Fundamental Structural Analysis (2)

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FUNDAMENTAL STRUCTURAL

Zubizuri Bridge Bilbao. Biscay, Euskadi, Spain

A N A L Y S I S

Jaroon Rungamornrat Faculty of Engineering Chulalongkorn University

FUNDAMENTAL STRUCTURAL ANALYSIS

J. Rungamornrat, Ph.D. Department of Civil Engineering Faculty of Engineering Chulalongkorn University

Copyright © 2011 J. Rungamornrat

Dedication To My parents, my wife and my beloved son

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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Preface

PREFACE

The book entitled Fundamental Structural Analysis is prepared with the primary objective to provide complete materials for a fundamental structural analysis course (i.e., 2101310 Structural Analysis I) in civil engineering at Chulalongkorn University. This basic course is offered every semester and is a requisite for the third year undergraduate students with a major in civil engineering. Materials contained in this book are organized into several chapters which are arranged in an appropriate sequence easy to follow. In addition, for each analysis technique presented, underlying theories and key assumptions are considered very crucial and generally outlined at the very beginning of the chapter, so readers can deeply understand its derivation, capability and limitations. To clearly demonstrate the step-by-step analysis procedure involved in each technique, various example problems supplemented by full discussion are presented. Structural analysis has been recognized as an essential component in the design of civil engineering and other types of structures such as buildings, bridges, dams, factories, airports, vehicle parts, machine components, aerospace structures, artificial human organs, etc. It concerns primarily the methodology to construct an exact or approximate solution of an existing or newly developed mathematical model, i.e., a representative of the real structure known as an idealized structure. A process to construct an appropriate model or idealized structure, commonly known as the structural idealization, is considered very crucial in the structural modeling (due to its significant influence on the accuracy of the representative solution to describe responses of the real structure) and must be carried out before the structural analysis can be applied. However, this process is out of scope of this text; a brief discussion is provided in the first chapter only to emphasize its importance and remind readers about the difference between idealized and real structures. A term “structure” used throughout this text therefore means, unless state otherwise, the idealized structure. Nowadays, many young engineers have exposed to various user-friendly, commercial software packages that are capable of performing comprehensive analysis of complex and largescale structures. Most of them have started to ignore or even forget the basic background of structural analysis since classical hand-based calculations have almost been replaced by computerbased analyses. Due to highly advances in computing devices and software technology, those available tools have been well-designed and supplemented by user-friendly interfaces and easy-tofollow user-manuals to draw attention from engineers. In the analysis, users are only required to provide complete information of (idealized) structures to be analyzed through the input channel and to properly interpret output results generated by the programs. The analysis procedure to determine such solutions has been implemented internally and generally blinded to the users. Upon the existence of powerful analysis packages, an important question concerning the necessity to study the foundation of structural analysis arises. Is only learning how to use available commercial programs really sufficient? If not, to what extent should the analysis course cover? Answers to above questions are still disputable and depends primarily on the individual perspective. In the author’s view, having the background of structural analysis is still essential for structural engineers although, in this era, powerful computer-aided tools have been dominated. The key objective is not to train engineers to understand the internal mechanism of the available codes or to implement the procedure into a code themselves as a programmer, but to understand fundamental theories and key assumptions underlying each analysis technique; the latter is considered crucial to deeply recognize Copyright © 2011 J. Rungamornrat


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Preface

their capability and limitations. In addition, during the learning process, students can gradually accumulate and finally develop sense of engineers through the problem solving strategy. An engineer fully equipped with knowledge and engineering sense should be able to recognize obviously wrong or unreasonable results, identify sources of errors, and verify results obtained from the analysis. Fully trusting results generated by commercial analysis packages without sufficient verification and check of human errors can lead to dramatic catastrophe if such information is further used in the design. The author has attempted to gather materials from various valuable and reliable resources (including his own experience accumulated from the undergraduate study at Chulalongkorn University, the graduate study at the University of Texas at Austin, and, more importantly, a series of lectures in structural analysis classes at Chulalongkorn University for several years) and put them together in a fashion, hopefully, easy to digest for both the beginners and ones who would like to review what they have learned before. The author anticipate that this book should be valuable and useful, to some extent, for civil engineering students, as supplemented materials to those covered in their classes and a source full with challenging exercise problems. The ultimate goal of writing this book is not only to transfer the basic knowledge from generations to generations but also to provide the motive for students, when start reading it, to gradually change their perspective of the subject from “very tough” to “not as tough as they thought”. Surprisingly, from the informal interview of several students in the past, their first impression about this subject is quite negative (this may result from various sources including exaggerated stories or scary legends told by their seniors) and this can significantly discourage their interest since the first day of the class. This book is organized into eleven chapters and the outline of each chapter is presented here to help readers understand its overall picture. The first chapter provides a brief introduction and basic components essential for structural modeling and analysis such as structural idealization, basic quantities and basic governing equations, classification of structures, degree of static and kinematical indeterminacy, stability of structures, etc. The second chapter devotes entirely to the static analysis for support reactions and internal forces of statically determinate structures. Three major types of idealized structures including plane trusses, beams, and plane frames are the main focus of this chapter. Chapter 3 presents a technique, called the direct integration method, to determine the exact solution of beams (e.g. deflections, rotations, shear forces, and bending moment as a function of position along the beam) under various end conditions and loading conditions. Chapter 4 presents a graphical-based technique, commonly known as the moment or curvature area method, to perform displacement and deformation analysis (i.e. determination of displacements and rotations) of statically determinate beams and frames. Chapter 5 introduces another method, called the conjugate structure analogy, which is based on the same set of equations derived in the Chapter 4 but such equations are interpreted differently in a fashion well-suited for analysis of beams and frames of complex geometry. Chapter 6 is considered fundamental and essential for the development of modern structural analysis techniques. It contains various principles and theorems formulated in terms of works and energies and having direct applications to structural analysis. The chapter starts by defining some essential quantities such as work and virtual work, complimentary work and complimentary virtual work, strain energy and virtual strain energy, complimentary strain energy and complimentary virtual strain energy, etc., and then outlines important work and energy theorems, e.g., conservation of work and energy, the principle of virtual work, the principle of complementary virtual work, the principle of stationary total potential energy, the principle of stationary total complementary potential energy, reciprocal theorem, and Castigliano’s 1st and 2nd theorems. Chapter 7 presents applications of the conservation of work and energy, or known as the method of real work, to the displacement and deformation analysis of statically determinate structures. Chapter 8 clearly demonstrates applications of the principle of complimentary virtual work, commonly recognized as the unit load method, to the displacement and deformation analysis of statically determinate trusses, beams and frames. Chapter 9 consists of two parts; the first part Copyright © 2011 J. Rungamornrat


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Preface

involves the application of Castigliano’s 2nd theorem to determine displacements and rotations of statically determinate structures whereas the second part presents its applications to the analysis of statically indeterminate structures. Chapter 10 devotes entirely to the development of a general framework for a force method, here called the method of consistent deformation, for analysis of statically indeterminate structures. Full discussion on how to choose unknown redundants, obtain primary structures and set up a set of compatibility equations is provided. The final chapter introduces the concept of influence lines and their applications to the analysis for various responses of structures under the action of moving loads. Both a direct procedure approach and those based on the well-known Müller-Breslau principle are presented with various applications to both statically determinate and indeterminate structures such as beams, floor systems, trusses, and frames.

Jaroon Rungamornrat, Ph.D. Department of Civil Engineering Faculty of Engineering Chulalongkorn University Bangkok Thailand


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Table of Contents

TABLE OF CONTENTS

PREFACE

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TABLE OF CONTENTS

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Chapter 1 INTRODUCTION TO STRUCTURAL ANALYSIS 1.1 Structural Idealization 1.2 Continuous Structure versus Discrete Structure Models 1.3 Configurations of Structure 1.4 Reference Coordinate Systems 1.5 Basic Quantities of Interest 1.6 Basic Components for Structural Mechanics 1.7 Static Equilibrium 1.8 Classification of Structures 1.9 Degree of Static Indeterminacy 1.10 Investigation of Static Stability of Structures Exercises

Chapter 2 ANALYSIS OF DETERMINATE STRUCTURES 2.1 Static Quantities 2.2 Tools for Static Analysis 2.3 Determination of Support Reactions 2.4 Static Analysis of Trusses 2.5 Static Analysis of Beams 2.6 Static Analysis of Frames Exercises

1 1 10 10 11 14 19 21 24 31 41 45

49 49 50 54 60 78 113 139

Chapter 3 DIRECT INTEGRATION METHOD

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3.1 Basic Equations 3.2 Governing Differential Equations 3.3 Boundary Conditions 3.4 Boundary Value Problem 3.5 Solution Procedure 3.6 Treatment of Discontinuity 3.7 Treatment of Statically Indeterminate Beams Exercises

143 148 149 154 156 176 198 207


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Table of Contents

Chapter 4 METHOD OF CURVATURE (MOMENT) AREA 4.1 Basic Assumptions 4.2 Derivation of Curvature Area Equations 4.3 Interpretation of Curvature Area Equations 4.4 Applications of Curvature Area Equations 4.5 Treatment of Axial Deformation Exercises

Chapter 5 CONJUGATE STRUCTURE ANALOGY 5.1 Conjugate Structure Analogy for Horizontal Segment 5.2 Conjugate Structure Analogy for Horizontal Segment with Hinges 5.3 Conjugate Structure Analogy for Inclined Segment 5.4 Conjugate Structure Analogy for General Segment Exercises

211 211 213 215 218 249 256

261 261 271 278 287 300

Chapter 6 INTRODUCTION TO WORK AND ENERGY THEOREMS

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6.1 Work and Complimentary Work 6.2 Virtual Work and Complimentary Virtual Work 6.3 Strain Energy and Complimentary Strain Energy 6.4 Virtual Strain Energy and Complimentary Virtual Strain Energy 6.5 Conservation of Work and Energy 6.6 Principle of Virtual Work (PVW) 6.7 Principle of Complimentary Virtual Work (PCVW) 6.8 Principle of Stationary Total Potential Energy (PSTPE) 6.9 Principle of Stationary Total Complimentary Potential Energy (PSTCPE) 6.10 Reciprocal Theorem 6.11 Castigliano’s 1st and 2nd Theorems Exercises

303 307 309 312 313 315 321 323 330 337 338 344

Chapter 7 DEFORMATION/DISPLACEMENT ANALYSIS BY PRW 7.1 Real Work Equation 7.2 Strain Energy for Various Effects 7.3 Applications of Real Work Equation 7.4 Limitations of PRW Exercises

Chapter 8 DEFORMATION/DISPLACEMENT ANALYSIS BY PCVW 8.1 8.2 8.3

PCVW with Single Virtual Concentrated Load Applications to Trusses Applications to Flexure-Dominating Structures


347 347 348 352 361 362

365 365 367 376

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Table of Contents

Exercises

403

Chapter 9 APPLICATIONS OF CASTIGLIANO’S 2nd THEOREM

407

9.1 Castigliano’s 2nd Theorem for Linearly Elastic Structures 9.2 Applications to Statically Determinate Structures 9.3 Applications to Statically Indeterminate Structures Exercises

407 409 422 432

Chapter 10 METHOD OF CONSIST DEFORMATION

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10.1 Basic Concept 10.2 Choice of Released Structures 10.3 Compatibility Equations for General Case Exercises

Chapter 11 INFLUENCE LINES 11.1 Introduction to Concept of Influence Lines 11.2 Influence Lines for Determinate Beams by Direct Method 11.3 Influence Lines by Müller-Breslau Principle 11.4 Influence Lines for Beams with Loading Panels 11.5 Influence Lines for Determinate Floor Systems 11.6 Influence Lines for Determinate Trusses 11.7 Influence Lines for Statically Indeterminate Structures Exercises

437 441 443 475

479 479 485 501 516 524 536 563 594

REFERENCE

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INDEX

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ACKNOWLEDGEMENT

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1

Introduction to Structural Analysis

CHAPTER 1 INTRODUCTION TO STRUCTURAL ANALYSIS This first chapter provides a brief introduction of basic components essential for structural analysis. First, the concept of structural modeling or structural idealization is introduced. This process involves the construction of a mathematical model or idealized structure to represent a real structure under consideration. The structural analysis is in fact a subsequent process that is employed to solve a set of mathematical equations governing the resulting mathematical model to obtain a mathematical solution. Such solution is subsequently employed to characterize or approximate responses of the real structure to a certain level of accuracy. Conservation of linear and angular momentum of a body in equilibrium is also reviewed and a well known set of equilibrium equations that is fundamentally important to structural analysis is also established. Finally, certain classifications of idealized structures are addressed.

1.1 Structural Idealization A real structure is an assemblage of components and parts that are integrated purposely to serve certain functions while withstanding all external actions or excitations (e.g. applied loads, environmental conditions such as temperature change and moisture penetration, and movement of its certain parts such as foundation, etc.) exerted by surrounding environments. Examples of real structures mostly encountered in civil engineering application include buildings, bridges, airports, factories, dams, etc as shown in Figure 1.1. The key characteristic of the real structure is that its responses under actions exerted by environments are often very complex and inaccessible to human in the sense that the real behavior cannot be known exactly. Laws of physics governing such physical or real phenomena are not truly known; most of available theories and conjectures are based primarily on various assumptions and, as a consequence, their validity is still disputable and dependent on experimental evidences.

Figure 1.1: Schematics of some real structures Copyright © 2011 J. Rungamornrat

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Since behavior of the real structure is extremely complex and inaccessible, it necessitates the development of a simplified or approximate structure termed as an idealized structure. To be more precise, an idealized structure is a mathematical model or a mathematical object that can be used to approximate behavior or responses of the real structure to certain degree of accuracy. The main characteristic of the idealized structure is that its responses are accessible, solvable, and can be completely determined using available laws of physics and mathematics. The process for obtaining the idealized structure is called structural idealization or structural modeling. This process generally involves imposing various assumptions and simplifying the complexity embedded in the real structure. The idealized structure of a given real structure is in general not unique and many different idealized structures can be established via use of different assumptions and simplifications. The level of idealization considered in the process of modeling depends primarily on the required degree of accuracy of (approximate) responses of the idealized structure in comparison with those of the real structure. The idealization error is an indicator that is employed to measure the discrepancy between a particular response of the real structure and the idealized solution obtained by solving the corresponding idealized structure. The acceptable idealization error is an important factor influences the level of idealization and a choice of the idealized structure. While a more complex idealized structure can characterize the real structure to higher accuracy, it at the same time consumes more computational time and effort in the analysis. The schematic indicating the process of structural idealization is shown in Figure 1.2. For brevity and convenience, the term “structure” throughout this text signifies the “idealized structure” unless stated otherwise. Some useful guidelines for constructing the idealized structure well-suited for structural analysis procedure are discussed as follows. Response interpretation Idealized solution

Structural analysis

Idealization error Assumptions + Simplification Real structure

Idealized structure Governing Physics

Complex & Inaccessible

Simplified & Solvable

Figure 1.2: Diagram indicating the process of structural idealization

1.1.1 Geometry of structure It is known that geometry of the real structure is very complex and, in fact, occupies space. However, for certain classes of real structures, several assumptions can be posed to obtain an idealized structure possessing a simplified geometry. A structural component with its length much larger than dimensions of its cross section can be modeled as a one-dimensional or line member, e.g. truss, beam, frame and arch shown in Figure 1.3. A structural component with its thickness much smaller than the other two dimensions can properly be modeled as a two-dimensional or surface member, e.g. plate and shell structures. For the case where all three dimensions of the Copyright © 2011 J. Rungamornrat


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structure are comparable, it may be obligatory to be modeled as a three-dimensional member, e.g. dam and a local region surrounding the connections or joints.

Beam

Arch

Truss

Frame

Figure 1.3: Schematics of idealized structures consisting of one-dimensional members

1.1.2 Displacement and deformation Hereby, the term deformation is defined as the distortion of the structure while the term displacement is defined as the movement of points within the structure. These two quantities have a fundamental difference, i.e., the former is a relative quantity that measures the change in shape or distortion of any part of the structure due to any action while the latter is a total quantity that measures the change in position of individual points resulting from any action. It is worth noting that the structure undergoing the displacement may possess no deformation; for instance, there is no change in shape or distortion of the structure if it is subjected to rigid translation or rigid rotation. This special type of displacement is known as the rigid body displacement. On the contrary, the deformation of any structure must follow by the displacement; i.e. it is impossible to introduce nonzero deformation to the structure with the displacement vanishing everywhere. For typical structures in civil engineering applications, the displacement and deformation due to external actions are in general infinitesimal in comparison with a characteristic dimension of the structure. The kinematics of the structure, i.e. a relationship between the displacement and the deformation, can therefore be simplified or approximated by linear relationship; for instance, the linear relationship between the elongation and the displacement of the axial member, the linear relationship between the curvature and the deflection of a beam, the linear relationship between the rate of twist and the angle of twist of a torsion member, etc. In addition, the small discrepancy between the undeformed and deformed configurations allows the (known) geometry of the undeformed configuration to be employed throughout instead of using the (unknown) geometry of the deformed configuration. It is important to remark that there are various practical situations where the small displacement and deformation assumption is not well-suited in the prediction of structural responses; for instance, structures undergoing large displacement and deformation near their collapse state, very flexible structures whose configuration is sensitive to applied loads, buckling and post-buckling behavior of axially dominated components, etc. Various investigations concerning structures undergoing large displacement and rotation can be found in the literature (e.g. Rungamornrat et al, 2008; Tangnovarad, 2008; Tangnovarad and Rungamornrat, 2008; Tangnovarad and Rungamornrat, 2009; Danmongkoltip, 2009; Danmomgkoltip and Rungamornrat, 2009; Rungamornrat and Tangnovarad, 2011; Douanevanh, 2011; Douanevanh et al, 2011). Copyright © 2011 J. Rungamornrat

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1.1.3 Material behavior The behavior of a constituting material in real structures is extremely complex (i.e. it is generally nonlinear, nonhomogeneous, anisotropic and time and history dependent) and, as a consequence, construction of a suitable constitutive model is both theoretically and computationally challenging. In constitutive modeling, the behavior of materials is generally modeled or approximated via the relationship between the internal force measure (e.g. axial force, torque, bending moment, shear force, and stress) and the deformation (e.g. elongation, rate of twist, curvature, and strain). Most of materials encountering in civil engineering applications (e.g. steel and concrete) are often modeled as an idealized, simple material behavior called an isotropic and linearly elastic material. The key characteristics of this class of materials are that the material properties are directional independent, its behavior is independent of both time and history, and stress and strain are related through a linear function. Only two material parameters are required to completely describe the material behavior; one is the so-called Young’s modulus denoted by E and the other is the Poisson’s ratio denoted by . Other material parameters can always be expressed in terms of these two parameters; for instance, the shear modulus, denoted by G, is given by

G

E 2(1  )

(1.1)

The Young’s modulus E can readily be obtained from a standard uniaxial tensile test while G is the elastic shear modulus obtained by conducting a direct shear test or a torsion test. The Poisson’s ratio can then be computed by the relation (1.1). Both E and G can be interpreted graphically as a slope of the uniaxial stress-strain curve (- curve) and a slope of the shear stress-strain curve (- curve), respectively, as indicated in Figure 1.4. The Poisson’s ratio  is a parameter that measures the degree of contraction or expansion of the material in the direction normal to the direction of the normal stress.





E

G

1

1

 Uniaxial stress-strain curve

 Shear stress-strain curve

Figure 1.4: Uniaxial and shear stress-strain diagrams

1.1.4 Excitations All actions or excitations exerted by surrounding environments are generally modeled by vector quantities such as forces and moments. The excitations can be divided into two different classes depending on the nature of their application; one called the contact force and the other called the remote force. The contact force results from the idealization of actions introduced by a direct Copyright © 2011 J. Rungamornrat


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contact between the structure and surrounding environments such as loads from occupants and wind while the remote force results from the idealization of actions introduced by remote environments such as gravitational force. The contact or remote force that acts on a small area of the structure can be modeled by a concentrated force or a concentrated moment while the contact or remote force that acts over a large area can properly be modeled by a distributed force or a distributed moment. Figure 1.5 shows an example of an idealized structure subjected to two concentrated forces, a distributed force and a concentrated moment.

Figure 1.5: Schematic of a two-dimensional, idealized structure subjected to idealized loads

1.1.5 Movement constraints Interaction between the structure and surrounding environments to maintain its stability while resisting external excitations (e.g. interaction between the structure and the foundation) can mathematically be modeled in terms of idealized supports. The key function of the idealized support is to prevent or constrain the movement of the structure in certain directions by means of reactive forces called support reactions. The support reactions are introduced in the direction where the movement is constrained and they are unknown a priori; such unknown reactions can generally be computed by enforcing static equilibrium conditions and other necessary kinematical conditions. Several types of idealized supports mostly found in two-dimensional idealized structures are summarized as follows.

1.1.5.1 Roller support A roller support is a support that can prevent movement of a point only in one direction while provide no rotational constraint. The corresponding unknown support reaction then possesses only one component of force in the constraint direction. Typical symbols used to represent the roller support and support reaction are shown schematically in Figure 1.6.

Figure 1.6: Schematic of a roller support and the corresponding support reaction Copyright © 2011 J. Rungamornrat

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1.1.5.2 Pinned or hinged support A pinned or hinged support is a support that can prevent movement of a point in both directions while provide no rotational constraint. The corresponding unknown support reaction then possesses two components of force in each direction of the constraint. Typical symbols used to represent the pinned or hinged support and the support reactions are shown schematically in Figure 1.7.

Figure 1.7: Schematic of a pinned or hinged support and the corresponding support reactions.

1.1.5.3 Fixed support A fixed support is a support that can prevent movement of a point in both directions and provide a full rotational constraint. The corresponding unknown support reaction then possesses two components of force in each direction of the translational constraint and one component of moment in the direction of rotational constraint. Typical symbols used to represent the fixed support and the support reactions are shown schematically in Figure 1.8.

Figure 1.8: Schematic of a fixed support and the corresponding support reactions

1.1.5.4 Guided support A guided support is a support that can prevent movement of a point in one direction and provide a full rotational constraint. The corresponding unknown support reaction then possesses one component of force in the direction of the translational constraint and one component of moment in the direction of rotational constraint. Typical symbols used to represent the guided support and the support reactions are shown schematically in Figure 1.9.

Figure 1.9: Schematic of a guided support and the corresponding support reactions Copyright © 2011 J. Rungamornrat


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1.1.5.5 Flexible support A flexible support is a support that can partially prevent translation and/or rotational constraints. The corresponding unknown support reaction is related to the unknown displacement and/or rotation of the support. Typical symbols used to represent the flexible support and the support reactions are shown schematically in Figure 1.10.

Figure 1.10: Schematic of a flexible support and the corresponding support reactions

1.1.6 Connections Behavior of a local region where the structural components are connected is very complicated and this complexity depends primarily on the type and details of the connection used. To extensively investigate the behavior of the connection, a three dimensional model is necessarily used to gain accurate results. For a standard, linear structural analysis, the connection is only modeled as a point called node or joint and the behavior of the node or joint depends mainly on the degree of force and moment transfer across the connection.

1.1.6.1 Rigid joint A rigid joint is a connection that allows the complete transfer of force and moment across the joint. Both the displacement and rotation are continuous at the rigid joint. This idealized connection is usually found in the beam or frame structures as shown schematically in Figure 1.11.

Figure 1.11: Schematic of a real connection and the idealized rigid joint

1.1.6.2 Hinge joint A hinge joint is a connection that allows the complete transfer of force across the joint but does not allow the transfer of the bending moment. Thus, the displacement is continuous at the hinge joint while the rotation is not since each end of the member connecting at the hinge joint can rotate freely from each other. This idealized connection is usually found in the truss structures as shown schematically in Figure 1.12. Copyright © 2011 J. Rungamornrat

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Figure 1.12: Schematic of a real connection and the idealized hinge joint.

1.1.6.3 Partially rigid joint A partially rigid joint is a connection that allows the complete transfer of force and a partial transfer of moment across the joint. For this particular case, both the displacement is continuous at the joint while rotation is not. The behavior of the flexible joint is more complex than the rigid joint and the hinge joint but it can better represent the real behavior of the connection in the real structure. The schematic of the partially rigid joint is shown in Figure 1.13.

Figure 1.13: Schematic of an idealized partially rigid joint

1.1.7 Idealized structures In this text, it is focused attention on a particular class of idealized structures that consist of onedimensional and straight components, is contained in a plane, and is subjected only to in-plane loadings; these structures are sometimes called “two-dimensional” or “plane” structures. Three specific types of structures in this class that are main focus of this text include truss, beam and frame.

Figure 1.14: Schematic of idealized trusses Copyright © 2011 J. Rungamornrat


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1.1.7.1 Truss Truss is an idealized structure consisting of one-dimensional, straight structural components that are connected by hinge joints. Applied loads are assumed to act only at the joints and all members possess only one component of internal forces, i.e. the axial force. Examples of truss structures are shown in the Figure 1.14.

1.1.7.2 Beam Beam is an idealized structure consisting of one-dimensional, straight members that are connected in a series either by hinge joints or rigid joints; thus, the geometry of the entire beam must be onedimensional. Loads acting on the beam must be transverse loadings (loads including forces normal to the axis of the beam and moments directing normal to the plane containing the beam) and they can act at any location within the beam. The internal forces at a particular cross section consist of only two components, i.e., the shear force and the bending moment. Examples of beams are shown in Figure 1.15.

Figure 1.15: Schematic of idealized beams

1.1.7.3 Frame Frame is an idealized structure consisting of one-dimensional, straight members that are connected either by hinge joints or rigid joints. Loads acting on the frame can be either transverse loadings or longitudinal loadings (loads acting in the direction parallel to the axis of the members) and they can act at any location within the structure. The internal forces at a particular cross section consist of three components: the axial force, the shear force and the bending moment. It can be remarked that when the internal axial force identically vanishes for all members and the geometry of the structure is one dimensional, the frame simply reduces to the beam. Examples of frame structures are shown in Figure 1.16.

Figure 1.16: Schematic of idealized frames Copyright © 2011 J. Rungamornrat

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1.2 Continuous Structure versus Discrete Structure Models A continuous structure is defined as an idealized structure where its responses at all points are unknown a priori and must be determined as a function of position (i.e. be determined at all points of the structure) in order to completely describe behavior of the entire structure. The primary unknowns of the continuous structure are in terms of response functions and, as a result, the number of unknowns counted at all points of the structure is infinite. Analysis of such continuous structure is quite complex and generally involves solving a set of governing differential equations. In the other hand, a discrete structure is a simplified idealized structure where the responses of the entire structure can completely be described by a finite set of quantities. This type of structures typically arises from a continuous structure furnishing with additional assumptions or constraints on the behavior of the structures to reduce the infinite number of unknowns to a finite number. A typical example of discrete structures is the one that consists of a collection of a finite number of structural components called members or elements and a finite number of points connecting those structural components to make the structure as a whole called nodes or nodal points. All unknowns are forced to be located only at the nodes by assuming that behavior of each member can be completely determined in terms of the nodal quantities – quantities associated with the nodes. An example of a discrete structure consisting of three members and four nodes is shown in Figure 1.17. Node 2

Node 3 Member 2 Member 3 Member 1

Node 4

Node 1

Figure 1.17: An example of a discrete structure comprising three members and four nodes

1.3 Configurations of Structure There are two configurations involve in the analysis of a deformable structure. An undeformed configuration is used to refer to the geometry of a structure at the reference state that is free of any disturbances and excitations. A deformed configuration is used to refer to a subsequent configuration of the structure after experiencing any disturbances or excitations. Figure 1.18 shows both the undeformed configuration and the deformed configuration of a rigid frame. 

Undeformed configuration Y

v

u Deformed configuration X

Figure 1.18: Undeformed and deformed configurations of a rigid frame under applied loads Copyright © 2011 J. Rungamornrat


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1.4 Reference Coordinate Systems In structural analysis, a reference coordinate system is an indispensable tool that is commonly used to conveniently represent quantities of interest such as displacements and rotations, applied loads, support reactions, etc. Following subsections provide a clear notion of global and local coordinate systems and a law of coordinate transformation that is essential for further development.

1.4.1 Global and local coordinate systems There are two types of reference coordinate systems used throughout the development presented further in this book. A global coordinate system is a single coordinate system that is used to reference geometry or involved quantities for the entire structure. A choice of the global coordinate system is not unique; in particular, an orientation of the reference axes and a location of its origin can be chosen arbitrarily. The global reference axes are labeled by X, Y and Z with their directions strictly following the right-handed rule. For a two-dimensional structure, the commonly used, global coordinate system is one with the Z-axis directing normal to the plane of the structure. A local coordinate system is a coordinate system that is used to reference geometry or involved quantities of an individual member. The local reference axes are labeled by x, y and z. This coordinate system is defined locally for each member and, generally, based on the geometry and orientation of the member itself. For plane structures, it is typical to orient the local coordinate system for each member in the way that its origin locates at one of its end, the x-axis directs along the axis of the member, the z-axis directs normal to the plane of the structure, and the y-axis follows the right-handed rule. An example of the global and local coordinate systems of a plane structure consisting of three members is shown in Figure 1.19. y

x y

Y

x

y X x Figure 1.19: Global and local coordinate systems of a plane structure

1.4.2 Coordinate transformation In this section, we briefly present a basic law of coordinate transformation for both scalar quantities and vector quantities. To clearly demonstrate the law, let introduce two reference coordinate systems that possess the same origin: one, denoted by {x1, y1, z1}, with the unit base vectors {i1, j1, k1} and the other, denoted by {x2, y2, z2}, with the unit base vectors {i2, j2, k2} as indicated in Figure 1.20. Now, let define a matrix R such that


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 i1  i 2 R   i1  j2 i1  k 2

j1  i 2 k 1  i 2  cos 11 cos 21 cos 31  j1  j2 k1  j2   cos 12 cos 22 cos 32  j1  k 2 k1  k 2  cos 13 cos 23 cos 33 

(1.2)

where {11, 21, 31} are angles between the unit vector i2 and the unit vectors {i1, j1, k1}, respectively; {12, 22, 32} are angles between the unit vector j2 and the unit vectors {i1, j1, k1}, respectively; and {13, 23, 33} are angles between the unit vector k2 and the unit vectors {i1, j1, k1}, respectively. z1

z2

k1

k2 i1 i2

j2

y2

j1

x1

y1

x2 Figure 1.20: Schematic of two reference coordinate systems with the same origin

1.4.2.1 Coordinate transformation for scalar quantities Let  be a scalar quantity whose values measured in the coordinate system {x1, y1, z1} and to the coordinate system {x2, y2, z2} are denoted by 1 and 2, respectively. Since a scalar quantity possesses only a magnitude, its values are invariant of the change of reference coordinate systems and this implies that 1   2

(1.3)

1.4.2.2 Coordinate transformation for vector quantities Let v be a vector whose representations with respect to the coordinate system {x1, y1, z1} and the coordinate system {x2, y2, z2} are given by v  v1x i1  v1y j1  v1zk 1  v 2x i 2  v 2y j2  v 2zk 2

(1.4)

where { v1x , v1y , v1z } and { v 2x , v 2y , v 2z } are components of a vector v with respect to the coordinate systems {x1, y1, z1} and {x2, y2, z2}, respectively. To determine the component v 2x in terms of the components { v1x , v1y , v1z }, we take an inner product between a vector v given by (1.4) and a unit vector i2 to obtain v 2x  v1x (i1  i 2 )  v1y ( j1  i 2 )  v1z (k 1  i 2 )

(1.5) Copyright © 2011 J. Rungamornrat


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Similarly, by taking an inner product between a vector v given by (1.4) and a unit vector j2 and k2, it leads to v 2y  v1x (i1  j2 )  v1y ( j1  j2 )  v1z (k 1  j2 )

(1.6)

v 2z  v1x (i1  k 2 )  v1y ( j1  k 2 )  v1z (k1  k 2 )

(1.7)

With use of the definition of the transformation matrix R given by (1.2), equations (1.5)-(1.7) can be expressed in a more concise form as v 2x   i1  i 2  2  v y    i1  j2  v 2  i  k  z  1 2

j1  i 2 j1  j2 j1  k 2

v1x  k1  i 2  v1x      k1  j2  v1y   R v1y   v1  k 1  k 2   v1z   z

(1.8)

The expression of the components { v1x , v1y , v1z } in terms of the components { v 2x , v 2y , v 2z } can readily be obtained in a similar fashion by taking a vector inner product of the vector v given by (1.4) and the unit base vectors {i1, j1, k1}. The final results are given by v1x   i 2  i1  1  v y    i 2  j1  v1  i  k  z  2 1

j2  i1 j2  j1 j2  k1

v 2x  k 2  i1  v 2x      k 2  j1  v 2y   R T v 2y  v2  k 2  k 1   v 2z   z

(1.9)

where RT is a transpose of the matrix R. Note that the matrix R is commonly termed a transformation matrix.

1.4.2.3 Special case Let consider a special case where the reference coordinate system {x2, y2, z2} is simply obtained by rotating the z1-axis of the reference coordinate system {x1, y1, z1} by an angle . The transformation matrix R possesses a special form given by  cos  sin  0 R   sin  cos  0  0 0 1

(1.10)

The coordinate transformation formula (1.8) and (1.9) therefore reduce to v 2x   cos  sin  0 v1x   2   1  v y    sin  cos  0 v y  v 2   0 0 1  v1z   z 

(1.11)

v1x  cos   sin  0 v 2x   1   2  v y    sin  cos  0 v y   v1   0 0 1  v 2z   z 

(1.12)


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This clearly indicates that the component along the axis of rotation is unchanged and is independent of the other two components. The laws of transformation (1.11) and (1.12) can also be applied to the case of two vectors v and w where v is contained in the x1-y1 plane (and the x2-y2 plane) and w is perpendicular to the x1-y1 plane (and the x2-y2 plane). More precisely, components of both vectors v and w in the {x1, y1, z1} coordinate system and in the {x2, y2, z2} coordinate system are related by  v 2x   cos  sin  0   v1x   2   1   v y     sin  cos  0   v y  w 2   0 0 1   w1z   z 

(1.13)

 v1x  cos   sin  0   v 2x   1   2   v y    sin  cos  0   v y   w1   0 0 1   w 2z   z 

(1.14)

1.5 Basic Quantities of Interest This section devotes to describe two different classes of basic quantities that are involved in structural analysis, one is termed kinematical quantities and the other is termed static quantities.

1.5.1 Kinematical quantities Kinematical quantities describe geometry of both the undeformed and deformed configurations of the structure. Within the context of static structural analysis, kinematical quantities can be categorized into two different sets: one associated with quantities used to measure the movement or change in position of the structure and the other is associated with quantities used to measure the change in shape or distortion of the structure. Displacement at any point within the structure is a quantity representing the change in position of that point in the deformed configuration measured relative to the undeformed configuration. Rotation at any point within the structure is a quantity representing the change in orientation of that point in the deformed configuration measured relative to the undeformed configuration. For a plane structure shown in Figure 1.18, the displacement at any point is fully described by a two-component vector (u, v) where u is a component of the displacement in Xdirection and v is a component of the displacement in Y-direction while the rotation at any point is fully described by an angle  measured from a local tangent line in the undeformed configuration to a local tangent line at the same point in the deformed configuration. It is important to emphasize that the rotation is not an independent quantity but its value at any point can be computed when the displacement at that point and all its neighboring points is known. A degree of freedom, denoted by DOF, is defined as a component of the displacement or the rotation at any node (of the discrete structure) essential for describing the displacement of the entire structure. There are two types of the degree of freedom, one termed as a prescribed degree of freedom and the other termed as a free or unknown degree of freedom. The former is the degree of freedom that is known a priori, for instance, the degree of freedom at nodes located at supports where components of the displacement or rotation are known while the latter is the degree of freedom that is unknown a priori. The number of degrees of freedom at each node depends primarily on the type of nodes and structures and also the internal releases and constraints present within the structure. In general, it is equal to the number of independent degrees of freedom at that node essential for describing the displacement of the entire structure. For beams, plane trusses, Copyright © 2011 J. Rungamornrat


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space truss, plane frames, and space frames containing no internal release and constraint, the number of degrees of freedom per node are 2 (a vertical displacement and a rotation), 2 (two components of the displacement), 3 (three components of the displacement), 3 (two components of the displacement and a rotation) and 6 (three components of the displacement and three components of the rotation), respectively. Figure 1.21 shows examples of both prescribed degrees of freedom and free degrees of freedom of beam, plane truss and plane frames. The number of degrees of freedom of a structure is defined as the number of all independent degrees of freedom sufficient for describing the displacement of the entire structure or, equivalently, it is equal to the sum of numbers of degrees of freedom at all nodes. For instance, a beam shown in Figure 1.21(a) has 6 DOFs {v1, 1, v2, 2, v3, 3} consisting of 3 prescribed DOFs {v1, 1, v3} and 3 free DOFs {v2, 2, 3}; a plane truss shown in Figure 1.21(b) has 6 DOFs {u1, v1, u2, v2, u3, v3} consisting of 3 prescribed DOFs {u1, v1, v2} and 3 free DOFs {u2, u3, v3}; and a plane frame shown in Figure 1.21(c) has 9 DOFs {u1, v1, 1, u2, v2, 2, u3, v3, 3} consisting of 3 prescribed DOFs { u1, v1, v3} and 6 free DOFs {1, u2, v2, 2, u3, 3}. It is evident that the number of degrees of freedom of a given structure is not unique but depending primarily on how the structure is discretized. As the number of nodes in the discrete structure increases, the number of the degrees of freedom of the structure increases. v3 Node 3

Y

u3

Y

v1 = 0 1 = 0

Node 3 3 X v3 = 0

Node 2

Node 1 v2

Node 1 u1=0 v1=0

2

Node 2

u2

X

v2=0 (b)

(a) Y Node 2

Node 3 v3=0 u3

u2

3

v2 2

1 Node 1 u1=0 v1=0

X (c)

Figure 1.21: (a) Degrees of freedom of a beam, (b) degrees of freedom of a plane truss, and (c) degrees of freedom of a plane frame Copyright © 2011 J. Rungamornrat

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Deformation is a quantity used to measure the change in shape or the distortion of a structure (i.e. elongation, rate of twist, curvature, strain, etc.) due to disturbances and excitations. The deformation is a relative quantity and a primary source that produces the internal forces or stresses within the structure. For continuous structures, the deformation is said to be completely described if and only if the deformation is known at all points or is given as a function of position while, for discrete structures, the deformation of the entire structure is said to be completely described if and only if the deformation of all members constituting the structure are known. The deformation for each member of a discrete structure can be described by a finite number of quantities called the member deformation (this, however, must be furnished by certain assumptions on kinematics of the member to ensure that the deformation at every point within the member can be determined in terms of the member deformation). The quantities selected to be the member deformation depend primarily on the type and behavior of such member. For instance, the elongation, e, or a measure of the change in length of a member is commonly chosen as the member deformation of a truss member as shown in Figure 1.22(a); the relative end rotations {s, e} where s and e denotes the rotations at both ends of the member measured relative to a chord connecting both end points as shown in Figure 1.22(b) are commonly chosen as the member deformation of a beam member; and the elongation and two relative end rotations {e, s, e} as shown in Figure 1.22(c) are commonly chosen as the member deformation of a frame member. It is remarked that the deformation of the entire discrete structure can fully be described by a finite set containing all member deformation.

y

y

L´= L s

L´= L + e x

L

L

e x

(b)

(a) y

L´= L+ e s

e

L (c)

x

Figure 1.22: Member deformation for different types of members: (a) truss member, (b) beam member, and (c) frame member A Rigid body motion is a particular type of displacement that produces no deformation at any point within the structure. The rigid body motion can be decomposed into two parts: a rigid translation and a rigid rotation. The rigid translation produces the same displacement at all points while the rigid rotation produces the displacement that is a linear function of position. Figure 1.23 shows a plane structure undergoing a series of rigid body motions starting from a rigid translation in the X-direction, then a rigid translation in the Y-direction, and finally a rigid rotation about a point A´. Copyright © 2011 J. Rungamornrat


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Within the context of static structural analysis, the structure under consideration must sufficiently be constrained to prevent both the rigid body motion of the entire structure and the rigid body motion of any part of the structure. The former is prevented by providing a sufficient number of supports and proper directions against movement and the latter is prevented by the proper arrangement of members and their connections. A structure shown in Figure 1.24(a) is a structure in Figure 1.23 after prevented all possible rigid body motions by introducing a pinned support at a point A and a roller support at a point B. A structure shown in Figure 1.24(b) indicates that although many supports are provided but in improper manner, the structure can still experience the rigid body motion; for this particular structure, the rigid translation can still occur in the Xdirection.

Y

A´ A

B X

Figure 1.23: An unconstrained plane structure undergoing a series of rigid body motions

Y

Y

X

X

(a)

(b)

Figure 1.24: (a) A structure with sufficient constraints preventing all possible rigid body motions and (b) a structure with improper constraints

1.5.2 Static quantities Quantities such as external actions and reactions in terms of forces and moments exerted to the structure by surrounding environments and the intensity of forces (e.g. stresses and pressure) and theirs resultants (e.g. axial force, bending moment, shear force, and torque, etc.) induced internally at any point within the structure are termed as static quantities. Applied load is one of static quantities referring to the prescribed force or moment acting to the structure. Support reaction is a term referring to an unknown force or moment exerted to the structure by idealized supports (representatives of surrounding environments) in order to prevent its movement or to maintain its Copyright © 2011 J. Rungamornrat

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stability. Support reactions are generally unknown a priori. There are two types of applied loads; one called a nodal load is an applied load acting to the node of the structure and the other called a member loads is an applied load acting to the member. An example of applied loads (both nodal loads and member loads) and support reactions of a plane frame is depicted in Figure 1.25. Stress is a static quantity used to describe the intensity of force (force per unit area) at any plane passing through a point. Internal force is a term used to represent the force or moment resultant of stress components on a particular surface such as a cross section of a member. Note again that a major source that produces the stress and the internal force within the structure is the deformation. The distribution of both stress and internal force within the member depends primarily on characteristics or types of that member. For standard one-dimensional members in a plane structure such as an axial member, a flexural member, and a frame member, the internal force is typically defined in terms of the force and moment resultants of all stress components over the cross section of the member – a plane normal to the axis of the member. Node 2 Node 3

Node 4 Node 1 Figure 1.25: Schematic of a plane frame subjected to external applied loads An axial member is a member in which only one component of the internal force, termed as an axial force and denoted by f – a force resultant normal to the cross section, is present. The axial force f is considered positive if it results from a tensile stress present at the cross section; otherwise, it is considered negative. Figure 1.26 shows an axial member subjected to two forces {fx1, fx2} at its ends where fx1 and fx2 are considered positive if their directions are along the positive local x-axis. The axial force f at any cross section of the member can readily be related to the two end forces {fx1, fx2} by enforcing static equilibrium of both parts of the member resulting from an imaginary cut; this gives rise to f = – fx1 = fx2. Such obtained relation implies that {f, fx1, fx2} are not all independent but only one of these three quantities can equivalently be chosen to fully represent the internal force of the axial member. y

y

fx1

fx2

x

fx1

f

f

fx2

x

Figure 1.26: An axial member subjected to two end forces A flexural member is a member in which only two components of the internal force, termed as a shear force denoted by V – a resultant force of the shear stress component and a bending moment denoted by M – a resultant moment of the normal stress component, are present. The shear Copyright © 2011 J. Rungamornrat


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force V and the bending moment M are considered positive if their directions are as shown in Figure 1.27; otherwise, they are considered negative. Figure 1.27 illustrates a flexural member subjected to forces and moments {fy1, m1, fy2, m2} at its ends where fy1 and fy2 are considered positive if their directions are along the positive local y-axis and m1 and m2 are considered positive if their directions are along the positive local z-axis. The shear force V and the bending moment M at any cross section of the member can readily be related to the end forces and moments {fy1, m1, fy2, m2} by enforcing static equilibrium of both parts of the member resulting from an imaginary cut. It can be verified that only two quantities from a set {fy1, m1, fy2, m2} are independent and the rest can be obtained from equilibrium of the entire member. This implies in addition that two independent quantities from {fy1, m1, fy2, m2} can be chosen to fully represent the internal force of the flexural member; for instance, {m1, m2} is a common choice for the internal force of the flexural member. y

y m1 fy1

m2

m1

x

M M

V

m2

V

fy1

fy2

x

fy2

Figure 1.27: A flexural member subjected to end forces and end moments. A frame member is a member in which three components of the internal force (i.e. an axial force f, a shear force V, and a bending moment M) are present. The axial force f, the shear force V and the bending moment M are considered positive if their directions are as indicated in Figure 1.28; otherwise, they are considered negative. Figure 1.28 shows a frame member subjected to a set of forces and moments {fx1, fy1, m1, fx2, fy2, m2} at its ends where fx1 and fx2 are considered positive if their directions are along the positive local x-axis, fy1 and fy2 are considered positive if their directions are along the positive local y-axis and m1 and m2 are considered positive if their directions are along the positive local z-axis. The axial force can readily be related to the end forces {fx1, fx2} by a relation f = – fx1 = fx2 and the internal forces {V, M} at any cross section of the member can be related to the end forces and end moments {fy1, m1, fy2, m2} by enforcing static equilibrium to both parts of the member resulting from a cut. It can also be verified that only three quantities from a set {fx1, fy1, m1, fx2, fy2, m2} are independent and the rest can be obtained from equilibrium of the entire member. This implies that two independent quantities from {fy1, m1, fy2, m2} along with one quantity from {f, fx1, fx2} can be chosen to fully represent the internal forces of the frame member; for instance, {f, m1, m2} is a common choice for the internal force of the frame member. y

y fx1

m1

m2 fx2

fy1

fy2

x

fx1

m1

V

M

M f

V

m2

f

fy1

fx2

x

fy2

Figure 1.28: A frame member subjected to a set of end forces and end moments.

1.6 Basic Components for Structural Mechanics There are four key quantities involved in the procedure of structural analysis: 1) displacements and rotations, 2) deformation, 3) internal forces, and 4) applied loads and support reactions. The first two quantities are kinematical quantities describing the change of position and change of shape or Copyright © 2011 J. Rungamornrat

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distortion of the structure under external actions while the last two quantities are static quantities describing the external actions and the intensity of force introduced within the structure. It is evident that the displacement and rotation at any constraint points (supports) and the applied loads are known a priori while the rest are unknown a priori. As a means to solve such unknowns, three fundamental laws are invoked to establish a set of sufficient governing equations.

1.6.1 Static equilibrium Static equilibrium is a fundamental principle essential for linear structural analysis. The principle is based upon a postulate: “the structure is in equilibrium if and only if both the linear momentum and the angular momentum conserve”. This postulate is conveniently enforced in terms of mathematical equations called equilibrium equations – equations that relate the static quantities such as applied loads, support reactions, and the internal force. Note that equilibrium equations can be established in several forms; for instance, equilibrium of the entire structure gives rise to a relation between support reactions and applied loads; equilibrium of a part of the structure resulting from sectioning leads to a relation between applied loads, support reactions appearing in that part, and the internal force at locations arising from sectioning; and equilibrium of an infinitesimal element of the structure resulting from the sectioning results in a differential relation between applied loads and the internal force.

1.6.2 Kinematics Kinematics is a basic ingredient essential for the analysis of deformable structures. The principle is based primarily upon the geometric consideration of both the undeformed configuration and the deformed configuration of the structure. The resulting equations obtained relate the kinematical quantities such as the displacement and rotation and the deformation such as elongation, rate of twist, curvature, and strain.

1.6.3 Constitutive law A constitutive law is a mathematical expression used to characterize the behavior of a material. It relates the deformation (a kinematical quantity that measures the change in shape or distortion of the material) and the internal force (a static quantity that measures the intensity of forces and their resultants). To be able to represent behavior of real materials, all parameters involved in the constitutive modeling or in the material model must be carried out by conducting proper experiments.

1.6.4 Relation between static and kinematical quantities Figure 1.29 indicates relations between the four key quantities (i.e. displacement and rotation, deformation, internal force, and applied loads and support reactions) by means of the three basic ingredients (i.e. static equilibrium, kinematics, and constitutive law). This diagram offers an overall picture of the ingredients necessitating the development of a complete set of governing equations sufficient for determining all involved unknowns. It is worth noting that while there are only three basic principles to be enforced, numerous analysis techniques arise in accordance with the fashion they apply and quantities chosen as primary unknowns. Methods of analysis can be categorized, by the type of primary unknowns, into two central classes: the force method and the displacement method. The former is a method that employs static quantities such as support reactions and internal forces as primary unknowns while the latter is a method that employs the displacement and rotation as primary unknowns. Copyright © 2011 J. Rungamornrat


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Applied Loads & Support Reactions (Known and unknown)

FORCE METHOD

Internal Forces (Unknown)

Constitutive Law

Deformation (Unknown)

Kinematics

DISPLACEMENT METHOD

Static Equilibrium

Displacement & Rotation (Known and unknown)

Figure 1.29: Diagram indicating relations between static quantities and kinematical quantities

1.7 Static Equilibrium Equilibrium equations are of fundamental importance and necessary as a basic tool for structural analysis. Equilibrium equations relate three basic static quantities, i.e. applied loads, support reactions, and the internal force, by means of the conservation of the linear momentum and the angular momentum of the structure that is in equilibrium. The necessary and sufficient condition for the structure to be in equilibrium is that the resultant of all forces and moments acting on the entire structure and any part of the structure vanishes. For three-dimensional structures, this condition generates six independent equilibrium equations for each part of the structure considered: three equations associated with the vanishing of force resultants in each coordinate direction and the other three equations corresponding to the vanishing of moment resultants in each coordinate direction. These six equilibrium equations can be expressed in a mathematical form as ΣFX  0

; ΣFY  0

; ΣFZ  0

ΣM AX  0 ; ΣM AY  0 ; ΣM AZ  0

(1.15) (1.16)

where {O; X, Y, Z} denotes the reference Cartesian coordinate system with origin at a point O and A denotes a reference point used for computing the moment resultants. Copyright © 2011 J. Rungamornrat


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For two-dimensional or plane structures (which are the main focus of this text), there are only three independent equilibrium equations: two equations associated with the vanishing of force resultants in two directions defining the plane of the structure and one associated with the vanishing of moment resultants in the direction normal to the plane of the structure. The other three equilibrium equations are satisfied automatically. If the X-Y plane is the plane of the structure, such three equilibrium equations can be expressed as ΣFX  0

; ΣFY  0

; ΣM AZ  0

(1.17)

It is important to emphasize that the reference point A can be chosen arbitrarily and it can be either within or outside the structure. According to this aspect, it seems that moment equilibrium equations can be generated as many as we need by changing only the reference point A. But the fact is these generated equilibrium equations are not independent of (1.15) and (1.16) and they can in fact be expressed in terms of a linear combination of (1.15) and (1.16). As a result, this set of additional moment equilibrium equations cannot be considered as a new set of equations and the number of independent equilibrium equations is still six and three for three-dimensional and two-dimensional cases, respectively. It can be noted, however, that selection of a suitable reference point A can significantly be useful in several situation; for instance, it can offer an alternative form of equilibrium equations that is well-suited for mathematical operations or simplify the solution procedures. To clearly demonstrate the above argument, let consider a plane frame under external loads as shown in Figure 1.30. For this particular structure, there are three unknown support reactions {RA, RBX, RBY}, as indicated in the figure, and three independent equilibrium equations (1.17) that provide a sufficient set of equations to solve for all unknown reactions. It is evident that if a point A is used as the reference point, all three equations FX = 0, FY = 0 and MAZ = 0 must be solved simultaneously in order to obtain {RA, RBX, RBY}. To avoid solving such a system of linear equations, a better choice of the reference point may be used. For instance, by using point B as the reference point, the moment equilibrium equation MBZ = 0 contains only one unknown RA and it can then be solved. Next, by taking moment about a point C, the reaction RBX can be obtained from MCZ = 0. Finally the reaction RBY can be obtained from equilibrium of forces in Y-direction, i.e. FY = 0. It can be noted, for this particular example, that the three equilibrium equations MBZ = 0, MCZ = 0 and FY = 0 are all independent and are alternative equilibrium equations to be used instead of (1.17). Note in addition that an alternative set of equilibrium equations is not unique and such a choice is a matter of taste and preference; for instance, {MBZ = 0,FX = 0,FY = 0}, {MBZ = 0,FY = 0,MDZ = 0}, {MBZ = 0,MCZ = 0,MDZ = 0} are also valid sets.

D

C Y

A RA

RBX

B

X

RBY Figure 1.30: Schematic of a plane frame indicating both applied loads and support reactions Copyright © 2011 J. Rungamornrat


23


The number of independent equilibrium equations can further be reduced for certain types of structures. This is due primarily to that some equilibrium equations are satisfied automatically as a result of the nature of applied loads. Here, we summarize certain special systems of applied loads that often encounter in the analysis of plane structures.

1.7.1 A system of forces with the same line of action Consider a body subjected to a special set of forces that have the same line of action as shown schematically in Figure 1.31. For this particular case, there is only one independent equilibrium equation, i.e. equilibrium of forces in the direction parallel to the line of action. The other two equilibrium equations are satisfied automatically since there is no component of forces normal to the line of action and the moment about any point located on the line of action identically vanishes. Truss members and axial members are examples of structures that are subjected to this type of loadings.

F1

F2

Line of action

F3

Figure 1.31: Schematic of a body subjected to a system of forces with the same line of action

1.7.2 A system of concurrent forces Consider the body subjected to a system of forces that pass through the same point as shown in Figure 1.32. For this particular case, there are only two independent equilibrium equations (equilibrium of forces in two directions defining the plane containing the body, i.e. FX = 0 and FY = 0). The moment equilibrium equation is satisfied automatically when the two force equilibrium equations are satisfied; this can readily be verified by simply taking the concurrent point as the reference point for computing the moment resultant. An example of structures or theirs part that are subjected to this type of loading is the joint of the truss when it is considered separately from the structure.

F1

Y X

F4

F2

F3

Figure 1.32: Schematic of a body subjected to a system of concurrent forces

1.7.3 A system of transverse loads Consider the body subjected to a system of transverse loads (loads consisting of forces where their lines of action are parallel and moments that direct perpendicular to the plane containing the body) Copyright © 2011 J. Rungamornrat


24


as shown schematically in Figure 1.33. For this particular case, there are only two independent equilibrium equations (equilibrium of forces in the direction parallel to any line of actions and equilibrium of moment in the direction normal to the plane containing the body, i.e. FY = 0 and MAZ = 0). It is evident that equilibrium of forces in the direction perpendicular to the line of action is satisfied automatically since there is no component of forces in that direction. Examples of structures that are subjected to this type of loading are beams.

F2

Y

M1 F3

M2 F4

F1

X

Figure 1.33: Schematic of a body subjected to a system of transverse loads An initial step that is important and significantly useful for establishing the correct equilibrium equations for the entire structure or any part of the structure (resulting from the sectioning) is to sketch the free body diagram (FBD). The free body diagram simply means the diagram showing the configuration of the structure or part of the structure under consideration and all forces and moments acting on it. If the supports are involved, they must be removed and replaced by corresponding support reactions, likewise, if the part of the structure resulting from the sectioning is considered, all the internal forces appearing along the cut must be included in the FBD. Figure 1.34(b) shows the FBD of the entire structure shown in Figure 1.34(a) and Figure 1.34(c) shows the FBD of two parts of the same structure resulting from the sectioning at a point B. In particular, the fixed support at A and the roller support at C are removed and then replaced by the support reactions {RAX, RAY, RAM, RCY}. For the FBD shown in Figure 1.34(c), the internal forces {FB, VB, MB} are included at the point B of both the FBDs.

1.8 Classification of Structures Idealized structures can be categorized into various classes depending primarily on criteria used for classification; for instance, they can be categorized based on their geometry into one-dimensional, two-dimensional, and three-dimensional structures or they can be categorized based on the dominant behavior of constituting members into truss, beam, arch, and frame structures, etc. In this section, we present the classification of structures based upon the following three well-known criteria: static stability, static indeterminacy, and kinematical indeterminacy. Knowledge of the structural type is useful and helpful in the selection of appropriate structural analysis techniques.

1.8.1 Classification by static stability criteria Static stability refers to the ability of the structure to maintain its function (no collapse occurs at the entire structure and at any of its parts) while resisting external actions. Using this criteria, idealized structures can be divided into several classes as follows.

1.8.1.1 Statically stable structures A statically stable structure is a structure that can resist any actions (or applied loads) without loss of stability. Loss of stability means the mechanism or the rigid body displacement (rigid translation Copyright © 2011 J. Rungamornrat


25


and rigid rotation) develops on the entire structure or any of its parts. To maintain static stability, the structure must be properly constrained by a sufficient number of supports to prevent all possible rigid body displacements. In addition, members constituting the structure must be arranged properly to prevent the development of mechanics within any part of the structure or, in the other word, to provide sufficient internal constraints. All “desirable” idealized structures considered in the static structural analysis must fall into this category. Examples of statically stable structures are shown in Figures 1.3, 1.5 and 1.14-1.16.

P

M

P

C

M RCY

Y B A

RAX

X

RAY

RAM

(a)

(b) M

P

RCY

MB

VB

F FB B MB VB RAX RAY

RAM

(c) Figure 1.34: (a) A plane frame subjected to external loads, (b) FBD of the entire structure, and (c) FBD of two parts of the structure resulting from sectioning at B.

1.8.1.2 Statically unstable structures A statically unstable structure is a structure that the mechanism or the rigid body displacement develops on the entire structure or any of its parts when subjected to applied loads. Loss of stability in this type of structures may be due to i) an insufficient number of supports as shown in Figure 1.35(a), ii) inappropriate directions of constraints as shown in Figure 1.35(b), iii) inappropriate Copyright © 2011 J. Rungamornrat


26


arrangement of member as shown in Figure 1.35(c), and iv) too many internal releases such as hinges as shown in Figure 1.35(d). This class of structures can be divided into three sub-classes based on how the rigid body displacement develops. 1.8.1.2.1 Externally, statically unstable structures An externally, statically unstable structure is a statically unstable structure that the mechanism or the rigid body displacement develops only on the entire structure when subjected to applied loads. Loss of stability of this type structure is due to an insufficient number of supports provided or an insufficient number of constraint directions. Examples of externally, statically unstable structures are shown in Figure 1.35(a) and 1.35(b). 1.8.1.2.2 Internally, statically unstable structures An internally, statically unstable structure is a statically unstable structure that the mechanism or the rigid body displacement develops only on a certain part of the structure when subjected to applied loads. Loss of stability of this type of structure is due to inappropriate arrangement of member and too many internal releases. Examples of internally, statically unstable structures are shown in Figure 1.35(c) and 1.35(d).

