (a + 2b)/2l

pa2/2l

b

Fig. 9.8

and uy and the rotation  of each joint. Only ®ve of them, i.e. uy …B†, …B†, ux …C†, uy …C†, …C†, are not prevented by the applied constraints. However, only the force P2 is applied directly to a joint. It is therefore necessary to ®nd out what the e€ects of the other loads are with respect to the degrees of freedom of the system. In order to do so we can restrain the structure against joint displacement by introducing imaginary joint restraints where necessary, as shown in Fig. 9.10. Next, the constraint reactions produced by the actual member loads (i.e. the loads that are not directly applied to the joints) are calculated. These reactions constitute the required set of ®xed end actions. By reversing these ®xed end actions in direction we obtain a set of forces and couples that we can consider equivalent to the member loads, that is the equivalent joint loads. This step is shown in Fig. 9.11.

258

Energy methods in structural mechanics q A

h

C B P2

P1 l/2

l/2

Fig. 9.9

The combined joint loads vector is obtained by adding the actual joint loads, which in this case is only the force P2 , to the equivalent joint loads, as shown in Fig. 9.12. It is straightforward to show that, once we have applied the direct sti€ness method to solve the situation pictured in Fig. 9.12, everything is known about the structure subject to the actual loads in Fig. 9.9. In fact, the superposition of the loads in Fig. 9.11(a) and the combined loads in Fig. 9.12 produces the loads in Fig. 9.9. At the same time the superposition of the displacement ®elds does not a€ect the joint displacements given by the sti€ness analysis of Figs 9.10 and 9.11(a), as the joint displacements of the frame shown in Fig. 9.10 must be zero.

Fig. 9.10

Matrix analysis of rigid-jointed framed structures qh2 12

259

qh 2 qh2 12

qh 2

q P1

P1

2

2

qh2 12

qh 2 qh2 12

P1l 8

P1l 8

P1

Pl

– 81

P1l 8

qh 2 P1

P1

2

2

(a)

(b)

Fig. 9.11

qh 2 qh2 12

qh2 12

qh 2

–

P1l 8

P1l 8

P1

P1

2

2

+ P2

Fig. 9.12

9.6

Restraint displacements

The e€ects of restraint displacements can be readily incorporated into the analysis of a framed structure by the direct sti€ness method. Generally, restraint displacements consist of known translations or rotation of the restraints. In order to illustrate this point with an example, we again refer to the framed structure of Fig. 9.1 and apply a vertical displacement u, at the ®xed node B. Such a displacement could be due to an inelastic yielding of the built-in support, as shown in Fig. 9.13. The analysis of such a condition can be e€ected easily by recalling the partitioned form, equation (8.50), of the general system equation

260

Energy methods in structural mechanics D

C

A

B u

P

Fig. 9.13

(8.49), i.e.   ‰KI Š ‰KI;II ŠT

‰KI;II Š ‰KII Š



‰DI Š ‰DII Š



 ˆ

‰PI Š



‰PII Š

This is an approach we have already considered in section 8.2. We recall that we can always rearrange the full set of equations of our problem to include in the sub-array ‰DI Š the degrees of freedom of those joints that are not restrained, and the degrees of freedom of the restrained joints are placed in the sub-array ‰DII Š. ‰PI Š and ‰PII Š are the sub-arrays in which the applied loads and constraints reactions are included, respectively. In the present example we have 2

ux …A†

3

6 7 7 ‰DI Š ˆ 6 4 uy …A† 5

…9:56†

…A† T

‰DII Š ˆ ‰ux …B† uy …B† …B† ux …C† uy …C† …C† ux …D† uy …D† …D†Š

and all the displacements of the nodes C and D are set to zero, while for the node B we have ux ˆ 0, uy ˆ ÿ u and  ˆ 0. Consequently, the system in equation (8.50) can be divided into the following two expressions ‰KI Š‰DI Š ˆ ‰PI Š ÿ ‰KI;II Š‰DII Š …9:57† ‰PII Š ˆ ‰KI;II ŠT ‰DI Š ‡ ‰KII Š‰DII Š

…9:58†

Matrix analysis of rigid-jointed framed structures

261

ui

–Ri

J

Fig. 9.14

The system (9.57) is, as before, disjoint from equation (9.58), as the sub-array ‰DII Š comprises known values of the displacement. Therefore, the problem of elastic equilibrium for the structure shown in Fig. 9.13 is reduced to the system, equation (9.57), of three equations for the three unknown degrees of freedom contained in the sub-array ‰DI Š, which are the components of displacement and the rotation of the loaded joint A. The substitution of these values in the relationships in equation (9.58) yields the value of the restraint reactions ‰PII Š. We can also take into consideration a linearly elastic displacement of the restraints, that is a displacement of the restraint that is proportional to the restraint reaction, for example ui …J† ˆ ÿcRi …J†

…9:59†

where c is a constant and Ri is the reaction of the restraint of a generic node J in the direction of the i-th axis, as shown in Fig. 9.14. This case can also be treated in a straightforward manner. In fact, it suces to notice that the reaction Ri …J† is represented in the array ‰PŠ as corresponding to the i-th component of the node J and perform the substitution expressed by equation (9.59). In this manner the i-th component of displacement of the restrained node J becomes an active degree of freedom of the structure and can be readily incorporated in the sub-array ‰DI Š in the partitioning of the general system equation (8.50). The rest of the procedure remains identically the same as presented above.

9.7

Some ®nal remarks on the matrix analysis of structures

The direct sti€ness method, introduced in chapter 8 and in this chapter for the analysis of trusses and frames, is highly suitable for automated computer procedures. Generally, a computer program can be conveniently divided into several steps, some requiring the input of the structural data from the operator, others concerning the manipulation of data and the computation of several quantities until the ®nal results

262

Energy methods in structural mechanics

are obtained. This book is intended to be a general introduction to energy methods that lead to the matrix methods, and therefore we do not deal with the subject of computer programming, not even with a simple example (plenty of excellent textbooks are totally devoted to this important topic). However, as a conclusion to the topic of frame and truss analysis it may be instructive to summarise the steps which are followed in an automatic procedure, as they are substantially the same as we will encounter later in our presentation of the ®nite element method. These steps are summarised as .

.

.

.

.

identi®cation of structural data. The geometrical and mechanical properties of the structure must be clearly identi®ed. In particular it is necessary to record the number of joints and their geometric coordinates, the number of members and their cross-section properties, the elastic properties of the material, the applied constraints and the inferred number of degrees of freedom. identi®cation of load related data. The joint loads and member loads acting on the structure under analysis must be clearly identi®ed. The member loads are subsequently modi®ed to obtain the value of the equivalent joint loads. These latter are then added to the directly applied joint loads and the problem is reduced to the equilibrium of a structure subject to nodal forces only. construction of sti€ness matrix. First, the individual member sti€ness matrices are calculated in the local reference frame of any element. Subsequently, they are transformed to the general reference frame and the contribution of each element is assembled in the combined sti€ness matrix of the problem. constraint application. The degrees of freedom that are suppressed by the constraints are eliminated in the general system of equations of equilibrium by deleting and rearranging the a€ected rows and columns in the matrices. calculation of results. The displacements of the joints are computed from the general system of equilibrium equations. The constraint reactions can then be simply obtained. Successively, the evaluation of the element strains and forces induced in the members proceeds on an individual basis: in particular, the member end-actions can be directly computed by means of the member sti€ness matrix, once the displacements of the joints are known, and the eventual member forces can be taken into account.

Matrix analysis of rigid-jointed framed structures 2m

263

1m 100 kN

A

B

C

1·5 m

D

Fig. 9.15

Suggested exercise problems 1.

Calculate the forces in the frame shown in Fig. 9.15 and the vertical de¯ection at B caused by the applied load if (a) (b)

the frame is considered ®rst as having pin-joint connections at A, B and C, see Fig. 9.16, and analysed as having all the members rigidly connected at these nodes, see Fig. 9.15. 2m

1m

B A

C

100 kN 1·5 m

D

Fig. 9.16

264

Energy methods in structural mechanics A

P

2P B

E

2m

C

F

2m

2m

2m

3m

D

Fig. 9.17

2.

3.

4.

All members are fabricated using steel tubes 300 mm outside diameter and 10 mm wall thickness. The modulus for steel is 205 kN/ mm2 . [Ans: pin-jointed
Av ˆ ÿ0:303 mm, NAB ˆ 33:6 kN, NBC ˆ 67.2 kN, rigid joints
Av ˆ ÿ0:193 mm, NAB ˆ 42 kN, NBC ˆ 84 kN, Mab ˆ 1:97 kNm] The continuous beam shown in Fig. 9.17 has two spans, AB with a ¯exural rigidity of I and BC with a ¯exural rigidity of 2I. It is welded to a rigid foundation at A and C and supported by an axially in®nitely rigid strut at B. The strut provides no rotational restraint to the horizontal beams at B. Calculate the de¯ection at the mid-length of both spans and calculate the maximum strain in the beam span AB if that member is composed of a length of hollow square channel, depth 200 mm and thickness 15 mm. The member BC is a hollow square section, depth 300 mm and uniform thickness. The modulus for steel is 205 kN/mm2 . Calculate the maximum allowable value of the load P if the maximum strain in the span AB is not to exceed the material yield strain, "0 ˆ 0:2% with a safety factor of 0:7. [Ans:
Ev ˆ 14:1 mm, "AB ˆ 8:6 mm, P ˆ 209 kN] Calculate the allowable load in the frame in Fig. 9.17 if the member DB has the same cross-section dimensions as member AB and cannot be considered to provide a rigid foundation at B. The member BD is rigidly connected at B to the horizontal members. [Ans: P ˆ 243 kN] An arch is constructed from straight steel members, each of length l and ¯exural rigidity EI, as shown in Fig. 9.18. The arch is attached to a rigid foundation at A and G and all the members are welded to each other so that it should be considered as a rigid-jointed frame. Calculate, taking account of symmetry as

Matrix analysis of rigid-jointed framed structures

265

P l

C

E D

B

30°

l/2

l/2

30°

F

l

A

60°

60°

G

Fig. 9.18

]

appropriate, the vertical de¯ection
D at the position of the applied load P. Calculate the magnitude and position of the maximum moment Mmax in the arch. Compare the moment and de¯ection with the corresponding value Mmax1 ,
D1 obtained from the Cotterill±Castigliano approach, chapter 6, considering the arch as a continuous curved structure. [Ans:
D ˆ 0:034…Pl 3 =EI†, Mmax ˆ 0:21Pl, Mmax1 ˆ 0:105Pl,
D1 ˆ 0:105…Pl 3 =EI†

10. Analysis of thin plates Introduction

Up to now in this book we have considered structures that are composed of one-dimensional linear elements. That is, the variation of de¯ections and forces are related only to the axis along the length of the elements. Particularly, the real three-dimensional nature of a beam is reduced to the one-dimensional variation by the assumption of linear strain through the depth of the beam section. There is, of course, a wider class of elements that are common in practical structures, that is plates and shells. Essentially these elements have two dimensions much greater than their third, which is called the thickness. The di€erence between plates and shells is that the former has an initially ¯at surface de®ned by the two main dimensions, lx and ly whereas shells have a surface that may be singly- or doubly-curved. In this introductory text, we consider only plates. Just as the analytical model of a beam is reduced to essentially a onedimensional model by a simplifying assumption so also a plate is reduced to a two-dimensional body with the aim of facilitating mathematical analysis. The assumption leading to this simpli®cation is identical to that for a beam; that is, the strain is assumed to vary linearly through the thickness of the plate. This is called the Kirchho€ model of plate deformation, named after the famous natural philosopher. It must be appreciated that this is an approximation to the real behaviour of plates, but within a restricted class of structural elements the model provides an adequate level of accuracy. The model will have reduced validity if, for example, the plate is composed of layers of material with di€erent moduli through the thickness, or the forces imposed on the plate in, or normal to its plane are very localised and the analysis is concerned with determining the strain distribution in the region of these loads. However, when we are concerned with the calculation of the general de¯ections of plates composed of homogeneous material and with a thickness much less than the other dimensions, then the Kirchho€ model serves very well and provides practical

268

Energy methods in structural mechanics

information at a decidedly acceptable level of accuracy, when compared with corresponding experimental results. The presentation in this chapter follows closely the pattern of the presentation of the Bernoulli beam theory in chapter 4. First, we will analyse the equilibrium of the plate subject to stretching loads, i.e. loads whose resultants always act in the mid-plane of the plate. Then we shall concern ourselves initially with the derivation of the expressions for the strain through the thickness, in terms of the de¯ection of the mid-plane surface of the plate. The expression for the strain energy in the plate material is developed for the condition where the plate mid-surface is deformed laterally in response to loads that generally are applied normal to the surface of the plate. We have two choices for attempting to determine the actual shape and magnitude of these de¯ections for speci®c values and patterns of loading. One approach is to apply the methods of calculus of variations to the energy functional to derive the di€erential equation of equilibrium, and then to attempt to ®nd a solution to this equation for speci®c forms of loading. This was the approach followed by some 18th and 19th century mathematicians and there is an interesting history of the search for the solutions that are reported in many textbooks. This approach o€ers also the possibility of elucidating the boundary conditions relevant to plate bending, which are not so obvious as in the case of bending beams. Inevitably, the solution of a fairly complex di€erential equation is dicult and in practice many of the solutions actually achieved relate to idealised plate shapes, boundary conditions and forms of loading. The practical cases of plate bending in engineering tend not to be so well conditioned and the solution of the di€erential equations of equilibrium is not feasible and so an alternative, and more widely applicable approach is to be found through the use of approximate energy methods, like the Rayleigh±Ritz procedure or the ®nite element method, which is developed later in this book. Once again, it is emphasised that the exact solutions to plate bending, albeit for idealised conditions, have a very valuable attribute in that they provide a basis for the calibration and validation for the more approximate solutions from the energy methods and the numerically based ®nite element models used in engineering practice. Of course, as in the case of beams, it is possible to obtain approximate solutions to the di€erential equation of equilibrium by using the method of ®nite di€erences. However, experience has shown that this method is not so conducive to automatic generation of the plate modelling and the boundary conditions as the ®nite element method so that this latter method is now the primary analysis tool available to structural engineers for plate analysis. For this

Analysis of thin plates

269

reason the ®nite di€erence method is not exempli®ed here and the main emphasis is on the approximate solution of equilibrium using the method of trial functions, including ®nite elements, applied to the energy functional. Nevertheless, for the sake of completeness, the differential equations for equilibrium of plates for in-plane and lateral loading, i.e. bending, are derived here.

10.1

Kirchho€ 's model of plate deformation

Let us start by establishing a mathematical model of the plate shown in Fig. 10.1. As was stated in the introduction, the plate is characterised by the fact that two of its dimensions, lx and ly , are generally much greater than its thickness hz . The mid surface of the plate is planar, and therefore we can think of it as being contained in the x±y plane. Actually, the Kirchho€ model of plate is rather intuitive and its correspondence with Bernoulli's model of bending beam is to some extent straightforward. In chapter 4 we assumed that irrespective of the deformation of the Bernoulli beam model, its cross sections must remain plane and normal to the curve representing the deformed axis of the beam. Now, as we are dealing with a two-dimensional solid, we can assume that, whatever is the deformation of the plate, the segments normal to its mid plane must remain straight and normal to the surface representing the deformed mid-plane of the plate. Thus, Kirchho€ 's model of a plate essentially incorporates a kinematic assumption regarding the manner in which the plate will deform. Of course, it must be emphasised once more that real plates

S Γ x y

z

Fig. 10.1

270

Energy methods in structural mechanics ux

φy x

z

Fig. 10.2

do not necessarily deform exactly in accordance with Kirchho€ 's assumptions. However, experience has shown that in most practical cases for isotropic materials the behaviour of practical plates is not far from the hypotheses stated above. Thus the results that we are going to obtain are usually acceptably accurate from a practical engineering viewpoint. As we could have anticipated, even if the basic assumptions of the model are very similar to those adopted for the bending beam, the fact that we are operating over two dimensions instead of one considerably complicates the development of the basic theory and obtaining practical solutions. Let us start from the analysis of the strain ®eld that occurs in the Kirchho€ model of plate deformation. The fact that the edges of segments normal to the mid plane, x±y, must remain straight and normal to the surface representing the deformed mid-plane yields ux …x; y; z† ˆ zy ‡ ux …x; y; z0 †

…10:1†

uy …x; y; z† ˆ zx ‡ uy …x; y; z0 †

…10:2†

uz …x; y; z† ˆ uz …x; y; z0 †

…10:3†

where ux …x; y; z0 † and uy …x; y; z0 † are the x and y components of displacement of the points lying on the mid-plane, z0  …z ˆ 0† and x and y are the angles of rotation along the x and y axes, respectively. The geometrical meaning of the term zy in equation (10.1) is depicted in Fig. 10.2.

10.2

Equilibrium of stretched plates

We start our analysis of the equilibrium of thin plates from the case of simply stretched plates, i.e. under the assumption that the resultants of the surface and boundary loads are contained in the mid-plane of the plate, as shown in Fig.10.3.

Analysis of thin plates

qx

qy q

271

Γ

x y S

Fx

Fy F

z

Fig. 10.3

The deformation of the plate, described by equations (10.1)±(10.2), can be simpli®ed in the following manner ux …x; y; z† ˆ ux …x; y; 0†

…10:4†

uy …x; y; z† ˆ uy …x; y; 0†

…10:5†

uz …x; y; z† ˆ 0

…10:6†

so that the strain ®eld of the deformed plate can be completely described by the relationships qux qx   1 qux quy ‡ ˆ 2 qy qx

"xx ˆ

…10:7†

"xy

…10:8†

"yy ˆ

quy qy

…10:9†

where "xx and "yy are the direct strains in the x and y directions, respectively, and "xy is the shear strain in the xy plane. Let us now consider an in®nitesimal element of plate, exempli®ed in Fig. 10.4. This slice has been obtained by means of four in®nitesimally close conceptual cuts: two in the z±x plane and two in the z±y plane. Resulting from the general theory of strain for linear elasticity, stated in chapter 3, the resultants of the internal forces per unit

272

Energy methods in structural mechanics

dx dy

z y x

Fig. 10.4

length are given by the relationships   E …" ‡ " † Nx ˆ h xx yy 1 ÿ 2   E Ny ˆ h …"yy ‡ "xx † 1 ÿ 2

…10:10† …10:11†

Nxy ˆ Nyx ˆ h…2G"xy †

…10:12†

These are a normal force Nx per unit length acting on the face of normal x, a normal force Ny per unit length acting on the face of normal y and ®nally the forces Nxy per unit length acting on both these faces. The situation is depicted in Fig. 10.5. On account of the above relationships the internal energy U stored in our model of stretched plate has the following expression … 1 Uˆ ‰Nx "xx ‡ Ny "yy ‡ 2Nxy "xy Š dS …10:13† 2 S

Nx Nxy Nyx Ny

Fig. 10.5

Analysis of thin plates

273

where the integral is performed over the surface S of the mid plane. At this point, in order to study the equilibrium of the plate, we can simply follow the usual procedure by writing the general condition (equation (1.28)) stated in chapter 1, that is U ˆ W To simplify the presentation here, we suppose the loading is a distributed load q…x; y†, acting in the mid-plane of the plate with components qx …x; y† and qy …x; y†, and a set of forces F per unit length on the boundary ÿ of the mid-surface, as shown in Fig. 10.3. Thus, the work done by the external forces is … … W ˆ …qx ux ‡ qy uy † dS ‡ …Fx ux ‡ Fy uy † dÿ …10:14† S

ÿ

and the condition of equilibrium (equation (1.28)) can be written as … ‰Nx "xx ‡ Ny "yy ‡ 2Nxy "xy Š dS S

…

…

S

ÿ

ˆ …qx ux ‡ qy uy † dS ‡ …Fx ux ‡ Fy uy † dÿ

…10:15†

On account of the general strain±displacement relationships equations (10.7)±(10.9), the left hand side of the above equation can be written as … ‰Nx "xx ‡ Ny "yy ‡ 2Nxy "xy Š dS S

… ˆ S

… ˆ S

  quy qux qux quy ‡ Ny ‡ Nxy ‡ dS Nx qx qy qx qy  q q …N u ‡ Nxy uy † ‡ …Ny uy ‡ Nxy ux † dS qx x x qy

    …  qNy qNxy qNx qNxy ‡ ‡ ux ‡ uy dS …10:16† ÿ qx qy qy qx S

and recalling the divergence theorem, i.e.  … … qfx qfy ‡ dS ˆ … fx x ‡ fy y † dÿ qx qy S

ÿ

…10:17†

274

Energy methods in structural mechanics

where x and y are the direction cosines of the normal to the boundary ÿ, we can write … ‰Nx "xx ‡ Ny "yy ‡ 2Nxy "xy Š dS S

… ˆ ‰…Nx ux ‡ Nxy uy †x ‡ …Ny uy ‡ Nxy ux †y Š dÿ ÿ

    …  qNy qNxy qNx qNxy ‡ ux ‡ ‡ uy dS …10:18† ÿ qx qy qy qx S

so that the equilibrium condition (equation (10.15)) becomes     …  qNy qNxy qNx qNxy ‡ ‡ qx ux ‡ ‡ ‡ qy uy dS qx qy qy qx S … ‡ ‰…Fx ÿ Nx x ÿ Nxy y †ux ÿ

‡ …Fy ÿ Ny y ÿ Nxy y †uy Š dÿ ˆ 0

…10:19†

As the virtual displacements ux and uy are arbitrary by de®nition, the vanishing of the above equation requires that qNx qNxy ‡ ‡ qx ˆ 0 qx qy Nx x ‡ Nxy y ˆ Fx

and

and

qNy qNxy ‡ ‡ qy ˆ 0 on S qy qx …10:20† Ny y ‡ Nxy y ˆ Fy

on ÿ

…10:21†

These are the equilibrium equations which relate the internal forces Nx , Ny and Nxy to the applied loads qx …x; y†, qy …x; y† and Fx , Fy in the case of simply stretched thin plates. It is worth noticing that the equations (10.20) and (10.21) are formally independent from the deformation of the plate.

10.3

Equilibrium of bent plates

As we have already done in the case of axially and laterally loaded beams, we again keep the stretching and the bending behaviour of the thin plates distinct. Therefore in the present section we deal with the pure bending of thin plates and suppose that the mid-plane of the plate does not undergo any stretching.

Analysis of thin plates

275

If we assume that no stretching of the mid-plane will occur, we must have ux …x; y; 0† ˆ uy …x; y; 0† ˆ 0

…10:22†

and, remembering the line of reasoning which led to equation (4.53), we can write quz qy qu y …x; y† ˆ ÿ z qx x …x; y† ˆ ÿ

…10:23† …10:24†

Equations (10.1) and (10.2) now become quz qx quz uy …x; y; z† ˆ ÿz qy ux …x; y; z† ˆ ÿz

…10:25† …10:26†

Finally, the strain ®eld of the deformed plate can be completely described by the relationships qux q2 u ˆ ÿz 2z qx qx   1 qux quy q2 uz ‡ ˆ ÿz ˆ 2 qy qx qx qy

"xx ˆ

…10:27†

"xy

…10:28†

"yy ˆ

quy q2 u ˆ ÿz 2z qy qy

…10:29†

where "xx and "yy are the direct strains in the x and y directions, respectively, and "xy is the shear strain in the xy plane. Let us now consider again the in®nitesimal element of the plate in Fig. 10.4. Resulting from the general theory of strain for linear elasticity, stated in chapter 3, the resultants of the internal forces per unit length are given by the relationships 

h=2 …

Mx ˆ ÿh=2

 E …"xx ‡ "yy † z dz 1 ÿ 2

h=2 …

ˆÿ ÿh=2



 2  E q uz q2 uz ‡  2 z dz qy 1 ÿ  2 qx2

…10:30†

276

Energy methods in structural mechanics h=2 …



My ˆ ÿh=2

 E …" ‡ " † yy xx z dz 1 ÿ 2 

h=2 …

ˆÿ ÿh=2

 2  E q uz q2 uz z dz ‡  qx2 1 ÿ  2 qy2

…10:31†

h=2 …

Mxy ˆ ÿMyx ˆ ÿ

…2G"xy †z dz ÿh=2

h=2 …

 2Gz

ˆ ÿh=2

 q2 uz z dz qx qy

…10:32†

These are a bending moment Mx acting on the face of normal x direction, a bending moment My acting on the face of normal y and ®nally what we will call a twisting moment Mxy on both these faces. The situation is depicted in Fig. 10.6.

My

Myx

Mx

Mxy

Mx y

x z

Mxy

Fig. 10.6

Analysis of thin plates

277

After a little mathematical manipulation, thanks to the assumption that the strain through the thickness of the plate varies linearly, equations (10.30)±(10.32) become  2  q uz q2 uz Mx ˆ ÿD ‡ 2 …10:33† qx2 qy  2  q uz q2 uz My ˆ ÿD ‡ 2 …10:34† qy2 qx Mxy ˆ D…1 ÿ †

q2 uz qx qy

…10:35†

The term D is called the bending sti€ness of the plate, that is D

Eh3 12…1 ÿ  2 †

…10:36†

It is worth noting that in the case of the Kirchho€ plate model the term D plays a role similar to that played by the term EIx for the Bernoulli beam model. We can now tackle the derivation of the equilibrium of the model. With reference to Fig. 10.6, it is clear that the internal moments Mx , My and Mxy are responsible for the energy stored in the in®nitesimal slice of plate on account of the work done by the e€ects of the curvatures x , y and xy . As we know from calculus (see also chapter 4), in the theory of in®nitesimal displacements, curvatures can be related to the normal de¯ection of the mid-plane by x ˆ ÿ

qx 1 q2 u ˆ ˆ ÿ 2z ry qy qy   1 qx qy q2 uz ‡ ˆ ˆ 2 qy qx qx qy

y ˆ ÿ xy

qy 1 q2 u ˆ ˆ ÿ 2z rx qx qx

…10:37† …10:38† …10:39†

As an example, the geometrical meaning of equation (10.38) is depicted in Fig. 10.7. On account of equations (10.37)±(10.39), the internal energy U stored in the Kirchho€ 's model of bent plate has the following expression … 1 Uˆ ‰Mx x ‡ My y ‡ 2Mxy xy Š dS …10:40† 2 S

278

Energy methods in structural mechanics

φy

rx

x

y z

Fig. 10.7

where the integral is performed over the surface S of the mid-plane. At this point, in order to study the equilibrium of the plate, we can simply follow the usual procedure by writing the general condition stated in chapter 1, that is U ˆ W In order to simplify the presentation here, we suppose that the loading is a distributed load qz …x; y†, which is directed normally to the plane x±y of the plate, and has a set of vertical forces F and bending couples  per unit length on the boundary ÿ of the mid surface, as shown in M Fig. 10.8. The work W done by the external forces is  … … quz quz    W ˆ qz uz dS ‡ ‡ Msn dÿ …10:41† Fuz ÿ Mn qn qs S

ÿ

 sn are the components of  n and M where, with reference to Fig. 10.9, M the generic bending couple along the normal n and the tangent s to the boundary ÿ, respectively.