(a)

(b)

(d) (c) Figure 1.35: Schematics of statically unstable structures 1.8.1.2.3 Mixed, statically unstable structures A mixed, statically unstable structure is a statically unstable structure that the mechanism or the rigid body displacement can develop on both the entire structure and any part of the structure when subjected to applied loads. Examples of mixed, statically unstable structures are shown in Figure 1.36. Copyright © 2011 J. Rungamornrat


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Figure 1.36: Schematics of mixed, statically unstable structures Figure 1.37 clearly demonstrates the classification of idealized structures based upon the static stability criteria.

Idealized structures

Statically unstable structures

Yes

Rigid body displacement of entire structure?

Development of rigid body displacement?

No

Statically stable structures

No Internally statically unstable structures

Yes

Rigid body displacement of part of structure?

No Externally statically unstable structures

Yes Mixed statically unstable structures

Figure 1.37: Diagram indicating classification of structures by static stability criteria Copyright © 2011 J. Rungamornrat


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1.8.2 Classification by static indeterminacy criteria Static indeterminacy refers to an ability or inability to determine static quantities (support reactions and internal force) at any point within a structure by means of static equilibrium. Using this criteria, statically stable idealized structures can be divided into several classes as follow.

1.8.2.1 Externally statically determinate structures An externally, statically determinate structure is a structure that all support reactions can be determined from static equilibrium. The internal force at any point within the structure can or cannot be obtained from static equilibrium. Examples of externally, statically determinate structures are shown in Figure 1.38.

(a)

(b)

(c)

Figure 1.38: Schematics of externally, statically determinate structures.

1.8.2.2 Externally statically indeterminate structures An externally, statically indeterminate structure is a structure that there exists at least one component of all support reactions that cannot be determined from static equilibrium. Note that there is no externally, statically indeterminate structure that the internal force at all points can be determined from static equilibrium. Examples of externally, statically indeterminate structures are shown in Figure 1.39.

1.8.2.3 Statically determinate structures A statically determinate structure is a structure that all support reactions and the internal force at all points within the structure can be determined from static equilibrium. It is evident that a statically determinate structure must also be an externally, statically determinate structure. Examples of statically determinate structures are shown in Figures 1.38(a) and 1.38(b).

1.8.2.4 Statically indeterminate structures A statically indeterminate structure is a structure that there exists at least one component of support reactions or the internal force at certain points within the structure that cannot be determined from static equilibrium. This definition implies that all statically stable structures must be either statically determinate or statically indeterminate. It can be noted that an externally, statically indeterminate structure must be a statically indeterminate structure. Examples of statically indeterminate structures are shown in Figures 1.38(c) and 1.39. Copyright © 2011 J. Rungamornrat


(a)

29


(b)

(c)

Figure 1.39: Schematics of statically indeterminate structures.

1.8.2.5 Internally statically indeterminate structures An internally statically indeterminate structure is a structure that is externally, statically determinate and, at the same time, statically indeterminate. This implies that all support reactions of an internally, statically indeterminate can be determined from static equilibrium while there exists the internal force at certain points within the structure that cannot be determined from static equilibrium. Examples of internally, statically indeterminate structures are shown in Figures 1.38(c) and 1.39(a). Figure 1.40 clearly demonstrates the classification of structures based on the static indeterminacy criteria.

1.8.3 Classification by kinematical indeterminacy criteria Kinematical indeterminacy referring to the ability or inability to determine kinematical quantities associated with a structure by means of kinematics or geometric consideration is utilized as a criterion for classification. A (discrete) structure can therefore be categorized as follows.

1.8.3.1 Kinematically determinate structures A kinematically determinate structure is a structure in which all degrees of freedom are prescribed degrees of freedom. With use of additional assumptions on kinematics of a member, the displacement and deformation at any point within a kinematically determinate structure are known. An example of this type of structures is given in Figure 1.41(a); all nine degrees of freedom are prescribed degrees of freedom.

1.8.3.2 Kinematically indeterminate structures A kinematically indeterminate structure is a structure in which there exists at least one free degree of freedom. As a result, the displacement and deformation of a kinematically indeterminate structure are not completely known. The example of this type of structures is given in Figure 1.41(b); for this particular discrete structure, there exist three free degrees of freedom, i.e. {u2, v2, 2}. Next, we define a term called degree of kinematical indeterminacy of a structure as a total number of free or unknown degrees of freedom present within that structure. Consistent with this definition, the degree of kinematical indeterminacy of a kinematically determinate structure is equal to zero while the degree of kinematical indeterminacy of a kinematically indeterminate structure is always greater than zero. Copyright © 2011 J. Rungamornrat


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Statically stable structures

Externally, statically determinate structure

Determination of internal force from static equilibrium?

Yes

Determination of reactions from static equilibrium?

No

Externally, statically indeterminate structure

Determination of internal force from static equilibrium?

Yes Statically determinate structure

No

No

Internally statically indeterminate structure

Statically indeterminate structure

Statically indeterminate structure

Figure 1.40: Diagram indicating classification of structures by static indeterminacy criteria



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v2

Node 2

Node 2 2

u2=0 v2=0 2=0

u1=0 v1=0 1=0

Node 3

Node 3 u3=0 v3=0 3=0

Node 1

u2

u3=0 v3=0 3=0

Node 1 u1=0 v1=0 1=0 (b)

(a)

Figure 1.41: (a) Kinematically determinate structure and (b) kinematically indeterminate structure

1.9 Degree of Static Indeterminacy The degree of static indeterminacy of a structure, denoted by DI, is defined as a number of independent static quantities (i.e. support reactions and the internal force) that must be prescribed in addition to available static equilibrium equations in order to completely describe a static state of the entire structure (a state where all support reactions and internal forces at any locations within the structure are known) or, equivalently, to render the structure statically determinate. From this definition, the degree of static indeterminacy is equal to the number of independent static unknowns subtracted by the number of independent static equilibrium equations. Thus, the degree of static indeterminacy of a statically determinate structure is equal to zero while the degree of static indeterminacy of a statically indeterminate structure is always greater than zero. The degree of static indeterminacy is also known as the degree of static redundancy and the corresponding extra, static unknowns exceeding the number of static equilibrium equations are termed as the redundants.

1.9.1 General formula for computing DI The degree of static indeterminacy of a statically stable structure can be computed from the general formula: DI  ra  n m  n j  n c

(1.18)

where ra is the number of all components of the support reactions, nm is the number of components of the internal member force, nj is the number of independent equilibrium equations at all nodes or joints, and nc is the number of static conditions associated with all internal releases present within the structure. It is evident that the term ra + nm represents the number of all static unknowns while the term nj + nc represents the number of all available equilibrium equations (including static conditions at the internal releases).

1.9.1.1 Number of support reactions The number of all components of support reaction at a given structure can be obtained using the following steps: 1) identify all supports within the structures, 2) identify the type and a number of Copyright © 2011 J. Rungamornrat


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components of the support reaction at each support (see section 1.1.5), and 3) sum the number of components of support reactions over all supports. It is emphasized here that for a beam structure, the component of the support reaction in the direction of the beam axis must not be counted in the calculation of ra since the beam is subjected only to transverse loads and there is no internal axial force at any cross section. For instance, the number of support reactions of the structure shown in Figure 1.42(a), Figure 1.42(b) and Figure 1.42(c) is 3, 4, and 8, respectively.

(b)

(a)

(c)

Figure 1.42: Schematics indicating all components of support reaction

1.9.1.2 Number of internal member forces As clearly demonstrated in subsection 1.5.2, the number of independent components of the internal force for an axial member, a flexural member, and a two-dimensional frame member are equal to 1, 2 and 3, respectively. Thus, the number of components of the internal forces for the entire structure (nm) can simply be obtained by summing the number of components of the internal forces for all individual members. It is worth noting that nm depends primarily on both the number and the type of constituting members of the structure. For instance, nm for the structure shown in Figure 1.42(a) is equal to 14(1) = 14 since it consists of 14 axial members; nm for the structure shown in Figure 1.42(b) is equal to 2(2) = 4 since it consists of 2 flexural members (by considering all supports as joints or nodes); and nm for the structure shown in Figure 1.42(c) is equal to 20(3) = 60 since it consists of 20 frame members (by considering supports and connections between columns and beams as joints or nodes).

1.9.1.3 Number of joint equilibrium equations To compute nj, it is required to know both the number and the type of joints present in the structure. The number of independent equilibrium equations at each joint depends primarily on the type of the joint. Here, we summarize standard joints found in the idealized structures. 1.9.1.3.1 Truss joints A truss joint is an idealized joint used for modeling connections of a truss structure. The truss joint behaves as a hinge joint so it cannot resist any moment and allows all members joining the joint to rotate freely relative to each other. Since the truss member possesses only the internal axial force, when the truss joint is separated from the structure to sketch the FBD, all forces acting to the joint are concurrent forces as shown in Figure 1.43; in particular, P1 and P2 are external loads and P1, P2 and P3 are internal axial forces from the truss members. As a consequence, the number of Copyright © 2011 J. Rungamornrat


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independent equilibrium equations per one truss joint is equal to 2 (i.e. FX = 0 and FY = 0; see also subsection 1.7.2).

P2 F1

Y

P1

X

F3

F2

Figure 1.43: FBD of the truss joint 1.9.1.3.2 Beam joints A beam joint is an idealized joint used for modeling connections of a flexural or beam structure. The beam joint behaves as a rigid joint so it can resist the external applied moment and can also transfer the moment among ends of members joining that joint. Since the flexural or beam member possesses only two components of the internal force, i.e. the shear force and the bending moment, when the beam joint is separated from the structure to sketch the FBD, all forces and moments acting to the joint form a set of transverse loads as shown in Figure 1.44; in particular, P and Mo are external loads and V1, M1, V2 and M2 are internal forces from the beam members. As a consequence, the number of independent equilibrium equations per one beam joint is equal to 2 (i.e. FY = 0 and MZ = 0; see also subsection 1.7.3).

P M1

V1

Y

Mo

M2 V2

X

Figure 1.44: FBD of the beam joint 1.9.1.3.3 Frame joints A frame joint is an idealized joint used for modeling connections of a frame structure. The frame joint behaves as a rigid joint so it can resist the external applied moment and can also transfer the moment among ends of the members joining the joint. Since the frame member possesses three components of the internal force, i.e. the axial force, the shear force and the bending moment, when the frame joint is separated from the structure to sketch the FBD, all forces and moments acting to the joint form a set of general 2D loads as shown in Figure 1.45; in particular, P1, P2 and Mo are external loads and F1, V1, M1, F2, V2, M2, F3, V3 and M3 are internal forces from the frame members. As a result, the number of independent equilibrium equations per one frame joint is equal to 3 (i.e. FX = 0, FY = 0 and MZ = 0). 1.9.1.3.4 Compound joints A compound joint is an idealized joint used for modeling connections where more than one types of members are connected. When the compound joint is separated from the structure to sketch the Copyright © 2011 J. Rungamornrat


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FBD, all forces and moments acting to the joint can form a set of general 2D loads as shown in Figure 1.46. As a result, the number of independent equilibrium equations per one compound joint is generally equal to 3 (i.e. FX = 0, FY = 0 and MZ = 0).

F1 Y

Mo

M1

P2

V1

P1 V3 M3

M2 V2

F2

F3

X

Figure 1.45: FBD of the frame joint

Frame member Y

Truss member

X Figure 1.46: FBD of the compound joint The number of independent joint equilibrium equations of the structure (nj) can simply be obtained by summing the number of independent equilibrium equations available at each joint.

1.9.1.4 Internal releases An internal release is a point within the structure where certain components of the internal force such as axial force, shear force and bending moment are prescribed. Presence of the internal releases within the structure provides extra equations in addition to those obtained from static equilibrium. Here, we summarize various types of internal releases that can be found in the idealized structure. 1.9.1.4.1 Moment release or hinge A moment release or hinge is an internal release where the bending moment is prescribed equal to zero or, in the other word, the bending moment cannot be transferred across this point (see Figure 1.47). At the moment release, the displacement is continuous while the rotation or slope is not. For this particular type of internal releases, it provides 1 additional equation per one hinge, i.e. M = 0 at the hinge point.



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Figure 1.47: Schematics of moment releases or hinges 1.9.1.4.2 Axial release An axial release is an internal release where the axial force is prescribed equal to zero or, in the other word, the axial force cannot be transferred across this point (see Figure 1.48). At the axial release, the longitudinal component of the displacement is discontinuous while the transverse component and the rotation are still continuous. For this particular type of internal releases, it provides 1 additional equation per one release, i.e. F = 0 at the axial release.

Figure 1.48: Schematic of axial release 1.9.1.4.3 Shear release A shear release is an internal release where the shear force is prescribed equal to zero or, in the other word, the shear force cannot be transferred across this point (see Figure 1.49). At the shear release, the transverse component of the displacement is discontinuous while the longitudinal component of the displacement and the rotation are still continuous. For this particular type of internal releases, it provides 1 additional equation per one release, i.e. V = 0 at the shear release.

Figure 1.49: Schematic of shear release 1.9.1.4.4 Combined release A combined release is an internal release where two or more components of the internal force are prescribed equal to zero (see Figure 1.50). Behavior of the combined release is the combination of behavior of the moment release, axial release, or shear release. For this particular type of internal releases, it provides two or more additional equations per one release depending on the number of prescribed components of the internal force.

Figure 1.50: Schematics of combined release Copyright © 2011 J. Rungamornrat


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1.9.1.4.5 Full moment release joint A joint or node where the bending moment at the end of all members jointing that joint is prescribed equal to zero is termed as a full moment release joint (see Figure 1.51). This joint has the same behavior and characteristic as the hinge joint. For truss structures, while all joints are full moment release joints, they provide no additional equation since presence of such joints has been considered in the reduction of the number of internal forces per member from three to one (i.e. only axial force is present). For beam or frame structures, presence of a full moment release joint provides n – 1 additional equations where n is the number of member joining the joint; for instance, a full moment release joint shown in Figure 1.51 provides 4 – 1 = 3 additional equations.

Figure 1.51: Schematic of full moment release joint 1.9.1.4.6 Partial moment release joint A joint or node where the bending moment at the end of certain but not all members jointing that joint is prescribed equal to zero is termed as a partial moment release joint (see Figure 1.52). This type of releases can be found in beam and frame structures. A partial moment release joint provides n additional equations if the bending moment at the end of n members are prescribed equal to zero; for instance, a partial moment release joint shown in Figure 1.52 provides 2 additional equations.

Figure 1.52: Schematic of partial moment release joint The number of static conditions associated with all internal releases present in the structure (nc) can simply be obtained by summing the number of additional equations provided by each internal release. Example 1.1 Determine the degree of static indeterminacy (DI) of the following structures

     

ra = 2 + 1 = 3 25 truss members  nm = 25(1) = 25 14 truss joints  nj = 14(2) = 28 No internal release  nc = 0 DI = 3 + 25 – 28 – 0 = 0 Statically determinate structure


37



     

ra = 2 + 1 + 1 + 1 = 5 3 beam members  nm = 3(2) = 6 4 beam joints  nj = 4(2) = 8 1 moment release  nc = 1 DI = 5 + 6 – 8 – 1 = 2 Statically indeterminate structure

     

ra = 3(3) + 2 + 1 = 12 28 frame members  nm = 28(3) = 84 21 frame joints  nj = 21(3) = 63 No internal release  nc = 0 DI = 12 + 84 – 63 – 0 = 33 Statically indeterminate structure

 ra = 4(3) = 12  28 frame members and 12 truss members  nm = 28(3) + 12(1) = 96  6 frame joints and 14 compound joints  nj = 6(3) + 14(3) = 60  No internal release  nc = 0  DI = 12 + 96 – 60 – 0 = 48  Statically indeterminate structure

1.9.2 Check of external static indeterminacy For a given statically stable structure, let ra be the number of all components of the support reactions, net be the number of independent equilibrium equations available for the entire structure and ncr be the number of additional static conditions that can be set up without introducing new static unknowns. The structure is externally, statically determinate if and only if ra  n et  n cr

(1.19)

and the structure is externally, statically indeterminate if and only if (1.20)

ra  n et  n cr

This check of external static indeterminacy is essential when the support reactions of the structure are to be determined. Copyright © 2011 J. Rungamornrat


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In general, for plane structures, the number of independent equilibrium equations that can be set up for the entire structure (net) is equal to 3, except for beam structures where the number of independent equilibrium equations reduces to 2 (the equilibrium of forces in the direction along the beam axis is automatically satisfied). Additional static conditions are typically the conditions associated with internal releases present within the structure; for instance, points where components of internal forces are prescribed such as “moment release or hinge”, “shear release”, and “axial release”. It is important to note that not all the static conditions can be incorporated in the counting of ncr but ones that introduce no additional unknowns other than the support reactions can be counted. These additional equations can be set up in terms of equilibrium equations of certain parts of the structure resulting from proper sectioning the structure at the internal releases. To clearly demonstrate the check of external static indeterminacy, let consider a frame structure as shown in Figure 1.53. For this structure, we obtain ra = 2(2) = 4, nm = 6(3) = 18, nj = 6(3) = 18, nc = 2(1) = 2  DI = 4 +18 – 18 – 2 = 2; thus the structure is statically determinate. In addition, net = 3 for frame structure and ncr = 1 since one additional equation (without introducing additional unknowns other than support reactions) can be set up by sectioning at the hinge A and then enforcing moment equilibrium about the point A of one part of the structure. The static condition associated with the hinge B cannot be included in ncr since no new equation can be set up without introducing additional unknown internal forces along the cut. It is evident that ra = 4 = net + ncr  the structure is externally, statically determinate and, therefore, all support reactions can be determined from static equilibrium. Since the structure is also statically indeterminate, from the definition provided above, this implies that the structure is internally, statically indeterminate.

B

A

Figure 1.53: Schematic of externally statically determinate structure

1.9.3 DI of truss structures Consider a statically stable truss structure that consists of m members and n joints. For this particular structure, we obtain nm = m(1) = m, nj = n(2) = 2n and nc = 0. Upon using the general formula (1.18), the degree of static indeterminacy of a truss is given by DI  ra  m  2n

(1.21)

It is important to emphasize that there cannot be an internal release at interior points of all truss members since each member possesses only one component of internal forces; presence of the (axial) internal release will render the structure statically unstable.


39



Example 1.2 Determine the degree of static indeterminacy (DI) of the following statically stable truss structures         

ra = 2 + 2 = 4 m = 35 n = 18 DI = 4 + 35 – 2(18) = 3 Statically indeterminate net = 3 ncr = 0 ra = 4 > net + ncr = 3 Externally statically indeterminate

        

ra = 2 + 1 = 3 m = 14 n=8 DI = 3 + 14 – 2(8) = 1 Statically indeterminate net = 3 ncr = 0 ra = 3 = net + ncr Externally statically determinate

        

ra = 2 + 2 = 4 m = 10 n=7 DI = 4 + 10 – 2(7) = 0 Statically indeterminate net = 3 ncr = 1 ra = 4 = net + ncr Externally statically determinate implies from DI as well)

(or

1.9.4 DI of beam structures Consider a statically stable beam structure that consists of m members and n joint. For this particular structure, we obtain nm = m(2) = 2m and nj = n(2) = 2n. Upon using the general formula (1.18), the degree of static indeterminacy of a beam is given by DI  ra  2(m  n)  n c

(1.22)

It is important to emphasize that in the determination of ra, components of the support reactions in the direction parallel to the beam axis must be ignored since there is no internal axial force in any beam members. In addition, the number of members of a given beam is not unique but it depends Copyright © 2011 J. Rungamornrat


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primarily on the choice of joints or nodes considered; in general, joints are located at the supports and free ends. However, the choice of joints and members does not affect the final value of DI. Example 1.3 Determine the degree of static indeterminacy (DI) of the following statically stable beam structures          

ra = 2(2) + 3(1) = 7 m=4 n=5 nc = 2 DI = 7 + 2(4 – 5) – 2 = 3 Statically indeterminate net = 2 ncr = 2 ra = 7 > net + ncr = 4 Externally statically indeterminate

         

ra = 2 + 2(1) = 4 m=3 n=4 nc = 2 DI = 4 + 2(3 – 4) – 2 = 0 Statically determinate net = 2 ncr = 2 ra = 4 = net + ncr Externally statically determinate

         

ra = 2 + 3(1) = 5 m=4 n=5 nc = 0 DI = 5 + 2(4 – 5) – 0 = 3 Statically indeterminate net = 2 ncr = 0 ra = 5 > net + ncr= 2 Externally statically indeterminate

1.9.5 DI of frame structures Consider a statically stable frame structure that consists of m members and n joints. For this particular structure, we obtain nm = m(3) = 3m and nj = n(3) = 3n. Upon using the general formula (1.18), the degree of static indeterminacy of a frame is given by DI  ra  3(m  n)  n c

(1.23)

Note that the free end must be treated as a joint or node. Copyright © 2011 J. Rungamornrat

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Example 1.4 Determine the degree of static indeterminacy (DI) of the following statically stable frame structures          

ra = 2(3) + 2 = 8 m=7 n=8 nc = 3 – 1 = 2 DI = 8 + 3(7 – 8) – 2 = 3 Statically indeterminate net = 3 ncr = 2 ra = 8 > net + ncr = 5 Externally statically indeterminate

         

ra = 2 + 1 = 3 m=7 n=6 nc = 0 DI = 3 + 3(7 – 6) – 0 = 6 Statically indeterminate net = 3 ncr = 0 ra = 3 = net + ncr Externally statically determinate

         

ra = 3 + 1 = 4 m=3 n=4 nc = 1 DI = 4 + 3(3 – 4) – 1 = 0 Statically determinate net = 3 ncr = 1 ra = 4 = net + ncr = 4 Externally statically determinate

1.10 Investigation of Static Stability of Structures Static stability of the real structure is essential and must extensively be investigated to ascertain that the structure can maintain its functions and purposes under external actions and excitations without excessive movement or collapse of the entire structure and its parts. In the structural modeling or structural idealization, the stability assurance can be achieved by requiring that all suitable idealized structures must be statically stable. This requirement is also essential in the sense that the subsequent process of static structural analysis can be performed. As previously mentioned, loss of stability of the structure can occur either on the entire structure or on the certain parts. The primary sources of instability are due to an insufficient number Copyright © 2011 J. Rungamornrat


42


of supports provided, inappropriate directions of constraints provided, inappropriate arrangement of constituting members that forms insufficient internal constraints, or presence of too many of internal releases. Here, we summarize three basic lemmas that can be used to investigate the static stability of a given idealized structure.

1.10.1 Lemma 1 From section 1.9, it can be deduced that “if the structure is statically stable, it must be either statically determinate (DI = 0) or statically indeterminate (DI > 0)”; thus DI of the structure is nonnegative if the structure is statically stable. This statement is mathematically equivalent to “if DI < 0, then the structure is statically unstable”. This lemma is simple and can be used to deduce the instability of the structure by the knowledge of negative DI. It is important to emphasize that, for any structure possessing DI ≥ 0, the lemma fails to provide information about their stability. Example 1.5 Use lemma 1 to check instability of the following structures.      

Truss structure ra = 1 + 1 = 2 m=9 n=6 DI = 2 + 9 – 2(6) = –1 < 0 Lemma 1  Statically unstable

      

Frame structure ra = 1 + 1 +1 = 3 m=3 n=4 nc = 0 DI = 3 + 3(3 – 4) – 0 = 0 Lemma 1  No conclusion about its stability

      

Beam structure ra = 2 + 1 +1 = 4 m=2 n=3 nc = 2 DI = 4 + 2(2 – 3) – 2 = 0 Lemma 1  No conclusion about its stability


43



      

Frame structure ra = 3 + 3 +1 = 7 m=5 n=6 nc = 1 DI = 7 + 3(5 – 6) – 1 = 3 Lemma 1  No conclusion about its stability

1.6.2 Lemma 2 The second lemma comes from the definition of the static instability condition of a structure: a structure is statically unstable if and only if there exists at least one pattern of a rigid body displacement developed within the structure under a particular action. This lemma can be used to conclude the instability of the structure by identifying one mechanism or rigid body displacement.

1.6.3 Lemma 3 The third lemma comes from the definition of the static stability condition of a structure: a structure is statically stable if and only if there is no development of a rigid body displacement in any part of the structure under any action. This lemma can be employed to conclude the stability of the structure by investigating all possible mechanisms or rigid body displacement. Example 1.6 Investigate the static stability of truss structures shown in the figures below. The structure I is obtained by adding the truss members a and b to the structure I and the structure III is obtained by adding the truss member c to the structure I.

a

b

Structure II

Structure I c

Structure III Solution Structure I: the degree of static indeterminacy (DI) is computed as follows: ra = 2 + 1 +1 = 4, nm = 11(1) = 11, nj = 8(2) = 16, nc = 0  DI = 4 + 11 – 16 – 0 = –1 < 0. Thus, from lemma 1, it can be Copyright © 2011 J. Rungamornrat


44


concluded that the structure I is statically unstable. Note that one can also use lemma 2 to conclude this instability by sketching the mechanism as shown in the figure below.

Structure II: the degree of static indeterminacy (DI) is computed as follows: ra = 2 + 1 +1 = 4, nm = 13(1) = 13, nj = 8(2) = 16, nc = 0  DI = 4 + 13 – 16 – 0 = 1 > 0. Thus, static stability of the structure cannot be concluded from the lemma 1. However, by investigating all parts of this structure, there exists a pattern of rigid body displacement as shown in the figure below. Therefore, the lemma 2 deduces that the structure is statically unstable.

Structure III: the degree of static indeterminacy (DI) is computed as follows: ra = 2 + 1 +1 = 4, nm = 12(1) = 12, nj = 8(2) = 16, nc = 0  DI = 4 + 12 – 16 – 0 = 0. Thus, static stability of the structure cannot be concluded from lemma 1. However, by investigating all parts of this structure, there is no development of rigid body displacement within any parts of the structure. Therefore, lemma 3 deduces that the structure is statically stable and, in addition, the structure is statically determinate since DI = 0. Example 1.7 Investigate the static stability of frame structures shown in the figure below. The structure II and structure III are obtained by adding the horizontal member to the structure I with the location of the hinge a being above or below the added member.

a

Structure I

a

Structure II Copyright © 2011 J. Rungamornrat

a

Structure III

45



Solution Structure I: the degree of static indeterminacy (DI) is computed as follows: ra = 3 + 3 = 6, nm = 4(3) = 12, nj = 5(3) = 15, nc = 4  DI = 6 + 12 – 15 – 4 = –1 < 0. Thus, from lemma 1, it can be concluded that the structure I is statically unstable. Note that one can also use lemma 2 to conclude this instability by sketching the mechanism as shown in the figure below.

a

a

Structure I

Structure II

Structure II: the degree of static indeterminacy (DI) is computed as follows: ra = 3 + 3 = 6, nm = 7(3) = 21, nj = 7(3) = 21, nc = 4  DI = 6 + 21 – 21 – 4 = 2 > 0. Thus, static stability of the structure cannot be concluded from the lemma 1. However, by investigating all parts of this structure, there exists a pattern of rigid body displacement as shown in the figure below. Therefore, the lemma 2 deduces that the structure is statically unstable. Structure III: the degree of static indeterminacy (DI) is computed as follows: ra = 3 + 3 = 6, nm = 7(3) = 21, nj = 7(3) = 21, nc = 4  DI = 6 + 21 – 21 – 4 = 2 > 0. Thus, static stability of the structure cannot be concluded from lemma 1. However, by investigating all parts of this structure, there is no development of rigid body displacement within any parts of the structure. Therefore, lemma 3 deduces that the structure is statically stable and, in addition, the structure is statically indeterminate since DI > 0.

Exercises 1. For each statically stable truss shown below, determine its degree of static indeterminacy and also use static indeterminacy criteria to identify whether it belongs to following classes (class 1: externally, statically determinate structures; class 2: externally, statically indeterminate structures; class 3: internally, statically indeterminate structures; class 4: statically determinate structures; class 5: statically indeterminate structures). Note that each structure may belong to several classes.



46


2. For each statically stable beam shown below, determine its degree of static indeterminacy and also use static indeterminacy criteria to identify whether it belongs to following classes (class 1: externally, statically determinate structures; class 2: externally, statically indeterminate structures; class 3: internally, statically indeterminate structures; class 4: statically determinate structures; class 5: statically indeterminate structures). Note that each structure may belong to several classes.

3. For each statically stable frame shown below, determine its degree of static indeterminacy and also use static indeterminacy criteria to identify whether it belongs to following classes (class 1: externally, statically determinate structures; class 2: externally, statically indeterminate structures; class 3: internally, statically indeterminate structures; class 4: statically determinate structures; class 5: statically indeterminate structures). Note that each structure may belong to several classes. Copyright © 2011 J. Rungamornrat

47





48


4. Investigate the static stability of the structures shown below using Lemma 1, Lemma 2 and Lemma 3. If the structure is statically unstable, show the sketch of possible rigid body displacements.



49

Analysis of Determinate Structures

CHAPTER 2 ANALYSIS OF DETERMINATE STRUCTURES This chapter focuses primarily on analysis of statically determinate structures. The key objective of the analysis is to determine unknown static quantities such as support reactions and internal forces resulting from external applied loads. As already discussed in the previous chapter, statically determinate structures belong to a special class of structures that all support reactions and internal forces at any location of the structure can completely be determined from equilibrium equations. In following sections, we first emphasize the definition of unknown static quantities (i.e. support reactions and internal forces) and a notion of applied loads, and then discuss essential tools for performing static analysis of statically determinate structures, e.g. equilibrium equations, method of structure partitioning, and free body diagram (FBD). Next, we demonstrate applications of equilibrium equations to determine support reactions of externally, statically determinate structures. Finally, analysis of the internal forces for certain classes of structures such as trusses, beams and rigid frames are presented. In addition, for the case of beams and rigid frames, the sketch of their qualitative elastic curve (or deformed shape) is also discussed.

2.1 Static Quantities Static quantities are the quantities associated with forces, intensity of forces (e.g. pressure, traction, and stress), or resultants of forces (e.g. moment and torque). Three basic static quantities interested in the analysis of structures are applied loads, support reactions, and internal forces as shown in Figure 2.1. Applied loads

Internal forces Support reactions Figure 2.1: Schematic indicating applied loads, support reactions and internal forces Applied loads represent prescribed forces, intensity of forces, resultants of forces, or in combination that are exerted to the structure by surrounding environments. It is necessary that applied loads must be known a priori (from idealization of excitations) before the analysis procedure is carried out. Support reactions represent unknown forces, intensity of forces, resultants of forces, or in combination that are exerted to the structure by (idealized) supports to maintain its equilibrium and stability under the action of applied loads or other excitations. The support reactions are naturally unknown a priori which can be obtained from static analysis. Copyright © 2011 J. Rungamornrat


50


Internal forces represent intensity of forces (e.g. stress) or resultant forces (e.g. axial force, shear force, bending moment, torque) induced at a material point or over a particular section of the structure under the action of applied loads or other excitations. Similar to the support reactions, the internal forces are unknown a priori which can be obtained from static analysis. The choice of the internal forces used to characterize the behavior of any structure depends primarily on the type of structures and the nature of excitations. This will be discussed further below.

2.2 Tools for Static Analysis Two key methodological components essential for determining support reactions and internal forces at any location of a structure are the static equilibrium equations and the method of structure partitioning. The first component is utilized generally to construct a necessary and sufficient set of equations to solve for unknown support reactions and internal forces, while the latter component accommodates the construction of equilibrium equations over certain portions of the structure in addition to those associated with the entire structure in order to supply adequate number of equations.

2.2.1 Static equilibrium Equilibrium condition of a body is a statement of conservation of linear momentum and angular momentum of the body. More precisely, the body is in equilibrium if and only if the linear momentum and angular momentum are conserved for any part of the body (the entire body can also be considered as a part of its body). For a two-dimensional body occupied a region on the X-Y plane as shown schematically in Figure 2.2, equilibrium of this body implies that all forces and moments applied to the body must satisfy the following three equations: FX = 0

(2.1)

FY = 0

(2.2)

MAZ = 0

(2.3)

where A is an arbitrary point used for computing the moment about the Z-axis and {X, Y, Z; O} denotes a reference Cartesian coordinate system. In particular, equations (2.1) and (2.2) indicate that the sum of components of all forces in X-direction and in Y-direction must vanish while the last equation (2.3) requires that the sum of moments about a point A in Z-direction must also vanish. As already pointed out in chapter 1, equilibrium condition of a two-dimensional body subjected to a system of general forces and moments provides exactly three independent equations (for a body subjected to special systems of applied loads, the number of independent equations can be less than three; readers are suggested to consult the section 1.7 for extensive discussion). It is important to emphasize that a set of three independent equations resulting from the equilibrium condition of a two-dimensional body is not unique. This non-uniqueness is due primarily to that the point for taking moment (i.e. point A) can be chosen arbitrarily (see discussion in section 1.7 of chapter 1). Here, we present four different but equivalent sets of three equilibrium equations that can be employed in static analysis of two-dimensional structures.

2.2.1.1 Set 1 A set consists of the equilibrium of forces in X-direction (2.1), the equilibrium of forces in Ydirection (2.2), and the equilibrium of moments about an arbitrarily selected point A (2.3). Copyright © 2011 J. Rungamornrat


51


Y

X

A

Figure 2.2: Schematic of two-dimensional body subjected to applied loads

2.2.1.2 Set 2 A set consists of the equilibrium of forces in X-direction (2.1), the equilibrium of moments about an arbitrarily selected point A (2.3), and the equilibrium of moments about another arbitrary selected point B, i.e. MBZ = 0

(2.4)

The only constraint placed on the choice of points A and B to render (2.1), (2.3) and (2.4) all independent is that the straight line connecting A and B must not parallel to the Y-axis.

2.2.1.3 Set 3 A set consists of the equilibrium of forces in Y-direction (2.2), the equilibrium of moments about an arbitrarily selected point A (2.3), and the equilibrium of moments about another arbitrary selected point B (2.4). Again, the only constraint placed on the choice of points A and B to render (2.2), (2.3) and (2.4) all independent is that the straight line connecting A and B must not parallel to the X-axis.

2.2.1.4 Set 4 A set consists of the equilibrium of moments about an arbitrarily selected point A (2.3), the equilibrium of moments about an arbitrary selected point B (2.4), and the equilibrium of moments about an arbitrary selected point C, i.e. MCZ = 0

(2.5)

The only constraint placed on the choice of points A, B and C to render (2.3), (2.4) and (2.5) all independent is that the A, B and C must not belong to the same straight line. To prove the equivalence among the above four sets, let’s assume that the body is subjected to a system of forces P1, P2, …, and PN acting at points (X1, Y1), (X2, Y2), …, (XN, YN), respectively, and a system of moments M1, M2, …, and MK as shown in Figure 2.3. Let A, B, and C be three arbitrarily selected points with coordinates (XA, YA), (XB, YB), and (XC, YC), respectively. The equilibrium of forces in the X-direction and Y-direction takes the form N

F

iX

(2.6)

= 0

i=1



52


FN

MK M1

C

Y

M2 B F1

A

X

F2

Figure 2.3: Schematic of two-dimensional body subjected to a system of forces and moments N

F

iY

(2.7)

= 0

i=1

where the subscripts “X” and “Y” indicate the X-component and Y-component of the force Fi, respectively. Similarly, equilibrium of moments about point A, B and C can readily be obtained as N

F

iX

(YA  Yi ) +

i=1

N

F

iX

N

iX

iY

(X i  X A ) +

i=1

(YB  Yi ) +

i=1

F

N

F N

F

iY

i=1

N

F

iY

i

= 0

(2.8)

i

= 0

(2.9)

i

= 0

(2.10)

i=1

(X i  X B ) +

i=1

(YC  Yi ) +

K

M K

M i=1

(X i  X C ) +

i=1

K

M i=1

Upon simple manipulations, equation (2.9) can be expressed as N

 FiX (YA  Yi ) + i=1

N

 FiY (Xi  X A ) + i=1

K

N

N

i=1

i=1

i=1

N

N

i=1

i=1

 Mi + (YB  YA ) FiX + (XA  XB ) FiY = 0

(2.11)

Similarly, equation (2.10) can also be written as N

F

iX

i=1

(YA  Yi ) +

N

F

iY

i=1

(X i  X A ) +

K

 M + (Y i

i=1

C

 YA ) FiX + (X A  X C ) FiY = 0

(2.12)

It is evident from (2.11) and (2.12) that equations (2.9) and (2.10) are not independent of the three equations (2.6), (2.7) and (2.8) but, in fact, they are the linear combinations of those three. Thus, the Set 1 is equivalent to the Set 2 provided that XA – XB does not vanish or, in the other word, a straight line connecting points A and B is not parallel to the Y-axis. It can readily be verified by employing equations (2.6) and (2.8) and then dividing the final result by XA – XB that (2.11) can be reduced to equation (2.7). This similar argument can also be used to prove the equivalence between Set 3 and Set 1. Finally, the equivalence between Set 4 and Set 1 can be verified by first substituting (2.8) into (2.11) and (2.12) to obtain Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat N

N

i=1

i=1

N

N

i=1

i=1

53


(YB  YA ) FiX + (X A  X B ) FiY = 0

(2.13)

(YC  YA ) FiX + (X A  X C ) FiY = 0

(2.14)

Equations (2.13) and (2.14) are equivalent to equations (2.6) and (2.7) if and only if (YB – YA)( XA – XC) – (YC – YA)( XA – XB) ≠ 0 or, equivalently, the three points A, B, C do not belong to the same straight line. There is no strong evidence to support and decide the best choice from the four sets given before. In general, the choice is a matter of taste and preference of an individual and, sometimes, it is problem dependent. The most reasonable choice is the one that allows all unknowns appearing in all three equations be solved in an easy manner as much as possible. In addition, after one of the four sets is already chosen, the order of three equations in the set to be employed and the choice of points used for taking moments are generally selected to eliminate the unknowns as many as possible in order to avoid solving a large system of linear equations. This strategy becomes more apparent in examples 2.1-2.3.

2.2.2 Method of structure partitioning To determine the internal forces at any location of the structure or to compute some components of the support reactions for certain structures, the consideration of equilibrium of certain parts of the structure is required in addition to that of the entire structure. From the fact that the structure is in equilibrium if and only if any part of the structure is in equilibrium, any portion of the structure resulting from partitioning of the structure must be in equilibrium with applied loads acting to that portion (i.e. reactions at supports present in that portion and internal forces exerted to that portion by the rest of the structure). Structure partitioning is simply a process to decompose the structure into two or several parts by introducing a sufficient number of fictitious or imaginary cuts at certain locations of the structures. For instance, a rigid frame shown in Figure 2.4 is partitioned into two parts by a fictitious cut at point B. One crucial function of the fictitious cut is to allow us to access and see the internal forces at the location of the cut. As can be observed from free body diagrams of the two parts in Figure 2.4, the internal forces {FB, VB, MB} at point B appear in both FBDs. M

P M P

RCY

C MB

Y

FB

B A

VB FB

MB X

VB RAX

RAM RAY

Figure 2.4: Schematic of the entire structure and two parts resulting from partitioning at point B Copyright © 2011 J. Rungamornrat


54


Since each portion of the structure resulting from partitioning must be in equilibrium, several equilibrium equations are therefore provided in addition to the equilibrium equations set up on the entire structure. However, the number of unknowns associated with the internal forces appearing at the cuts also increases at the same time. Hence, locations of the cuts and the number of the cuts are important and must properly be chosen in order to ensure that the number of all unknowns does not exceed the number of available equilibrium equations. In addition, the imaginary cut must be made at the point where the internal forces are of interest. For instance, the cut must be made at point B of the rigid frame in Figure 2.4 if the internal forces {FB, VB, MB} are to be determined.

2.3 Determination of Support Reactions This section demonstrates the application of static equilibrium to compute all support reactions of externally, statically determinate structures (note that statically determinate structures are also contained in this class of structures). Recalling the definition provided in subsection 1.8.2 of chapter 1, all support reactions of externally, statically determinate structures can completely be obtained by solving static equilibrium equations. A brief summary of guidelines for determining support reactions is given below:  Identify type of all supports  Determine the number of unknown support reactions (ra)  Determine the number of independent equilibrium equations that can be set up for the entire structure (net)  Determine the number of additional static conditions that can be set up without introducing new static unknowns (ncr); consult subsection 1.9.2 of chapter 1 for extensive discussion  If ra > net + ncr, the structure is externally statically indeterminate and support reactions cannot completely be obtained by using only equilibrium equations. If ra = net + ncr, the structure is externally statically determinate and support reactions can be determined as described below. It is worth noting that if a given structure is known to be statically determinate (i.e. DI = 0), it automatically implies that ra = net + ncr.  If ra = net, all support reactions can be obtained from equilibrium of the entire structures and the following steps are suggested: (i) sketch a free body diagram (FBD) of the entire structure, (ii) write down all independent equilibrium equations, and (iii) solve for unknown support reactions  If ra > net, all support reactions cannot be obtained from equilibrium of the entire structures alone and the subsequent steps are as follows: (i) introduce suitable fictitious cuts; in general, fictitious cuts are made at the internal releases present within the structure (e.g. hinges, shear releases, axial releases), (ii) ensure that the number of fictitious cuts is sufficient to supply adequate number of equations to solve for all unknowns (i.e. all support reactions and extra unknowns associated with the internal forces induced at the cuts), (iii) sketch a free body diagram of the entire structure and free body diagrams of parts resulting from the cuts, (iv) write down all independent equilibrium equations for the entire structure and for parts resulting from the cuts, and (v) solve for unknown support reactions. It is important to emphasize again that after the free body diagram(s) is sketched, a careful choice of reference points for computing moment and the order of equilibrium equations employed can significantly reduce the computational effort associated with solving linear equations. This becomes apparent in following examples. Copyright © 2011 J. Rungamornrat


55


Example 2.1 Determine all support reactions of a rigid frame shown below P

PL D

C

L A

B 4P L

L

Solution The given structure is constrained at points A and B by a pinned support and a roller support, respectively; thus, the total number of support reactions is ra = 2 + 1 = 3. The number of independent equilibrium equations for a two-dimensional rigid frame is net = 3. Since ra = net, the structure is externally, statically determinate and all support reactions can be obtained by considering equilibrium of the entire structure. FBD of the entire structure is given below where {RAX, RAY} and RBY are unknown support reactions at point A and point B, respectively. To demonstrate the equivalence among four sets of equilibrium equations mentioned in section 2.2.1, we apply each set separately and then compare the final results. P

PL D

C

Y

RAX

A

B 4P

RAY

X RBY

Option I: Use equilibrium equations from the Set 1 [FX = 0]  

:

RAX + P = 0 RAX = –P

[MA = 0]  

:

(RBY)(2L) – PL – (P)(L) – (4P)(L) = 0 RBY = 3P

[FY = 0]  

:

Leftward Upward

RAY + RBY – 4P = 0 RAY = P

Upward

Option II: Use equilibrium equations from the Set 2 [FX = 0]  

:

RAX + P = 0 Copyright © 2011 J. Rungamornrat


56


RAX = –P [MA = 0]  

:

Leftward

(RBY)(2L) – PL – (P)(L) – (4P)(L) = 0 RBY = 3P

[MB = 0]  

:

Upward

–(RAY)(2L) + (4P)(L) – (P)(L) – PL = 0 RAY = P

Upward

Option III: Use equilibrium equations from the Set 3 [MA = 0]  

:

(RBY)(2L) – PL – (P)(L) – (4P)(L) = 0 RBY = 3P

[FY = 0]  

:

Upward

RAY + RBY – 4P = 0 RAY = P

[MC = 0]  

:

Upward

(RAX)(L) + (RBY)(2L) – (4P)(L) – PL = 0 RAX = –P

Leftward

Option IV: Use equilibrium equations from the Set 4 [MA = 0]  

:

(RBY)(2L) – PL – (P)(L) – (4P)(L) = 0 RBY = 3P

[MB = 0]  

:

Upward

–(RAY)(2L) + (4P)(L) – (P)(L) – PL = 0 RAY = P

[MC = 0]  

:

Upward

(RAX)(L) + (RBY)(2L) – (4P)(L) – PL = 0 RAX = –P

Leftward

It is evident that use of any set of equilibrium equations leads to the same results. While the order of equilibrium equations employed does not matter, those containing only one unknown are considered first. This allows the unknown be easily obtained without solving any system of linear equations. Example 2.2 Determine all support reactions of a beam shown below q

P = qL C

A

B L

E

D L

L

L

Solution Since ra = 2 + 1 + 1 = 4, m = 2, n = 3, nc = 2, then DI = 4 + 2(2) – 2(3) – 2 = 0. Thus, the structure is statically determinate and all support reactions can be determined form static equilibrium. However, the number of independent equilibrium equations that can be set up for a Copyright © 2011 J. Rungamornrat


57


beam is net = 2 < ra; thus, the support reactions cannot be obtained by considering only equilibrium of the entire structure. To clarify this, let us sketch the FBD of the entire beam as shown below.

q C

RAM

A

Y

P = qL B

E

D

RAY

X

REY

RCY

Equilibrium of the entire beam requires that [MA = 0]  

:

RAM + (RCY)(2L) + (REY)(4L) – (qL)(L/2) – (qL)(3L) = 0 RAM + 2RCYL + 4REYL = 7qL2/2

[FY = 0]  

:

RAY + RCY + REY – qL – qL = 0 RAY + RCY + REY = 2qL

It is emphasized that there are only two independent equilibrium equations and they are insufficient to be solved for four unknown support reactions RAM, RAY, RCY and REY. To overcome this problem, two additional equations associated with the presence of two moment releases or hinges at points B and D, i.e. M = 0 at B and M = 0 at D, must be employed. By introducing two cuts, one at point B and the other at point just to the right of point D called DR, the original structure is decomposed into three parts and the FBD of each part is shown below. Note that the cut is not made exactly at point D due to the application of a concentrated load at point D and we choose to avoid the question on how to distribute this concentrated load to the left and right parts of the point D. For this particular choice of the cut (cut at point DR), the concentrated load P = qL appears at the point DR of the FBD of the middle part. q RAM

A RAY

VB B VB

P = qL C

B

D RCY

Y VDR E

VDR

REY

X

The left portion contains three unknowns, i.e., {RAM, RAY, VB}; the middle portion also contains three unknowns, i.e., {RCY, VB, VDR}; and the right portion contains only two unknowns, i.e., {VDR, REY}. The total number of unknowns (including all support reactions and the shear forces appearing at the cuts) now becomes six, i.e., {RAM, RAY, RCY, REY, VB, VDR}. Two independent equilibrium equations can be set up for each individual portion and this leads to a set of six independent linear equations sufficient for determining all unknowns. To avoid solving a system of six linear equations, we first consider the right portion in which the number of unknowns is equal to the number of independent equilibrium equations. By first applying equilibrium of moments about point E and then considering equilibrium of forces in Y-direction, we obtain Copyright © 2011 J. Rungamornrat


58


[ME = 0]  

:

–VDRL = 0 VDR = 0

[FY = 0]  

:

REY + VDR = 0 REY = 0

Since the shear force VDR is already known, the number of unknowns for the middle portion now reduces to two, i.e., {RCY, VB}, and they can be solved by considering equilibrium of this portion as follows: [MB = 0]  

:

(RCY)(L) – (qL)(2L) – (VDR)(2L) = 0 RCY = 2qL

[FY = 0]  

:

Upward

VB +RCY – qL – VDR = 0 VB = –qL

Downward

Since the shear force VB is already known, the number of unknowns for the left portion now reduces to two, i.e., {RAM, RAY}, and they can be solved by considering equilibrium of this portion as follows: [MA = 0]  

:

RAM – (qL)(L/2) – (VB)(L) = 0 RAM = –qL2/2

[FY = 0]  

:

Clockwise

RAY – qL – VB = 0 RAY = 0

Example 2.3 Determine all support reactions of a truss shown below 4P 2P C

0.4L 0.2L 0.4L B

A

0.4L 0.2L 0.4L 0.4L 0.2L 0.4L Solution Since ra = 2 + 2 = 4, m = 10, n = 7, nc = 0, then DI = 4 + 10(1) – 7(2) – 0 = 0. Thus, the structure is statically determinate and all support reactions can be determined from static Copyright © 2011 J. Rungamornrat


59


equilibrium. However, the number of independent equilibrium equations that can be set up for a beam is net = 3 < ra; thus, the support reactions cannot be obtained by considering only the equilibrium of the entire structure. Partitioning of the structure to construct additional equations is then required. 4P Y

2P

FX

FY C

C X RAX

RBX

A

B RBY

RAY

RBX

B RBY

First, let us consider the equilibrium of the entire structure (its FBD is shown above). By applying equilibrium of moments about point A and B, the two support reactions {RAY, RBY} can readily be determined: [MA = 0]  

:

(RBY)(2L) – (2P)(L) – (4P)(L) = 0 RBY = 3P

[MB = 0]  

:

Upward

–(RAY)(2L) – (2P)(L) + (4P)(L) = 0 RAY = P

Upward

The remaining equilibrium equation associated with equilibrium of forces in X-direction only provides a relation between the two unknowns {RAX, RBX}: [FA = 0]  

:

RAX + RBX + 2P = 0

To obtain an additional equation, we introduce a cut at point just to the right of point C and then consider the right portion resulting from that cut; the FBD is shown in the figure below. Since point C is a pinned or hinge joint, only two new unknowns {FX, FY} are introduced at the cut. In addition, the applied loads acting at point C do not appear in the FBD of the right portion due to that the cut is not made exactly at point C but at point just to the right of point C. By considering equilibrium of moments about point C and then using the known value of RBY, we obtain [MC+ = 0]  

:

(RBX)(L) + (RBY)(L) = 0 RBX = –3P

Leftward

Once RBX is known, the reaction RAX can readily be obtained from above equation, i.e., RAX = –2P –RBX = P (Rightward).



60


2.4 Static Analysis of Truss In this section, we focus our attention on the analysis of statically determinate trusses with the primary objective to determine all support reactions and internal forces due to external applied loads. This section starts with a brief introduction to characteristics of trusses, notations and terms used throughout. Next, we present two standard methods typically employed to determine the internal force of each member of the truss. Various examples are then presented to demonstrate the principle and procedure of each method.

2.4.1 Characteristics of truss An idealized structure is termed a truss if and only if (i) all members are straight, (ii) all members are connected by pinned (or hinge or truss) joints, (iii) all applied loads are in terms of concentrated forces and they act only at joints, and (iv) all supports provide only constraints against translations. Examples of trusses are shown schematically in Figure 2.5. Note that concentrated moments are not allowed to be applied to any joint of the truss since they provide no rotational constraint. From above definition, it can readily be verified that any member of the truss possesses only one component of the internal force (i.e. the axial force) and this axial force is constant throughout the member. In addition, the angle between any two members joining the same pinned joint is not preserved; the angle measured before and after application of applied loads is generally not identical. To demonstrate this feature, let us consider a truss shown in Figure 2.6. After subjected to applied loads, this truss is deformed to a new state and the configuration associated with this state is termed the deformed configuration (represented by a dash line). Clearly, angles between any two members before and after movement are not necessarily the same, e.g. 0 ≠  and 0 ≠ .

Figure 2.5: Example of truss structures Copyright © 2011 J. Rungamornrat


0

61


 0 

Figure 2.6: Schematic of undeformed and deformed configurations of a truss

2.4.2 Sign and convention For support reactions of a given truss, there is no specific notation for their label. Typically, they are named based on the label of the joint where the support is located and their direction with respective to a reference coordinate system. For example (see Figure 2.7), support reactions induced at a pinned support located at a point A can be labeled RAX and RAY where the first one denotes the reaction at the point A in the X-direction and the last one denotes the reaction at the point A in the Y-direction, and, similarly, the support reaction induced at a roller support located at a point B can be labeled RBY. Note that the upper case R is used only to distinguish the reaction from the other two static quantities, applied loads and internal forces. The sign convention of the support reaction is defined based on the reference coordinate system; specifically, it is positive if it directs along the positive coordinate direction otherwise it is negative. In the analysis for support reactions, it is typical to assume a priori that all components of support reactions are positive or direct along the coordinate directions (as shown in Figure 2.7) in order to prevent any confusion and error. The actual direction of all support reactions can be known once the analysis is complete. If the value of the support reaction obtained is positive, the assumed direction of such reaction is correct, and if its value is negative, the assumed direction of such reaction is wrong and must be reversed.

Y

RAX

A

B

RAY

RBY

X

Figure 2.7: Labels of support reactions and their sign convention For the internal force or member force of trusses, it is standard to follow notations and sign convention as described below:  A member joining joint i and joint j is called “member ij” or equivalently “member ji”. Copyright © 2011 J. Rungamornrat


62


 The internal force (or the axial force or the member force) of a member ij is denoted by Fij. The internal force Fij is positive if and only if the member ij is in tension and negative if and only if the member ij is in compression. Figure 2.8 shows the schematic of the internal force at both ends of the member ij and at the joints i and j for a member in tension and a member in compression. It is worth pointing out that for a member in tension, the internal force directs outward from the joint in the FBD of that joint and directs outward from the member end in the FBD of that member and this observation is reverse if the member is in compression. i

Fij Fij

Fij Fij

j

Member in tension i

Fij Fij

Fij Fij

j

Member in compression Figure 2.8: Schematic of the internal force in FBD of joints and member

2.4.3 Determination of support reactions In the analysis of statically determinate trusses, all support reactions have been generally computed before the process in determining the internal or member forces starts. The procedure to obtain these quantities for truss structure follows exactly that given in the section 2.3. For some trusses such as those shown in Figure 2.5, the number of support reactions is equal to 3 and, hence, the consideration of equilibrium of the entire structure provides a sufficient number of equations to solve for those unknowns. For instance, the truss shown in Figure 2.7 has three unknown support reactions {RAX, RAY, RBY} and they can readily be computed as follow:  the reaction RBY is obtained from equilibrium of moment about point A of the entire structure;  the reaction RAY is obtained from equilibrium of forces in Y-direction of the entire structure; and  the reaction RAX is obtained from equilibrium of forces in X-direction of the entire structure. Y C

X RAX

A D

RCX

B

RBY

RAY

Figure 2.9: Schematic of a truss that a number of support reactions is more than three Copyright © 2011 J. Rungamornrat


63


For some trusses, the number of support reactions may exceed three, while they are still statically determinate (see for example a truss shown in Figure 2.9). In this case, it requires, in addition, the consideration of equilibrium of certain parts of the structure that results from fictitious cuts at proper locations such as a joint connecting between two sub-trusses (see for example a joint D of a truss shown in Figure 2.9). The need for introducing cuts is to supplement extra equations sufficient for, when combined with a set of equilibrium equations for the entire structure, solving all unknown reactions. To clearly demonstrate the above argument, consider, for example, a truss shown in Figure 2.9. This structure is obviously statically determinate (i.e., ra = 4, nm = 32(1) = 32, nj = 18(2) = 36, nc = 0 → DI = 4 + 32 – 36 – 0 = 0) and this therefore ensures that all support reactions can be obtained from static equilibrium. However, it is evident that all four support reactions {RAX, RAY, RBY, RCX} cannot be obtained by considering only equilibrium of the entire structure (it yields only three independent equations). To overcome such degeneracy, we can introduce a cut at the point D and then separate the truss into two parts as shown in Figure 2.10. While we introduce two extra unknowns {FX, FY} at the cut, the total number of unknowns (4 + 2 = 6) is now equal to the number of equilibrium equations that can be set up for the two parts (3 + 3 = 6). To obtain all support reactions without solving a system of six linear equations, equilibrium of both the entire structure and its parts may be considered together as follow:  the reaction RAY is obtained from equilibrium of moments about the point D of the left part;  the reaction RBY is obtained from equilibrium of forces in the Y-direction of the entire structure;  the reaction RCX is obtained from equilibrium of moments about the point D of the right part; and  the reaction RAX is obtained from equilibrium of forces in the X-direction of the entire structure.