Analysis of thin plates

279

S

qz

Γ x

y

M F

z

Fig. 10.8

Γ

s

n

Fig. 10.9

In the present case the general condition of equilibrium, see equation (1.28), can be therefore written as … ‰Mx x ‡ My y ‡ 2Mxy xy Š dS S

…

…

ˆ qz uz dS ‡ S

ÿ

 quz quz    ‡ Msn  dÿ …10:42† Fuz ÿ Mn  qn qs

Now, in order to illustrate what this condition of equilibrium really means in terms of the displacement ®eld uz …x; y†, some mathematics is required. Even if the procedure can appear lengthy, it is, however, worth paying attention to each step, because it constitutes a standard

280

Energy methods in structural mechanics

technique in the theory of deformable bodies and gives us a clear example of the diculties that can underlie even a simpli®ed model, like the Kirchho€ model for plate deformation. In this spirit, let us now proceed along our way. First of all, let us substitute the expressions, equation (10.37)± (10.39), for the curvatures into the left hand side of equation (10.42). Thus we obtain … ÿ S

 2   2   2  q uz q uz q uz dS Mx  ‡ My  ÿ 2Mxy  2 2 qxqy qx qy …

… ˆ qz uz dS ‡ S

ÿ

 quz quz    ‡ Msn  dÿ …10:43† Fuz ÿ Mn  qn qs

We now replace the variational operators by the di€erential ones, and vice versa, and get … ÿ

 q2 …uz † q2 …uz † q2 …uz † dS ‡ M ÿ 2M y xy qx qy qx2 qy2  … …  sn  quz dÿ …10:44†  n  quz ‡ M  zÿM ˆ qz uz dS ‡ Fu qn qs

Mx S

S

ÿ

Let us now consider the following relationships   q2 …uz † q q…uz † qMx q2 Mx M ÿ u Mx ‡ u ˆ x z z qx qx qx qx2 qx2

…10:45†

  q2 My q2 …uz † q q…uz † qMx M ÿ u ‡ u ˆ y z z qy qy qy qy2 qy2

…10:46†

My

    qMxy q2 …uz † q q…uz † q ˆ Mxy ÿ uz Mxy qxqy qx qy qy qx ‡ uz

q2 Mxy qx qy

…10:47†

Bearing in mind that in equation (10.47) the x and the y coordinate can be substituted one for the other, on account of the symmetry, we can substitute the relationships in equations (10.45)±(10.47) in the left

Analysis of thin plates

281

hand side of equation (10.44) and obtain  … q2 …uz † q2 …uz † q2 …uz † dS ÿ Mx ‡ My ÿ 2Mxy qxqy qx2 qy2 S

…

ˆÿ S

… ‡ S

 q2 Mxy q2 My q2 Mx ‡ uz dS ÿ 2 qx qy qx2 qy2    q qMx qMxy uz ÿ qx qx qy

   qMxy qMy q uz ÿ ‡ dS ‡ qy qx qy  …  q q…uz † q…uz † Mx ÿ Mxy ÿ qx qx qy S

  q q…uz † q…uz † ÿMxy ‡ My dS ‡ qy qx qy Recalling the divergence theorem, i.e.  … … qfx qfy ‡ dS ˆ … fx x ‡ fy y † dÿ qx qy S

…10:48†

…10:49†

ÿ

where x and y are the direction cosines of the normal to the boundary ÿ, we can apply it to the last two integrals of equation (10.48). Thus  … q2 …uz † q2 …uz † q2 …uz † ÿ Mx dS ‡ M ÿ 2M y xy qx qy qx2 qy2 S

…

ˆÿ S

 q2 Mxy q2 My q2 Mx ‡ uz dS ÿ2 qx qy qx2 qy2 

…

‡ uz ÿ

    qMxy qMy qMx qMxy ÿ x ‡ ÿ ‡ y dÿ qx qy qx qy

…  Mx

ÿ ÿ

 q…uz † q…uz † ÿ Mxy x qx qy

   q…uz † q…uz † ‡ My y dÿ ‡ ÿMxy qx qy

…10:50†

282

Energy methods in structural mechanics

This expression can still look quite complicated, however, if we de®ne     qMxy qMy qMx qMxy …10:51† ÿ x ‡ ÿ ‡ y Sn ˆ qx qy qx qy and remember from calculus that quz quz qu ˆ  ÿ z qx qn x qs y quz quz qu ˆ  ‡ z qy qn y qs x we have  …  q…uz † q…uz † ÿ Mxy x Mx qx qy ÿ    q…uz † q…uz † ‡ My y dÿ ‡ ÿMxy qx qy … q…uz † q…uz † …Mx 2x ÿ 2Mxy x y ‡ My 2y † ‡ ˆ qn qs ÿ  2 2
‰…My ÿ Mx †x y ÿ Mxy …x ÿ y †Š dÿ  … q…uz † q…uz † ÿ Msn dÿ Mn ˆ qn qs

…10:52† …10:53†

…10:54†

ÿ

On account of the transformation rules that link the moments Mn , Msn , Mx , My and Mxy (see Fig. 10.10), i.e. Mn ˆ Mx 2x ÿ 2Mxy x y ‡ My 2y Msn ˆ …Mx ÿ My †x y ‡ Mxy …2x ÿ 2y †

…10:55† …10:56†

we can ®nally write the condition of equilibrium, equation (10.42), in the form  … 2 q2 Mxy q2 My q Mx ÿ ‡ uz dS ÿ2 qxqy qx2 qy2 S  … q…uz † q…uz † ‡ Msn dÿ Sn uz ÿ Mn ‡ qn qs ÿ  … … quz quz    ˆ qz uz dS ‡ ‡ Msn  dÿ …10:57† Fuz ÿ Mn  qn qs S

ÿ

Analysis of thin plates

283

Myx

0 Mx

αx

My

αy

x

Mxy Msn Mn n y

Fig. 10.10

This expression is more appropriate to our purposes and the ®nal step in this development consists of grouping the terms, so that  … 2 q2 Mxy q2 My q Mx ‡ ÿ2 ‡ qz uz dS qxqy qx2 qy2 S

… …  sn ÿ Msn † q…uz † dÿ ‡ …F ÿ Sn †uz dÿ ‡ …M qs ÿ

…

ÿ

 n † q…uz † dÿ ˆ 0 ‡ …Mn ÿ M qn

…10:58†

ÿ

As the virtual displacement ®eld uz …x; y† is arbitrary, by de®nition, the vanishing of this variational expression implies some clearly de®ned conditions. First of all we must have q2 Mxy q2 My q2 Mx ‡ ÿ2 ‡ qz ˆ 0 qx qy qx2 qy2

on S

which, by means of equation (10.33)±(10.35), becomes  4  q uz q4 uz q4 uz D ‡ 2 2 2 ‡ 4 ˆ qz on S qx4 qx qy qy

…10:59†

…10:60†

The requirement of equilibrium of the plate under the assigned loads for the displacement ®eld uz …x; y†which de®nes the deformed con®guration of the Kirchho€ model means the satisfaction of this di€erential

284

Energy methods in structural mechanics

equation . As may be noted, there is a strong resemblance between this equilibrium condition, equation (10.60), and the corresponding equilibrium condition for the Bernoulli beam model, i.e. Euler's equation of the deformed beam, equation (4.43). Both are fourth-order di€erential equations (even if, of course, equation (10.60) involves partial derivatives as a consequence of the two-dimensional character of the plate model) and involve the ¯exural sti€ness EIx and D, respectively, for the beam and the plate bending. This is not surprising, since the kinematic hypotheses for the models, i.e. linear variation of strain through the depth of the plate and beam, are essentially the same in both cases. However, in contrast to the case of Bernoulli's beam, not all the boundary conditions associated to the di€erential equation (10.60) are obvious from a mechanical point of view. In fact, the vanishing of the general condition of equilibrium in the form of equation (10.58) requires also the vanishing of its last integral. This happens if quz q…uz †  n on ÿ ˆ 0 or Mn ˆ M ˆ n ) …10:61† qn qn This requires that the rotation quz =qn on the boundary ÿ coincides with the imposed value  or, alternatively, the bending moment Mn  n . Moreover, the vanishing of the coincides with the imposed value M second and third integral in equation (10.58) must be treated in conjunction. In fact the derivative q…uz †=qs along the tangent to the boundary ÿ is linked to the variation uz along the boundary itself. Integrating the third integral by parts we get … …  sn ÿ Msn †uz Š dÿ  sn ÿ Msn † q…uz † dÿ ˆ q ‰…M …M qs qs ÿ

ÿ

ÿ

 …  qMsn qMsn ÿ uz dÿ ÿ qs qs

…10:62†

ÿ

where the ®rst integral at the right hand side is obviously zero. In fact, this term is the line integral along a closed path (i.e. the edge of the plate) of a perfect di€erential. Combining this result with the second integral of equation (10.58), we get …  sn qMsn  qM  ‡ uz dÿ ˆ 0 F ÿ Sn ÿ …10:63† qs qs ÿ

This statement leads to the boundary conditions  sn qMsn qM ˆ F ÿ on ÿ uz ˆ uz or Sn ÿ qs qs

…10:64†

Analysis of thin plates [ Msn +

∂Msn ∂s ]

285

ds

Msn ds ∂M

Msn + ∂ssn Msn ds ds

ds ds ∂M

–Sn = ∂ssn

ds ds

Fig. 10.11

The meaning of the de®nition in equation (10.51) is now evident: the term Sn represents the shear forces normal to the mid-plane x±y. Its existence on the boundary ÿ is strictly related to the twisting moment Msn , as shown in Fig. 10.11, and this explains the mechanical condition expressed by the second alternative of equation (10.64). In other words, if we represent the distribution of twisting moments Msn along the edge of the plate by means of a continuous set of twisting couples, we are led to the conclusion that the gradient of the distribution of twisting moments Msn is statically equivalent to a distribution of shear forces of intensity Sn ˆ qMsn =qs. In closing this presentation it is useful to summarise all the equations that have been yielded by the general condition of equilibrium, equation (10.42), that must be ful®lled by the displacement ®eld uz …x; y†  4  q uz q4 uz q4 uz Field equation on S D ‡ 2 2 2 ‡ 4 ˆ qz qx4 qx qy qy Boundary condition on ÿ

uz ˆ uz

or Sn ÿ

 sn qMsn qM ˆ F ÿ qs qs

286

Energy methods in structural mechanics

Boundary condition on ÿ

ÿ

quz ˆ n qn

n or Mn ˆ M

Of course the mechanical boundary conditions can also be expressed in terms of the displacement ®eld. Thus, because of equations (10.33)± (10.35), (10.51), (10.55) and (10.56), we can write  3  3    q uz q3 uz q uz q3 uz Sn ˆ ÿD ‡ ‡ y x ‡ …10:65† qx3 qx qy2 qy3 qx2 qy  2  q uz q2 uz 2 q2 uz Mn ˆ ÿD   ‡  ‡ 2…1 ÿ †  x qx qy x y qx2 qy2    2 q uz q2 uz 2 y …10:66† ‡ ‡  qy2 qx2  2   2  q uz q2 uz q uz q2 uz Msn ˆ ÿD ‡ 2 ÿ ‡  2 x y qx2 qy qy2 qx ‡ D…1 ÿ †

q2 uz 2 … ÿ 2y † qx qy x

…10:67†

It is now clear that the problem of the equilibrium of the plate bending model is not simple, even for the example of an evenly distributed load qz …x; y†. Several special solutions have been developed for speci®c problems of plate bending. However, the study of these solutions is not really the purpose of this book and the interested reader can make reference to specialised texts. We consider that what has been said is sucient to elucidate the fundamentals of the problem. In the next section we will deal with a more practical and general procedure that is developed by making use of the energy approach with the method of trial functions.

10.4

An example of the method of trial function for the solution of bending plates.

It is apparent that the solution of the derived di€erential equation describing the elastic equilibrium does not constitute a simple task even for the simpli®ed model of bending plates. From an engineering point of view it therefore appears very worthwhile to make reference to all those methods capable of providing approximate solutions. Among the ways we have already studied is the method of trial function, which we will now apply to the evaluation of the response of the example plate shown in Fig. 10.12 to the speci®ed applied loading.

Analysis of thin plates

287

l

x

y

l

Fig. 10.12

The ®gure shows a square plate simply supported along two edges and rigidly clamped along the other two, subject to an evenly distributed load qz …x; y†. Its thickness is constant, as well as its bending sti€ness D. To be in a position to apply the trial function method, ®rst of all we must write down the expression for the total potential energy of the system in terms of the displacement ®eld, uz …x; y†. This is simply achieved by means of equations (10.40) and (10.41), and taking into account equations (10.33)±(10.35) and (10.37)±(10.39). We thus have …uz † ˆ U ÿ W …  2 2  2 2 1 q uz q uz q2 uz q2 uz ˆ D ‡ ‡ 2 2 qx2 qy2 qx2 qy2 S

 2 2  q uz dS ‡ 2…1 ÿ † qx qy  … … quz quz    ‡ Msn dÿ ÿ qz uz dS ÿ Fuz ÿ Mn qn qs S

…10:68†

ÿ

For the example plate being analysed here, the boundary conditions are essentially geometrical     l l …10:69† uz ; y ˆ 0 and uz ÿ ; y ˆ 0 2 2     l l uz x; ˆ 0 and uz x; ÿ ˆ0 …10:70† 2 2     quz quz ˆ 0 and ˆ0 …10:71† qy x; y ˆ l=2 qy x; y ˆ ÿl=2

288

Energy methods in structural mechanics

Since there are no forces applied on the edges, the last term of equation (10.68) is zero, so we have …uz † ˆ U ÿ W …  2 2  2 2 1 q uz q uz q2 uz q2 uz ˆ D ‡ ‡ 2 2 qx2 qy2 qx2 qy2 S

 2 2  … q uz dS ÿ qz uz dS ‡ 2…1 ÿ † qx qy

…10:72†

S

Moreover, the fact that the boundary conditions, equations (10.69)± (10.71) are not dependent on Poisson's ratio  (as they would have been in the case of expressions (10.66) and (10.67) for the bending and twisting moments were involved on the edge) leads us to conclude that the solution cannot be dependent on this ratio, as D is also constant. Therefore, if we rearrange equation (10.72) in the following manner …uz † ˆ U ÿ W 1 ˆ D 2

… S

…
S

q2 uz q2 uz ‡ 2 qx2 qy

2

dS ÿ D…1 ÿ †

 2 2  … q2 uz q2 uz q uz dS ÿ ÿ qz uz dS qx qy qx2 qy2

…10:73†

S

we can establish that the term  2 2  … 2 q uz q2 uz q uz dS ÿ 2 2 qx qy qx qy

…10:74†

S

will vanish and we have 1 …uz † ˆ U ÿ W ˆ D 2

… S

 … q2 uz q2 uz 2 ‡ dS ÿ qz uz dS qx2 qy2

…10:75†

S

It can be rigorously demonstrated that this simpli®cation holds for all the plates rigidly supported on the boundary and for all polygonal plates whose boundary conditions are either uz ˆ 0 or quz =qn ˆ 0. We now assume a shape trial function that satis®es the geometrical boundary conditions equations (10.69)±(10.71). A general format for a

Analysis of thin plates

289

suitable trial function is uz ˆ a1 f …x†g…y†

…10:76†

with the following expressions for …x† and g…y† f …x† ˆ 1 ÿ

4x2 l2

…10:77†

8y2 16y4 g…y† ˆ 1 ÿ 2 ‡ 4 l l Thus we have    4x2 8y2 16y4  uz ˆ a1 1 ÿ 2 1ÿ 2 ‡ 4 l l l

…10:78†

This function satis®es the geometrical boundary conditions. We can now substitute it in the expression for the total potential energy functional, equation (10.75). Thus   q2 uz 8 8y2 32y4 ˆ ÿa1 2 1 ÿ 2 ‡ 4 …10:79† qx2 l l l    q2 uz 4x2 16 192y2 1 ÿ ÿ …10:80† ˆ a ‡ 1 qy2 l2 l4 l2 and …uz †

1 ˆ D 2

l=2 …

l=2 …

ÿl=2 ÿl=2

64a21 l 12

4 2 2 2 2 2 2 4 2
…ÿ3l ‡ 8l x ‡ 32l y ÿ 96x y ÿ 16y † dx dy l=2 …

l=2 …

ÿ ÿl=2 ÿl=2

‡

  4x2 8y2 32x2 y2 q a1 1 ÿ 2 ÿ 2 ‡ l l l4

16y4 64x2 y4 ÿ l4 l6

 dx dy

…10:81†

After evaluation of the integrals equation (10.81) is 1 32768 D 16 ÿ a1 ql 2 …a1 † ˆ a21 2 175 l 2 45

…10:82†

Imposing the stationary condition on this expression of the total

290

Energy methods in structural mechanics

potential energy, to satisfy equilibrium, we ®nally achieve q ˆ0 qa1

)

a1 ˆ

35 ql 4 18432 D

…10:83†

The de¯ection of the plate can be obtained by substituting equation (10.83) in equation (10.78). At the centre point of the plate, we have 4

ql uz …0; 0† ˆ 0:00190 …10:84† D This value is quite accurate. In fact the analytical solution1 to the di€erential equation of equilibrium yields 4

ql uz …0; 0† ˆ 0:00192 …10:85† D We therefore have a solution from the method of trial function with an error of only 1:1%.

10.5

More examples of application of simple trial functions to plate bending

The previous section has shown an example of the application of the method of trial functions for calculating the magnitude of the de¯ections of plate deformed by loads normal to their plane. The present section expands on that example and exempli®es the simplicity of the method by application to a number of types of boundary conditions and loading. With respect to the general de®nition of the plate geometry shown in Fig. 10.13, the energy formulation equation (10.75) is written D …uz † ˆ 2

…a …b  0 0

q2 uz q2 uz ‡ 2 qx2 qy

2

…a …b dx dy ÿ

q…x; y†uz dx dy …10:86† 0 0

Plates subjected to uniform pressure

Three examples of boundary conditions are shown in Fig. 10.14, varying from the situation where all the edges are simple supported to the condition where all the edges are encastreÂ. The applied loading is a uniform pressure, i.e. q…x; y† ˆ q0

1

…10:87†

Timoshenko, S. P. and Woinowsky-Kreiger, S. (1959), Theory of plates and shells, McGraw-Hill, chapter 6.

Analysis of thin plates

291

a

x

y b

Fig. 10.13

Case (a): All edges simple supported The boundary conditions for this case are uz ˆ 0

at

x ˆ 0;

x ˆ a;

y ˆ 0;

yˆb

…10:88†

From section 7.1, it is reasonable to employ the one-degree-of-freedom trial function in the form     x y  sin …10:89† uz …x; y† ˆ u0 sin a b This function satis®es the geometry boundary conditions along all edges. Substituting equation (10.89) into equation (10.86) and performing the mathematical operations we have …u0 † ˆ

4 D 4 4q ‰a ‡ 2a2 b2 ‡ b4 Šu20 ÿ 20 abu0 3 3  8a b

…10:90†

Applying the condition of equilibrium, i.e. q…u0 † ˆ0 qu0

…10:91†

we have the relationship between the magnitude of the de¯ection and the applied pressure as   16q0 a4 b4 …10:92† u0 ˆ 6  D a4 ‡ 2a2 b2 ‡ b4

292

Energy methods in structural mechanics

Fig. 10.14

Case (b): Two edges simple supported and two edges encastre The boundary conditions for this case, see Fig. 10.14(b), are uz ˆ 0

at x ˆ 0;

x ˆ a;

quz ˆ0 qy

at y ˆ 0;

yˆb

y ˆ 0;

yˆb …10:94†

An appropriate trial function satisfying the boundary conditions for this case is    u x 2y 1 ÿ cos …10:95† uz …x; y† ˆ 0 sin a b 2

Analysis of thin plates

293

Substituting equation (10.95) into equation (10.86), carrying out the mathematical operations and applying the equilibrium condition, equation (10.91), we have   16q0 a4 b4 u0 ˆ 3 …10:96†  D 16a4 ‡ 8a2 b2 ‡ 3b4 Case (c): All edges encastre The boundary conditions for this case, see Fig. 10.14(c), are uz ˆ 0

at x ˆ 0;

x ˆ a;

y ˆ 0;

quz ˆ0 qx

at x ˆ 0;

x ˆ a;

quz ˆ0 qy

yˆb at y ˆ 0;

yˆb …10:97†

An appropriate trial function for this case is    u0 2x 2y  uz …x; y† ˆ 1 ÿ cos 1 ÿ cos a b 4

…10:98†

Substituting equation (10.98) into equation (10.86), carrying out the mathematical operations and applying the equilibrium condition, equation (10.91) we have   q0 a4 b4 u0 ˆ 4 …10:99†  D 3a4 ‡ 2a2 b2 ‡ 3b4 The accuracy of these solutions can be assessed by comparison with the exact values for the de¯ection for a square plate, i.e. a ˆ b ˆ l. The general formulation is u0 ˆ 

q0 l 4 D

…10:100†

and Case a b c

exact 0:00406 0:00192 0:00126

trial function 0:00416 0:00191 0:00128

As the table shows the solutions from the one degree of freedom trial function analysis are within a few percent of the corresponding exact solutions.

294

Energy methods in structural mechanics

P

a/2 a

x

y b/2 b

Fig. 10.15

Plates subjected to central concentrated load

A particular merit of the method of trial functions is that the analysis can very easily accommodate a range of types of loading without requiring extensive re-analysis. This is exempli®ed here by considering the condition where the uniform pressure of the previous section is replaced by a concentrated load P acting at the centre of the plate normal to its plane. This is shown in Fig. 10.15. The potential energy functional, equation (10.75), is for this form of loading D …uz † ˆ 2

…a …b  0 0

q2 uz q2 uz ‡ 2 qx2 qy

2 dx dy ÿ Pu0

…10:101†

where u0 is the de¯ection at the position of application of the load. Evidently the strain energy component of the potential energy is unchanged by the type of loading. Consider now a plate with all edges simply supported (Case a) in which the trial function is described by equation (10.89) and the energy functional in terms of the de¯ection u0 is …u0 † ˆ

4 D 4 ‰a ‡ 2a2 b2 ‡ b4 Šu20 ÿ Pu0 8a3 b3

…10:102†

Thus applying the equilibrium condition equation (10.91) to equation (10.102)   4P a4 b4 u0 ˆ 4 …10:103†  D a4 ‡ 2a2 b2 ‡ b4

Analysis of thin plates

295

The corresponding relationships for the boundary conditions described in Cases b and c are   16P a4 b4 Case b u0 ˆ 4  D 16a4 ‡ 8a2 b2 ‡ 3b4 …10:104†   4P a4 b4 Case c u0 ˆ 4  D 3a4 ‡ 2a2 b2 ‡ 3b4 The conditions for a square plate, a ˆ b ˆ l, are simply obtained from equations (10.103) and (10.104) and are u0 ˆ 

Pl 2 D

…10:105†

where Case a b c

10.6

exact 0.0116 0.0056

 0:0103 0:00608 0:00513

Localised Rayleigh±Ritz method applied to plate bending

The method of localised trial functions was introduced in chapter 7 and the concept of these trial functions was applied to the calculation of the deformation of beams. It is intuitively obvious that this method can be extended to the analysis of plate bending by the derivation of localised trial functions in the two dimensions of the plane of a plate. The method is described in this section. The localised trial function that ensures continuity of displacement and slope at the boundaries of the elements of a beam is described by equation (7.78) and is de®ned here as uny …i ÿ 1 † ˆ ui ÿ 1 …1 ÿ 3i2ÿ 1 ‡ 2i3ÿ 1 † ‡ i ÿ 1 …ÿi ÿ 1 ‡ 2i2ÿ 1 ÿ i3ÿ 1 † ‡ ui …3i2ÿ 1 ÿ 2i3ÿ 1 † ‡ i …i2ÿ 1 ÿ i3ÿ 1 †

…10:106†

where, as shown in Fig. 10.16, the non-dimensional coordinate is i ÿ 1  zi ÿ 1 =ln , ln is the length of the n-th element, uz is the de¯ected shape normal to the coordinate , and u i ÿ1 and ui are the de¯ections at the nodes bounding the n-th element. Again, we de®ne i ÿ 1  i ÿ 1 ln and i  i ln .

296

Energy methods in structural mechanics ln

z ui–1 ui

φi–1 φi zi–1 uy

Fig. 10.16

The one-dimensional function can be extended to provide a set of two-dimensional functions that can be used as localised trial functions to analyse plate bending for a variety of conditions of loading and boundary conditions. The plate is sub-divided into zones, or elements, each of which can have di€erent lengths in the x and y directions, as shown in Fig. 10.17. The orthogonal functions describing the deformations of an element n, shown in Fig. 10.18, are:

x

Element n i + 1, j i + 1, j + 1 lxn i, j

i, j + 1

y lyn uz

Fig. 10.17

Analysis of thin plates i + 1, j + 1

i + 1, j

ξn

297

Element n

i, j

i, j + 1

ηn

Fig. 10.18

In the x direction unz …n † ˆ ui …1 ÿ 3n2 ‡ 2n3 † ‡ i …ÿn ‡ 2n2 ÿ n3 † ‡ ui‡1 …3n2 ÿ 2n3 † ‡ i‡1 …n2 ÿ n3 †

…10:107†

and in the y direction unz …n † ˆ uj …1 ÿ 3n2 ‡ 2n3 † ‡ j …ÿn ‡ 2n2 ÿ n3 † ‡ uj ‡ 1 …3n2 ÿ 2n3 † ‡ j ‡ 1 …n2 ÿ n3 †

…10:108†

where n 

x lxn

n 

y lyn

…10:109†

The displacement function for the plate element is obtained by combining the functions in equations (10.107) and (10.108). This is exempli®ed for the displacement and slopes at node, i, j. The function associated with the displacement, ui;j is F1 …n ; n † ˆ ui; j …1 ÿ 3n2 ‡ 2n3 †…1 ÿ 3n2 ‡ 2n3 †

…10:110†

and for the slope in the x direction F2 …n ; n † ˆ xi; j …ÿ ‡ 2n2 ÿ n3 †…1 ÿ 3n2 ‡ 2n3 †

…10:111†

similarly for the slope in the y direction F3 …n ; n † ˆ yi; j …1 ÿ 3n2 ‡ 2n3 †…ÿn ‡ 2n2 ÿ n3 †

…10:112†

and for the twist term F4 …n ; n † ˆ xyi; j …ÿ ‡ 2n2 ÿ n3 †…ÿn ‡ 2n2 ÿ n3 †

…10:113†

The graphical representation of these deformation formulations is shown in Fig. 10.19. This process of identifying the functions is completed for

298

Energy methods in structural mechanics x

x

φx

u b

b a

a

c

i, j

c

i, j d

d a z

b a

b

z

y

y

(a)

(b) x

x

φxy

φy

b

b a

c

i, j

a

i, j d

d z

c

z y

y

(c)

(d)

Fig. 10.19

all the 16 degrees of freedom associated with element n, i.e. ui; j ; x ; y ; xy i; j

ui ‡ 1; j ; ui; j ‡ 1 ;

xi ‡ 1; j ; x ; i; j ‡ 1

ui ‡ 1; j ‡ 1 ; xi ‡ 1; j ‡ 1 ;

i; j

yi ‡ 1; j ; y ; i; j ‡ 1

i; j

xyi ‡ 1; j xy

…10:114†

i; j ‡ 1

yi ‡ 1; j ‡ 1 ; xyi ‡ 1; j ‡ 1

The complete function can be written succinctly as unz …n ; n † ˆ

16 X rˆ1

Dnr Fr …n ; n †

…10:115†

where Dnr are the elements of the vector collecting the nodal degrees of freedom of the n-th element, i.e. 2 3 ui; j 6


7 ‰Dn Š ˆ 4


5 …10:116†  xyi ‡ 1; j ‡ 1

Analysis of thin plates

299

The formulation of the sti€ness matrix is simpli®ed and made more amenable to computer programming by rede®ning new generalised coordinates. The de¯ection function, equation (10.115), can be described as unz ˆ A0 ‡ A1 n ‡ A2 n ‡ A3 n2 ‡ A4 n n ‡ A5 n2 ‡ A6 n3 ‡ A7 n2 n ‡ A8 n n2 ‡ A9 n3 ‡ A10 n3 n ‡ A11 n2 n2 ‡ A12 n n3 ‡ A13 n3 n2 ‡ A14 n2 n3 ‡ A15 n3 n3

…10:117†

The generalised coordinates are related to the nodal coordinates Dnr , see equation (10.115), by ‰AŠ ˆ ‰BŠ‰Dn Š with

…10:118†

2

3 A0 6


7 6 7 ‰AŠ ˆ 6 7 4


5

…10:119†

A15 The transformation matrix ‰BŠ is 2

1 6 0 6 6 6 0 6 6 ÿ3 6 6 6 0 6 6 3 6 6 6 ÿ2 6 6 0 6 ‰BŠ ˆ 6 6 0 6 6 ÿ2 6 6 6 0 6 6 9 6 6 6 0 6 6 6 6 6 4 6 4

0 0 0 0 1 0 0 0 0 1 0 0 2 0 0 3 0 0 1 0 0 ÿ2 0 0 1 0 0 2 0 3 ÿ2 0 3 0 ÿ2 0 0 1 0 0 0 2 ÿ1 0 ÿ6 ÿ6 4 ÿ9 ÿ2 0 1 0 ÿ3 ÿ4 ÿ4 ÿ3 ÿ2 ÿ2

2 ÿ6 2 ÿ6 1 ÿ4

3 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 07 7 7 0 0 0 0 0 0 0 0 0 0 07 7 ÿ1 0 0 0 0 0 0 0 0 0 07 7 7 0 0 0 0 0 0 0 0 0 0 07 7 0 0 0 ÿ3 0 ÿ1 0 0 0 0 07 7 7 1 0 0 0 0 0 0 0 0 0 07 7 0 ÿ3 ÿ1 0 0 0 0 0 0 0 07 7 7 0 0 0 0 ÿ3 0 ÿ1 0 0 0 0 7 7 0 0 0 2 0 1 0 0 0 0 07 7 7 0 ÿ2 ÿ1 0 0 0 0 0 0 0 07 7 ÿ3 6 2 ÿ9 6 ÿ3 2 9 3 3 17 7 7 0 0 0 0 2 0 1 0 0 0 07 7 ÿ3 4 2 ÿ6 3 ÿ2 1 6 3 2 17 7 7 ÿ2 3 1 ÿ6 4 ÿ3 2 6 2 3 15 ÿ2 2 1 ÿ4 2 ÿ2 1 4 2 2 1

(10.120) The formulation for the strain energy of deformation of the element of plate bending Un is obtained from the ®rst term in equation (10.73).