Y C

X FY

A RAX

FX

D RAY

RCX

B

D FY

RBY

Figure 2.10: Schematic of two parts of the truss resulting from a cut at point D

2.4.4 Method of joints For a truss consisting of ra support reactions, n joints and m member, the total number of unknowns is ra + m and the total number of independent equilibrium equations that can be set up for all joints is 2n (two independent equilibrium equations can be set up for each joint). Since the degree of static indeterminacy DI = (ra + m) – 2n = 0 for statically determinate trusses, the number of equilibrium equations at all joints is therefore sufficient for determining all unknown member forces and also support reactions. Copyright © 2011 J. Rungamornrat


64


The above idea constitutes a basis for the development of a well-known technique, called the method of joints, for determining all member forces of statically determinate trusses. In this technique, each joint of the truss is first isolated from the structure and its corresponding free body diagram is then sketched (see Figure 2.11 for examples of FBDs for isolated joints). P3 P2

P4

5

P6

P5 7

6

8

P5 7

F67

2 3

4

F78

P7 F27

1

P6 8

P1

F23

F37

3

F38

F48

F38

F34

P1

4

(b)

(a)

P7

R4Y

Figure 2.11: (a) Schematic of a 2D truss and (b) FBDs of joints 3, 4, 7 and 8 Since the internal force of any truss member consists of only the axial force, all forces acting to each joint (member forces and applied loads) constitute a system of concurrent forces. For twodimensional truss, two independent equilibrium equations can be set up for each joint, one corresponding to equilibrium of forces in the X-direction (FX = 0) and the other corresponding to equilibrium of forces in the Y-direction (FY = 0). Once equilibrium equations of all joints are set up, such a system of linear equation can, in principle, be solved to obtain all member forces and support reactions. To reduce computational effort especially when manual calculation is performed, we typically choose to avoid solving a large system of linear equations. Following guidelines can be useful when applied along with the method of joints.  To prevent confusion and accidental errors, the member force is assumed a priori to be in tension in the sketch of joint FBD.  Support reactions should be computed first by using a procedure stated in the section 2.4.3 in order to reduce the number of unknowns. However, this is not a must since a set of equations constructed at all joint is sufficient for determining both member forces and support reactions.  Joint that consists of only two unknowns should be considered first since such unknowns can be solved by considering only equilibrium of that joint. Consider for example a truss shown in Figure 2.11. Once the support reactions {R1X, R1Y, R4Y} are computed, either joint 4 or joint 5 can be considered first since they contain only two unknowns {F15, F56} and {F34, F48}, respectively. Let’s say that we start with the joint 4. Once the member forces {F34, F48} are obtained, joint 8 now becomes a good candidate for the next step since it contains only two unknowns {F38, F78}. This process can be repeated until there is no joint containing two unknowns. Determination of all support reactions before application of the method of joints, while is not necessary, increases the possibility to find joints that contain only two unknowns.  If there exists a joint such that (1) it contains only two members, (2) the angle between those two members is not equal to 180 degrees, and (3) there is no applied load in both Copyright © 2011 J. Rungamornrat


65


directions, then the member forces of the two members vanish. This argument can readily be proved by considering equilibrium of that joint and the fact that a pair of nonparallel forces can be in equilibrium if and only if they vanish identically. For instance, a joint F in a truss shown in Figure 2.12 satisfies above conditions thus rendering the member forces FAF and FFG vanish (see also the FBD of the joint F in Figure 2.13 for clarity). P3

P1 F

H

G

A

C

B

J

I

P4

E

D

P2 Figure 2.12: Schematic of truss containing members of zero member forces F

FAF = 0 FBG = 0

FAB

B

FBC FCD

J

FIJ

FFG = 0

P4

FEJ = 0 FDI = 0

D

FGH

FDE

H

FHI

FCH = 0

Figure 2.13: FBDs of joints B, D, F, H and J of truss shown in Figure 2.12 



If there exists a joint such that (1) it contains only two members, (2) the angle between those two members is not equal to 180 degrees, and (3) there is only one applied load or one component of the support reaction in the direction parallel to one member, then the member force of the other member vanishes. This argument can readily be proved by considering equilibrium of forces of that joint in the direction perpendicular to the applied load. For instance, a joint J in the truss shown in Figure 2.12 satisfies above conditions and this gives FEJ = 0 (see also the FBD of the joint J in Figure 2.13 for clarity). If there exists a joint such that (1) it contains only three members, (2) two of the three members are parallel, and (3) there is no applied load at that joint, then the member force of the third member that are not parallel to the other two vanishes. Again, this argument can be verified by considering equilibrium of forces of that joint in the direction perpendicular to the two parallel members. For instance, joints B, D, and H in the truss shown in Figure 2.12 satisfies above conditions and this yields FBG = FDI = FCH = 0 (see also the FBDs of the joint B, D, and H in Figure 2.13 for clarity). Copyright © 2011 J. Rungamornrat


66




For a joint that all forces acting to that joint can be represented by a set of three forces, equilibrium of such joint implies that vectors of the three forces must form sides of a triangle. This feature allows magnitude of two of the three forces be obtained by the law of sine provided that magnitude of one force is known. From equilibrium of the joint, it is implied in addition that the direction of each vector (as indicated by an arrow) must ensure a zero sum of the three vectors (i.e. the three arrows form a closed loop triangle). This latter property is sufficient for identifying direction of two forces provided that the direction of the other force is known. 2P

P 4

5 60o

o

30

30

60o

R1X = 2P

6

o

60o

1

2P

3

2 R1Y = P

R3Y = 4P F14

2P 4

F45

6

F56

2P

1

F12 P

F14

F14

F24

30o

60o

F45

–F24

F36

-F36

30o

2P

60o

F56

F14 F12

30o

P

2P – F12

o

60

2P

Figure 2.14: Schematic of a 2D truss and FBDs of joints 1, 4 and 6 For instance, consider a truss shown in Figure 2.14. A joint 6 contains only three forces (i.e. one applied load and two member forces) as shown in the FBD of this joint. From the law of sine, we obtain  F36 F56 2P = = o o sin90 sin30 sin60o

From above equation, the two unknowns {F36, F56} can readily be solved since the applied load is known. In addition, the members 36 and 56 must be in compression and tension, respectively, for the joint to be in equilibrium with the applied load 2P (see a triangle representing those three forces in Figure 2.14). Next, consider a joint 1 containing a pinned supports. While there are four forces acting at this joint (i.e. two Copyright © 2011 J. Rungamornrat


67


reactions and two member forces), the support reaction 2P and the member force F12 are parallel and can be combined into one force as shown in Figure 2.15. Since reactions are known, the two unknown member forces {F12, F14} can be obtained from the law of sine: F14 2P  F12 P = = o o sin90 sin30 sin60o

From the diagram of forces, it is evident that both members 12 and 14 are in tension. Finally, let us consider a joint 4 that contains three member forces {F14, F24, F45} as shown by its FBD in Figure 2.14. Since the magnitude and direction of the member force F14 are already known from the joint 1, the members 24 and 45 must be in compression and in tension, respectively, and the magnitude of {F24, F45} can be computed from the law of sine: F45 F14  F24 = = o o sin60 sin30 sin90o



Note that the negative sign appearing in –F36, –F12, and –F24 is due to that the length of each side of the triangle must be non-negative and the member force is always positive in tension. Note also that the consideration of equilibrium of joints containing only three forces by representing them by sides of the triangle and then applying the law of sine provides an attractive alternative to that by solving the following two equilibrium equations of forces in X- and Y-directions, i.e. FX = 0 and FY = 0. If all support reactions are determined before the method of joints is applied, there will be ra equations exceeding the number of member forces. These equations can be used as a final check of support reactions and member forces already computed. Specially, if all support reactions and member forces are computed correctly, these ra equations must be satisfied automatically; on the contrary, if some equilibrium equations are not satisfied, either support reactions or member forces are wrong.

As a final remark, the method of joints is a good candidate if all member forces are to be determined but, if only certain member forces are of interest, the method leads to a significant amount of effort associated with determination of other member forces. For instance, if only the member force F37 of the truss shown in Figure 2.11(a) is of interest, the joint 4 and the joint 8 must be considered first before the joint 3 can be treated. Another drawback of the method of joints becomes evident when the technique is applied to statically determinate trusses with all of their joints containing at least three members (e.g. trusses shown in Figure 2.15). For these particular structures, the method of joints leads to a large system of linear equations to be solved.

Figure 2.15: Example of determinate trusses with all joints containing at least three members Copyright © 2011 J. Rungamornrat


68


Example 2.4 Determine all support reactions and then compute all member forces for a truss shown below by the method of joints 2P

3P 2

4

P 8

6

Y L 45o

45o 5

3

1 L

X

45o

L

7 L

Solution The structure given above is statically determinate (i.e. ra = 2 + 1 = 3, m = 13, n = 8, nc = 0, then DI = 3 + 13(1) – 8(2) – 0 = 0); thus all support reactions and member forces can be determined from static equilibrium. Since the number of support reactions is equal to 3, they can be computed by the consideration of equilibrium of the entire structure as shown below (the FBD of the entire structure is also shown below).

R2Y 2

R2X

2P

3P 4

6

P 8 Y X

R1X

[FY = 0]  

: : :

Upward

(R1X)(L) – (3P)(L) – (2P)(2L) – (P)(3L) = 0 R1X = 10P

[FX = 0]  

7

R2Y – 3P – 2P – P = 0 R2Y = 6P

[M2 = 0]  

5

3

1

Rightward

R1X + R2X = 0 R2X = –10P

Leftward

The member forces are then determined by the method of joints. Since the joint 7 contains only two non-parallel members and there is no applied load, it can be deduced that F57 = F78 = 0. Based on the known member forces {F57, F78} and the known support reactions {R1X, R2X, R2Y}, only joints 2 and 8 that contain only two unknowns. Let us start with the joint 8. The member forces {F58, F68} can be obtained as follow: Copyright © 2011 J. Rungamornrat


[FY = 0]  

:

69


F58 = –√2 P [FX = 0]  

:

P

–F58cos45o – F78 – P = 0 (Compression)

F68

8

o

–F58sin45 – F68 = 0 F68 = P

45o

F58

(Tension)

F78 = 0

Since member force F58 and F68 are negative and positive, respectively, the members 58 and 68 are therefore in compression and in tension, respectively. Note that since the member force F78 = 0, the joint 8 contains only three non-zero forces and this allows the law of sine be alternatively used to determine the two unknown member forces {F58, F68} as follow: F68  F58 P P = =  F58  sin90o   2P sin90o sin45o sin45o sin45o P  F68  sin45o P sin45o

(Compression)

45o

-F58

P

45o

(Tension)

F68

Once the member force F58 is determined, the joint 5 now contains only two unknown {F35, F56}. Since the member force F57 = 0, the joint 5 contains only three non-zero forces and they are shown in the diagram below. The unknowns {F35, F56} are obtained as follow:  F35 F56  F58 2P = =  F35  sin45o  P o o o sin45 sin45 sin90 sin90o 2P  F56  sin45o P sin90o

(Compression)

45o

-F58

(Tension)

F56

45o

-F35

Once the member forces F56 and F68 are determined, the next joint to be considered is the joint 6 since it contains only two unknowns {F36, F46}. By considering a FBD of the joint 6 and then setting two equilibrium equations, we obtain 2P o [FY = 0]   : –F36cos45 – F56 – 2P = 0 6 F68 F46 F36 = –3√2 P (Compression) [FX = 0]  

:

–F36sin45o – F46 + F68 = 0 F46 = 4P

F36

(Tension)

45o F56

Once the member forces F35 and F36 are determined, the next joint to be considered is the joint 3 since it contains only two unknowns {F36, F46}. By considering a FBD of the joint 3 and then setting two equilibrium equations, we obtain [FX = 0]  

:

F36cos45o + F35 – F13 = 0 F13 = –4P

[FY = 0]  

:

F36

(Compression)

F36sin45o + F34 = 0 F34 = 3P

F34

(Tension)


F13

45o 3

F35


70


Once the member forces F34 and F46 are determined, the next joint to be considered is the joint 4 since it contains only two unknowns {F14, F24}. By considering a FBD of the joint 4 and then setting two equilibrium equations, we obtain 3P o [FY = 0]   : –F14cos45 – F34 – 3P = 0 4 F24 F46 F = –6√2 P (Compression) 14

[FX = 0]  

:

–F14sin45o – F24 + F46 = 0 F24 = 10P

(Tension)

F14

45o F34

Once the member forces F13 and F14 are determined, the next joint to be considered is the joint 1 since it contains only one unknown {F12}. By considering a FBD of the joint 1 and then setting two equilibrium equations, we obtain [FY = 0]  

:

F12

F14sin45o + F12 = 0

F14

F12 = 6P (Tension) [FX = 0]  

:

F14cos45o + F13 + R1X = 0 ?

R1X

(–6√2P)(1/√2) – 4P+ 10P = 0 OK

45o 1

F13

It is evident that the second equation is not needed in the calculation of member forces and, in addition, there are still two equations left at joint 2. This is not a surprise since three equilibrium equations associated with the entire structure were already employed in the calculation of support reactions. As a result, the three equilibrium equations (one at joint 1 and the other two at joint 2) constitute no new independent equation but, in fact, they are linear combinations of other equilibrium equations established above. Although they are not required in the calculation of member forces, such extra equations are still useful as a part of verification of calculated results; if all member forces are computed accurately, they must automatically satisfy those equations. For instance, two equilibrium equations at joint 2 must also be satisfied as follow: [FY = 0]  

[FX = 0]  

R2Y

:

R2y – F12 = 0

?

:

6P – 6P = 0

OK

:

R2x + F24 = 0 ?

:

–10P + 10P = 0

OK

R2X

2

F24

F12

2.4.5 Method of sections As already pointed out in the previous section, the method of joints becomes inefficient if the internal forces of certain members are of interest. To further enhance the efficiency of computation of member forces of statically determinate trusses, another technique called a method of sections is introduced. This method simply employs static equilibrium conditions along with the method of structure partitioning. To outline and clearly demonstrate the method of sections, let us consider a truss shown in Figure 2.16(a). Assume that the member forces {F23, F27, F67} are of interest. As a first step, all support reactions {R1X, R1Y, R4Y} are determined via the consideration of equilibrium of the entire truss. To access and see the member forces {F23, F27, F67}, we need to introduce a fictitious cut or a Copyright © 2011 J. Rungamornrat


71


section passing through the members 23, 27 and 67 as shown by a dash line in Figure 2.16(a). This cut not only exposes the member forces {F23, F27, F67} but also divides the structure into two parts with the FBD of each part shown in Figure 2.16(b) and 2.16(c). As is evident, either the FBD of the right part or the FBD of the left part contains exactly three unknowns {F23, F27, F67}. Thus, the consideration of equilibrium of either one of those two parts yields a sufficient number of equations for solving the three unknown member forces. For instance, enforcing equilibrium of moments about joint 7 of the right part yields the member force F23; enforcing equilibrium of moments about joint 2 of the right part yields the member force F67; and enforcing equilibrium of forces in the Ydirection of the right part yields the member force F27. It is worth noting that other equivalent sets of three equilibrium equations can also be used and that the left part of the truss can also be employed to determine {F23, F27, F67}. Note, however, that the consideration of equilibrium of those two parts, while given totally six equations, still yields only three independent equations since the other three equations are equivalent to those employed for computing the support reactions {R1X, R1Y, R4Y}.

6

5

8

7

4

1

2

3 (a)

6

5

R1X

1 2 R1Y

F27

F67

F23

(b)

F67

F27 F23

8

7

4 3 (c)

R4Y

Figure 2.16: (a) Schematic of a truss and location of fictitious cut (b) & (c) FBDs of parts resulting from partitioning If the member force F37 is of interest, we may introduce a fictitious cut as shown in Figure 2.17(a). This cut passes through the member 37 and divides the structure into two parts with the corresponding FBDs shown in Figure 2.17(b) and (c). It is obvious that both FBDs contain exactly three unknown member forces {F23, F37, F78} and they can in principle be determined from equilibrium equations set up for one of these two parts. To determine the member force F37, we simply enforce equilibrium of forces in Y-direction of the left part. If the member forces {F23, F78}



72


are also of interest, they can be obtained by enforcing equilibrium of moments about joint 7 and joint 3, respectively.

6

5

8

7

4

1

3

2 (a)

5

R1X

6

F78

7

F78 F37

8

F37

1 2

F23

F23

4 3

R1Y

R4Y (b)

(c)

Figure 2.17: (a) Schematic of a truss and location of fictitious cut and (b) & (c) FBDs of parts resulting from partitioning To apply the method of sections in an efficient manner, following remarks may be taken into account to reduce computational effort:     

Since the internal force is constant throughout the member, the location through which the section passes does not affect the results. This therefore provides flexibility for choosing a path for sectioning. To prevent confusion and errors, it is generally assumed a positive sign convention for all unknown member forces (i.e. members are assumed in tension). If the negative member force is obtained, the member is therefore in compression. All support reactions should be determined before the method of sections is applied in order to reduce the number of unknowns appearing in any parts resulting from the sectioning. If a section introduces only three unknown member forces along the cut, such unknown forces can be determined from equilibrium of either one of the two parts provided that all forces are not concurrent forces, e.g. sections shown in Figure 2.16 and Figure 2.17. If a section and a reference point used for taking the moment are chosen appropriately, the member force can readily be obtained from equilibrium of moments. For instance, if the member force F12 of trusses shown in Figure 2.18 is to be computed, the section and reference point A may be chosen as shown below. Copyright © 2011 J. Rungamornrat


73


2 F12 1

A

2 A 1

F12

A

1

F12

2

Figure 2.18: Schematic of trusses, sections, FBD of parts of truss, and reference point A The method of sections has been found more efficient than the method of joints when the internal forces are to be determined for certain members. However, for certain trusses, both methods may be used together to increase the efficiency. For instance, in the analysis of the last two trusses shown in Figure 2.18 (every joint of these truss contains at least three members), the method of sections is applied first to determine some member forces. The method of joints can subsequently be used to compute all remaining member forces in a simple fashion since it is always possible to find joints containing only two unknowns. Example 2.5 Determine all support reactions and then use the method of sections to compute the member forces F23, F25, F35, and F56 of a truss shown below 4L 1

P

3L 2

5

3L 3

6

P

3L 4

7

3P 3P


Y X


74


Solution The structure given above is statically determinate (i.e. ra = 2 + 1 = 3, m = 11, n = 7, nc = 0, then DI = 3 + 11(1) – 7(2) – 0 = 0); thus all support reactions and member forces can be determined from static equilibrium. Since the number of support reactions is equal to 3, they can be computed by considering equilibrium of the entire structure as shown below. P

1

P

1

S1

1 F15 F15

2

5

R3X

2

F25

5

F25 5

F23

R3X

F23 3

R4X

6

4 7

3P

X

:

4

: :

7

3P 3P

R4Y – 3P = 0 Upward

–(R3X)(6L) – (P)(9L) – (P)(3L) – (3P)(4L) = 0 R3X = –4P

[FX = 0]  

R4X

P

6

R4Y

R4Y = 3P [M4 = 0]  

3

Y

3P

R4Y [FY = 0]  

P

Leftward

R4X + R3X + P + P + 3P = 0 R4X = –P

Leftward

Next, by introducing a fictitious cut S1 and then considering equilibrium of the top part of the truss, the member forces F23 and F25 can be obtained as follow: [M5 = 0]  

:

(F23)(4L) – (P)(3L) = 0 F23 = 3P/4

[M1 = 0]  

:

(Tension)

(F25)(3L) = 0 F25 = 0

Alternatively, the member forces F23 and F25 can also be obtained by considering equilibrium of the bottom part of the truss as shown below: [M5 = 0]  

:

P(3L) + (3P + R4X)(6L) – (F23 + R4Y)(4L) = 0 F23 = 3P/4

[M1 = 0]  

:

(Tension)

(R3X –F25)(3L) + (3P + R4X)(9L) + P(6L) – (3P)(4L) = 0 F25 = 0 Copyright © 2011 J. Rungamornrat


1

75


1

P

2

5

2

R3X

R4X

6

4 7

5

F23

S2 3

P

P

Y

3P

R3X

F56 F35

F35 3

F56

3

R4X

X

3P

R4Y

F23

6

4 7 R4Y

P

3P

3P

Next, by introducing another fictitious cut S2 and again considering equilibrium of the top part of the truss, the member forces F35 and F56 can then be obtained as follow: [M3 = 0]  

:

–(F56)(4L) – (R3X)(3L) – (P)(6L) = 0 F56 = 3P/2

[FX = 0]  

:

(Tension)

–(F35)(4/5) + P + R3X = 0 F35 = –15P/4 (Compression)

Similar to the previous case, the member forces F35 and F56 can equivalently be obtained by considering equilibrium of the bottom part of the truss as shown below: [M3 = 0]  

:

(F56 – 3P)(4L) + (3P + R4X)(3L) = 0 F56 = 3P/2

[FX = 0]  

:

(Tension)

(F35)(4/5) + P + 3P + R4X = 0 F35 = –15P/4 (Compression)

Example 2.6 Determine all support reactions and all member forces for a truss shown below a1 Y

a8

a3

a2 a6

a9

a10

a4

a5

a7

a11

2@3L

a12

2P 6@4L Copyright © 2011 J. Rungamornrat

a13

a14

X


76


Solution The structure given above is statically determinate (i.e. ra = 2 + 1 = 3, m = 25, n = 14, nc = 0, then DI = 3 + 25(1) – 14(2) – 0 = 0); thus, all support reactions and member forces can be determined from static equilibrium. Since the number of support reactions is equal to 3, they can be computed by considering equilibrium of the entire structure as shown below. S1 a1

Y

a2

a3

a5

a12

a13

a7

a6 Ra8X

a4

a8 a9

a11

a10

2P

Ra8Y

X

a14 Ra14Y

[FX = 0]  

:

Ra8X = 0

[Ma8 = 0]  

:

(Ra14Y)(24L) – (2P)(12L) = 0 Ra14Y = P

[FY = 0]  

Ra14Y + Ra8Y – 2P = 0 Ra8Y = P

From geometry and loading of a given truss, it can readily be verified that the following member forces vanish: Fa2a6 = Fa3a11 = Fa4a7 = Fa6a10 = Fa6a9 = Fa1a9 = Fa7a12 = Fa7a13 = Fa5a13 = 0 From symmetry of geometry and loading, it can be deduced that Fa1a8 = Fa5a14

; Fa8a9 = Fa13a14 ; Fa1a2 = Fa4a5

Fa9a10 = Fa12a13 ; Fa2a3 = Fa3a4

; Fa1a6 = Fa5a7

; Fa6a11 = Fa7a11 ; Fa10a11 = Fa11a12

Next, by applying the method of joints to joints a2, a9, a6 and a10, it leads to Joint a2:

[FX = 0]



Fa1a2 = Fa2a3

Joint a9:

[FX = 0]



Fa8a9 = Fa9a10

Joint a6:

[FX = 0]



Fa1a6 = Fa6a11

Joint a10:

[FX = 0]



Fa9a10 = Fa10a11

Now, it still remains to determine the member forces Fa2a3, Fa6a11, Fa10a11 and Fa1a8. To compute the member forces Fa2a3, Fa6a11 and Fa10a11, we apply the method of sections along with introducing a fictitious cut S1 that passes through the members a2a3, a6a11 and a10a11 as indicated in the above figure and detail calculations are given below. Copyright © 2011 J. Rungamornrat


77


a1

a2

Y

Fa2a3 Fa1a8

a6 X Ra8X = 0

Fa6a11

a8

a10 Fa10a11 a11

a9

Ra8X = 0

Ra8Y = P [Ma1 = 0]  

:

a8



Fa8a9

Ra8Y = P

(Fa10a11)(6L) – (P)(4L) = 0 Fa10a11 = 2P/3 (Tension)

[Ma11 = 0]  

:

–(Fa2a3)(6L) – (P)(12L) = 0 Fa2a3 = –2P

[FY = 0]  

:

(Compression)

–(Fa6a11)(3/5) + P = 0 Fa6a11 = 5P/3 (Tension)

Finally, the member force Fa1a8 is obtained by applying the method of joints to joint a8 as shown below (sin = 3/√13 and cos = 2/√13). [FY = 0]  

:

Fa1a8 sin + P = 0 Fa1a8 = –√13P/3

[FX = 0]  

:

(Compression)

Fa1a8 cos + Fa8a9 = 0 ? (–√13P/3)(2/√13) + 2P/3 = 0 OK

Note that the second equation is just an extra equation that can be used to partially verify computed results. All member forces are summarized below: Fa1a8 = Fa5a14 = –√13P/3 (Compression) Fa8a9 = Fa13a14 = Fa9a10 = Fa10a11 = Fa11a12 = Fa12a13 = 2P/3 (Tension) Fa1a2 = Fa2a3

= Fa3a4 = Fa4a5

= –2P (Compression)

Fa1a6 = Fa6a11 = Fa5a7 = Fa7a11 = 5P/3 (Tension) Fa1a9 = Fa2a6

= Fa6a9 = Fa6a10 = Fa3a11 = Fa4a7 = Fa7a12 = Fa5a13 = Fa7a13 = 0

As a final remark, the method of joints and method of sections described above is applied under the assumption that the displacement at all joints is infinitesimal allowing equilibrium to be enforced in the undeformed state (whose geometry is considered known a priori). For a truss undergoing large displacement, equilibrium must be checked based on the (unknown) geometry in the deformed state and, for this particular case, only equilibrium equations are not sufficient to determine support reactions and internal forces (see Rungamornrat et al (2008) for the analysis of truss with consideration of geometric nonlinearity). Copyright © 2011 J. Rungamornrat


78


2.5 Static Analysis of Beams This section devotes to the static analysis of another type of structures termed beams. The primary objective is to present basic techniques commonly used in the determination of support reactions and the internal forces at any location, i.e. the shear force diagram (SFD) and the bending moment diagram (BMD). In addition, sketch of a qualitative elastic or deformed shape of beam under external applied loads is also discussed. The section starts with a brief introduction on characteristics of beams and standard notations and sign convention commonly used for beam. Next, a basic technique based upon the method of sections to determine the internal forces at a particular location of interest is presented. A more general technique based on the differential and integral formula is further introduced to construct the SFD and BMD. Finally, guidelines useful for sketching the elastic curve are summarized. Various examples are also presented to demonstrate the principle and details of each technique.

2.5.1 Characteristics of beams An idealized structure is called a beam if and only if (i) all members are straight and form a straight-line-configuration structure, (ii) all members are generally connected by beam joints (full moment release is allowed for certain joints), and (iii) all applied loads must form a system of transverse loads. Examples of beams are shown schematically in Figure 2.19. Note that there is no restriction on the type of supports present in beams, i.e. roller supports, pinned supports, guide supports and fixed supports are allowed. Note, however, that the component of support reactions in the direction of the beam axis, if exists, can be ignored since all applied loads are transverse loads. From above definition, it can readily be verified that the internal forces at any location of the beam can be represented by two components called the shear force and the bending moment. The shear force is the resultant force in the direction perpendicular to the axis of the beam and the bending moment is the resultant moment in the direction perpendicular to the plane of transverse loads, see Figure 2.20. Unlike the truss member, the shear force and bending moment are in general not constant throughout the member. Another different feature is that the angle between the two members connecting to any beam joint is preserved before and after undergoing deformation provided that such a joint contains no moment release.

Figure 2.19: Schematic of some statically determinate beams Copyright © 2011 J. Rungamornrat


79


Shear force

Bending moment

Figure 2.20: Schematic indicating two components of the internal force in beam

2.5.2 Sign and convention A reference Cartesian coordinate system commonly used for beam, denoted by {O; X, Y, Z} where O is the origin and {X, Y, Z} represents three mutually perpendicular axes following a standard right handed rule, is defined such that the X-axis directs along the axis of the beam and the X-Y plane is a plane of transverse loads. Note that there is no restriction on the location of the origin O. An example of the reference coordinate system for a beam is shown in Figure 2.21. The sign convention and notations for support reactions of beams can be defined in a similar fashion to those for support reactions of trusses. For instance, reactions at the fixed support of the beam shown in Figure 2.21 are denoted by RAY and RAM where the former stands for a force reaction in the Y-direction and the latter stands for a moment reaction in the Z-direction, and a force reaction in the Y-direction of the roller support located at a point B is denoted by RBY. Since all support reactions are unknown a priori, it is common in the analysis to assume that they possess a positive direction, i.e. they direct along the positive coordinate axes (see examples of positive sign convention for reactions in Figure 2.21). Once results are obtained, the actual direction of each reaction can be decided from their sign; specifically, if the computed reaction is positive, the assumed direction is correct but, if the computed reaction is negative, the actual direction is opposite to the assumed direction. Y RAM

A

B

RAY

X

RBY

Figure 2.21: Schematic showing a reference coordinate system and sign convention and notations of support reactions of a beam For the shear force and bending moment, it is standard to follow notations and sign convention given below.  The shear force at a particular point A is denoted by a symbol VA and the shear force as a function of position x along the beam is denoted by V(x). The shear force at any point is Copyright © 2011 J. Rungamornrat


80


considered to be positive if and only if it tends to produce a clockwise rotation of the infinitesimal beam element in the neighborhood of that point; otherwise it is negative. The positive and negative shear forces are shown in Figure 2.22. A traditional strategy commonly used to memorize the sign convention for the shear force is that “the shear force at any point is considered positive if and only if, when a cut is made at that point, the shear force directs upward in the FBD of the right part of a beam and directs downward in the FBD of the left part of a beam”.

Positive shear force

Negative shear force

Figure 2.22: Schematic indicating positive and negative sign convention for shear force  The bending moment at a particular point A is denoted by a symbol MA and the bending moment as a function of position x along the beam is denoted by M(x). The bending moment at any point is considered to be positive if and only if it produces a compressive stress at the top and produces a tensile stress at the bottom; otherwise it is negative. The positive and negative bending moments are shown in Figure 2.23. A traditional strategy commonly used to memorize the sign convention for the bending moment is that “the positive bending moment produces a concave upward curve or a smile shape while the negative moment produces a concave downward curve or a sad shape”.

Top fiber

Top fiber

Bottom fiber

Bottom fiber

Positive bending moment

Negative bending moment

Figure 2.23: Schematic indicating positive and negative sign convention for bending moment

2.5.3 Determination of support reactions A standard procedure for determining support reactions of statically determinate beams follows exactly that given in the section 2.3. For a beam containing only two components of support reactions, consideration of equilibrium of the entire beam is sufficient for solving all unknown reactions. For instance, support reactions {RAY, RBY} of a simply-supported beam shown in Figure 2.24 can be obtained by solving two equilibrium equations set up on the entire beam as follows:  the reaction RAY is obtained from equilibrium of moments of the entire structure about a point A, and  the reaction RBY is obtained either from equilibrium of forces in Y-direction of the entire structure or from equilibrium of moments of the entire structure about a point B. Copyright © 2011 J. Rungamornrat


81


B

A

Y B

A

X

RBY

RAY

Figure 2.24: Schematic of a simply-supported beam and its FBD For various beams, they may contain more than two components of support reactions while still statically determinate, for instance, the third beam shown in Figure 2.19, the beam shown in Figure 2.21 and the beam shown in Figure 2.25. For this particular case, only the consideration of equilibrium of the entire beam is insufficient to determine all support reactions. Additional conditions related to the presence of internal releases within the beam must be employed to supply adequate number of equations. These extra equations can equivalently be viewed as equilibrium equations written for some parts of the beam that resulting from suitable cuts (e.g. cuts passing through the location of the internal releases). Y RAM

C A RAY

B

D RCY

E

X

REY

Figure 2.25: Schematic of a two-span beam containing four support reactions To clearly demonstrate how to determine support reactions for this particular case, consider, for example, the beam shown in Figure 2.25. This structure is obviously statically determinate (i.e., ra = 4, nm = 2(2) = 4, nj = 3(2) = 6, nc = 2 → DI = 4 + 4 – 6 – 2 = 0) and this therefore ensures that all support reactions can be obtained only from static equilibrium. Since there are four unknowns reactions {RAM, RAY, RCY, REY}, we still need to construct two additional equations, when used together with those two constructed on the entire beam, to render a sufficient number of equations. To achieve this task, let us first introduce a cut at a hinge D and consider the right part of the beam (see FBD in Figure 2.26(a)) and next introduce a cut at a point just to the right of a hinge B and consider the right part of the beam (see FBD in Figure 2.27(b)). Note that we intend not to make a cut exactly at the hinge B since we want to avoid an argument related to how to distribute a concentrated load acting at the hinge B to the left and right parts of the beam. The unknown reactions {RAM, RAY, RCY, REY} can then be computed as follow:  the reaction REY is obtained from equilibrium of moments about a point D of the right part of the beam shown in Figure 2.26(a),  the reaction RCY is obtained from equilibrium of moments about a point BR of the right part of the beam shown in Figure 2.26(b), Copyright © 2011 J. Rungamornrat


82


 the reaction RAM is obtained from equilibrium of moments about a point A of the entire beam shown in Figure 2.26(c), and  the reaction RAY is obtained from equilibrium of forces in Y-direction of the entire structure. C VD

VBR

E

D

BR

E

D RCY

REY

REY

(b)

(a)

Y C RAM

A

B RCY

RAY

X

E

D

REY

(c) Figure 2.26: (a) FBD of right part of beam when a cut is made at hinge D, (b) FBD of right part of the beam when a cut is made at point just to the right of hinge B, and (c) FBD of entire beam Alternatively, let us introduce two cuts simultaneously, one at point just to the right of a hinge B and the other at a hinge D. With these two cuts, we can sketch corresponding FBDs of three parts of the beam as shown in Figure 2.27. While we introduce two extra unknowns {VBR, VD} at the cut, the total number of unknowns (4 + 2 = 6) is now equal to the number of equilibrium equations that can be set up for the three parts (2 + 2 + 2 = 6). To obtain all reactions without solving a system of six linear equations, we can consider equilibrium of each part as follow: Y RAM

VBR

A B VBR

RAY (a)

VD

C D RCY

VD

(b)

E

D

X

REY (c)

Figure 2.27: Free body diagrams of three parts of the beam resulting from two cuts at point just to the right of hinge B and at hinge D  the reaction REY is obtained from equilibrium of moments about a point D of the right part of the beam shown in Figure 2.27(c),  the shear force VD is obtained from equilibrium of forces in Y-direction of the right part of the beam shown in Figure 2.27(c),  the reaction RCY is obtained from equilibrium of moments about a point BR of the middle part of the beam shown in Figure 2.27(b),  the shear force VBR is obtained from equilibrium of forces in Y-direction of the middle part of the beam shown in Figure 2.27(b), Copyright © 2011 J. Rungamornrat


83


 the reaction RAM is obtained from equilibrium of moments about a point A of the left part of the beam shown in Figure 2.27(a), and  the reaction RAY is obtained from equilibrium of forces in Y-direction of the left part of the beam shown in Figure 2.27(a). It is remarked that while the both solution strategies yield identical results, the intermediate unknowns {VBR, VD} must be solved in the second strategy in order to obtain all support reactions.

2.5.4 Method of sections In various situations, the shear force and bending moment at some specific locations are of interest. The method of sections similar to that employed in the analysis of member forces in trusses can efficiently be applied here. The procedure starts with introducing a cut at a location where the shear force and bending moment are to be determined in order to access and see those unknown internal forces and then follows by applying static equilibrium equations to solve for such unknowns. If all support reactions are determined before the method of sections is applied, only two unknowns (one corresponding to the shear force and the other corresponding to the bending moment) are introduced at the cut and appear in the FBD of both parts of the beam resulting from the cut. Consideration of equilibrium of either part provides two independent equilibrium equations and this is sufficient for solving the two unknown internal forces at a particular location. Procedures for determining shear force and bending moment at a particular point P by the method of sections can be summarized as follow (see for example a beam shown in Figure 2.28):    

Determine all support reactions following guideline given in section 2.5.4 Introduce a cut at point P and then separate the beam into two parts Choose one of the two parts that seems to involve less computation Sketch the FBD of a selected part; both shear force and bending moment are assumed a priori to be positive.  Apply equilibrium of forces of the selected part in Y-direction; this yields the shear force at point P (VP). If the computed shear force is positive, the assumed direction is correct; otherwise, the actual direction is opposite to the assumed direction.  Apply equilibrium of moments of the selected part about a point P; this yields the bending moment at point P (MP). If the computed bending moment is positive, the assumed direction is correct; otherwise, the actual direction is opposite to the assumed direction. Y RAM

A

B

P RAY

RAM

X

RBY VP

A P

MP

RAY

MP VP

P

B RBY

Figure 2.28: Schematic indicating a cut used to access the shear force and bending moment at point P and FBDs of the two parts resulting from the cut Copyright © 2011 J. Rungamornrat


84


It is remarked that the method of sections can be used not only to determine shear force and bending moment at a particular point but also to determine the shear force and bending moment at any location of the beam, i.e. V(x) and M(x). Procedures to obtain V(x) and M(x) are similar to those described above except that a cut must be made at any point x instead of a specific location. In addition, the entire beam needs to be separated into several subintervals due to the discontinuity induced by the presence of supports, concentrated applied loads, and locations where distributed loads change their distribution. The function forms of V(x) or M(x) for those subintervals are generally different and need to be constructed separately. Once the shear force and bending moment as a function of position x along the beam are determined, graphs of V(x) and M(x) can be plotted with the x-axis directing along the axis of the beam. These two graphical representations are known as a shear force diagram (SFD) and a bending moment diagram (BMD), respectively. Example 2.7 and Example 2.8 demonstrate applications of the method of sections to determine shear force and bending moment at some specific locations and to construct V(x) and M(x) and sketch the corresponding SFD and BMD, respectively. Example 2.7 Determine shear force and bending moment at points AR, BL, BR, CL, CR, DL, DR of a beam shown below. Subscripts L and R is used to indicate a point just to the left and a point just to the right of the indicated point, e.g. CL, CR are point just to the left and point just to the right of the point C. Y q

Po = qL Mo = qL2

A

B L

C L

D L

E

X

L

Solution The given beam is statically determinate (i.e. ra = 2, nm = 1(2) = 2, nj = 2(2) = 4, nc = 0, then DI = 2 + 2 – 4 – 0 = 0); thus all support reactions and the internal forces at any location can be determined from static equilibrium. Since the number of support reactions is equal to 2, they can be obtained from equilibrium of the entire structure as shown below. Y q

Po = qL RAM

Mo = qL2

A B

C

D

E

RAY [FY = 0]  

:

RAY – qL – qL = 0 RAY = 2qL

[MA = 0]  

:

Upward

RAm – (qL)(L) + qL2 – (qL)(3L+L/2) = 0 RAm = 7qL2/2 CCW Copyright © 2011 J. Rungamornrat

X


85


The shear force and bending moment at point AR are obtained by making a cut at point just to the right of point A and then considering equilibrium of the left part: [FY = 0]  

:

RAY – VAR = 0 VAR = 2qL

[MA = 0]  

:

RAM VAR MAR

RAM + MAR = 0

RAY

MAR = –7qL2/2

Similarly, the shear force and bending moment at point BL are obtained by making a cut at point just to the left of point B and then considering equilibrium of the left part: [FY = 0]  

:

VBL = 2qL [MBL = 0]  

:

L–

RAY – VBL = 0 RAM

RAM – RAYL+ MBL = 0 MBL = –3qL2/2

VBL A

BL

MBL

RAY

Note that the distance between the point A and point BL is denoted by L– and, in the limit as BL approaches the point B, L– = L. Next, the shear force and bending moment at point BR are obtained by making a cut at point just to the right of point B and then considering equilibrium of the left part: [FY = 0]  

:

VBR = qL [MBR = 0]  

:

L+

RAY – VBR – qL = 0 RAM

RAM – RAYL+ MBR = 0 MBR = –3qL2/2

qL VBR A

BR

MBR

RAY

Similarly, in the limit as BR approaches the point B, L+ = L. Next, the shear force and bending moment at point CL are obtained by making a cut at point just to the left of point C and then considering equilibrium of the left part: [FY = 0]  

:

RAY – VCL – qL = 0 VCL = qL

[MCL = 0]  

:

qL

RAM

RAM – 2RAYL + qL2+ MCL = 0 MCL = –qL2/2

L–

L

VCL

B

A

CL

MCL

RAY

The shear force and bending moment at point CR can be obtained by making a cut at point just to the right of point C and then considering equilibrium of the left part: [FY = 0]  

:

VCR = qL [MCR = 0]  

:

RAM – 2RAYL + 2qL2 + MCR = 0 MCR = –3qL2/2


L+

L

RAY – VCR – qL = 0

qL

RAM

qL2 A

RAY

B

VCR MCR CR


86


The shear force and bending moment at point DL can be obtained by making a cut at point just to the left of point C and then considering equilibrium of the right part: [FY = 0]  

:

L+

VDL – qL = 0 VDL = qL

[MDL = 0]  

:

q

VDL

–MDL – (qL)(L/2) = 0 MDL = –qL2/2

MDL

DL

E

Finally, the shear force and bending moment at point DR can be obtained by making a cut at point just to the right of point D and then considering equilibrium of the right part: [FY = 0]  

:

VDR – qL = 0

L–

VDR = qL [MDR = 0]  

:

–MDR – (qL)(L/2) = 0 MDR = –qL2/2

q

VDR MDL

DR

E

From results obtained, it is worth noting that at a point where a concentrated force is applied, there is a jump of the shear force equal to the magnitude of the concentrated force while there is no jump of the bending moment. For instance, at the point B, we have VBR – VBL = –qL MBR – MBL = 0 Next, at a point where a concentrated moment is applied, there is a jump of the bending moment equal to the magnitude of the concentrated moment while there is no jump of the shear force. For instance, at the point C, we have VCR – VCL = 0 MCR – MCL = –qL2 Finally, at a point where the distributed load is discontinuous, there is no jump of both the bending moment and the shear force. For instance, at the point D, we have VDR – VDL = 0 MDR – MDL = 0 Example 2.8 Determine all support reactions and then use the method of sections to construct the SFD and BMD of a beam shown below. Y 2

2q

Po = 3qL

Mo = qL

A

B L

C L

D L


E L

X


87


Solution The given beam is statically determinate (i.e. ra = 2, nm = 2(2) = 4, nj = 3(2) = 6, nc = 0, then DI = 2 + 4 – 6 – 0 = 0); thus all support reactions and the internal forces at any location can be determined from static equilibrium. Since the number of support reactions is equal to 2, they can be obtained from equilibrium of the entire structure as shown below. Y 2

Mo = qL

A

C

B

: :

X

E

3RDYL + qL2 – (3qL)(2L) – (2qL)(3L+L/2) = 0 RDY = 4qL

[FY = 0]  

D RDY

RAY [MA = 0]  

2q

Po = 3qL

Upward

RAY + RDY – 3qL – 2qL = 0 RAY = qL

Upward

Since there are three points of discontinuity within the beam (i.e. points B, C, and D), the beam is then divided into four subintervals (i.e. subintervals AB, BC, CD and DE) and the shear force V(x) and bending moment M(x) are to be obtained for each subinterval as shown below. Starting with the subinterval AB, a cut is made at any point x  (0, L) and equilibrium of the left part of the beam is considered: [FY = 0]  

:

x

RAY –V(x) = 0

V(x)

V(x) = qL [Mx = 0]  

:

M(x) – (RAY)(x) = 0

A

M(x) = qLx

M(x) RAY

Next, the shear force V(x) and bending moment M(x) within the subinterval BC are obtained by introducing a cut at any point x  (L, 2L) and considering equilibrium of the left part of the beam: [FY = 0]   [Mx = 0]  

: :

RAY –V(x) = 0

x

V(x) = qL

qL2

M(x) – (RAY)(x) + qL2 = 0 M(x) = qLx – qL2

A

B

V(x) M(x)

RAY

Next, the shear force V(x) and bending moment M(x) within the subinterval CD are obtained by introducing a cut at any point x  (2L, 3L) and considering equilibrium of the left part of the beam: [FY = 0]  

:

RAY –V(x) – 3qL = 0 V(x) = –2qL

[Mx = 0]  

:

M(x) – (RAY)(x) + qL2 + (3qL)(x-2L) = 0 M(x) = –2qLx + 5qL2 Copyright © 2011 J. Rungamornrat


88


x 3qL

V(x)

2

qL

A

B

M(x)

C

RAY Finally, the shear force V(x) and bending moment M(x) within the last subinterval DE are obtained by making a cut at any point x  (3L, 4L) and equilibrium of the right part of the beam is considered as follows: [FY = 0]  

:

4L – x

V(x) – 2q(4L– x) = 0 V(x) = 8qL – 2qx

[Mx = 0]  

:

V(x)

–M(x) – (2q)(4L– x)(4L– x)/2 = 0

2q

M(x)

E

2

M(x) = – q(4L– x)

The shear force V(x) and the bending moment M(x) for the entire beam are summarized and the corresponding shear force diagram (SFD) and bending moment diagram (BMD) are sketched as shown below. 

Shear force V(x) = qL V(x) = qL V(x) = –2qL V(x) = 8qL–2qx



Y , , , ,

x  (0, L) x  (L, 2L) x  (2L, 3L) x  (3L, 4L)

Po = 3qL

, , , ,

x  (0, L) x  (L, 2L) x  (2L, 3L) x  (3L, 4L)

2q

Mo = qL

A

B

C

D RDY

RAY

Bending moment M(x) = qLx M(x) = qLx–qL2 M(x) = –2qLx + 5qL2 M(x) = –q(4L–x)2

2

E

X

2qL qL

qL

SFD

2

qL

–2qL qL2 BMD –qL2

From above SFD and BMD, the maximum positive shear force is equal to 2qL occurring at point just to the right of the point D; the maximum negative shear force is equal to 2qL occurring at the entire subinterval CD; the maximum positive bending moment is equal to qL2 occurring at points C and a point just to the left of the point B; and the maximum negative bending moment is equal to qL2 occurring at point D. Copyright © 2011 J. Rungamornrat


89


2.5.5 Method of differential and integral formula It has been apparent from the previous section that the method of sections can become inefficient when applied to construct SFD and BMD of beams. This is due to the need to obtain the shear force and bending moment as a function of position along the entire beam, i.e. V(x) and M(x); the construction of these two functions is somewhat cumbersome especially when there are many points of loading discontinuity (e.g. points where supports are present, concentrated forces and moments are applied, distributed loads change their distribution, etc.) thus requiring to form V(x) and M(x) in several subinterval separately. To overcome this tedious task, another technique called “a method of differential and integral formula” is introduced. This technique is still based primarily on static equilibrium but the key equilibrium equations employed are expressed in both differential form and integral form. The special feature of these equations is that they relate the shear force and bending moment to the external applied loads in both local and global senses and, therefore, allow the shear force and bending moment be obtained as a function of position without introducing any cut along the beam. More explanation about this technique is presented further below.

2.5.5.1 Equilibrium equations in differential form Consider a beam that is in equilibrium with applied transverse loads as shown schematically in Figure 2.29(a). Let’s introduce two cuts, one at the coordinate x and the other at the coordinate x + dx, and then sketch the FBD of an infinitesimal element dx as shown in Figure 2.29(b). The shear force and bending moment at x are denoted by V(x) and M(x), respectively, and the shear force and bending moment at x + dx are denoted by V(x) + dV and M(x) + dM, respectively, where dV and dM are increments of shear force and bending moment. Note that the positive sign conventions for the shear force, bending moment, and the distributed load q (distributed load is considered to be positive if its direction is along the Y-axis) are assumed. Y q

q V(x) M(x)

X x

V(x) + dV dx

M(x) + dM

dx (b)

(a)

Figure 2.29: (a) Schematic of a beam subjected to transverse loads and (b) FBD of an infinitesimal element dx By enforcing static equilibrium of the infinitesimal element dx shown in Figure 2.29(b) and then taking appropriate limit process, we obtain the following two equilibrium equations in a differential form: 

dV(x) dx

ΣFY  0



V(x) + qdx  V(x)  dV = 0

ΣM Z  0



M(x) + dM  M(x)  V(x)dx  qdx(dx/2) = 0



(2.15)

q(x)




dM(x) dx



V(x)

(2.16)


90


It is important to emphasize that the equation (2.15) is valid at any point x such that the distributed load q is continuous and it is free of any concentrated force while the equation (2.15) is valid at any point x such that the shear force is continuous and it is free of any concentrated moment. The first equation (2.15) states that the spatial rate of change of the shear force or “the slope of the shear force diagram” at any point is equal to the value of the distributed load q applied at that point. Based on this argument, following cases can be deduced:  A segment of the beam that is free of distributed load and concentrated force (i.e. q = 0) must have a constant shear or, equivalently, a portion of the SFD corresponding to that segment possesses a zero slope or, in the other word, assumes a horizontal straight line; for instance, segments AB, CD, FG, and IJ of a beam shown in Figure 2.30.  The shear force of a beam segment that is subjected only to uniformly distributed load q possesses a linear distribution across that segment or, equivalently, a portion of the SFD corresponding to that beam segment assumes a straight line (e.g. segments BC and JK of a beam shown in Figure 2.30). The slope of this straight line depends on the direction of q; the slope is positive when q directs upwards otherwise it is negative.  If the distributed load q directs upward (i.e. q > 0) with monotonically increasing magnitude over a segment of the beam, a portion of the SFD corresponding to that segment assumes a rising and concave upward curve (e.g. segment GH shown in Figure 2.30). On the contrary, if the magnitude of the distributed load decreases monotonically (while its direction is still upward), the corresponding portion of the SFD assumes a rising and concave downward curve (e.g. segment HI shown in Figure 2.30).  If the distributed load q directs downward (i.e. q < 0) with monotonically increasing magnitude over a segment of the beam, a portion of the SFD corresponding to that segment assumes a dropping and concave downward curve (e.g. segment DE shown in Figure 2.30). On the contrary, if the magnitude of the distributed load decreases monotonically (while its direction is still downward), the corresponding portion of the SFD assumes a dropping and concave upward curve (e.g. segment EF shown in Figure 2.30). Y

A

B

C

D

E

F

G

H

I

J

K

X

SFD

Figure 2.30: SFD of a beam subjected to different types of distributed load Copyright © 2011 J. Rungamornrat


91


The second equation (2.16) states that the spatial rate of change of the bending moment or “the slope of the bending moment diagram” at any point is equal to the value of shear force at that point. Based on this argument, following cases can be deduced:  A segment of the beam that the shear force identically vanishes (i.e. V = 0) must have a constant shear or, equivalently, a portion of the BMD corresponding to that segment possesses a zero slope or, in the other word, assumes a horizontal straight line (e.g. segment AB shown in Figure 2.31).  For a segment of the beam that possesses a constant positive shear force, the bending moment increases linearly or, equivalently, a portion of the BMD corresponding to that segment assumes a rising straight line (e.g. segment BC shown in Figure 2.31).  For a segment of the beam that possesses a constant negative shear force, the bending moment decreases linearly or, equivalently, a portion of the BMD corresponding to that segment assumes a dropping straight line (e.g. segment GH shown in Figure 2.31).  If the shear force is positive and increases monotonically in magnitude over a segment of the beam, a portion of the BMD corresponding to that segment assumes a rising and concave upward curve (e.g. segment EF shown in Figure 2.31).  If the shear force is positive and decreases monotonically in magnitude over a segment of the beam, a portion of the BMD corresponding to that segment assumes a rising and concave downward curve (e.g. segment FG shown in Figure 2.31).  If the shear force is negative and increases monotonically in magnitude over a segment of the beam, a portion of the BMD corresponding to that segment assumes a dropping and concave downward curve (e.g. segment CD shown in Figure 2.31).  If the shear force is negative and decreases monotonically in magnitude over a segment of the beam, a portion of the BMD corresponding to that segment assumes a dropping and concave upward curve (e.g. segment DE shown in Figure 2.31).

A

B

C

D

E

F

G

H

SFD

BMD

Figure 2.31: SFD and the corresponding BMD of a beam Copyright © 2011 J. Rungamornrat


92


2.5.5.2 Equilibrium equations in integral form Now, let A and B be two points within the beam and xA and xB be their X-coordinates; without loss of generality, let’s assume further that the point B is on the right of point A, i.e., xB > xA. By directly integrating equation (2.15) from the point A to the point B, we then obtain the integral formula xB

VB  VA   q dx  VA  Q AB

(2.17)

xA

where VA and VB are shear forces at the points A and B, respectively, and QAB is the sum of all distributed load over the segment AB. This equation implies that the shear force at the point B can be obtained by adding the total load over the segment AB to the shear force at the point A. It is important to emphasize that equation (2.17) applies only to the case that there is no concentrated force acting to the segment AB. Note that the sign convention of the total load QAB follows that of the distributed load q. Similarly, by directly integrating equation (2.16) from the point A to the point B, we obtain the integral formula xB

M B  M A   V dx  M A  AreaVAB

(2.18)

xA

where MA and MB are the bending moment at the point A and point B, respectively, and AreaVAB is the area of the shear force diagram over the segment AB. This equation implies that the bending moment at the point B can be obtained by adding the area of the shear force diagram over the segment AB to the bending moment at the point A. It is important to emphasize that the relation (2.18) applies to the case that there is no concentrated moment acting to the portion AB. The sign convention of the area AreaVAB follows directly from that of the shear force V; it can therefore be either positive or negative. Both the relations (2.17) and (2.18) can be used to determine the shear force and bending moment at a particular point when there exists at least one point that the shear force and the bending moment are known. If the relation (2.18) is to be employed in the construction of BMD, the SFD must be known a priori.

2.5.5.3 Discontinuity of shear force and bending moment It has been found in various situations that actual applied loads are suitable to be modeled by concentrated forces or concentrated moments in the idealized structure. Presence of such concentrated loads within the beam may introduce the discontinuity of certain components of the internal force at the location where the concentrated loads are applied. Here, we employ static equilibrium to establish the discontinuity conditions of the shear force and bending moment at points where the concentrated loads are applied. First, let us investigate the discontinuity condition at the location where a concentrate force is applied. Let Po be a concentrated force applied to a point A of the beam (this force is considered to be positive if it directs upward in Y-direction otherwise it is negative). By introducing two cuts at a point just to left and a point just to the right of the point A and then considering equilibrium of an infinitesimal element containing a point A (its free body diagram is shown in Figure 2.32) along with taking appropriate limit process, we obtain the discontinuity conditions of the shear force and bending moment: Copyright © 2011 J. Rungamornrat


93


VAL MAL

Po

VAR

A

MAR

Figure 2.32: FBD of infinitesimal element containing a point where concentrated force is applied VAR  VAL  Po

(2.19)

M AR  M AL

(2.20)

where VAR and VAL are the shear force at a point just to the right and a point just to the left of the point A, respectively, and MAR and MAL are the bending moment at a point just to the right and a point just to the left of the point A, respectively. It is evident from (2.19) and (2.20) that, at the location where a concentrated force is applied, the shear force is not defined and is discontinuous with the magnitude of the jump equal to the magnitude of the concentrated force while the bending moment is still continuous. In addition, the shear force experiences a positive jump if the concentrated force is positive (or directs upward in the Y-direction) and it experiences a negative jump if the concentrated force is negative (or directs downward in the opposite Y-direction). Next, let us consider the discontinuity condition at a location where a concentrated moment Mo is applied. Let Mo be a concentrated moment applied to a point A (this moment is considered to be positive if it directs in a counter clockwise direction or in the Z-direction otherwise it is negative). By introducing two cuts at a point just to left and a point just to the right of the point A and then considering equilibrium of an infinitesimal element containing a point A (its free body diagram is shown in Figure 2.33) along with taking appropriate limit process, we obtain the discontinuity conditions of the shear force and bending moment:

MAL

VAL VAR Mo A

MAR

Figure 2.33: FBD of infinitesimal element containing a point where concentrated moment is applied VAR  VAL

(2.21)

M AR  M AL  M o

(2.22)

Equations (2.21) and (2.22) imply that at a point where the concentrated moment is applied, the bending moment is discontinuous with the magnitude of the jump equal to the magnitude of the concentrated moment while the shear force experiences no jump but it is not defined at this point. We emphasize in addition that the bending moment experiences a positive jump if the concentrated moment is negative (or directs in a clockwise direction or Z-direction) and it experiences a negative jump if the concentrated moment is positive (or directs in a counter clockwise direction or opposite Z-direction). Finally, let us investigate the location where the distributed load q is discontinuous, say a point A. By introducing two cuts at a point just to left and a point just to the right of the point A and then considering equilibrium of an infinitesimal element containing a point A (its free body diagram is shown in Figure 2.34) along with taking appropriate limit process, we obtain the discontinuity conditions of the shear force and bending moment: Copyright © 2011 J. Rungamornrat


94


q VAR

VAL MAL

A

MAR

Figure 2.34: FBD of infinitesimal element containing point where distributed load is discontinuous VAR  VAL

(2.23)

M AR  M AL

(2.24)

This implies that both the shear force and bending moment are continuous at a location where the distributed load is discontinuous.