300

Energy methods in structural mechanics

In non-dimensional notations D Un ˆ 2lxn lyn

…1 …1  0 0

b q2 unz a q2 unz ‡ a qn2 b qn2



 2 n 2 n  2 n 2  q uz q uz q uz ÿ 2…1 ÿ † dn dn ÿ qn qn qn2 qn2

…10:121†

The strain energy for the complete plate can be obtained by summing the energy contributions of all the elements, i.e. X Uˆ Un …10:122† n

However, for a typical element the substitution of equation (10.115) into equation (10.121) and carrying out the mathematical operation results in replacement of the energy functional by a function of the generalised coordinates Ai de®ned in equation (10.118). That is a quadratic function, F, Un ) F…Ai †

…10:123†

This is equivalent to the strain energy being in the form D Un ˆ ‰AŠT ‰CŠ‰AŠ 2lxn lyn ˆ

D ‰Dn ŠT ‰BŠT ‰CŠ‰Dn Š‰BŠ 2lxn lyn

…10:124†

Application of the extremum condition, i.e. qUn ˆ0 qDnr

…10:125†

provides the condition D ‰BŠT ‰CŠ‰BŠ‰Dn Š ˆ 0 lxn lyn

…10:126†

The sti€ness matrix for a typical element is ‰K n Š ˆ ‰BŠT ‰CŠ‰BŠ

…10:127†

 is obtained by adding and the sti€ness matrix for the whole plate, ‰KŠ, the sti€ness matrices of the all the elements according to equation (10.122). The expression for total of potential energy is obtained by addition of the loss of potential energy of the load to the strain energy equation

Analysis of thin plates

301

(10.122). Take as an example the bending of a plate in response to a concentrated load P applied normal to the plane of the plate. The loss of potential energy of the load is ÿP u0

…10:128†

where u0 is the de¯ection at the position of application of the load. In vector form we can write ÿ‰PŠ

…10:129†

where the loading vector, ‰PŠ has zeros except at the position corresponding to u0 . The equilibrium condition for the whole plate is  ‰KŠ‰DŠ ÿ ‰PŠ ˆ 0 …10:130† The boundary conditions are incorporated into equation (10.130) by eliminating appropriate rows and columns. The displacements and slopes at the nodes of the elements are obtained as  ÿ1 ‰PŠ ‰DŠ ˆ ‰KŠ

…10:131†

The accuracy of the results from a localised trial function analysis generally increases as the number of elements included in the model is increased. This is exempli®ed by application of the method described in this section to a square plate with edge length, l, subjected to a concentrated load at the centre of the plate as shown in Fig. 10.20. The analysis of the plate is run with the mesh geometries shown in Fig. 10.21. Note that symmetry conditions of loading and deformations allow that only one quarter of the area of the plates needs to be modelled.

l

P l

x

y

Fig. 10.20

302

Energy methods in structural mechanics Symmetry

m=1

m=2

m=4

Fig. 10.21

The formulation relating the de¯ection at the centre of the plate, u0 , to the applied load P is Pl 2 D The solutions are u0 ˆ 

…10:132†

Case a: Centrally loaded plate with all edges simply supported Mesh size 1 2 4

 0:0111 0:0115 0:0116

Case b: Centrally loaded plate with all edges encastre Mesh size 1 2 4

 0:00530 0:00548 0:00557

It is evident that even with only a few elements the accuracy of the results, see section 10.5, is very good. Again, the great merit of localised trial function approach is that once the element derivation is programmed for a computer it is very simple to alter the boundary conditions and to incorporate a variety of types of loading in the numerical model.

11. The theory of ®nite elements Introduction

Preceding chapters in this book have examined various energy methods of analysis of elastic solids; particularly bars, beams, trusses, frames, plates. In the case of plane and space trusses and frames we have established the direct sti€ness method, a procedure that is especially suitable for computer programming and therefore for the automatic analysis of structures, even involving thousands of members. Moreover, we have seen that the treatment of structures involving plates, even with the simpli®cations allowed by Kirchho€'s model, remains a fairly complicated matter. In this form of structural component approximate methods, and particularly the method of trial functions, can still turn out to be a dicult approach. Indeed, the method of trial functions requires considerable skill in the choice of suitable approximating functions and substantial e€ort in the construction of the algebraic functions. In fact, the choice of the trial functions depends on the geometry of the structural component under analysis and on its boundary conditions and form of loading, therefore the procedure cannot be automated and there are ever-present risks associated with the possibility of human errors that make embarking on lengthy calculations somewhat hazardous. This diculty is common to all those structures that are not composed simply of a ®nite number of well-de®ned components, such as bars and beams, whose behaviour is unequivocally known. However, we have seen that most of these diculties can be overcome by the method of localised trial functions. In this chapter we introduce a generalisation of the method of localised trial functions, i.e. the ®nite element method, which is without any doubt the most powerful tool for the analysis of even very complex structures and the major analytical development in structural engineering in the last 40 years. The method was created by structural engineers, but its versatility, easy of use and accuracy have rendered this methodology a standard tool of analysis even for problems which may seem very

304

Energy methods in structural mechanics

far from the structural ones, like heat di€usion, ¯uid dynamics and magnetic ®eld analyses. Most of the high technology resources available to us, from supersonic aeroplanes to mobile telephones, owe their actual present state of sophistication to the ®nite element method. Thanks to this method, engineers and physicists can now model mathematically and obtain solutions to a very wide range of problems that would indeed be virtually impossible to handle in any other way. Alternative approaches are the numerical integration of the di€erential equations of equilibrium, using the ®nite di€erence method and perhaps by very costly testing at full- or small-scale. But these alternatives require special skills in numerical calculus and model-making that in most cases are obviated by the ®nite element programs now available. These programs can be accessed and operated very simply, often by engineers who, unfortunately, sometimes have little understanding of the fundamental theory of the method they are using. Despite these impressive capabilities, the underlying idea of the ®nite element method is very simple. It consists in a sub-division of the structure conceptually into smaller zones or areas. Once this sub-division has been carried out, trial functions are given within each zone, usually called an element, and the behaviour of each element can be analysed separately and independently from all the others. The boundary conditions between these elements can be imposed quite easily, provided the choice of the trial functions are simple polynomials that provide the continuity between the element under consideration and those contiguous with it. Then, exactly as we have done in the case of pin-jointed structures and frames, the sti€ness matrix of each element can be assembled with the sti€ness matrices of all the other elements to produce the combined sti€ness matrix of the problem. The forces applied to the structure are reduced once again to equivalent nodal loads and the boundary conditions can be imposed very straightforwardly along the edges of the constrained elements. It may be easily inferred that the fundamental basis of the ®nite element is in fact the localised Rayleigh±Ritz method introduced and exempli®ed in earlier chapters. The procedure is very general and requires only limited data to particularise the analysis to any structural form, i.e. the geometry of the structure, the shape of the elements, the material properties. The trial functions are often the same for each element, thus an analysis can be fully automated on a computer by means of a standard set of instructions. By following the standardised method even large-scale ®nite element problems can be treated in a very straightforward and almost routine manner.

The theory of ®nite elements

305

It seems reasonable that a better approximation may generally be obtained by increasing the number of sub-divisions, which is certainly more simple than including more and more complex trial functions by the classical Ritz method. However, despite the fact than the subdivision into simple elements and the equations of compatibility and equilibrium between the elements have a sound physical meaning, the problem of the convergence of the ®nite element solution to the exact mathematical one may be quite complex and has given origin to many signi®cant studies during the last 30 years. Apart from some brief notes regarding the physical interpretation of the ®nite element method, we will not deal with the topic of a formal proof of convergence, i.e. the application of precise mathematical theorems to the ®nite element method or the estimation of accuracy inherent in the ®nite element approximation, nor will we give any detail beyond a very simple outline of the basic procedure. Excellent textbooks exist on this topic and the interested reader can make reference to them in order to gain a more mathematical understanding of this basic tool in current engineering practice.

11.1

Sub-division of a structure into ®nite elements

We illustrate the basic steps of the ®nite element method by referring to the example of a simple structure, i.e. a retaining wall with lateral pressure, shown in Fig. 11.1. As we have already seen in chapter 3, if the dimension of the structure is very large in the z-direction and the lateral pressure is perpendicular to the longitudinal axis z and does not vary along the length, we may assume that all the cross sections are in the same state. In fact, we can suppose that the end sections are con®ned between P

Fig. 11.1

306

Energy methods in structural mechanics

Fig. 11.2

smooth rigid planes, which implies that displacement in the axial direction is prevented at the ends and, by symmetry, also at the mid-section. Consequently, we can assume that displacement in the axial direction is prevented at any cross section. Thus, the deformation of the wall can be assumed to be the same at any cross section and can be de®ned completely by the non-zero components of displacement ux and uy , which are assumed to be functions of x and y only. Hence, as already noticed in chapter 3, we have ux ˆ ux …x; y†

uy ˆ uy …x; y†

quy qux "yy …x; y† ˆ qx qy   1 qux quy "xy …x; y† ˆ ‡ 2 qy qx   1 qux quz ‡ ˆ0 "xz …x; y† ˆ 2 qz qx   1 quy quz ‡ ˆ0 "yz …x; y† ˆ 2 qz qy "xx …x; y† ˆ

uz ˆ 0 "zz ˆ

quz ˆ0 qz

There are several important engineering problems of this kind, like, for example, tunnels or cylindrical tubes subject to internal pressure, as illustrated in Fig. 11.2. Despite the fact that the problem of elastic equilibrium is reduced to a two-dimensional model in the x±y plane, nonetheless it remains a quite complex problem and, apart from some cases characterised by a very simple geometry of the boundary, no closed-form mathematical solutions are readily available to engineers. We make resort to the ®nite element procedure and start by conceptually sub-dividing the structure into zones, i.e. elements. This can be done, for example, by means of four-noded plane elements as shown in Fig. 11.3.

The theory of ®nite elements 14

15

16

13 7

8

4

11 6 7

6

1

1

12

5

5

y

9

10

9

8

2

2

307

3

3

4

x

Fig. 11.3

In this way we substitute our structure with an assembly of zones, whose deformations must be characterised by a ®nite number of parameters to permit the replacement of the continuum problem by a discrete one. It is clear that in the vast majority of structural problems these parameters may be assumed to be the displacements of the nodes. If we reduce also the applied loads to equivalent joint loads, the solution of the whole problem follows exactly the same rules as those applicable to standard discrete systems, like the trusses and frames examined in chapters 8 and 9. Of course, the key to the whole procedure lies in the de®nition of the behaviour of each individual element in terms of the displacements of its nodes. From a kinematic point of view this means that the displacements of the internal points of the element must be functions of the displacements of the nodes and satisfy basic requirements of compatibility at the boundaries. Thus, for example, the displacement along contiguous sides of adjacent elements may be required to be identical for any speci®ed displacement of the nodes and, in certain cases, this requirement of continuity may be extended to the slope of the displacement. The de®nition and the employment of such interpolating functions is the subject of the following sections.

11.2

Displacement functions and shape functions

We have already encountered the concept of displacement functions in the presentation of the Rayleigh±Ritz procedure, where trial functions

308

Energy methods in structural mechanics y

uy

uy L

H

ux

ux

l uy

uy I ux

αl

J

ux

x

Fig. 11.4

dependent on a certain number of parameters were employed to represent the displacement ®eld of the elastic body under analysis. The values of these parameters were determined by means of a minimisation of the total potential energy of the system. However, in that case, trial functions extended over the whole structure, whereas in the ®nite element method the displacement functions extend over a single sub-domain of the region under analysis. This concept was introduced in section 7.6 under the heading localised Rayleigh±Ritz. The localised nature of the trial functions in the ®nite element method is further examined in this section. If we restrict ourselves to the plane strain problem introduced in the previous section, we can take into consideration the four-noded rectangle shown in Fig. 11.4 and express the displacement ®eld u…x; y† over this sub-region as a function of the displacements of its nodes. For example, we can assume the components along x and y of the displacement ®eld are described by polynomial expressions of the type ux …x; y† ˆ a1 ‡ a2 x ‡ a3 y ‡ a4 xy

…11:2†

uy …x; y† ˆ b1 ‡ b2 x ‡ b3 y ‡ b4 xy

…11:3†

By inserting the coordinates of the four nodes of the element in equations (11.2) and (11.3) we obtain a system of linear equations in the eight unknowns a1 ; . . . ; b4 , that is ux …I† ˆ a1

ux …H† ˆ a1 ‡ a2 l ‡ a3 l ‡ a4 l 2

uy …I† ˆ b1

uy …H† ˆ b1 ‡ b2 l ‡ b3 l ‡ b4 l 2

ux …J† ˆ a1 ‡ a2 l uy …J† ˆ b1 ‡ b2 l

ux …L† ˆ a1 ‡ a3 l ux …L† ˆ b1 ‡ b3 l

…11:4†

The theory of ®nite elements

309

Solving this system and substituting the results in equations (11.2) and (11.3), we have      x y x y ux …x; y† ˆ 1 ÿ 1 ÿ ux …I† ‡ 1 ÿ ux …J† l l l l   xy y x 1ÿ u …L† …11:5† ‡ 2 ux …H† ‡ l l x l      x y x y 1 ÿ uy …I† ‡ 1 ÿ uy …J† uy …x; y† ˆ 1 ÿ l l l l   xy y x 1ÿ u …L† …11:6† ‡ 2 uy …H† ‡ l l y l Therefore the displacement functions, equations (11.5) and (11.6), can be taken to represent the displacement ®eld inside the ®nite element under consideration with respect to the displacement of its nodes. At this point it must be stressed once again that the displacement functions, similar to the situation in the Raleigh±Ritz method, are not exactly the displacement ®eld that would be provided by the solution of the elastic equilibrium equation for the particular structure under consideration, but instead, represent an approximation. However, as the solution of the elastic equilibrium problem cannot be obtained in closed form for the vast majority of engineering cases, this approximation is very often the only one available to engineers and in most cases may indeed be made very accurate. For the sake of generality we can represent the displacement functions, equations (11.5) and (11.6), in matrix form. This is easily obtained by writing u    0 NL 0 NI 0 NJ 0 NH x ˆ ‰Di Š uy 0 NI 0 NJ 0 NH 0 NL ˆ ‰NŠ‰Di Š

…11:7†

i

where ‰D Š is the array containing the displacement of the nodes of the i-th element, i.e. ‰Di ŠT ˆ ‰ux …I† uy …I† ux …J† uy …J† ux …H† uy …H† ux …L† uy …L†Š …11:8† and Nj …x; y† are the so-called shape functions. The shape functions have a crucial importance in the ®nite element method. By reference to equation (11.7) it seems clear that in general they have to be chosen such as to provide appropriate

310

Energy methods in structural mechanics

nodal displacements when the coordinates of the corresponding nodes are inserted in the shape function. In other words, we must have NI …x…I†; y…I†† ˆ 1

…11:9†

NI …x…J†; y…J†† ˆ 0

…11:10†

In the case at hand the shape functions take the form    x y 1ÿ NI …x; y† ˆ 1 ÿ l l   x y NJ …x; y† ˆ 1ÿ l l xy NH …x; y† ˆ 2 l   y x 1ÿ NL …x; y† ˆ l l

…11:11† …11:12† …11:13† …11:14†

As both the components of displacement are interpolated in an identical manner, four shape functions are sucient to represent the displacement functions, equations (11.5) and (11.6). The graphical form of the shape functions NI ±NL over the rectangular sub-domain under consideration are shown in Fig. 11.5. It is worth noting that both the components of displacement ux and uy on the four sides of the rectangle, i.e. the straight lines of equation x ˆ 0, y ˆ 0, x ˆ l and y ˆ l, respectively, depend only on the displacement of the nodes that belong to the sides themselves. For example, the displacement of the side connecting the nodes J and H is given by the expressions   y y ux …x ˆ l; y† ˆ 1 ÿ ux …J† ‡ ux …H† …11:15† l l   y y …11:16† uy …x ˆ l; y† ˆ 1 ÿ uy …J† ‡ uy …H† l l and similar expressions hold true for the remaining sides. This means that two adjacent rectangular elements, like those represented, for example, by the pairs 1 and 4 or 4 and 7, in Fig. 11.3, are characterised by the same displacement on contiguous sides, provided, of course, these latter are interpolated by means of equations (11.5) and (11.6) on both the sub-domains. Now let us turn our attention to the trapezoid elements no. 2, 5 and 8. A generic element of this set is shown in Fig. 11.6. Following the same line of reasoning adopted for the rectangular elements, we can

The theory of ®nite elements

L

311

L

H

H

I

I

J

NI

J

NJ L

L

H

H

I

I

J

NH

J

NL

Fig. 11.5

βl y uy

uy L

H

ux

ux

l uy

uy I ux

Fig. 11.6

αl

J

ux

x

312

Energy methods in structural mechanics

again consider the components of displacement in the form given by equations (11.2) and (11.3). In order to obtain appropriate nodal displacements corresponding to the coordinates of the nodes, we are now led to the following system of equations in the eight unknowns a1 ; . . . ; b4 ux …I† ˆ a1

ux …H† ˆ a1 ‡ a2 l ‡ a3 l ‡ a4 l 2

uy …I† ˆ b1 ux …J† ˆ a1 ‡ a2 l

uy …H† ˆ b1 ‡ b2 l ‡ b3 l ‡ b4 l 2 …11:17† ux …L† ˆ a1 ‡ a3 l

uy …J† ˆ b1 ‡ b2 l

ux …L† ˆ b1 ‡ b3 l

The substitution of the solution of the above system in the expressions, equations (11.2) and (11.3), yields      x y x y 1 ÿ ux …I† ‡ 1 ÿ ux …J† ux …x; y† ˆ 1 ÿ l l l l   xy y x 1ÿ u …L† ‡ 2 ux …H† ‡ …11:18† l l x l      x y x y 1ÿ uy …I† ‡ 1ÿ uy …J† uy …x; y† ˆ 1 ÿ l l l l   xy y x 1ÿ u …L† …11:19† ‡ 2 uy …H† ‡ l l y l which we may take as the expressions of the components of the displacement ®eld in the trapezoidal sub-domain, Fig. 11.6. Again, we can express the displacement function in matrix form, equation (11.7), provided the shape functions Nj …x; y† have the following expressions    x y 1ÿ …11:20† NI …x; y† ˆ 1 ÿ l l   x y NJ …x; y† ˆ 1ÿ …11:21† l l xy NH …x; y† ˆ 2 …11:22† l   y x 1ÿ …11:23† NL …x; y† ˆ l l At this point of the presentation we can anticipate that this way of proceeding may present serious drawbacks. For example, even if we assume the same general polynomial expression for the displacement

The theory of ®nite elements

313

functions (in the present example, equations (11.2) and (11.3)) and the meshing of the structure is achieved using quadrilateral elements only, it is evident that the resulting expressions for the displacement function varies according to the shape of the element and, in general, cannot be derived as a single function for all elements. Moreover, the conditions ful®lled by the displacement functions may vary according to the shape of the element as well. This factor can be easily explained by reference to the property of two adjacent rectangular elements to present the same displacement on contiguous sides. This property does not hold true for the trapezoidal element, Fig. 11.6. In fact, the side connecting the nodes J and H is a straight line of equation x ˆ l ÿ y… ÿ †

…11:24†

and the displacements of its points are given by the expressions   y ux …x ˆ l ÿ y… ÿ †; y† ˆ 1ÿ u …I† l  x     y y 1ÿ 1 ÿ ux …J† ‡ 1‡ l  l   2 y y  ÿ ÿ 1 ux …H† ‡ l l 2     y  ÿ 1 ux …L† ‡ 1ÿ ‡ l …11:25†   y 1ÿ u …I† uy …x ˆ l ÿ y… ÿ †; y† ˆ l  y     y y 1ÿ 1ÿ uy …J† ‡ 1‡ l  l    y y2  ÿ 2 ÿ 1 uy …H† ‡ l l     y  ÿ 1 uy …L† ‡ 1ÿ ‡ l …11:26† These expressions involve the components of displacement of all the four nodes of the element. Therefore, the displacement along contiguous sides of adjacent elements is not governed only by the behaviour of the nodes associated with the sides themselves and, in general, it will not be the same for both of elements. The elimination of such a

314

Energy methods in structural mechanics

pathological behaviour would imply a di€erent choice of the polynomial displacement functions, equations (11.2) and (11.3). Generally speaking, this means that an appropriate analysis of the displacement functions is required, at least for any di€erent set of elements. However, in order to obtain a procedure with the greatest degree of automation, it is desirable that a small amount of data can be used to represent the various types of elements that may be required in the analysis of a real engineering problem. This can be obtained by making resort to the use of intrinsic coordinates and appropriate transformation rules, as we will see in the next section.

11.3

Element mapping and intrinsic coordinates Ð isoparametric elements

Up to this point, we have discussed some diculties that arise in the direct generation of the displacement functions for the few elements employed for the simple meshing of the retaining wall, Fig. 11.3. If we are to deal with more complex practical engineering structures, which are characterised by quite complicated geometrical boundaries, it would be valuable to ®nd a way to construct ®nite element models using only a fairly restricted library of elements. This can be done by mapping a simple element (the so-called parent element), such as a rectangle in the local coordinate system ; , into a more complex shape in the general coordinate system x±y, as shown in Fig. 11.7. In the present context, mapping means a unique, one-to-one relationship between the local system of coordinates ,  and the general reference system x±y. It should be noted that in this case the relationship between the local system of coordinates and the general one is somewhat broader than the relationship between the local and L

H L

η

H

ξ I I

J J

y

x

Fig. 11.7

The theory of ®nite elements φ

φ2

φ1

y

L

H

H

L r2

r1 I

I

φ1

r

r2

J

φ2

J r1

315

x

Fig. 11.8

the general reference frames encountered in the matrix analysis of pinjointed and rigid-jointed structures. In fact, in the previous situation we dealt with a transformation rule that allowed us to establish a correspondence between systems of Cartesian orthogonal axes that di€ered by simple translations or rotations in the three-dimensional space. Here we have relationships that allow a continuous distortion of the element, too. For instance, we can consider the relationship between cylindrical polar and Cartesian coordinates, i.e. x ˆ r cos 

y ˆ r sin 

…11:27†

that may imply the mapping of a rectangular domain in the local system r ÿ  into a distorted element in the x±y space. This situation is illustrated in Fig. 11.8. For the sake of convenience the local system ± is generally assumed to be a system of natural or intrinsic coordinates, i.e. a system of non-dimensional coordinates varying between ÿ1 and 1, see Fig. 11.9. In the case of four noded elements, this means that they can be derived from a parent element which is a square whose sides measure 2. With regards to the transformation rules between the coordinate

η 2

ξ

2

Fig. 11.9

316

Energy methods in structural mechanics

systems, a very convenient method of establishing them is to make use of the same shape functions that we use to represent the displacement function. This means that the array ‰ci Š of the coordinates of the nodes of the element in the general reference frame is used to de®ne the global coordinates …x; y; z† of any point in the element itself. For instance, we can perform this by deriving a set of shape functions Nj …; † in the local reference frame ± according to the general procedure introduced in section 11.3 (that is, they must have a unit value at the reference node and zero at all the others) and by writing for each element the following transformation rules x…; † ˆ NI …; †x…I† ‡ NJ …; †x…J† ‡ NH …; †x…H† ‡ NL …; †x…L†

…11:28†

y…; † ˆ NI …; †y…I† ‡ NJ …; †y…J† ‡ NH …; †y…H† ‡ NL …; †y…L† In matrix form we can write   x…; † ˆ ‰N…; †Š‰ci Š y…; †

…11:29†

…11:30†

where ‰N…; †Š is the shape function matrix and ‰ci Š is the array containing the global coordinates of the nodes of the generic element, i.e. ‰ci ŠT ˆ ‰x…I† y…I† x…J† y…J† x…H† y…H† x…L† y…L†Š …11:31† In this manner we obtain a set which is readily suitable for mapping. In fact, it is evident that the locations of coordinates …I†, …I†; . . . ; …L†; …L†, i.e. the four nodes of the parent element, will automatically assume the required coordinates x…I†; y…I†; . . . ; x…L†; y…L† in the mapping. Continuity and uniqueness requirements are generally satis®ed by such a type of transformation, with the exception of excessive and pathological distortions. The feature can be checked by making reference to the Jacobian matrix (after Jacobi1 ) of the transformation, that is 3 2 qx qy 6 q q 7 7 6 …11:32† ‰JŠ ˆ 6 7 4 qx qy 5 q 1

q

Jacobi, Karl Gustav Jacob (Potsdam, 1804 ± Berlin, 1851), German mathematician.

The theory of ®nite elements

317

The Jacobian matrix of the transformation can be trivially calculated from equation (11.30), as 2 3 qNI qNL 0 ... 0 6 q 7 i q 7‰c Š ‰JŠ ˆ 6 …11:33† 4 qNI qNL 5 ... 0 0 q q From calculus we know that the condition for a one-to-one mapping is that the sign of the determinant of equation (11.33) remains unchanged at all the points of the domain under mapping. When such a transformation rule is adopted, the elements will be called isoparametric. One of the most signi®cant advantages of the isoparametric formulation is that if two adjacent elements are generated from the same parent element, whose shape functions make the displacement of any side dependent on the behaviour of only the nodes belonging to the side itself, then the distorted elements will be always contiguous, before and after deformation has occurred. This can be demonstrated very easily by making reference to Fig. 11.10. It is evident that along a common edge, such as the line I±J, elements A and B have the same tangent coordinate, being A ˆ B . Moreover, as the mapping into the distorted elements is governed along the common edge by the same functions of , i.e. the shape functions Nj , J

ηA

ηB

ξA

ξB

I

J

B A

I

Fig. 11.10

318

Energy methods in structural mechanics

and the sets of global nodal coordinates are the same, it follows that the inter-element boundary curve I±J will also be the same. The same line of reasoning applied to the displacement functions, equation (11.7), demonstrates that displacement is also the same along a common edge. It goes without saying that the continuity of the displacement functions is preserved in the mapping and that isoparametric elements will be invariant as a consequence of the adoption of the intrinsic coordinate system ±. Finally, it must be underlined that, although we have limited our development here to two-dimensional structures, nevertheless all the comments and results derived above are equally valid and applicable to the analysis of three-dimensional structures and components.

11.4

Convergence criteria for the displacement functions Ð the patch test

Now it is appropriate to consider some conditions to which displacement functions should conform so that it is likely that the exact mathematical solution of the problem of elastic equilibrium will be approached by the ®nite element treatment as more and more elements are used to mesh a certain structure. From a rigorous mathematical standpoint this is quite a complex matter and in the past 30 years many studies have been devoted to underwriting the intrinsically approximate ®nite element method with a formal mathematical basis. A review of such studies and the mathematical developments are outside the purpose of this textbook and we limit the following presentation to self-evident and heuristic physical convergence requirements. These requirements are . .

the displacement ®eld must be continuous within every subdomain. This is a fairly simple requirement and is met by any polynomial displacement function. inter-element compatibility should be respected, i.e. there should be no gaps or overlaps between elements. Moreover, bending elements (like, for instance, those associated with the Kircho€ plate model or the Bernouilli beam model) should not allow discontinuities in slope between elements. The ®rst point of this statement is physically evident and we have seen that the use of the isoparametric formulation provides a straightforward way of meeting such a condition. The second point is physically evident, too. However, it is rather more complicated to meet such a requirement and there are many element formulations of common use that violate the requirement in a coarse mesh arrangement. Nevertheless, these elements may be able to

The theory of ®nite elements

.

.

.