2.5.5.4 Procedures for constructing SFD and BMD The differential formula (2.15) and (2.16), the integral formula (2.17) and (2.18) and the discontinuity conditions (2.19)-(2.24) constitutes basic components for constructing the SFD and BMD of a beam. In particular, the two integral formula (2.17) and (2.18) are employed to obtain the shear force and the bending moment at the right end of any segment when values at the left end of those quantities are known and there is no point of loading discontinuity within the segment (e.g. concentrated forces and moments). The two differential formulae (2.15) and (2.16) are then used to identify the type of a curve that connects a part of the SFD and BMD over a segment where values of the shear force and bending moment are already known at its ends. The discontinuity conditions are used to dictate the jump of the shear force and bending moment in the SFD and BMD due to the presence of concentrated forces and moments. Here, we summarize standard procedures or guidelines for constructing the SFD and BMD of a beam.  Determine all support reactions  Identify and mark points of loading discontinuity, e.g. supports, points where concentrated forces and moments are applied, and points where distributed load changes its distribution  Identify all possible segments such that points of loading discontinuity must be at the ends of each segment  Identify a point that both the shear force and bending moment are known as a starting point (in general, the left end of the beam is chosen since all forces and moments are known at both ends of the beam once the reactions are already determined.)  Start drawing the SFD as follow: (i) start with the first segment on the left of the beam since the shear force at the left end of this segment is already known, (ii) use the integral formula (2.17) to compute the shear force at the right end of the segment, (iii) use the differential formula (2.15) to identify the type of a curve and then use that curve to draw the SFD over the segment, (iv) choose the segment just to the left of the previous segment as a current segment to be considered, and (v) use the jump conditions (2.19), (2.21) and (2.23) along with the value of the shear force at the right end of the previous segment to obtain the shear force at left end of the current segment. Repeat from steps (ii) until all segments are considered. Note that, for the last segment, the shear force at the right end of the segment must be consistent with the force applied at that end.  Once the SFD is obtained, the BMD can be constructed as follow: (i) start with the first segment on the left of the beam since the bending moment at the left end of this segment Copyright © 2011 J. Rungamornrat


95


is already known, (ii) use the integral formula (2.18) to compute the bending moment at the right end of the segment, (iii) use the differential formula (2.16) to identify the type of a curve and then use that curve to draw the BMD over the segment, (iv) choose the segment just to the left of the previous segment as a current segment to be considered, and (v) use the jump conditions (2.20), (2.22) and (2.24) along with the value of the bending moment at the right end of the previous segment to obtain the bending moment at the left end of the current segment. Repeat from the step (ii) until all segments are considered. Note that, for the last segment, the bending moment at the right end of the segment must be consistent with the moment applied at that end. Once the SFD and BMD are completed, one can identify both the magnitude and location of the maximum shear force and maximum bending moment. In general, the maximum shear force can occur at the supports, the locations where the distributed load q vanishes, and locations where the concentrated forces are applied. Similarly, the maximum bending moment can occur at the supports, the locations where the shear force vanishes, the locations where the shear force changes its sign, and the locations where the concentrated moments are applied. Example 2.9 Draw the SFD and BMD of a beam shown in the example 2.8 using the method of differential and integral formula Y 2

2q

Po = 3qL

Mo = qL

A

B L

C L

D L

E

X

L

Solution All support reactions of this beam were already determined in the example 2.8 and the FBD diagram of the entire beam is shown again below. Y 2

2q

Po = 3qL

Mo = qL

A

B RAY = qL

C

D

E

X

RDY = 4qL

From above FBD, there are three points of loading discontinuity within the beam (excluding the two ends of the beam), e.g. points B, C and D. Specifically, the point B is a point where the concentrated moment is applied, the point C is a point where the concentrated force is applied, and the point D is a point where the support reaction (viewed as the concentrated force) is applied and the distributed load is discontinuous. Thus, we can divide the entire beam into four segments: AB, BC, CD and DE. The shear force and the bending moment at the left end of the beam (i.e. point A) are already known, i.e. VA = RAY = qL and MA = 0. First, let us construct the SFD. The differential formula (2.15), the integral formula (2.17), and the discontinuity conditions (2.19), (2.21) and (2.23) are utilized for each segment as follow: Copyright © 2011 J. Rungamornrat


96


 Segment AB - VA = qL - There is no distributed load over the segment + equation (2.17)  VBL = VA + 0 = qL - There is no distributed load over the segment + equation (2.15)  SFD over the segment is a horizontal straight line  Segment BC - The concentrated moment is applied at point B + equation (2.21)  there is no jump of the shear force at point B  VBR = VBL = qL - There is no distributed load over the segment + equation (2.17)  VCL = VBR + 0 = qL - There is no distributed load over the segment + equation (2.15)  SFD over the segment is a horizontal straight line  Segment CD - The negative concentrated force equal to 3qL is applied at point C + equation (2.19)  there is a jump of the shear force at point C  VCR = VCL – 3qL = –2qL - There is no distributed load over the segment + equation (2.17)  VDL = VCR + 0 = –2qL - There is no distributed load over the segment + equation (2.15)  SFD over the segment is a horizontal straight line  Segment DE - The positive concentrated force equal to 4qL is applied at point D  there is a jump of the shear force at point D  VDR = VDL + 4qL = 2qL - A negative uniform distributed load –2q is applied over the segment + equation (2.17)  VE = VDR + (–2q)(L) = 0  consistent with condition at the right end of the beam - A negative uniform distributed load –2q is applied over the segment + equation (2.15)  SFD over the segment is a dropping straight line Y 2

2q

Po = 3qL

Mo = qL

A

C

B

D E RDY = 4qL

RAY = qL

X

2qL qL

qL SFD –2qL

Once the SFD is obtained, the BMD can then be constructed. The differential formula (2.16), the integral formula (2.18), and the discontinuity conditions (2.20), (2.22) and (2.24) are utilized for each segment as follow: Copyright © 2011 J. Rungamornrat


97


 Segment AB - MA = 0 - Area of the SFD over the segment is (qL)(L) + equation (2.18)  MBL = MA + (qL)(L) = qL2 - The shear force is constant and positive over the segment + equation (2.16)  BMD over the segment is a rising straight line  Segment BC - The positive concentrated moment equal to qL2 is applied at point B + equation (2.22)  there is a jump of the bending moment at point B  MBR = MBL – qL2 = 0 - Area of the SFD over the segment is (qL)(L) + equation (2.18)  MCL = MBR + (qL)(L) = qL2 - The shear force is constant and positive over the segment + equation (2.16)  BMD over the segment is a rising straight line  Segment CD - The concentrated force is applied at point C + equation (2.20)  there is no discontinuity of the bending moment at point C  MCR = MCL = qL2 - Area of the SFD over the segment is (–2qL)(L) + equation (2.18)  MDL = MCR + (–2qL)(L) = –qL2 - The shear force is constant and negative over the segment + equation (2.16)  BMD over the segment is a dropping straight line  Segment DE - The concentrated force is applied at point D + equation (2.20)  there is no discontinuity of the bending moment at point D  MDR = MDL = –qL2 - Area of the SFD over the segment is (2qL)(L)/2 + equation (2.18)  MEL = MDR + (2qL)(L)/2 = 0  consistent with condition at the right end of the beam - The shear force is positive and decreases monotonically in magnitude over a segment + equation (2.16)  BMD over the segment is a rising and concave downward curve Y 2

Po = 3qL

2q

Mo = qL

A

B

C

D E RDY = 4qL

RAY = qL

X

2qL qL

qL SFD –2qL qL2

qL2 BMD –qL2



98


Example 2.10 Draw the SFD and BMD of a beam shown below using the method of differential and integral formula. Y Po = qL

2q 2

Mo = qL C

B

A L

L

E

D L

F L

X

L

Solution The given beam is statically determinate (i.e. ra = 2 + 1 = 3, nm = 2(2) = 4, nj = 3(2) = 6, nc = 1, then DI = 3 + 4 – 6 – 1 = 0); thus, all support reactions and the internal forces at any location can be determined from static equilibrium. Since the number of support reactions is equal to 3, they cannot be obtained from equilibrium of the entire structure alone. By making a cut at point just to the left of the hinge B and then considering equilibrium of the right part (part BF), one of the reactions can be determined. The rest of the reactions can be computed from equilibrium of the entire beam. Details of calculation are shown below: Po = qL

2q 2

Mo = qL

VBR

BR

C

E

D

F

X

REY FBD of part BF Y Po = qL

2q 2

Mo = qL

RAM A

B

C

E

D

F REY

RAY FBD of entire beam Equilibrium of part BF: [MBR = 0]  

:

3REYL – (2q)(L)(L/2) – qL2 – (qL)(4L) = 0 REY = 2qL

Upward

Equilibrium of entire beam: [MA = 0]  

:

RAM + 4REYL – (2q)(2L)(L) – qL2 – (qL)(5L) = 0 RAM = 2qL2

[FY = 0]  

:

CCW

RAY + REY – (2q)(2L) – qL = 0 RAY = 3qL

Upward


X


99


From the FBD of the entire structure, there are three points of loading discontinuity within the beam (excluding the two ends of the beam), e.g. points C, D and E. Specifically, the point C is a point where the distributed load is discontinuous, the point D is a point where the concentrated moment is applied, and the point E is a point where the support reaction (viewed as the concentrated force) is applied. Note that the point B, while it is an internal release, is not considered as a point of loading discontinuity since, at this point, the distributed load is continuous and there is no applied concentrated load. Therefore, we can divide the entire beam into four segments: AC, CD, DE and EF. Once the support reactions are determined, the shear force and the bending moment at the left end of the beam (i.e. point A) are known, i.e. VA = RAY = 3qL and MA = –RAM = –2qL2. First, let us construct the SFD. The differential formula (2.15), the integral formula (2.17), and the discontinuity conditions (2.19), (2.21) and (2.23) are utilized for each segment as described below:  Segment AC - VA = 3qL - A negative uniform distributed load –2q is applied over the segment + equation (2.17)  VCL = VA + (–2q)(2L) = –qL - A negative uniform distributed load –2q is applied over the segment + equation (2.15)  SFD over the segment is a dropping straight line  Segment CD - The distributed load is discontinuous at point C + equation (2.23)  there is no discontinuity of the shear force at point C  VCR = VCL = –qL - There is no distributed load over the segment + equation (2.17)  VDL = VCR + 0 = –qL - There is no distributed load over the segment + equation (2.15)  SFD over the segment is a horizontal straight line  Segment DE - The negative concentrated moment is applied at point C + equation (2.21)  there is no discontinuity of the shear force at point D  VDR = VDL = –qL - There is no distributed load over the segment + equation (2.17)  VEL = VDR + 0 = –qL - There is no distributed load over the segment + equation (2.15)  SFD over the segment is a horizontal straight line  Segment EF - The positive concentrated force equal to 2qL is applied at point E + equation (2.19)  there is a jump of the shear force at point E  VER = VEL + 2qL = qL - There is no distributed load over the segment + equation (2.17)  VFL = VER + 0 = qL  consistent with condition at the right end of the beam - There is no distributed load over the segment + equation (2.15)  SFD over the segment is a horizontal straight line The SFD is shown in the figure below. It is evident that there exists a point within the segment AC, say a point G, such that the shear force vanishes. In particular, the point G is located, with a distance L/2, on the right of the hinge B. The segment AC can further be separated into two subsegments AG and GC; the shear force is positive for the first sub-segment while it is negative for the second sub-segment. In the construction of the BMD, the differential formula (2.16), the integral formula (2.18), and the discontinuity conditions (2.20), (2.22) and (2.24) are utilized for each segment as described below: Copyright © 2011 J. Rungamornrat


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Y Po = qL

2q 2

Mo = qL

RAM

A

G C

B

D

RAY 3qL

E

X

F REY

L/2 qL

qL

qL SFD

–qL

–qL

–qL

 Segment AG - MA = –RAM = –2qL2 - Area of the SFD over the segment is (3qL)(3L/2)/2 + equation (2.18)  MGL = MA + (3qL)(3L/2)/2 = qL2/4 - The shear force is positive and decreases monotonically in magnitude over a segment + equation (2.16)  BMD over the segment is a rising and concave downward curve with a zero slope at point G  Segment GC - No loading discontinuity at point G  there is no discontinuity of the bending moment at point G  MGR = MGL = qL2/4 - Area of the SFD over the segment is (–qL)(L/2)/2 + equation (2.18)  MCL = MGR + (–qL)(L/2)/2 = 0 - The shear force is negative and increases monotonically in magnitude over a segment + equation (2.16)  BMD over the segment is a dropping and concave downward curve  Segment CD - The distributed load is discontinuous at point C + equation (2.24)  there is no discontinuity of the bending moment at point C  MCR = MCL = 0 - Area of the SFD over the segment is (–qL)(L) + equation (2.18)  MDL = MCR + (– qL)(L) = –qL2 - The shear force is constant and negative over the segment + equation (2.16)  BMD over the segment is a dropping straight line  Segment DE - The negative concentrated moment –qL2 is applied at point D + equation (2.22)  there is a jump of the bending moment at point D  MDR = MDL – (–qL2) = 0 - Area of the SFD over the segment is (–qL)(L) + equation (2.18)  MEL = MDR + (– qL)(L) = –qL2 - The shear force is constant and negative over the segment + equation (2.16)  BMD over the segment is a dropping straight line  Segment EF - The positive concentrated force 2qL is applied at point E + equation (2.20)  there is no discontinuity of the bending moment at point E  MER = MEL = –qL2 Copyright © 2011 J. Rungamornrat


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Area of the SFD over the segment is (–qL)(L) + equation (2.18)  MFL = MER + (qL)(L) = 0  consistent with the condition at the right end of the beam The shear force is constant and positive over the segment + equation (2.16)  BMD over the segment is a rising straight line

-

Y Po = qL

2q 2

Mo = qL

RAM

A

G C

B

E

D

X

F REY

RAY L/2

3qL

qL

qL

qL SFD

–qL

–qL

–qL qL2/4

BMD –qL2

–qL2

–2qL2

From above SFD and BMD, the maximum positive shear force is equal to 3qL occurring at point A; the maximum negative shear force is equal to qL occurring at the entire subinterval CE; the maximum positive bending moment is equal to qL2/4 occurring at point G; and the maximum negative bending moment is equal to 2qL2 occurring at point A. Example 2.11 Draw the SFD and BMD of a beam shown below using the method of differential and integral formula Y qL

q

qL2/3 A

B

X

D C

L

L

E L


L


102


Solution Since ra = 2 + 1 + 1 = 4, nm = 2(2) = 4, nj = 3(2) = 6, nc = 2, then DI = 4 + 4 – 6 – 2 = 0. Thus, the structure is statically determinate and all support reactions can be determined form static equilibrium. However, the number of independent equilibrium equations that can be set up for beams is net = 2 < ra; thus, the support reactions cannot be obtained by considering only equilibrium of the entire structure. To overcome this problem, two additional equations associated with the presence of two moment releases or hinges at points B and D, i.e. M = 0 at B and M = 0 at D, must be employed. By introducing a cut at the point D and considering moment equilibrium of the right part of the beam, the reaction REY can be determined; by introducing a cut at the point just to the right of the point B and considering moment equilibrium of the right part of the beam, the reaction RCY can be determined; finally, by considering equilibrium of the entire beam, the rest of reactions can readily be determined. Details of calculations are shown below: VD

q

D

E

X

FBD of a part DE

X

FBD of a part BE

X

FBD of entire beam

REY VBR BR Y

q

qL2/3 C RCY

D

REY

qL RAM

q

qL2/3 A

B

RAY

C RCY

E

D

E REY

Equilibrium of part DE [MDR = 0]  

:

REYL – (q)(L/2)(2L/3) = 0 REY = qL/3

Upward

Equilibrium of part BE [MDR = 0]  

:

RCYL + 3REYL – qL2/3 – (q)(L/2)(2L+2L/3) = 0 RCY = 2qL/3 Upward

Equilibrium of entire beam [MA = 0]  

:

RAM + 2RCYL + 4REYL – (qL)(L) – qL2/3– (qL/2)(3L+2L/3) = 0 RAM = qL2/2 CCW

[FY = 0]  

:

RAY + RCY +REY – qL – (q)(L/2) = 0 RAY = qL/2

Upward



103


From the FBD of the entire structure, there are three points of loading discontinuity within the beam (excluding the two ends of the beam), e.g. points B, C and D. Specifically, the point B is a point where the concentrated force is applied, the point C is a point where the concentrated moment and the support reaction (viewed as the concentrated force) are applied, and the point D is a point where the distributed load changes its distribution but still continuous. Therefore, we can divide the entire beam into four segments: AB, BC, CD and DE. Once the support reactions are determined, the shear force and the bending moment at the left end of the beam (i.e. point A) are known, i.e. VA = RAY = qL/2 and MA = –RAM = –qL2/2. First, let us construct the SFD. The differential formula (2.15), the integral formula (2.17), and the discontinuity conditions (2.19), (2.21) and (2.23) are utilized for each segment as described below:  Segment AB - VA = qL/2 - There is no distributed load over the segment + equation (2.17)  VBL = VA + 0 = qL/2 - There is no distributed load over the segment + equation (2.15)  SFD over the segment is a horizontal straight line  Segment BC - The negative concentrated force –qL is applied at point B + equation (2.19)  there is a jump of the shear force at point B  VBR = VBL + (–qL) = –qL/2 - There is no distributed load over the segment + equation (2.17)  VCL = VBR + 0 = –qL/2 - There is no distributed load over the segment + equation (2.15)  SFD over the segment is a horizontal straight line  Segment CD - The positive concentrated force 2qL/3 is applied at point C + equation (2.19)  there is a jump of the shear force at point C  VCR = VCL + (2qL/3) = qL/6 - There is no distributed load over the segment + equation (2.17)  VDL = VCR + 0 = qL/6 - There is no distributed load over the segment + equation (2.15)  SFD over the segment is a horizontal straight line  Segment DE - The distributed load changes its distribution at point D  there is no discontinuity of the shear force at point D  VDR = VDL = qL/6 - The distributed load is negative and increases monotonically in magnitude over the segment + equation (2.17)  VEL = VDR – (q)(L)/2 = –qL/3  consistent with the condition at the right end of the beam - The distributed load is negative and increases monotonically in magnitude over the segment + equation (2.15)  SFD over the segment is a dropping and concave downward curve From the SFD shown above, it is evident that there exists a point within the segment DE, say a point F, such that the shear force vanishes. The exact location of the point G is obtained as (qs/L)(s/2) = qL/6



s = L/√3

where s is the distance between the point D and point F. In addition, the value of the distributed at the point F is q/√3. The segment DE can now be separated into two sub-segments DF and FE; the shear force is positive for the first sub-segment while it is negative for the second sub-segment. Copyright © 2011 J. Rungamornrat


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Y qL RAM

q

qL2/3 A

B

F

D

C

REY

RCY

RAY

X

E

s qL/2 qL/6 SFD –qL/3 –qL/2 In the construction of the BMD, the differential formula (2.16), the integral formula (2.18), and the discontinuity conditions (2.20), (2.22) and (2.24) are utilized for each segment of the beam as described below:  Segment AB - MA = –RAM = –qL2/2 - Area of the SFD over the segment is (qL/2)(L) + equation (2.18)  MBL = MA + (qL/2)(L) = 0 - The shear force is constant and positive over a segment + equation (2.16)  BMD over the segment is a rising straight line  Segment BC - The negative concentrated force –qL is applied at point B + equation (2.20)  there is no discontinuity of the bending moment at point B  MBR = MBL = 0 - Area of the SFD over the segment is (–qL/2)(L) + equation (2.18)  MCL = MBR + (–qL/2)(L) = –qL2/2 - The shear force is constant and negative over a segment + equation (2.16)  BMD over the segment is a dropping straight line  Segment CD - The negative concentrated moment –qL2/3 is applied at point C + equation (2.22) 2  there is a jump of the bending moment at point C  MCR = MCL – (–qL /3) = – 2 qL /6 - Area of the SFD over the segment is (qL/6)(L) + equation (2.18)  MDL = MCR + (qL/6)(L) = 0 - The shear force is constant and positive over the segment + equation (2.16)  BMD over the segment is a rising straight line  Segment DF - The distributed load changes its distribution at point D but it is still continuous  there is no discontinuity of the bending moment at point D  MDR = MDL = 0 - Area of the SFD over the segment is 2(qL/6)( L/√3)/3 + equation (2.18)  MFL = MDR + qL2/9√3 = qL2/9√3 - The shear force is positive and decreases monotonically in magnitude over a segment + equation (2.16)  BMD over the segment is a rising and concave downward curve with a zero slope at point F Copyright © 2011 J. Rungamornrat


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 Segment FE - There is no loading discontinuity at point F  there is no discontinuity of the bending moment at point F  MFR = MFL = qL2/9√3 - Area of the SFD over the segment is –qL2/9√3 + equation (2.18)  MEL = MFR – qL2/9√3 = 0  consistent with the condition at the right end of the beam - The shear force is negative and increases monotonically in magnitude over a segment + equation (2.16)  BMD over the segment is a dropping and concave downward curve Y qL RAM

q

qL2/3 A

B

RCY

RAY

X

E

F

D

C

REY s

qL/2 qL/6 SFD –qL/3 –qL/2 qL2/9√3 BMD 2

–qL /6 –qL2/2

–qL2/2

From above SFD and BMD, the maximum positive shear force is equal to qL/2 occurring at the entire segment AB; the maximum negative shear force is equal to –qL/2 occurring at the entire segment BC; the maximum positive bending moment is equal to qL2/9√3 occurring at point F; and the maximum negative bending moment is equal to qL2/2 occurring at points A and C.

2.5.6 Qualitative elastic curve In this section, we provide basic idea for sketching qualitative elastic curve of a beam (a curve represented the deformed configuration of the neutral axis of the beam) under applied loads once the bending moment diagram (BMD) is determined. The key assumptions employed are those associated with Euler-Bernoulli beam theory; i.e. (i) beam is made from a linearly elastic material; (ii) plane section remains plane before and after undergoing deformation; (iii) no shear deformation; and (iv) no internal axial force and no axial deformation of the neutral axis. Schematics of EulerBernoulli beam before and after undergoing deformation are indicated in Figure 2.35. According to the kinematics assumption of deformation of the cross section, it is standard to represent the entire beam by its neutral axis. The elastic or deformed curve is therefore the deformed configuration of the neutral axis. Copyright © 2011 J. Rungamornrat


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NA NA Undeformed state

Deformed state

Figure 2.35: Schematics of undeformed and deformed configurations of a beam under bending Let v(x) and (x) be the deflection and the rotation at any point x as shown schematically in Figure 2.36. The deflection v(x) is considered to be positive if it directs upward in the Y-direction and the rotation (x) is considered to be positive if it directs in a counter clockwise direction or in the Zdirection. By assuming that the deflection and the rotation are infinitesimally small in comparison with the length of the beam, the deflection v(x), the rotation (x), and the curvature (x) are related through the following relations

θ(x) 

dv dx

(2.25)

κ(x) 

dθ d 2 v  dx dx 2

(2.26)

Y (x) v(x) X x Figure 2.36: Schematics indicating the deflection and rotation at any point x It is evident from the relation (2.26) that the curvature (x) is positive if the deflection is concave upward (i.e. d 2 v/dx 2  0 ), negative if the deflection is concave downward (i.e. d 2 v/dx 2  0 ), and zero if it is an inflection point (i.e. d 2 v/dx 2  0 ). In addition, a direct consequence of the infinitesimal displacement and rotation assumption leads to zero displacement in the longitudinal direction of the beam or, equivalently, the preservation of the projected length of the beam onto its undeformed axis. This behavior implies that any point of the beam displaces only in a vertical direction or, more precisely, in a direction perpendicular to the axis of the beam. By exploiting the kinematics assumption of the beam cross section, utilizing material constitutive, and computing the moment resultant of the cross section, it leads to a well-known moment-curvature relationship: κ(x) 

M(x) EI



107


where E is the Young’s modulus and I is the moment of inertia of the cross section. It can be deduced from the relations (2.26) and (2.27) that  A segment of a beam possessing the positive bending moment undergoes a positive curvature and, as a result, leading to a concave upward elastic curve (see Figure 2.37);

Figure 2.37: Schematics of concave upward elastic curve  A segment of a beam possessing the negative bending moment undergoes a negative curvature and, as a result, leading to a concave downward elastic curve (see Figure 2.38);

Figure 2.38: Schematics of concave downward elastic curve  A segment of a beam possessing the zero bending moment undergoes a zero curvature and, as a result, leading to a straight-line elastic curve; and  A point within the beam where the bending moment changes sign at that point is an inflection point on the elastic curve. To sketch the qualitative elastic curve, the following procedures are suggested:    

Construct BMD for the entire beam Use equation (2.27) to identify the shape of elastic curve at any segment of the beam Patch all segments of elastic curve together Check compatibility with all supports and internal releases.

It is important to emphasize that the deflection of the beam is continuous everywhere except at the shear releases and the rotation of the beam is continuous everywhere except at the moment releases or hinges. Figure 2.39 shown below indicate the discontinuity occurs at the shear release and the moment release.

Figure 2.39: Schematics of deformed shape in the neighborhood of the shear and moment releases Copyright © 2011 J. Rungamornrat


108


Example 2.12 Compute all support reactions, sketch the SFD and BMD, and then sketch the elastic curve of a beam shown below Y q

2qL D

A

B L

X

C L

L

Solution The given beam is statically determinate (i.e. ra = 2 + 1 = 3, nm = 1(2) = 2, nj = 2(2) = 6, nc = 1, then DI = 3 + 2 – 4 – 1 = 0); thus, all support reactions and the internal forces at any location can be determined from static equilibrium. Since the number of support reactions is equal to 3, they cannot be obtained from equilibrium of the entire structure alone. By making a cut at the hinge B and then considering equilibrium of the right part (part BD), one of the reactions can be determined. The rest of the reactions can be computed from equilibrium of the entire beam. Details of calculation are shown below:

2qL

VB

D C

B

FBD of part BD

X

FBD of entire beam

RDY

Y q RAM

X

2qL D

A

C

B

RAY

RDY

Equilibrium of a part BD [MBR = 0]  

:

2RDYL – (2qL)(L) = 0 RDY = qL

Upward

Equilibrium of entire beam [MA = 0]  

:

RAM + 3RDYL – (q)(L)(L/2) – (2qL)(2L) = 0 RAM = 3qL2/2 CCW

[FY = 0]  

:

RAY + RDY – (q)(L) – 2qL = 0 RAY = 2qL

Upward



109


From the FBD of the entire structure, there are two points of loading discontinuity within the beam (excluding the two ends of the beam), e.g. points B and C. Specifically, the point B is a point where the distributed load is discontinuous and the point C is a point where the concentrated forced is applied. Therefore, we can divide the entire beam into three segments: AB, BC, and CD. Once the support reactions are determined, the shear force and the bending moment at the left end of the beam (i.e. point A) are known, i.e. VA = RAY = 2qL and MA = –RAM = –3qL2/2. First, let us construct the SFD. The differential formula (2.15), the integral formula (2.17), and the discontinuity conditions (2.19), (2.21) and (2.23) are utilized for each segment as shown below:  Segment AB - VA = 2qL - A negative uniform distributed load –q is applied over the segment + equation (2.17)  VBL = VA + (–q)(L) = qL - A negative uniform distributed load –q is applied over the segment + equation (2.15)  SFD over the segment is a dropping straight line  Segment BC - The distributed load is discontinuous at point B + equation (2.23)  there is no discontinuity of the shear force at point B  VBR = VBL = qL - There is no distributed load over the segment + equation (2.17)  VCL = VBR + 0 = qL - There is no distributed load over the segment + equation (2.15)  SFD over the segment is a horizontal straight line  Segment CD - The negative concentrated force –2qL is applied at point C + equation (2.19)  there is a jump of the shear force at point C  VCR = VCL + (–2qL) = –qL - There is no distributed load over the segment + equation (2.17)  VDL = VDR + 0 = –qL  consistent with condition at the right end of the beam - There is no distributed load over the segment + equation (2.15)  SFD over the segment is a horizontal straight line Once the SFD is obtained, the BMD can then be constructed. The differential formula (2.16), the integral formula (2.18), and the discontinuity conditions (2.20), (2.22) and (2.24) are utilized for each segment as described below:  Segment AB - MA = –3qL2/2 - Area of the SFD over the segment is (2qL + qL)(L)/2 + equation (2.18)  MBL = MA + 3qL2/2 = 0 - The shear force is positive and decreases monotonically in magnitude over the segment + equation (2.16)  BMD over the segment is a rising and concave downward curve  Segment BC - The distributed load is discontinuous at point B + equation (2.24)  there is no discontinuity of the bending moment at point B  MBR = MBL = 0 - Area of the SFD over the segment is (qL)(L) + equation (2.18)  MCL = MBR + (qL)(L) = qL2 - The shear force is constant and positive over the segment + equation (2.16)  BMD over the segment is a rising straight line  Segment CD Copyright © 2011 J. Rungamornrat


110


-

The concentrated force is applied at point C + equation (2.20)  there is no discontinuity of the bending moment at point C  MCR = MCL = qL2 Area of the SFD over the segment is (–qL)(L) + equation (2.18)  MDL = MCR + (– qL)(L) = 0  consistent with condition at the right end of the beam The shear force is constant and negative over the segment + equation (2.16)  BMD over the segment is a dropping straight line

From movement constraints provided by roller and fixed supports, a moment release, and the BMD shown below, we obtain the following information that is useful for sketching an elastic curve:  Point A: fixed support  there is no rotation and deflection at this point  Point B: hinge  the rotation is discontinuous at this point while displacement is still continuous  Point D: roller support  there is no vertical displacement at this point while the rotation is allowed  Segment AB: bending moment is negative  the elastic curve of this segment must be concave downward  Segment BD: bending moment is positive  the elastic curve of this segment must be concave upward Y q RAM

2qL D

A

X

C

B

RAY

RDY

2qL qL SFD –qL 2

qL

BMD

–3qL2/2 Elastic curve Hinge



111


Example 2.13 Sketch the elastic curve of the beam shown in example 2.9 Y 2

2q

Po = 3qL

Mo = qL

A

C

B L

L

D L

X

E L

Solution From movement constraints provided by roller and pinned supports and the BMD obtained in example 2.9, we obtain the following information that is useful for sketching an elastic curve:  Point A: pinned support  there is no vertical displacement at this point while the rotation is allowed  Point D: roller support  there is no vertical displacement at this point while the rotation is allowed  Point F: change sign of bending moment  inflection point  Segment AB: bending moment is positive  the elastic curve of this segment must be concave upward  Segment BC: bending moment is positive  the elastic curve of this segment must be concave upward  Segment CF: bending moment is positive  the elastic curve of this segment must be concave upward  Segment FD: bending moment is negative  the elastic curve of this segment must be concave downward  Segment DE: Bending moment is negative  the elastic curve of this segment must be concave downward The resulting elastic curve is shown below. Y A

2

Mo = qL B

F C

RAY = qL qL2

2q

Po = 3qL

D E RDY = 4qL

X

qL2 BMD –qL2 Elastic curve



112


Example 2.14 Sketch the elastic curve of the beam shown in example 2.11 Y qL

q

qL2/3 A

B

X

D C L

L

E L

L

Solution From movement constraints provided by roller and fixed supports, moment releases, and the BMD obtained in example 2.11, we obtain the following information that is useful for sketching an elastic curve:  Point A: fixed support  there is no rotation and deflection at this point  Point C: roller support  there is no vertical displacement at this point while the rotation is allowed  Point E: roller support  there is no vertical displacement at this point while the rotation is allowed  Point B: hinge  the rotation is discontinuous at this point while displacement is still continuous  Point D: hinge  the rotation is discontinuous at this point while displacement is still continuous  Segments AB, BC, CD: bending moment is negative  the elastic curve of these segments must be concave downward  Segment DE: bending moment is positive  the elastic curve of this segment must be concave upward The resulting elastic curve is shown below. Y qL RAM

q

qL2/3 A RAY

B

D

C RCY

X

E

F

REY qL2/9√3 BMD 2

–qL /6 –qL2/2

–qL2/2

Elastic curve



113


2.6 Static Analysis of Frames The primary objective of this section is to generalize the concept and techniques presented in the previous section to analyze a more general class of structures called frames. Basic quantities of interest from the static analysis are still the support reactions and the internal forces at any location. However, as will be clear in later discussion, the internal force of a frame is relatively more complex than that of a truss and a beam since it consists of all three components known as the axial force, the shear force and the bending moment. This section begins with a brief introduction on characteristics of frames and standard notations and sign convention commonly used. A brief overview on how to determine support reactions of statically determinate frames is addressed along with some useful remarks. In the analysis for the internal forces, both the method of sections and the method of differential and integral formula are outlined. While these two methods are similar in accord to those employed in the analysis of beams, they possess an additional feature capable of treating structures whose internal force fully containing the axial force, the shear force and the bending moment. Finally, some guidelines for sketching a qualitative elastic curve for frames are summarized. At the end of this section, some examples are also presented to clearly demonstrate all techniques outlined.

2.6.1 Characteristics of frames An idealized structure is called a planar frame if and only if (i) all members form a twodimensional structure, (ii) members are connected by rigid or frame joints (full or partial moment releases are allowed for certain joints and this can be viewed as rigid joints supplying by moment releases), and (iii) applied loads form a system of general two-dimensional forces and moments (i.e. it includes both transverse and longitudinal loads). Examples of planar frames following the above definition are shown in Figure 2.40. Note that there is no restriction on the type of supports present in frames, i.e. roller supports, pinned supports, guide supports and fixed supports are allowed. The pinned support and fixed support in frames contain two and three components of the reaction, respectively; the restriction on the number of reactions as in the case of beams is now removed. Note in addition that a one-dimensional structure shown in Figure 2.40 is also classified as a frame since it is subjected to both transverse and longitudinal loads.

Figure 2.40: Schematic of some statically determinate frames Copyright © 2011 J. Rungamornrat


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As a consequence of the above definition, the internal forces at any location of the beam can be represented by three components of resultants called the axial force, the shear force and the bending moment. The last two components of the internal force are defined in the same fashion as those for the beams and the first component, i.e. the axial force, is the resultant force in the direction parallel to the axis of the member, see Figure 2.41 for clarity. Like a beam member, the axial force, the shear force and the bending moment along a member of the frame are in general not constant but vary as a function of position. Note also that the displacement and rotation at any point within the frame that contain no internal release are always continuous; for instance, the angle between any two members connected by a rigid joint is preserved and there is no gap and overlapping at any point containing no internal release after undergoing deformation.

Bending moment

Axial force Shear force

Figure 2.41: Schematic indicating three components of the internal force in planar frame

2.6.2 Sign and convention Since members of a frame can possess different orientations, it is generally impossible to find a single reference Cartesian coordinate system with one of its axes directing along the axis of all members as in the case of beams. As a result, it is common to employ two different types of reference coordinate systems, one termed a global coordinate system and the other termed a local coordinate system. The global coordinate system is a single coordinate system used as a reference of the entire structure. A choice of the global coordinate system is not unique; in particular, an orientation of the reference axes and a location of its origin can be chosen arbitrarily or for convenience. The global reference axes are labeled by X, Y, and Z with their directions following the right-hand rule. For a two-dimensional structure, the global coordinate system is typically oriented such that the Z-axis directs normal to the plane of the structure. The local coordinate system is a coordinate system defined for an individual member. The local reference axes are Copyright © 2011 J. Rungamornrat


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labeled by x, y, and z. The local coordinate system for each member is commonly defined based on the geometry and orientation of that member. Specifically, the origin is taken at one end of the member; the x-axis directs along the axis of the member; the z-axis directs normal to the plane of the structure (align with the Z-axis), and the y-axis follows from the right-hand rule. An example of global and local coordinate systems of a planar frame is shown in Figure 2.42. y

x y

Y

x

y X x Figure 2.42: Global and local coordinate systems of a planar frame. The sign convention and notations for support reactions of frames are defined in a similar fashion as those for trusses and beams with reference to the global coordinate system. For instance, reactions at the fixed support of a frame shown in Figure 2.43 are denoted by RAX, RAY and RAM where the first two symbols stand for a force reaction in the X-direction and a force reaction in the Y-direction and the last symbol stands for a moment reaction in the Z-direction, and a force reaction in the Y-direction of the roller support located at a point B is denoted by RBY. In the analysis for support reactions, it is common to assume positive support reactions in a sketch of the FBD. Once results are obtained, the actual direction of each reaction can be decided from their sign; specifically, if the computed reaction is positive, the assumed direction is correct but, if the computed reaction is negative, the actual direction is opposite to the assumed direction. Y

B RAX

A

X RAM

RBY

RAY Figure 2.43: Schematic showing sign convention and notations of support reactions of a frame The sign convention for the shear force and the bending moment for a frame member depend primarily on a choice of the local coordinate system. Once the local coordinate system is selected, the sign convention is defined in the same way as that for a beam (the local coordinate system x-y of a frame member can be viewed as the X-Y axis of a beam member): Copyright © 2011 J. Rungamornrat


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 The shear force at a particular point A is denoted by a symbol VA and the shear force as a function of position x along the member is denoted by V(x). The shear force at any point is considered to be positive if and only if it tends to rotate an infinitesimal element in the neighborhood of that point in the negative z-direction otherwise it is negative. The positive and negative shear forces are shown in Figure 2.44. y

x y

Positive shear force

x Negative shear force Figure 2.44: Schematic indicating positive and negative sign convention for shear force  The bending moment at a particular point A is denoted by a symbol MA and the bending moment as a function of position x along the member is denoted by M(x). The bending moment at any point is considered to be positive if and only if it produces a compressive stress at the top and produces a tensile stress at the bottom; otherwise it is negative. Note that the top and bottom sides of the member are defined based on the local coordinate system as shown in Figure 2.45. y Top fiber x Bottom fiber Positive bending moment y Top fiber x Bottom fiber Negative bending moment Figure 2.45: Schematic indicating positive and negative sign convention for bending moment Similar to the axial force in a truss member, the sign convention for the axial force in a frame member is defined based primarily upon the characteristic of the axial deformation, thus rendering it independent of the local coordinate system (or the member orientation). In particular, Copyright © 2011 J. Rungamornrat


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the axial force at a particular point A is denoted by a symbol FA and the axial force as a function of position x along the member is denoted by F(x). The axial force at any point is considered to be positive if and only if it produces an elongation in the neighborhood of that point or it is in tension otherwise it is negative. The positive axial force and the negative axial force are shown schematically in Figure 2.46. y x y

Positive axial force

x Negative axial force Figure 2.46: Schematic indicating positive and negative sign convention for axial force

2.6.3 Determination of support reactions Determination of support reactions of statically determinate frame follows the same procedures described in the section 2.3. For frames containing only three components of support reactions, such unknown reactions can readily be determined from equilibrium of the entire structure. For example, support reactions {RAX, RAY, RBY} of a frame shown in Figure 2.47 can be obtained as follows:  the reaction RAX is obtained from equilibrium of forces in X-direction;  the reaction RBY is obtained from equilibrium of moments about a point A; and  the reaction RAY is obtained from equilibrium of forces in Y-direction.

B

B

RBY Y

A

RAX

A RAY

Figure 2.47: Schematic of a planar frame and its FBD Copyright © 2011 J. Rungamornrat

X


118


For various statically determinate frames, more than three components of support reactions may be present; for instance, both the frame shown in Figure 2.43 and a frame shown in Figure 2.48 contain four and five components of support reactions, respectively. For this particular case, the consideration of equilibrium of the entire structure provides only three independent equations and this is insufficient to determine all unknown reactions. Since the structure is statically determinate, there must be some internal releases that supply additional conditions, when combined with existing equilibrium equations, adequate for resolving all unknowns. Such extra or additional conditions available at the internal releases are commonly set up in a form of equilibrium equations of parts of the structure resulting from introducing proper fictitious cuts.

B

C

E

D

RCY

RBY Y

RAX

A

X

RAM RAY

Figure 2.48: Schematic of a statically determinate frame containing five support reactions To clearly demonstrate the procedures, let us consider a frame shown in Figure 2.48. This structure is obviously statically determinate (i.e., ra = 5, nm = 3(3) = 9, nj = 4(3) = 12, nc = 2 → DI = 5 + 9 – 12 – 2 = 0) and this therefore ensures that all five support reactions can be obtained only from static equilibrium. To solve for all reactions {RAM, RAX, RAY, RBY, RCY}, two strategies may be used. The first strategy employs additional equilibrium equations of parts of the structure along with equilibrium of the entire structure. Specifically, we first introduce a cut at a hinge E and consider equilibrium of the right part of the structure (see FBD in Figure 2.49(a)); next, we introduce a cut at a point just to the right of a hinge D and consider equilibrium of the right part of the structure (see FBD in Figure 2.49(b)); and, finally, we consider equilibrium of the entire structure. Details of equilibrium equations employed are shown below:  the reaction RCY is obtained from equilibrium of moments about the point E of the right part of the frame shown in Figure 2.49(a);  the reaction RBY is obtained from equilibrium of moments about a point DR of the right part of the frame shown in Figure 2.49(b);  the reaction RAM is obtained from equilibrium of moments about a point A of the entire frame shown in Figure 2.49(c);  the reaction RAX is obtained from equilibrium of forces in the X-direction of the entire structure; and  the reaction RAY is obtained from equilibrium of forces in the Y-direction of the entire structure. Copyright © 2011 J. Rungamornrat


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VDR

VE FE

C

E

B

FDR

DR

D RBY

RCY (a)

C RCY

(b)

B E

D RBY

C RCY

Y RAX

A

X

RAM

RAY

(c) Figure 2.49: (a) FBD of the right part of the frame when a cut is made at hinge E, (b) FBD of the right part of the frame when a cut is made at the point just to the right of hinge D, and (c) FBD of the entire frame The second strategy employs equilibrium equations set up for all parts of the structure. Specifically, we first introduce two cuts simultaneously, one at the hinge E and the other at point just to the right of the hinge D. With these two cuts, the structure is divided into three parts whose the FBDs are shown in Figure 2.50. While we introduce four extra unknowns {FDR, VDR, FE, VE} at the cuts and the total number of unknowns becomes 5 + 4 = 9, it is equal to the number of equilibrium equations that can be set up for the three parts (3 + 3 + 3 = 9). To obtain all reactions without solving a system of nine linear equations, we can consider equilibrium of each part as follow:  the reaction RCY is obtained from equilibrium of moments about the point E of the right part of the frame shown in Figure 2.50(c), and  the axial force FE is obtained from equilibrium of forces in X-direction of the right part of the frame shown in Figure 2.50(c), and  the shear force VE is obtained from equilibrium of forces in Y-direction of the right part of the frame shown in Figure 2.50(c), and  the reaction RBY is obtained from equilibrium of moments about a point BR of the middle part of the frame shown in Figure 2.50(b), and  the axial force FDR is obtained from equilibrium of forces in X-direction of the middle part of the frame shown in Figure 2.50(b), and  the shear force VDR is obtained from equilibrium of forces in Y-direction of the middle part of the frame shown in Figure 2.50(b), and  the reaction RAM is obtained from equilibrium of moments about a point A of the left part of the frame shown in Figure 2.50(a), and  the reaction RAX is obtained from equilibrium of forces in X-direction of the left part of the frame shown in Figure 2.50(a). Copyright © 2011 J. Rungamornrat


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 the reaction RAY is obtained from equilibrium of forces in Y-direction of the left part of the frame shown in Figure 2.50(a). As be apparent, while both strategies yield identical results, one may prefer the first strategy since it is not required to solve for the intermediate unknowns {FDR, VDR, FE, VE} introduced at the cuts. VDR B D

FDR FDR VDR

DR

A

E

FE FE

C

E

RBY (b)

Y

RAX

VE

VE

RCY (c)

X RAM

RAY (a) Figure 2.50: Free body diagrams of three parts of the frame resulting from two cuts at point just to the right of hinge D and at hinge E

2.6.4 Method of sections To determine the internal forces (i.e. axial force, shear force, and bending moment) at a particular location of a statically determinate frame, the method of sections is similar to that used in the case of beams can be employed. A fictitious cut is properly made to access the internal forces at a specific location of interest and equilibrium of parts of the structure resulting from that cut is then enforced to determine all unknown internal forces. Note in particular that three unknown internal forces (i.e. axial force, shear force, and bending moment) are introduced at each cut except at the internal releases where certain components of the internal force vanish, and that three independent equilibrium equations (e.g. FX = 0; FY = 0; and MAZ = 0 or other equivalent sets) can be set up for each part resulting from the cut. Procedures for obtaining the internal force at a particular location of a frame can be summarized below (see also a frame shown in Figure 2.51 to clarify such procedures):  Determine all support reactions  Introduce a fictitious cut at a point P (point where the internal force is determined) and then separate the frame into two parts  Choose one of the two parts that seems to involve less computation  Sketch the FBD of a selected part; the positive sign convention of the internal forces follows the local coordinate system of a member containing the point P  Consider equilibrium of the selected part and this yields three independent equations. For instance, the axial force FP can directly be obtained from equilibrium of forces in the Xdirection; the shear force VP can directly be obtained from equilibrium of forces in the Ydirection; and the bending moment MP can be obtained from equilibrium of moment about the point P. Copyright © 2011 J. Rungamornrat


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y x

P

Y X B A

VP P VP

MP

FP

FP MP

P

B RAX

A RAM RAY

RBY

Figure 2.51: Schematic indicating a cut made to access the axial force, shear force and bending moment at point P and FBDs of the two parts resulting from the cut Similar to the case of beams, the method of sections can also be used to determine the axial force, the shear force and the bending moment at any point x of each frame member, i.e. F(x), V(x) and M(x). Procedures to obtain F(x), V(x) and M(x) are similar to those described above except that a cut must be made at any point x instead of a specific location. Similar to the SFD and BMD, a graph of F(x) plotted along the local x-axis of each member is known as an axial force diagram (AFD). Note however that the construction of the AFD, SFD and BMD by the method of sections is somewhat cumbersome especially when there are many points of loading discontinuity and points where members change their direction.

2.6.5 Method of differential and integral formula An alternative technique to the method of sections for the construction of the AFD, SFD and BMD of a frame is the method of differential and integral formula. This technique is based mainly upon two sets of equations, one associated with equilibrium equations in a differential form and the other corresponding to equilibrium equations in an integral form, and some special discontinuity conditions at points of loading discontinuity. Note that this technique is similar in accord to that presented in the section 2.5.5 for beams except that one equation associated with equilibrium of forces in the direction of the member axis is added due to the presence of the axial force. Copyright © 2011 J. Rungamornrat


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2.6.5.1 Equilibrium equations in differential form Consider a frame that is in equilibrium with applied loads as shown schematically in Figure 2.52(a). Let us focus on a particular member (with a local coordinate system x-y) and introduce following two cuts, one at the local coordinate x and the other at the local coordinate x + dx. An infinitesimal element dx is then separated from the structure and its FBD is shown in Figure 2.52(b). The axial force, shear force and bending moment at x are denoted by F(x), V(x) and M(x), respectively; the axial, shear force and bending moment at x + dx are denoted by F(x) + dF, V(x) + dV and M(x) + dM, respectively, where dF, dV and dM are increments of axial force, shear force and bending moment; and the distributed transverse and longitudinal loads are denoted by qy and qx, respectively. Note that all components of the internal forces follow the sign convention defined in the section 2.6.2 while the distributed loads qy and qx are considered to be positive if their direction is along the y-axis and x-axis, respectively. x

qy

qx

y

F+dF

qy M

dx

qx V

F

x

M+dM V+dV dx

(b)

(a)

Figure 2.52: (a) Schematic of a frame subjected to applied loads and (b) FBD of an infinitesimal element dx By enforcing static equilibrium of the infinitesimal element dx shown in Figure 2.52(b) and then taking appropriate limit process, we obtain the following three equilibrium equations in a differential form: ΣFx  0



dF(x) dx



 q x (x)

(2.28)

ΣFy  0



dV(x) dx



q y (x)

(2.29)

ΣM z  0



dM(x) dx



V(x)

(2.30)

It is important to emphasize that the equation (2.28) is valid at any point x where the distributed longitudinal load qx is continuous and it is free of a longitudinal concentrated force; the equation (2.29) is valid at any point x where the distributed transverse load qy is continuous and it is free of a transverse concentrated force; and the equation (2.30) is valid at any point x such that the shear force is continuous and it is free of a concentrated moment. Copyright © 2011 J. Rungamornrat


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The first equation (2.28) implies that the spatial rate of change of the axial force or the “slope of the axial force diagram” at any point is equal to the negative distributed longitudinal load –qx acting to that point. The second equation (2.29) implies that the spatial rate of change of the shear force or the “slope of the shear force diagram” at any point is equal to the distributed transverse load qy acting that point. The last equation (2.30) implies that the spatial rate of change of the bending moment or the “slope of the bending moment diagram” at any point is equal to the shear force at that point. Similar to the case of beams, the three differential relations (2.28)-(2.30) are very useful for identifying the type of curve connecting any two points in the AFD, the SFD and the BMD (see the section 2.5.5.1 for more details on some specific types of curves).

2.6.5.2 Equilibrium equations in integral form Now, let A and B be any two points within a frame member and let xA and xB be their x-coordinates with respect to the local coordinate system of the member. By directly integrating equation (2.28) from the point A to the point B, we obtain the integral formula FB  FA 

xB

q

x

dx  FA  Qlong AB

(2.31)

xA

where FA and FB are the axial force at the point A and the point B, respectively, and Q long AB is the sum of distributed longitudinal load qx over the segment AB. This equation implies that the axial force at the point B can be obtained by subtracting the total longitudinal load over the segment AB to the axial force at the point A. Note that the equation (2.31) is valid if the segment AB is free of concentrated longitudinal force and that the sign convention of the total load Q long AB follows that of the distributed load qx. Similarly, by directly integrating equation (2.29) from the point A to the point B, we obtain the integral formula VB  VA 

xB

q

y

tran dx  VA  Q AB

(2.32)

xA

tran where VA and VB are the shear force at the point A and the point B, respectively and Q AB is the sum of all distributed transverse load qy over the segment AB. This equation implies that the shear force at the point B can be obtained by adding the total transverse load over the segment AB to the shear force at the point A. Note that the equation (2.32) is valid if the segment AB is free of concentrated tran transverse force and that the sign convention of the total load Q AB follows that of the distributed load qy. Finally, by directly integrating equation (2.30) from the point A to the point B, we obtain the integral formula

MB  MA 

xB

 V dx  M

A

 AreaVAB

(2.33)

xA

where MA and MB are the bending moment at the point A and the point B, respectively and AreaVAB denotes the area of the shear force diagram over the segment AB. This equation implies that the bending moment at the point B can be obtained by adding the area of the shear force diagram over the portion AB to the bending moment at the point A. Note that the equation (2.33) is Copyright © 2011 J. Rungamornrat


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valid if the segment AB is free of concentrated moment and that the sign convention of the area AreaVAB follows that of the shear force.

2.6.5.3 Discontinuity of axial force, shear force and bending moment As already pointed out, the differential formula (2.28)-(2.30) and the integral formula (2.31)-(2.33) are not valid for points of loading discontinuity or segments that contain those points, e.g. points where the concentrated force is applied, points where the concentrated moment is applied, and points where the distributed load is discontinuous. Knowledge of discontinuity conditions at such points is required in the construction of the AFD, SFD, and BMD. First, let us investigate the discontinuity condition at the location where a concentrated longitudinal force is applied. Let Po be such a concentrated force applied to a point A of a particular frame member (it is emphasized again that this force is considered to be positive if it directs in the local x-direction of the member otherwise it is negative). By introducing two cuts at a point just to left and a point just to the right of the point A and then considering equilibrium of a resulting infinitesimal element (its free body diagram is shown in Figure 2.53 along with taking appropriate limit process, we obtain the discontinuity conditions of the axial force, shear force and bending moment: x

y

FAL

FAR MAR Po A VAR VAL MAL

Figure 2.53: FBD of infinitesimal element containing a point where concentrated longitudinal force is applied FAR  FAL  Po

(2.34)

VAR  VAL

(2.35)

M AR  M AL

(2.36)

where {FAR, VAR, MAR} and {FAL, VAL, MAL} are the axial force, shear force and bending moment at a point just to the right and a point just to the left of the point A, respectively. It is evident from (2.34)-(2.36) that, at the location where a concentrated longitudinal force is applied, the axial force is discontinuous with the magnitude of the jump equal to the magnitude of the concentrated longitudinal force while the shear force and the bending moment are still continuous. In particular, the axial force experiences a positive jump if the concentrated longitudinal force is negative (or directs along the opposite local x-direction) and it experiences a negative jump if the concentrated longitudinal force is positive (or directs along the local x-direction). Next, let us consider the discontinuity conditions at a location where a concentrated transverse force is applied. Let Po be such a concentrated force applied to a point A of a particular frame member (it is emphasized again that this force is considered to be positive if it directs in the local y-direction of the member otherwise it is negative). By introducing two cuts at a point just to the left and a point just to the right of the point A and then considering equilibrium of a resulting Copyright © 2011 J. Rungamornrat


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infinitesimal element (its free body diagram is shown in Figure 2.54) along with taking appropriate limit process, we obtain the discontinuity conditions of the axial force, shear force and bending moment: x

Po

y

A VAL

FAL

FAR MAR VAR

MAL

Figure 2.54: FBD of infinitesimal element containing a point where concentrated transverse force is applied FAR  FAL

(2.37)

VAR  VAL  Po

(2.38)

M AR  M AL

(2.39)

Equations (2.37)-(2.39) imply that, at the location where a concentrated transverse force is applied, the shear force is discontinuous with the magnitude of the jump equal to the magnitude of the concentrated transverse force while the axial force and the bending moment are still continuous. Unlike the previous case, the shear force experiences a positive jump if the concentrated transverse force is positive (or directs along the local y-direction) and it experiences a negative jump if the concentrated transverse force is negative (or directs along the opposite local y-direction). Next, let us consider the discontinuity conditions at a location where a concentrated moment Mo is applied. Let Mo be a concentrated moment applied to a point A (this moment is considered to be positive if it directs to a counter clockwise direction or to the local z-direction of the member otherwise it is negative). By introducing two cuts at a point just to the left and a point just to the right of the point A and then considering equilibrium of a resulting infinitesimal element (its free body diagram is shown in Figure 2.55) along with taking appropriate limit process, we obtain the discontinuity conditions of the axial force, shear force and bending moment: FAR  FAL

(2.40)

VAR  VAL

(2.41)

M AR  M AL  M o

(2.42)

Equations (2.40)-(2.42) imply that at a location where the concentrated moment is applied, the bending moment is discontinuous with the magnitude of the jump equal to the magnitude of the concentrated moment while the axial force and the shear force are still continuous. In particular, the bending moment experiences a positive jump if the concentrated moment is negative (or directs to a clockwise direction or the local z-direction of the member) and it experiences a negative jump if the concentrated moment is positive (or directs to a counter clockwise direction or opposite Zdirection). Copyright © 2011 J. Rungamornrat


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x

Mo y

A VAL

FAR MAR VAR

MAL FAL

Figure 2.55: FBD of infinitesimal element containing a point where concentrated moment is applied Finally, we state without proof that points of loading discontinuity such as points where the distributed longitudinal or transverse load is discontinuous and points where both the distributed loads are continuous but change their distribution do not produce the discontinuity in the axial force, the shear force and the bending moment, i.e. FAR  FAL

(2.43)

VAR  VAL

(2.44)

M AR  M AL

(2.45)

where A denotes a point where the distributed load is discontinuous or change its distribution.

2.6.5.4 Procedures for constructing AFD, SFD and BMD The differential formula (2.28)-(2.30), the integral formula (2.31)-(2.33) and the discontinuity conditions (2.34)-(2.45) are basic components essential for constructing the AFD, SFD and BMD of a frame. In particular, the three integral formula (2.31)-(2.33) are employed to obtain the axial force, the shear force and the bending moment at the right end of any segment when values of those quantities at the left end are known and there is no point of loading discontinuity within the segment. The three differential formula (2.28)-(2.30) are then used to identify the type of a curve that connects a part of the AFD, SFD and BMD over a segment where values of the axial force, shear force and bending moment are already known at its ends. The discontinuity conditions are used to dictate the jump of the shear force and bending moment in the AFD, SFD and BMD where the concentrated forces and moments are present. Here, we summarize standard procedures or guidelines for constructing the AFD, SFD and BMD of a frame.  Determine all support reactions  Identify and mark points of loading discontinuity, e.g. supports, points where concentrated forces and moments are applied, and points where distributed load changes its distribution  Identify and mark points where members change their orientation  Separate a given frame into several members using points where members change their orientation  Identify all possible segments within each member such that points of loading discontinuity must be at the ends of each segment  Identify a point that the axial force, the shear force and the bending moment are known (in general, a point containing only one member is chosen since all forces and moments at that point are always known once the reactions are already determined.) Copyright © 2011 J. Rungamornrat


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 Draw the SFD as follow: (i) start with a member where the shear force is known at one of its ends, (ii) use the differential formula (2.29), the integral formula (2.32) and the discontinuity conditions (2.35), (2.38), (2.41) and (2.44) to construct the SFD over each segment in the selected member (procedures are similar to those used for the case of beams), (iii) choose the next member where the shear force is known at one end and then follow step (ii), and (iv) repeat step (iii) until all members are considered  Draw the BFD as follow: (i) start with a member where the bending moment is known at one of its ends, (ii) use the differential formula (2.30), the integral formula (2.33) and the discontinuity conditions (2.36), (2.39), (2.42) and (2.45) to construct the BMD over each segment in the selected frame member (procedures are similar to those used for the case of beams), (iii) choose the next frame member where the bending moment is known at one end and then follow step (ii), and (iv) repeat step (iii) until all members are considered  Draw the AFD as follow: (i) start with a member where the axial force is known at one of its ends, (ii) use the differential formula (2.28), the integral formula (2.31) and the discontinuity conditions (2.34), (2.37), (2.40) and (2.43) to construct the AFD over each segment in the selected frame member, (iii) choose the next frame member where the axial force is known at one end and then follow step (ii), and (iv) repeat step (iii) until all members are considered Once the AFD, SFD and BMD are completed, one can identify both the magnitude and location of the maximum axial force, maximum shear force and maximum bending moment. In general, the maximum axial force and shear force can occur at the supports, the locations where the distributed load q vanishes, the locations where the concentrated forces are applied, and the locations where members change their orientation. Similarly, the maximum bending moment can occur at supports, locations where the shear force vanishes, locations where the shear force changes its sign, locations where the concentrated moments are applied, and locations where members change their orientation.

2.6.6 Qualitative Elastic Curve In this section, we demonstrate how to sketch a qualitative elastic curve or deformed curve of a frame under applied loads once the bending moment diagram is constructed. The key assumptions employed are those associated with Euler-Bernoulli beam theory utilized in the sketch of an elastic curve of beams; i.e. (i) frame is made of a linearly elastic material; (ii) plane section remains plane before and after undergoing deformation; (iii) no shear deformation; and (iv) no axial deformation. According to a kinematics assumption of deformation of the cross section, it is sufficient to represent any frame member by their axis and, therefore, the elastic or deformed curve is the deformed configuration of such axis. Let u(x), v(x) and (x) be the longitudinal component of the displacement, transverse component of the displacement, and the rotation at any point x within a frame member as shown in Figure 2.56. The displacement u(x) and v(x) are considered to be positive if they direct to the positive local x-direction and local y-direction, respectively, and the rotation (x) is considered to be positive if it directs to a counter clockwise direction or the local z-direction. By assuming that the displacement and the rotation are infinitesimally small in comparison with the characteristic length of the frame, the transverse displacement v(x), the rotation (x), and the curvature (x) are related through the relations θ(x) 

dv dx



128


(x) y

v(x)

x

u(x)

x

Figure 2.56: Schematics indicating the longitudinal and transverse displacement and the rotation at any point x within a frame member κ(x) 

dθ d 2 v  dx dx 2

(2.47)

It is evident from the definition (2.47) that the curvature (x) is positive if the transverse displacement is concave upward with respect to the local coordinate system {x, y, z}, negative if the displacement is concave downward, and zero if it is an inflection point. In addition, a direct consequence of the small displacement and rotation assumption and the kinematic assumption (iv) leads to that the longitudinal displacement is constant for the entire member or, equivalently, the projection of the deformed curve to the undeformed axis possesses the same length as that of the undeformed member. By considering the deformation of the cross section, employing material constitutive, and computing the moment resultant of the cross section, it leads to a well-known moment-curvature relationship κ(x) 

M(x) EI

(2.48)

where E is Young’s modulus and I is the moment of inertia of the cross section. It can be deduced from the relations (2.47) and (2.48) that  A segment of a frame possessing the positive bending moment undergoes a positive curvature and, as a result, leading to a concave upward elastic curve;  A segment of a frame possessing the negative bending moment undergoes a negative curvature and, as a result, leading to a concave downward elastic curve;  A segment of a frame possessing the zero bending moment undergoes a zero curvature and, as a result, leading to a straight-line elastic curve; and  A point within a frame where the bending moment changes sign at that point is an inflection point on the elastic curve. To sketch the qualitative elastic curve, the following procedures can be used:  Construct BMD for the entire beam  Use equation (2.48) to identify the shape of elastic curve at any segment of the frame Copyright © 2011 J. Rungamornrat


129


 Patch all segments of elastic curve together  Check compatibility with all supports and internal releases. It is important to emphasize that the longitudinal displacement at any point is continuous except at the axial release, that the transverse displacement at any point is continuous except at the shear release, and that the rotation at any point is continuous except at the moment release (hinge). Schematics shown in Figure 2.57 indicate the deformation of a segment with no internal release and the discontinuity induced at the axial release, shear release and the moment release.