319

yield a good convergence towards the exact mathematical solution because incompatibilities in slope tend to disappear with increasing mesh re®nement. Also, incompatible elements are very often deliberately adopted because they tend to `soften' the behaviour of the model and counteract the natural oversti€ness of the displacement functions. In fact it must be borne in mind that by substituting the real solid by a ®nite element model we seek the solution of the elastic equilibrium (which for linear problems corresponds to a minimum of the total potential energy) in a class of functions which is a subset of the class of all the kinematically admissible functions and therefore provides displacements that are generally smaller than the real ones. the elements should be geometrically invariant or isotropic. This requirement is once again physically evident and means that the element behaviour should have no preferred direction. It implies that the polynomial expression adopted to interpolate the displacement ®eld over the element sub-domain should be symmetric with respect to any coordinate. Of course obtaining a set of isotropic elements is quite easy in an isoparametric formulation, provided the assumed displacement functions for the parent element retain symmetry with respect to all the intrinsic coordinates. the elements should be able to represent rigid-body motions exactly. Of course, a reliable ®nite element model must be able to represent correctly this situation, which may occur in the whole structure or only in a part of it. An obvious implication of this requirement is the following: when the vector of nodal displacements ‰dŠ represents a rigid body motion, the element must exhibit a zero strain ®eld. From a mathematical standpoint, this implies that the polynomial adopted for the representation of the displacement ®eld must possess appropriate and adequate constant terms. the elements should be able to represent constant strain ®elds. From a physical standpoint this last requirement may seem less obvious than the previous ones, but it is nevertheless a very signi®cant requirement. We can imagine any continuous structure as a system characterised by an in®nite number of elements, each one displaying a constant strain, which is the value of the strain at each point of the whole domain. Thus, it seems appropriate that a ®nite element is able to reproduce constant strain ®elds so that, as the mesh is further and further re®ned and the sub-division of the domain becomes ®ner and

320

Energy methods in structural mechanics qz (z)

z y l

εz (z)

Fig. 11.11

®ner, elements will approach the state of constant strain in the neighbourhood of each point. In order to give a simple example of this situation, we can make reference to an axially loaded bar, Fig. 11.11. The form of the axial strain "z …z† is displayed below the bar. We may now sub-divide the bar in a speci®ed number of ®nite elements, say n, as shown in Fig. 11.12. If the elements are able to represent a constant strain ®eld, the actual strain ®eld in Fig. 11.12 can be approximated by the piecewise diagram shown in Fig. 11.13 and as n ! 1 the real strain ®eld will be approached by the ®nite element solution. qz (z)

z l/n

y l

Fig. 11.12

εzz (z)

Fig. 11.13

The theory of ®nite elements

321

Fig. 11.14

If, however, the elements are not able to represent a constant strain ®eld, it may be that even when the dimension l/n of each element becomes very small, the form of the axial strain resulting from the ®nite element modelling will still di€er signi®cantly from the exact solution, as shown in Fig. 11.14. We recall that in the linear theory of elasticity the strain ®eld within the structure is derived from the displacement ®eld by means of the well-known relationships qu "xx ˆ x qx ...   …11:34† 1 quy qux "xy ˆ ‡ 2 qx qy ... and it follows that a complete linear polynomial is required in order to represent constant strain states. A complete quadratic polynomial is required in order to represent constant curvatures in models for structural ¯exure, such as Kircho€'s plates. As a generality, we can say that it is desirable that polynomial expressions of displacement ®elds possess the highest order of a complete polynomial for the assigned number of parameters to be determined. The whole set of complete polynomials of any order can be shown graphically by means of the Pascal2 triangle, Fig. 11.15. However, it must be underlined that this condition, like each of the previous requirements, if taken alone, is neither a necessary nor a suf®cient condition to guarantee the convergence of the ®nite element model to the exact mathematical solution. On the whole, it can be armed that displacement functions that satisfy the above requirements show a good convergence to an accurate solution and do not display strange or pathologic behaviours even for a relatively coarse discretisation of the structure. A practical test of validity for a generic non-conforming element (i.e. those elements that do not respect the second of the conditions above, 2

Pascal, Blaise (Clermont-Ferrand, 1623±Paris 1662), French mathematician, physicist and religious philosopher.

322

Energy methods in structural mechanics 1

ξ

ξ2

ξ3

ξ4

η

ξη

ξ2 η

ξ3 η

η2

ξ η2

ξ2 η2

η3

ξ η3

η4

Fig. 11.15

which is normally the case for elements used to model plates and shells, as discussed earlier) is the so-called patch test. This test consists of assembling an arbitrary patch of elements in such a way that at least one node is completely surrounded by elements, see Fig. 11.16. Then, a boundary nodal displacement that corresponds to a state of constant strain is applied. Once the equations of elastic equilibrium have been solved for the ®nite element model, the test is considered to be passed if resulting strains agree exactly with exact values at every point in any sub-domain. Of course, in order to be signi®cant, the test must be repeated for more than one geometry, mesh arrangement and state of strain. In fact, some elements may pass the test in some conditions and fail it in others. It can be shown that, under some precise requirements, the patch test is a necessary and sucient condition for convergence. In any case and apart from precise

Fig. 11.16

The theory of ®nite elements

323

mathematical reasons, the patch test is also the most e€ective and practical benchmark for ®nite element programming and checking.

11.5

Some general families of shape functions

In section 11.2 we started from simple polynomial expressions for the components of the displacement function in the x±y plane, i.e. equations (11.2) and (11.3). The choice of a four-term polynomial expression in x and y was obviously suggested by the fact that we were dealing with a four-noded element and thus we had four conditions to impose for each component of displacement in order to make the value of the displacement function at the position of each node equal to the displacement of the node itself. In general this approach can be repeated for any parent element in intrinsic coordinates, regardless of the number of nodes. As an example of this, let us consider the eight-noded element shown in Fig. 11.17. The components of displacement at eight points must determine uniquely the variation of the displacement function exempli®ed by a polynomial expression in  and  with eight free coecients, such as ui ˆ a1 ‡ a2  ‡ a3  ‡ a4  ‡ a5 2 ‡ a6 2 ‡ a7 2  ‡ a8 3 …11:35† The substitution of the coordinates of the nodes I; . . . ; P will provide a system of eight linear equations in the eight unknowns a1 ; . . . ; a8 , i.e. 2 3 2 32 3 1 …I† . . . 2 …I†…I† 3 …I† ui …I† a1 4 ... 5 ˆ 4... ... ... ... . . . 54 . . . 5 a8 ui …P† 1 …P† . . . 2 …P†…P† 3 …P† …11:36† O

N

M

η

L

P

ξ

I

Fig. 11.17

J

H

324

Energy methods in structural mechanics

or, in more symbolic form ‰d i Š ˆ ‰CŠ‰aŠ

…11:37†

The solution of this system provides the unknown coecients a1 ; . . . ; a8 , that is ‰aŠ ˆ ‰CŠÿ1 ‰d i Š

…11:38†

Thus, equation (11.35) may be written ‰ui …; †Š ˆ ‰Š‰aŠ ˆ ‰Š‰CŠÿ1 ‰d i Š

…11:39†

where ‰Š  ‰ 1







2

2

2 

3 Š

…11:40†

Comparison with the general expression, equation (11.7), allows us to conclude that in this case the required expression of the shape functions is ‰NŠ ˆ ‰ NI

. . . . . . NP Š ˆ ‰Š‰CŠÿ1

…11:41†

However, there are some serious drawbacks to this procedure. First, the existence of the inverse of ‰CŠÿ1 is not always guaranteed. Moreover, even when the inversion of ‰CŠ is possible, it is still a quite complicated and lengthy formal procedure for many element geometries. Finally, the choice of the lowest possible power for each term of the polynomial expression does not necessarily lead to expressions that are symmetric with respect to  and . This is the case in equation (11.35), which is a polynomial of order 3 that does not include the terms 3 and 2 , as can be easily observed from the Pascal triangle, Fig. 11.15. In this manner the element shown in Fig. 11.17 has preferential directions, allowing a fully cubic variation of the components of displacement only with respect to the  coordinate and thus violates the third of the conditions stated in the previous section. Therefore it appears advantageous to seek to establish a procedure to specify the shape functions directly and systematically with full compliance with the required conditions. Actually, there are several ways to ful®l this goal and each one results in a precise family of shape functions. Here we will consider two basic and popular `families' of shape functions: the Lagrange family and the serendipity family. The Lagrange family of shape functions stems directly from the expression of Lagrange polynomials in one coordinate. It is known from elementary calculus that these polynomials have the following remarkable property: given n points on the  axis, it is always possible

The theory of ®nite elements

325 1

1

k

2

n

ξ

Fig. 11.18

to construct a Lagrange polynomial that has unit value at one point, say the k-th, and zero value at all the other n ÿ 1. The situation is illustrated in Fig. 11.18. The general expression of any Lagrange polynomial in one coordinate is given by the relationship Lnk …† ˆ

… ÿ 1 †… ÿ 2 † . . . … ÿ k ÿ 1 †… ÿ k ‡ 1 † . . . … ÿ n † …k ÿ 1 †…k ÿ 2 † . . . …k ÿ k ÿ 1 †…k ÿ k ‡ 1 † . . . …k ÿ n † …11:42†

which immediately yields Lnk …k † ˆ 1

and Lnk … j † ˆ 0

for j 6ˆ k

…11:43†

It follows from equation (11.42) that a Lagrange polynomial through n points is a polynomial of order n ÿ 1. If we take into consideration a generic rectangular element with r  s nodes, as shown in Fig. 11.19, a whole family of shape functions can be written down directly from the expression of Lagrange polynomials.

s· r

η ξ j

1

1

r 1

326

Energy methods in structural mechanics

In fact, by labelling the node by its column and row number, say h and k, we have Nj …; † ˆ Lrh …†Lsk …†

…11:44†

It is clear from equation (11.44) that the fundamental advantage of such a family of shape functions lies in the extreme simplicity of its general expression. Additionally, as for a rectangular element both r and s result  2, it follows that the expression, equation (11.44), will always include a complete polynomial of order 1, thus allowing the representation of rigid displacements and constant strains. Also, if we assume r ˆ s, as is usually the case for parent elements in intrinsic coordinates, geometrical isotropy is automatically assured. However, from Fig. 11.19 it is also clear that the Lagrange family may imply a large number of internal nodes for each element. Despite the fact that the behaviour of such nodes can be linked to the displacements of the nodes on the edges by means of a standard substructuring procedure (according to the same technique by which, for instance, we can make the displacement of the internal node M of the truss of Fig. 11.20 an explicit function of the displacements of the end nodes I, J, H, L), nevertheless the presence of such degrees of freedom can adversely a€ect the convergence of the ®nite element model. Moreover, if we take into consideration the generic shape function which is derived from equation (11.44) for a 3  3 nodes element, i.e. Nj …; † ˆ a1 ‡ a2  ‡ a3  ‡ a4 2 ‡ a5  ‡ a6 2 ‡ a7 2  ‡ a8 2 ‡ a9 2 2

…11:45† u (H)

u (L) L

H

M

u (I) I

Fig. 11.20

J

u (J)

The theory of ®nite elements L

327

H

η ξ

I

J

Fig. 11.21

we see that it contains some high-order terms and it omits some lower ones. In fact equation (11.45) contains, together with the complete second order expansion, the higher order terms 2 , 2 and 2 2 , while it omits the terms 3 and 3 (needed for a complete third-order expansion) and the terms 4 , 3 , 3 and 4 (needed for a complete fourth-order expansion). This holds true for every element of the Lagrange family, which therefore does not o€set the welcome property that polynomial expressions of displacement ®elds possess the highest order of a complete polynomial for the assigned number of parameters to be determined. The serendipity family of shape functions is somewhat slightly less simple to generate, but it o€ers substantial advantages, too. The name serendipity was given to this class of shape functions after the princes of Serendip, who were reported by Horace Walpole3 to be remarkable for their chance discovery. The basis for the name stems evidently from the way in which we derive the serendipity shape functions. In fact, let us consider the four noded square element in Fig. 11.21. The equations of the lines J±H and L±H are, respectively, 1ÿ ˆ0

1ÿ ˆ0

…11:46†

Therefore the function P…; † ˆ …1 ÿ †…1 ÿ †

…11:47†

will give zero value for the coordinates of the nodes J, H and L. Thus, we can write NI …; † ˆ a…1 ÿ †…1 ÿ †

…11:48†

and impose that NI …ÿ1; ÿ1† ˆ 1, which yields a ˆ 1=4. Consequently, we have NI …; † ˆ 14 …1 ÿ †…1 ÿ † 3

Walpole, Horace (London, 1717±1797), English writer.

…11:49†

328

Energy methods in structural mechanics

O

N

M

η

L

P

ξ

I

J

H

Fig. 11.22

By repeating the same reasoning for each of the four nodes of the element in Fig. 11.21, we are led to the general formula Nj …; † ˆ 14 …1 ‡ j †…1 ‡ j †

…11:50†

It is clear that however simple it seems, this procedure requires some ingenuity for any element we may wish to consider. In fact, if we shift our attention to the eight-noded element shown in Fig. 11.17, we have to identify for each node a certain number of lines that pass, in turn, through all the other nodes and these may not be simply the sides which do not contain the node in question. As an example, consider a corner node, such as the node I, for which the required lines are J±P, H±M and O±M, as shown in Fig. 11.22. Their respective equations are 1‡‡ ˆ0

1ÿ ˆ0

1ÿ ˆ0

…11:51†

and the appropriate expression for the shape function NI …; † is NI …; † ˆ a…1 ‡  ‡ †…1 ÿ †…1 ÿ †

…11:52†

Then, the condition NI …ÿ1; ÿ1† ˆ 1 yields NI …; † ˆ ÿ 14 …1 ‡  ‡ †…1 ÿ †…1 ÿ †

…11:53†

For a side node, such as the node J, the required lines are H±M, O±M and I±O, as shown in Fig. 11.23. Their respective equations are 1ÿ ˆ0

1ÿ ˆ0

ÿ1ÿ ˆ0

…11:54†

The theory of ®nite elements

O

N

329

M

η

L

P

ξ

I

J

H

Fig. 11.23

and the expression of the shape function NJ …; † is NJ …; † ˆ a…1 ÿ †…1 ÿ †…ÿ1 ÿ †

…11:55†

Once again, the condition NJ …0; ÿ1† ˆ 1 yields NJ …; † ˆ ÿ 12 …1 ÿ †…1 ÿ †…ÿ1 ÿ †

…11:56†

By iterating the same line of reasoning, for the example element we obtain the following general formulae corner nodes:

Nj …; † ˆ 14 …j  ‡ j  ÿ 1†…1 ‡ j †…1 ‡ j † …11:57†

mid-side nodes: Nj …; † ˆ 12 …1 ÿ 2 †…1 ‡ j † Nj …; † ˆ 12 …1 ‡ j †…1 ÿ 2 †

for j ˆ 0 for

j ˆ 0 …11:58†

In general the serendipity shape functions require fewer degrees of freedom for a certain complete polynomial expansion than the Lagrangian family. Moreover, they do not require the presence of internal nodes a priori, but this can become necessary in cases where the presence of complete polynomials beyond the cubic for mid-side nodes is required for particular analyses.

330

Energy methods in structural mechanics

11.6

Strain energy, work of applied loads and equilibrium equations

Once the kinematic behaviour of any element has been interpolated in terms of shape and displacement functions, the study of the elastic equilibrium of the ®nite element model becomes simple and straightforward. For the sake of simplicity and clarity of presentation, we continue with the example of the two-dimensional plane strain problem introduced in section 11.1. From a conceptual point of view the development of the theory for a generic three-dimensional problem is identical to that for the two-dimensional case and the extension should present no real diculties. The simplest way to study the problem consists once again in writing down the total potential energy of the ®nite element model and seeking the values of the degrees of freedom of the system, i.e. the displacements of its nodes, corresponding to the stationary value of this expression. This is exactly what we have already described in chapters 8 and 9 for the systems of bars and beams and most of the observations made at that stage remain valid also in the theory for ®nite elements. Therefore, let us recall the general expression of the total potential energy, that is equation (6.2) ˆUÿW Also in the present case the strain energy of the whole structure is given by the sum of the energy stored in each of its elements, i.e. X i Uˆ U …11:59† i

where, according to chapter 3, in the global reference system x±y we can write equation (3.107) in a more compact form … 1 Ui ˆ " …x; y†Cijhk "hk …x; y† dA …11:60† 2 ij Si

i

S is the domain of the generic element and Cijhk are the constants which de®ne the elastic properties of the material. Remember that we are examining a system of unit depth, so that the integrals can be performed over the area A of the elements. For the sake of compactness, in the case at hand the components of strain "ij …x; y† can be ordered in a three-element array, i.e. 3 2 "xx …x; y† 7 6 …11:61† ‰"…x; y†Š ˆ 4 "yy …x; y† 5 2"xy …x; y†

The theory of ®nite elements

331

and the elastic constants Cijhk can be set in a 3  3 matrix, that is 2 3 …1 ÿ †  0 6  E …1 ÿ † 0 7 6 7 …11:62† ‰CŠ ˆ 4 …1 ‡ †…1 ÿ 2† 1 ÿ 2 5 0 0 2 Notice that the elastic constants matrix (11.62) can be straightforwardly derived from the relationships (3.93)±(3.95) and (3.99) by setting "zz ˆ 0 in equation (3.95) and then substituting the resulting value of zz in the preceding two equations. The strain±displacement relationships can also be expressed in matrix form, 2 q 3 3 6 qx 0 7 2 "xx 6 7" # q 7 ux 6 " 7 6 7 ‰"Š ˆ 4 yy 5 ˆ 6 …11:63† 6 0 qy 7 u y 6 7 2"xy 4 q q 5 qy

qx

or, more compact form, ‰"Š ˆ ‰SŠ‰uŠ

…11:64†

where ‰SŠ formally represents a di€erential operator. Thus, the strain energy of the generic element can be written as … … 1 1 i T U ˆ ‰"Š ‰CŠ‰"Š dA ˆ ‰uŠT ‰SŠT ‰CŠ‰SŠ‰uŠ dA …11:65† 2 2 Si

Si

By substituting the equation (11.7) for the displacement function, in equation (11.65) i.e. ‰u…x; y†Š ˆ ‰N…x; y†Š‰Di Š, we get, ®nally … 1 i U ˆ ‰Di ŠT ‰NŠT ‰SŠT ‰CŠ‰SŠ‰NŠ‰Di Š dA 2 Si

1 ˆ ‰Di ŠT 2

…

‰NŠT ‰SŠT ‰CŠ‰SŠ‰NŠ dA‰Di Š

…11:66†

Si

The above formula can be put into a form identical to equations (8.40)±(8.42), obtained in the case of frame of structures, by setting … …11:67† ‰K i Š ˆ ‰NŠT ‰SŠT ‰CŠ‰SŠ‰NŠ dA Si

332

Energy methods in structural mechanics

so that equation (11.66) can be written as U i ˆ 12 ‰Di ŠT ‰K i Š‰Di Š

…11:68†

We remember that ‰Di Š is the array that contains the displacements of the nodes of the generic four-node element in the global reference frame. Thus for the present example, see equation (11.8) ‰Di ŠT ˆ ‰ux …I† uy …I† ux …J† uy …J† ux …H† uy …H† ux …L† uy …L†Š ‰K i Š is therefore an 8  8 matrix. In order to make the formulae as general as possible, we take into consideration the vector ‰DŠ, which contains all the nodal displacements of the ®nite element model of Fig. 11.3. We have ‰DŠT ˆ ‰ux …1† uy …1†

...

uy …9† ux …9†

...

ux …16† uy …16†Š …11:69†

At this point, exactly as we have done in the case of framed structures, we can identify the joints I, J, H and L of any element in the general numbering system. For example, for element 7 in Fig. 11.3, we have I

!

9

J

! 10

H

! 14

L

! 13

…11:70†

On the basis of this identi®cation, we can enlarge the matrix ‰K i Š from an 8  8 array to a 32  32 array simply by setting its terms in the appropriate position and leaving all the others as zero. Thus, we can write U i ˆ 12 ‰DŠT ‰K i Š‰DŠ

…11:71†

and the strain energy of the whole structure is X i X1 T i ‰DŠ ‰K Š‰DŠ U ˆ Uˆ 2 i i ˆ

1 T X i 1  ‰DŠ ‰K Š‰DŠ ˆ ‰DŠT ‰KŠ‰DŠ 2 2 i

where  ˆ ‰KŠ

X i

‰K i Š

…11:72†

…11:73†

The theory of ®nite elements

333

Let us now take into account the body forces b that act on the domain S of the structure, and the tractions f that act on its boundary, ÿ. If we order their components along the axes of the global reference frame as two arrays, i.e.     bx …x; y† fx …x; y† ‰b…x; y†Š ˆ ‰ f …x; y†Š ˆ …11:74† by …x; y† fy …x; y† we can write the work of the applied loads as … … W ˆ bj uj dA ‡ fj uj ds S

ÿ

… ˆ ‰u…x; y†ŠT ‰b…x; y†Š dA ‡ ‰u…x; y†ŠT ‰ f …x; y†Š ds … S

…11:75†

ÿ

Given the property of the addition operator for integrals, the above expression can also be written as X… T X… T ‰uŠ ‰bŠ dA ‡ ‰uŠ ‰ f Š ds …11:76† Wˆ i

i

Si

ÿi

where S i is the sub-domain of the generic element and ÿi is its boundary. By substituting the equation (11.7) for the displacement function into equation (11.76), that is ‰uŠ ˆ ‰NŠ‰Di Š, we obtain X… i T T X… i T T Wˆ ‰D Š ‰NŠ ‰bŠ dA ‡ ‰D Š ‰NŠ ‰ f Š ds i

ˆ

X i

i

Si

ÿi

9 = ‰D Š ‰NŠ ‰bŠ dA ‡ ‰NŠ ‰ f Š ds : i ; i i

8… T<

…

T

S

T

…11:77†

ÿ

Again, in order to obtain a formulation formally identical to the one presented in chapters 8 and 9 for the framed structures, we can set 8… 9 … < = i T T …11:78† ‰P Š ˆ ‰NŠ ‰bŠ dA ‡ ‰NŠ ‰ f Š ds : i ; i S

ÿ

In the present example the vector ‰P i Š is given by the integration of the product of an 2  8 matrix, ‰NŠT , by a two-element vector, i.e. ‰bŠ and ‰ f Š, which gives an eight-element column array.

334

Energy methods in structural mechanics

Thus, we have X iT i ‰D Š ‰P Š Wˆ

…11:79†

i

Similarly to what has been done before, identi®cation of the joints I, J, H and L of any element in the general numbering system allows us to enlarge the eight-element vectors ‰P i Š to 32-element arrays. This is done by setting the terms of ‰P i Š in the appropriate positions and leaving all the other elements as zero. In this manner we can write X i X T i Wˆ ‰DŠ ‰P Š ˆ ‰DŠT ‰P Š ˆ ‰DŠT ‰PŠ …11:80† i

where ‰PŠ ˆ

X

i

‰P i Š

…11:81†

i

At this point the total potential energy of the ®nite element model is   ˆ U ÿ W ˆ 12 ‰DŠT ‰KŠ‰DŠ ÿ ‰DŠT ‰PŠ

…11:82†

and, from a formal point of view, appears exactly the same as equation (8.47) obtained in the case of framed structures, as we set out to show. The solution of the elastic equilibrium problem can be pursued by seeking the stationary value of the total potential energy expression equation (11.82) with respect to the degrees of freedom set in the displacement vector ‰DŠ. This means that we have to impose that q ˆ0 qDj

…11:83†

where Dj is the generic component of the displacement vector ‰DŠ. In the case of the retaining wall, Fig. 11.3, equation (11.83) is a linear system of 32 equations in 32 unknowns, namely the components of the vector ‰DŠ de®ned by equation (11.69). In matrix notation it is  ‰KŠ‰DŠ ÿ ‰PŠ ˆ 0 …11:84†  de®ned by equations (11.73) and It is evident that the matrix ‰KŠ, (11.67), represents the combined sti€ness matrix for the structure, while the vector ‰PŠ, de®ned by equations (11.81) and (11.78), represents the equivalent combined loads vector. As was the case with framed structures, the system equation (11.84) is singular and not solvable before the e€ects of restraints are taken into account. Imposition of the respect of the kinematic constraints at the nodes 1, 2, 3 and 4 of the ®nite element model in Fig. 11.3 follows the same procedure as in section 8.2 and is therefore not repeated here. In the

The theory of ®nite elements

335

present example introducing the boundary constraints yields a reduced set of 24 linear equations in 24 unknowns, that are the active degrees of freedom of the structure, i.e. the components of displacement of the nodes 5±16. The system can be written as ‰KI Š‰DI Š ˆ ‰PI Š …11:85† and admits a unique solution. The reader can also verify that, once the e€ect of the constraints has been imposed, the joint sti€ness matrix of the problem ‰KI Š is positive de®nite. Restraint displacements, too, can be taken into account following the same line of reasoning as presented in section 9.6. Again, as in the case of the framed structures of chapters 8 and 9, the construction of the equilibrium equations (11.85) can be pursued directly, without the necessity to repeat every time the underlying reasoning in terms of energy. This constitutes the direct sti€ness method previously introduced. In fact, once the actual structure has been conceptually sub-divided into ®nite elements, the only components required to produce the equilibrium equations (11.84) are the expressions for the sti€ness matrices ‰K i Š, given by equation (11.67), and of the equivalent nodal loads ‰P i Š, given by equation (11.78). However, both equations (11.67) and (11.78) are expressed in the global reference frame x±y, while we generally establish the expressions of the shape and displacement functions for the generic ®nite element in the intrinsic reference frame ±, as seen in sections 11.3 and 11.5. An appropriate manipulation of data involving the transformation rules is therefore required. Moreover, the computation of the sti€ness matrices ‰K i Š and of the equivalent joint loads ‰P i Š involves integration of several functions. In dealing with real largescale complex structures it is clear that these integration operations must be performed in an automated way by means of suitable numerical procedures. Both these points will be clari®ed in the next section.

11.7

Evaluation of sti€ness matrices and equivalent nodal loads Ð numerical integration

At the conclusion of the previous section is was noted that we generally map the elements from the intrinsic system of coordinates ± to the global system x±y and therefore we possess the expression of the shape functions matrix ‰N…; †Š in the intrinsic system ±, while the integrations in the expressions equations (11.67) and (11.78), are to be performed in the global system x±y. Some simple mathematical manipulation is therefore required at this stage. We start from formula

336

Energy methods in structural mechanics

equation (11.67), i.e. … ‰K i Š ˆ ‰N…x; y†ŠT ‰SŠT ‰CŠ‰SŠ‰N…x; y†Š dA Si

Basically, in order to perform this integration we require the expression of the matrix ‰B…x; y†Š ˆ ‰SŠ‰N…x; y†Š 2 qN I 0 6 qx 6 6 qNI 6 ˆ6 0 qy 6 6 4 qN qN I

qy

I

qx

... ...

qNL qx

... ...

0

... ...

qNL qy

0

3

7 7 qNL 7 7 7 qy 7 7 qNL 5 qx

…11:86†

A way to attain this target is obviously by means of the explicit expression of the shape function matrix ‰NŠ in the global reference frame x±y. However, as said above, in the isoparametric formulation we hold the expression of the shape functions matrix ‰N…; †Š in the intrinsic system ± and the inversion of the transformation rules, equation (11.30), may turn out to be quite a dicult task. Fortunately, some very simple manipulation can provide the matrix ‰B…x; y†Š without requiring the inversion of equation (11.30). If we consider that the mapping correspondence equation (11.30) yields x and y as functions of  and , i.e. x ˆ x…; † and y ˆ y…; †, the chain rule of derivatives allows us to write qNj qNj qx qNj qy ‡ ˆ q qx q qy q qNj qNj qx qNj qy ‡ ˆ q qx q qy q or, in matrix form, 3 3 2 2 qNj qNj 6 qx 7 6 q 7 7 7 6 6 ˆ ‰JŠ 7 7 6 6 4 qNj 5 4 qNj 5 qy q

…11:87†

…11:88†

where ‰JŠ is the Jacobian matrix of the transformation, given by equation (11.33). Consequently, the terms qNj =qx and qNj =qy required for

The theory of ®nite elements

337

the expression of the matrix ‰B…x; y†Š are yielded by the formula 2 3 2 3 qNj qNj 6 q 7 6 qx 7 7 6 7 ÿ1 6 …11:89† 7 6 7 ˆ ‰JŠ 6 4 qNj 5 4 qNj 5 qy q Of course, the existence of the inverse of the Jacobian matrix of the transformation is guaranteed under the condition of uniqueness on the mapping discussed in section 11.3. At this point, computation of the sti€ness matrices ‰K i Š involves the integration of known functions in the global reference frame x±y over the sub-domain of the ®nite element, Si . In fact, we have … i  ‰K Š ˆ ‰B…x; y†ŠT ‰CŠ‰B…x; y†Š dA …11:90† Si

Also, this integral can be transformed back to the intrinsic system of coordinates ±. It is known from calculus that …

…1 …1 f …x; y† dA ˆ

j f~…; † d d

…11:91†

ÿ1 ÿ1

Si

with j ˆ det‰JŠ

…11:92†

f~…; † ˆ f …x…; †; y…; ††

…11:93†

and Thus the integration of equation (11.90) can be performed directly in the intrinsic system of coordinates of the parent element, given that the expression of ‰B…x; y†Š is already obtained in terms of  and  by virtue of equation (11.89). Then, we have ‰K i Š ˆ

…1 …1

j‰B…; †ŠT ‰CŠ‰B…; †Šd d

…11:94†

ÿ1 ÿ1

This integration technique applies also to the computation of the equivalent nodal loads, equation (11.78) 8… 9 … < = i T T ‰P Š ˆ ‰N…x; y†Š ‰b…x; y†Š dA ‡ ‰N…x; y†Š ‰ f …x; y†Š ds : i ; i S

ÿ

338

Energy methods in structural mechanics

In this case we have the expression of the shape functions matrix ‰N…; †Š and we can obtain the expressions ‰b…x…; †; y…; ††Š and ‰ f …x…; †; y…; ††Š of the applied loads directly from substitution of the transformation rules, equation (11.30). Consequently, we have 81 1 …1 <… … i T ‰P Š ˆ j‰N…; †Š ‰b…; †Š d d ‡ ‰N…; ÿ1†ŠT ‰ f  …; ÿ1†Š d : ÿ1 ÿ1

…1

ÿ1

T

…1



‰N…ÿ1; †Š ‰ f …ÿ1; †Š d ‡

‡ ÿ1

…1 ‡ ÿ1

where

9 = ‰N…1; †ŠT ‰ f  …1; †Š d ;



fJ f ˆ s 11 fn J11 

ÿ fn J12 ÿ fs J12

 f



‰N…; 1†ŠT ‰ f  …; 1†Š d

ÿ1

…11:95† 

fJ ˆ s 21 fn J21

ÿ fn J22 ÿ fs J22



and fs , fn are the components of the tractions f tangent and normal to the boundary, respectively. With regards to the e€ective evaluation of the integral equations (11.94) and (11.95) in the intrinsic system of coordinates, it must be noted that, however possible in a small and mostly academic number of cases, the exact integration of these expressions is quite dicult and complicated and generally not feasible in the vast majority of practical applications that may involve complex distorted elements. Making resort to the techniques of numerical integration is therefore the standard procedure in almost all ®nite element analyses. Given the introductory purposes of this textbook, we will only summarise here a few elementary principles regarding these numerical integration operations, referring the interested reader to one of the many valuable treatises readily available on the subject. One of the most e€ective methods of evaluating a de®nite integral numerically is the Gauss4 quadrature. The basic idea is well-known: to approximate a certain one-dimensional integral …1 Iˆ

f …† d

…11:96†

ÿ1 4

Gauss, Karl Friedrich (Brunswick, 1777 ± GoÈttingen, 1855), German mathematician and physicist, one of the greatest mathematicians of all times.

The theory of ®nite elements

339

f

–1

0

1

ξ

Fig. 11.24 f

ξi

ξ

Δ

Fig. 11.25

over the interval ‰ÿ1; 1Š we can take the value f …0† at the mid-point of the interval and multiply it by the interval length, i.e. 2. The result is I  2f …0† …11:97† This is exact only in the case where f …† represents the equation of a straight line, as shown in Fig. 11.24. It is clear that a better approximation can be achieved by sub-dividing the interval in many parts, Fig. 11.25, and by assuming X I f …i †
…11:98† i

where i are the mid-points of each sub-interval and
their common length. However, it may turn out that a better approximation can be achieved with the same number of sampling points for the function f …† if they are chosen at special locations i . Thus, we can write X f …i †Wi …11:99† I i

where Wi are appropriate weighting coecients, that may vary according to the sampling point i . Gauss quadrature, on the basis of the properties of Legendre5 polynomials, speci®es the position of the sampling points so that for any number n of them a very high 5

Legendre, Adrien-Marie (Paris, 1752 ± Auteuil, 1833), French mathematician.

340

Energy methods in structural mechanics Table 11.1 Sampling points and relative weights for Gauss quadrature n

i

Wi

1

0:00000 0:57735

2:00000 1:00000

0:77459 0:00000 0:86113 0:33998

0:55555 0:88888 0:34785 0:65214

2 3 4

accuracy is attained. In particular it allows us to integrate exactly a polynomial of degree 2n ÿ 1 by means of n-point sampling. Sampling points are placed symmetrically with respect to the centre of the interval and each pair of symmetric points is characterised by the same weighting coecient Wi . In Table 11.1 some examples of data for one-dimensional Gauss quadrature are presented. Extension to the case of n-dimensional integrals is very straightforward; for instance, in two-dimensional problems we can evaluate the de®nite integral …1 …1 Iˆ

f …; † d d

…11:100†

ÿ1 ÿ1

by integrating ®rst with respect to  and then with respect to , i.e.  …1 …1 …1 X  Iˆ f …; † d d  f …i ; †Wi d ÿ1 ÿ1



X X j

i

ÿ1



i

f …i ; j †Wi Wj ˆ

XX i

j

f …i ; j †Wi Wj

…11:101†

Of course, as numerical integration always introduces additional approximations in the ®nite element analyses, the following question arises naturally: what is the minimum number of sampling points necessary for convergence of the problem towards the exact solution? The answer is strictly linked to the types of element employed and appropriate tables have been developed and made available in literature. Generally, we must bear in mind that the computational cost of numerical integration is quite signi®cant, but poor treatment and lack of attention to detail at this phase of the preparation of a ®nite element program always leads to inaccurate and poorly reliability for practical structural analyses.

The theory of ®nite elements

341

p = 300 kN/m2

1·5 m

3·00 m

Fig. 11.26

11.8

Numerical example: analysis of a compressed foundation

The example concerns the analysis of a very long concrete foundation subject to a vertical pressure p ˆ 300 kN=m2 on the top, as shown in Fig. 11.26. On account of its length, we can consider the foundation to be in a state of plane strain and therefore we will limit ourselves to analysing a cross section of unit depth in the longitudinal direction, Fig. 11.27.

p

l y x

l

Fig. 11.27

l

342

Energy methods in structural mechanics p 1

2

l

y 4

x

3 l

Fig. 11.28

Moreover, on account of the symmetry about the vertical plane which divides the cross section in two mirrored parts, we can further simplify our analysis and take into account the scheme of Fig. 11.28, where the points coincident with the symmetry axis 2±4 are allowed to experience vertical displacements only. Given the simplicity of the problem, we employ triangular elements and derive the shape functions directly, without making resort to the concept of parent elements and intrinsic coordinates. Thus, we assume that the ®nite element model for the structure in Fig. 11.28 is composed of two triangular elements, as shown in Fig. 11.29. As we have seen in the previous sections, the ®rst step consists in deriving the expression of the shape functions Nj …x; y† which represent the displacement ®eld in the element as function of the displacements of its nodes.

1

2

1 l 2 3

4

l

Fig. 11.29

The theory of ®nite elements

343

uy (H)

y H

ux (H)

uy (J) uy (I)

ux (J)

J

I

ux (I)

x

Fig. 11.30

With reference to the generic triangular element of Fig. 11.30, we have     ux …x; y† NI 0 NJ 0 NH 0 ˆ ‰Di Š ˆ ‰NŠ‰Di Š uy …x; y† 0 NI 0 NJ 0 NH with

2

ux …I†

3

6 u …I† 7 7 6 y 7 6 7 6 u …J† x 7 ‰Di Š ˆ 6 6 u …J† 7 7 6 y 7 6 4 ux …H† 5 uy …H† In this example we use the simplest assumption for the deformed shape of the element, that is, the components along x and y of the displacement ®eld are described by ®rst-order polynomial expressions of the type ux …x; y† ˆ a1 ‡ a2 x ‡ a3 y uy …x; y† ˆ b1 ‡ b2 x ‡ b3 y

344

Energy methods in structural mechanics

Notice that this very simple expression has the advantage that it reduces the complexity of the algebra to obtain the sti€ness matrix for the element but the assumed polynomial function also restricts the variation of the strain in the plate to be piecewise constant. Thus for the strain "x , in the x direction, qux ˆ a2 qx and, similarly, for the other strain components. This simplicity of representation, i.e. piecewise, of a continuous strain ®eld may have implications for the number of elements that are required to obtain an accurate solution using the ®nite element method. By inserting in the above expressions the coordinates of the three nodes of the element we have the following systems of linear equations in the unknowns a1 ; a2 ; a3 and b1 ; b2 ; b3 , respectively 8 > < ux …I† ˆ a1 ‡ a2 x…I† ‡ a3 y…I† ux …J† ˆ a1 ‡ a2 x…J† ‡ a3 y…J† > : ux …H† ˆ a1 ‡ a2 x…H† ‡ a3 y…H† 8 > < uy …I† ˆ b1 ‡ b2 x…I† ‡ b3 y…I† uy …J† ˆ b1 ‡ b2 x…J† ‡ b3 y…J† > : uy …H† ˆ b1 ‡ b2 x…H† ‡ b3 y…H† "xx ˆ

From linear algebra, the solution of these systems is readily found in the form 2 3 ux …I† x…I† y…I† 6 7 a1 ˆ det4 ux …J† x…J† y…J† 5=2A ux …H† x…H†

y…H† 3 1 ux …I† y…I† 7 6 a2 ˆ det4 1 ux …J† y…J† 5=2A 1 ux …H† y…H† 3 2 1 x…I† ux …I† 7 6 a3 ˆ det4 1 x…J† ux …J† 5=2A 1 x…H† ux …H† 2 3 uy …I† x…I† y…I† 6 7 b1 ˆ det4 uy …J† x…J† y…J† 5=2A uy …H† x…H† y…H† 2

The theory of ®nite elements

2

1 61 b2 ˆ det4 1 2 1 61 b3 ˆ det4 1

345

3 y…I† y…J† 7 5=2A uy …H† y…H† 3 x…I† uy …I† x…J† uy …J† 7 5=2A x…H† uy …H† uy …I† uy …J†

where A is the area of the triangular element, and 3 2 1 x…I† y…I† 7 6 2A ˆ det4 1 x…J† y…J† 5 1 x…H† y…H† On account of these expressions, with simple calculations we ®nd that the shape functions Nj …x; y† take the form NI …x; y† ˆ f‰x…J†y…H† ÿ x…H†y…J†Š ‡ ‰y…J† ÿ y…H†Šx ‡‰x…H† ÿ x…J†Šyg=2A NJ …x; y† ˆ f‰x…H†y…I† ÿ x…I†y…H†Š ‡ ‰y…H† ÿ y…I†Šx ‡‰x…I† ÿ x…H†Šyg=2A NH …x; y† ˆ f‰x…I†y…J† ÿ x…J†y…I†Š ‡ ‰y…I† ÿ y…J†Šx ‡‰x…J† ÿ x…I†Šyg=2A At this point we can evaluate the sti€ness matrix ‰K i Š of the generic element, as given by equation (11.90) … i  ‰K Š ˆ ‰BŠT ‰CŠ‰BŠ dA Si

where ‰BŠ and ‰CŠ are, see equations (11.86) and (11.62), 2 qN

I

6 qx 6 6 ‰BŠ ˆ 6 6 0 6 4 qN

I

qy 2 6 ˆ4

0 qNI qy qNI qx

qNJ qx 0 qNJ qy

0 qNJ qy qNJ qx

qNH qx 0 qNH qy

3

0

7 7 qNH 7 7 qy 7 7 qN 5 H

qx

‰y…J† ÿ y…H†Š 0

0 ‰x…H† ÿ x…J†Š

‰y…H† ÿ y…I†Š 0

0 ‰x…I† ÿ x…H†Š

‰y…I† ÿ y…J†Š 0

3 0 7 ‰x…J† ÿ x…I†Š 5

‰x…H† ÿ x…J†Š

‰y…J† ÿ y…H†Š

‰x…I† ÿ x…H†Š

‰y…H† ÿ y…I†Š

‰x…J† ÿ x…I†Š

‰y…I† ÿ y…J†Š

346

Energy methods in structural mechanics

2 ‰CŠ ˆ

6 E 6 6 …1 ‡ †…1 ÿ 2† 4

…1 ÿ †



0



…1 ÿ †

0

0

0

1 ÿ 2 2

3 7 7 7 5

It is evident that for the assumed choice of displacement polynomials the deformation matrix ‰BŠ actually does not depend on the coordinates x and y of the points inside the domain of the element. Thus, the integration over the domain of the element is simple and straightforward, and the required sti€ness matrix ‰K i Š can be written as 1 E…1 ÿ † 4A …1 ‡ †…1 ÿ 2† 2 2 c4 ‡ c21 c4 c1 ‡ c4 c1 6 6 c21 ‡ c24 6 6 6 6
6 6 6 6 6 sym 4

‰K i Š ˆ

c4 c5 ‡ c1 c2

c4 c2 ‡ c1 c5

c6 c4 ‡ c1 c3

c1 c5 ‡ c4 c2

c1 c2 ‡ c4 c5

c1 c6 ‡ c4 c3

c25 ‡ c22

c5 c2 ‡ c5 c2

c5 c6 ‡ c2 c3

c22 ‡ c25

c2 c6 ‡ c5 c3 c26 ‡ c23

c4 c3 ‡ c1 c6

3

7 c1 c3 ‡ c4 c6 7 7 7 c3 c5 ‡ c2 c6 7 7 7 c2 c3 ‡ c5 c6 7 7 7 c6 c3 ‡ c6 c3 7 5 c23 ‡ c26

with  

 1ÿ

 

1 ÿ 2 2…1 ÿ †

and c1  x…H† ÿ x…J†

c2  x…I† ÿ x…H†

c3  x…J† ÿ x…I†

c4  y…J† ÿ y…H†

c5  y…H† ÿ y…I†

c6  y…I† ÿ y…J†

For the plane analysis being exempli®ed here, the sti€ness matrix of the element 1 (which is de®ned by the coordinates I ˆ 4, J ˆ 2, H ˆ 1), ‰K 1 Š, is readily obtained from the general expression by setting c1 ˆ ÿl

c2 ˆ 0

c3 ˆ l

c4 ˆ 0

c5 ˆ l

c6 ˆ ÿl

while the sti€ness matrix of the element 2 (which is de®ned by the coordinates I ˆ 4, J ˆ 3, H ˆ 2), ‰K 2 Š, is obtained by setting c1 ˆ 0

c2 ˆ ÿl

c3 ˆ l

c4 ˆ ÿl

c5 ˆ l

c6 ˆ 0

The theory of ®nite elements

347

Before we impose any constraints, the nodal displacement vector ‰DŠ for the model in Fig. 11. 29 is represented by the following eightelement array ‰DŠT ˆ ‰ ux …1†

uy …1†

...

. . . ux …4† uy …4† Š

so that the corresponding combined sti€ness matrix for the structure,  is an 8  8 array. This matrix can be easily obtained by adding the ‰KŠ, contributions of the sti€ness matrices of the elements 1 and 2, i.e. ‰K 1 Š and ‰K 2 Š in the appropriate positions, as formally indicated by equation (11.73). However, the presence of the constraints at the nodes 2, 3 and 4 allows only the horizontal and the vertical displacement of the node 1 and the vertical displacement of the node 2. Therefore, the nodal displacement vector ‰DŠ reduces to 3 2 ux …1† 7 6 ‰DI Š ˆ 4 uy …1† 5 uy …2† and the joint sti€ness matrix of the structure, ‰KI Š, becomes a 3  3 array. Therefore the joint sti€ness matrix is seen to be 2 3 1 1 1 K55 K56 K54 6 7 1 1 7 ‰KI Š ˆ 6 K66 K64 4 5 1 2 sym …K44 ‡ K66 † 1 E…1 ÿ † 4A …1 ‡ †…1 ÿ 2† 2 3  l 2 ÿ l 2 † l 2 …l 2 ‡ l 2 † …ÿ 6 7 …l 2 ‡ l 2 † ÿl 2 7
6 4 5 2 2 sym …l ‡ l † 3 2 …1 ‡ † …ÿ  ÿ †  E…1 ÿ † 7 6 ˆ …1 ‡ † ÿ 5 4 2…1 ‡ †…1 ÿ 2† sym …1 ‡ †

ˆ

given that A ˆ l 2 =2. The analytical method described above is applied here to a concrete foundation which is composed of material with the following properties: Young's modulus and Poisson's ratio: E ˆ 3:11  107 kN=m2 and  ˆ 0:15, respectively. The weight per unit volume is ˆ 22 kN=m3 .

348

Energy methods in structural mechanics

Thus, we have  ˆ 0:176

 ˆ 0:412 2 : 3 1 412 ÿ0:588 0:412 1:412 ÿ0:412 5 ‰KI Š ˆ 1:642  107 4 sym 1:412

The vector of the equivalent nodal loads is given by equations (11.78) and (11.81), i.e. 8… 9 … < = i T T ‰NŠ ‰bŠ dA ‡ ‰NŠ ‰ f Š ds ‰P Š ˆ : i ; S ÿi X i ‰PŠ ˆ ‰P Š i

with

 ‰bŠ ˆ

0



 ‰fŠ ˆ

ÿ22

0

 on the side 1ÿ2

ÿ300

For element 1 we have N4 ˆ …l 2 ÿ ly†=l 2 ˆ …l ÿ y†=l N2 ˆ lx=l 2 ˆ x=l N1 ˆ …ÿlx ‡ ly†=l 2 ˆ …y ÿ x†=l and for element 2 N4 ˆ …l 2 ÿ lx†=l 2 ˆ …l ÿ x†=l N3 ˆ …lx ÿ ly†=l 2 ˆ …x ÿ y†=l N2 ˆ ly=l 2 ˆ y=l so that the vector of the nodal forces corresponding to the active degrees of freedom is seen to be 2 3 2 3 0 0 6 7 6 7 …l …l …l 6 7 6 7 6 7 6 yÿx lÿx 7 6 7 6 ÿ300 ÿ22 dx dy dx 7 6 7 6 7 l l 7‡6 7 ‰PI Š ˆ 6 x 0 0 6 7 6 7 0l 1 6 7 6 7 … …l …l …l …x 6 7 6 7 6 7 6 7 x y x 4 ÿ22@ dx dy ‡ dx dyA 5 4 ÿ300 dx 5 l l l 0

x

0

0

0

The theory of ®nite elements

349

Notice that for the sake of clarity we kept the contribution of the body forces and that of the applied surface load p separate. With a fairly simple integration we obtain 3 3 2 2 0 0 6 6 l2 7 l 7 7 7 6 6 6 ÿ22 7 6 ÿ300 7 ‰PI Š ˆ6 6 7‡6 27 7 7 6 6 5 5 4 4 l2 l ÿ300 ÿ22 2 3 3 3 2 2 3 2 0 0 0 7 7 6 6 7 6 : 7 : 7 6 7 6 ˆ6 4 ÿ8 25 5 ‡ 4 ÿ225 5 ˆ 4 ÿ233 25 5 ÿ241:50 ÿ225 ÿ16:5 It is straightforward to verify that the sum of the nodal forces corresponding to the body forces due to the self-weight of the concrete are equal to half of the weight of the whole section, i.e.

 A ˆ 22  2:25 ˆ 49:5 kN (remember that we are dealing with a plane strain problem and therefore we are examining a slice of unit depth). Only half the weight is included in the analysis of the deformation because the remaining part of the total weight would be applied to the constrained nodes 3 and 4 with no e€ect on the deformation of the foundation. Conversely, the resultant of the surface load p, i.e. p  l ˆ 300  1:5 ˆ 450 kN/m, is applied to the nodes 1 and 2 in the form of two forces of equal value. Finally, the equilibrium equations of the problem is ‰KI Š‰DI Š ˆ ‰PI Š )

32 3 ux …1† 1:412 ÿ0:588 0:412 76 7 6 1:412 ÿ0:412 54 uy …1† 5 ˆ 1:642  107 4 uy …2† sym 1:412 2 3 0 6 :25 7 ÿ233 4 5 2

ÿ241:50

the solution of which is 2 3 ÿ2:14418  10ÿ6 m 6 7 ‰DI Š ˆ 4 ÿ0:150952  10ÿ4 m 5 ÿ0:141951  10ÿ4 m

350

Energy methods in structural mechanics

It is worth noting that if we neglect the e€ect of the self-weight, we can write 3 32 3 2 2 ux …1† 0 1:412 ÿ0:588 0:412 7 76 7 6 6 1:642  107 4 1:412 ÿ0:412 54 uy …1† 5 ˆ 4 ÿ225 5 sym and get

1:412

uy …2†

ÿ225

2

3 ÿ2:148710  10ÿ6 m 6 7 ‰DI Š ˆ 4 ÿ0:144809  10ÿ4 m 5 ÿ0:133029  10ÿ4 m

The strain for each element can be evaluated by means of equations (11.64), (11.7) and (11.86), that is 3 2 "xx 6 " 7 i i 4 yy 5 ˆ ‰SŠ‰NŠ‰D Š ˆ ‰BŠ‰D Š 2"xy Thus, we have 2 3 2 3 "xx 0:145038  10ÿ4 6 " 7 6 : 7 4 yy 5 ˆ 4 ÿ0 977461  10ÿ4 5 2"xy ÿ0:655216  10ÿ5 3 2 3 2 "xx 0 6 " 7 6 : ÿ4 7 4 yy 5 ˆ 4 ÿ0 897944  10 5 2"xy 0 for elements 1 and 2, respectively. In order to validate the quality of the solution obtained, we can refer to the exact value of the strain component "y which is ÿ0:964630  10ÿ4 . It is found that even with the extremely coarse mesh adopted, the ®nite element solution di€ers from the exact one by no more than 7:5%. If a more exact solution were required for the ®nite element modelling there are two possible ways forward. One is simply to re®ne the mesh geometry de®ning the elements, that is to include more elements in the analysis. This is a simple and straight forward operation once the element sti€ness matrix has been formulated in a general manner. Indeed this approach is the one that most engineers would follow when they use commercial programs for ®nite elements. They would carry out a sensitivity analysis to determine, by progressively re®ning

The theory of ®nite elements

351

the element mesh geometry, the geometry that would give an adequately accurate answer and ensure that the answer is not sensitive to the change of mesh geometry. An alternative approach would be to assume a higher order polynomial to describe the variation of the deformations ux …x; y†, uy …x; y†, in place of the simple linear variation assumed in this example. This alternative approach requires the re-formulation of the sti€ness matrix for a generic element. The potential advantage for this approach is that possibly the number of elements required to converge to an adequately accurate solution may be reduced.

12. Stability of equilibrium and non-linear deformations of beam±columns Introduction

The great majority of this text is devoted to the presentation of the linear response of systems of bars and plates to applied loading. That is, the structural deformations can be modelled by a linear relationship between the loading and the strains induced by the loads. A basic assumption is that the deformations of the structural elements are suciently small that the points of application and directions of the loads can be considered as unchanged during the deformations and always acting on the undeformed structure. Despite the increased complexity of incorporating non-linearity into the analysis even for linearly elastic materials, we feel that some introduction to these types of phenomena and solution techniques is a valuable complement to the linear analysis methods presented in earlier chapters. This chapter therefore is concerned with a development of the analysis of some of the simplest structural elements that exhibit geometric non-linearity, mainly the beam-column. The deformations of this type of element have a signi®cant e€ect on the geometry of the applied loading. In this presentation we shall restrict the theoretical development to linear elastic materials.

12.1

The concept of stability. Initial buckling of columns

Most of us are familiar with the concept of a mechanical system which is capable of exhibiting stable and unstable con®gurations of equilibrium. Perhaps the most simple of these systems is that already introduced in chapter 6, i.e. the ball subject to a gravitational ®eld in the landscape of hills and valleys of Fig. 6.1.

354

Energy methods in structural mechanics

P P

Fig 12.1

In section 6.1 we stated that the total potential energy of the system coincides with the load potential ÿPh…s†, which represents the capability of the gravitational force P to perform work on account of the height h…s† of the ball. If no work tends to be performed by the gravitational force, the principle of conservation of energy assures that no kinetic energy can be gained by the system and it is in static equilibrium. However, it is intuitively clear that the state of equilibrium attained at the top of a hill is qualitatively very di€erent from the equilibrium attained at the bottom of a valley. In fact, let us consider the static forces, acting as a result of a very small displacement about an equilibrium state, as a means of inferring the likely dynamic response. If the static forces acting upon the system after such a small displacement about an equilibrium state (i.e. the gravitational force and the frictionless reaction of the support) are in a direction which tends to restore the system to its original equilibrium state we consider that the original position is a stable equilibrium state, but if the static forces are such that they tend to upset the system then the original position is considered to be an unstable equilibrium state. In the case under consideration, if the ball is at the bottom of a valley then the gravitational force P will always tend to restore the ball to its original position of equilibrium after any small displacement about this point, as shown in Fig. 12.1. On the contrary, if the ball is at the top of a hill then the gravitational force P will always tend to move the ball away from its original position of equilibrium for any small displacement about this point, see Fig. 12.2.

P P

Fig 12.2

Stability of equilibrium and non-linear deformations

P

355

P

Fig 12.3

Of course, from a theoretical point of view we may also encounter cases of neutral equilibrium. With reference to the present example this may happen when the ball is on a perfectly horizontal plane surface and therefore it may be placed in any displaced con®guration and still be in a state of equilibrium, Fig. 12.3. From a mathematical standpoint a situation of neutral equilibrium can also be imagined as a limiting case dividing stable from unstable equilibrium, but from a practical point of view in the analysis of structures this is not a signi®cant case. The simplest problem of elastic instability in the theory of structures is that concerning lateral buckling of compressed members. With reference to Fig. 12.4, which shows a slender bar subject to a compressive axial load N, it is easy to ®nd experimentally that for a certain range of values of the axial force, say from 0 to Nc , the axial shortening of the bar is directly proportional to the applied load. This happens according to the deformation relationship, equation (4.5), stated in section 4.2 for an axially loaded bar in the linear theory of elasticity. However, once the axial load has reached the N Δl

N

Nc

Δl

N

Fig 12.4

356

Energy methods in structural mechanics N

u (l/2)

N

Fig 12.5

critical value, Nc , the bar de¯ects laterally and the value of the de¯ection is found not to be directly proportional to the value of applied load, even if the material is perfectly linearly elastic. At the onset of the lateral de¯ection we say that the bar experiences initial buckling. Of course, as long as we remain in the realm of the linear theory of elasticity and under the hypotheses formulated in chapter 4, such a behaviour cannot be mathematically described, given that the deformation relationship, equation (4.5), holds for any value of the axial load. Still, the reason for the discrepancy between the predictions of the model introduced in chapter 4 and the observations in tests on slender columns is quite simple. Essentially, if we perturb the equilibrium of a slender axially loaded bar, for example by slightly displacing its mid-span, the axial load will cause a bending moment given by the product of the force and the distance between its direction and the deformed axis of the beam, Fig. 12.5. As long as the elastic sti€ness of the beam is capable of restoring the original con®guration of equilibrium, the equilibrium is stable and the only equilibrium con®guration is the straight one. However, when the magnitude of the axial force is such that the bending moment yielded by the perturbation exceeds the restoring capacity of the elastic sti€ness of the beam, other displaced con®gurations of equilibrium become possible and the column will bend laterally. In this case the straight con®guration of equilibrium is classi®ed as unstable, because, as in the case of the ball at the crest of the hill of

Stability of equilibrium and non-linear deformations

357

Fig. 12.2, the column always departs from the crest as a result of any small displacement of its position. From a practical point of view it must also be said that it is experimentally very dicult to keep the column in a straight con®guration of equilibrium once the critical value of the axial load Nc has been attained because this implies the absence of any imperfection in the specimen or external perturbation of the column. It is now clear that the modelling of axially loaded bar illustrated in chapter 4 ignores the additional moment induced by perturbations or imperfections of the axis of the column and is therefore incapable of describing the buckling phenomena. Intuitively it is anticipated that initial buckling a€ects only slender bars, being dependent on the length and on the bending sti€ness of the element. In fact, for a squat structural member it is found that yielding and plastic deformation of the material occurs before any lateral initial buckling takes place. However, there is a large range of practical compressed elements that can be analysed using the theory of linear elasticity. The mathematical problem of the lateral instability of compressed members was solved in the 18th century by the Swiss mathematician Leonhard Euler, who also formulated the equation of the equilibrium for laterally loaded beams presented in chapter 4. In the realm of large displacements this theory leads to the formulation of what is generally known as the Euler's equation of the elastica. We will present the Euler theory of compressed bars, i.e. struts, in section 12.4, limiting our treatment to relatively small de¯ections. First, however, in the next section, we present an energy criterion for stability and give some elementary examples.

12.2

Energy criterion of stability

According to the simple example of the ball discussed previously, the understanding of the dynamic response of mechanical systems to disturbances is the basis of stability studies. However, from our point of view, that is the analysis of structures, this route is probably not the most convenient one to follow and we shall now introduce a very useful and intuitive alternative approach based on the potential energy functional. We will assume that if the potential energy is a local minimum the structure is stable with respect to small disturbances, and if the potential energy is a relative maximum the structure is correspondingly unstable. In order to attempt a rigorous proof of the link between dynamical response to disturbances and the potential energy, a great deal of work, which is outside the scope of this introductory textbook, would be

358

Energy methods in structural mechanics

required. It must also be noted that in recent times some scienti®c objections, related to very special physical systems, have been directed towards this classical energy criterion. However, these special systems do not impinge on the structural models considered here. Therefore, given its intuitive basis, we will assume the criterion above to be a statement which is con®rmed by correspondence between experimental results and analytical predictions made using this assumption. Alternatively, the following simple line of reasoning can be considered. In chapter 6 we stated the principle of stationary value of the total potential energy, that is the relationship  ˆ 0 holds in the equilibrium state of conservative systems (which in statics of structures is always the speci®c state for which the stability is investigated). Let us assume that the potential functional  can be made dependent on a system parameter that speci®es the position, i.e.  ˆ …q†. As a consequence, …q0 † ˆ 0 holds for the value q0 which corresponds to equilibrium. For a small variation of q, we have q 1 q2  …q0 ‡ q† ˆ …q0 † ‡ q ‡ q2 ‡


qq q0 2 qq2 q0 ˆ …q0 † ‡ …q0 † ‡ 12 2 …q0 † ‡


ˆ …q0 † ‡ 12 2 …q0 † ‡




…12:1†

Thus, in the vicinity of the equilibrium position de®ned by q0 the potential energy  is a relative maximum if 2  < 0. This means that potential energy is released and kinetic energy gained when q is varied from q0 and the system will tend to move away from the equilibrium position that is therefore unstable. On the contrary, if 2  > 0, potential energy is gained when q is varied from q0 and the system cannot therefore leave the position of equilibrium by itself, that is the equilibrium is stable. The stability boundary is given by 2  ˆ 0, which corresponds to neutral equilibrium. As an historical remark, it should be noticed that these results make the energy criterion formally identical to Dirichlet's theorem for a point system.