 

 

Figure 2.57: Deformation of a segment with no internal release and the discontinuity induced at the axial release, shear release and the moment release Example 2.15 Determine all support reactions and draw AFD, SFD, BMD and elastic curve of a frame shown below Y q 2

3qL qL

D

C

L qL

B

L A

X 2L

Solution The given frame is statically determinate (i.e. ra = 2 + 1 = 3, nm = 2(3) = 6, nj = 3(3) = 9, nc = 0, then DI = 3 + 6 – 9 – 0 = 0); thus, all support reactions and the internal forces at any location can be determined from static equilibrium. Since the number of support reactions is equal to 3, they Copyright © 2011 J. Rungamornrat


130


can be obtained from equilibrium of the entire structure. The FBD of the entire structure and details of calculation are shown below: x

Y

FC MC

q 2

3qL qL

C

D

C

VC

RDY qL

B

RAX

A

y

X

RAY [MA = 0]  

:

RAX

A

2RDYL – (qL)(L) – (qL)(2L) – (2qL)(L) + 3qL2 = 0

: :

Upward

RAX + qL + qL = 0 RAX = –2qL

[FY = 0]  

B

RAY

RDY = qL [FX = 0]  

qL

Leftward

RAY + RDY – 2qL = 0 RAY = qL

Upward

For the given frame, there is only one point where members change their orientation, i.e. a point C; therefore, only two frame members, a member AC and a member CD, are considered. For the member AC, there is only one point of loading discontinuity (excluding the two ends of the member), i.e. a point B where a concentrated transverse force is applied, while the member CD contains no point of loading discontinuity. Since all support reactions are already determined, the axial force, shear force and bending moment at the point A are already known. First, let us construct the AFD, SFD and BMD of the member AC. The differential formula (2.28)-(2.30), the integral formula (2.31)-(2.33) and the discontinuity conditions (2.34)-(2.45) are utilized. The local coordinate system for this particular member is shown in above figure.  AFD - No point where the concentrated longitudinal force is applied + no longitudinal distributed load being applied for the entire member  it is sufficient to consider only one segment AC - FA = –RAY = –qL - No longitudinal distributed load being applied for the entire member + equation (2.31)  FC = FA + 0 = –qL - No longitudinal distributed load being applied for the entire member + equation (2.28)  AFD over the member is a horizontal straight line Copyright © 2011 J. Rungamornrat


131


 SFD - A concentrated transverse force is applied at point B  member AC is divided into two segments, AB and BC - VA = –RAX = 2qL - There is no transverse distributed load over the segment AB + equation (2.32)  VBL = VA + 0 = 2qL - There is no transverse distributed load over the segment AB + equation (2.29)  SFD over the segment AB is a horizontal straight line - The negative concentrated transverse force is applied at point B + equation (2.38)  there is a jump of the shear force at point B  VBR = VBL – qL = qL - There is no transverse distributed load over the segment BC + equation (2.32)  VC = VBR + 0 = qL - There is no transverse distributed load over the segment BC + equation (2.29)  SFD over the segment BC is a horizontal straight line  BMD - No point where the concentrated moment is applied + SFD over the member AC  it is sufficient to consider only two segments, AB and BC - MA = 0 - Area of the SFD over the segment AB is (2qL)(L) + equation (2.33)  MBL = MA + (2qL)(L) = 2qL2 - The shear force is constant and positive over the segment AB + equation (2.30)  BMD over the segment is a rising straight line - There is no concentrated moment applied at point B  there is no jump of the bending moment at point B  MBR = MBL = 2qL2 - Area of the SFD over the segment BC is (qL)(L) + equation (2.33)  MC = MBR + (qL)(L) = 3qL2 - The shear force is constant and positive over the segment BC + equation (2.30)  BMD over the segment is a rising straight line Next, let us consider the member CD. Once the axial force, shear force and bending moment at the point C of the member AC are determined, the axial force, shear force and bending moment at the point C of the member CD can readily be obtained. The FBD of the member CD and the corresponding local coordinate system are shown in the figure below. y q 2

3qL qL C qL 3qL2

D

x

RDY

qL  AFD - No point where the concentrated longitudinal force is applied + no longitudinal distributed load being applied for the entire member  it is sufficient to consider only one segment CD - FC = qL – qL = 0 - No longitudinal distributed load being applied for the entire member + equation (2.31)  FD = FC + 0 = 0  consistent with condition at the point D Copyright © 2011 J. Rungamornrat


132


-

No longitudinal distributed load being applied for the entire member + equation (2.28)  AFD over the member is a horizontal straight line  SFD - No point where the concentrated transverse force is applied + the distributed transverse load is continuous for the entire member  it is sufficient to consider only one segment CD - VC = qL - A negative uniform distributed transverse load is applied over the segment CD + equation (2.32)  VD = VC + (–q)(2L) = –qL  consistent with condition at the point D - A negative uniform distributed load is applied over the segment CD + equation (2.29)  SFD over the segment AB is a dropping straight line  BMD - No point where the concentrated moment is applied + SFD over the member CD  it is sufficient to consider only two segments CE and ED where E is the mid point of the segment CD - MC = 3qL2 – 3qL2 = 0 - Area of the SFD over the segment CE is (qL)(L)/2 + equation (2.33)  MEL = MC + (qL)(L)/2 = qL2/2 - The shear force is positive and decreases monotonically in magnitude over a segment CE + equation (2.30)  BMD over the segment is a rising and concave downward curve - There is no concentrated moment applied at point E  there is a jump of the bending moment at point E  MER = MEL = qL2/2 - Area of the SFD over the segment ED is (–qL)(L)/2 + equation (2.33)  MD = MER + (–qL)(L)/2 = 0  consistent with condition at the point D - The shear force is negative and increases monotonically in magnitude over a segment ED + equation (2.30)  BMD over the segment is a dropping and concave downward curve The AFD, SFD, and BMD of the entire frame are shown in the figure below. The maximum axial force, shear force and maximum bending moment and the location where they occur are summarized as follow: for a member AC, the maximum negative axial force is equal to qL occurring at the entire segment AC, the maximum positive shear is equal to 2qL occurring at the entire segment AB, and the maximum positive bending moment is equal to 3qL2 occurring at a point C; for the member CD, the maximum positive shear force is equal to qL occurring at a point C, the maximum negative shear force is equal to qL occurring at point D, and the maximum positive bending moment is equal to qL2/2 occurring at a point E. From movement constraints provided by roller and pinned supports and the BMD shown below, we obtain following information that is useful for sketching an elastic curve:  Point A: pinned support  there is no vertical and horizontal displacements at this point  Point C: rigid joint  both the displacement and rotation are continuous at this point  Point D: roller support  there is no vertical displacement at this point while the horizontal displacement and rotation are allowed  Segment AB: bending moment is positive  the elastic curve of this segment must be concave upward  Segment BC: bending moment is positive  the elastic curve of this segment must be concave upward  Segment CD: bending moment is positive  the elastic curve of this segment must be concave upward Copyright © 2011 J. Rungamornrat


133


 Length constraint of member AC: the vertical displacement at point C must vanish, i.e. vC = 0  Length constraint of member CD: the horizontal displacement at point C and point D must be identical, i.e. uC = uD

0

AFD

qL SFD qL2/2

–qL BMD

q 2

3qL qL

3qL2 –qL

D

C

RDY

qL 2qL2

qL

B

RAX

A

2qL

AFD

SFD

BMD

RAY uC C

A


uD C

D

D


134


Example 2.16 Determine all support reactions and draw AFD, SFD, BMD and elastic curve of the statically determinate frame structure shown below. q D

C L/2

L/3

2

qL qL

E

B

Y

2L/3

L/2 A

F L

qL

X

L

Solution The given frame is statically determinate (i.e. ra = 3 + 1 = 4, nm = 3(3) = 9, nj = 4(3) = 12, nc = 1, then DI = 4 + 9 – 12 – 1 = 0); thus, all support reactions and the internal forces at any location can be determined from static equilibrium. However, the number of independent equilibrium equations that can be set up for the entire frame is net = 3 < ra; thus, the support reactions cannot be obtained from equilibrium of the entire structure. To overcome this problem, an additional equation associated with the presence of a hinge at point C, i.e. M = 0 at C, must be employed. By introducing a cut at the point C and employing moment equilibrium of the right part of the frame, the reaction RFY can readily be determined and, by considering equilibrium of the entire frame, the rest of reactions can be computed. Details of calculations are shown below: q

q Y

FC VC

D

C

2

qL

qL

E

F

X qL

RFY

RAX RAM

qL

F

RAY FBD of entire frame

RFYL + (qL)(L) + qL2 – (qL)(L/2) = 0 RFY = –3qL/2 Downward Copyright © 2011 J. Rungamornrat

qL RFY

Equilibrium of portion CDF :

E

B

A

FBD of portion CDF

[MC = 0]  

D

C

2


135


Equilibrium of entire frame [FX = 0]  

:

RAX + qL + qL = 0 RAX = –2qL

[MA = 0]  

:

RAM + 2RFYL + qL2 – (qL)(3L/2) – (qL)(L/2) = 0 RAM = 4qL2

[FY = 0]  

:

Leftward CCW

RAY + RFY – qL = 0 RAY = 5qL/2 Upward

For the given frame, there are only two points where members change their orientation, i.e. a points C and D; therefore, only three frame members, a member AC, a member CD and a member DF, are considered. In particular, the member AC contains one point of loading discontinuity (i.e. a point B where a concentrated force is applied); the member CD contains no point of loading discontinuity; and the member DF one point of loading discontinuity (i.e. a point E where a concentrated moment is applied). Since all support reactions are already determined, the axial force, shear force and bending moment at the point A are already known. First, let us construct the AFD, SFD and BMD of the member AC. The local coordinate system and the FBD for this particular member are shown in the figure below. x C

y RAX RAM

qL

FC MC VC

B

A RAY

 AFD - A concentrated longitudinal force is applied at point B  member AC is divided into two segments, AB and BC - FA = –RAXcos45o – RAYsin45o = –√2qL/4 - No longitudinal distributed load being applied to the segment AB + equation (2.31)  FBL = FA + 0 = –√2qL/4 - No longitudinal distributed load being applied to the segment AB + equation (2.28)  AFD over the segment AB is a horizontal straight line - The positive concentrated longitudinal force is applied at point B + equation (2.34) o  there is a jump of the axial force at point B  FBR = FBL – qLcos45 = –3√2qL/4 - No longitudinal distributed load being applied to the segment BC + equation (2.31)  FC = FBR + 0 = –3√2qL/4 - No longitudinal distributed load being applied to the segment BC + equation (2.28)  AFD over the segment BC is a horizontal straight line Copyright © 2011 J. Rungamornrat


136


 SFD - A concentrated transverse force is applied at point B  member AC is divided into two segments, AB and BC - VA = –RAXsin45o + RAYcos45o = 9√2qL/4 - There is no transverse distributed load over the segment AB + equation (2.32)  VBL = VA + 0 = 9√2qL/4 - There is no transverse distributed load over the segment AB + equation (2.29)  SFD over the segment AB is a horizontal straight line - The negative concentrated transverse force is applied at point B + equation (2.38)  there is a jump of the shear force at point B  VBR = VBL – qLsin45o = 7√2qL/4 - There is no transverse distributed load over the segment BC + equation (2.32)  VC = VBR + 0 = 7√2qL/4 - There is no transverse distributed load over the segment BC + equation (2.29)  SFD over the segment BC is a horizontal straight line  BMD - No point where the concentrated moment is applied + SFD over the member AC  it is sufficient to consider only two segments, AB and BC - MA = –RAM = –4qL2 - Area of the SFD over the segment AB is (9√2qL/4)(√2L/2) + equation (2.33)  MBL = MA + (9√2qL/4)(√2L/2) = –7qL2/4 - The shear force is constant and positive over the segment AB + equation (2.30)  BMD over the segment is a rising straight line - There is no concentrated moment applied at point B  there is no jump of the bending moment at point B  MBR = MBL = –7qL2/4 - Area of the SFD over the segment BC is (7√2qL/4)(√2L/2) + equation (2.33)  MC = MBR + (7√2qL/4)(√2L/2) = 0 - The shear force is constant and positive over the segment BC + equation (2.30)  BMD over the segment is a rising straight line Next, let us construct the AFD, SFD and BMD of the member CD. The local coordinate system and the FBD for this particular member are shown in the figure below. y

7√2qL/4

–3√2qL/4

C

D

0

FD VD

x

MD

 AFD - No point of loading discontinuity within the member  it is sufficient to consider only one segment - FC = (7√2qL/4)cos45o – (3√2qL/4)sin45o = qL - No longitudinal distributed load being applied to the member CD + equation (2.31)  FD = FC + 0 = qL - No longitudinal distributed load being applied to the member CD + equation (2.28)  AFD over the segment CD is a horizontal straight line Copyright © 2011 J. Rungamornrat


137


 SFD - No point of loading discontinuity within the member  it is sufficient to consider only one segment - VC = (7√2qL/4)sin45o + (3√2qL/4)cos45o = 5qL/2 - A negative uniform distributed transverse load is applied over the member CD + equation (2.32)  VD = VC – (q)(L) = 3qL/2 - A negative uniform distributed transverse load is applied over the member CD + equation (2.29)  SFD over the member CD is a dropping straight line  BMD - No point of loading discontinuity within the member + SFD over the member CD  it is sufficient to consider only one segment - MC = 0 - Area of the SFD over the segment AB is (5qL/2+3qL/2)(L/2) + equation (2.33)  MD = MC + (5qL/2+3qL/2)(L/2) = 2qL2 - The shear force is positive and decreases monotonically in magnitude over the member CD + equation (2.30)  BMD over the segment is a rising and concave downward curve Finally, let us construct the AFD, SFD and BMD of the member DF. The local coordinate system and the FBD for this particular member are shown in the figure below. 3qL/2 D

qL

y

2qL2 qL2

E

qL RFY x  AFD - No point of longitudinal loading discontinuity within the member  it is sufficient to consider only one segment - FD = 3qL/2 - No longitudinal distributed load being applied to the member DF + equation (2.31)  FF = FD + 0 = 3qL/2  consistent with condition at the point F - No longitudinal distributed load being applied to the member DF + equation (2.28)  AFD over the segment DF is a horizontal straight line  SFD - No point of transverse loading discontinuity within the member  it is sufficient to consider only one segment - VD = –qL - No transverse distributed load being applied to the member DF + equation (2.32)  VF = VD + 0 = –qL  consistent with condition at the point F Copyright © 2011 J. Rungamornrat


138


No transverse distributed load being applied to the member DF + equation (2.29)  SFD over the member DF is a horizontal straight line  BMD - A concentrated moment is applied at point E + SFD over the member DF  the member DF is divided into two segments, DE and EF - MD = 2qL2 - Area of the SFD over the segment DE is (–qL)(L/3) + equation (2.33)  MEL = MD + (–qL)(L/3) = 5qL2/3 - The shear force is constant and negative over the segment DE + equation (2.30)  BMD over the segment is a dropping straight line - A positive concentrated moment is applied at point E  there is a jump of the bending moment at point E  MER = MEL – qL2 = 2qL2/3 - Area of the SFD over the segment EF is (–qL)(2L/3) + equation (2.33)  MF = MER + (–qL)(2L/3) = 0  consistent with condition at the point F - The shear force is constant and negative over the segment EF + equation (2.30)  BMD over the segment is a dropping straight line -

qL

AFD

A FD

5qL/2

3qL/2

BMD

2q

7

2q L/ 4

BM D

2q 3 L/ 4

2qL2

SF D

L/ 4

SFD

2q



L/ 4

q D

C

7

9

qL 2 /4

2qL2 2

B

E

3qL/2

5qL2/3 2qL2/3

4

qL 2

qL

qL

–qL

RAX RAM

F qL

A

RFY

RAY

BMD

SFD AFD

From movement constraints provided by roller and fixed supports, a moment release and the BMD shown below, we obtain following information that is useful for sketching an elastic curve:  Point A: fixed support  there is no vertical and horizontal displacements and rotation at this point  Point C: hinge joint  the rotation is discontinuous at this point while the displacement is continuous  Point D: rigid joint  both the displacement and rotation are continuous at this point  Point F: roller support  there is no vertical displacement at this point while the horizontal displacement and rotation are allowed Copyright © 2011 J. Rungamornrat


139


 Segment ABC: bending moment is negative  the elastic curve of this segment must be concave downward  Segment CD: bending moment is positive  the elastic curve of this segment must be concave upward  Segment DEF: bending moment is positive  the elastic curve of this segment must be concave upward  Length constraint of member AC: the vertical displacement and the horizontal displacement at point C must be identical, i.e. uC = vC  Length constraint of member CD: the horizontal displacement at point C and point D must be identical, i.e. uC = uD  Length constraint of member DF: the vertical displacement at point D must vanish, i.e. vD = 0

vC

uC

uD

C

D D C

A

F

F

Exercises 1. Show that structures shown below are externally statically determinate. Sketch free body diagram (FBD) of these structures and then apply static equilibrium equations to determine all support reactions. 2P

P

L/2

2P

L

L/2 P

3P 2P L/2

L/2 P

L

L/2

L/2 2P

L

P

2PL

L

L Copyright © 2011 J. Rungamornrat

L/2

L

L/2

L/2 q

L

2L


140


q 2qL

D

L

L

2q C

B

qL

qL L

L A

L

L

L

L

2. For truss structures shown below, you are required to (i) show that they are statically determinate, (ii) sketch FBD and then determine support reactions, (iii) identify zero member forces (if they exist), and (iv) determine the remaining forces using either the method of joints or method of sections. P

2P

2P L

L

4P L

L

L/2

L

L/2 L/2

L/2

L/2

L/2

4P

P L

3L 4P

2P

P L

L

L

3L L

4P 3L

L 2P

3L

L 3P L

L

L

3P L

4L Copyright © 2011 J. Rungamornrat

4L


141


3. For beam structures shown below, you are required to (i) show that they are statically determinate, (ii) sketch FBD and then determine all support reactions, (iii) sketch shear force and bending moment diagrams, (iv) identify the maximum shear force and bending moment, and (v) sketch qualitative elastic curve. q

qL2

L

L

L/2

qL

q

qL

qL2/2

L

L/2

L

L

qL 2qL

q

L

L

qL qL2

L

L

L/2

q

L

L/2

2qL

2q

L/2 L/2

L

qL

3qL2

L

L

2qL 2q

qL

2

2qL

L

L

L

L

L

4. For frame structures shown below, you are required to (i) show that they are statically determinate, (ii) sketch FBD and then determine all support reactions, (iii) sketch axial force, shear force and bending moment diagrams, (iv) identify the maximum axial force, shear force and bending moment, and (v) sketch qualitative elastic curve. q 2qL L

L

qL qL2

L

qL

qL2

2qL L

q

L


2L


142


q

qL 2qL q

qL

2L

2L

3L

L

qL2

L

L

qL 2qL L 3L

qL2

qL

L

2qL2 q

L

L

L

2L

2L

2L

2L

P P 2L

2L

L

L

2qL

2L

2L 3qL2

P

P 2L

2L P

2L


143


Direct Integration Method

CHAPTER 3 DIRECT INTEGRATION METHOD This chapter devotes to a classical method, called a direct integration method, commonly used to determine both the deflection and rotation of beams. The key idea of this approach is to combine three basic components in structural mechanics (i.e. kinematics, constitutive law, and equilibrium equation) to form a differential equation governing behavior of the beam. This equation is in fact an equilibrium equation formulated in terms of the deflection and rotation of the beam. This differential equation, when supplied by proper and sufficient boundary conditions at both ends of the beam, constitutes a complete boundary value problem for a particular beam. Due to a special form of the governing differential equation, the rotation and deflection of the beam can simply be obtained via a direct integration technique (the name direct integration method results directly from this solution strategy). Boundary conditions prescribed at both ends of the beam are then imposed to uniquely determine arbitrary constants resulting from the integration. Following sections present the development of governing differential equations, treatment of various boundary conditions, treatment of data discontinuity, and finally applications of the direct integration method to analyze various beam problems.

3.1 Basic Equations In this chapter, we focus attention on a beam structure that is made from a linearly elastic material whose constitutive behavior is completely characterized by a single material parameter termed the Young’s modulus E (this parameter can be obtained via conducting proper laboratory experiments). In the development of differential equations governing behavior of such beam, we follow EulerBernoulli beam theory. More precisely, this theory is based on following key assumptions: (i) beam is made from a linearly elastic material whose properties are uniform across the section, (ii) a plane section remains plane after undergoing deformation, (iii) shear deformation is negligible, (iv) the rotation of the beam is relatively small, and (v) equilibrium equation is set up in the undeformed state. These assumptions play a central role in derivation presented below. Y dx* q = q(x) Mo a

A x

B

C

Po D

X

dx

Figure 3.1: Schematic of a beam subjected to a set of transverse loads Consider a beam of length L occupying a line defined by x = 0 and y  [a, a + L] as shown schematically in Figure 3.1. Note that a constant a, which defines a coordinate of the left end of the beam, can be chosen arbitrarily as a matter of preference. Without loss of generality, we may choose a = 0 and, as a result, the left end of the beam is located at x = 0 while its right end is located at x = L. The beam is subjected to a set of transverse loads as illustrated in Figure 3.1; this set of Copyright © 2011 J. Rungamornrat

144



applied loads may consist of a distributed transverse force q = q(x) and concentrated forces and moments acting at certain points; q is positive if it directs upward or in Y-direction, otherwise it is negative. Under these actions, the beam moves to a new deformed state as indicated by a red dash line. More specifically, a point occupying the coordinate (x, 0) in the undeformed configuration displaces to a point occupying the coordinate (x + u, v) in the deformed configuration where u = u(x) and v = v(x) denote the longitudinal displacement and transverse displacement at point x, respectively. Let V = V(x) and M = M(x) denote the shear force and bending moment at the cross section located at any point x.

3.1.1 Kinematics Y

(x + u, v)

dx* 

(x + dx u + du, v + dv)

dx (x, 0)

(x + dx, 0)

X

Figure 3.2: Schematic of undeformed and deformed infinitesimal elements Let ds be an infinitesimal element connecting a point (x, 0) to a neighboring point (x + dx, 0) in the undeformed configuration and dx* be the same infinitesimal element in the deformed configuration as shown in Figure 3.2. In particular, dx* is a curve element connecting a point (x + u, v) to a point (x + dx + u + du, v + dv) in the deformed configuration as shown in Figure 1(a). From geometric consideration of the element dx* and the fact that there is no axial deformation for the entire beam (i.e. dx* = dx), components of the displacement u and v can readily be related to the rotation  = (x) at any point (x, 0) by sinθ 

dv dx

cosθ  1 

(3.1) du dx

(3.2)

From the definition of the curvature (a quantity that is commonly used to represent the deformation of flexural members) and the inextensibility condition dx* = dx, we then obtain a relation between the curvature  =  (x) and the rotation  = (x) by 

d dx

(3.3)

From the assumption (iv), following approximations are sound and commonly recognized sin    

3 5     6 120


145


cos   1 


2 4   1 2 24

(3.5)

With the approximations (3.4) and (3.5), the relations (3.1) and (3.2) simply reduce to dv dx

(3.6)

du 0 dx

(3.7)

θ

Equation (3.7) simply implies that the longitudinal displacement must be constant throughout the beam and, when supplied by a proper constraint to prevent rigid translation in the longitudinal direction, u must vanish identically for the entire beam. As a result, based on a small rotation assumption, any point of the beam undergoes only a transverse displacement v and it is commonly termed a deflection. Further, by combining (3.3) and (3.6), it leads to a linear relation between the deflection v and the curvature : d2v (x)  2 dx

(3.8)

Equation (3.8) is commonly termed a linearized kinematics of a beam that undergoes a small rotation. It is noted by passing that for beams undergoing large displacement and rotation, the approximations (3.4) and (3.5) cannot be employed to accurately capture responses of those structures. Various investigations of such problems using exact kinematics (i.e., exact relationship between the deflection and the curvature) can be found in the literature (e.g., Tangnovarad, 2008; Tangnovarad and Rungamornrat, 2008; Tangnovarad and Rungamornrat, 2009; Rungamornrat and Tangnovarad, 2011; Douanevanh, 2011; Douanevanh et al, 2011). Since the beam is represented by a one dimensional line model or, equivalently, a cross section is represented by a single point, the deformation at any point within a cross section cannot completely be characterized only by the curvature  but requiring additional assumption on kinematics of the cross section. To investigate this issue, let us consider an infinitesimal element of length dx of the beam in the undeformed state and the corresponding element in the deformed state as shown in Figure 3.3. From the assumptions (ii) and (iii), the geometry of the deformed element must be a sector of hollow circular cross section. Next, let us define a neutral axis (NA) which is a locus of points that undergo no deformation in the deformed state or, equivalently, there is no change in length in the deformed state, i.e. dx* = dx, and let  denote the radius of curvature of the neutral axis in the deformed state. Note by passing that the elastic curve of a beam (as shown by a red dash line in Figure 3.1) is in fact the neutral axis of the deformed beam. To obtain a formula for a normal strain at any point within the cross section, let us consider a fiber of length dx located at the distance y from the neutral axis. This fiber deforms to a curve fiber of length ds and, again from the assumption (ii) and (iii), this deformed fiber is in fact an arc segment of a circle of radius  – y. From the definition of the engineering strain, the normal strain at a point with a distance y from the neutral axis is given by (x, y) 

ds  dx (  y)d  d d   y dx dx dx

(3.9)

where x denotes the coordinate of the cross section. Upon using (3.3), we finally obtain the normal strain at any point within the cross section in terms of the curvature of that cross section as: (x, y)   y(x)


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Equation (3.10) implies that the normal strain at any point within the cross section varies linearly with respect to the distance from the neutral axis. The negative sign simply indicates that the positive curvature produces a compressive strain at all points above the neutral axis.

d 

Neutral axis

ds

y dx

–y Neutral axis

dx* = dx

Figure 3.3: Schematic of undeformed infinitesimal element dx and corresponding deformed element

3.1.2 Constitutive law From the assumption (i), the normal stress at any point within the cross section is related to the normal strain at the same point via a linear stress-strain relation: (x, y)  E(x)(x, y)

(3.11)

where E = E(x) is Young’s modulus at any cross section. From (3.11) along with (3.10), it can be deduced that the normal strain at any point within the cross section also varies linearly with respect to the distance from the neutral axis.

3.1.3 Equilibrium equations From assumption (v), we obtain following two equilibrium equations in differential form (see derivation in subsection 2.5.5.1) dV dx



q(x)

(3.12)

dM dx



V(x)

(3.13)

where V = V(x) and M = M(x) are the shear force and bending moment at any point x. Validity of equations (3.12) and (3.13) depends primarily on the smoothness of loading data at point x as elaborated in details below. For a point A that is free of concentrated force and moment and q is continuous, V, M, dV/dx and dM/dx are well-defined at this point and, as a result, both (3.12) and (3.13) are valid at this point. Also, it can readily be verified that there is no jump of the shear force and bending moment at point A: [V]A  limV(x A  )  V(x A  )  lim 0

0

x A 

[M]A  limM(x A  )  M(x A  )  lim  0

 q(x)dx  0

(3.14)

x A 

 0

x A 

 V(x)dx  0

x A 


(3.15)

147



where (xA, 0) is a coordinate of a point A. For a point that is free of concentrated force and concentrated moment but q is discontinuous (such as a point B in Figure 3.1), V, M and dM/dx are well-defined while dV/dx is not defined at this point; as a result, only the relation (3.13) is valid at this point. In addition, by considering equilibrium at this point (see subsection 2.5.5.3), it can be shown that there is no jump of the shear force and no jump of the bending moment at point B, i.e. [V]B  0

(3.16)

[M]B  0

(3.17)

For a point that is subjected to a concentrated force Po (such as a point C in Figure 3.1), M is welldefined while V, dV/dx and dM/dx are not defined at this point and, similarly, it can be deduced from equilibrium at this point that the jump of the shear force and the bending moment satisfy [V]C  P0

(3.18)

[M]C  0

(3.19)

For a point that is subjected to a concentrated moment Mo (such as a point D in Figure 3.1), V, M, dV/dx and dM/dx are not defined at this point. Again, by considering equilibrium at this point, we can conclude following jump conditions: [V]D  0

(3.20)

[M]D   M 0

(3.21)

In addition, force and moment resultants of the normal stress  over the entire cross section yield the axial force F(x) and the bending moment M(x) as follows: F(x)   (x,y)dA

(3.22)

M(x)    y(x,y)dA

(3.23)

A

A

By substituting (3.10) and (3.11) into (3.22) and (3.23), it leads to F(x)  E(x)κ(x)  ydA  E(x)κ(x)y

(3.24)

M(x)  E(x)κ(x)  y 2 dA  E(x)I(x)κ(x)

(3.25)

A

A

where y is the distance from the centroid of the cross section to the neutral axis and I is the moment of inertia of the cross section. From (3.24) and the fact that the axial force vanishes for the entire beam (i.e. F(x) = 0), it implies that the neutral axis is located at the centroid of the cross section. Equation (3.25) that can be viewed as a constitutive relation in the cross section level is also known as a moment-curvature relationship. It is evident that the moment-curvature relationship (3.25) is linear; this results directly from the linear stress-strain relation (3.11). For nonlinear elastic and inelastic materials, the relation between the bending moment and the curvature is, in general, nonlinear. Analysis of beams by taking material nonlinearity into account can be found, for examples, in the work of Danmongkoltip (2009), Danmomgkoltip and Rungamornrat (2009) and Pinyochotiwong et al (2009). Copyright © 2011 J. Rungamornrat

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3.2 Governing Differential Equations A set of differential equations governing behavior of the entire beam can be obtained by combining kinematics, constitutive laws and equilibrium equations established in section 3.1. In this section, we focus only on the case that the loading data q and the flexural rigidity EI are continuous everywhere and, in addition, it is free of concentrated forces and moments. The treatment of the discontinuity induced by discontinuous loading q, abrupt change of flexural rigidity EI, and presence of concentrated forces and moments are deferred to later sections. A complete set of differential equations governing the deflection v for an Euler-Bernoulli beam consists of four basic equations: 1 kinematics (3.8), 2 equilibrium equations (3.12) and (3.13), and 1 moment-curvature relation (3.25) as summarized again below (x) 

d2v dx 2

(3.26a)

dV  q(x) dx

(3.26b)

dM  V(x) dx

(3.26c)

M(x)  EI(x)κ(x)

(3.26d)

where EI(x) = E(x)I(x) is termed the flexural rigidity of the beam. Note that the highest order of differential equations appearing in this set is equal to 2 (i.e. equation (3.26a)). Instead of using a system of four differential equations (3.26a)-(3.26d), it is common and more convenient in the solution procedure to introduce a single differential equation that can represent all four equations. This equation can be obtained via simple substitution as follows: q(x) 

dV d  dM  d 2 d2  d2 v      2  EI(x)κ(x)   2  EI(x) 2  dx dx  dx  dx dx  dx 

(3.27)

Clearly, the curvature , the shear force V and the bending moment M are eliminated and (3.27) involves only the unknown deflection v and prescribed loading data q and flexural rigidity EI. This fourth-order differential equation is well-recognized as a governing equation for a beam undergoing small rotation. Another form of the governing equations that is also widely used in the analysis for deflection of beams consists of three differential equations, 1 equation resulting from combining (3.26a) and (3.26d) and 2 equilibrium equations (3.26b) and (3.26c): EI(x)

d2v  M(x) dx 2

(3.28a)

dV  q(x) dx

(3.28b)

dM  V(x) dx

(3.28c)

This set involves differential equations of order less than or equal to 2. For statically determinate beams, equations (3.28b) and (3.28c) can be solved independently of equation (3.28a) to obtain the Copyright © 2011 J. Rungamornrat

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shear force V and bending moment M. In steady of directly solving the differential equations (3.28b) and (3.28c), the shear force and bending moment can also be obtained via the method of sections and the method of differential and integral formula as discussed in subsection 2.5.4 and 2.5.5. Once the bending moment is completely determined, equation (3.28a) is now a key differential equation governing the deflection of the beam. For statically indeterminate beams, the bending moment M cannot be fully obtained from (3.28b) and (3.28c) but it will involve extra static unknowns (e.g. reactions and internal forces). Such extra unknowns can be determined once (3.28a) is solved and extra kinematical conditions are imposed (see examples for clarification). The differential equation (3.28a) along with the known bending moment M forms a second-order differential equation for beams. While it has not received the same level of popularity, another form of the governing equations has proven more convenient than those stated above when it is applied to certain situations. This set of governing equations consists of following two differential equations, 1 equation resulting from combining (3.26a), (3.26c) and (3.26d) and 1 equilibrium equation (3.26b): d  d2 v   EI(x) 2   V(x) dx  dx 

(3.29a)

dV  q(x) dx

(3.29b)

This set involves differential equations of order less than or equal to 3. For statically determinate beams, equations (3.29b) can be solved independently of equation (3.29a) to obtain the shear force. Again, this can also be achieved by using the method of sections and the method of differential and integral formula as discussed in subsections 2.5.4 and 2.5.5. Once the shear force V is completely determined, equation (3.29a) becomes a key governing equation for the deflection v. For statically indeterminate beam, the shear force V may not completely be obtained from (3.29b) but it will contain extra static unknowns. Procedure for determining such extra unknowns is similar to the previous case (also see examples for clarification). The differential equation (3.29a) along with the known shear force V forms a third-order differential equation for beams. Once the deflection of the beam is already solved, other quantities can be obtained as follows: (i) if (3.27) is chosen as the key governing equation, the rotation is obtained from (3.6) and the shear force and bending moment are obtained from (3.29a) and (3.28a), respectively; (ii) if (3.28a)-(3.28c) are chosen as the key governing equations, the rotation is obtained from (3.6) and the shear force is obtained from (3.29a); and (iii) if (3.29a) and (3.29b) are chosen as the key governing equations, the rotation is obtained from (3.6) and the bending moment is obtained from (3.28a). Finally, we remark that all four sets of ordinary differential equations stated above are mathematically equivalent and either one of them can be chosen in the analysis of beams. Equation (3.27) is often termed the full-order differential equation while equations (3.28a) and (3.29a) are known as the reduced-order differential equations. There is no strong evidence to support and decide the best choice from these four sets; in general, the choice is a matter of test and preference and, frequently, problem-dependent. This will become more apparent when they are applied to solve various beam problems.

3.3 Boundary Conditions To obtain a unique solution for a particular beam, it is required to specify sufficient end conditions termed boundary conditions in addition to loading data and flexural rigidity. These boundary


150



conditions play an important role in the determination of arbitrary constants resulting from integration of the differential equation in the solution procedure. From fundamental theorems on differential equations, it is required to specify n boundary conditions to render the two-point, elliptic, ordinary differential equation of order n in terms of a variable v to be well-posed (the term two-point is used to emphasize that a domain has two end points). All such boundary conditions can only involve quantities associated with v and its derivatives of order lower than n, i.e. dv/dx, d2v/dx2, …, dn-1v/dxn-1. For instance, a two-point, elliptic, ordinary differential equation of order 4 in terms of a variable v requires 4 boundary conditions in terms of v, dv/dx, d2v/dx2 or d3v/dx3. For a special class of two-point, elliptic, ordinary differential equations of even order (i.e. n is even), boundary conditions can be divided into two categories: essential boundary conditions and natural boundary conditions. The first category involves quantities such as v and its derivative of order less than n/2 (i.e. v, dv/dx, d2v/dx2, …, dn/2-1v/dxn/2-1) while the other involves the remaining derivatives (i.e. dn/2v/dxn/2, dn/2+1v/dxn/2+1, …, dn-1v/dxn-1). In addition, exactly half of boundary conditions must be specified at each end point. As is evident from the previous subsection, the full-order differential equation governing the beam deflection, i.e. equation (3.27), is elliptic and is of order 4. As a result, boundary conditions at both ends of the beam involve prescribed values of either the deflection v, the rotation dv/dx, the bending moment EId2v/dx2, or the shear force d(EId2v/dx2)/dx. The first two are essential boundary conditions and the last two are natural boundary conditions. For each end of the beam, exactly two boundary conditions must be prescribed. To specify proper boundary conditions for each end of a particular beam, following two guidelines are useful:  Both the deflection and the shear force cannot be prescribed independently at the end of the beam. One boundary condition can be deduced from following three cases: (i) the deflection is prescribed while the shear force is unknown a priori (e.g. roller support and pinned-end support); (ii) the shear force is prescribed while the deflection is unknown a priori (e.g. free end and guided support); and both the deflection and shear force are unknown a priori but there exists a relation (generally obtained from considering force equilibrium at the beam end) relating both quantities (e.g. beam end with a translational spring).  Both the rotation and the bending moment cannot be prescribed independently at the end of the beam. One boundary condition can be deduced from following three cases: (i) the rotation is prescribed while the bending moment is unknown a priori (e.g. fixed-end support and guided support); (ii) the bending moment is prescribed while the rotation is unknown a priori (e.g. free end, roller support and pinned-end support); and both the rotation and bending moment are unknown a priori but there exists a relation (generally obtained from considering moment equilibrium at the beam end) relating both quantities (e.g. beam end with a rotational spring). With the above guidelines, we obtain boundary conditions for certain types of beam ends that are mostly found in beam as demonstrated below. Comprehensive summary of Boundary conditions for various beam ends can be found in Table 3.1.

3.3.1 Fixed-end support For a beam end with a fixed-end support, both the deflection and rotation at this point are fully prevented while the bending moment and the shear force are unknown a priori. Two boundary conditions are therefore given by v(0)  0

and v (0)  0 if a support is at the left end Copyright © 2011 J. Rungamornrat

(3.30a)

151


v(L)  0 and

v (L)  0


if a support is at the right end

(3.30b)

where v (x) denotes the first derivative of v(x), i.e. v (x)  dv/dx . If the vertical and rotational settlements occur at this support, the prescribed values of the deflection and the rotation are now non-zero and must be set equal to the settlements.

3.3.2 Pinned-end support or Roller support From the small rotation assumption and a proper constraint against the rigid translation, there is no component of the displacement parallel to the beam axis and this renders no difference between kinematical constraints provided by a pinned-end support and a roller support. These two supports provide a full constraint to beam end against the movement in the direction normal to the beam axis (i.e. no deflection) while the bending moment is fully prescribed. Two boundary conditions at these two supports are therefore given by v(0)  0

and EI

v(L)  0 and

EI

d2v (0)   M 0 dx 2

if a support is at the left end

(3.31a)

d2v (L)  M 0 dx 2

if a support is at the right end

(3.31b)

where M0 is a moment acting at the supports; this applied moment is considered positive if it directs in the Z-direction or counterclockwise, otherwise it is negative. The negative sign appearing only in (3.31a) is due to that the counterclockwise applied moment acting at the left end produces a negative bending moment at that point (following the sign convention defined in subsection 2.5.2 in chapter 2) while the counterclockwise applied moment acting at the right end produces a positive bending moment at that point. Boundary conditions, for a special case when there is no applied moment M0, can readily be obtained by substituting M0 = 0 into (3.31a) and (3.31b). In addition, if the vertical settlement occurs at this support, the prescribed value of the deflection is now non-zero and must be set equal to the settlement.

3.3.3 Guided support For a beam end with a guided support, the rotation is fully prevented at this point while the shear force is prescribed. Two boundary conditions are therefore given by dv (0)  0 dx

and

d  d2v   EI  (0)  P0 dx  dx 2 

dv (L)  0 dx

and

d  d2 v   EI  (L)   P0 dx  dx 2 

if a support is at the left end if a support is at the right end

(3.32a) (3.32b)

where P0 is a concentrated force acting at the support; this applied force is considered positive if it directs in the Y-direction or upward, otherwise it is negative. Again, the negative sign appearing only in (3.32b) is due to that the upward applied force acting at the left end produces a positive shear force at that point (following the sign convention defined in subsection 2.5.2 in chapter 2) while the upward applied force acting at the right end produces a negative shear force at that point. Boundary conditions, for a special case when there is no applied force P0, can readily be obtained by substituting P0 = 0 into (3.32a) and (3.32b). In addition, if the rotational settlement occurs at this support, the prescribed value of the rotation is now non-zero and must be set equal to the settlement. Copyright © 2011 J. Rungamornrat

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Table 3.1: Boundary conditions for various types of beam end Left end (x = 0)

Boundary conditions

Right end (x = L)

v(0)  0 v(0)  0

v(L)  0 v(L)  0

v(0)  0 EIv(0)  0

v(L)  0 EIv(L)  0

M0 M0

P0

P0

M0 v(0)  0 EIv(0)  M 0

M0

v(L)  0

(EIv) (0)  0

(EIv) (L)  0

P0

(EIv) (L)  P0

EIv(0)  0

EIv(L)  0

(EIv) (0)  0

(EIv) (L)  0

EIv(0)  M 0

P0

(EIv) (0)  P0

M0

EIv(L)  M 0 (EIv) (L)  P0

v(L)  0 ks

(EIv) (0)  k s v(0)  0

(EIv) (L)  k s v(L)  0

v(L)  0

EIv(0)  0 ks

v(L)  0

(EIv) (0)  P0

EIv(0)  0 ks

v(L)  0 EIv(L)  M 0

v(0)  0

v(0)  0

M0

Boundary conditions

(EIv) (0)  k s v(0)  0

ks

EIv(0)  k  v(0)  0

(EIv) (L)  k s v(L)  0 EIv(L)  k  v(L)  0

(EIv) (0)  0

k

(EIv) (L)  0

k

v(0)  0 EIv(0)  k  v(0)  0

k

v(L)  0 EIv(L)  k  v(L)  0

k

EIv(0)  k  v(0)  0

k

EIv(L)  k  v(L)  0

k

ks

(EIv) (0)  k s v(0)  0


ks

(EIv) (L)  k s v(L)  0

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3.3.4 Free end For a free end of a beam, there is no constraint on both the deflection and the rotation. As a result, both the bending moment and the shear force are fully prescribed and this leads to following two boundary conditions EI

d2v (0)   M 0 dx 2

EI

d2v (L)  M 0 dx 2

and and

d  d2v   EI  (0)  P0 dx  dx 2  d  d2v   EI  (L)   P0 dx  dx 2 

for a left free end

(3.33a)

for a right free end

(3.33b)

where P0 and M0 are applied force and applied moment at the free end; sign convention of P0 and M0 are the same as stated in the two previous cases. Boundary conditions, for a special case when there is no applied force and applied moment, can readily be obtained by substituting P0 = 0 and M0 = 0 into (3.33a) and (3.33b).

3.3.5 Beam end with a translational spring Consider a beam end connected to a linear translational spring with a spring constant ks. For this particular case, both the shear force and the deflection at this end are unknown a priori; therefore, neither of these two quantities can be treated as a boundary condition. Presence of a translational spring generally induces a force proportional to the deflection of the beam end (in fact it is equal to the product of the deflection and the spring constant) and in the direction opposite to the deflection. To construct a proper boundary condition, we consider force equilibrium of an infinitesimal element containing the end point as shown schematically in Figure 3.4 and this leads to one boundary condition: V(0)  k s v(0)  P0 

d  d2 v   EI  (0)  k s v(0)  P0  0 dx  dx 2 

V(L)  k s v(L)  P0 

d  d2v   EI  (L)  k s v(0)  P0  0 dx  dx 2 

for a translational spring at the left end

(3.34a)

for a translational spring at the right end (3.34b)

where P0 is an applied force at the beam end. The other boundary condition can be deduced from the rotational constraint. P0

V(L) P0

V(0) M(0)

M(L)

ksv(0) Left end (x = 0)

ksv(L) Right end (x = L)

Figure 3.4: FBD of beam end containing translational spring and subjected to force P0 Copyright © 2011 J. Rungamornrat

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3.3.6 Beam end with a rotational spring Consider a beam end connected to a linear rotational spring with a spring constant k. For this particular case, both the bending moment and the rotation at this end are unknown a priori; therefore, neither of these two quantities can be treated as a boundary condition. Presence of a rotational spring generally induces a moment proportional to the rotation of the beam end (in fact it is equal to the product of the rotation and the spring constant) and in the direction opposite to the rotation. To construct a proper boundary condition, we consider moment equilibrium of an infinitesimal element containing the end point as shown schematically in Figure 3.5 and this leads to one boundary condition: M(0)  k 

dv d2 v dv (0)  M 0  EI 2 (0)  k  (0)  M 0  0 dx dx dx

for a rotational spring at the left end

M(L)  k 

dv d2 v dv (L)  M 0  EI 2 (0)  k  (0)  M 0  0 dx dx dx

for a rotational spring at the right end (3.35b)

(3.35a)

where M0 is an applied moment at the beam end. The other boundary condition can be deduced from the translational constraint. V(L)

V(0) M0

M(0)

M(L)

M0 dv (L) dx

dv (0) dx

kθ

Left end (x = 0)

Right end (x = L)

kθ

Figure 3.5: FBD of beam end containing rotational spring and subjected to moment M0 If the reduced-order differential equation (i.e. second-order and third-order differential equations) is chosen as a key governing equation, consideration of boundary conditions as described above still applies. However, it is important to emphasize that not all four boundary conditions at both ends of the beam apply to the reduced-order differential equations since some of them are used in the construction of either the shear force or the bending moment. If the second-order differential equation is employed, only two boundary conditions are needed for determining constants from the integration and they involve only the deflection and the rotation. If the third-order differential equation is employed, only three boundary conditions are needed to determine constants from the integration and they involve only the deflection, the rotation, and the bending moment.

3.4 Boundary Value Problem For a beam of length L that is fully supplied by data such as flexural rigidity, applied loads and kinematical constraints, a set of differential equations that completely characterizes behavior of the entire beam (e.g. equations (3.26a)-(3.26d), or equation (3.27), or equations (3.28a)-(3.28c), or equations (3.29a)-(3.29b)) furnished by proper and sufficient boundary conditions at both ends forms a boundary value problem for the given beam. Copyright © 2011 J. Rungamornrat

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Y q EI

X

L Figure 3.6: Schematic of a cantilever beam subjected to uniformly distributed load q For instance, let us consider a cantilever beam of length L and constant flexural rigidity EI and subjected to a uniformly distributed load q for the entire beam as shown in Figure 3.6. If equations (3.26a)-(3.28d) are used as the key governing equations, a boundary value problem for this particular beam becomes (x) 

d2v dx 2

dV  q dx

for x  (0, L)

(3.36a)

for x  (0, L)

(3.36b)

dM  V(x) for x  (0, L) dx

(3.36c)

M(x)  EIκ(x)

(3.36d)

for x  (0, L)

v(0)  0

(3.36e)

v(0)  0

(3.36f)

EIv(L)  0

(3.36g)

EIv(L)  0

(3.36h)

If a fourth-order differential equation (3.27) is employed as a key governing equation, a boundary value problem for this particular beam becomes EI

d4v  q dx 4

for x  (0, L)

(3.37a)

v(0)  0

(3.37b)

v(0)  0

(3.37c)

EIv(L)  0

(3.37d)

EIv(L)  0

(3.37e)

If equations (3.28a)-(3.28c) are chosen as the key governing equations, we may first apply force and moment equilibrium equations (3.28b) and (3.28c) to obtain the bending moment M(x) = –q(L – x)2/2. A boundary value problem for this particular beam can then be formulated in terms of the second-order differential equation: Copyright © 2011 J. Rungamornrat

156



EI

d2v q(L  x) 2  2 dx 2

for x  (0, L)

(3.38a)

v(0)  0

(3.38b)

v(0)  0

(3.38c)

If equations (3.29a)-(3.29b) are chosen as the key governing equations, we may first apply force equilibrium equation (3.29b) to obtain the shear force for the entire beam V(x) = q(L – x). A boundary value problem for this particular beam can then be formulated in terms of the third-order differential equation: EI

d3v  q L  x dx 3

for x  (0, L)

(3.39a)

v(0)  0

(3.39b)

v(0)  0

(3.39c)

EIv(L)  0

(3.39d)

3.5 Solution Procedure Since the ordinary differential equation(s) involved in the boundary value problem formulated above contains only a single derivative term, a direct integration technique can therefore be exploited to solve such differential equations. Arbitrary constants resulting from the integration can then be obtained by enforcing boundary conditions for a particular beam treated. To demonstrate the procedure, let us consider first a boundary value problem formulated in terms of the fourth-order differential equation (3.27). A direct integration of this equation yields the shear force V(x): V(x) 

d  d2 v  EI(x)    q(x)dx  C1 dx  dx 2  

(3.40)

where C1 is an arbitrary constant resulting from the integration. By performing one more integration of (3.40), it leads to the bending moment M(x): M(x)  EI(x)

d2v  q(x)dxdx  C1 x  C 2 dx 2  

(3.41)

where C2 is again another arbitrary constant resulting from the second integration. By dividing both sides of (3.41) by EI(x) and then performing a direct integration, it results in the rotation (x): (x) 

dv 1 1  q(x)dxdxdx    C1x  C2 dx  C3   dx EI(x) EI(x)

(3.42)

where C3 is an arbitrary constant resulting from the third integration. Finally, by taking the last integration of (3.42), it yield the deflection v(x): v(x)   

1 1 q(x)dxdxdxdx     C1x  C2 dxdx  C3 x  C4   EI(x) EI(x) Copyright © 2011 J. Rungamornrat

(3.43)

157



where C4 is an arbitrary constant resulting from the last integration. For the special case of beam with constant flexural rigidity, (3.40)-(3.43) simply reduce to V(x)  EI

d3v  q(x)dx  C1 dx 3 

(3.44a)

M(x)  EI

d2v  q(x)dxdx  C1 x  C 2 dx 2  

(3.44b)

EI(x)  EI

dv 1     q(x)dxdxdx  C1x 2  C 2 x  C3 dx 2

1 1 EIv(x)      q(x)dxdxdxdx  C1x 3  C 2 x 2  C3 x  C 4 6 2

(3.44c) (3.44d)

Four arbitrary constants {C1, C2, C3, C4} can uniquely be obtained from four boundary conditions available at both ends of the beams. Next, let us consider a boundary value problem formulated in terms of the second-order differential equation (3.28a). By dividing both sides of this equation by EI(x) and then performing a direct integration, it results in the rotation (x): (x) 

dv 1  M(x)dx  C1 dx  EI(x)

(3.45)

By performing one more integration, it leads to the deflection v(x): v(x)   

1 M(x)dxdx  C1 x  C 2 EI(x)

(3.46)

Similarly, if the flexural rigidity is constant throughout the beam, (3.45) and (3.46) become EI(x)  EI

dv  M(x)dx  C1 dx 

(3.47a)

EIv(x)    M(x)dxdx  C1x  C 2

(3.47b)

Two arbitrary constants {C1, C2} resulting from the integrations can uniquely be obtained from two boundary conditions available at both ends of the beams. Finally, let us consider a boundary value problem formulated in terms of the third-order differential equation (3.29a). A direct integration of this equation yields the bending moment M(x): M(x)  EI(x)

d2v  V(x)dx  C1 dx 2 

(3.48)

By dividing both sides of (3.48) by EI(x) and then performing another direct integration, it results in the rotation (x): (x) 

dv 1 1  V(x)dxdx   C1dx  C 2  dx EI(x) EI(x)


(3.49)

158



Finally, by performing the last integration, it leads to the deflection v(x): v(x)   

1 1 V(x)dxdxdx    C1dxdx  C 2 x  C3  EI(x) EI(x)

(3.50)

For the special case of beam with constant flexural rigidity, (3.48)-(3.50) reduce to M(x)  EI

d2 v  V(x)dx  C1 dx 2 

EI(x)  EI

(3.51a)

dv  V(x)dxdx  C1x  C 2 dx  

(3.51b)

1 EIv(x)     V(x)dxdxdx  C1x 2  C 2 x  C3 2

(3.51c)

Again, three arbitrary constants {C1, C2, C3} resulting from the integrations can uniquely be obtained from three boundary conditions available at both ends of the beams. Example 3.1 Consider a cantilever beam of length L and constant flexural rigidity EI as shown below. The beam is subjected to a distributed load q = q0(x/L)n where q0 and n are non-negative constants. Determine the shear force, bending moment, deflection and the rotation of the beam.

Y q = q0(x/L)n EI

q0 X

L Solution To clearly demonstrate the application of the full-order and the reduced-order differential equations, this problem is solved in three different ways as indicated below. Option I: Use fourth-order differential equation The beam is fully fixed at the left end while the shear force and the bending moment at the right end vanish. A boundary value problem formulated in terms of the fourth-order differential equation for this particular beam is given by EI

d4v x  q0   4 dx L

n

for x  (0, L)

(e3.1.1)

v(0)  0

(e3.1.2)

v(0)  0

(e3.1.3)

EIv(L)  0

(e3.1.4)

EIv(L)  0

(e3.1.5) Copyright © 2011 J. Rungamornrat

159



Performing direction integrations of (3.1.1) yields n

n+1

EI

q L x  d3v x   q 0   dx  C1  0   3 dx n+1  L  L

EI

q0L  x  q 0 L2 d2v x  dx  C x  C  1 2     2  dx n+1  L  (n+1)(n+2)  L 

EI

q 0 L2 q 0 L3 dv 1 x x 2  dx  C x  C x  C  1 2 3     dx (n+1)(n+2)  L  2 (n+1)(n+2)(n+3)  L 

 C1

(e3.1.6)

n+1

n+2

 C1x  C 2

n+2

q 0 L4 x EIv    (n+1)(n+2)(n+3)(n+4)  L 

n+4

(e3.1.7) n+3

1  C1x 2  C 2 x  C3 (e3.1.8) 2

1 1  C1x 3  C 2 x 2  C3 x  C 4 6 2

(e3.1.9)

Four constants of integration {C1, C2, C3, C4} can be obtained as follows: v(0)  0



C4  0

v(0)  0



C3  0

EIv(L)  0



q0L q L  C1   P0  C1   0 n+1 n+1

EIv(L)  0



q 0 L2 q 0 L2 q L2  C1L  C 2  0  C 2    C1L  0 (n+1)(n+2) (n+1)(n+2) n+2

By substituting {C1, C2, C3, C4} into (e3.1.6)-(e3.1.9), we obtain the shear force, the bending moment, the deflection and the rotation for the given beam: V(x)  EI

n+1  d 3 v q 0 L  x      1 3 dx n+1  L  

 x  n+2  q 0 L2 d2v x M(x)  EI 2     (n+2)  n +1 dx (n+1)(n+2)  L  L  EI(x)  EI

EIv(x) 

n+3 2 q 0 L3 dv 1 x   x  x     (n+2)(n+3)    (n+1)(n+3)  dx (n+1)(n+2)(n+3)  L  2 L  L

 x  q 0 L4   (n+1)(n+2)(n+3)(n+4)  L 

n+4

3

1 x 1 x  (n+2)(n+3)(n+4)    (n+1)(n+3)(n+4)   6 L 2 L

2

  

Option II: Use second-order differential equation Since the given beam is statically determinate (i.e. ra = 2, nm = 1(2) = 2, nj = 2(2) = 4, nc = 0, then DI = 2 + 2 – 4 – 0 = 0), all support reactions and the bending moment can readily be obtained from static equilibrium as demonstrated below.