12.3

Some illustrative examples of stability conditions. Branching points and limit points.

To illustrate the application of the energy criterion, let us start by considering the elementary model in Fig. 12.6. This is a rigid cantilever elastically hinged at the bottom end and subject to a vertical force P at the upper end. The rotational spring is an idealised device that is capable of resisting rotations by a reactive moment. A familiar example might be a common door spring that restores the door to

Stability of equilibrium and non-linear deformations

359

P

l

l cos φ

φ

M = –kφ

Fig 12.6

its closed state. The relationship between rotation and the reactive moment M is M ˆ ÿk

…12:2†

 being the rotation about the hinge axis and k the rotational sti€ness of the spring. By virtue of Clapeyron's theorem, the strain energy stored in the spring upon rotation is U…† ˆ 12 k2

…12:3†

In order to compute the work done by the vertical load P, we must remove the hypothesis of small displacements. The total potential energy of the system becomes, consequently …† ˆ 12 k2 ÿ P
ˆ 12 k2 ÿ Pl…1 ÿ cos †

…12:4†

and the equation of equilibrium is  ˆ 0

)

q ˆ k ÿ Pl sin  ˆ 0 q

…12:5†

This equilibrium equation is always satis®ed by the trivial solution  ˆ 0, which corresponds to the perfectly vertical position of the bar. If we wish to examine the stability of such a con®guration, we

360

Energy methods in structural mechanics P

Pc

φ

Fig 12.7

must examine the term 1 2 1 q2  2  …0† ˆ  2 2 q2 0 Since 2 is always positive, the above term reduces to q2  ˆ k ÿ Pl cos…0† q2 0

…12:6†

…12:7†

This means that for P < k=l the con®guration of equilibrium  ˆ 0 is stable, while for P > k=l the equilibrium  ˆ 0 is unstable. The value of the applied load P ˆ k=l constitutes the boundary between stable and unstable equilibrium and that load is called the critical load, Pc . Apart from the trivial solution, equation (12.5) is also satis®ed by the values of P and  satisfying the relationship   k  Pˆ …12:8† l sin  In the space P ÿ  this equilibrium path has the form shown in Fig. 2.7 and intersects the trivial or fundamental path,  ˆ 0 coincident with the load axis at the point Pc ˆ k=l. Actually, this point, apart from being the boundary between stable and unstable con®gurations on the fundamental path, constitutes a branching point in the equilibrium path. Thus the critical load Pc makes a bifurcation in the deformation state. The rising post-buckling path, i.e. increasing deformation with increasing load, given by the equation (12.8) is easily found to be

Stability of equilibrium and non-linear deformations

everywhere stable, given that q2  ˆ k ‡ Pl cos  > 0 q2 

for

 6ˆ 0

and P >

k l

361

…12:9†

In other words, if we perturb the equilibrium of the cantilever in its fundamental position of equilibrium, it will tend to return to it as long as P < k=l. On the contrary, if P > k=l any small disturbance will cause the cantilever to move away from the fundamental path  ˆ 0 and it will tend to become on the rising equilibrium branch, which is always stable. Of course a perfectly straight cantilever loaded by a perfectly axial force is a pure idealisation and in reality some imperfections will always a€ect this simple model. In order to examine such a situation, let us consider the cantilever in Fig. 12.8, where the vertical load P is slightly o€set from the vertical axis of the bar. In such a case the total potential energy of the system becomes …† ˆ 12 k2 ÿ P
ˆ 12 k2 ÿ Pl…1 ÿ cos † ‡ Pe sin 

…12:10†

e

P

P e sin φ

l cos φ

l

φ

M = –kφ

Fig 12.8

362

Energy methods in structural mechanics

on account of the presence of the small eccentricity e. Thus, the condition of equilibrium is  ˆ 0

)

q ˆ k ÿ Pl sin  ‡ Pe cos  ˆ 0 q

…12:11†

By using the ratio e " l

…12:12†

as a measure of the loading imperfection, the equation of equilibrium (12.11) becomes   k  Pˆ …12:13† l sin  ÿ " cos  Plotting the equilibrium paths yielded by equation (12.13), as was done in Fig. 12.9, we notice that there is no longer a bifurcation condition. This is due to the fact that, on account of the loading eccentricity, the trivial equilibrium path has disappeared. Nevertheless, the value of the critical load Pc ˆ k=l for the corresponding perfect system retains all its importance because it represents the magnitude of the load at which the system starts to present three possible and di€erent con®gurations of equilibrium. Moreover, if we proceed to examine the stability of the equilibrium paths by studying the sign of the parameter q2  ˆ k ‡ Pl cos  ÿ Pe sin  q2

…12:14†

P

3

2

1

φ

Fig 12.9

Stability of equilibrium and non-linear deformations

363

P l cos α

φ

α l

l

Fig 12.10

we discover that the equilibrium paths 1 and 3 are stable, while the equilibrium path 2 is unstable. To summarise the results, we may say that the imperfect system of Fig. 12.8, loaded from its unloaded state, will follow a constantly rising stable path so that no instability is encountered. The de¯ections will simply grow more rapidly as the critical load of the perfect system is passed. In addition to this natural equilibrium path, the system will also exhibit a complementary equilibrium path in the other quadrant of the P ÿ  space, which is partly stable and partly unstable and cannot be normally encountered in a natural loading process from P ˆ 0. To complete our presentation of simple examples, we include a di€erent type of instability, which may occur also in perfect systems without the presence of bifurcation of the equilibrium path. To this purpose, we can consider the shallow arch of Fig. 12.10. The arch comprises two axially deformable bars pinned to each other and to rigid foundations. It is evident from the simple equilibrium of rigid bodies that for any position, de®ned by the rotation angle , the bars are always subject to axial loading only. The deformation law (4.5) allows us to write
l ˆ

N l N ˆ EA cos  k

…12:15†

where k is the axial sti€ness of the bar. By de®ning the rotation angle  as the parameter describing the con®guration of the system, we can write the following expression for the strain energy of the system   2  1 l l Uˆ2 k ÿ …12:16† 2 cos  cos… ÿ †

364

Energy methods in structural mechanics

The corresponding work of the applied load P is W ˆ P‰l tan  ÿ l tan… ÿ †Š

…12:17†

and the total potential energy of the system is therefore 2    1 l l ÿ …† ˆ 2 k ÿ P‰l tan  ÿ l tan… ÿ †Š 2 cos  cos… ÿ † …12:18† The present example is simpli®ed if we restrict it to a shallow arch, that is we can take both  and  to be small and write cos   1 ÿ tan   ;

2 ; 2

cos… ÿ †  1 ÿ

… ÿ †2 ; 2

…12:19†

tan… ÿ †   ÿ 

so that equation (12.18) becomes …† ˆ 14 kl 2 …42 2 ÿ 43 ‡ 4 † ÿ Pl

…12:20†

Once again, the condition of equilibrium is  ˆ 0

)

q ˆ kl 2 …22  ÿ 32 ‡ 3 † ÿ Pl ˆ 0 q

…12:21†

and yields P ˆ kl…22 ÿ 3 ‡ 2 †

…12:22†

By plotting this relationship in the P ÿ  space, we can visualise the equilibrium path, see Fig. 12.11. P

E

A

D

B

0

2α

α

C

Fig 12.11

φ

Stability of equilibrium and non-linear deformations

365

First of all we notice that there are three equilibrium positions for P ˆ 0, i.e.  ˆ 0,  ˆ  and  ˆ 2. In order to investigate the stability of the equilibrium path, we again make reference to the sign of the second derivative q2  ˆ kl 2 …22 ÿ 6 ‡ 32 † q2

…12:23†

In this manner we ®nd that the two unloaded equilibrium con®gurations at  ˆ 0 and  ˆ 2 are both stable, while the unloaded equilibrium con®guration at  ˆ  is unstable. Moreover, the path 0±A is stable, as well as the rising path C±E, while the path A±B±C is unstable. In other words, starting from the unloaded origin of the co-ordinates, the arch equilibrium will be described by the stable rising path 0±A. At this point the equilibrium would require a diminishing load along the path A±B until zero magnitude is attained at B. From B to C the magnitude of the load would be required to increase again but with opposite direction from the initial one, until point C is reached. In reality, on attaining the point A under the prescribed value of the load P, the arch will move dynamically to the state D, this is termed snap buckling. In a practical case the energy released in the rapid transition from the geometry at A to that at D will eventually be dissipated by vibrations due to the presence of some damping and the system will come to rest in this position. Subsequent increments of the load will cause the inverted arch equilibrium states to follow the stable rising path D±E. In conclusion, in this example there are no branching points and the quality of the equilibrium changes from stable to unstable at some points, namely A and C, which we will call limit points. In other words, this is a form of progressive instability in the sense that the structure tends to lose its sti€ness gradually as an e€ect of the change of con®guration due to increasing values of loading. At the end of the presentation of these simple examples, it seems clear that when we have to investigate the stability of equilibrium we must start by removing one of the fundamental assumptions of the linear theory of elasticity, that is on account of the smallness of the displacement ®eld the forces acting on the body can be always referred to the undeformed con®guration. In the next section we will see what this implies for Euler's equation of beam deformation.

12.4

Euler's equation of beam deformation in the presence of axial loads

We have already encountered Euler's equation of beam deformation when dealing with laterally loaded beams in chapter 4. Euler was

366

Energy methods in structural mechanics P B

Δ

l

z A

y

Fig 12.12

also the ®rst mathematician to provide a complete and correct solution to the problem of an axially loaded elastic bar. Thus, at this point of the presentation we wish to illustrate how it is possible to write the total potential energy functional for a beam subject to axial compressive loading and establish, exactly in the same manner as we did in chapter 4 for the laterally loaded beams, the di€erential equation of the problem. We start by considering the simply supported beam±column of Fig. 12.12. However, the procedure is absolutely general and can account, as we will see, for any kind of boundary conditions. For the sake of simplicity, we consider the beam±column to have uniform bending sti€ness EIx . If we neglect the axial extensibility, the only component of the strain energy is that due to bending, is once more given by equation (4.33) …l 1 Uˆ EIx u002 y dz 2 0

Since the load P is axially applied, its potential is related to the vertical displacement of the sliding support of the structure
. Given that we have neglected any axial extensibility, this displacement is dependent only on the lateral de¯ection of the beam±column. However, in

Stability of equilibrium and non-linear deformations

367

chapter 4 and in the realm of the theory of very small displacements, we assumed that u02 y 1 and therefore this term was discarded in order to obtain the simpli®ed expression of the curvature, equation (4.25), from the exact relationship, equation (4.23). This also implied that the axis of the de¯ected cantilever of Fig. 4.8 had, according to equations (4.27) and (4.28), the same length of its projection on the z-axis so that the horizontal displacement of its right end could be considered zero under bending. We now need to remove this hypothesis in order to evaluate the displacement of the sliding support. However, it must be stressed that we will not remove the assumption u02 y  1 with reference to the simpli®cation of the curvature relationship equation (4.23) into equation (4.25). In other words, we assume to have small strains but, with respect to the applied loading, not negligibly small displacements. This fact may appear incongruous from a mathematical point of view, but it must be borne in mind that we are always working with idealised models of real systems and that in order to describe a phenomenon a model must be simpli®ed enough to allow a solvable mathematical treatment. Formulating a problem is, in fact, asking questions whereas the solution is answering those questions. As long as we obtain a solution which gives acceptable correspondence with experimental observations, we may be satis®ed that we have grasped the essence of the phenomenon under investigation. Of course, every simpli®cation introduces a degree of approximation and we are going to see how the present treatment will allow only the evaluation of the initial buckling load, inasmuch as it is not able to provide any information about the post-buckling equilibrium states. Thus, let us consider a generic element of the beam±column in its displaced con®guration, as shown in Fig. 12.13. From the theorem of Pythagoras1 we see that …dz ÿ duz †2 ˆ dz2 ÿ du2y

…12:24†

so that we have q dz ÿ duz ˆ dz2 ÿ du2y

0

)

1 s  2 duy A dz duz ˆ @1 ÿ 1 ÿ dz …12:25†

1

Pythagoras (Samos, 571±479

BC),

Greek philosopher and mathematician.

368

Energy methods in structural mechanics duy

duz

dz dz

z

y

Fig 12.13

Also, from calculus we know that, by means of a truncated series expansion, we can write p x 1ÿx1ÿ …12:26† 2 thus equation (12.25) becomes duz ˆ 12 u02 y dz

…12:27†

As a result, the vertical displacement of the sliding roller of the beam± column of Fig. 12.12 can be computed as …l 1 02
ˆ duz ˆ u dz 2 y …l 0

…12:28†

0

and the total potential energy functional of the system can be written as …l …l 1 1 02 002 EIx uy dz ÿ P u dz …uy † ˆ U ÿ P
ˆ 2 2 y 0

0

…12:29†

Stability of equilibrium and non-linear deformations

369

Once again, the stationary value of the functional …uy † gives the equilibrium states of the system. Thus …l  ˆ

EIx u00y u00y

…l

dz ÿ P u0y u0y dz ˆ 0

0

…12:30†

0

By repeatedly integrating by parts, and taking into account the same boundary conditions exempli®ed in section 4.4 (and in a more general way, in section 4.5), we get …l  ˆ

EIx u0000 y uy

…l

dz ‡ P u00y uy dz ˆ 0

0

…12:31†

0

which, since the virtual displacement ®eld uy is arbitrary, is equivalent to the fourth-order di€erential equation 00 EIx u0000 y ‡ Puy ˆ 0

…12:32†

This equation is called Euler's equation of beam buckling, and its solution represents the deformed con®guration of the elastic beam±column under the speci®ed axial load P. By dividing both terms by the bending sti€ness EIx and setting s P  …12:33† EIx we can write 2 00 u0000 y ‡  uy ˆ 0

…12:34†

The general integral of such equation is uy …z† ˆ A sin z ‡ B cos z ‡ Cz ‡ D

…12:35†

where A, B, C and D are constants which need to be determined by means of the geometrical and mechanical conditions at the supports A and B. These constraints require that u…z† ˆ 0

for

zˆ0

M…z† ˆ ÿEIx u00y ˆ 0

and z ˆ l for

zˆ0

…12:36† and z ˆ l

…12:37†

370

Energy methods in structural mechanics

Therefore we have an homogeneous linear system of four equations in the four unknown constants A, B, C and D. The system is 8 uy …0† ˆ B ‡ D ˆ 0 > > > > > < ÿEIx u00y …0† ˆ EIx 2 B ˆ 0 …12:38† > uy …l† ˆ A sin l ‡ B cos l ‡ Cl ‡ D ˆ 0 > > > > : ÿEIx u00y …l† ˆ EIx A2 sin l ‡ EIx B2 cos l ˆ 0 or, in matrix form 2 0 0 6 0 2 6 6 4 sin l cos l 2 2  sin l  cos l

1 0 l 0

2 3 0 6 7 6 7 0 76 B 7 6 0 7 7 76 7 ˆ 6 7 1 54 C 5 4 0 5 1

32

A

D

0

3

…12:39†

0

A homogeneous linear system of equations always admits the trivial solution A ˆ B ˆ C ˆ D ˆ 0, which corresponds to the fundamental equilibrium of the beam±column, i.e. its straight con®guration. From linear algebra we also know that other solutions are possible if, and only if, 3 2 0 0 1 1 6 0 2 0 07 7 6 det6 …12:40† 7ˆ0 4 sin l cos l l 15 2 sin l

2 cos l

0

0

Therefore we are led to a transcendental equation (it involves trigonometric functions) in the parameter . The solution of equation (12.40) is found to be PE ˆ n2

2 EIx l2

…12:41†

where n is a generic integer. For n ˆ 1 we get the minimum magnitude of the axial load for which a de¯ected con®guration of equilibrium is possible. This is usually called the Euler buckling load, PE , or the critical load. It must be noted that in correspondence to this buckling load the matrix equation (12.39) will admit a solution which is di€erent from the trivial one, but the solution is modi®ed by an arbitrary constant. Thus the buckled con®guration represented by the equation (12.35)

Stability of equilibrium and non-linear deformations

371

can be obtained in shape, but not in magnitude. In Fig. 12.14 the buckled shapes for n ˆ 1, n ˆ 2 and n ˆ 3 are represented. As we anticipated before, the fact that we are left only with the critical buckling load and not with a relationship between the load and the corresponding de¯ection is to be ascribed to neglecting the e€ect of higher-order terms in the de¯ection±curvature relationship, equation (4.23). A fully non-linear equilibrium equation enabling the study of post-buckling equilibrium paths can be obtained quite easily by formulating the exact expression of the curvature, but it is not integrable in an elementary manner and is therefore outside the scope of this introductory book.2 It is however interesting to notice that a similar behaviour to the Euler initial buckling formulation might be found also in the previously examined example of the elastically hinged cantilever of Fig. 12.6. In fact, if we expand the total potential energy expression (equation (12.4)) with reference to the term cos  and retain only up to the second 2 Nevertheless, the analysis of the post-buckling behaviour of the simply supported beam of Fig. 12.12 can be tackled in an approximate way by means of the trial function method. In fact, with a line of reasoning similar to equation (4.23), we can write the curvature of the de¯ected column as

u00y 1 d ˆ ˆ ÿ q r ds 1 ÿ u02 y

where, however, this time the derivatives are taken with respect to the curvilinear coordinate of the undeformed axis of the beam, s. Moreover, we can write  2 u002 1 y 02 ˆ  u002 y …1 ‡ uy † r 1 ÿ u02 y Thus, the total potential energy is ˆ

…l …l 1 P 02 02 EIx ‰u002 …1 ‡ u †Š ds ÿ u ds y y 2 2 y 0

0

By assuming a trial function of the type   ns uy ˆ A sin l and performing the substitution in the total potential energy expression, setting d=dA ˆ 0 gives Pˆ

n2 2 ‰8EIx l 2 n ‡ 4A2 EIx n3 3 ÿ 4EIx l 2 sin…2n† ÿ A2 EIx n2 2 sin…4n†Š 4l 4 ‰2n ‡ sin…2n†Š

which constitutes the load±displacement relationship in post-buckling.

372

Energy methods in structural mechanics

a2

a3

a1

PE = π

2

EI l2

PE = 4π 2 EI 2

l

PE = 9π 2 EI 2

l

Fig 12.14

order terms, we have 1 1 2 …† ˆ k2 ÿ Pl…1 ÿ cos † ˆ k2 ÿ Pl 2 2 2

…12:42†

and the equation of equilibrium results  ˆ 0

)

q ˆ k ÿ Pl ˆ 0 q

…12:43†

Equation (12.43) gives the value of the critical load Pc ˆ k=l as the value of the load at which a non-trivial con®guration of equilibrium becomes possible, but it fails to give information about the bifurcated equilibrium paths. However, from a practical point of view engineers are mostly interested in designing structures with a degree of safety such that the critical load is not attained and therefore the knowledge only of the magnitude of such a load is generally sucient for design analysis purposes together with an understanding of the e€ects of imperfections in the structure loading or geometry.

12.5

Analytical treatment of some examples of columns

The analysis in this section will apply quite systematically the method introduced above for obtaining the load at which the onset of lateral deformations are initiated. In the previous section we have seen that the load at which the perfect column becomes unstable and develops lateral deformations is called the Euler, or critical load. The form of

Stability of equilibrium and non-linear deformations

373

buckling considered in that section is thus called Euler buckling or sometimes bar buckling to distinguish the phenomenon from the other forms of buckling exhibited by plates and shells. It has also been noted that a perfect column is an idealisation of reality and implies that . . .

the column is perfectly straight before the application of the perfectly axial loading the material is perfectly linearly elastic, homogeneous and isotropic the column cross section is perfectly uniform along its length.

Therefore, in the following we will take into account other very common examples of beam±columns, these are the rotational restrained column, the column with a lateral load and the column with initial out-of-straightness. The analytical treatment in this section reduces the degree of idealisation and the e€ect of lateral loads on the column behaviour eliminates the condition of the unique critical load behaviour. Nevertheless, the later analysis emphasises the value of the Euler load as an important parameter characterising the behaviour of a column subjected to increasing load levels

Column with various degrees of rotational restraint

A column very seldom exists as a separate entity, it is usually a component in a frame and as such it has a ¯exural restraint at its ends from the contiguous members of the frame. The e€ect of this restraint, which is a function of the relative ¯exural sti€ness of the column and the members attached to it are considered here. Figure 12.15(a) shows the deformed shape of an initially perfect column subjected to a load P applied along the initial axis. The moments m at the end of the column are applied restraints from other members attached to the column and in the present example the restraints are equal at both ends. Note that these restraining moments have zero value until the column begins to de¯ect and that they will increase in magnitude as the magnitude of the column lateral de¯ection increases. We refer to Euler's equation of initial buckling (12.32) 00 EIx u0000 y ‡ Puy ˆ 0

and again, by setting s P  EIx

374

Energy methods in structural mechanics P

P m Q z

φ uy (z)

φ

φ Q m

m

P

P (a)

(b)

Fig 12.15

we obtain 2 00 u0000 y ‡  uy ˆ 0

for which the general solution is given by equation (12.35) uy …z† ˆ A sin z ‡ B cos z ‡ Cz ‡ D where A, B, C and D are constants determined from the boundary conditions as follows u…z† ˆ 0 Mˆ

EIx u00y

for ˆ ÿm

zˆ0

and z ˆ l

for z ˆ 0 and

…12:44† zˆl

Thus, we can write the following system 8 uy …0† ˆ B ‡ D ˆ 0 > > > > < EI u00 …0† ˆ ÿEI B2 ˆ m x y x > u …l† ˆ A sin l ‡ B cos l ‡ Cl ‡ D ˆ 0 y > > > : EIx u00y …l† ˆ ÿEIx A2 sin l ÿ EIx B2 cos l ˆ m

…12:45†

…12:46†

Stability of equilibrium and non-linear deformations

375

which, contrary to the linear system equation (12.38) found in the previous section, is not homogeneous in the present case. Its solution yields   m 1 ÿ cos l m l Aˆÿ ˆÿ tan EIx sin l EIx 2 …12:47† m m Bˆÿ C ˆ 0 D ˆ EIx 2 EIx 2 so that uy …z† ˆ ÿ

m l m m tan sin z ÿ cos z ‡ 2 EIx 2 EIx  EIx 2

Now, let us de®ne the coecient of restraint  as m  

…12:48†

…12:49†

then, since we have u0y …0† ˆ …0† ˆ ÿ

m l tan EIx 2

…12:50†

we get tan

l 2EIx ˆÿ 2 l

…12:51†

l 2 l ˆÿ 2 2

…12:52†

or also tan

with  l=EIx . The degree of ¯exural restraint applied to the column from a contiguous member provides the speci®c value of  in equation (12.49) and hence in equation (12.52) which can then be solved to provide the appropriate value of the Euler load. The Euler load obtained from the solution of equation (12.52) can be written as

2 EI …12:53† l2 where is the particular coecient related to the value of . An alternative approach and convenient to describing the Euler load is to de®ne an e€ective length le such that PE ˆ

PE ˆ

2 EIx le2

…12:54†

376

Energy methods in structural mechanics 1·0

P 0·9 le

le /l

0·8 P

(b )

P m

0·7

le 0·6

m P

(a )

0·5 0

5

10

15 β = αl/EI

20

25

30

β

Fig 12.16

where

l le  p

…12:55†

Curve (a) in Fig. 12.16 shows the variation of the e€ective length with increasing values of . Curve (b) show the corresponding variation for columns in which one end is restrained and the other is simply supported. The free body diagram for this case is shown in Fig. 12.15(b) in which the asymmetry of the constraint moments requires that a lateral load Q be applied at the ends of the column. This load, as in the case of the restraint moments ÿm is zero until the column experiences lateral de¯ections at the Euler load. The physical interpretation of the e€ective length le is made more clear in Fig. 12.17 where the de¯ected shape of the columns at the respective Euler loads are shown. The analysis shows that for a column simply supported at each end, the e€ective length is le ˆ l, and for a column encastre at both ends, le ˆ l=2. The corresponding condition for a column encastre at one end and simply supported at the other has le ˆ 0:71l. These are shown graphically in Fig. 12.17(a) to (c) respectively. The e€ective length is in fact the distance between the points of in¯ection, i.e. zero moment, in the de¯ected shape of the columns. The case shown in

Stability of equilibrium and non-linear deformations P

P

P

377

P

l l e = 0·71l l e = 0·5l

le = l

(a)

l e = 2l

(b)

(c)

(d)

Fig 12.17

Fig. 12.17(d) is for a column encastre at one end and totally free from constraint at the other. The analysis shows that le ˆ 2l, that is the Euler load for this column is 25% of that for a column simply supported at both ends. This form of buckling mode is called sway buckling and is of particular importance to the design of frames since the buckling load is so low.

Column with a lateral load

In practice a column very seldom is in a condition where it experiences only a load aligned with the axis. More often, the loading is a combination of axial and lateral loading, as shown in Fig. 12.18. The following analysis considers the de¯ections imposed on a column due to the loading. This is for simplicity of presentation, but the method of analysis and, more importantly, the results can easily be applied to the other forms of boundary constraint considered up to now. The lateral load Q is applied at mid-length of the beam, whose behaviour is therefore symmetrical with respect to this cross section. Given the presence of the lateral load Q, the total potential energy of the system is given by the equation (12.29) plus the addition of the potential relative to this lateral load, that is …uy † ˆ U ÿ P
1 ÿ Q
2 …l …l 1 1 02 002 ˆ EIx uy dz ÿ P u dz ÿ Q
2 2 2 y 0

0

…12:56†

378

Energy methods in structural mechanics P Δ1

Q

Δ2 z

Fig 12.18

On account of the symmetry of the de¯ection with respect to the midspan section, we can express the lateral de¯ection at the mid-span
2 as follows l=2 …


2 ˆ 0

u0y

…l 1 0 dz ˆ u dz 2 y

…12:57†

0

so that …l …l …l 1 1 02 1 0 002 ˆ EIx uy dz ÿ P u dz ÿ Q u dz 2 2 y 2 y 0

0

…12:58†

0

As usual, the equilibrium is given by …l  ˆ

EIx u00y u00y

…l dz ÿ P

0

u0y u0y

0

…l 1 dzÿQ u0y dz ˆ 0 2

…12:59†

0

and by integrating by parts and taking into account the boundary conditions, we get …l ÿ 0

0 EIx u000 y uy

…l dz ÿ P 0

u0y u0y

…l 1 dzÿQ u0y dz ˆ 0 2 0

…12:60†

Stability of equilibrium and non-linear deformations

379

which, since the virtual ®eld of rotations u0y is arbitrary, is equivalent to the third-order di€erential equation 0 EIx u000 y ‡ Puy ˆ ÿ

Q 2

…12:61†

By integrating both the left and the right hand terms, we obtain EIx u00y ‡ Puy ˆ ÿ

Q z ‡ const: 2

…12:62†

and by evaluating the above equality at z ˆ 0, it is immediatly possible to see that the constant of integration is zero. Thus, using the de®nition in equation (12.33) once again, we have the following equilibrium equation u00y ‡ 2 uy ˆ ÿ

Q z 2EIx

…12:63†

Note that also in this case the equilibrium equation is no longer a homogeneous equation, as it was the case for the perfect column, and the general solution of equation (12.63) is uy …z† ˆ A sin z ‡ B cos z ÿ

Q z 2EIx 2

…12:64†

where A and B are constants evaluated using the boundary conditions, that are, on account of the symmetry with respect to the mid-span uy …0† ˆ 0

and

u0y …l=2† ˆ 0

…12:65†

Finally, the de¯ected shape of the column is 0 1 Q sin z uy ˆ ÿ zA @ 2EIx 3 cos z 2

…12:66†

and the de¯ection at mid-length of the column is   Q l l ÿ uy …l=2† ˆ tan 2 2 2EIx 3

…12:67†

Note that as P ! PE

)

l  ! 2 2

)

uy …l=2† ! 1

…12:68†

that is, as the axial load increases close to the magnitude of the Euler load for the column, the de¯ection becomes very large. The magnitude

380

Energy methods in structural mechanics

of the de¯ections of the column are clari®ed by using the approximation    2  3  l  l ÿ ÿ1 l 4 2 2 12 tan  …12:69†  2  2 2  l ÿ 2 2 This approximation enables equation (12.67) to be written as 0 1 Ql 3 1 …12:70† uy …l=2†  PA 48EIx @ 1ÿ PE Thus it can be seen that when the axial load, P, is zero the initial de¯ection at the mid-length of the beam corresponds to that due to the lateral load Q only. This is uy …l=2† ˆ

Ql 3 48EIx

…12:71†

As the axial load is increased in magnitude, the de¯ection increases in a non-linear manner until, as stated above, when the axial load approaches the magnitude of the Euler load the de¯ections become very large. If the initial de¯ection uy …l=2† is de®ned as the de¯ection at zero value of P, then the variation of de¯ection with load is uy …l=2† ˆ

uy 1ÿ

P PE

…12:72†

It is evident from equation (12.72) that the Euler load is the governing parameter in determining the increase in the de¯ection as the axial load is applied. The e€ect of the ampli®cation factor is shown in Fig. 12.19.

Column with initial out-of-straightness

The previous example has shown how a column with a lateral load Q initially will have a lateral de¯ection and that de¯ection will grow nonlinearly as the axial load P is imposed. It was assumed in section 12.4, where the perfect column was analysed, that the initial form of the column was perfectly straight. That is an idealisation of the actual initial condition of a practical column which due to the manufacturing process will have some degree of out-of-straightness along its length. The present example deals with the evaluation of de¯ections from

Stability of equilibrium and non-linear deformations

381

P

PE

uy (l/2)

uy

Fig 12.19

that initial condition as the axial load P is imposed on such an imperfect column. The initial deformations of the column are de®ned as u0y …z†, as shown in Fig. 12.20(a). According to equation (4.25), they de®ne the curvature of the axis as 1 ˆ ÿu000y r0

…12:73† P

u0y (z)

uy (z)

z

P (a)

Fig 12.20

(b)

382

Energy methods in structural mechanics

so that the total potential energy of the system is …l …l 1 1 00 00 2 02 EIx …uy ÿ u0y † dz ÿ P …u02 …uy † ˆ y ÿ u0y † dz 2 2 0

…12:74†

0

The equilibrium equation is therefore …l  ˆ

EIx …u00y

ÿ

u000y †u00y

…l

dz ÿ P u0y u0y dz ˆ 0

0

…12:75†

0

and by integrating by parts and taking into account the boundary conditions, we get …l

EIx …u00y

ÿ

u000y †u00y

…l

dz ‡ P uy u00y dz ˆ 0

0

…12:76†

0

which, since the virtual ®eld of curvatures u00y is arbitrary, is equivalent to the second order di€erential equation …u00y ÿ u000y † ‡ uy ˆ 0

…12:77†

with  given by equation (12.33). The initial de¯ected shape of a column will be dependant on the manufacturing process and the treatment the column experiences during fabrication. The analysis is simpli®ed if the initial de¯ection, u0y …z†, is represented as a trigonometric series u0y …z† ˆ b1 sin

z 2z 3z ‡ b2 sin ‡ b3 sin ‡


l l l

…12:78†

The components, b1 , etc., could be evaluated if an actual column was being analysed and its initial de¯ected shape had been measured. However, for the moment it suces that the trigonometric series representation is feasible. Substitution of equation (12.78) into equation (12.77) gives  2  2  z 2 2z u00y ‡ uy ˆ ÿ ÿ


b1 sin ÿ b2 sin l l l l

…12:79†

Stability of equilibrium and non-linear deformations

The general solution of equation (12.79) is  2  z l b1 sin uy …z† ˆ A sin z ‡ B cos z ‡  2 l  ÿ 2 l  2  4 2z l ‡


‡  2 b2 sin l  2 4 ÿ l

383

…12:80†

where the constants A and B are determined by the following boundary conditions uˆ0

for

zˆ0

and z ˆ l

…12:81†

Hence Bˆ0

and

A sin l ˆ 0

…12:82†

If sin kl ˆ 0, then at distinct values of the load, that is the eigenvalues l ˆ n, the magnitude of A can have any value. But at a value of loading di€erent from the eigenvalues, sin kl 6ˆ 0 and therefore A ˆ 0. Thus, provided the consideration is restricted to load levels less than the Euler load, it is permissible to substitute A ˆ 0 in equation (10.80). Consequently uy ˆ

b1

P 1ÿ PE

sin

z ‡ l

b2

P 1ÿ 4PE

sin

2z ‡ l

b3

P 1ÿ 9PE

sin

3z ‡


l …12:83†

It is clear from equation (10.83) that as the axial load P increases, the denominators in the right hand terms will decrease non-linearly. Thus, provided the magnitudes of the coecients b1 , b2 , etc. are fairly similar, the contribution to the magnitude of the de¯ection of the column will increase and the rate of increase with the load will be greater for the term with b1 compared to the others. Actually, generally it is b3 > b3 > b3 > . . ., so that as P ! PE the contribution of the other terms becomes negligible and without introducing much error, equation (12.83) can be approximated as uy 

b1 z sin l …1 ÿ P=PE †

…12:84†

384

Energy methods in structural mechanics

and the de¯ection at mid-length of the column is uy …l=2† ˆ

b1 1ÿ

…12:85†

P PE

Note the similarity between equation (12.85) and equation (12.72). In the latter the initial de¯ection is speci®ed by the actions of the lateral load Q, and in the former the initial de¯ections occur due to manufacturing e€ects. In this second case the practical levels of initial de¯ection are determined by measurements made on a wide range of practical column geometries. In both cases, the initial de¯ections are ampli®ed by the action of the axial load P and it is evident that the Euler load PE , although derived on the basis of idealised column geometry and loading, is a very important parameter in determining the practical behaviour of columns.

12.6

Method of trial functions for the initial buckling of columns

The method for applying trial functions to the evaluation of the Euler loads for columns is very similar to that considered in chapter 7 for beams. The method is exempli®ed ®rst by application to the perfect columns analysed above, with uniform cross-section geometry and with simple support and encastre boundary conditions, and then by showing the power of the method by considering a perfect column in which one half has a ¯exural rigidity di€erent from the other half. For a perfect column with uniform cross-section geometry, the total potential energy of the system, equation (12.29), is …l …l 1 1 02 002 …uy † ˆ U ÿ P
ˆ EIx uy dz ÿ P u dz 2 2 y 0

0

As usual, the application of the method of trial functions is simpli®ed if the energy formulation and the trial functions are non-dimensionalised. Thus we will assume EI  ˆ 3x 2l

…1  2 …1  2 2 d uy duy P dÿ d 2 2l d d 0

Simply supported column

0

with  

z l

…12:86†

The ®rst example concerns the derivation of the Euler load for the column simply supported at both ends and subjected to an axial

Stability of equilibrium and non-linear deformations

385

load P, shown in Fig. 12.12. The boundary conditions, in non-dimensional form, are uy …0† ˆ uy …1† ˆ 0

…12:87†

The simplest function satisfying these conditions is uy ˆ uy sin 

…12:88†

Substituting equation (12.88) into equation (12.86) and performing the integration gives ˆ

EIx 4 2 P2 2 uy ÿ u 4l y 4l 3

…12:89†

Thus the functional relationship, equation (12.86), has been replaced by an approximate but very simple algebraic expression. The equilibrium condition is q ˆ0 q uy

…12:90†

Hence   EIx 2 ÿ P uy ˆ 0 l2

…12:91†

Two solutions exist for equation (12.91), either uy ˆ 0 or PE ˆ

2 EIx l2

…12:92†

Thus we have obtained exactly the same results for the Euler load which were obtained from the exact analytical approach presented in section 12.4. However, in this case it is evident that the choice of trial function coincided with the exact mode of de¯ection of the column at the Euler load.

Encastre column

The boundary conditions in non-dimensional form for this column (Fig. 12.17(b)) are uy ˆ 0 and

duy ˆ0 d

at  ˆ 0

and  ˆ 1

…12:93†

A simple function that satis®es these conditions is uy ˆ 12 uy …1 ÿ cos 2†

…12:94†

386

Energy methods in structural mechanics

Substitution of this into the energy functional, equation (12.86), and performing the integration gives ˆ

EIx 4 2 P2 2 u uy ÿ 4l y l3

…12:95†

from which by means of the equilibrium condition the Euler load is obtained as PE ˆ

42 EIx l2

…12:96†

which again corresponds exactly to the expression we would have found by the analytical approach.

Column with step-change in ¯exural rigidity

The previous two examples of the application of the method of trial functions to the evaluation of the Euler load have shown how simple the method is, but have not provided information beyond that available using the exact analytical approach. However, we know that the merit in the method of trial functions is that it can provide practically useful approximate solutions in cases where the exact analysis is quite complex and sometimes not feasible. This advantage is exempli®ed here by applying the trial functions to the evaluation of the Euler load for a column with a step change of ¯exural rigidity. Figure 12.21 shows a perfect column that has half its length with a ¯exural rigidity of I1 and the other half with I2 . The total potential energy functional for this condition is 2 3 1=2 …1  2 …1  2 2 …  2 2 d uy d uy duy EI1 6 7 P d‡I d5 ÿ d ˆ 34 2l d 2l d 2 d 2 0

where I

I2 I1

1=2

0

(12.97)

…12:98†

We start the analysis by means of a one-term trial function. The column is simply supported at both ends therefore a suitable trial function is uy ˆ uy sin 

…12:99†

Stability of equilibrium and non-linear deformations

387

P

ζ EI1

l/2

EI2 l/2

Fig 12.21

which substituted in equation (12.97) gives, after integration   EI1 2 4 4 2 P2 2 ‡I u ÿ u …12:100† ˆ 4 4 y 4l y 2l 3 Applying the equilibrium conditions, the Euler load is obtained as …1 ‡ I† 2 EIx …12:101† 2 l2 Evidently when I ˆ 1, i.e. the column has a uniform ¯exural rigidity along its length, the Euler load corresponds to the exact value. The accuracy of the one-term trial function can be tested using more extensive trial functions. A two-term function can be, for example, PE ˆ

uy ˆ u1 sin  ‡ u2 sin 2

…12:102†

Substituting this function in the total potential energy functional, performing the integrations and applying the equilibrium conditions, i.e. q q ˆ 0; ˆ0 …12:103† q u1 q u2

388

Energy methods in structural mechanics

gives the equations …1 ‡ I† u1 ‡

32 …1 ÿ I† u2 ÿ p u1 ˆ 0 3

…12:104†

32 …1 ÿ I† u1 ‡ 16…1 ‡ I† u2 ÿ 4p u2 ˆ 0 3 where p

2Pl 2 2 EI1

…12:105†

The value of the critical loads are obtained from the zero value of the determinant of equation (12.104), that is 32 ‰…1 ‡ I† ÿ pŠ …1 ÿ I† 3 …12:106† ˆ0 32 …1 ÿ I† ‰16…1 ‡ I† ÿ 4pŠ 3 Expanding the determinant gives the quadratic equation 2    32 2 2 4p ÿ 20p…1 ‡ I† ‡ 16…1 ‡ I† ÿ …1 ÿ I† ˆ 0 …12:107† 3 Figure 12.22 shows a plot of the Euler load obtained from the solution of equation (12.107) for varying values of I, compared to the corresponding values from equation (12.101). It may be seen that for I ! 1 the two solutions converge and give the Euler load that agrees with the exact analytical value. However, as the degree of step-change of the ¯exural rigidity along the column 2·5 p1 one degree of freedom

p1, p2

2·0

p2 two degrees of freedom

1·5

1·0

0·5 0·2

0·4

0·6

0·8 I

Fig 12.22

1·0

1·2

Stability of equilibrium and non-linear deformations

389

increases the one-term solution diverges from the supposedly more accurate two-term solution.

Column with step-change in ¯exural rigidity and a lateral load

We now apply the method of trial functions to the case where the column with a step-change in ¯exural rigidity has also an initial lateral load Q. This load is applied at mid-length, as shown in Fig. 12.23. The non-dimensional expression of the total potential energy functional for this condition is …1  2 …1  2 2 d uy duy EIx P ˆ 3 d ÿ d ÿ Q
2 …12:108† 2 2l d 2l d 0

0

With reference to Fig. 12.23, the last term in this functional is relative to the potential energy of the lateral load during the increase in the de¯ection
2 at the point of application of the load. The two-term trial functions described in equation (12.102) are applied to the energy functional, equation (12.116) and after the appropriate P

ζ EI1

Δ2

l/2

Q

EI2 l/2

Fig 12.23

390

Energy methods in structural mechanics

integration operations, the equilibrium conditions equation (12.103) q q ˆ 0; ˆ0 q u1 q u2 are applied. Note that in this example, and for the trial functions chosen, the de¯ection at the point of application of the lateral load is speci®ed by
2  u1 , so that the potential energy term associated with the lateral load is Q
2  Q u1 . Thus, we get 32 u2 ˆ q ‰…1 ‡ I† ÿ pŠ u1 ‡ …1 ÿ I† 3 …12:109† 32 …1 ÿ I† u1 ‡ ‰16…1 ‡ I† ÿ 4pŠ u2 ˆ 0 3 where p

2Pl 2 2 EI1

and q 

2Ql 3 4 EI1

…12:110†

Once more, these equations can be put in matrix format 3 2 32     …1 ÿ I† ‰…1 ÿ I† ÿ pŠ q 7 u1 6 3 ˆ …12:111† 5 4 32 u2 0 …1 ÿ I† ‰16…1 ‡ I† ÿ 4pŠ 3 The solution of equations (12.111) to evaluate the development of the lateral de¯ections are exempli®ed in Fig. 12.24 for the conditions I ˆ 0:2, q ˆ 0:1. Figure 12.22 shows that the non-dimensional Euler load is found to be p ˆ 0:745 by means of the same two-degrees of freedom trial function, 20

u1 u2

u1 (p), u2 (p)

15

10

5

0 0

Fig 12.24

0·1

0·2

0·3

0·4 p

0·5

0·6

0·7

0·8

Stability of equilibrium and non-linear deformations

391

2·5 2·0

uy (ζ)

1·5 1·0 0·5 0 –0·5

0·2

0·4

0·6 ζ

0·8

1·0

1·2

Fig 12.25

equation (12.102). Figure 10.24 shows that as the value of p increases to close to the Euler load the de¯ections of the column increase in a nondimensional manner. The de¯ected shape is a combination of the two terms in the trial function. The shape is exempli®ed for p ˆ 0:7, where Fig. 12.25 shows the de¯ection shape is asymmetrical about the midlength of the column due to the step-change of the ¯exural rigidity.

12.7

Localised trial functions applied to the initial buckling of struts

Section 7.6 introduced the concept of localised trial functions that in e€ect are an automated way of choosing appropriate functions. This approach actually can be considered the basis for the ®nite element method and chapter 11 has presented in some detail the more formalised aspect of the ®nite element application in practice. The treatment in this section starts once again by considering the socalled perfect column, that is the condition in which the column is initially perfectly straight and the load P is applied perfectly co-axial with the column, then it takes into account the column with a lateral load and the column with a step-change of ¯exural rigidity. The development follows closely that of section 7.6.

Perfect column

The total potential energy of the system is given by equation (12.29), i.e. …l …l 1 1 02 002 EIx uy dz ÿ P u dz …uy † ˆ U ÿ P
ˆ 2 2 y 0

0

The practical application of the method is exempli®ed with respect to a

392

Energy methods in structural mechanics P

Node 1

Element 1

Node 2

Element 2

Node 3 P

Fig 12.26

column built-in at both ends, Fig. 12.26. For the sake of simplicity we treat the column as having two elements. The strain energy can be particularised for an element, that is a length of the column ln between nodes, as a function of the nondimensional length . Thus we can write …1  2 n 2 d uy EI Un ˆ 3 d …12:112† d 2 2ln 0

Exactly in the same manner we can evaluate the shortening of the vertical projection of each element as …1  n 2 duy 1 d …12:113†
n ˆ 2ln d 0

so that the total potential energy of the system can be written as 2 3 2 1 3 …1  2 n 2 …  n 2 n n X X d uy duy EI 1 4 4 d 5ÿP d 5 …12:114† n ˆ 3 2 2l d d 2l n n iˆ1 iˆ1 0

0

In the particular case considered here, n ˆ 2 and ln ˆ l=2.

Stability of equilibrium and non-linear deformations

393

The development in section 7.6 demonstrated that for laterally loaded beams an e€ective trial function, yet maintaining simplicity, is uny …† ˆ ui ÿ 1 …1 ÿ 3 2 ‡ 2 3 † ‡ i ÿ 1 … ÿ 2 2 ‡  3 † ‡ ui …3 2 ÿ 2 3 † ‡ i …ÿ 2 ‡  3 †

…12:113†

where ui ÿ 1 , ui , are the displacements at nodes i ÿ 1 and i, respectively. As before, for the slopes we set n  n ln . Substitution of equation (12.113) into equation (12.114) and performance of the stationary conditions for equilibrium, i.e. qn ˆ 0; q ui ÿ 1

qn ˆ 0; qi ÿ 1

qn ˆ 0; q ui

qn ˆ0 qi

…12:114†

yields, in the present case 2 32 3 u1 24 12 ÿ24 12 0 0 6 12 7 6 8 ÿ12 4 0 0 76 1 7 6 7 6 76 7 ÿ24 ÿ12 48 0 ÿ24 12 76 u2 7 4EI 6 6 76 7 6 7 l3 6 4 0 16 ÿ12 47 6 12 76 2 7 6 76 7 4 0 0 ÿ24 ÿ12 24 ÿ12 54 u3 5 0 0 12 4 ÿ12 8 3 32 3 2 u1 36 3 ÿ36 3 0 0 7 6 3 6 4 ÿ3 ÿ1 0 0 76 1 7 6 7 76 7 6 36 ÿ3 72 0 ÿ36 3 76 u2 7 P 6 76 7 ˆ 0 6 ÿ 6 7 15l 6 0 8 ÿ3 ÿ1 7 76 2 7 6 3 ÿ1 76 7 6 …12:115† 4 0 0 ÿ36 ÿ3 36 ÿ3 54 u3 5 0 0 3 ÿ1 ÿ3 4 3 The boundary conditions correspond to u1 ˆ 1 ˆ u3 ˆ 3 ˆ 0 thus, the system equation (12.115) reduces to       4EI 48 0 u2 P 72 0 u2 ÿ ˆ0 15l 0 8 2 l3 0 16 2

…12:116†

…12:117†

De®ning a non-dimensional load as p

Pl 2 60EI

…12:118†

394

Energy methods in structural mechanics

equation (12.117) can be written as       72 0 u2 48 0 u2 ÿp ˆ0 ) 0 8 2 0 16 2    u2 …48 ÿ 72p† 0 ˆ0 0 8…2 ÿ p† 2

…12:119†

As usual in the case of perfect columns, the system equation (12.119) admits the trivial solution, i.e. u2 ˆ 2 ˆ 0, for any value of p. Values of p that allows the system equation (12.119) to have a solution di€erent from the trivial one correspond to the zero value of the determinant on the matrix. That is …48 ÿ 72p† 0 ˆ0 …12:120† 0 8…2 ÿ p† Expanding this determinant, we have …0:776 ÿ p†…2 ÿ p† ˆ 0

…12:121†

Evidently the roots of this equation are p ˆ 0:667 or 2. Intuitively the lesser of these values is the one that is practically relevant since the lowest buckling load of a column will govern the onset of the buckling deformations. Thus, in this case p ˆ 0:667 is the relevant value and gives the value of the load at which instability of the column occurs as PE ˆ …40EI†=l 2 . This compares closely with the analytical value PE ˆ …42 EI†=l 2 even if, of course, the choice of the trial function, equation (12.113), cannot represent the trigonometric shape yielded by the Euler di€erential equation. It is interesting to notice that in the development shown above it is possible, and trivial, to expand the determinant since there are only two elements and therefore with the boundary conditions this reduces the problem to a 2  2 matrix. When the number of elements included in the modelling is increased the order of the matrix increases also and the e€ort to expand the determinant becomes prohibitive. Apart from this fact, algebraic equations of an order greater than four cannot be solved algebraically. The approach that is used to obtain the lowest buckling load in these circumstances is therefore di€erent. If we put equation (12.119) in a totally symbolic form, we can write ‰AŠ‰uŠ ÿ p‰BŠ‰uŠ ˆ …‰AŠ ÿ p‰BŠ†‰uŠ ˆ …‰BŠÿ1 ‰AŠ ÿ p‰IŠ†‰uŠ ˆ 0 …12:122†

Stability of equilibrium and non-linear deformations

395

where ‰uŠ is the vector of the degrees of freedom of the system and ‰IŠ is the identity matrix. Thus, the required values of the critical loads p are provided by the eigenvalues of the matrix ‰BŠÿ1 ‰AŠ, which is a standard algebraic eigenvalue problem which can be solved by means of a classical Jacobi procedure. Of course, the least of these values is the relevant value for the onset of large deformations. The method of localised Rayleigh functions, or ®nite elements, is further exempli®ed here by considering the analysis of a column simply supported at its ends. In such a case the boundary conditions for the two-element model are u1 ˆ u3 ˆ 0

…12:123†

so that eliminating the corresponding rows and columns from equation (12.115) we can arrange the matrix equation in the form (12.122) ‰AŠ‰uŠ ÿ p‰BŠ‰uŠ ˆ 0 where

2

Pl 2 ; p 60EI 2

4

6 6 ÿ3 6 ‰BŠ ˆ 6 6 ÿ1 4 0

8

6 6 ÿ12 6 ‰AŠ ˆ 6 6 4 4 ÿ3

ÿ1

72

0

0

8

3

ÿ1

ÿ12

4

0

3

7 0 12 7 7 7; 0 16 4 7 5

48

0 12 4 8 3 2 3 0 1 7 6 7 6 u2 7 37 7 6 7 ; ‰uŠ ˆ 7 6 7 7 6 2 7 ÿ1 5 4 5 4

…12:124†

3

The result from the calculations are that the eigenvalues of the matrix ‰BŠÿ1 ‰AŠ are 0:166, 0:894, 2:145 and 3:36. The lowest of these values is the one of practical signi®cance and gives the buckling load as PE ˆ …10EI†=l 2 , which as in the previous example, compares very well with the analytical value PE ˆ …2 EI†=l 2 .

Column with lateral load

In this section the formatting of the localised trial functions method is considered for the case of a column with a lateral load. The method is exempli®ed by application to a column with uniform ¯exural rigidity and simple support conditions at both ends. This is shown in Fig. 12.27(a).

396

Energy methods in structural mechanics P Node 1

Element 1

Δ2

Node 2 Q

Element 2

Node 3

(a)

(b)

Fig 12.27

The energy formulation, in non-dimensional terms and for n elements, is 2 3 …1  2 n 2 n X d uy 4 EI n ˆ d 5 3 d 2 2l n iˆ1 0

n X

2

41 ÿP 2ln iˆ1

3 …1  n 2 duy d 5 ÿ Q
2 d

…12:125†

0

where
2 is the lateral displacement of the node at which Q is applied. By dividing the column in two elements, Fig. 12.27(b), we have
2  u2 . With the employment of the trial function equation (12.113), the formulation of the sti€ness matrix ‰AŠ and of the axial loading matrix ‰BŠ follows as in the development presented for the perfect column. Thus the equilibrium equations, after imposition of the boundary conditions, can be written in the form ‰AŠ‰uŠ ÿ p‰BŠ‰uŠ ˆ q‰EŠ

…12:126†

Stability of equilibrium and non-linear deformations

397

1·0 0·8

P/PE

0·6 0·4 0·2 0 0

9

18

Δ2/Δ02

27

36

45

Fig 12.28

where p

Pl 2 ; 60EI

q

Ql 3 ; 48EI

2 3 0 617 7 ‰EŠ ˆ 6 405 0

…12:128†

The matrices ‰AŠ, ‰BŠ and the vector ‰uŠ are as de®ned in equation (12.124). The system equation (12.126) is no longer, as may be expected given the presence of the lateral load, an homogeneous one. The displacement vector can be obtained for a speci®c value of the applied axial load P and of the initial lateral load Q from ‰uŠ ˆ ‰A ÿ pBŠÿ1 q‰EŠ

…12:129†

Figure 12.28 shows the variation of the de¯ection at mid-length of the column, i.e. where the lateral load is applied, with the increase in the axial load. The de¯ection is plotted as the ratio of the total de¯ection
2 to the initial de¯ection caused by the lateral load alone
02 . The variation of the axial load is plotted as the ratio of the load to the Euler load PE . Not surprisingly, the localised trial function method replicates the non-linear increase in the de¯ection that is predicted by the exact analytical method.

Column with step-change of ¯exural rigidity

In this example the localised trial function method is applied to a column that has a step-change of ¯exural rigidity form EI1 to EI2 at mid-length, see section 12.6. The matrix that is a€ected by this change of rigidity is ‰AŠ and using the de®nition I  I2 =I1 , the matrix

398

Energy methods in structural mechanics

for the complete beam for a two-element model can be written as 2 3 24 12 ÿ24 12 0 0 6 12 8 ÿ12 4 0 0 7 6 7 6 ÿ24 ÿ12 24…I ‡ 1† 12…I ÿ 1† ÿ24I 7 12I 6 7 ‰AŠ ˆ 6 4 12…I ÿ 1† 8…I ‡ 1† ÿ12I 4I 7 6 12 7 4 0 0 ÿ24I ÿ12I 24I ÿ12I 5 0 0 12I 4I ÿ12I 8I …12:130† Applying the boundary conditions relevant to simple support conditions, u1 ˆ u3 ˆ 0

…12:131†

and eliminating the appropriate rows and columns in the matrix above, the resulting system is again in the form shown in equation (12.122) ‰AŠ‰uŠ ÿ p‰BŠ‰uŠ ˆ 0 with

2

p

Pl 2 ; 60EI1 2

4 6 ÿ3 ‰BŠ ˆ 6 4 ÿ1 0

8 ÿ12 4 6 ÿ12 24…I ‡ 1† 12…I ÿ 1† ‰AŠ ˆ 6 4 4 12…I ÿ 1† 8…I ‡ 1† 0 12I 4I 2 3 3 1 ÿ3 ÿ1 0 6 7 72 0 37 6 u2 7 7; ‰uŠ ˆ 6 7 0 8 ÿ1 5 4 2 5 3 ÿ1 4 3

3 0 12I 7 7 4I 5 8I …12:132†

This equation can be solved for speci®c values of I and the results compared to the corresponding values from section 12.6. The following typical results are obtained PE ˆ 

2 EI1 l2

Trial function One-term Two-term Localised (two elements)

I ˆ1 1 1 0.99

 I ˆ 0:5 0:75 0:67 0:64

I ˆ 0:2 0:60 0:37 0:30

Stability of equilibrium and non-linear deformations

399

Thus the localised trial function agrees fairly well with the two-term trial functions described earlier with the greater advantage that the localised method can be automated and the numbers of elements increases very easily.

12.8

Method of trial functions for the initial buckling of frames

The previous sections, namely section 12.6 and section 12.7, have presented the application of the method of trial functions to the calculation of the critical load of columns with various end geometric constraints. A column is, of course, usually a component in a frame structure and it has the important attribute that when it attains its critical load the frame, or perhaps a region of the frame, will experience the onset of buckling. The ultimate load analysis of frames is the subject of many papers and textbooks and it is not the purpose of the presentation in this section to emulate the scope of these publications. Instead this section introduces, by means of a simple example portal frame, the application of the method of trial functions to the calculation of the critical loads for frame structures. The presentation starts by considering the analysis of the portal frame shown in Fig. 12.29, when subjected to the loads P=2 aligned co-axially with the vertical members of the frame. Both the vertical members have the same ¯exural sti€ness EI1 and the horizontal member B±C has a ¯exural sti€ness EI2 . As in all applications of the trial function method, we must ®rst consider the likely modes in which the frame will deform at the critical loads. Figure 12.30 shows two possible modes: (a) sway and (b) symmetric. There may be other modes of deformation, and each will have its corresponding critical load. However, observations of simple tests show that the two modes in Fig. 12.30 are related to the two lowest critical loads. The trial functions we choose to represent these modes are developed and incorporate the compatibility condition that the frame is ¯exurally rigid at joints B and C. Thus the basic assumption is that the changes of slope of the vertical and horizontal members have a common value at these joints.

Sway mode

With reference to Fig. 12.31, and using the non-dimensional variables 1 

z1 l

2 

z2 l

…12:133†

400

Energy methods in structural mechanics l

P/2

P/2 B

C EI2

EI1

EI1

l

A

D

Fig 12.29

The appropriate mode shapes are Member A±B uAB y



ˆ a1



1 1 ÿ cos 2

 …12:134†

This function satis®es the geometric constraints of zero slope and zero

P/2

P/2

(a)

Fig 12.30

P/2

P/2

(b)

Stability of equilibrium and non-linear deformations

401

φB P/2 B

Δl

a 2 = a 1/4

φB B

C u (z 2)

a1 z2 u (z 1)

l

(b) z1

A (a)

Fig 12.31

de¯ection at 1 ˆ 0. The displacement and slope at 1 ˆ 1 are AB 1 duy  AB AB uy …B† ˆ a1 and  …B† ˆ ÿ ˆ ÿ a1 l d 2l 1

Member B±C uBC y ˆ a2 sin…22 †

1 ˆ 1

(12.135) …12:136†

The slope of member B±C at the joint with the vertical member, i.e. at 2 ˆ 0, is BC 1 duy 2 ˆ ÿ a2 …12:137† BC …B† ˆ ÿ l d2 2 ˆ 0 l Thus from the condition of compatibility of slope at the joint, a …12:138† a2 ˆ 1 4 Hence a1 uBC sin…22 † …12:139† y ˆ 4 It is evident that in this manner the compatibility condition at the joint C with the column C±D is automatically satis®ed, provided we assume for the column C±D the same shape function of the member A±B, equation (12.134).