[FY = 0]  

n

:

L qL x R AY   q 0   dx  0  R AY   0 n+1 L 0


160



Y q = q0(x/L)n

q0

A

MA

X

RAY L

Y

q = q0(x/L)n

V(x)

A

MA

M(x)

X

RAY x L

n

[MA = 0]  

:

q L2 x M A   q 0   xdx  0  M A   0 n+2 L 0

[Mx = 0]  

:

 M(x)  M A  R AY x   q 0   (x  )d  0 L 0

n

x

 x  q 0 L2  M(x)    (n+1)(n+2)  L 

n+2

 (n+2)

 x  n +1 L 

Once, the bending moment M(x) is obtained, we can formulate a boundary value problem in terms of the second-order differential equation for this particular beam as follows: EI

 x  n+2  q 0 L2 d2v x     (n+2)  n +1 2 dx (n+1)(n+2)  L  L 

for x  (0, L)

(e3.1.10)

v(0)  0

(e3.1.11)

v(0)  0

(e3.1.12)

Performing direction integrations of (3.1.10) yields  x  n+2  q 0 L2 dv x EI     (n+2)  n +1dx  C1 dx (n+1)(n+2)  L  L  

q 0 L3  x    (n+1)(n+2)(n+3)  L 

 x  q 0 L3 EIv     (n+1)(n+2)(n+3)  L  

n+3

n+3

 x  q 0 L4  (n+1)(n+2)(n+3)(n+4)  L 

2 1 x  x  (n+2)(n+3)    (n+1)(n+3)   C1 2 L  L

(e3.1.13)

2 1 x  x  (n+2)(n+3)    (n+1)(n+3) dx  C1 x  C 2 2 L  L

n+4

3

1 x 1 x  (n+2)(n+3)(n+4)    (n+1)(n+3)(n+4)   6 L 2 L Copyright © 2011 J. Rungamornrat

2

   C1 x  C 2 

(e3.1.14)

161



Two constants of integration {C1, C2} can be obtained as follows: v(0)  0



C2  0

v(0)  0



C1  0

By substituting {C1, C2} into (e3.1.13)-(e3.1.14), we obtain the same deflection and the same rotation as those obtained in option I. Once the deflection v is solved, the shear force can readily be obtained using the relation (3.29a) and, again, the final result is identical to that obtained in option I. Option III: Use third-order differential equation By considering force equilibrium of the left portion of the beam (whose its FBD is indicated above), we then obtain the shear force at any point x, V(x), as

[FY = 0]  

n

x

:

R AY

x   q 0   dx  V(x)  0 L 0

 V(x) 

q 0 L  x    n+1  L 

n+1

  1 

Once, the shear force V(x) is obtained, we can formulate a boundary value problem in terms of the third-order differential equation for this particular beam as follows: n+1  d 3 v q 0 L  x  EI 3     1 dx n+1  L  

for x  (0, L)

(e3.1.15)

v(0)  0

(e3.1.16)

v(0)  0

(e3.1.17)

EIv(L)  0

(e3.1.18)

Performing direction integrations of (3.1.15) yields EI

n+1 n+2 q 0 L  x  q 0 L2 d2v x    x  1 dx C        1      (n+2)   C1 2  dx n+1  L  (n+1)(n+2)  L  L  

EI

 x  n+2 q 0 L2 dv x      (n+2) dx  C1 x  C 2 dx (n+1)(n+2)  L  L   x  q 0 L3    (n+1)(n+2)(n+3)  L 

EIv  

 x  q 0 L3   (n+1)(n+2)(n+3)  L 

n+3

n+3

1 x  (n+2)(n+3)   2 L 1 x  (n+2)(n+3)   2 L

 x  q 0 L4    (n+1)(n+2)(n+3)(n+4)  L 

n+4

2

2

   C1 x  C 2 

(e3.1.19)

(e3.1.20)

 1 2 dx  C1 x  C 2 x  C3 2 

1 x  (n+2)(n+3)(n+4)   6 L

3

 1 2   C1x  C 2 x  C3  2


(e3.1.21)

162



Three constants of integration {C1, C2, C3} can be obtained as follows: v(0)  0



C3  0

v(0)  0



C2  0

EIv(L)  0





q 0 L2 q L2  C1  0  C1  0 n+2 n+2

By substituting {C1, C2, C3} into (e3.1.19)-(e3.1.21), we obtain the deflection, the rotation, and the bending moment identical to those obtained in option I. The shear force V(x), the bending moment M(x), the rotation (x), and the deflection v(x) obtained above can be specialized to a beam subjected to different distributed loads: (i) a uniformly distributed load (n = 0), (ii) a linearly distributed load (n = 1) and (iii) a parabolic distributed load (n = 2) as shown in a table below.

n

EIv, EI, M, V

Type of distributed load EIv(x) 

Y q = q0 0

EI, L

3 2  x  4 x  x      4    6    L  L    L 

2  x 3 x  x 3  3      L  L  L  2 q L2  x   x   M(x)  0    2    1 2  L   L  

EI(x) 

X

q 0 L4 24

q 0 L3 6

 x   V(x)  q 0 L    1  L   EIv(x) 

Y q = q0(x/L) 1

EI, L

V(x) 

q = q0(x/L)2 2

EI, L

X

q 0 L3 24

2  q 0 L  x     1 2  L  

6 3 2  x  x  x      20    45    L  L    L  5 2 q L3  x  x  x   EI(x)  0    10    15    60  L  L  L   4 q L2  x   x   M(x)  0    4    3 12  L   L  

EIv(x) 

Y

5 3 2  x  x  x      10    20    L  L    L 

4 2  x  x  x      6    8    L  L    L  3  q L2  x  x M(x)  0    3    2  6  L  L 

EI(x) 

X

q 0 L4 120

V(x) 

q 0 L4 360

3  q 0 L  x     1 3  L  



163


Example 3.2 Consider a simply-supported beam of length L and constant flexural rigidity EI as shown below. The beam is subjected to a distributed load q = q0(x/L)n where q0 and n are nonnegative constants. Determine the shear force, bending moment, deflection and the rotation of the beam. Y q = q0(x/L)n

q0 X

EI L

Solution As another example for demonstrating the use of the second-order, third-order, and fourthorder differential equations, the problem is again solved in three different ways similar to example 3.1. Option I: Use fourth-order differential equation Since the left end and right end of the beam are supported by pinned-end and roller supports, respectively, the deflection and bending moment vanish at both ends of the beam. A boundary value problem formulated in terms of the fourth-order differential equation for this particular beam is therefore given by d4 v x EI 4  q 0   dx L

n

for x  (0, L)

(e3.2.1)

v(0)  0

(e3.2.2)

EIv(0)  0

(e3.2.3)

v(L)  0

(e3.2.4)

EIv(L)  0

(e3.2.5)

Performing direction integrations of (3.2.1) yields n

n+1

EI

q L x  d3v x dx  C1  0    q0 dx 3   L  n+1  L 

EI

q0L  x  q 0 L2 d2v x  dx  C x  C  1 2     2  dx n+1  L  (n+1)(n+2)  L 

EI

q 0 L2 q 0 L3 dv 1 x x 2  dx  C x  C x  C  1 2 3     dx (n+1)(n+2)  L  2 (n+1)(n+2)(n+3)  L 

n+1

 C1

(e3.2.6) n+2

 C1x  C 2

n+2

EIv 

q 0 L4 x   (n+1)(n+2)(n+3)(n+4)  L 

n+4

1 1  C1 x 3  C 2 x 2  C3 x  C 4 6 2


(e3.2.7) n+3

1  C1x 2  C 2 x  C3 (e3.2.8) 2

(e3.2.9)


164


It is not surprising that (e3.2.6)-(e3.2.9) are identical to (e3.1.6)-(e3.3.9) since the loading data q(x) for both cases is identical. Four constants of integration {C1, C2, C3, C4} can then be obtained by using boundary conditions (e3.2.2)-(e3.2.5): v(0)  0



C4  0

EIv(0)  0



C2  0

EIv(L)  0



q 0 L2 q0L  C1L  C 2  0  C1   (n+1)(n+2) (n+1)(n+2)

v(L)  0



q 0 L4 1 1  C1L3  C 2 L2  C 3 L  C 4  0 (n+1)(n+2)(n+3)(n+4) 6 2



C3 

(n+6)q 0 L3 6(n+2)(n+3)(n+4)


n+1 d 3 v q 0 L  x  1       3 dx n+1  L  n+2 

M(x)  EI

n+2 q 0 L2 d2 v  x   x          2 dx (n+1)(n+2)  L   L  

EI(x)  EI

n+3 2 q 0 L3 dv (n+3)  x  (n+1)(n+6)   x        dx (n+1)(n+2)(n+3)  L  2  L  6(n+4) 

 x  q 0 L4 EIv(x)   (n+1)(n+2)(n+3)(n+4)  L 

n+4

3 (n+3)(n+4)  x  (n+1)(n+6)  x    L   L  6 6     

Option II: Use second-order differential equation Since the given beam is statically determinate (i.e. ra = 2, nm = 1(2) = 2, nj = 2(2) = 4, nc = 0, then DI = 2 + 2 – 4 – 0 = 0), all support reactions can be obtained by considering equilibrium of the entire beam and the bending moment can readily be computed from moment equilibrium of the left portion as demonstrated below. n

[MA = 0]  

:

L q L x R BY L   q 0   xdx  0  R BY   0 n+2 L 0

[FY = 0]  

:

L q L q0L x R AY  R BY   q 0   dx  0  R AY   0  R BY   L n+1 (n+1)(n+2)   0

n

[Mx = 0]  

x

:

n

 M(x)  R AY x   q 0   (x  )d  0 L 0  M(x) 

 x  q 0 L2  (n+1)(n+2)  L 

n+2

 x       L  



165


Y q = q0(x/L)n

q0

A

B

RAY

X

RBY L

Y

q = q0(x/L)n A

V(x) M(x)

X

RAY x Once, the bending moment M(x) is obtained, we can formulate a boundary value problem in terms of the second-order differential equation for this particular beam as follows:  x  n+2  x   q 0 L2 d2v EI 2      dx (n+1)(n+2)  L   L  

for x  (0, L)

(e3.2.10)

v(0)  0

(e3.2.11)

v(L)  0

(e3.2.12)

Performing direction integrations of (3.2.10) yields  x  n+2  x   q 0 L2 dv EI     dx  C1  dx  (n+1)(n+2)  L   L      C1  n+3 2  x  q 0 L3 (n+3)  x   EIv      dx  C1 x  C 2 (n+1)(n+2)(n+3)  L  2  L   



 x  q 0 L3   (n+1)(n+2)(n+3)  L 

n+3



q 0 L4  x   (n+1)(n+2)(n+3)(n+4)  L 

(n+3)  x  2  L 

n+4

2

3 1  x    (n+3)(n+4)     C1 x  C 2 6  L  

(e3.2.13)

(e3.2.14)

Two constants of integration {C1, C2} can be obtained from boundary conditions (e3.2.11) and (e3.2.12) as follows: v(0)  0



C2  0

v(L)  0





(n+6)q 0 L4 (n+6)q 0 L3  C1L  C 2  0  C1  6(n+2)(n+3)(n+4) 6(n+2)(n+3)(n+4) Copyright © 2011 J. Rungamornrat


166


By substituting {C1, C2} into (e3.2.13)-(e3.2.14), we obtain the same deflection and the same rotation as those obtained in option I. Once the deflection v is solved, the shear force V(x) can readily be obtained using the relation (3.29a) and, again, the final result is identical to that obtained in option I. Option III: Use third-order differential equation By considering force equilibrium of the left portion of the beam (whose its FBD is indicated above), we then obtain the shear force at any point x, V(x), as [FY = 0]  

n

x

x R AY   q 0   dx  V(x)  0 L 0

:

 V(x) 

q 0 L  x   n+1  L 

n+1



1   n+2 

Once, the shear force V(x) is obtained, we can formulate a boundary value problem in terms of the third-order differential equation for this particular beam as follows: EI

n+1 d 3 v q 0 L  x  1       3 dx n+1  L  n+2 

for x  (0, L)

(e3.2.15)

v(0)  0

(e3.2.16)

v(L)  0

(e3.2.17)

EIv(0)  0 or EIv(L)  0

(e3.2.18)

Note that either one of the two conditions stated in (e3.2.18) can be treated as a boundary condition since the condition M(0)  M(L)  EIv(0)  EIv(L)  0 has already been employed in the determination of the shear force (i.e. it is used to determine the support reactions). Performing direction integrations of (3.2.15) yields EI

n+1  x  n+2  x   q 0 L  x  q 0 L2 d2v 1           C1 dx C    1 dx 2  n+1  L  n+2  (n+1)(n+2)  L   L  

(e3.2.19)

 x  n+2  x   q 0 L2 dv EI     dx  C1 x  C 2  dx  (n+1)(n+2)  L   L   

 x  q 0 L3   (n+1)(n+2)(n+3)  L 

EIv  

 x  q 0 L3   (n+1)(n+2)(n+3)  L 



n+3



(n+3)  x  2  L 



(n+3)  x  2  L 

n+3

q 0 L4  x   (n+1)(n+2)(n+3)(n+4)  L 

n+4

2

2

   C1x  C 2 

(e3.2.20)

 1 2 dx  C1 x  C 2 x  C3 2 

1 x  (n+3)(n+4)   6 L

3

 1 2   C1 x  C 2 x  C3  2


(e3.2.21)


167


Three constants of integration {C1, C2, C3} can be obtained from (e3.2.16)-(e3.2.18) as follows: EIv(0)  0



C1  0

v(0)  0



C3  0

v(L)  0





(n+6)q 0 L4 (n+6)q 0 L3 1  C1 x 2  C 2 L  C3  0  C 2  6(n+2)(n+3)(n+4) 2 6(n+2)(n+3)(n+4)

By substituting {C1, C2, C3} into (e3.2.19)-(e3.2.21), we obtain the deflection, the rotation, and the bending moment identical to those obtained in option I. The shear force V(x), the bending moment M(x), the rotation (x), and the deflection v(x) obtained above can be specialized to a beam subjected to different distributed loads: (i) a uniformly distributed load (n = 0), (ii) a linearly distributed load (n = 1) and (iii) a parabolic distributed load (n = 2) as shown in a table below.

n

Type of distributed load

EIv, EI, M, V EIv(x) 

Y q = q0 0

EI, L

 x 3 3  x  2 1          L  2  L  4  2 q L2  x   x   M(x)  0       2  L   L  

EI(x) 

X

4 3 q 0 L4  x   x   x      2       24  L   L   L  

q 0 L3 6

 x  1  V(x)  q 0 L      L  2  EIv(x) 

Y q = q0(x/L) 1

EI, L

4 2 q 0 L3  x  7  x    2     24  L  L 15    3 2  q L  x   x   M(x)  0       6  L   L   2 q L  x  1  V(x)  0     2  L  3 

X

EIv(x) 

q = q0(x/L)2 EI, L

 x 5 10  x 3 7  x            3  L  3  L    L 

EI(x) 

Y

2

q 0 L4 120

q 0 L4 360

5 2 q 0 L3  x  5  x  2         60  L  2  L  3  4 q L2  x   x   M(x)  0       12  L   L   3 q L  x  1  V(x)  0     3  L  4 

EI(x) 

X

3  x 6 x  x     4   5     L  L    L 



168


Example 3.3 Consider a beam of length L and constant flexural rigidity EI as shown below. The beam is subjected to a moment M0 at the left end and a force P0 at the right end. Determine the shear force, bending moment, deflection and the rotation of the beam. Y P0 M0

EI

X

L Solution In this example, we show how to treat boundary conditions associated with concentrated force and concentrated moment acting to both ends of the beam. Again, a solution procedure for three different approaches using second-order, third-order, or fourth-order differential equations is demonstrated. Option I: Use fourth-order differential equation At the left end of the beam, the deflection is fully prevented and the bending moment is prescribed equal to applied moment M0 (this applied moment produces a positive bending moment at the left end) while, at the right end, the rotation is fully prevented and the shear force is prescribed equal to P0 (this applied force produces a positive shear force at the right end). In addition, there is no distributed load (i.e. q = 0) acting to this beam. Therefore, a boundary value problem formulated in terms of the fourth-order differential equation for this particular beam is given by EI

d4v 0 dx 4

for x  (0, L)

(e3.3.1)

v(0)  0

(e3.3.2)

EIv(0)  M 0

(e3.3.3)

v(L)  0

(e3.3.4)

EIv(L)  P0

(e3.3.5)

By performing four direction integrations of (3.3.1), it simply leads to following results: EI

d3v  C1 dx 3

(e3.3.6)

EI

d2 v  C1 x  C 2 dx 2

(e3.3.7)

EI

dv 1  C1x 2  C 2 x  C3 dx 2

(e3.3.8)

1 1 EIv  C1 x 3  C 2 x 2  C3 x  C 4 6 2

(e3.3.9)



169


Four constants of integration {C1, C2, C3, C4} can then be obtained by using boundary conditions (e3.3.2)-(e3.3.5): v(0)  0



C4  0

EIv(0)  M 0



C2  M 0

EIv(L)  P0



C1  P0

v(L)  0



1 1 C1L2  C 2 L  C 3  0  C 3   M 0 L  P0 L2 2 2


d3 v  P0 dx 3

M(x)  EI

d2 v  P0 x  M 0 dx 2

EI(x)  EI

EIv(x) 

2  x   dv 1  x    P0 L2    1  M 0 L    1 dx 2  L     L 

3 2 1  x   x   x   1  x   P0 L3    3     M 0 L2    2    6  L   2  L    L   L 

Option II: Use second-order differential equation Since the given beam is statically determinate (i.e. ra = 2, nm = 1(2) = 2, nj = 2(2) = 4, nc = 0, then DI = 2 + 2 – 4 – 0 = 0), all support reactions can be obtained by considering equilibrium of the entire beam and the bending moment can readily be computed from moment equilibrium of the left portion as demonstrated below. [MA = 0]  

M B  M 0  P0 L  0  M B  M 0  P0 L

: Y

P0 M0

A

B

MB

X

RAY L

Y

V(x) M0

A

M(x)

RAY x Copyright © 2011 J. Rungamornrat

X


170


[FY = 0]  

:

R AY  P0  0  R AY  P0

[Mx = 0]  

:

M(x)  R AY x  M 0  0  M(x)  P0 x  M 0


d2 v  P0 x  M 0 dx 2

for x  (0, L)

(e3.3.10)

v(0)  0

(e3.3.11)

v(L)  0

(e3.3.12)

Performing two direction integrations of (3.3.10) yields dv 1    P0 x  M 0 dx  C1  P0 x 2  M 0 x  C1 dx 2 1 1 1   EIv    P0 x 2  M 0 x dx  C1x  C 2  P0 x 3  M 0 x 2  C1x  C 2 6 2 2 

EI

(e3.3.13) (e3.3.14)




C2  0

v(L)  0



1 1 P0 L2  M 0 L  C1  0  C1   P0 L2  M 0 L 2 2


:

R AY  V(x)  0  V(x)  P0

Once, the shear force V(x) is obtained, we can formulate a boundary value problem in terms of the third-order differential equation for this particular beam as follows: EI

d3v  P0 dx 3

for x  (0, L)

(e3.3.15)

v(0)  0

(e3.3.16)

EIv(0)  M 0

(e3.3.17)

v(L)  0



171


By performing direction integrations of (3.3.15), it yields following results: EI

d2v  P0 dx  C1  P0 x  C1 dx 2 

(e3.3.19)

EI

dv 1   P0 xdx  C1x  C 2  P0 x 2  C1x  C 2 dx 2

(e3.3.20)

1 1 1 1 EIv   P0 x 2 dx  C1 x 2  C 2 x  C3  P0 x 3  C1x 2  C 2 x  C3 2 2 6 2

(e3.3.21)

Three constants of integration {C1, C2, C3} can be obtained from (e3.3.16)-(e3.3.18) as follows: v(0)  0



C3  0

EIv(0)  M 0



C1  M 0

v(L)  0



1 1 P0 L2  C1L  C 2  0  C 2   P0 L2  M 0 L 2 2

By substituting {C1, C2, C3} into (e3.3.19)-(e3.3.21), we obtain the deflection, the rotation, and the bending moment identical to those obtained in option I. Example 3.4 Consider a beam of length L and constant flexural rigidity EI and constrained by a roller support at the left end and a linear translational spring of constant ks at the right end as shown below. The beam is subjected to a downward linear distributed load q = q0(1 – x/L) and a clockwise moment M0 at the right end. Determine the shear force, bending moment, deflection and the rotation of the beam. Y q = q0(1 – x/L) q0

M0 EI

X

ks

L Solution In this example, we demonstrate how to treat a boundary condition at the beam end connected by a linear translational spring. Option I: Use fourth-order differential equation At the left end of the beam, the deflection is fully prevented by a pinned-end support and the bending moment vanishes while, at the right end, the deflection is partially constrained by a linear translational spring of spring constant ks and the bending moment is prescribed equal to –M0 (the applied moment produces a positive bending moment at the right end). In addition, there is a downward linear distributed load q = q0(1 – x/L) acting to this beam. Therefore, a boundary value problem formulated in terms of the fourth-order differential equation for this particular beam is given by Copyright © 2011 J. Rungamornrat


172


EI

d4 v x   q 0  1   4 dx  L

for x  (0, L)

(e3.4.1)

v(0)  0

(e3.4.2)

EIv(0)  0

(e3.4.3)

EIv(L)  k s v(L)  0

(e3.4.4)

EIv(L)   M 0

(e3.4.5)

By performing four direction integrations of (3.4.1), it simply leads to following results: EI

2 q 0 L  x  d3v x   x     q 1  dx  C     2     C1 0 1  3  dx 2  L   L  L  

(e3.4.6)

EI

2 3 2 q 0 L  x  q 0 L2  x  d2v  x    x     2 dx  C x  C   3   1 2  L   L    C1x  C 2 dx 2  2  L  6  L       

(e3.4.7)

EI

q L2 dv  0 dx 6 

q 0 L3 24

q L3 EIv   0 24 q L4  0 120

2  x 3 1  x   2  3   dx  C1 x  C 2 x  C3   L L 2      

4 3  x   x   1 2    4     C1x  C 2 x  C3 L L 2      

(e3.4.8)

3  x  4 1 1  x   3 2    4   dx  C1 x  C 2 x  C3 x  C 4 6 2  L    L 

4  x 5 1  x   1 3 2    5     C1x  C 2 x  C3 x  C 4 2  L   6  L 

(e3.4.9)

Four constants of integration {C1, C2, C3, C4} can then be obtained by using boundary conditions (e3.4.2)-(e3.4.5): v(0)  0



C4  0

EIv(0)  0



C2  0

EIv(L)   M 0 



q 0 L2 q L M  C1L  C 2   M 0  C1  0  0 3 3 L

EIv(L)  k s v(L)  0  



 q0L k  q L4 1 1  C1  s   0  C1L3  C 2 L2  C3 L  C 4   0 2 EI  30 6 2 

C3 

M0L  6EI  q 0 L3  15EI   1  1   3  6  k s L  45  2k s L3 

By substituting {C1, C2, C3, C4} into (e3.3.6)-(e3.3.9), we obtain the shear force, the bending moment, the deflection and the rotation for the given beam: Copyright © 2011 J. Rungamornrat


V(x)  EI

2 d 3 v q 0 L  x  x   2      3 dx 2  L  L

173


2  M 0  3  L

3 2 d 2 v q 0 L2  x  x  x   x M(x)  EI 2     3    2     M 0   dx 6  L  L  L   L

EI(x)  EI

EIv(x) 

4 3 2 2 dv q 0 L3  x  8 15EI   M 0 L  x  1  6EI   x x 4 4 1             1        3  dx 24  L  2  L  3  k s L3   L  L  15  2k s L  

q 0 L4 120

4 3 3  x 5 15EI   x   M 0 L  x   6EI   x    x  20  x  8  5 1             1          3  3  L  3  2k s L   L   6  L   k s L3   L   L  L 

For a special case associated with k s   , a translational spring now acts as a roller support. Above results, when specialized to this particular case, are given by V(x)  EI

2 d 3 v q 0 L  x  x     2    3 dx 2  L  L

2  M 0  3  L

M(x)  EI

3 2 d 2 v q 0 L2  x  x  x   x   3  2    M 0        2 dx 6  L  L  L   L

4 3 2 2 dv q 0 L3  x  8  M 0 L  x  1  x x EI(x)  EI   4   4         dx 24  L  2  L  3  L  L  15 

q L4 EIv(x)  0 120

4 3 3  x 5  x  20  x  8  x   M 0 L  x   x       5             3  L  3  L   6  L   L   L  L 

Option II: Use second-order differential equation Since the given beam is statically determinate, all support reactions can be obtained by considering equilibrium of the entire beam and the bending moment can readily be computed from moment equilibrium of the left portion as demonstrated below. Y q = q0(1 – x/L) B M0

A RAY

X

RBY L

Y

q = q0(1 – x/L)

RBY V(x)

A

M(x) RAY x Copyright © 2011 J. Rungamornrat

X


174


[MA = 0]  

:

R BY L  M 0 

q0L  L  q0L M0     0  R BY  2 3 6 L

[FY = 0]  

:

R AY  R BY 

q0L q L M  0  R AY  0  0 2 3 L

[Mx = 0]  

:

M(x)  R AY x 

q0 x 2  x  1  0 2  3L 

q L2  M(x)  0 6

2  x 3 x  x   x    3    2     M 0   L  L   L  L 


3 2 d 2 v q 0 L2  x  x  x   x   3  2    M0        2 dx 6  L  L  L   L

for x  (0, L)

v(0)  0

R BY 

(e3.4.10) (e3.4.11)

q0L M0    k s v(L) 6 L

(e3.4.12)

Performing two direction integrations of (3.4.10) yields EI

q L2 dv  0 dx 6

3 2 2  x  4 x  x   M 0 L  x   C1    4    4     2  L  L  L    L 

q L3  0 24 EIv  



q 0 L3 24

q 0 L4 120

2  x 3 x  x   x  3  2    dx   M 0  dx  C1     L L L      L  

(e3.4.13)

4 3 2 2 M0L  x   x  x  x   dx  C1 x  C 2    4    4    dx   2  L  L  L    L 

4 3 3 2  x 5  x  20  x   M 0 L  x   5    C1 x  C 2    L 3  L   6  L     L 

(e3.4.14)




C2  0

q0L M 0    k s v(L)  6 L 

 q0L M0 k  q L4 M L2    s  0  0  C1L  C 2  6 L EI  45 6  C1 

M 0L  6EI  q 0 L3  15EI   1  1   3  6  k s L  45  2k s L3 



175



R AY  V(x) 

:

2 q0 x  q 0 L  x  x x 2   0  V(x)   2    2  L  2  L  L

2  M 0  3  L

Once, the shear force V(x) is obtained, we can formulate a boundary value problem in terms of the third-order differential equation for this particular beam as follows: 2 d 3 v q 0 L  x  x EI 3     2    dx 2  L  L

2  M 0  3  L

for x  (0, L)

(e3.4.15)

v(0)  0

(e3.4.16)

EIv(0)  0

(e3.4.17)

q0L M0    k s v(L) 6 L

R BY 

(e3.4.18)

Note that both (e3.4.3) and (e3.4.5) cannot be treated as boundary conditions at the same time since a sum of end moments was already used in the determination of support reactions. By performing direct integrations of (3.4.15), it yields the following results: q L  x  d2v x   0    2    2 dx 2  L  L 2

EI



2  x 3 x  x   x  3  2     M 0    C1     L L L      L  

q 0 L2 6

q L2 dv EI  0 dx 6 

q 0 L3 24

EIv  

q 0 L3 24



q 0 L4 120

M0 2  dx  C1 dx   3  L

(e3.4.19)

2  x 3 x  x   x    3    2   dx   M 0  dx  C1 x  C 2 L  L   L  L 

3 2 2  x  4 x  x   M 0 L  x   C1 x  C 2    4    4     2  L  L  L    L 

(e3.4.20)

3 2 2  x  4 M0L  x  1 x  x   2  4  4 dx    L L   L  dx  2 C1 x  C 2 x  C3  L 2         

5 4 3 3 2  x   x  20  x   M 0 L  x  1  C1 x 2  C 2 x  C3    5          3  L   6 L 2 L  L 


(e3.4.21)


176


Three constants of integration {C1, C2, C3} can be obtained from (e3.4.16)-(e3.4.18) as follows: v(0)  0



C3  0

EIv(0)  0



C1  0

q0L M 0    k s v(L)  6 L 

 q0L M0 k  q L4 M L2 1    s  0  0  C1L2  C 2 L  C 3  6 L EI  45 6 2  C2 

M 0L  6EI  q 0 L3  15EI  1   1   6  k s L3  45  2k s L3 

By substituting {C1, C2, C3} into (e3.4.19)-(e3.4.21), we obtain the deflection, the rotation, and the bending moment identical to those obtained in option I.

3.6 Treatment of Discontinuity In the previous section, we focused only on beams that have constant flexural rigidity and are subjected to either continuous distributed load on the entire span or concentrated force and moment at their ends. Here, we enhance the capability of the direct integration technique to treat various types of interior discontinuity mostly found in beams, for instance, a point of abrupt change in flexural rigidity (point D in Figure 3.7), a point where q is discontinuous (points B, C, J and K in Figure 3.7), a point where a concentrated force is applied (point F in Figure 3.7), a point where a concentrated moment is applied (point E in Figure 3.7), a hinge point (point I in Figure 3.7), and a shear release (point L in Figure 3.7). Two commonly used techniques, a domain decomposition technique and a discontinuity-function technique, are proposed to handle above situations. Y q(x)

Po

Mo A

B

C

D

E

F

G

X

Y q(x) H

I

J

K

X L

M

Figure 3.7: Schematic of beams subjected to various types of interior discontinuity

3.6.1 Domain decomposition technique A direct integration method, when supplied by a domain decomposition technique, has a capability to treat discontinuity within a beam. The basic idea is to identify locations of discontinuity and use this information to decompose a beam into several segments containing only continuous data. Each segment resulting from the decomposition can therefore be solved independently using the same procedure as described in section 3.5. The total number of unknown constants resulting from the integration depends primarily on the number of segments and the order of governing differential equation employed, and they can be determined from boundary conditions at the beam ends and Copyright © 2011 J. Rungamornrat


177


continuity conditions at the inter-boundary of all segments. For instance, a cantilever beam shown in Figure 3.7 must be decomposed into at least six segments, i.e. segments AB, BC, CD, DE, EF and FG, and constants from integration can be obtained by imposing boundary conditions at points A and G and continuity conditions at points B, C, D, E and F. An important step in the solution procedure is to properly and sufficiently set up continuity conditions at the segment inter-boundary. Following guidelines are useful for setting up continuity conditions for various types of discontinuity.  A point of abrupt change in flexural rigidity: the shear force, the bending moment, the deflection, and the rotation are continuous at this point  A point of discontinuous distributed load: the shear force, the bending moment, the deflection, and the rotation are continuous at this point  A point where a concentrated force is applied: the deflection and the rotation are continuous at this point while the shear force and the bending moment must satisfy the jump conditions (3.18) and (3.19)  A point where a concentrated moment is applied: the deflection and the rotation are continuous at this point while the shear force and the bending moment must satisfy the jump conditions (3.20) and (3.21)  A hinge point: the deflection and the shear force are continuous at this point while the bending moment vanishes and the rotation is discontinuous  A shear release: the rotation and the bending moment are continuous at this point while the shear force vanishes and the deflection is discontinuous Note that if two types of discontinuity or more are present simultaneously at one point, the continuity conditions stated above must be properly combined. For instance, at a point where both the concentrated force and moment are applied, the deflection and rotation are continuous while the shear force and bending moment satisfy the jump conditions (3.18) and (3.21), respectively; at a hinge point subjected to a concentrated force, the deflection is continuous, the shear force satisfies the jump condition (3.18), the rotation is discontinuous, and the bending moment vanishes; and, at a point where the distributed load is discontinuous and a concentrated moment is applied, the deflection and rotation are continuous while the shear force and bending moment satisfy the jump conditions (3.20) and (3.21). Another crucial remark is that the number of continuity conditions at a point of discontinuity must be matching to the choice of a key governing differential equation employed. In general, at each point of discontinuity, the second-order, the third-order, and the fourth-order differential equations require two continuity conditions in terms of the deflection and rotation, three continuity conditions in terms of the deflection, rotation and bending moment, and four continuity conditions in terms of the deflection, rotation, shear force and bending moment, respectively. To clearly demonstrate the domain decomposition technique, let us consider following examples. Example 3.5 Consider a cantilever beam subjected to distributed load and concentrated forces as shown below. The flexural rigidity of the segment AB and segment BC are given by 2EI and EI, respectively. Determine the shear force, bending moment, deflection and the rotation of the beam. Y q0L q0 A

2EI L/2

2q0L B

EI L/2


C

X


178


Solution For this particular beam, the flexural rigidity and loading data are discontinuous at point B; in particular, at this point, the flexural rigidity changes abruptly from 2EI to EI, the distributed load changes abruptly from q0 to 0, and a concentrated load q0L is applied. Therefore, the beam must be decomposed into two segments (i.e. a segment AB and a segment BC) and, clearly, data associated with these two segments are continuous. At point B, the displacement, the rotation, and the bending moment are continuous while the shear force experiences a jump given by (3.18). Here, we demonstrate the domain decomposition technique when either a second-order, third-order or the fourth-order differential equations is employed as a key governing equation. Option I: Use fourth-order differential equation A boundary value problem formulated in terms of the fourth-order differential equation for this particular beam is given by Segment AB:

2EI

Segment BC:

EI

d4v  q 0 dx 4

d4v 0 dx 4

for x  (0, L/2)

for x  (L/2, L)

Boundary conditions: v(0)  0

(e3.5.2) (e3.5.3a) (e3.5.3b) (e3.5.3c) (e3.5.3d)

v(0)  0 EIv(L)  0 EIv(L)  2q 0 L

Continuity conditions: v(L/2 )  v(L/2 ) v(L/2 )  v(L/2 ) 2EIv(L/2 )  EIv(L/2 ) 

(e3.5.1)



2EIv(L/2 )  EIv(L/2 )  q 0 L

(e3.5.4a) (e3.5.4b) (e3.5.4c) (e3.5.4d)

where the symbols x  and x  denote the left limiting point and the right limiting point of a point x, respectively. By performing direct integrations of (3.5.1) and (3.5.2), it leads to following results: d3v  q 0 x  C1 for x  (0, L/2) dx 3 d3 v EI 3  C5 for x  (L/2,L) dx d2v 1 2EI 2   q 0 x 2  C1 x  C 2 for x  (0, L/2) dx 2 2 d v EI 2  C5 x  C6 for x  (L/2,L) dx dv 1 1 2EI   q 0 x 3  C1 x 2  C 2 x  C3 for x  (0, L/2) dx 6 2 dv 1 EI  C5 x 2  C 6 x  C 7 for x  (L/2,L) dx 2 1 1 1 2EIv   q 0 x 4  C1x 3  C 2 x 2  C3 x  C 4 for x  (0, L/2) 24 6 2 1 1 EIv  C5 x 3  C6 x 2  C7 x  C8 for x  (L/2,L) 6 2

2EI


(e3.5.5a)

(e3.5.5b)

(e3.5.5c)

(e3.5.5d)


179


Eight constants resulting from the integration {C1, C2, C3, C4} and {C5, C6, C7, C8} can be obtained from boundary conditions (e3.5.3a)-(e3.5.3d) and continuity conditions (e3.5.4a)-(e3.5.4d) as follows: v(0)  0  v(0)  0  EIv(L)  2q 0 L  EIv(L)  0 

C4  0 C3  0 C5  2q 0 L C5 L  C6  0  C6  2q 0 L2 q0L 7q L  C1  C5  q 0 L  C1  0 2 2

2EIv(L/2 )  EIv(L/2 )  q 0 L 



2EIv(L/2 )  EIv(L/2 )



21q 0 L2 1 1 1  q 0 L2  C1L  C 2  C5 L  C6  C 2   8 2 2 8

v(L/2 )  v(L/2 )

1  1 1 1 1  1 1  3 2 2   q 0 L  C1L  C 2 L    C5 L  C6 L  C7  2EI  48 8 2 2  EI  8  29q 0 L3 C7  96

 

v(L/2 )  v(L/2 )

 

1  1 1 1 1 1  1  1   q 0 L4  C1L3  C 2 L2    C 5 L3  C 6 L2  C 7 L  C8  2EI  384 48 8 EI 48 8 2    55q 0 L4 C8   768

By substituting {C1, C2, C3, C4} and {C5, C6, C7, C8} into (e3.5.5a)-(e3.5.5d), we obtain the shear force, the bending moment, the deflection and the rotation for the given beam:   x  7  d3 v 2EI   q 0 L      dx 3   L  2  V(x)   3  dv  EI dx 3  2q 0 L

for x  (0, L/2) for x  (L/2,L)

2  q L2  x  d2v  x  21   2EI 2   0    7     for x  (0, L/2) dx 2  L   L  4   M(x)     d2v 2  x  for x  (0, L/2)  EI dx 2  2q 0 L  L   1    

 q L3  x 3 21  x  2 63  x     0          2 L 4  L    12EI  L  (x)   2  q 0 L3  x   x  29    EI  L   2  L   96       

for x  (0, L/2) for x  (L/2,L)

3 2  q L4  x  4  x  63  x   for x  (0, L/2)   0    14       2  L   L  48EI  L  v(x)   3 2  q 0 L4  x   x  29  x  55  3   L   32  L   256  for x  (L/2,L)  3EI  L         



180


Option II: Use second-order differential equation Since the given beam is statically determinate, we use a method of sections to determine the bending moment at any point within the segment AB and segment BC as demonstrated below: Y 2q0L

V(x) FBD 1

A

C

M(x) x

L–x q0L

Y

FBD 2

X

q0

V(x) M(x)

2q0L B

A x

C

X

L/2

L/2 – x 1 2

FBD 2: M(x)  2q 0 L(L  x)  q 0 L(L/2  x)  q 0 (L/2  x) 2  

q 0 L2 2

 x  2  x  21     7      L  4   L 

 x  FBD 1: M(x)  2q 0 L(L  x)  2q 0 L2    1  L 



Now, a boundary value problem formulated in terms of the fourth-order differential equation for this particular beam is given by Segment AB:

2EI

Segment BC:

EI

q 0 L2 d2v   dx 2 2

 x  2  x  21     7      L  4   L 

 x   d2 v  2q 0 L2    1 2 dx  L  

for x  (0, L/2)

for x  (L/2, L)


(e3.5.7) (e3.5.8a) (e3.5.8b)

v(0)  0

Continuity conditions: v(L/2 )  v(L/2 )

(e3.5.9a) (e3.5.9b)

v(L/2 )  v(L/2 ) 

(e3.5.6)



Performing direct integrations of (3.5.6) and (3.5.7) yields following results: 2EI

q L3 dv  0 dx 6

3 2  x  21  x  63  x             C1 2 L 4  L    L 

 x  2 dv  x   EI  q 0 L3    2     C3 dx  L    L 

for x  (0, L/2)

(e3.5.10a) for x  (L/2,L)



2EIv  

q 0 L4 24

q L4 EIv  0 3

181


4 3 2  x   x  63  x      14        C1x  C 2 2  L   L  L 

2  x 3  x   3        C3 x  C 4  L    L 

for x  (0, L/2)

(e3.5.10b) for x  (L/2,L)

Four constants resulting from the integration {C1, C2} and {C3, C4} can be obtained from boundary conditions (e3.5.8a)-(e3.5.8b) and continuity conditions (e3.5.9a)-(e3.5.9b) as follows: v(0)  0 v(0)  0

 

C1  0 C2  0

v(L/2 )  v(L/2 )



v(L/2 )  v(L/2 )



 1  3q 0 L3  29q 0 L3 1  43q 0 L3   C    C  C    1 3 3 2EI  48 4 96  EI    1  5q L4 1  55q 0 L4 1  33q 0 L4 1  C1L  C 2     0  C3 L  C 4   C 4    2EI  128 2 24 2 768  EI  

By substituting {C1, C2} and {C3, C4} into (e3.5.10a)-(e3.5.10b), we obtain the same deflection and the same rotation as those obtained in option I. Once the deflection v is solved, the shear force V(x) can readily be obtained using the relation (3.29a) and the final result is identical to that obtained in option I. Option III: Use third-order differential equation Again, by considering force equilibrium of portions of the beam whose FBD indicated above, we can readily obtain the shear force at any point x, V(x), as  x 7 Segment AB: V(x)  2q 0 L  q 0 L  q 0 (L/2  x)  q 0 L      L 

2

Segment BC: V(x)  2q 0 L Now, a boundary value problem formulated in terms of the fourth-order differential equation for this particular beam is given by Segment AB:

2EI

Segment BC:

EI

 x  7  d3v   q 0 L     dx 3  L  2 

d3v  2q 0 L dx 3

for x  (0, L/2)

for x  (L/2, L)


(e3.5.12) (e3.5.13a) (e3.5.13b) (e3.5.13c)

v(0)  0 EIv(L)  0

Continuity conditions: v(L/2 )  v(L/2 ) v(L/2 )  v(L/2 ) 2EIv(L/2 )  EIv(L/2 ) 

(e3.5.11)



Performing three direct integrations of (3.5.11) and (3.5.12) leads to following results: Copyright © 2011 J. Rungamornrat

(e3.5.14a) (e3.5.14b) (e3.5.14c)


182


2EI

q 0 L2 d2v   dx 2 2

2  x   x      7     C1  L    L 

for x  (0, L/2)

(e3.5.15a)

2

d v  2q 0 Lx  C 4 for x  (L/2,L) dx 2 3 2 q L3  x  21  x   dv 2EI   0        C1x  C 2 for x  (0, L/2) dx 6  L  2  L   EI

(e3.5.15b)

2

dv x  q 0 L2    C 4 x  C5 EI dx L 2EIv   EIv 

q 0 L4 24 4

for x  (L/2,L)

3  x  4  x   1 2  14       C1 x  C 2 x  C3 L L 2     

for x  (0, L/2)

(e3.5.15c)

3

q0L  x  1 2    C 4 x  C5 x  C6 3 L 2

for x  (L/2,L)

Six constants resulting from the integration {C1, C2, C3} and {C4, C5, C6} can be obtained from boundary conditions (e3.5.13a)-(e3.5.13c) and continuity conditions (e3.5.14a)-(e3.5.14c) as follows: v(0)  0 v(0)  0 EIv(L)  0

 

C3  0 C2  0



C 4  2q 0 L2

2EIv(L/2 )  EIv(L/2 ) 

v(L/2 )  v(L/2 )



v(L/2 )  v(L/2 )



13q 0 L2 21q 0 L2  C1  q 0 L2  C 4  C1   8 8  1  q 0 L2 1  29q 0 L3 1  5q 0 L3 1       C L C L C C   1  4 5 5 2EI  12 2 2 96  EI  4   55q 0 L4 1  9q 0 L4 1 1  q 0 L4 1 1 2 2  C L   C L  C L  C  C      1 4 5 6 6 2EI  128 8 2 768  EI  24 8 

By substituting {C1, C2, C3} and {C4, C5, C6} into (e3.5.15a)-(e3.5.15c), we obtain the deflection, the rotation, and the bending moment identical to those obtained in option I. Example 3.6 Determine the deflection and rotation of a prismatic beam of constant flexural rigidity EI as shown below by using a second-order differential equation Y q0

q0L2 A

C

B L/2

L/4

D

X

L/4

Solution For a given beam, the prescribed data is discontinuous at points B and C; in particular, at point B, a concentrated moment is applied and, at point C, the distributed load changes abruptly from 0 to q0 and a concentrated load (in terms of a support reaction) is applied. Therefore, the beam Copyright © 2011 J. Rungamornrat


183


must be decomposed into at least three segments (e.g. a segment AB, a segment BC and a segment CD). To formulate a boundary value problem in terms of a second-order differential equation, we first determine all support reactions and the bending moment M(x) for all three segments by using static equilibrium along with the method of sections as demonstrated below: Y q0

2

q0L

A

FBD 1

C

B

RAY L/2

Y

D

RCY

L/4

X

L/4

V(x) A

FBD 2

RAY x

Y

q0L2

A

FBD 3

X

M(x)

V(x)

RAY

X

M(x)

B x

Y q0

2

q0 L

A

FBD 4

B

RAY

C

V(x)

RCY

M(x)

X

x 13q 0 L 1 q 0 L2  q 0 (L/4)(7L/8)  3L/4 8 11q 0 L  q 0 (L/4)  R CY   8

FBD 1: R CY  R AY



FBD 2: M(x)  R AY x  



11q 0 L2  x    8 L

FBD 3: M(x)  R AY x  q 0 L2  

11q 0 L2 8

 x  8       L  11  1 2

FBD 4: M(x)  R AY x  q 0 L2  R CY (x  3L/4)  q 0 (x  3L/4) 2  

q 0 L2 2

 x  2  x      2    1  L    L 

Since both the deflection and rotation at point B and C must be continuous, a boundary value problem formulated in terms of the second-order differential equation for this beam is given by Copyright © 2011 J. Rungamornrat


184


Segment AB:

EI

11q 0 L2  x  d2v     dx 2 8 L

Segment BC:

EI

11q 0 L2 d2v   dx 2 8

Segment BC:

EI

q 0 L2 d2v   dx 2 2

for x  (0, L/2)

 x  8       L  11 

(e3.6.1)

for x  (L/2, L/4)

 x  2  x      2    1  L    L 

(e3.6.2)

for x  (L/4, L)


(e3.6.3) (e3.6.4)

Continuity conditions: v(L/2 )  v(L/2 )

(e3.6.5a) (e3.6.5b) (e3.6.5c)

v(L/2 )  v(L/2 ) v(3L/4 )  v(3L/4 ) 



Interior support & continuity condition: v(3L/4 )  0

(e3.6.6a) (e3.6.6b)



v(3L/4 )  0

Note that conditions (e3.6.6a) and (e3.6.6b) result from the presence of a roller support at point C and the continuity of the deflection. By performing direct integrations of (e3.6.1)-(e3.6.3), it leads to following results: 2  3   11q 0 L  x   C for x  (0, L/2) 1  16  L   2 dv  11q 0 L3  x  16  x     EI for x  (L/2,3L/4)        C 3 dx  16  L  11  L    3 2 3 x  x    q 0 L  x  3      3     C5 for x  (3L/4, L)  6  L  L  L     3  4   11q 0 L  x   C x  C for x  (0, L/2) 1 2    48  L   3 2  11q 0 L4  x  24  x   EIv    for x  (L/2,3L/4)        C3 x  C 4 48  L  11  L     4 3 2 4 x  x    q 0 L  x  4 6          C5 x  C6 for x  (3L/4, L)  24  L  L   L     

(e3.6.7a)

(e3.6.7b)

Six constants resulting from the integration {C1, C2}, {C3, C4} and {C5, C6} can be obtained from boundary and continuity conditions (e3.6.4)-(e3.6.6b) as follows: v(0)  0



v(L/2 )  v(L/2 )

v(3L/4 )  v(3L/4 ) v(3L/4 )  0



C2  0

(e3.6.8a) 3

3

3

11q 0 L 21q 0 L q L  C1   C3  C1  C3  0 64 64 2 93q 0 L3 21q 0 L3 135q 0 L3   C3    C5  C3  C 5   256 128 256 189q 0 L4 3 189q 0 L4  C3 L  C 4  0  3C3 L  4C 4   1024 4 256






(e3.6.8b) (e3.6.8c) (e3.6.8d)


v(3L/4 )  0



v(L/2 )  v(L/2 )

185


171q 0 L4 3 171q 0 L4  C5 L  C6  0  3C5 L  4C6  2048 4 512 11q 0 L4 1 37q 0 L4 1    C1L  C 2   C3 L  C 4 384 2 384 2 q 0 L4  C1L  C3 L  2C 4  4 

(e3.6.8e)

(e3.6.8f)

By solving a system of linear equations (e3.6.8b)-(e3.6.8f), we obtain C1 

67q 0 L3 317q 0 L3 11q 0 L3 q L4 5q L4 , C3   , C5  , C 4  0 , C6   0 768 768 96 8 2048

By substituting {C1, C2}, {C3, C4} and {C5, C6} into (e3.6.7a)-(e3.6.7b), we obtain the deflection and the rotation for the given beam:  11q L3  x  2 67  0       16EI  L  528   2  11q 0 L3  x  16  x  317  (x)            16EI  L  11  L  528   3 2 3 x  x  11   q 0 L  x    3    3     6EI  L  L  L  16   

for x  (0, L/2) for x  (L/2,3L/4) for x  (3L/4, L)

 11q L4  x 3 67  x   0        48EI  L  176  L    3 2  11q 0 L4  x  24  x  317  x  6  v(x)              48EI  L  11  L  176  L  11   4 3 2 4 x  x  11  x  15   q 0 L  x  4 6             24EI  L  4  L  256  L L  

for x  (0, L/2) for x  (L/2,3L/4) for x  (3L/4, L)

Example 3.7 Determine the deflection and rotation of a beam shown below by using a second-order differential equation Y q0 A

3EI

B

2L/3

EI

C

X

L/3

Solution For a given beam, the prescribed data is discontinuous at point B; in particular, at this point, the flexural rigidity changes abruptly from 3EI to EI, the distributed load changes abruptly from 0 to q0, and a hinge is present. From this discontinuity information, we decompose the beam into two segments (i.e. a segment AB and a segment BC). Since the beam is statically determinate, Copyright © 2011 J. Rungamornrat


186


all support reactions and the bending moment M(x) for both segments can be obtained by using static equilibrium along with the method of sections as demonstrated below: Y q0

VB FBD 1

B Y

FBD 2

MA

A B 2L/3

Y

MA

X

C RCY

X

L/3 q0

RAY

FBD 3

C RCY

L/3 V(x) M(x)

A

X

RAY x Y V(x) M(x)

FBD 4

q0 EI

x

C RCY

X

L–x

q L 1  q 0 (L/3)(L/6)   0 L/3 6 q0L q0L q L2 ; M A  q 0 (L/3)(5L/6)  R CY L  0 FBD 2: R AY   R CY  3 6 9 2 q L  x 2 FBD 3: M(x)  R AY x  M A  0     6  L  3 

FBD 1: R CY 

1 2

FBD 4: M(x)  R CY (L  x)  q 0 (L  x) 2 

q 0 L2 2

2  x  5  x  1          L  6  L  3 

Since the deflection is continuous at hinge B while the rotation is discontinuous, a boundary value problem formulated in terms of the second-order differential equation for this beam is given by Segment AB:

EI

d 2 v q 0 L2  x  2      dx 2 6  L  3 

Segment BC:

EI

2 d 2 v q 0 L2  x  5  x  1       dx 2 2  L  6  L  3 

for x  (0, 2L/3) for x  (2L/3, L)


(e3.7.1) (e3.7.2)


187



(e3.7.3a) (e3.7.3b) (e3.7.3c) (e3.7.4)

v(0)  0 v(L)  0

Continuity conditions: v(2L/3 )  v(2L/3 ) By performing direct integrations of (e3.7.1)-(e3.7.2), it leads to following results: 3EI

2 dv q 0 L2  x  4  x           C1 dx 12  L  3  L  

for x  (0, 2L/3)

3 2 dv q 0 L2  x  5  x   x   EI            C3 dx 6  L  4  L   L  

3EIv 

q 0 L2 36

q L2 EIv  0 24

(e3.7.5a) for x  (2L/3, L)

2  x 3  x    2       C1 x  C 2  L    L 

for x  (0, 2L/3)

2  x  4 5  x 3  x   2           C3 x  C 4  L    L  3  L 

(e3.7.5b) for x  (2L/3, L)

Four constants resulting from the integration {C1, C2} and {C3, C4} can be obtained from boundary conditions (e3.7.3a)-(e3.7.3c) and continuity condition (e3.7.4) as follows: v(0)  0 v(0)  0

 

C2  0 C1  0

q 0 L2 q L2  C3 L  C 4  0  C3 L  C 4   0 36 36 2  1  2q L2 2  2q L2 1  4q 0 L 2 v(2L/3 )  v(2L/3 )   C1L  C 2     0  C3 L  C 4   2C3 L  3C 4  0  3EI  243 3 3 243  EI  243  v(L)  0



By solving the last two linear equations simultaneously, we obtain C3  

89q 0 L3 31q 0 L4 , C4  972 486

By substituting {C1, C2} and {C3, C4} into (e3.7.5a)-(e3.7.5b), we then obtain  q L2  x  2 4  x    0        36EI  L  3  L   (x)   3 2  q 0 L2  x  5  x   x  89     6EI  L  4  L    L   162         

for x  (0, 2L/3) for x  (2L/3, L)

2  q L2  x 3  x    0    2     L   108EI  L  v(x)   4 3 2  q 0 L2  x  5  x   x  178  x  124  2        L   81  L   81         24EI  L  3  L 

for x  (0, 2L/3) for x  (2L/3, L)



188


3.6.2 Discontinuity-function technique As evident from the previous subsection, the number of integration constants involved in the domain decomposition technique becomes vast as the number of segments resulting from the decomposition increases. This therefore renders the technique losing popularity and impractical for analysis of beams under complex loading conditions; more precisely, when several points of discontinuity are introduced. To reduce such cumbersome determination of integration constants, a solution technique based upon a single domain formulation is more preferable. Loading discontinuities present within a beam may be represented in terms of discontinuous functions and, as a result, the distributed load q, the shear force V and the bending moment M (appearing as input data in the fourth-order, third-order, and second-order differential equations, respectively) can be expressed in terms of a single function for the entire beam. Before demonstrating application of this technique to handle loading discontinuity, we first introduce certain special functions and their properties essential for further use. Let a be a real number and H(x – a) be a real-value function defined by 0 H (x  a )   1

for x  a

(3.52)

for x  a

This special function is commonly termed as unit-step or Heaviside function due to its graphical interpretation as shown in Figure 3.8. It is evident that H is continuous everywhere except at x = a where it experiences a unit jump, i.e. [ H ](a)  limH (a  )  H (a  )  1

(3.53)

0

Due to the finite jump at point a, it can readily be verified that a definite integral of the Heaviside function over an interval of infinitesimal length and containing a point a always vanishes, i.e. a 

lim  0

a 

a 

 H (x  a)dx  lim  H (x  a)dx  lim  1dx  lim   0

a 

 0

a

 0

a

 0

(3.54)

Next, let (x – a) and(x – a) be defined in terms of the first and second derivatives of H(x – a) as follows: (x  a ) 

dH (x  a ) dx

(3.55)

 (x  a ) 

d(x  a ) d 2 H (x  a )  dx dx 2

(3.56)

Note that both (x – a) and (x – a) defined by (3.55) and (3.56) are not functions defined in an ordinary sense but, in fact, they are commonly termed the distribution functional (comprehensive details for this topic can easily be found in textbooks of mathematics). Both (x – a) and (x – a) vanish everywhere except at x = a where both distributions are not well-defined; graphical interpretation of these two distributions is shown in Figure 3.8. By integrating (3.55) and (3.56) over an interval containing the point a, it leads to following useful results: a 

 (x  a)dx  H (a  )  H (a  )  1

(3.57)

a 



189


H(x – a) 1

x=a

x

(x – a)

x=a

x

(x – a)

x=a

x

Figure 3.8: Graphical interpretation of functions H(x – a), (x – a) and (x – a) a 

 (x  a)dx  (a  )  (a  )  0

(3.58)

a 

In addition, it can also be verified that

 (x  a)dx  H (x  a)  C

(3.59)

 (x  a)dx  (x  a)  C

(3.60)

1

2

where C1 and C2 are arbitrary constants of integration. Now, let f be a continuous function and define a cut-off function of the function f at point a, denoted by f a , by 0 f a (x)  f (x) H (x  a )    f (x)

for x < a for x > a

(3.61)

The graphical interpretation of the cut-off function f a is shown in Figure 3.9. It is obvious either from the definition or from its graph, the cut-off function f a takes the value of the function f for x > a while vanishing for x < a. We state without proof following two properties regarding to the differentiation and integration of the cut-off function: df a (x) df (x)  H (x  a )  f (a )(x  a ) dx dx



(3.62)

x

f a (x)dx  H (x  a )  f (x)dx + C

(3.63)

xA



190


f(x)

f a (x)

x

x=a Figure 3.9: Graphical interpretation of function f and it cut-off function f a Y

q(x) P

MO

xA

M xB

xC

X

RO

Figure 3.10: Graphical interpretation of function f and it cut-off function f a where C is an arbitrary constant of integration. For the special case when f vanishes at x = a, i.e. f(a) = 0, the formula (3.62) simply reduces to df a (x) df (x)  H (x  a ) dx dx

(3.64)

To clearly demonstrate how to treat loading discontinuities (e.g. discontinuous distributed loads, concentrated forces and concentrated moments) within a beam using special functions stated above, let us consider a model problem as shown in Figure 3.10. The prismatic cantilever beam is assumed to be subjected to a concentrated load P at x = xA, a concentrated moment M at x = xB and a distributed load q0(x) from x = xC to x = L. The primary objective here is to represent all applied loads in terms of a single representation of the distributed load q(x) valid for the entire beam.  The discontinuity of the distributed load can readily be treated by using special features of the cut-off function. For this particular case, the distributed load shown in Figure 3.10 can simply be represented by q(x)  q 0 (x) H (x  x A )

(3.65)

Obviously, this single representation provides the right behavior and complete information of the distributed load for the entire beam; more precisely, it vanishes for x < xC and is equal to q0(x) for x > xC.  For a concentrated load P, it is motivated by its localized nature (i.e. it is applied only at x = xA) and the jump conditions (3.18) and (3.19) to represent a concentrated load in terms of a distributed load of a form q(x)  P(x  x A )



191


To verify the representation (3.66), let us explore behavior of the shear force, bending moment and the jump conditions at x = xA. By substituting (3.66) into (3.12) and then performing a direct integration, it leads to V(x) =  P(x  x A )dx = PH (x  x A )  C1

(3.67)

where C1 is an arbitrary constant of integration. The shear force given by (3.67) possesses the right behavior in that (i) the jump condition (3.18) is satisfied and (ii) its function form is correct (i.e. the concentrated force can produce only constant shear force). Now, let investigate the bending moment. By performing an integration of (3.13) with use of (3.67), we obtain x

M(x) =  PH (x  x A )dx  C1 x = H (x  x A )  Pdx  C1 x  C 2  (x  x A )PH (x  x A )  C1 x  C 2 (3.68) xA

where C2 is another arbitrary constant of integration. The bending moment given by (3.68) also possesses the right behavior since (i) the jump condition (3.19) is satisfied and (ii) its function form is correct (i.e. the concentrated force can produce up to linear bending moment).  For a concentrated moment M, it is suggested by its localized nature (i.e. it is applied only at x = xA) along with the jump conditions (3.20) and (3.21) to represent a concentrated moment in terms of the distribution (x – a) as follows: q(x)  M (x  x A )

(3.69)

Substituting (3.69) into (3.12) and then performing a direct integration yield the shear force V(x) =  M (x  x A )dx = M(x  x A )  C1

(3.70)

where C1 is an arbitrary constant of integration. The shear force given by (3.70) possesses the right behavior in that (i) the jump condition (3.21) is satisfied and (ii) the function form is correct (i.e. the concentrated moment can produce only constant shear force). Next, by integrating (3.13) along with the result (3.70), we obtain the bending moment within a beam: M(x) =  M(x  x A )dx  C1 x = MH (x  x A )  C1 x  C 2

(3.71)

where C2 is an arbitrary constant of integration. The bending moment obtained from (3.71) also possesses the right behavior since (i) the jump condition (3.21) is satisfied and (ii) the function form is correct (i.e. the concentrated moment can produce up to linear bending moment). By incorporating (3.65), (3.66) and (3.69), all applied loads to the cantilever beam shown in Figure 3.10 can be represented by a single representation: q(x)  q 0 (x) H (x  x A )  P(x  x A )  M (x  x A )


(3.72)


192


With the representation (3.72), this beam can now be solved using the fourth-order differential equation without decomposing the beam into four separate segments. It is important to emphasize that direct integrations involved in the solution procedure must carefully be treated via properties (3.59), (3.60) and (3.63). By substituting (3.72) into (3.12) and then integrating the result, we obtain x

V(x) = H (x  x A )  q 0 (x)dx  PH (x  x A )  M(x  x A )  R O

(3.73)

xA

where a constant of integration Ro is obtained by using the boundary condition V(0) = Ro. Again, the shear force given by (3.73) can also be used along with the third-order differential equation to solve this particular problem without domain decomposition. Alternatively, by substituting (3.73) into (3.13) and then integrating the result, it leads to x

M(x) = H (x  x A ) 

x

q

0

(3.74)