402

Energy methods in structural mechanics

The expression of the total potential energy for the frame results   …1  2 AB 2 …1  2 BC 2 d uy d uy EI1 EI2 ˆ2 d1 ‡ 3 d2 2l 3 d12 2l d22 0

0

  …1  AB 2 duy P ÿ d1 2l d1

…12:140†

0

Substituting the chosen trial functions, we have ˆ

EI1 4 2 EI2 4 2 2 2 a a a ‡ ÿ P 1 1 32 1 l 3 32 l3 2

…12:141†

The equilibrium condition is obtained from the stationary value of the total potential energy, thus   4 d  EI1 4 EI2 2 P a ˆ0 ˆ ‡ ÿ …12:142† da1 8l 1 16l 3 2l 3 and, consequently, we have PE ˆ

2 EI1 ‰1 ‡ 8IŠ 2l 2

…12:143†

where I  I2 =I1 . When the member B±C has a very small value of ¯exural sti€ness compared to the vertical members, A±B and C±D, i.e. I ! 0, it is evident that the critical load for the frame approaches the critical load for initial buckling of a simple cantilever encastre at the bottom end, i.e. PE ˆ …2 EI1 †=4l 2 , by virtue of the fact that the load P is applied equally to the two vertical members.

Symmetric mode

We assume now that the frame experiences the form of buckling shown in Fig. 12.30(b). The deformation conditions for the members of the frame are shown in Fig. 12.32. Members A±B and D±C As by the nature of the method we can assume any trial function which satis®es the appropriate geometric constraints, we propose a function in the form 2 5 3 2 4 uAB y …1 † ˆ ÿb1 …1 ÿ 3 1 ‡ 3 1 †

…12:144†

Stability of equilibrium and non-linear deformations φB

403

z2

P/2

b2 = b1 /3π

φB B

C

B u (z 2) u (z1)

(b)

z1

A (a)

Fig 12.32

which evidently satis®es the following boundary conditions for the symmetrical mode duAB y ˆ0 d1

uAB y ˆ 0

and

uAB y

at 1 ˆ 1

ˆ0

at 1 ˆ 0

…12:145†

The slope of this member at the joint with B±C, that is at 1 ˆ 1 AB 1 duy b AB …12:146† ˆÿ 1  …B† ˆ ÿ l d1 1 ˆ 1 3l Member B±C A suitable deformation trial function is uBC y …2 † ˆ b2 sin…2 † The slope of this member at joint B is BC 1 duy b ˆÿ 2 BC …B† ˆ ÿ l d2 2 ˆ 0 l

…12:147†

…12:148†

Hence from compatibility of the slope at joint B b2 ˆ

b1 3

…12:149†

404

Energy methods in structural mechanics

Substituting the trial functions in equation (12.140) gives the equilibrium condition as   d EI1 8 EI2 2 P 4 ‡ 3 b ˆ0 ÿ ˆ …12:150† db1 l 105 1 l3 5 l 9 That provides the critical load for the symmetric mode of frame buckling as PE ˆ

4:262 EI1 ‰1 ‡ 0:343IŠ l2

…12:151†

where, as in the sway mode, we have here I  I2 =I1 . In this case therefore the ¯exural sti€ness of the member B±C has a much less e€ect in increasing the critical load of the frame than is the case for the sway buckling mode. Notice as I becomes very small the critical load approaches the value for the strut pinned at one end and encastre at the other end, taking into account that the load P is applied equally to the two vertical members of the frame. The method of analysis presented above required a separate identi®cation of the two signi®cant modes of initial buckling associated with the critical loads of the portal frame. However, it goes without saying that the choice of deformation mode does not have to be made at the outset of the analysis. If sucient degrees of freedom are included in the trial functions, then the calculation of stationary values of the potential energy will identify the buckling modes automatically. This is simply attained by incorporating the two buckling modes, i.e. the sway and symmetric, as one trial function with two degrees of freedom. Thus, the trial function for the members A±B and C±D becomes     1 5 3 2 4 AB 2 uy …1 † ˆ a1 1 ÿ cos ÿ b 1 1 ÿ 1 ‡ 1 …12:152† 3 3 2 and by reason of compatibility of rotation at the joint B, the trial function for member B±C is uBC y …2 † ˆ

12.9

a1 b sin 22 ‡ 1 sin 2 4 3

Analysis of a frame loaded by vertical and lateral loads

…12:153†

In this section, we shall consider the situation in which a load P is applied to the frame at mid-length along the horizontal member as shown in Fig. 12.33. An additional horizontal load Q is applied at B. This implies that there will be some initial bending in the frame and this modi®es the

Stability of equilibrium and non-linear deformations

405

l/2

P B

C

Q EI2

EI1

EI1 l

A

D

Fig 12.33

response to the applied load from that of the frame analysed in the previous section in the same manner as for a column, which was shown in section 12.5. Let us assume the combined trial functions given by equations (12.152) and (12.153) where, by means of equation (12.149), we express b1 in terms of b2 . The total potential energy functional is   …1  2 AB 2 …1  2 BC 2 d uy d uy EI1 EI2 ˆ2 d1 ‡ 3 d2 2l 3 d12 2l d22 0

0

  …1  AB 2 duy P ÿ d1 ÿ Pb2 ÿ Qa1 2l d1

…12:154†

0

Substituting equations (12.152) and (12.153) in the above expression and performing the indicated derivatives and integrals we substitute the functional equation (12.154) with a simple function …a1 ; b2 † of the parameters a1 and b2 . Thus, the equilibrium conditions are simply given by q ˆ 0; qa1

q ˆ0 qb2

…12:155†

Finally, we have ‰…1 ‡ 8I† ÿ 4pŠa1 ‡ ‰0:0611 ‡ 0:2443pŠb2 ˆ 0:405q ‰0:0611 ‡ 0:2443pŠa1 ‡ ‰0:263…1 ‡ 0:343I† ÿ 0:1235pŠb2 ˆ 0:405p

…12:156†

406

Energy methods in structural mechanics

where a1 

a1 ; l

b2 

b2 ; l

I

I2 ; I1

p

P ; PES

q

Q PES

…12:157†

PES ˆ …22 EI1 †=l 2 is the critical load for a column built-in at one end and simply supported at the other. Putting equation (12.156) in matrix notation, we get ‰AŠ‰uŠ ÿ p‰BŠ‰uŠ ˆ ‰qŠ where

"

‰AŠ ˆ " ‰BŠ ˆ

…1 ‡ 8I† 0:0611 4:00

…12:158† 0:0611

0:263…1 ‡ 0:343I† # 0:2443

0:2443 0:1235 " # " # q a1 :405 ‰uŠ ˆ ‰qŠ ˆ 0 p b2

#

…12:159†

Therefore, the system equation (12.158) is not a homogenous one and this happens not only on account of the transverse load Q but also if we set Q ˆ 0, which implies q ˆ 0, as the vector ‰qŠ does not become zero. This happens on account of the mid-span position of the vertical load P which always produces a deformation of the member B±C that is transmitted to the columns A±B and C±D by means of the slope at the junctions B and C. We come therefore to the conclusion that the trivial solution ‰uŠ ˆ 0 is not admissible for any value of the loading. In order to quantify this e€ect, we consider ®rst the case Q ˆ 0. Figure 12.34(a) shows the de¯ection amplitudes a1 and b2 for I ˆ 0 plotted against a range of values of p. Note that the de¯ections remain quite small until the load p approaches the symmetric critical value, p ˆ 17:67. Thus the e€ect of the positioning of the load P is to introduce some de¯ections in the frame at loads less than the critical value and as the load is increased to the critical value the de¯ections grow very rapidly. This is a very similar behaviour to that shown by the individual struts considered in section 12.5. Notice also that even if the bending sti€ness of the horizontal member is set to zero, the corresponding de¯ection b2 does not tend to in®nity because the member is described by means of the trial shape function equation (12.153) and b1 ˆ 3b2 on account of equation (12.149). Therefore the de¯ection b2 is related to the de¯ection of the columns by means of the equation (12.152).

Stability of equilibrium and non-linear deformations

407

1000

p q

800

a1*, b*2

600

400

a*1 b*2

200

20

15

10 p

5

0

0

(a) 20

a*1 16

a1*

12

8

4

2·0

1·5

1·0 p

0·5

0

0

(b)

Fig 12.34

However, the more important result from this analysis is that the frame responded to the application of the load P as if there was no sway buckling mode. This mode has an eigenvalue much less that the symmetric mode and therefore it would have been anticipated that de¯ections in the sway mode would have developed at quite lower values of p. However, if a small lateral load Q is applied, then

408

Energy methods in structural mechanics

the picture changes quite dramatically. Considering the condition in which the lateral load has a value of 1% of the vertical load P so that q ˆ 0:01p, it may be seen from Fig. 12.34(b) that the magnitude of the de¯ections in the sway mode, i.e. a1 gradually increases with increasing values of p, until at the eigenvalue for the sway mode, i.e. p ˆ 2, the de¯ections grow very rapidly. This simple example illustrates one of the most dicult aspects of the non-linear buckling analysis of structures which is that unless all the potential buckling modes are activated by the form of loading there is a danger that the lowest buckling mode of the structure will be missed and an engineer designing a structure can easily overestimate the load carrying capacity of the structure.

12.10

Localised trial function analysis of frame buckling

The method of localised trial functions, the progenitor of the ®nite element method, has been applied to the analysis of a frame with a linear response to the applied loading. The method has also been shown earlier in this chapter being applied to the analysis of columns with various boundary conditions and step changes of section geometry. In the present analysis the method is applied to the evaluation of the initial buckling load for the simple frame previously considered. The geometry of the frame is shown in Fig. 12.35. Once again the analysis is carried out under the hypothesis that the axial deformation

l z2

P

y2

y3 3

4

5

EI2 EI1

l

z1

Fig 12.35

y1

z3

EI1

2

6

1

7

Stability of equilibrium and non-linear deformations

409

of the elements, due to their axial sti€ness, is negligible with respect to the e€ects of the lateral de¯ection. The potential of the vertical load P is related both to the bending de¯ection of the horizontal beam and to the vertical displacements of the nodes 3 and 5. Thus, the potential energy of the system  in terms of the de¯ection of its members is …uyn † ˆ U ÿ P
EI ˆ 1 2

…l  0

EI ‡ 1 2

q2 uy1 qz21

…l  0

2

EI dz1 ‡ 2 2

q2 uy3 qz23

2

…l  0

q2 uy2 qz22

2

dz2

dz3

8 2l 39 2 2 … …l  < = quy3 1 1 4 quy1 ÿ P dz1 ‡ dz3 5 ; 2 :2 qz1 qz3 0

0

ÿ Puy2 …l=2†

…12:160†

The members are each composed of two elements with the nodes numbered 1 to 7 as shown in Fig. 12.36. The interaction between the elements, i.e. their boundary conditions, is clari®ed in Fig. 12.36 where the nodes at the corners of the frame are separated into the 3"

Fig 12.36

4

5'

3'

5"

2

6

1

7

410

Energy methods in structural mechanics

nodes 30 and 500 , associated with the vertical members of the frame, and 300 and 50 , associated with the horizontal members of the frame. The generic displacement function for each element is obtained from equation (12.113). The boundary conditions at corner nodes are u300 ˆ 0;

30 ˆ 300 ;

u50 ˆ 0;

50 ˆ 500

…12:161†

The boundary conditions at the encastre base of the vertical members are 1 ˆ 0;

u1 ˆ 0;

7 ˆ 0

u7 ˆ 0;

…12:162†

Taking these conditions into account, the equilibrium equations for the frame, which come from the stationary value of the total potential energy functional, equation (12.160), are  ‰KŠ‰DŠ ÿ

Pl 2 Pl 3 ‰L1 Š‰DŠ ÿ ‰L2 Š ˆ 0 120 4

…12:163†

where the matrices are 2

48

0 ÿ24

12

0

6 0 16 ÿ12 4 0 6 6 6 ÿ24 ÿ12 24 ÿ12 0 6 6 4 ÿ12 …8 ‡ 8I† ÿ12I 6 12 6 6 0 0 0 ÿ12I 48I  ˆ EI1 6 ‰KŠ 6 6 0 0 0 4I 0 6 6 0 0 0 0 0 6 6 6 0 0 0 0 12I 6 6 4 0 0 0 0 0 0 2

72

0

0

0

0

0

0

0

0

0

0

0

0

0

0

4I

0

0

0

0

0

12I

0

16I

0

4I

0

0

24

12

ÿ24

4I

12 …8 ‡ 8I†

0

ÿ24

0

12

0 ÿ36

3 0 0

0

0

0

ÿ3

ÿ1 0 0

0

0

0

36

ÿ3 0 0

0

0

0

6 0 8 6 6 6 ÿ36 ÿ3 6 6 6 3 ÿ1 6 6 0 0 6 ‰L1 Š ˆ 6 6 0 0 6 6 0 0 6 6 6 0 0 6 6 4 0 0 0

0

0

0

ÿ12 4 0

3

0

0 0 0

ÿ36 ÿ3

72

07 7 7 07 7 7 07 7 07 7 7 07 7 37 7 7 ÿ1 7 7 7 05

0

0 0 0

3 ÿ1

0

4

ÿ3

4 0 0

0

0

0

0

0 0 0

0

0

0

0

0 0 0

0

0

0

0

0 0 0

36

0

0 0 0

3

3 ÿ36 4

ÿ3

ÿ12 48

0

3

07 7 7 07 7 7 07 7 07 7 7 07 7 12 7 7 7 47 7 7 05

0 16

Stability of equilibrium and non-linear deformations

411

2 3 0 607 6 7 6 7 607 6 7 607 6 7 6 7 617 7 ‰L2 Š ˆ 6 607 6 7 6 7 607 6 7 607 6 7 6 7 405 0 As usual, ‰DŠ is the vector containing the nodal degrees of freedom, i.e. 3 2 u2 6  7 6 27 7 6 6 u30 7 7 6 6  0 7 6 3 7 7 6 6 u4 7 6 ‰DŠ ˆ 6  7 7 6 4 7 7 6 6 u500 7 7 6 6  00 7 6 5 7 7 6 4 u6 5 6 and I  I2 =I1 . Notice that the two rows and columns in ‰L1 Š associated with u4 and 4 are completely null. This is because the vertical load P plays its role through two distinct terms: the ®rst one is related to the shortening of the vertical members on account of their lateral de¯ection, the second one is related to the vertical displacement of the mid-span of the horizontal beam on account of its bending. However, these rows and columns are required to maintain the correct order of matrix ‰L1 Š to be compatible with the vector ‰DŠ. It is evident that the ®rst two terms in equation (12.163) determine the eigenvalues and corresponding modes of the initial buckling of the frame while the third term re¯ects the bending de¯ection of the horizontal member in the frame. Eliminating for the moment the

412

Energy methods in structural mechanics P/2

P/2

EI2

EI1

EI1

Fig 12.37

third term, equation (12.163) becomes  ÿ ‰L Š†‰DŠ ˆ 0 …‰KŠ 1

…12:164†

where 

Pl 2 ; 120EI1

  ‰KŠ

1  ‰KŠ EI1

The formulation in equation (12.164) is equivalent to divide the load P such that it is aligned with the two vertical members of the frame as shown in Fig. 12.37.  gives ten eigenvalues, whose number is equal to Setting I ˆ 0 in ‰KŠ the number of degrees of freedom in the ®nite element model, of which the lowest is c1 ˆ 0:0425

i:e:

PE1 ˆ

0:5172 EI1 l2

…12:165a†

with the eigenmode de¯ected shape shown in Fig. 12.38(a). The initial buckling load PE1 is evidently associated with a sway form of buckling. The next lowest eigenvalue is c2 ˆ 0:348

i:e:

PE2 ˆ

4:232 EI1 l2

…12:165b†

with the corresponding de¯ected shape shown in Fig. 12.38(b), i.e. a symmetric de¯ected buckling mode. These initial buckling loads and modes correspond to those obtained in section 12.5 for columns encastre at one end and with zero ¯exural restraint, i.e., simply supported, at the other, as indeed would be the case for I ˆ 0.

Stability of equilibrium and non-linear deformations

λ = 0·0425 PE1 2

=

P/2

P/2

P/2

P/2

λ = 0·348

I=0

0·258 π 2 EI1 l2

PE2 2

=

(a)

λ = 0·1772 PE1 2

=

2·12 π 2 EI1 l2

(b)

P/2

P/2

P/2

P/2

413

λ = 0·6667

I=∞

1·08 π 2 EI1 l2

PE2 2

(c)

=

4·05 π 2 EI1 l2

(d)

Fig 12.38

 the lowest two eigenvalues from equation Applying I ˆ 1 in ‰KŠ, (12.164) are c1 ˆ 0:1772

i:e:

c2 ˆ 0:6667

i:e:

Pc1 ˆ Pc2 ˆ

2:152 EI1 l2 8:112 EI1

…12:166†

l2 with the corresponding deformed shapes shown in Figs 12.38(c) and (d). These two initial buckling loads also correspond to sway and symmetric modes of buckling but now for the condition where the horizontal member of the frame has an in®nite ¯exural sti€ness. The value of the lowest eigenvalue is dependent on the value of I, that is, the ratio of the ¯exural sti€ness of the horizontal and vertical members of the frame. This was shown also in section 12.8 for the application of the method of trial functions for the initial buckling

414

Energy methods in structural mechanics

Coefficient of initial buckling load

4 3 Variation of the initial buckling load with I (2 term trial function)

2 1 0

0

0·5

1·0

1·5

Value of I (a) Coefficient of initial buckling load

1·2 1·0 0·8

Variation of the initial buckling load with I (finite element)

0·6 0·4 0·2 0 0

0·5

1·0

1·5

Value of I (b)

Fig 12.39

of frames. The analysis in section 12.8 involved two degrees of freedom whereas the present analysis using the localised trial function method has ten degrees of freedom. Figure 12.39(a) shows the variation of the lowest initial buckling loads, as I is increased. Figure 12.39(b) shows the corresponding variation of the initial buckling loads obtained from the two degree of freedom analysis in section 12.8. It is evident that in the case at hand incorporating a larger number of degrees of freedom predicts lower buckling loads.

Example analysis for I ˆ 2:68

Section 6.5 has shown the linear analysis of a plane frame with a vertical load at the mid-span of the horizontal member. The ratio of the ¯exural sti€ness of the horizontal and vertical member is I ˆ 2:68. The same frame geometry is considered here from the viewpoint of the e€ects of the vertical load to cause initial buckling of the frame.  and obtaining the lowest eigenvalue from Putting I ˆ 2:68 in ‰KŠ, equation (12.164) gives c1 ˆ 0:1264 and equation (12.163) can be written  ÿ ‰L Š†‰D0 Š ˆ 30‰L Š …‰KŠ …12:167† 1

2

Stability of equilibrium and non-linear deformations

415

where 1 ‰D0 Š  ‰DŠ; l

 ˆ ‰KŠ

1  ‰KŠ EI1

thus  ÿ ‰L Š†ÿ1 ‰L Š ‰D0 Š ˆ 30…‰KŠ 1 2

…12:168†

Substituting various values of  into equation (12.168) the corresponding values of the de¯ections of the frame at the nodes can be evaluated. It is evident from the symmetrical form of the frame that if the vertical load is applied to the mid-span of the horizontal member then the sway mode of de¯ection may not be detected. This was shown in section 12.8 where it was clear that in order to ensure that the lowest buckling mode was initiated in the analysis, a lateral load has to be applied to the frame. Similarly, in the present analysis, a small non-dimensional lateral load is required to ensure that the deformations calculated using equation (12.168) correspond to the lowest value of the initial buckling load, i.e. the sway mode of buckling. Figure 12.40(a) shows the arrangement in which the lateral load, i.e. 0:01 is classi®ed as an initial imperfection. This loading is incorporated in equation (12.168) by modifying the vector ‰L2 Š to be 3 2 0 60 7 7 6 6 : 7 6 0 01 7 7 6 60 7 7 6 7 6 61 7 7 ‰L2 Š ˆ 6 60 7 7 6 7 6 60 7 7 6 60 7 7 6 7 6 40 5 0 Figure 12.40(b) shows the variation of the horizontal de¯ection at node 3 with increasing values of . The applied load , has been normalised by dividing by the lowest eigenvalue c1 . It is relatively simple in the present example to ensure that the initial imperfection, i.e. the lateral load, will indeed induce deformation that correspond to the lowest initial buckling load. In other cases of ®nite

416

Energy methods in structural mechanics 30 λ 0·3 λ

Normalized vertical load

(a) Variation of lateral deflection with increasing vertical load (1% lateral load) 1·0 0·8 0·6 0·4 0·2 0

0

0·5 1·0 1·5 Non-dimensional lateral deflection

2·0

(b)

Fig 12.40

element modelling, where there may be thousands of degrees of freedom, the application of an initial loading system may be much more complex. In such cases the eigenvalues can be used to introduce a system of initial geometric imperfections that will indeed ensure that the lowest initial buckling deformations are not overlooked in the analysis. The procedure is straightforward. Equation (12.164) is used to calculate the eigenvalues of the system and the corresponding eigen~n Š. These eigenmodes are then introduced into equation modes, ‰D (12.167) as follows  ÿ ‰L Š†‰D0 Š ˆ 30‰L Š ‡ a ‰D ~ 0 Š ‡ a ‰D ~0 Š ‡


…‰KŠ …12:168† 1

2

1

1

2

2

Stability of equilibrium and non-linear deformations

417

Normalized vertical load

Variation of lateral deflection with increasing vertical load (initial geometric imperfection) 1·0 0·8 0·6 0·4 0·2 0 0

1 Non-dimensional lateral deflection

2

Fig 12.41

~1 Š is the eigenmode for c1 , etc., and a1 ; a2 ; . . . are constants where ‰D that are generally chosen such that the form of the initial de¯ection of the structure is as general as possible and the magnitude of the initial de¯ection corresponds to measurements on practical structures. ~0i Š  ‰D ~i Š=l. Again, ‰D Using equation (12.168) the deformations at the nodes of the model can be calculated for increasing values of . Figure 12.41 shows the results from such a procedure in which a1 ˆ 0:1 and a2 ˆ a3 ˆ


ˆ 0. Comparison between Figs 12.41 and 12.40(b) show that the procedure of using the eigenmodes is very similar, in the present example, to applying a small arti®cial lateral load.

Suggested exercise problems 1.

The column shown in Fig. 12.42 is simply supported at both ends and has an initial de¯ection at mid-length of 50 mm. The cross-section is a cylindrical tube with 300 mm outside diameter and 10 mm wall thickness. Calculate the magnitude of the applied P

Δoc = 50 mm 15 m

Fig 12.42

418

Energy methods in structural mechanics P

Δoc = l

l 500

q

axial load, P, such that the maximum strain does not exceed the yield strain, "0 ˆ 0:0015. The modulus of the material is 205 kN/mm2 . [Ans: P ˆ 673 kN] The column shown in Fig. 12.43 has a length l, and is built-in at both ends. The column has an initial de¯ection of
0c ˆ l=500. The column is subjected concurrently to a lateral uniformly distributed load, q, and an axial load P. Derive an expression for the total de¯ection at mid-length,
c , of the column in terms of P, l, q and the ¯exural sti€ness, EI. " #   l 1:3ql 3 Pl 2 1‡
c ˆ 1‡ 2 Ans: 500 EI 4 EI

2.

3.

A column, pinned at its base, is attached to a cross-beam as shown in Fig. 12.44. The column has an initial de¯ection at mid-length of
0c ˆ l=300 and both the column and the cross-beam have the P

P

Mode a l

l

l Δoc =

l 300

P

Mode b

Fig 12.44

Stability of equilibrium and non-linear deformations

419

same ¯exural rigidity, EI. Show, using the method of localised trial functions, whether the rate of increase in the total strain, "t at midlength of the column is greater in mode a or mode b. [Ans: mode b has an eigenvalue 70% of the value of the eigenvalue of mode a, therefore the rate of growth of bending strain in mode b is greater than in mode a at the same applied load.]

Index ampli®cation factor, 380 anisotropic material properties, 63 bar buckling, 373 Beltrami, Eugenio, 81 Beltrami's equations, 81 Beltrami's yield criterion, 88 bending moment, 37 bending moment diagram, 39 Bernoulli, Daniel, 106 Bernoulli, Jacob, 94 Bernoulli's model of beam bending, 94 Betti's theorem, 148 bifurcation condition, 360 branching point, 360 buckling sway, 377 buckling, bifurcation, 360 buckling, snap, 365 bulk modulus, 90 calculus of variations, 108 Castigliano, Carlo, 155 Clapeyron, Benoit-Paul-EÂmile, 82 Clapeyron's theorem, 82 closed system, 5 combined joint load vector, 254 compatibility equation, 214 concatenated displacements, 28 conservative systems, 154 coordinate functions, 170, 188 Cotteril, James, 155 critical load, 360, 370 Croll, James, 106 degrees of freedom, 6, 174, 211

Descartes, ReneÂ, 12 direct sti€ness method, 221 Dirichlet's theorem, 358 displacement work, 147 ductility, 54 eigenvalue problem, 395 eigenwork, 148 elastic, 54 elastic energy, 3 element, 191 elements, non-conforming, 321 energy, 1 energy conservation, 5 engineering shear strain, 70 equilibrium, neutral, 355 equilibrium, unstable, 354 equilibrium, stable, 354 equivalent joint loads, 254 Euler buckling, 373 Euler buckling load, 370 Euler, Leonhard, 20 Euler's equation of beam buckling, 369 Euler's equation of beam deformation, 109 extensometer, 51 external constraints, 21 ®rst law of thermodynamics, 5 ¯exural sti€ness, 110 free body, 36 fundamental path, 360 Gauss, Karl, 338 generalised coordinates, 6

422

Index

generalised coordinates, 174 generalised displacement, 149 generalised mass, 9 Green, George, 66 Green's strains, 66

Newton, Isaac, 1,12 node, 234 nominal strain, 52 nominal stress, 52 non-conforming elements, 321

heat, 3 Hooke, Robert, 57 Hooke's law, 57

Pascal, Blaise, 321 patch test, 322 perfect column, 391 perfect system, 362 plane strain, 79 plate bending sti€ness, 277 plate deformation, 267 Poisson, SimeÂon-Denis, 61 Poisson's ratio, 61 principal components of strain, 77 principle of conservation of energy, 5 principle of stationary work, 12 principle of substitution of constraints, 33 principle of superpositon, 143 pure bending, 99 Pythagorus, Samar, 367

imperfections, 357 ine€ective constraint, 25 in®nitesimal displacements, 102 initial buckling, 356 initial imperfection, 415 initial out-of-straightness, 380 internal constraints, 21 intrinsic coordinates, 315 isoparametric, 317 isotropic material, 63 Jacobi, Karl, 316 Jacobian matrix, 316 kinetic energy, 3, 8 Kircho€, Gustav, 144 Kirkho€ model of plate deformation, 267 Lagrange family of shape functions, 324 Lagrange, Joseph-Louis, 12 Lagrangian coordinates, 12 lateral load, 377 Legendre, Adrien-Marie, 339 limit points, 365 linear elastic, 56 localised Rayleigh-Ritz method, 189 LuÈder's plateau, 54 material modulus, 57 Mises, Richard von, 90 mode, sway, 399 mode, symmetric, 402 modulus, shear, 85 moment of inertia, 101 neutral equilibrium, 355

Rayleigh, Lord, 174 relative centre of rotation, 27 rigid joints, 234 Ritz, Walther, 174 rotational restraint, 373 Saint-Venant, AdheÂmar-Jean-Claude, 89 Saint-Venant's yield criterion, 89 second moment of area, 101 serendipity family of shape functions, 324 shape function, 170 shape functions, 309 shape functions, Lagrange family, 324 shape functions, serendipity family, 324 shear force, 37 shear force diagram, 39 shear modulus, 85 shear strain, 69 Shute, Nevil, iv snap buckling, 365

Index static equilibrium, 9 statical determinacy, 35 statical indeterminacy, 35 sti€ness matrix, 215 strain compatibility, 80 strain energy, 9 strain energy density, 83 strain invariants, 76 superposition, 143 sway buckling, 377 sway mode, 399 symmetric mode, 402 theory of in®nitesimal displacements, 102 thermodynamics, 5 Timoshenko, Stephen, 290

423

trial displacements, 10 true strain, 52 true stress, 52 virtual displacements, 10 Von Mises yield criterion, 90 Walker, Alastair, 106 Walpole, Horace, 327 Woinowsky-Krieger, Stephen, 290 work, 1 yield plateau, 54 yield strain, 57 Young, Thomas, 57 Young's modulus, 57