(x)dxdx  (x  x A )PH (x  x A )  MH (x  x A )  R O x  M O

xA xA

where a constant of integration –Mo is obtained by using the boundary condition M(0) = –Mo. This known moment M(x) can also be used along with the second-order differential equation to solve this problem. Applications of the discontinuity-function technique along with the solution procedure by direct integration technique for all three types of the governing differential equations are shown in examples below. Example 3.8 Determine the deflection and rotation of a prismatic beam shown below by using a second-order, third-order and fourth-order differential equations Y q0

q0L q0L2

X

(c) L/3

L/6

L/6

L/3

Solution From prescribed loadings and boundary conditions provided by pinned and roller supports, we obtain q(x)  q 0  H (x  L/3)  1  q 0 L(x  L/2)  q 0 L2  (x  2L/3)

(e3.8.1)

v(0)  0

(e3.8.2a)

EIv(0)  0

(e3.8.2b)

v(L)  0

(e3.8.2c)

EIv(L)  0

(e3.8.2d)

Option I: Use fourth-order differential equation A boundary value problem for this particular beam formulated in terms of the fourth-order differential equation is given by Copyright © 2011 J. Rungamornrat


EI

193


d4v  q 0 H (x  L/3)  1  q 0 L(x  L/2)  q 0 L2  (x  2L/3) dx 4

for x  (0, L)

(e3.8.2)

v(0)  0

(e3.8.3a)

EIv(0)  0

(e3.8.3b)

v(L)  0

(e3.8.3c)

EIv(L)  0

(e3.8.3d)


d3v L L L 2L    q 0 (x  ) H (x  )  x   q 0 LH (x  )  q 0 L2 (x  )  C1 3 dx 3 3 2 3  

(e3.8.3)

EI

d2v L L 1  L L 2L 1  q 0  (x  ) 2 H (x  )  x 2   q 0 L(x  ) H (x  )  q 0 L2 H (x  )  C1 x  C 2 2 dx 3 3 2  2 2 3 2

(e3.8.4)

EI

dv L L 1  q L L L 2L 2L 1  q 0  (x  )3 H (x  )  x 3   0 (x  ) 2 H (x  )  q 0 L2 (x  ) H (x  ) dx 6 3 3 6 2 2 2 3 3   1  C1x 2  C 2 x  C3 2

L L 1 4  q0L L L q L2 2L 2 2L 1 EIv  q 0  (x  ) 4 H (x  )  x  (x  )3 H (x  )  0 (x  ) H (x  ) 24 3 3 24 6 2 2 2 3 3   1 1  C1x 3  C 2 x 2  C3 x  C 4 6 2

(e3.8.5)

(e3.8.6)

Four constants of integration {C1, C2, C3, C4} can then be obtained by using boundary conditions (e3.8.3a)-(e3.8.3d): v(0)  0



C4  0

EIv(0)  0



C2  0

EIv(L)  0



2q L 2 1 q L q 0 L2     0  q 0 L2  C1L  C 2  0  C1   0 9 2 2 9  

v(L)  0



1  q L4 q L4 1 1  2 q 0 L4     0  0  C1L3  C 2 L2  C3 L  C 4  0 18 6 2  243 24  48



C3 

2

139q 0 L3 3888

By substituting {C1, C2, C3, C4} into (e3.8.3)-(e3.8.6), we obtain the shear force, the bending moment, the deflection and the rotation for the given beam: L L L 2L 2q 0 L   V(x)  q 0 (x  ) H (x  )  x   q 0 LH (x  )  q 0 L2 (x  ) 3 3 2 3 9   L L 1  L L 2L 2q 0 Lx 1 M(x)  q 0  (x  ) 2 H (x  )  x 2   q 0 L(x  ) H (x  )  q 0 L2 H (x  ) 2 3 3 2 2 2 3 9   Copyright © 2011 J. Rungamornrat


194


L L 1  q L L L 2L 2L q 0 Lx 2 139q 0 L3 1 EI(x)  q 0  (x  )3 H (x  )  x 3   0 (x  ) 2 H (x  )  q 0 L2 (x  ) H (x  )  3 3 6  2 2 2 3 3 9 3888 6 q L2 L L 1 4  q0L L L 2L 2 2L q 0 Lx 3 139q 0 L3 x 1 EIv(x)  q 0  (x  ) 4 H (x  )  x  (x  )3 H (x  )  0 (x  ) H (x  )  3 3 24  6 2 2 2 3 3 27 3888  24

Option II: Use third-order differential equation Y q0

q0L q0L2 EI

RAY

X RBY

By considering equilibrium of the entire beam, we then obtain support reactions as follows: [MA = 0]  

:

R BY L  q 0 (L/3)(L/6)  q 0 L(L/2)  q 0 L2  0  R BY 

[FY = 0]  

:

R AY  R BY 

14q 0 L 9

q0L 2q L  q 0 L  0  R AY   0 3 9

By first substituting (e3.8.1) into (3.12), then integrating the result and finally determining a constant of integration from V(0) = RAY = –2q0L/9, we obtain the shear force V(x) L L L 2L 2q 0 L   V(x)  q 0 (x  ) H (x  )  x   q 0 LH (x  )  q 0 L2 (x  ) 3 3 2 3 9  

(e3.8.7)

Once V(x) is obtained, we can formulate a boundary value problem in terms of the third-order differential equation for this particular beam as follows: EI

d3v L L L 2L 2q 0 L    q 0 (x  ) H (x  )  x   q 0 LH (x  )  q 0 L2 (x  ) 3 dx 3 3 2 3 9  

for x  (0, L)

(e3.8.9)

v(0)  0

(e3.8.10a)

EIv(0)  0

(e3.8.10b)

v(L)  0

(e3.8.10c)

Performing direction integrations of (e3.8.9) yields EI

d2v L L 1  L L 2L 2q 0 Lx 1 )  q 0  (x  ) 2 H (x  )  x 2   q 0 L(x  ) H (x  )  q 0 L2 H (x   C1 2 dx 3 3 2  2 2 3 9 2

EI

dv L L 1  q L L L 2L 2L 1 ) H (x  )  q 0  (x  )3 H (x  )  x 3   0 (x  ) 2 H (x  )  q 0 L2 (x  dx 3 3 6  2 2 2 3 3 6



q 0 Lx 2  C1x  C2 9

(e3.8.11)



195


L L 1 4  q0L L L q L2 2L 2 2L 1 EIv  q 0  (x  ) 4 H (x  )  x  (x  )3 H (x  )  0 (x  ) H (x  ) 3 3 24  6 2 2 2 3 3  24



q 0 Lx 3 1  C1x 2  C2 x  C3 27 2

(e3.8.13)

Three constants of integration {C1, C2, C3} can be obtained from boundary conditions (e3.8.10a) and (e3.8.10c) as follows: v(0)  0



C3  0

EIv(0)  0



C1  0

v(L)  0



139q 0 L3 1  q L4 q L2 q L4 1  2 q 0 L4     0  0  0  C1L2  C 2 L  C3  0  C 2  18 27 2 3888  243 24  48

By substituting {C1, C2, C3} into (e3.8.11)-(e3.8.13), we obtain the same shear force, deflection and rotation as those obtained in option I. Option III: Use second-order differential equation By substituting (e3.8.7) into (3.13), then integrating the result and finally determining the constant of integration from M(0) = 0, it yields L L 1  L L 2L 2q 0 Lx 1 M(x)  q 0  (x  ) 2 H (x  )  x 2   q 0 L(x  ) H (x  )  q 0 L2 H (x  ) 3 3 2  2 2 3 9 2

(e3.8.14)


d2v L L 1  L L 2L 2q 0 Lx 1 )  q 0  (x  ) 2 H (x  )  x 2   q 0 L(x  ) H (x  )  q 0 L2 H (x  2 dx 3 3 2  2 2 3 9 2

for x  (0, L)

(e3.8.15) v(0)  0

(e3.8.16)

v(L)  0

(e3.8.17)


dv L L 1  q L L L 2L 2L q 0 Lx 2 1  q 0  (x  )3 H (x  )  x 3   0 (x  ) 2 H (x  )  q 0 L2 (x  ) H (x  )  C1 dx 3 3 6  2 2 2 3 3 9 6

(e3.8.18) L L 1 4  q0L L L q L2 2L 2 2L q 0 Lx 3 1 EIv  q 0  (x  ) 4 H (x  )  x  (x  )3 H (x  )  0 (x  ) H (x  ) 3 3 24  6 2 2 2 3 3 27  24  C1x  C2 (e3.8.19)




C2  0 Copyright © 2011 J. Rungamornrat


196




v(L)  0

139q 0 L3 1  q L q L2 q L3  2 q 0 L4     0  0  0  C1L  C 2  0  C1  18 27 3888  243 24  48

By substituting {C1, C2} into (e3.8.18)-(e3.8.19), we obtain the same deflection and the same rotation as those obtained in option I. Once the deflection v is solved, the shear force can readily be obtained using the relation (3.29a) and, again, the final result is identical to that obtained in option I. Example 3.9 Determine the deflection and rotation of the beam shown below by using the secondorder differential equation along with the discontinuity-function technique Y q0 L A

C

B L/4

2q0L

q0

2

L/4

D

L/4

X

L/4

Solution Support reactions are obtained first by considering equilibrium of the entire structure as shown below Y 2q0L

q0

q0L2 A

B

C

D

X

RBY

RAY [MA = 0]  

:

R BY (3L/4)  q 0 L2  q 0 (L/4)(5L/8)  2q 0 L(L)  0  R BY 

[FY = 0]  

:

R AY  R BY 

37q 0 L 32

q0L 35q 0 L  2q 0 L  0  R AY  4 32

All applied loads within the beam can now be expressed in terms of the distributed load by q(x)  q 0  H (x  L/2)  H (x  3L/4) 

37q 0 L (x  3L/4)  q 0 L2  (x  L/4) 32

The bending moment is then obtained by first integrating (3.12) and then using the obtained result to integrate (3.13) and final results are given by V(x)  q 0 (x  L/2) H (x  L/2)  (x  3L/4) H (x  3L/4) 

37q 0 L 35q 0 L H (x  3L/4)  q 0 L2 (x  L/4)  32 32

q0 37q 0 L (x  L/2) 2 H (x  L/2)  (x  3L/4) 2 H (x  3L/4)  (x  3L/4) H (x  3L/4) 2 32 35q 0 Lx q 0 L2 H (x  L/4)  32

M(x)  







197


A boundary value problem for this particular beam formulated in terms of the second-order differential equation is given by EI

q 37q 0 L d2v   0 (x  L/2) 2 H (x  L/2)  (x  3L/4) 2 H (x  3L/4)  (x  3L/4) H (x  3L/4) 2 dx 2 32





q 0 L2 H (x  L/4) 

35q 0 Lx 32

for x  (0, L)

(e3.9.1)

v(0)  0

(e3.9.2a)

v(L)  0

(e3.9.2b)


q 37q 0 L dv   0 (x  L/2)3 H (x  L/2)  (x  3L/4)3 H (x  3L/4)  (x  3L/4) 2 H (x  3L/4) dx 6 64





q 0 L2 (x  L/4) H (x  L/4)  EIv  



35q 0 Lx 2  C1 64

(e3.9.3)

q0 37q 0 L (x  L/2) 4 H (x  L/2)  (x  3L/4) 4 H (x  3L/4)  (x  3L/4)3 H (x  3L/4) 24 192





q 0 L2 35q 0 Lx 3 (x  L/4) 2 H (x  L/4)   C1x  C2 2 192

(e3.9.4)




C2  0

v(L)  0





q 0 L4  1 621q 0 L3 1  37q 0 L4 q 0 L4 35q 0 L4    C1L  C 2  0  C1      24 16 256  12288 32 192 4096

By substituting {C1, C2} into (e3.9.3)-(e3.9.4), it leads to the deflection and rotation of the beam: q0 37q 0 L (x  L/2) 4 H (x  L/2)  (x  3L/4) 4 H (x  3L/4)  (x  3L/4)3 H (x  3L/4) 24 192 q L2 35q 0 Lx 3 621q 0 L3 x  0 (x  L/4) 2 H (x  L/4)   2 192 4096 q0 37q 0 L EI   (x  L/2)3 H (x  L/2)  (x  3L/4)3 H (x  3L/4)  (x  3L/4) 2 H (x  3L/4) 6 64 35q 0 Lx 2 621q 0 L3 q 0 L2 (x  L/4) H (x  L/4)   64 4096

EIv  









It is obvious from the above two example problems that the discontinuity-function technique significantly reduces effort associated with the determination of constants resulting from direct integrations. The number of constants is still identical to the case that the loading data is continuous everywhere. While solutions obtained from this technique are left in terms of H(x – a), (x – a) and (x – a), they can readily be expressed for each individual segments by recalling the definition of those special function and distributions. For instance, the deflection and rotation obtained above can also be expressed as Copyright © 2011 J. Rungamornrat


198


 35q 0 Lx 3 621q 0 L3 x for x  [0,L/4]   4096  192  q 0 L2 35q 0 Lx 3 621q 0 L3 x L (x  ) 2  for x  [L/4,L/2]   4 192 4096  2 EIv   2 3 3   q 0 (x  L ) 4  q 0 L (x  L ) 2  35q 0 Lx  621q 0 L x for x  [3L/4,L]  24 2 2 4 192 4096  q L2 35q 0 Lx 3 621q 0 L3 x q 37q L 3L L L 3L   0   0 (x  ) 4  (x  ) 4   (x  )3  0 (x  ) 2    24  4 2 4 192 4096 2 4  192

 35q 0 Lx 2 621q 0 L3 for x  [0,L/4]   4096  64  L 35q 0 Lx 2 621q 0 L3 2 for x  [L/4,L/2]   q 0 L (x  )  4 64 4096  EI   2 3   q 0 (x  L )3  q L2 (x  L )  35q 0 Lx  621q 0 L for x  [3L/4,L] 0  6 2 4 64 4096  q 37q L L 35q 0 Lx 2 621q 0 L3 3L 0   0 (x  L )3  (x  3L )3    (x  ) 2  q 0 L2 (x  )   6  4 64 4096 2 4  64 4

for x  [3L/4,L]

for x  [3L/4,L]

3.7 Treatment of Statically Indeterminate Beams In this section, we generalize the direct integration technique to have a capability to treat statically indeterminate beams. If the fourth-order differential equation is used in the formulation of the boundary value problem, the displacement is chosen as the primary unknown and the key governing equation is in fact the equilibrium equation expressed in terms of this primary unknown. As a result, a direct integration technique with this type of differential equation is classified as a displacement method. The crucial nature of the displacement method is that there is no prejudice on the static indeterminacy; equivalently, the technique can be applied equally to both statically determinate and statically indeterminate beams. Solution procedure stated above therefore requires no generalization when applying to statically indeterminate beam. If the reduced-order (i.e. second-order and third-order) differential equations are used in the formulation of the boundary value problem, the shear force V(x) and the bending moment M(x) must be determined first. The fact that the equilibrium equation is the only tool available for obtaining V(x) and M(x), limits the capability of the technique to statically determinate beams. More precisely, for statically determinate beams, both V(x) and M(x) can completely be obtained in terms of prescribed loading data merely by using equilibrium equations while, for statically indeterminate beams, available equilibrium equations are not sufficient to solve for these two quantities. To overcome such limitation and enhance the capability of the technique to treat statically indeterminate beams, following strategy is followed: (i) a (statically stable) primary structure is obtained by fictitiously removing static quantities (e.g. support reactions and internal forces at certain points) from the original statically indeterminate beam until it becomes statically determinate; (ii) the released static quantities, commonly termed redundants, are applied back to the locations where they are released but, now, they are viewed as (unknown) applied loads on the primary structure; (iii) either V(x) or M(x) of the primary structure with redundants in step (ii) can now be obtained from equilibrium equation; (iv) identify boundary conditions associated with the structure in step (ii); (v) identify kinematical conditions at released locations to make structure in step (ii) identical to the original structure; (vi) use information from steps (iv) and (v) to determine constants of integration and all redundants; and (vii) obtain quantities of interest. Copyright © 2011 J. Rungamornrat


199


Y q0

M (a)

A Y

(b)

C

D

RCY

RDY

B

q0

M A

B

X

C

D

X

Y q0

M (c)

A

B

C R1

D

X

R2

Figure 3.11: (a) Schematic of statically indeterminate beam of second degree, (b) primary structure resulting from removing support reactions at point C and D, and (c) primary structure after redundants are applied To clearly demonstrate the above strategy, let us consider a model problem shown in Figure 3.11(a). This beam is statically indeterminate of second degree (i.e. ra = 4, nm = 2(2) = 4, nj = 3(2) = 6, nc = 0, then DI = 4 + 4 – 6 – 0 = 2). First, we construct a primary structure by removing two support reactions RCY and RDY at points C and D; this primary structure is shown in Figure 3.11(b). Next, the redundants R1 and R2 are applied back to the primary structure as shown in Figure 3.11(c). It is evident that the beams shown in Figure 3.11(a) and Figure 3.12(c) are different in nature; more specifically, the former is statically indeterminate while the latter is statically determinate with applied redundants. Support reactions (at point A), shear force, and bending moment of the beam in Figure 3.11(c) can readily be obtained by static equilibrium. If the second-order differential equation is chosen as a key governing equation, two boundary conditions (of the structure in Figure 3.11(c)) are v(0) = 0 and v'(0) = 0. If the third-order differential equation is chosen as a key governing equation, three boundary conditions are v(0) = 0, v'(0) = 0 and EIv''(L) = 0. Next, by comparing beams in Figure 3.11(a) and Figure 3.12(c), the former beam can be viewed as a special case of the latter in the sense that the redundant R1 and R2 appearing in Figure 3.11(a) are unknowns and can still vary arbitrarily while both support reactions RCY and RDY in Figure 3.11(c) possess a single value to maintain zero deflection at points C and D. Thus, the necessary and sufficient conditions for turning the beam in Figure 3.11(a) to the original structure are v(xC) = 0 and v(xD) = 0. Steps (vi) and (vii) follow exactly the same procedure as discussed in previous subsections. It is worth noting that a primary structure constructed in step (i) is generally not unique. Any set of redundants that produces statically stable structure with DI = 0 is acceptable. There is no prejudice on a set of redundants chosen and no strong evidence to support the best choice. In general, the choice is often problem dependent and, most of the time, a matter of preference. Note also that boundary conditions specified in step (iv) are for structure in step (ii) not the original one. Copyright © 2011 J. Rungamornrat


200


Example 3.10 Determine the deflection and rotation of a statically indeterminate beam shown below by using a second-order, third-order and fourth-order differential equations Y q0 A

B

EI L/2

X

L/2

Solution Note that the given beam is statically indeterminate to the first degree (i.e. ra = 3, nm = 1(2) = 2, nj = 2(2) = 4, nc = 0, then DI = 3 + 2 – 4 – 0 = 1) and the applied load can be expressed as q(x)  q 0 H (x  L/2)

(e3.10.1)

Option I: Use fourth-order differential equation A boundary value problem for this particular beam formulated in terms of the fourth-order differential equation is given by EI

d4v  q 0 H (x  L/2) dx 4

(e3.10.2)

v(0)  0

(e3.10.3a)

EIv(0)  0

(e3.10.3b)

v(L)  0

(e3.10.3c)

EIv(L)  0

(e3.10.3d)

Performing direction integrations of (3.10.2) yields d3v  q 0 (x  L/2) H (x  L/2)  C1 dx 3 q d2v EI 2   0 (x  L/2) 2 H (x  L/2)  C1 x  C 2 dx 2 q dv 1   0 (x  L/2)3 H (x  L/2)  C1 x 2  C 2 x  C3 EI dx 6 2 q0 1 1 EIv   (x  L/2) 4 H (x  L/2)  C1x 3  C 2 x 2  C3 x  C 4 24 6 2 EI

(e3.10.4) (e3.10.5) (e3.10.6) (e3.10.7)

Four constants of integration {C1, C2, C3, C4} can then be obtained by using boundary conditions (e3.10.3a)-(e3.10.3d) as follows: v(0)  0



C4  0

v(0)  0



C3  0

EIv(L)  0





q 0 L2 q L2  C1L  C 2  0  C1L  C 2  0 8 8 Copyright © 2011 J. Rungamornrat

(e3.10.8)


v(L)  0





201


q 0 L4 1 q L2 1  C1L3  C 2 L2  C3 x  C 4  0  C1L  3C 2  0 384 6 2 64

(e3.10.9)

By solving (e3.10.8) and (e3.10.9) simultaneously, we obtain C1 

23q 0 L 7q L2 and C 2   0 128 128

By substituting {C1, C2, C3, C4} into (e3.10.6)-(e3.10.7), we obtain the deflection and the rotation for the given beam: q0 23q 0 Lx 2 7q 0 L2 x (x  L/2)3 H (x  L/2)   6 256 128 3 q 23q 0 Lx 7q L2 x 2 EIv(x)   0 (x  L/2) 4 H (x  L/2)   0 6 768 256 EI(x)  

Option II: Use third-order differential equation Since the given beam is statically indeterminate to the first degree, a primary structure is obtained by removing only one redundant. Here, we choose a support reaction at point B, RBY, as a redundant and the corresponding primary structure subjected to this redundant is shown below Y q0 A

B RBY

EI L/2

X

L/2

To obtain the shear force V(x), we first substitute (e3.10.1) into (3.12), then integrate the result, and finally determine a constant of integration by using the condition V(L) = –RBY. This leads to V(x)  q 0 (x  L/2) H (x  L/2)  q 0 L/2  R BY

(e3.10.10)

Boundary conditions associated with above primary structure are v(0) = v'(0) = EIv''(0) = 0 and a kinematical condition required to turn above structure to the original structure is v(L) = 0. Now, we can formulate a boundary value problem in terms of the third-order differential equation for this particular beam as follow: EI

d3v  q 0 (x  L/2) H (x  L/2)  q 0 L/2  R BY dx 3

for x  (0, L)

(e3.10.11)

v(0)  0

(e3.10.12a)

v(0)  0

(e3.10.12b)

EIv(L)  0

(e3.10.12c)

v(L)  0

(e3.10.12d) Copyright © 2011 J. Rungamornrat


202



q q Lx d2v   0 (x  L/2) 2 H (x  L/2)  0  R BY x  C1 2 dx 2 2

(e3.10.13)

EI

q q Lx 2 R BY x 2 dv   0 (x  L/2)3 H (x  L/2)  0   C1 x  C 2 dx 6 4 2

(e3.10.14)

EIv  

q0 q Lx 3 R BY x 3 1   C1 x 2  C 2 x  C3 (x  L/2) 4 H (x  L/2)  0 24 12 6 2

(e3.10.15)

Three constants of integration {C1, C2, C3} and the redundant RBY can be obtained from boundary conditions (e3.10.12a)-(e3.10.12d) as follow: v(0)  0



C3  0

v(0)  0



C2  0

EIv(L)  0





q 0 L2 q 0 L2 3q L2   R BY L  C1  0  R BY L  C1  0 8 2 8

v(L)  0





q 0 L4 q 0 L4 R BY L3 1 31q 0 L2 (e3.10.17)    C1L2  C 2 L  C3  0  R BY L  3C1  384 12 6 2 64

(e3.10.16)

By solving (e3.10.16) and (e3.10.17) simultaneously, we obtain R BY 

41q 0 L 7q L2 and C1   0 128 128

By substituting {C1, C2, C3} and RBY into (e3.10.14)-(e3.10.15), we obtain the same deflection and rotation as those obtained in option I. Option III: Use second-order differential equation The same primary structure as shown in option II is considered again here. To obtain the bending moment M(x), we first substitute (e3.10.10) into (3.13), then integrate the result, and finally determine a constant of integration by using the condition M(L) = 0. This leads to M(x)  

q0 q Lx 3q L2 (x  L/2) 2 H (x  L/2)  0  0  R BY (L  x) 2 2 8

(e3.10.18)

Two boundary conditions associated with above primary structure are v(0) = v'(0) = 0 and a kinematical condition required to turn above structure to the original structure is v(L) = 0. Now, we can formulate a boundary value problem in terms of the second-order differential equation for this particular beam as follow: EI

q0 q 0 Lx 3q 0 L2 d2v 2    H     R BY (L  x) (x L/2) (x L/2) dx 2 2 2 8

for x  (0, L)

(e3.10.19)

v(0)  0

(e3.10.20a)

v(0)  0

(e3.10.20b)

v(L)  0

(e3.10.20c) Copyright © 2011 J. Rungamornrat


203



q q Lx 2 3q 0 L2 x dv x2   0 (x  L/2)3 H (x  L/2)  0   R BY (Lx  )  C1 dx 6 4 8 2

EIv  

(e3.10.21)

q0 q Lx 3 3q 0 L2 x 2 Lx 2 x 3   R BY (  )  C1 x  C 2 (x  L/2) 4 H (x  L/2)  0 24 12 16 2 6

(e3.10.22)

Two constants of integration {C1, C2} and the redundant RBY can be obtained from boundary conditions (e3.10.20a)-(e3.10.20c) as follow: v(0)  0



C2  0

v(0)  0



C1  0

v(L)  0





q 0 L4 q 0 L4 3q 0 L4 R BY L3 41q 0 L     C1L  C 2  0  R BY  384 12 16 3 128

By substituting {C1, C2} and RBY into (e3.10.21)-(e3.10.22), we obtain the same deflection and rotation as those obtained in option I. Example 3.11 Determine the deflection and rotation of a statically indeterminate beam shown below by using a second-order differential equations Y q0 A

B

EI

EI

C

X

L

L

Solution Since the given beam is statically indeterminate to the first degree (i.e. ra = 3, nm = 2(2) = 4, nj = 3(2) = 6, nc = 0, then DI = 3 + 4 – 6 – 0 = 1), a primary structure is obtained by removing the roller support at point B. The primary structure subjected to the redundant RBY is shown below. Y q0 A

EI

B

RBY

EI

RAY

C

X

RCY L

L

By considering equilibrium of the entire structure, we can readily obtain support reactions at point A and C as RAY = RCY = q0L – RBY/2. All applied loads acting to above primary structure can then be expressed as q(x)  q 0  R BY (x  L)



204


First, the shear force V(x) is obtained by integrating (3.12) along with the use of (e3.11.1): V(x)  q 0 x  R BY H (x  L)  q 0 L 

R BY 2

(e3.11.2)

A constant of integration is obtained from V(0) = RAY. To obtain the bending moment M(x), we substitute (e3.11.2) into (3.13) and then perform a direct integration. Once a condition M(0) = 0 is employed to determine a constant of integration, we then obtain M(x)  

q0 x 2 R x  R BY (x  L) H (x  L)  q 0 Lx  BY 2 2

(e3.11.3)

Now, we can formulate a boundary value problem in terms of the second-order differential equation for this particular beam as follow: EI

q0 x 2 R x d2v    R BY (x  L) H (x  L)  q 0 Lx  BY 2 dx 2 2

for x  (0, 2L)

(e3.11.4)

v(0)  0

(e3.11.5a)

v(2L)  0

(e3.11.5b)

v(L)  0

(e3.11.5c)

Note that (e3.11.5a) and (e3.11.5b) are boundary conditions for the primary structure while (e3.11.5c) is a kinematical condition rendering the primary structure identical to the original structure. By performing direction integrations of (e3.11.4), it leads to EI

q x3 R q Lx 2 R BY x 2 dv   0  BY (x  L) 2 H (x  L)  0   C1 dx 6 2 2 4

EIv  

q 0 x 4 R BY q Lx 3 R BY x 3 (x  L)3 H (x  L)  0    C1 x  C 2 24 6 6 12

(e3.11.6) (e3.11.7)

Two constants of integration {C1, C2} and the redundant RBY can be obtained from boundary conditions (e3.11.5a)-(e3.11.5c) as follow: v(0)  0



C2  0

v(2L)  0





v(L)  0



2q 0 L4 R BY L3 4q 0 L4 2R BY L3 4q L3     2C1L  C 2  0  4C1  R BY L2   0 3 6 3 3 3 4 4 3 3 q L q L R L 3q L  0  0  BY  C1L  C 2  0  12C1  R BY L2   0 24 6 12 2

By solving above two linear equations simultaneously, we obtain R BY 

5q 0 L q L3 and C1   0 48 4

By substituting {C1, C2} and RBY into (e3.11.6)-(e3.11.7), we then obtain the deflection and rotation for the given beam: Copyright © 2011 J. Rungamornrat


205


EI(x)  

q 0 x 3 3q 0 Lx 2 q 0 L3 5q 0 L(x  L) 2    H (x  L) 6 16 48 8

EIv(x)  

q 0 x 4 q 0 Lx 3 q 0 L3 x 5q 0 L(x  L)3    H (x  L) 24 16 48 24

Example 3.12 Determine the deflection and rotation of a statically indeterminate beam shown below by using a second-order differential equations Y M0 A

B

EI L

EI

C

X

L

Solution Since the given beam is statically indeterminate to the second degree (i.e. ra = 4, nm = 2(2) = 4, nj = 3(2) = 6, nc = 0, then DI = 4 + 4 – 6 – 0 = 2), a primary structure is obtained by removing two roller supports at points B and C. The primary structure subjected to the two redundant RBY and RCY is shown below. Y M0 A

EI

B

L

RBY

EI

C RCY

X

L

Applied loads acting within the primary structure including the redundant RBY can be expressed as q(x)   M 0  (x  L)  R BY (x  L)

(e3.12.1)

First, the shear force V(x) is obtained by integrating (3.12) along with the use of (e3.12.1): V(x)   M 0 (x  L)  R BY H (x  L)  1  R CY

(e3.12.2)

A constant of integration is obtained from V(2L) = –RCY. To obtain the bending moment M(x), we substitute (e3.12.2) into (3.13) and then perform a direct integration. Once a condition M(2L) = 0 is employed to determine a constant of integration, we then obtain M(x)   M 0 H (x  L)  1  R BY (x  L) H (x  L)  x  L  R CY (x  2L)

(e3.12.3)

Now, we can formulate a boundary value problem in terms of the second-order differential equation for this particular beam as follow: Copyright © 2011 J. Rungamornrat


206


EI

d2v   M 0 H (x  L)  1  R BY (x  L) H (x  L)  x  L  R CY (x  2L) dx 2

for x  (0, 2L)

(e3.12.4)

v(0)  0

(e3.12.5a)

v(0)  0

(e3.12.5b)

v(L)  0

(e3.12.5c)

v(2L)  0

(e3.12.5d)

Note that (e3.12.5a) and (e3.12.5b) are boundary conditions for the primary structure while (e3.12.5c) and (e3.12.5d) are kinematical conditions rendering the primary structure identical to the original structure. By performing direction integrations of (e3.12.4), it leads to EI

 (x  L) 2  dv x2 x2   M 0 (x  L) H (x  L)  x  R BY  H (x  L)   Lx   R CY (  2Lx)  C1 dx 2 2  2 

(e3.12.6)

 (x  L) 2  (x  L)3 x2  x 3 Lx 2  x3 2  EIv   M 0  H (x  L)    R BY  H (x  L)    R CY (  Lx )  C1 x  C 2 (e3.12.7) 2 2 6 6 2 6    

Two constants of integration {C1, C2} and the redundants RBY and RCY can be obtained from boundary conditions (e3.12.5a)-(e3.12.5d) as follow: v(0)  0



C2  0

v(0)  0



C1  0

v(L)  0



M 0 L2 R BY L3 5R CY L3 3M 0    0  2R BY + 5R CY   2 3 6 L

v(2L)  0



3M 0 L2 5R BY L3 8R CY L3 9M 0    0  5R BY  16R CY   2 6 3 L

By solving above two linear equations simultaneously, we obtain R BY  

3M 0 3M and R CY   0 7L 7L

By substituting {C1, C2} and RBY and RCY into (e3.12.6)-(e3.12.7), we then obtain the deflection and rotation for the given beam: EI(x)   M 0 (x  L) H (x  L)  x 

 3M 0  (x  L) 2 H (x  L)  x 2  3Lx   7L  2 

 (x  L) 2 x 2  3M 0  (x  L)3 x 3 3Lx 2   EIv(x)   M 0  H (x  L)    H (x  L)    2 2  7L  6 3 2  



207


Exercises 1. Use second-order, third-order and fourth-order differential equations to solve for deflection and rotation of statically determinate beams shown below 4q0x(L – x)/L2 q0(1 – x/L) EI

EI

L

L q0(1 – x/L)

P M

EI

EI L

L

M

M

EI

EI

L

L

q0

q0(1 – x/L)

EI

EI

L

L

2. Apply domain decomposition technique along with the second-order differential equation to solve for deflection and rotation of statically determinate beams shown below q0

P EI

EI a

L–a

a

L–a q0

q0 EI

EI a

L–a Copyright © 2011 J. Rungamornrat

L–a

a


208


q0

q0 EI

EI a

a

L–a

L–a

M

M EI a

EI a

L–a

L–a q0

q0 EI

EI a

a

L–a

L–a

q0

P EI

EI a

a

L–a

L–a

q0

P EI

EI a

a

L–a

L–a q0

q0 EI

EI a

L–a

L–a

a

q0 M EI

EI a

L–a


a

L–a


209


3. Apply discontinuity-function technique along with the second-order differential equation to solve for deflection and rotation of statically determinate beams shown in problem 2 4. Determine the deflection and rotation of statically determinate beams containing internal releases and points where the flexural rigidity is discontinuous q0

q0 EI

a

L–a

a

L–a q0

q0

EI

EI

EI

a

L–a

L/4

EI

EI

EI

a

L–a

q0

q0

EI

EI

L/2

L/4

a

L–a

5. Use second-order, third-order and fourth-order differential equations to solve for deflection and rotation of statically indeterminate beams shown below q0

q0

EI

EI

L

L

q0

q0 EI

EI

EI

L/2

L/2

L P

q0 EI

EI a


L–a


210


P

P EI

EI a

L–a

a

L–a

M

M EI

EI a

L–a

a

L–a

q0 M EI a

EI L–a

a q0

q0

EI

EI a

L–a

L–a


a

L–a


211

Method of Curvature (Moment) Area

CHAPTER 4 METHOD OF CURVATURE (MOMENT) AREA A method of curvature area (or commonly known as a moment area method) is a classical technique that can be used to determine the displacement and rotation at any point of flexuredominating structures such as beams and rigid frames. As obvious from its name, key governing equations for this technique involve the area of the curvature diagram (or a bending moment diagram divided by the flexural rigidity EI) of each member; more precisely, these equations present the relationship among the curvature of a member (in terms of its area and first moment of area), the relative displacement, and the relative rotation of both ends of the member. It is worth noting that those relative quantities are measures of the member deformation due to the bending effect and, when combined with prescribed boundary conditions (e.g., known displacements and/or rotations at supports), it allows the displacement and rotation at any points be calculated. In the following sections, we present basic assumptions and limitations of the method, a complete derivation of the two curvature area equations, their physical interpretation and key remarks. At the end of the chapter, various example problems are solved to clearly demonstrate the technique.

4.1 Basic Assumptions Consider a two-dimensional, flexure-dominated structure consisting of several components as shown in Figure 4.1. Let So denote the undeformed configuration of a structure (a configuration corresponding to a state of the structure that is free of applied loads, internal forces and displacements) and let S denote the deformed configuration of the structure (a configuration corresponding to a state of the structure that undergoes deformation and change of position due to excitations from surrounding environment). Let further define u, v and  as the longitudinal component of the displacement (i.e. displacement parallel to the member axis), the transverse component of the displacement (i.e. displacement normal to the member axis), and the rotation at any point of the structure as shown schematically in Figure 4.2. In the development of the curvature area equations presented further below, following assumptions are employed: member

node

So S

(a)

(b)

Figure 4.1: (a) Schematic of two-dimensional rigid frame and (b) schematic of undeformed configuration So and deformed configuration S (i) Geometry: a structure consists of one-dimensional, straight segments termed members and they are connected by points termed nodes or joints. Members and nodes in both the undeformed configuration So and the deformed configuration S can be represented graphically by lines and points, respectively, as shown in Figure 4.1. Copyright © 2011 J. Rungamornrat


212


So

 S u v

Figure 4.2: Schematic indicating longitudinal displacement u, transverse displacement v and rotation  at any point (ii) Structure kinematics: the displacement components u, v and the rotation  at any point of the structure is relatively small, i.e. |u/L|, |v/L|, || << 1 where L represents the characteristic dimension of the structure. As a consequence of this assumption, following approximations are legitimate (ii.1) there is no difference between So and S and this therefore allows the known geometry of So be used in the analysis; (ii.2) the rotation is related to the transverse displacement by a linear relation  = dv/dx; and (ii.3) the curvature at any point, denoted by , is related to the transverse displacement by a linearized kinematics, i.e.,  = d2v/dx2. (iii) Member kinematics: only deformation due to bending moment is considered while axial and shear deformations are neglected. As a consequence of this assumption, following approximations are legitimate (iii.1) a plane section that is normal to the axis of the member in So always remains plane and normal to the axis of the member in S; (iii.2) the normal strain resulting from the bending moment varies linearly across the cross section normal to the axis of the member; and (iii.3) there is no change in length of any segment of the structure. (iv) Constitutive relation: material constituting any part of the structure is isotropic and linearly elastic. Following additional simplifications are also imposed (iv.1) the Young's modulus E is sufficient to completely characterize the material behavior; (iv.2) the Young's modulus E is uniform across the cross section normal to the axis of the member, but can be a function of position along the axis of the member; (iv.3) normal stresses in the transverse direction is small and neglected; and (iv.4) lateral deformation due to Poisson effect is neglected. (v) Static equilibrium: equilibrium of a structure is enforced in the undeformed state, i.e. all equilibrium equations are set up based on the known geometry of So. Additional idealizations resulting from combination of above assumptions should also be addressed. By combining assumptions (ii) and (iii.3), it leads to that the projection of any deformed segment onto its undeformed axis has the same length as that of its undeformed segment (see Figure 4.3, this assumption leads to L = Lo and therefore uA = uB). Combining assumptions (iii.2) and (iv.2) leads to the linear distribution of the bending stress across the cross section normal to the axis of a member. Copyright © 2011 J. Rungamornrat


213


B′ vB

S B A′ vA A

uA

So

uB

Lo L

Figure 4.3: Schematic indicating projection of deformed segment onto its original undeformed axis

4.2 Derivation of Curvature Area Equations Consider a straight segment AB which is a part of a structure as shown schematically in Figure 4.4. This segment is chosen such that there is no discontinuity of the displacement and rotation at any interior point, i.e. containing no internal releases such as hinges, shear releases and axial releases within the segment. A local coordinate system {o; x, y, z} of the segment in the undeformed configuration is defined such that its origin o is at point A, the x-coordinate direction is along the axis of the segment, the y-coordinate direction is perpendicular to axis of the segment, and the zcoordinate direction directs outward from the paper. According to this local coordinate system, the longitudinal component uP and transverse component vP of the displacement at any point P become the components of the displacement in the x-direction and the y-direction, respectively, and the rotation P at any point P becomes an angle measured in radian from the x-axis to the tangent line at that point in the deformed configuration. Both uP and vP are considered positive if they direct along the positive coordinate directions and the rotation P is considered positive if the tangent line at point P rotates in a counterclockwise direction (for instance, A andB of the segment AB in Figure 4.4 are negative and positive, respectively). We also introduce a single coordinate system {O; X, Y, Z}, termed the global coordinate system, for the entire structure. The Z-coordinate direction is taken to be same as the z-coordinate direction while the choice of X- and Y-coordinate directions is arbitrary and a matter of preference. Components of the displacement at any point P in the X- and Y-directions are denoted by UP and VP, respectively. Note that the rotation at any point P measured in the global coordinate system, denoted by P, is the same as that measured in the local coordinate system, i.e. P = P. The bending moment M at any cross sections of the segment AB is assumed to be known and the corresponding curvature diagram  = M/EI is shown in Figure 4.4 where EI is the flexural rigidity of the segment. Upon employing kinematics, constitutive relations, equilibrium of the cross section and assumptions indicated above, we obtain the relation among the bending moment M = M(x), the curvature (x), the rotation (x) and the transverse displacement v = v(x) at any point x of the segment AB as (also see section 3.1 in chapter 3 for derivation) κ=

dθ d2v M = = dx dx 2 EI

(4.1)



214


B B vB VB

S B

UA A y

A vA

VA

UB

So

x

uB Y

uA

A

X

I M/E

Lo

Figure 4.4: Schematic of segment AB undergoing deformation and associated curvature diagram By directly integrating equation (4.1) from x = 0 to x =   [0, Lo] and then using the boundary condition (0) = A, we obtain the rotation at any point  by ξ

θ() = A +

 0

M dx EI

(4.2)

By substituting  = Lo and using the boundary condition (Lo) = B into (4.2), it leads to the first curvature area equation B = A +

Lo

 0

M dx EI

(4.3)

By employing the relation = dv/d, integrating equation (4.2) from = 0 to = Lo, and then using displacement boundary conditions v(0) = vA and v(Lo) = vB, we obtain v B = v A + A L o +

Lo



0

0



M dxd = v A + A L o + EI

Lo Lo

  0

x

M ddx EI

(4.4)

Note that the last term is obtained by changing order of integration of the double integral. Since the bending moment M and the flexural rigidity EI are only functions of x, i.e. M = M(x) and EI = EI(x), the inner integral of the double integral in equation (4.4) can be integrated explicitly to obtain v B = v A + A L o +

Lo

 0

M EI

Lo

 x

d dx = v A + A L o +

Lo

 0

(Lo  x)

M dx EI


(4.5)


215


Equation (4.5) is known as the second curvature area equation. An alternative form of the second curvature area equation can be obtained by replacing A in (4.5) by that from (4.3). This leads to  v B = v A +  B   

Lo

 0

 M dx  Lo +  EI 

Lo

 0

(Lo  x)

M dx EI



v A = v B  B L o +

Lo

 0

x

M dx EI

(4.6)

From the derivation of (4.6), it is important to emphasize that the equations (4.3), (4.5) and (4.6) are not all independent but only two from these three equations are independent. Note further that the first curvature area equation (4.3) and the second curvature area equation (4.5) or (4.6) are valid for a segment containing no discontinuity of the displacement and rotation (i.e. presence of the internal releases within a segment is prohibited). It is worth noting that the derivation of both curvature area equations presented above also applies to the case of nonlinear moment-curvature relationship. Instead of replacing the curvature  by M/EI at the beginning of the derivation, all equations can be obtained directly in terms of the curvature k and the final general form of the first and second curvature area equations becomes B = A +

Lo



 dx

(4.7a)

0

v B = v A + A L o +

Lo



(Lo  x) dx

(4.7b)

0

The form of these two equations is clearly independent of the constitutive relation or, equivalently, the moment-curvature relationship. Applications of such pair of curvature area equations to the analysis of nonlinear elastic beams can be found in the work of Pinyochotiwong et al (2009). As be evident from (4.3) and (4.6) or (4.7), the first and second curvature area equations provide no information about the longitudinal component of the displacement at both ends of the segment. To relate those two quantities, i.e. uA and uB, we invoke the inextensibility assumption (no axial deformation) and this gives rise to a simple relation: (4.8)

uB = uA

This equation is known as the length constraint equation. As will be clear in later presentation, the first and second curvature area equations along with the length constraint equation form a sufficient set of equations to quantitatively determine the displacement and rotation at any point of the structure.

4.3 Interpretation of Curvature Area Equations In this section, the geometrical interpretation of the first and second curvature area equations is discussed. Such graphical demonstration may significantly be useful when these two equations are employed in the determination of the displacement and rotation of beams and rigid frames.

4.3.1 First curvature area equation Let us define a relative quantity B/A such that B/A = B – A. This quantity physically means an angle between a tangent line at the end A and a tangent line at the end B of the segment in the deformed configuration as shown in Figure 4.5 and is termed the relative end rotation of the segment AB. The sign convention of the relative angle B/A follows directly from that of the angle B and A; i.e. it is positive if the angle sweeping from the tangent line at the end A to the tangent



216


line at the end B is in a counterclockwise direction. Let further define Af,AB as the area under a graph of a function f between a point A and a point B, i.e. xB

A f,AB =

 f(x) dx

(4.9)

xA

The first curvature area equation simply states that the relative end rotation of the segment AB is equal to the area of the curvature-diagram (or M/EI-diagram) between the point A and the point B or, equivalently, B/A = A M/EI,AB

(4.10)

It is important to remark that the area of the curvature over the segment AB can either be positive or negative depending primarily on values of the bending moment; for instance, the negative bending moment over the entire segment produces the negative area of the curvature. B B S B/A B

A

x

A

y A

So

I M/E , AB

AM / E I

Lo

Figure 4.5: Schematic indicating relative end rotation B/A and area of curvature over segment AB

4.3.2 Second curvature area equation Let tB/A be a relative quantity defined by tB/A = vB – (vA + ALo). This quantity then has a graphical representation as the displacement deviation of the end B measured relative to a tangent line of the end A in the direction perpendicular to its original undeformed axis as shown in Figure 4.6. The deviation tB/A is positive if a vector connecting a point on the tangent line of the end A and the end B directs in the positive y-direction; otherwise, it is negative. The integral term appearing in the Copyright © 2011 J. Rungamornrat


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relation (4.6) can be interpreted graphically as the first moment of the area of the curvature-diagram (or M/EI-diagram) over the segment AB about the end B, i.e. Lo

 0

(Lo  x)

M dx = A M/EI,AB x B EI

(4.11)

where x B is the distance, measured along the undeformed axis of the segment, from the end B to the centroid of the curvature-diagram.

B tB/A vB

S

ALo x vAALo

B

A A

y

vA A

So

I M/E CG xB

Lo

Figure 4.6: Schematic indicating displacement deviation tB/A and centroid of curvature diagram The second curvature area equation (4.6) states that the displacement deviation tB/A of the segment AB due to the bending moment is equal to the first moment of area of the curvaturediagram (or M/EI-diagram) about the end B, i.e. (4.12)

t B/A = A M/EI,AB x B

The geometrical interpretation of the relation (4.7) can also be established in a similar fashion as shown in Figure 4.7. The displacement deviation of the end A measured relative to a tangent line of the end B in the direction perpendicular to its original undeformed axis, denoted by tA/B = vA – (vB – BLo), is equal to the first moment of area of the curvature-diagram (or M/EI-diagram) about the end A, i.e. (4.13)

t A/B = A M/EI,AB x A Copyright © 2011 J. Rungamornrat


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where x A is the distance, measured along the undeformed axis of the segment, from the end A to the centroid of curvature diagram. Note that the displacement deviation tA/B is positive if a vector connecting a point on the tangent line of the end B to the end A directs in the positive y-direction; otherwise, it is negative. B B vB

S BLo tA/B y

x B

A So

vA

A vBBLo

I M/E

CG Lo xA

Figure 4.7: Schematic indicating the displacement deviation tA/B and centroid of curvature diagram

4.4 Applications of Curvature Area Equations Let us consider a planar, flexure-dominating structure where the bending moment at any cross section is already known (either by means of static equilibrium for statically determinate structures or by certain structural analysis techniques for statically indeterminate structures). For any straight segment AB which is a part of this structure, there exists three kinematical degrees of freedom at each end of the segment; one is associated with the end rotation and the other two are associated AB with transverse and longitudinal components of the displacement. For instance, we have { u AB A , vA , AB AB AB AB A } at the end A and { u B , v B ,  B } at the end B. The superscript ‘AB’ is used only to emphasize that those kinematical quantities belong to the segment AB. The rotation and transverse AB AB AB components of the displacement at both ends { AB A ,  B , v A , v B } are related by the first curvature area equation (4.3) and the second curvature area equation (either equation (4.6) or (4.7)), while the AB longitudinal components of the displacement at both ends { u AB A , u B } are related by the length constraint equation (4.8). These four equations are summarized again below: B/A = θ BAB  θ AAB = A M/EI,AB

(4.14)

AB t B/A = v AB  (v AB B A +  A L AB ) = A M/EI,AB x B


(4.15a)


219


t A/B = v AB  (v AB  AB A B B L AB ) = A M/EI,AB x A u AB  u AB B A

(4.15b) (4.16)

From this set of equations, following remarks can be deduced: (1) Only three equations from this set are linearly independent. One of them is the length constraint equation (4.16) and the other two come from the three equations (4.14), (4.15a) and (4.15b). (2) There are six kinematical unknowns that are involved in the above equations, two end AB AB AB rotations { AB A ,  B }, two end transverse displacements { v A , v B } and two end AB longitudinal displacements { u AB A , u B }. AB (3) If one of the longitudinal displacements { u AB A , u B } is known, the other can readily be computed from the length constraint equation (4.16). AB AB (4) If one of the end rotations { AB A ,  B } and one of the transverse displacements { v A , v AB B } are known, the other end rotation can readily be obtained from the first curvature area equation (4.14) and the other transverse displacement can be obtained from the second curvature area equation (4.15a) or (4.15b). AB AB (5) If both transverse displacements { v AB A , v B } are known, one of the end rotations {  A , AB B } can be computed from the second curvature area equation (4.15a) or (4.15b) and the other end rotation can then be determined from the first curvature area equation (4.14). (6) If the end rotation and the longitudinal and transverse displacements at one end are known, the rotation, transverse displacement, longitudinal displacement at the other end can readily be obtained from (4.14), (4.15a) or (4.15b), and (4.16), respectively. AB (7) Both { AB A ,  B } cannot be prescribed arbitrarily and independently for a given bending moment diagram; they must satisfy the first curvature area equation (4.14). AB (8) Both { u AB A , u B } cannot be prescribed arbitrarily and independently; they must satisfy the length constraint equation (4.16). (9) For statically stable beams (with rigid body movement properly prevented), the longitudinal component of the displacement vanishes everywhere since there is no axial deformation. As a consequence, the length constraint equation (4.16) is satisfied automatically and is not necessary to be considered. The total number of kinematical unknowns per segment now reduces from six to four, i.e. two end rotations and two end transverse displacements, and only the first and second curvature area equations are available to solve two unknowns once the other two are known. In addition, the global and local coordinate systems for beams can be chosen such that they are coincident and, as a result, there is no need to distinguish between the local and global displacements and rotations. The above remarks are useful when the curvature area equations are applied to determine the displacement and rotation at any points within the structure. For instance, the remark (6) implies that if there exists a point in the structure where the displacement and the rotation are known, the displacement and rotation at all other points can then be readily obtained from (4.14)-(4.16). The key point is to choose a proper straight segment that contains a point where the displacement and rotation are already known. For example, consider a rigid frame shown in Figure 4.8. The displacement and rotation at a fixed support at point A (with respect to the reference global coordinate system) are known, i.e. U A  0 , VA  0 , and  A  0 . By using the relations U A =  v AAB ,



220


VA = u AAB , and  A = AAB , the transverse and longitudinal displacements and the rotation at end A of a

segment AB are known and then, by applying (4.14)-(4.16) to this segment, the longitudinal and AB AB transverse displacements and the rotation at end B of the segment AB { u AB B , v B ,  B } can be AB BC AB BC AB obtained. Next, by using the continuity relations u BC B = U B   v B , v B = VB  u B , and  B =  B   B , the displacement and rotation at point B with respect to the global coordinate system { U B , VB , Θ B } BC BC and with respect to the local coordinate system of a segment BC { u BC B , v B ,  B } are now known. By applying (4.14)-(4.16) again to the segment BC, the transverse and longitudinal displacement and the rotation at the end C { u CBC , vCBC , CBC } can readily be computed. Next, by using the continuity BC CD BC CD BC relations vCD C = U C  u C ,  u C = VC  v C , and C =  C  C , the displacement and rotation at the point C with respect to the global coordinate system { U C , VC , Θ C } and with respect to the local CD CD coordinate system of a segment CD { u CD C , v C , C } are now known. By applying (4.14)-(4.16) to CD CD the segment CD, the displacement and rotation at the end D { u CD D , v D ,  D } can be obtained. CD CD Finally, by using the relations U D  vCD D , VD   u D , and  D   D , the displacement and rotation at the point D with respect to the global coordinate system { U D , VD , Θ D } are known. x, y

C

B

x, y

D Y x y

A

X

Figure 4.8: Schematic of statically determinate frame containing a fixed-support. The shaded area represents the curvature diagram of the structure For certain structures, a point whose the displacement and the rotation are known a priori may not exist, but a straight segment can be properly chosen from the structure and then (4.14)(4.16) are applied to solve for the remaining components of the displacement and rotation at both ends. After the displacement and rotation at a particular point were resolved, calculation of the displacement and rotation at other points follows the same procedure as discussed in the previous case. To clearly demonstrate the idea, let us consider a statically determinate beam subjected to a concentrated load as shown in Figure 4.9. From the prescribed boundary conditions, the transverse displacement at points A and B vanishes and, from the inextensibility condition, the longitudinal displacement at any point identically vanishes. For a segment AB, it contains only two unknowns AB { AB A ,  B } and, by using the remark (5), these two unknowns can readily be determined. Once the deflection (or transverse displacement) and rotation at points A and B are completely determined, Copyright © 2011 J. Rungamornrat


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the deflection and rotation at any points within the beam can then be calculated. For instance, the deflection and rotation at point D can be obtained by considering a segment BD and using the BD AB BD continuity at point B, i.e. v AB B = v B , and  B =  B , and the deflection and rotation at any point C within the segment AB can be obtained by considering either a segment AC or a segment CB since the deflection and rotation at points A and B are already known. Y, y

A

B D

C

X, x

M/EI-diagram

Figure 4.9: Schematic of statically determinate beam subjected to concentrated load and the corresponding curvature diagram For structures with a general configuration, a sufficient number of segments must be considered to establish a sufficient number of equations to solve for unknown displacements and rotations at the ends of all segments. After these unknowns were determined, a similar procedure as previously described is employed to determine the displacement and rotation at other points. To illustrate such a procedure, let us consider a statically determinate rigid frame subjected to a horizontal concentrated force as shown in Figure 4.10. From boundary conditions at supports, two degrees of freedom at point A are prescribed, i.e. U A  VA  0 , and one degree of freedom at point D is prescribed, i.e. VD  0 . It is obvious that there is no a straight segment containing only three unknowns. As a result, it is impossible to completely determine the displacement and rotation at a point using only one straight segment. However, by separating the structure into three segments AB, BC and CD, the total number of kinematical degrees of freedom associated with both ends of the three segments is equal to 3x6 = 18 where three of them are prescribed and the remaining fifteen are unknown. The number of equations that can be established for the three segments is equal to 3x3 = 9. Six additional equations are obtained from the continuity of the displacement and rotation at the joint B and the joint C. Once the 15 equations are solved simultaneously, displacements and rotations at points A, B, C, and D are now known. The displacement and rotation at other points can readily be determined; for instance, the displacement and rotation at point E can be computed from a segment AE or a segment EB, and the displacement and rotation at a point F can be computed from a segment BF or a segment FC. For this particular structure, an alternative strategy may be AB employed to avoid solving a large system of linear equations. For the segment AB, { u AB A , v A } are AB AB AB AB can be prescribed and { AB A , u B , v B ,  B } are unknown. Toward using the remark (3), u B AB AB AB obtained from the length constraint equation (4.14) while { A , v B , B } still cannot be solved is from the first and second curvature area equations (4.15)-(4.16). For the segment CD, u CD D CD CD CD CD CD CD prescribed and { u C , vC , C , v D , D } are unknown. Again, from the remark (3), u C can be CD CD CD obtained from (4.14) while { v CD C , C , v D ,  D } still cannot be obtained by solving (4.15) and Copyright © 2011 J. Rungamornrat


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BC BC AB (4.16). For the segment BC, { v BC and B , v C } are known from the continuity relations, i.e. v B  u B BC CD AB CD vC  u C , along with the known u B from the segment AB and the known u C from the segment BC BC BC BC BC CD, while { u BC B ,  B , u C , C } are still unknown. By using the remark (5), {  B , C } can be BC still cannot be determined from (4.15) and (4.16) while the other two unknowns { u BC B , uC } solved. Now, we return to the segment AB. With results obtained for the segment BC and the BC AB AB continuity at point B (e.g. AB B   B ), the remaining two unknowns {  A , v B } can be obtained by BC solving (4.15) and (4.16). By returning to the segment BC, the two unknowns { u BC B , u C } can now AB be obtained from the continuity at the point B (e.g. u BC B   v B ) and the length constraint equation CD (4.14). By returning to the segment CD, the remaining two unknowns { vCD D ,  D } can now be CD computed from (4.15) and (4.16) since { v CD C , C } are already known from the continuity at point C. With a procedure utilized above, the displacements and the rotations at points A, B, C, and D can be determined without solving 15 linear equations simultaneously.

x, y

M/EI

B

F

C

x, y

M/EI E D

y

A x

Figure 4.10: Schematic of statically determinate frame subjected to concentrated force and the corresponding curvature diagram Note that, for a structure consisting of segments with an arbitrary orientation or containing supports with directions of constraint neither aligned with and normal to the axis of the connecting segment as shown schematically in Figure 4.11, it may not be possible to use above strategy to avoid solving a large system of linear algebraic equations. This results primarily from the fact that the length constraint equation and the first and second curvature area equations have been derived within the context of a straight segment and based on its local reference coordinate system. This limitation renders the method of curvature area efficient only for structures with simple configurations. To enhance their capability, equations (4.14)-(4.16) must be established in terms of quantities referring to a global coordinate system and to be valid for a part of structure consisting of several straight segments. Details of such development will be discussed within the context of a conjugate structure analogy presented in the chapter 5. Copyright © 2011 J. Rungamornrat


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Figure 4.11: Schematic of structure consisting of segments with arbitrary orientation and supports with a direction of constraint neither aligned with and normal to the axis of connecting segments As a final remark, for structures containing the moment releases or hinges as shown schematically in Figure 4.12, straight segments chosen in the calculation must not contain those hinges except at their ends. For instance, segment AC and segment CE cannot be used in the calculation while segments AB, BC, CD, DE and EF are allowed. This restriction results from the fact that both the first and second curvature area equations were derived based on a key assumption that the rotation at any point within the segment must be continuous.

C

D

E

B C A

B

D

A

F

Figure 4.12: Schematic of structures containing moment releases or hinges

4.4.1 Construction of BMD by method of superposition It is obvious that in applications of both the first and second curvature area equations to any segment, it is required to compute the area and the first moment of area of the curvature diagram over that segment. While this can be achieved, in principle, by performing integration of the total curvature of the segment, it can lead to substantial computational effort especially when the total curvature diagram is complex. To avoid such direct integration, the total curvature diagram over a given segment is often decomposed into several sub-curvature diagrams whose area and centroid can readily be computed. By linearity of the problem, the area of the total curvature diagram is equal to the sum of the area of all sub-curvature diagrams and, similarly, the first moment of area of the total curvature diagram is equal to the sum of the first moment of area of all sub-curvature diagrams. Before demonstrating the construction of sub-curvature diagrams, we review how to sketch the bending moment diagram of a cantilever beam subjected to various loading conditions. These useful results will be essential for decomposing the total bending moment diagram over a straight segment into several simple bending moment diagrams called sub-BMD (“simple” in this sense Copyright © 2011 J. Rungamornrat


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means its area and its centroid can readily be computed without performing direct integration) and these bending moment diagrams are used further to construct sub-curvature diagrams.

4.4.1.1 Cantilever beam subjected to a concentrated load Consider a cantilever beam subjected to a concentrated force P at a distance h from the fixed support. The bending moment diagram for this particular loading condition simply forms a triangle of base h and height +Ph or –Ph (depending on the direction of the concentrated force) as shown schematically in Figure 4.13. The centroid of the BMD is located at a distance h/3 from the fixed support and the area can simply be computed from a simple formulae Area = (height)(base)/2. Ph

Ph P

P

h/3

2h/3

2h/3 P

h/3

P

–Ph

–Ph

Figure 4.13: Cantilever beams subjected to a concentrated force and the corresponding BMD

4.4.1.2 Cantilever beam subjected to a concentrated load Consider a cantilever beam subjected to a concentrated moment M at a distance h from the fixed support. The bending moment diagram for this particular loading condition simply forms a rectangle of base h and height +M or –M (depending on the direction of the concentrated moment and the location of the fixed support) as shown schematically in Figure 4.14. The centroid of the BMD is located at a distance h/2 from the fixed support and the area can simply be computed from a simple formulae Area = (height)(base). M

M M

M

h/2

h/2

h/2

h/2

M

M

–M

–M

Figure 4.14: Cantilever beams subjected to a concentrated moment and the corresponding BMD Copyright © 2011 J. Rungamornrat


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4.4.1.3 Cantilever beam subjected to uniformly distributed load Consider a cantilever beam subjected to a uniformly distributed load q over a distance h from the fixed support. The bending moment diagram for this particular loading condition simply forms a parabola of base h and height +qh2/2 or –qh2/2 (depending on the direction of distributed load q) as shown schematically in Figure 4.15. The centroid of the BMD is located at a distance h/4 from the fixed support and the area can simply be computed from a simple formulae Area = (height)(base)/3. qh2/2

qh2/2 q

h/4

q

h/4

3h/4

3h/4

q

q

–qh2/2

–qh2/2

Figure 4.15: Cantilever beams subjected to uniformly distributed load and the corresponding BMD

4.4.1.4 Cantilever beam subjected to power-law distributed load Consider a cantilever beam subjected to distributed load q over a distance h from the fixed support. The distributed load possesses the following form q(x) = c(x/h)n where c is a constant representing the value of distributed load at the fixed support, n is a positive integer, and x is a distance measured from the end point of the loading zone toward the fixed support. The bending moment diagram for this particular loading condition simply forms a xn+2-curve of base h and height +ch2/(n + 1)(n + 2) as shown schematically in Figure 4.16. The centroid of the BMD is located at a distance h from the fixed support where  = 1/(n + 4) and the area can simply be computed from a simple formulae Area = (height)(base)/(n + 3). For n = 1, the distributed load varies linearly over the distance h and the height of the BMD is ch2/6. The centroid is located at a distance h/5 from the fixed support and the area can be computed from a formulae Area = (height)(base)/4. d

d c(x/h)n

c(x/h)n

x

x

h (1–)h

(1–)h

–c(x/h)n

h

–c(x/h)n

–d

–d

Figure 4.16: Cantilever beams subjected to power-law distributed load and the corresponding BMD Copyright © 2011 J. Rungamornrat


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To construct sub-curvature diagrams (sub-CDs) of a given prismatic straight segment, following procedures are recommended:  choose one end of the segment as a fixed support and treat it as a fictitious cantilever beam  identify all forces and moments acting to the segment (i.e. all external applied loads and the internal forces appearing at the non-fixed end)  sketch the sub-BMD for each loading condition using results from the sections 4.4.1.14.4.1.4  divide each sub-BMD by the flexural rigidity EI of the segment to obtain sub-curvature diagrams  the area of each sub-curvature diagram over the segment is equal to the area of the corresponding sub-BMD divided by the flexural rigidity EI  the centroid of each sub-curvature diagram is the same as the centroid of the corresponding sub-BMD For instance, consider a beam of length 3L subjected to a concentrate load 3qL and a distributed load 2q as shown schematically in Figure 4.17. The flexural rigidity of the segment AB and segment BD are 2EI and EI, respectively. We first separate the beam into three prismatic segments AB, BC and CD and sub-curvature diagrams for each segment are then constructed as follows: 3qL A

B

2EI, L

2q EI, L

C

qL

D

4qL

A qL

EI, L

B B

qL2

C 2q

2qL

D

C qL2

qL2

Sub-BMDs –qL2 –2qL2 2

qL2/EI

qL /2EI

Sub-CDs –qL2/EI –2qL2/EI

Figure 4.17: Sub-BMDs and sub-CDs of a beam with 3 segments Copyright © 2011 J. Rungamornrat


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

Segment AB: This segment is separated from the structure by making a cut at point just to the left of point B  point B is chosen as a fixed support  a cantilever segment AB is now subjected to a positive reaction qL at point A  bending moment diagram can readily be obtained using results from section 4.4.1.1.  Segment BC: This segment is separated from the structure by making two cuts, one at a point just to the right of point B and the other at a point just to left of point C  point C is chosen as a fixed support  a cantilever segment BC is now subjected to a negative shear force 2qL and a positive moment qL2 which appear at the cut at point B  bending moment diagrams can be obtained using results from section 4.4.1.1 and section 4.4.1.2.  Segment CD: This segment is separated from the structure by making a cut at a point just to the right of point C  point C is chosen as a fixed support  a cantilever segment CD is now subjected to a negative uniform distributed load 2q  bending moment diagrams can be obtained using results from section 4.4.1.3. It is important to emphasize that the choice of the fixed support can be chosen arbitrarily and, most of the time, it is a matter of preference. In general, one may choose the end of the segment to be a fixed support to avoid determining the internal forces along the cut as much as possible. While the sub-curvature diagrams can appear very different when different choices of the fixed support are employed, the total curvature diagrams for each segment resulting from any choice are identical. Figure 4.18 shows the sub-bending moment diagrams and sub-curvature diagrams of the same beam shown in Figure 4.17 but with different choice of the fixed support.  Segment AB: This segment is separated from the structure by making a cut at point just to the left of point B  point A is chosen as a fixed support  a cantilever segment AB is now subjected to a positive shear force qL and a positive moment qL2  bending moment diagrams can then be obtained using results from sections 4.4.1.1 and 4.4.1.2.  Segment BC: This segment is separated from the structure by making two cuts, one at point just to the right of point B and the other at point just to left of point C  point B is chosen as a fixed support  a cantilever segment BC is now subjected to a negative shear force 2qL and a negative moment qL2  bending moment diagrams can then be obtained using results from sections 4.4.1.1 and 4.4.1.2.  Segment CD: This segment is separated from the structure by making a cut at point just to the right of point C  point D is chosen as a fixed support  a cantilever segment CD is now subjected to negative uniform distributed load 2q, a positive shear force 2qL, and a negative moment qL2  bending moment diagrams can then be obtained using results from sections 4.4.1.1, 4.4.1.2, and 4.4.1.3. By comparing sub-curvature diagrams from Figure 4.17 and Figure 4.18, it is suggested that the segment AB should be fixed at point B, the segment BC can be fixed either at the point B or the point C, and the segment CD should be fixed at the point C in order to minimize the number of subcurvature diagrams and also reduce the calculation of internal forces (i.e. shear force and bending moment) at the cut. For instance, if point A is chosen as a fixed support for the segment AB, it is unavoidable to compute both the shear force an bending moment at point B before sub-bending moment diagrams can be constructed. Similarly, if point D is chosen as a fixed support for the segment CD, the shear force and bending moment at point C are needed. Instead of separating the beam into three segments, it is also possible to construct the subcurvature diagrams by considering only two segments. For example, the same beam shown in Figure 4.17 may be separated into a segment AB and a segment BD, as shown schematically in Figure 4.19. Sub-BMD and sub-curvature diagrams for these two segments can readily be obtained as follows:  Segment AB: This segment is separated from the structure by making a cut at a point just to the left of point B  point B is then chosen as a fixed support  a cantilever Copyright © 2011 J. Rungamornrat


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segment AB is now subjected to a positive reaction qL at point A  bending moment diagram can then be obtained using results from section 4.4.1.1.  Segment BD: This segment is separated from the structure by making a cut at point just to the right of point B  point B is then chosen as a fixed support  a cantilever segment BD is now subjected to a negative distributed load 2q over its entire span, a positive distributed load 2q over a half-span, and a positive reaction 4qL  bending moment diagram can then be obtained using results from sections 4.4.1.1 and 4.4.1.3. It is worth noting that the sub-curvature diagrams constructed in this fashion will not be suitable, when we need to apply the first and second moment area equations to the segment BC, since the area and centroid of the sub-curvature diagram associated with the distributed load over the segment BC cannot readily be computed. Another possible choice for constructing the sub-curvature diagrams is to separate the beam into two segments, a segment AC and a segment CD, as shown in Figure 4.20: 3qL A

2EI, L

2q

B

C

EI, L

qL

EI, L

D

4qL

A

B

qL2

qL

2qL C qL2

B

2q 2

qL

C 2qL

2qL2

D

2qL2

qL2

Sub-BMDs –qL2

–qL2 2qL2/EI

–qL2 –qL2 2qL2/EI

qL2/2EI

Sub-CDs 2

–qL /2EI

–qL2/EI

–qL2/EI –qL2/EI

Figure 4.18: Sub-BMDs and sub-CDs of beam shown in Figure 4.17 Copyright © 2011 J. Rungamornrat






229


Segment AC: This segment is separated from the structure by making a cut at a point just to the left of point C  point C is then chosen as a fixed support  a cantilever segment AC is now subjected to a positive reaction qL at point A and a negative concentrated force 3qL at point B  bending moment diagram can then be obtained using results from section 4.4.1.1. Segment CD: This segment is separated from the structure by making a cut at a point just to the right of point C  point C is then chosen as a fixed support  a cantilever segment CD is now subjected to a negative distributed load 2q over its entire span  bending moment diagram can then be obtained using results from section 4.4.1.3. 3qL A

B

2EI, L

2q EI, L

C

qL A

EI, L

D

4qL B

qL

2q B 2q

D 4qL

4qL2 qL2

Sub-BMDs

–4qL2 qL2

4qL2/EI qL2/2EI

Sub-CDs

–4qL2/EI qL2/EI

Figure 4.19: Sub-BMDs and sub-CDs of beam shown in Figure 4.17 Copyright © 2011 J. Rungamornrat


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For this particular case, special attention must be paid to the discontinuity of the flexural rigidity at point B. The sub-bending moment diagram associated with the reaction at the point A forms a single triangle while the corresponding sub-curvature diagram is discontinuous at point B. 3qL A

2EI, L

2q

B

EI, L

C

qL

D

EI, L 4qL

3qL A C

B

qL

2q C

D

2qL2

Sub-BMDs –qL

2

–3qL2 2qL2/EI 2

qL /EI qL2/2EI

Sub-CDs –qL2/EI

–3qL2/EI

Figure 4.20: Sub-BMDs and sub-CDs of beam shown in Figure 4.17 Example 4.1 Consider a cantilever beam of length 2L and subjected to concentrated force P and concentrated moment PL as shown below. Determine the tip deflection and tip rotation of a beam for the following two cases: 1) the flexural rigidity EI is constant throughout the beam and 2) the flexural rigidity of segments AB and BC is given by 2EI and EI, respectively. P PL

A L

B L


C


231


Solution Since the given structure is a beam, there are only two kinematical quantities at each point, i.e. the rotation and the vertical displacement (or deflection). From boundary conditions, the beam is fully fixed at point A and, therefore, the deflection and rotation vanish at that point, i.e. A  v A  0 . The tip deflection and the rotation can then be obtained by applying the first and second curvature area equations to the segment AC. Sub-curvature diagrams of this beam can be constructed by considering only one segment AC (with a fixed point A) with results shown below. P PL

A L

B L

C

PL

PL

Sub-BMDs –PL PL/EI

PL/EI

Sub-CDs – Case 1) –PL/EI PL/EI

PL/2EI

Sub-CDs – Case 2) –PL/2EI

From the bending moment diagram, a qualitative elastic curve of the beam can be sketched as shown below: C

B A

tC/A = vCAC C/A = CAC

B

C

AC AC and Case 1): consider the segment AC  AC A  0 and v C  0  there remain two unknowns C AC vC  first and second curvature area equations are sufficient to solve for these two unknowns as shown below

1st curvature area equation: C/A = CAC  AC A = A M/EI,AC 1  PL  3PL2  PL     2L L  CAC  0          2  EI  2EI  EI  2 3PL CCW  CAC  2EI AC 2nd curvature area equation: t C/A = v CAC  (v AC A   A L AC ) = A M/EI,AC x C Copyright © 2011 J. Rungamornrat


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1  PL  2L  7PL3  PL    vCAC  0   2L L   L L+         2  EI  3  6EI  EI   7PL3  vCAC  Upward 6EI AC AC Case 2): consider the segment AC  AC and A  0 and v C  0  there remain two unknowns C AC vC  first and second curvature area equations are sufficient to solve for these two unknowns as shown below

1st curvature area equation: C/A = CAC  AC A = A M/EI,AC 1  PL  5PL2  PL   PL  L L L      CAC  0              2  2EI  4EI  EI   2EI  2 5PL CCW  CAC  4EI AC 2nd curvature area equation: t C/A = v CAC  (v AC A   A L AC ) = A M/EI,AC x C 3  PL   L   PL   3L  1  PL   5L  5PL  vCAC  0   L  L   L          2   2EI   2  2  2EI   3  6EI  EI            3 5PL  vCAC  Upward 6EI

While sketch of a qualitative elastic curve is not necessary in the application of curvature area equations (at least within the context of the current development), it can still be useful to gain physical insight into kinematical quantities of interest. In addition, geometric consideration of qualitative elastic curve can be employed to retrieve relations such as C/A = CAC  AC and A AC AC AC t C/A = v C  (v A  A L AC ) without memorization. For instance, it is obvious from the above elastic curve that C/A = CAC and t C/A = v CAC . Example 4.2 For a prismatic beam of flexural rigidity EI and subjected to applied loads shown below, determine the deflection and rotation at point C and the maximum deflection within the span AB qL

2q A

EI 2L

B

EI

C

L

Solution From boundary conditions, the deflections at point A and point B vanish, i.e. v A  v B  0 . With these conditions, there exists only one segment (i.e. a segment AB) within the beam that AB contains only two kinematical unknowns, i.e. { θ AB A , θ B }. For mathematical convenience, we should apply the curvature area equations to the segment AB first in order to obtain the remaining unknowns at points A and B. Once those unknowns are resolved (i.e. the deflection and rotation at points A and B are completely known), any segments containing either the point A or the point B as one of its ends can be chosen to determine the deflection and rotation at any point of interest. For Copyright © 2011 J. Rungamornrat


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instance, the deflection and rotation at point C can be obtained by applying the curvature area equations to the segment BC. Let us start with the construction of sub-curvature diagrams for the entire beam. First, two support reactions are determined from equilibrium of the entire beam. Next, the beam is separated into two segments, i.e. segment AB and segment BC, with point B being chosen as a fixed point for both segments. The final sub-curvature diagrams are shown below. qL

2q A

B

EI 3qL/2

EI

7qL/2

2L

C

L

3qL2

Sub-BMDs

–qL2 –4qL2 3qL2/EI

Sub-CDs

–qL2/EI –4qL2/EI

AB AB Segment AB: v AB and AB A  0 and v B  0  there remain only two unknowns  A B  first and second curvature area equations are sufficient to obtain these two unknowns as shown below.

A

AB B 

AB A

B

C

B/A tB/A

AB 2nd curvature area equation: t B/A = v BAB  (v AB A   A L AB ) = A M/EI,AB x B

2 4 1  3qL2   2L  1  4qL   2L  2qL  AB    2L     2L   A  2L       2  EI   3  3  EI   4  3EI 3 qL  AB CW  A 3EI  AB 1st curvature area equation: B/A = AB B A = A M/EI,AB

 qL3  1  3qL2  1  4qL2  qL3    2L   2L   AB  AB 0           B B 3EI 2 EI 3 EI 3EI       Copyright © 2011 J. Rungamornrat


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Since there is no internal release at point B, the deflection and rotation at this particular point must AB BC AB be continuous. This leads to v BC B  v B  0 and  B   B  0 . To obtain the rotation and deflection at point C, we now consider a segment BC. BC BC Segment BC: v BC and CBC  first and B  0 and  B  0  there remain only two unknowns v C second curvature area equations are sufficient to obtain these two unknowns

BC B

B

A

C

tC/B

CBC

1st curvature area equation: C/B = CBC  BC B = A M/EI,BC 1  qL2  qL3  CBC  0     L   2  EI  2EI BC 2nd curvature area equation: t C/B = vCBC  (v BC B   B L BC ) = A M/EI,BC x C

CBC  

qL3 CW 2EI

1  qL2  qL4 qL4  2L  Downward   vCBC     vCBC   L  3EI 2  EI  3EI  3 

To determine the maximum deflection within the span AB, first we need to identify the location where it occurs. This can be achieved simply by using the necessary condition that the rotation vanishes at the point where the deflection is maximum. To proceed, let the maximum deflection be located at point D with the distance x from point A. Since the deflection and rotation at point A are already known, the segment AD now contains only two unknowns and they can be obtained from the first and second curvature area equations. Note however that the sub-curvature diagrams shown above are not well-suited for the segment AD since their area and centroid cannot be easily calculated. As a result, we reconstruct sub-curvature diagrams again, but only for the segment AD, as shown below (in this construction, the segment AD is fixed at point D). qL

2q A

EI

B

D

3qL/2

EI

C

7qL/2 x

2L – x

L

3qLx/2

Sub-BMDs –qx2 3qLx/2EI

Sub-CDs –qx2/EI



235


B

A

AD A



AD D

D/A

v

C

AD D

D tD/A

Now, the deflection and rotation at point D can be obtained from the first and second curvature area equations as shown below. 1st curvature area equation: D/A = DAD  AAD = A M/EI,AD  qL3  1  3qLx  1  qx 2  3qLx 2 qx 3    AD x    x     D  2  2EI  3  EI  4EI 3EI  3EI  3qLx 2 qx 3 qL3     AD D 4EI 3EI 3EI AD  (v AD 2nd curvature area equation: t D/A = v AD D A   A L AD ) = A M/EI,AD x D 2  qL3  1  3qLx   x  1  qx  x   x  x    v AD      x  D     2  2EI   3  3  EI  4  3EI  3 3 4 qL x qLx qx     v AD D 3EI 4EI 12EI

To obtain the location of point D or the distance x  (0, L), we simply set AD  0 and then solve D the corresponding equation to obtain x = 0.8431L. The maximum deflection at point D can now be obtained as v AD D =

qL3 (0.8431L) qL(0.8431L)3 q(0.8431L) 4 0.1733qL4 Downward   =  3EI 4EI 12EI EI

Example 4.3 Consider a statically determinate beam subjected to external applied load as shown below. The Young's modulus E is assumed to be constant throughout and the moment of inertia of the cross section for segments AB and BD is given by 2I and I, respectively. Determine the relative rotation at hinge B and the deflection at point D. 2qL

2q C A

2EI 2L

B

EI

EI

2L

L

D

Solution: From boundary conditions, the deflections and rotation at point A and the deflection at point C vanish, i.e. v A  A  v C  0 . Since all kinematical quantities are known at point A, the deflection and rotation at point just to the left of hinge B can readily be determined by applying curvature area equations to a segment AB. Upon enforcing the continuity of the deflection at the hinge B, a segment BC now contains only two kinematical unknowns (i.e. both end rotations) and they can be solved from the two curvature area equations. It is important to emphasize that the Copyright © 2011 J. Rungamornrat


236


rotation is discontinuous at the hinge B or, equivalently, the rotation at a point just to the left and a point just to the right of the hinge B are not necessary identical. The jump in the rotation at the hinge is generally termed the relative hinge rotation. Once the rotation at point C is determined from the segment BC, a segment CD is then considered to compute the deflection and rotation at point D. Let us start with the construction of sub-curvature diagrams for the entire beam. First, three support reactions (two at the fixed support and one at the roller support) are determined from static equilibrium. Next, the beam is divided into three segments AB, BC and CD with points A and C being chosen as the fixed point for the segment AB and the segments BC and CD, respectively. The final sub-curvature diagrams are shown below. 2qL

2q C

2qL2

A 3qL

2EI

EI

B

EI

D

3qL 2L

L

2L

2qL2

Sub-BMDs –2qL2 –4qL2 qL2/EI

Sub-CDs –2qL2/EI

–2qL2/EI

AB AB and AB Segment AB: v AB  first and second A   A  0  there remain only two unknowns v B B curvature area equations are sufficient to solve for these two unknowns

B

C

A B/A = AB B

D

tB/A = v AB B

1st curvature area equation: B/A = AB  AB B A = A M/EI,AB 1  qL2  1  2qL2  qL3  0  2L   2L    AB         B 2  EI  3  EI  3EI Copyright © 2011 J. Rungamornrat


237


qL3 CW 3EI AB  (v AB A   A L AB ) = A M/EI,AB x B

  AB B

2nd curvature area equation: t B/A = v BAB

1  qL2    v AB B 2  EI 2qL4    v AB B 3EI

2  2qL4  4L  1  2qL   6L    2L       2L       4  3EI  3  3  EI    

Downward

AB 4 From continuity of the deflection at point B, we then obtain v BC B  v B  2qL /3EI  . Next, we move to the segment BC. 4 BC BC and CBC  Segment BC: v BC B  2qL /3EI and v C  0  there remain only two unknowns  B first and second curvature area equations are applied to solve for these two unknowns as shown below

C/B tC/B B A AB B

v BC B

C  BC B Δθ B

BC C

D

BC 2nd curvature area equation: t C/B = vCBC  (v BC B   B L BC ) = A M/EI,BC x C

 2qL4  1  2qL2  4qL4  2L   BC   0      2L   B (2L)    3EI  3   3EI  2  EI  3 qL CCW   BC B EI 1st curvature area equation: C/B = CBC  BC B = A M/EI,BC qL3 1  2qL2  2qL3     2L    EI 2  EI  EI qL3 CW  EI

 CBC   CBC

The relative hinge rotation at point B, denoted by Δθ B , now becomes AB Δθ B  BC B  B 

qL3  qL3  4qL3  CCW  EI  3EI  3EI

BC 3 From continuity of the rotation at point C, we have CD C  C  qL /EI CW. Finally, we move to the segment CD to compute the displacement and rotation at point D. CD 3 CD and vCD  two Segment CD: vCD C  0 and C  qL /EI  there remain only two unknowns  D D curvature area equations are sufficient to obtain these two unknowns as follows



238


B

A

CD C C

D θ D/C

CD D

tD/C

1st curvature area equation: D/C = CD  CCD = A M/EI,CD D  qL3  1  2qL2  qL3     L    CD      D EI  EI  2  EI  2qL3 CW   CD D EI CD  (vCD 2nd curvature area equation: t B/A = v CD D C  C L CD ) = A M/EI,CD x D  qL3  1  2qL2  2qL4  2L  (L) L  vCD             D  3  2  EI  3EI    EI  4 5qL Downward   vCD D 3EI

Example 4.4: Given a statically determinate frame subjected to a concentrated force P as shown below. The Young's modulus E and the moment of inertia of the cross section I are assumed to be constant throughout the structure. Determine the displacement and rotation at point E. P

D

E Y

L C

A

X

B L

L

L

Solution: First, let us define local coordinate systems for following three segments AC, BD, and DE as shown in the figure below. Since the given structure is a rigid frame, for any proper segment chosen, there are three independent equations available in the analysis for the displacement and rotation (i.e. two curvature area equations and one length constraint equation). y D E

x,y

y B

A x


C

x


239


To obtain the displacement and rotation at point E, following strategy can be employed. From AC AC boundary conditions at point A and C (i.e. u AC A  v A = v C  0 ), there are only three kinematical AC AC unknowns left for the segment AC (i.e. AC A , u C , C ) and, as a result, this segment must be considered first to solve for those three unknowns. Next, either the segment AB or the segment BC can be considered to compute the displacement and rotation at point B since the displacement and rotation at point A and point C are completely known and these two segments contain exactly three kinematical unknowns. Once the displacement and rotation at point B are determined, the segment BD can then be considered to compute the displacement and rotation at point D. Finally, the displacement and rotation at point E can be obtained from the segment DE. Let us start with the construction of sub-curvature diagrams for the entire frame. First, three support reactions (two at the pinned support and one at the roller support) are determined from static equilibrium of the entire frame. Next, the frame is divided into four segments AB, BC, BD and DE with points B and D being chosen as fixed points for segments AB, BC and BD and a segment DE, respectively. It is emphasized that the sign convention of the sub-curvature diagrams must strictly follow the local coordinate system defined above. The final sub-curvature diagrams are given below. P D E –PL

–PL

4PL/3 PL/3

A

0

Sub-BMDs C

B 2P/3

P/3 D E –PL/EI

–PL/EI

4PL/3EI PL/3EI

A

Sub-CDs C

B AC AC AC AC and CAC  two Segment AC: u AC A  v A  v C  0  there remain only three unknowns  A , u C curvature area equations and one length constraint equation are sufficient to obtain these three unknowns as shown below.

E

D

y A

AC A

B C/A


CAC C

tC/A

x


240

Method of Curvature (Moment) Area AC 2nd curvature area equation: t C/A = v CAC  (v AC A   A L AC ) = A M/EI,AC x C

1  4PL  2L 1 PL 2L 7PL3  AC  2L   L       L     A  3L     2  3EI  3  2  3EI    3  3EI 2 7PL  AC CW  A 9EI 1st curvature area equation: C/A = CAC  AC A = A M/EI,AC  7PL2  1  4PL  1  PL  3PL2  CAC         2L    L  2  3EI  2  3EI  2EI  9EI  2 13PL CCW  CAC  18EI Length constraint equation: u CAC  u AC A

 u CAC  0 AB AB AB Segment AB: u AB  7PL2 /9EI  there remain only three unknowns u AB and A  v A  0 , A B , vB AB B  two curvature area equations and one length constraint equation are sufficient to obtain these three unknowns as follows.

E

D

y A

AB A

B 

C

AB B

B/A

x

tB/A

1st curvature area equation: B/A = AB  AB B A = A M/EI,AB  7PL2  1  4PL  4PL2     2L  AB     B   2  3EI  3EI  9EI  5PL2 CCW   AB B 9EI AB 2nd curvature area equation: t B/A = v BAB  (v AB A   A L AB ) = A M/EI,AB x B

 7PL2   v AB  B  9EI 2PL3    v AB B 3EI AB AB Length constraint equation: u B  u A

3  1  4PL   2L  8PL 2L 2L         3  9EI 2  3EI    

Downward

0  u AB B AB 3 Segment DB: continuity of displacement and rotation at point B  u DB B   v B  2PL /3EI , AB DB AB DB v DB  5PL2 /9EI  now there remain only three unknowns u DB and DB B  u B  0 , B  B D , vD D  two curvature area equations and one length constraint equation are sufficient to obtain these three unknowns as follows.



241


E

D tD/B

B/D

DB D

A

y

DB B

B

C

x  DB 1st curvature area equation: B/D = DB B D = A M/EI,DB 5PL2 PL2  PL   DB   D L   9EI EI  EI  2 14PL CCW   DB D 9EI DB  (v DB 2nd curvature area equation: t D/B = v DB D B   B L DB ) = A M/EI,DB x D



 5PL2  L PL3  PL   v DB  L        L    D   2EI  EI  2  9EI  3 19PL Leftward   v DB D 18EI  u DB Length constraint equation: u DB D B   u DB D

2PL3 Downward 3EI

DB 3 Segment ED: continuity of displacement and rotation at point D  u ED D  v D  19PL /18EI , DB 3 ED DB ED v ED  14PL2 /9EI  now there remain only three unknowns u ED D   u D  2PL /3EI ,  D   D E , vE and ED  two curvature area equations and one length constraint equation are sufficient to solve E for these three unknowns.

y E D/E tE/D A

ED E

D 

ED D

x

B

1st curvature area equation: D/E = DED  EED = A M/EI,ED 14PL2 1  PL  PL2  ED   E L   9EI 2  EI  2EI 2 37PL CCW   DB D 18EI




C


242

Method of Curvature (Moment) Area ED 2nd curvature area equation: t E/D = v ED  (v ED E D   D L ED ) = A M/EI,ED x E

 2PL3 14PL2  1  PL  2L PL3  (L)      v ED  L      E   9EI 2  EI  3EI  3   3EI  3 23PL Downward   v ED E 9EI  u ED Length constraint equation: u ED E D   u ED E

19PL3 Leftward 18EI

Example 4.5 Given a statically determinate frame subjected to external loads as shown below. The Young’s modulus E and moment of inertia of the cross section I are constant throughout the structure. Determine the displacements and rotations at points B and C. Y 2PL C

P

B D

2L

A

X 2L

L

Solution First, let us define the local coordinate systems for following two segments AB and CBD as shown in the figure below. y

x x

B

C

D

y

A

AB BD From boundary conditions at points A and D, we obtain u AB A  v A = v D  0 . It is evident for this particular structure that there is no straight segment that contains only three kinematical unknowns. AB AB AB AB AB For the segment AB, { u AB A , v A } are known and {  A , u B , v B ,  B } are unknown; for the BD BD BD BD BD segment BD, { v BD D } is known and { u B , v B ,  B , u D ,  D } are unknown; and, for the segment



243


CB CB CB CB CB CB, { u CB C , v C , C , u B , v B ,  B } are unknown. Following strategy can be employed to solve all is obtained from the length constraint equation of the those unknowns. The displacement u AB B BD segment AB. Next, the displacement v B is obtained from the continuity of the displacement at BD BD = u AB point B (i.e. v BD B B ). Then, {  B ,  D } can be determined from the curvature area equations of is obtained from the the segment BD. Now, by returning to the segment AB, the rotation AB B AB BD AB AB continuity of the rotation at point B (i.e. B = B ) and { A , v B } can be obtained from the is obtained curvature area equations. Next, by returning to the segment BD, the displacement u BD B BD AB BD from the continuity of the displacement (i.e. u B = – v B ) and u D can then be computed from the length constraint equation. Now, the displacement and rotation at points A, B and D are completely known. Finally, the displacement and rotation at point C can readily be obtained from the segment CB. Let us start with the construction of sub-curvature diagrams for the entire frame. First, three support reactions (two at the pinned support and one at the roller support) are determined from static equilibrium of the entire structure. Next, the frame is divided into three segments AB, BD and CB with point B being chosen as a fixed point for all segments. Again, the sign convention of the sub-curvature diagrams strictly follows the local coordinate system defined above. The final subcurvature diagrams are shown below.

4PL/EI

4PL 2PL/EI

2PL

2PL

D C

2PL

B

P

C 2PL/EI

D B

2P

P

A

A 2P

Sub-CDs

Sub-BMDs Segment AB: u AB A  0 and, from the length constraint equation, we have u AB  u AB B A

0  u AB B

Continuity of displacement at point B: v BD = u AB =0 B B BD Segment BD: v BD B  v D  0 and, from two curvature area equations, we obtain

2nd curvature area equation: t D/B = v DBD  (v BBD  BBD LBD ) = A M/EI,BD x D 3 8PL2 1  4PL   4L  16PL BD CW     BD  2L    B B  2L    3  3EI 2  EI  3EI  



244


BD B

B

D

C

BD D

D/B t D/B

A 1st curvature area equation: D/B = BD  BD D B = A M/EI,BD  8PL2  1  4PL  4PL2 4PL2 BD CCW      2L   BD      D D 3EI 2  EI  EI  3EI  8PL2 BD Continuity of rotation at point B: AB =  =  B B 3EI 2 8PL AB Segment BD: v AB and, from two curvature area equations, we obtain A  0 , B =  3EI

B C

tB/A

D

AB B

B/A AB A

A

1st curvature area equation: B/A = AB  AB B A = A M/EI,AB 8PL2 1  2PL  2PL2 14PL2 CW   AB    AB A A   2L   3EI 3EI 2  EI  EI AB 2nd curvature area equation: t B/A = v BAB  (v AB A   A L AB ) = A M/EI,AB x B

 

3  14PL2  1  2PL   2L  4PL  v AB   2L    B    2L    2  EI   3  3EI  3EI  3 8PL Rightward   v AB B EI 8PL2 Continuity of displacement at point B: u BD = – v AB = B B EI 2 8PL Segment BD: u BD = and, from the length constraint equation, we have B EI 8PL2 BD BD u BD  u Rightward u   D D B EI



245


Continuity of displacement and rotation at point B: 8PL2 8PL2 CB BD BD , vCB B  v B  0 and  B   B   EI 3EI 2 2 8PL 8PL Segment CB: u CB , CB and, from two curvature area equations, we obtain B  B  EI 3EI BD u CB B  uB 

CB C

C

B

CB B

D

tB/C

B/C

A 1st curvature area equation: B/C = CB  CB B C = A M/EI,CB 14PL2 8PL2 2PL2  2PL  CW   CB  CB  L  C C  3EI 3EI EI  EI  CB  (v CB 2nd curvature area equation: t B/C = vCB B C  C L CB ) = A M/EI,CB x B  

  14PL2 2PL  L PL3  0   vCB  L      L     C   3EI  2  EI    EI  3 11PL  vCB Upward  C 3EI  u CB Length constraint equation: u CB C B   u CB C

8PL2 Rightward EI

Example 4.6 Consider a statically determinate frame subjected to a concentrated load P as shown in the figure below. The Young's modulus E is assumed to be constant throughout and the moment of inertia of a segment AB and a segment BC are I and 2I, respectively. Determine the unknown displacements and rotations at points A, B, and C. Given that  =  = 45o. A

Y 

X L

C

B

L


P




246


Solution First, let us define the local coordinate systems for segments AB and BC as shown below. A

y

y B

C

x

x Let us start with the construction of sub-curvature diagrams of the entire frame. Three support reactions (two at the pinned support and one at the roller support) are determined from static equilibrium. Next, the frame is divided into two straight segments AB and BC with point B being chosen as a fixed point for both segments. The final sub-curvature diagrams referring to above local coordinate systems are shown below. 2P 2 2P

A

A

C

B

 2PL

P

B  2PL/2EI  2PL/EI

3P

 2PL

C

Sub-CDs

Sub-BMDs

According to the inclined orientation of the segment AB and the direction of the constraint at the roller support, the unknown displacements and rotations at points A, B and C cannot be obtained independently from each segment but requiring consideration of both segments simultaneously as follow. Segment AB: By applying two curvature area equations and one length constraint equation, it leads to following three linear equations  AB 1st curvature area equation: B/A = AB B A = A M/EI,AB

1 2PL   AB     AB  B A 2 EI 





2L  


PL2 EI

(e4.6.1)


A

247


AB A

B/A

tB/A AB B

C B AB 2nd curvature area equation: t B/A = v BAB  (v AB A   A L AB ) = A M/EI,AB x B

1 2PL  AB  v AB 2LAB    B  vA  A  2 EI   u AB Length constraint equation: u AB A B



 2L  2PL3 2L     3  3EI  



(e4.6.2) (e4.6.3)

Segment BC: By applying two curvature area equations and one length constraint equation, it leads to following three linear equations A

C/B

tC/B

CBC

BC B

C B 1st curvature area equation: C/B = CBC  BC B = A M/EI,BC 1 2PL  2PL2   2L    CBC  BC    B 2  EI  EI BC 2nd curvature area equation: t C/B = vCBC  (v BC B   B L BC ) = A M/EI,BC x C 1 2PL  2PL3  2L  BC   vCBC  v BC     L   B  L B   2  2EI  6EI  3   u CBC Length constraint equation: u BC B

(e4.6.4)

(e4.6.5) (e4.6.6)

Continuity of displacement and rotation at point B: u BC B 

u AB B

v BC B 

2



u AB B 2

v AB B

(e4.6.7)

2 

v AB B

(e4.6.8)

2 Copyright © 2011 J. Rungamornrat


248

Method of Curvature (Moment) Area AB BC B  B

(e4.6.9)

Boundary conditions at point A: u AB A 0

(e4.6.10)

v AB A 0

(e4.6.11)

Boundary conditions at point C: the displacement in the direction of the constraint provided by the roller support must vanish, i.e. vCBC 2



u CBC 2

0

(e4.6.12)

Note that a set of ten linear equations (e4.6.1)-(e4.6.9) and (e4.6.12) along with the two boundary AB AB AB BC conditions (e4.6.10) and (e4.6.11) is sufficient to solve for ten unknowns { AB A , u B , v B , B , u B , BC BC BC BC v BC B ,  B , u C , v C , C }. To solve this set of equations, the continuity equations (e4.6.7)-(e4.6.9) are first expressed, by using equations (e4.6.1)-(e4.6.6) and (e4.6.12), in terms of kinematical BC BC quantities at points A and C (i.e. { AB A , u C , C }) as





2PL2 4+2 2 4EI PL3 u CBC  Lθ AAB =  3EI 2PL3 u CBC  2Lθ AB 1+4 2 A = 6EI

BC AB A  C 



(e4.6.13) (e4.6.14)



(e4.6.15)

A system of three linear equations (e4.6.13)-(e4.6.15) can readily be solved simultaneously to obtain 2 u CBC  (4+ 2)PL3 / 6EI , AB CBC  5 2PL2 / 12EI A  (6+ 2)PL / 6EI ,

The other unknowns can then be determined using equations (e4.6.1)-(e4.6.6) and (e4.6.12); results are given by AB 3 AB BC 3 BC 3 u AB 2PL2 / 6EI , u BC B  0 , v B = (1+2 2)PL / 3EI ,  B   B  B  (4+ 2)PL / 6EI , v B  (4+ 2)PL / 6EI

Thus, the unknown displacements and rotations at points A, B, and C, referring to the global coordinate system, are given by U A =VA =0 , Θ A  (6+ 2)PL2 / 6EI CCW

U B  (4+ 2)PL3 / 6EI Rightward, VB  (4+ 2)PL3 / 6EI Upward, Θ B  2PL2 / 6EI CCW U C  (4+ 2)PL3 / 6EI Rightward, VC  (4+ 2)PL3 / 6EI Upward, ΘC  5 2PL2 / 12EI CW



249


4.5 Treatment of Axial Deformation Development presented above is based primarily upon the inextensibility assumption where the axial deformation is completely neglected. This implies that the projected length of any deformed segment onto its undeformed axis must be the same as the original length. In this section, we remove such restriction by taking the axial deformation into account. It is worth noting that the static equilibrium is still enforced in the undeformed configuration; i.e. there is no interaction between axial and bending. As a result, the first and second curvature area equations established above are still valid while the length constraint equation must be modified to incorporate the axial deformation. B θ AB B v AB B

S θ

A y

v AB A

A

x

AB A

B

u AB B

So

u AB A

F/EA

Lo

Figure 3.21: Schematic of straight segment AB undergoing deformation and its strain diagram To modify the length constraint equation (4.8), we recall the differential relation between the internal axial force F = F(x) and the longitudinal component of the displacement u = u(x): F = EA

du dx

(4.17)

where E = E(x) is the Young's modulus or modulus of elasticity and A = A(x) is the cross sectional area. Remark that the axial force F is considered to be positive if it is tension otherwise it is negative. By integrating the equation (4.17) from x = 0 to x = Lo and then using the end conditions AB u(0) = u AB A and u(Lo) = u B , we obtain Copyright © 2011 J. Rungamornrat


250


u

AB B

=u

AB A



Lo

 0

F dx EA

(4.18)

It is important to emphasize that the expression (4.18) is valid as long as the longitudinal component of the displacement u is continuous at any point within the segment AB. This relation is generally known as the axial strain area equation; the name follows from the fact that F/EA represents the axial strain at any cross section of the segment and its graphical interpretation of the integral term appearing in (4.18), i.e. Lo

 0

F dx = A F/EA,AB EA

(4.19)

The strain area equation (4.18) then states that the relative displacement in the longitudinal direction between the two end points of the segment is equal to the area of the axial strain diagram over that segment, i.e. u B/A  A F/EA,AB

(4.20)

 u AB denotes the relative end longitudinal displacement. This equation where u B/A  u AB B A constitutes the generalization of the length constraint equation (4.8) by taking the axial deformation into account; if the segment is axially rigid, equation (4.20) simply reduces to (4.8) since F/AE  0 . Now, the two displacement components and the rotation at both ends of a segment AB due to the bending and shear deformations are governed by the following three equations:

B/A = θ BAB  θ AAB = A M/EI,AB t B/A = v

AB B

 (v

AB A

t A/B = v

AB A

 (v

AB B

(4.21) (4.22a) (4.22b) (4.23)

+  L AB ) = A M/EI,AB x B AB A

 AB B L AB ) = A M/EI,AB x A

u B/A  A F/EA,AB

Note that the first two equations are, in fact, identical to (4.14) and (4.15) and that all useful remarks stated above still apply to above three equations. Certain example problems are presented further below to demonstrate applications of (4.21)-(4.23) and to investigate the influence of the axial deformation on the displacement and rotation. Example 4.7 Consider a statically determinate frame subjected to concentrated forces P and Q as shown below. Young's modulus, the cross sectional area and the moment of inertia of the cross section are assumed to be constant throughout the structure and denoted by E, A and I, respectively. Determine the displacement and rotation at points B and C. P Q

C L

A

B 3L Copyright © 2011 J. Rungamornrat


251


Solution First, let us define the local coordinate systems for two segments AB and BC as shown below. x C y y A

x

B

Since the structure is statically determinate, the axial force diagram and the bending moment diagram can readily be obtained from static equilibrium. The strain diagram F/EA and the curvature diagram M/EI are plotted for each segment as indicated in the figure below. C

A

B

–QL/EI

–P/EA

Q/EA

–QL/EI –3PL/EI

From boundary conditions (at the fixed support), the displacement and rotation at point A are fully AB AB prescribed, i.e. u AB A  v A =  A  0 . The displacement and rotation at other points can readily be obtained as follow. Segment AB: since the displacement and rotation at point A are already known, i.e. AB AB u AB A  v A =  A  0 , the displacement and rotation at point B can be determined from the two curvature area equations and the strain area equation (see also the elastic curve): C

uB/A A

B/A B


tB/A


252


1st curvature area equation: B/A = AB  AB B A = A M/EI,AB

3  3P+2Q  L 1  3PL  QL  AB 0    3L       3L    B  2  EI  2EI  EI  2 3  3P+2Q  L  CW  AB B 2EI AB 2nd curvature area equation: t B/A = v BAB  (v AB A   A L AB ) = A M/EI,AB x B

2

18P  9Q  L 1  3PL  QL 3L  0    v AB  3L  2L       3L      B  2EI 2  EI   EI   2  3 18P  9Q  L Downward   v AB B 2EI Axial strain area equation: u B/A = u BAB  u AB A = A F/EA,AB

3

3QL 3QL  Q   0    u AB  u AB Rightward B B   3L   EA EA EA  

Continuity of displacement and rotation at point B: 3QL EA 18P  9Q  L3

v BC   u AB  B B u BC  v BAB   B

BC  AB  B B

2EI 3  3P+2Q  L2

2EI

Segment BC: since the displacement and rotation at point B are already known from the continuity at point B, i.e. v BC  3QL/EA , u BC   18P  9Q  L3 /2EI , and BC  3  3P+2Q  L2 /2EI , the B B B displacement and rotation at point C can be determined from the two curvature area equations and the strain area equation (see also the elastic curve): C tC/B B/A A

B

CBC

BC B

1st curvature area equation: C/B = CBC  BC B = A M/EI,BC  CBC 

3  3P+2Q  L2

 CBC  

2nd curvature area equation: t C/B = vCBC

2EI  9P+7Q  L2

2

1  QL  QL    L    2  EI  2EI

CW

2EI BC  (v BC B   B L BC ) = A M/EI,BC x C



253


3QL 3  3P+2Q  L 1  QL  2L QL3     L       EA 2EI 2  EI  3EI  3  3 3QL  27P+20Q  L   Rightward EA 6EI 3

 vCBC 

 vCBC

Axial strain area equation: u C/B = u CBC  u BC B = A F/EA,BC

 u CBC

18P  9Q  L3

PL  P    L    2EI EA  EA  3 PL 9  2P  Q  L   Downward EA 2EI

 u CBC 

Example 4.8 Consider a statically determinate frame subjected to a concentrated force P at point C as shown below. The Young's modulus, the cross sectional area and the moment of inertia of the cross section are assumed to be constant throughout the structure and denoted by E, A and I, respectively. Determine the displacements and rotations at points A, B, C, and D due to the axial deformation. P B

C

L

A

D L

Solution First, let us define the local coordinate systems for three segments AB, BC, and CD as shown below. x, y B

y

A

C

D

x Copyright © 2011 J. Rungamornrat

x, y


254


Since the structure is statically determinate, the axial force diagram and the bending moment diagram can readily be obtained from static equilibrium. For this particular case, the M/EI vanishes identically for all segments and the F/EA vanishes for the segments AB and BC while F/EA for the segment CD is non-zero and given below. P B

C

A

0 0

D

–P/EA

P

From boundary conditions, the horizontal and vertical displacement at the point A and the vertical AB CD displacement at the point D are fully prevented, i.e. u AB A  v A = 0 and u D  0 . The displacement and rotation at other points can be obtained as follow. AB Segment AB: Only two kinematical quantities are already known, i.e. u AB A  v A = 0 . Some of the remaining unknown quantities can be determined and the rest of the unknowns are related as shown (see also the elastic curve):

C

B

A

D

1st curvature area equation: B/A = BAB  AB A = A M/EI,AB  AB  AB B A = 0

(e4.8.1)

AB 2nd curvature area equation: t B/A = v BAB  (v AB A  A L AB ) = A M/EI,AB x B

 AB  v AB B A L 0

Axial strain area equation: u B/A = u BAB  u AB A = A F/EA,AB  0  0  u AB 0  u AB B B Copyright © 2011 J. Rungamornrat

(e4.8.2)


255


Continuity of displacement and rotation at point B: u BC   v AB B B v



BC B BC B

u 

AB B AB B

(e4.8.3)

0

(e4.8.4)

Segment CD: Only one kinematic quantity is already known, i.e. u CD D  0 . Some of the remaining unknown quantities can be determined and the rest of the unknowns are related as shown:  CCD = A M/EI,CD 1st curvature area equation: D/C = CD D  CD  CD D C = 0 nd

2 curvature area equation: t D/C = v  v

CD D

 (v

 v

CD D CD D

Axial strain area equation: u D/C = u

CD C

 u

CD C

(e4.8.5)

  L) = A M/EI,CD x D CD C

 CD C L 0 CD C

(e4.8.6)

= A F/EA,CD

PL  P     0  u CD C  L   EA  EA  PL  Downward  u CD C EA

Continuity of displacement and rotation at point C: u CBC  vCD C vCBC  u CD C 

(e4.8.7) PL EA

CBC  CD C

(e4.8.8)

BC Segment BC: Two kinematic quantities are already known, i.e. v BC  PL/EA . Some of B  0 and v C the remaining unknown quantities can be determined and the rest of the unknowns are related as shown: BC 2nd curvature area equation: t C/B = vCBC  (v BC B  B L) = A M/EI,BC x C

PL P BC  0  BC CW B L  0  B   EA EA 1st curvature area equation: C/B = CBC  BC B = A M/EI,BC  

 CBC 

Axial strain area equation: u C/B = u CBC

P P = 0  CBC   CW EA EA  u BC B = A F/EA,BC

0  u CBC  u BC B

From BC B   P/EA and the continuity condition (e4.8.4), we then have AB  B

P CW EA

AB From AB B   P/EA , v A = 0 and equations (e4.8.1) and (e4.8.2), we obtain


(e4.8.9)


256


AB  A

P CW EA

 v AB B

,

PL Rightward EA

From v AB  PL/EA , the continuity condition (e4.8.3) and the relation (e4.8.9), we get B  u BC B

PL Rightward EA

u CBC 

,

PL Rightward EA

From u CBC  PL/EA , CBC  P/EA , and the continuity conditions (e4.8.7) and (e4.8.8), we get  vCD C

PL Rightward EA

CD  C

,

P CW EA

Finally, we return to the relations (e4.8.5) and (e4.8.6) and then obtain vCD 0 D

, CD  D

P CW EA

It is worth noting that the displacement and rotation of this particular structure are due only to the axial deformation. The curvature area method developed previously without consideration of the axial deformation predicts no displacement and rotation due to the vertical concentrated load P acting to the point C.

Exercises 1. Use the method of curvature (moment) area to determine the displacement and rotation at points indicated in the figure below. In the calculations, only bending deformation is considered and Young's modulus and moment of inertia of the cross section are assumed to be constant throughout the structures and denoted by E and I, respectively. qL

q B

A

D C

2L P

L

L

E

C

A

D

B L

L

L

L

2P

P C

B A

D L


L

PL


257


B A L P C

D L

A

L PL D

P

L

C

B 2L q B

A

C

C q

L

L A

D L

L

D

B

L

L

D L

q P

C

B

B C L

L A

A L

2L



258


2. Rework problem 1 by taking the axial deformation into account in the calculation of the displacement and rotation. 3. Determine the magnitude of the force P to introduce a gap o (measured in the direction of the applied force) between the two ends of structures shown below. Assume that the Young's modulus, the moment of inertia, and the cross sectional area are constant throughout the structure and denoted by E, I, and A, respectively. Compare results between following two cases: the first case considering only bending effect and the other considering both bending and axial effects. P

P P

P P

P P

2L

P

2L 60o

60o

L

L

L

L

L

L

60o

60o

4. Consider a statically determinate beam subjected to external applied load as shown below. The beam consists of three segments; the segments AB and CD are made of the same material of Young’s modulus E and the same moment of inertia I and the segment BC is assumed to be rigid, i.e. EI → . Determine the displacement and rotation at points A, B, C and D.

A

D B 2L

C L

L

5. Consider the following simply supported beams subjected to end moments as shown below. For each structure, develop the relation between the applied end moments and the end rotations in the matrix form:  1   f11 f12   M1      2  f 21 f 22  M 2 

where {M1, 1} are applied moment and rotation at the left end; {M2, 2} are applied moment and rotation at the right end; and f11, f12, f21, f22 are entries of the flexibility matrix. The bending moment and the rotation are considered to be positive when they direct in a counterclockwise direction. M 1 , 1

M 2 , 2 EI L Copyright © 2011 J. Rungamornrat


259


M 1 , 1

M 2 , 2 EI

EI

L/2

L/2

M 1 , 1

M 2 , 2 EI → 

EI

L/2

L/2

M 1 , 1

M 2 , 2 EI

EI

EI

L/3

L/3

L/3

M 1 , 1

M 2 , 2 EI

EI → 

EI

L/3

L/3

L/3

M 1 , 1

M 2 , 2 EI

EI

k

L/2

L/2

M 1 , 1

M 2 , 2 EI

k

L/3

EI

EI

k

L/3

L/3

k

M 1 , 1

M 2 , 2

EI

EI

L/2

L/2



260


6. Consider a beam of length L and moment of inertia I and made from an elastic material of Young’s modulus E. This beam is placed on a rigid and frictionless floor as shown in the figure below. Beside its uniform weight per length w, the beam is also subjected to end moment M. Determine the length of a contact region between the beam and the floor, the displacement and rotation at a point where the moment is applied, and the maximum moment M that reduces the contact length to zero.

EI, w L


M


261

Conjugate Structure Analogy

CHAPTER 5 CONJUGATE STRUCTURE ANALOGY As be apparent from the previous chapter, the method of curvature (moment) area experiences difficulty when it is applied to rigid frames containing members that are neither aligned with nor perpendicular to the global coordinate axes. This is due to that the curvature area and length constraint equations are expressed in terms of transverse and longitudinal components of the displacement and, as a result, the continuity of the displacement at nodes joining those members cannot be enforced easily but involves the coordinate transformation. Moreover, the curvature area equations derived previously are strictly for a straight segment containing no internal release (e.g., hinges); this restriction limits the choice of segments used in the analysis (i.e., only straight segments containing no interior internal release can be chosen) and, for various cases, it renders the calculation of quantities of interest inefficient. To remove such limitations, the method of curvature area can be generalized to be capable of to treat both a segment consisting of multiple, non-aligned sub-segments and a segment containing the internal moment releases (i.e., hinges). In addition to their enhanced features, an analogy can be established to interpret those equations in a simpler and more familiar fashion. Specifically, the curvature area equations and the length constraint equation can simply be viewed as static equilibrium equations of a fictitious structure, termed a conjugate structure, under a specific set of loadings. This method has been recognized from such analogy as the “method of conjugate structure analogy”. In following sections, we first present the conjugate analogy for a horizontal, straight segment that can be applied to solve beams, then the analogy is generalized for a straight segment of arbitrary orientation with a direct application to rigid frames, and finally, a segment consisting of multiple straight sub-segments and containing hinges are treated. To clearly demonstrate the technique and its capability, various examples involving both beams and rigid frames are included.

5.1 Conjugate Structure Analogy for Horizontal Segment Consider a horizontal, straight segment AB of length LAB and rigidity EI as shown schematically in AB Figure 5.1. Recalling results from the previous chapter, the end rotations { AB A ,  B }, the end AB AB AB longitudinal displacements { u AB A , u B }, and the end transverse displacements { v A , v B } of the segment are related through the first and second curvature area equations and the length constraint equation: AB  AB  A M/EI,AB  0 B A

(5.1)

AB v AB  v AB B A   A L AB  A M/EI,AB x B  0

(5.2)

u AB  u AB B A = 0

(5.3)

where AM/EI,AB denotes the area of curvature diagram (i.e., M/EI diagram) over the segment AB and x B is the distance, measured along the undeformed axis of the segment, from the end B to the centroid of the curvature diagram. To construct the analogy, let us consider the following thought process. Imagine that we have a fictitious segment AB which has identical geometry to that of the real segment AB and is subjected to the following set of fictitious loadings: (i) a force of value AB acting at the end A in A Copyright © 2011 J. Rungamornrat


262


the positive Z-direction and a force of value AB B acting at the end B in the negative Z-direction; (ii) AB a moment of value u A acting at the end A in the positive X-direction and a moment of value u AB B AB acting at the end B in the negative X-direction; (iii) a moment of value v A acting at the end A in acting at the end B in the negative Y-direction; the positive Y-direction and a moment of value v AB B and (iv) a force of value AM/EI,AB acting at the centroid of the curvature diagram in the positive Zdirection. The corresponding free body diagram of this segment is shown in Figure 5.2. Y

B

θ A

AB B

θ AB A

v AB B

v AB A AB A uA

B

X

u AB B

M/EI xA

xB LAB

Figure 5.1: Schematic of deformed and undeformed configurations of horizontal straight segment AB and its curvature diagram

Y

v AB B

v AB A

u AB A

A M/EI,AB

θ AB A P

A

AB B

θ B

u AB B

X

xP xA

xB LAB

Figure 5.2: Schematic of conjugate structure of horizontal straight segment AB shown in Figure 5.1 Copyright © 2011 J. Rungamornrat


263


By considering static equilibrium of the fictitious segment AB shown in Figure 5.2, it requires that ΣFZ = 0



AB  A M/EI,AB  AB A B = 0

(5.4)

ΣM YB = 0



AB AB v AB A   A L AB  A M/EI,AB x B  v B  0

(5.5)

ΣM XB = 0



u AB  u BAB = 0 A

(5.6)

It is important to emphasize that the sign convention of all quantities appearing in (5.4)-(5.6) follow exactly that employed in the method of curvature area. For instance, the end longitudinal displacement, the end transverse displacement and the end rotations are positive if they direct in the positive coordinate directions. By comparing a set of two curvature area equations and a length constraint equation (5.1)(5.3) of the real segment AB and a set of three equilibrium equations (5.4)-(5.6) of the fictitious segment AB, we can deduce the following three correspondences:  The first curvature area equation of the real segment is identical to the equilibrium equation of forces in the Z-direction of the fictitious segment.  The second curvature area equation of the real segment is identical to the equilibrium equation of moments in the Y-direction of the fictitious segment.  The length constraint equation of the real segment is identical to the equilibrium equation of moments in the X-direction of the fictitious segment.

These three correspondences form the well-known conjugate structure analogy and the fictitious segment subjected to a set of fictitious loadings is known as the conjugate structure. This conjugate structure analogy provides a useful, convenient and systematic means to set up three equations relating the longitudinal and transverse displacements and the rotation at both ends of the real segment to the curvature diagram without geometric consideration (no need to sketch the elastic or deformed curve of the structure). The key step of the analogy is to establish the correct conjugate structure and the remaining task only involves setting static equilibrium equations of that conjugate structure. It is worth noting that a reference point used to form the moment equilibrium equation in the Y-direction is not restricted to the end B. Other points (e.g. end A and any interior point of the segment) can be chosen as the reference point and the resulting moment equilibrium equation is in fact a linear combination of (5.4) and (5.5) or, equivalently, a linear combination of (5.1) and (5.2). To demonstrate this argument, let us form the moment equilibrium equation in the Y-direction about an arbitrary point P (see Figure 5.2). The resulting moment equilibrium equation, upon certain manipulations, is given by ΣM YP = 0



AB AB AB v AB A   A x P  A M/EI,AB (L AB  x B  x P )  v B   B (L AB  x P )  0



AB AB AB v AB  A M/EI,AB  BAB  0 A   A L AB  A M/EI,AB x B  v B  (x P  L AB ) A





(5.7)

This is obviously the linear combination of (5.4) and (5.5). As a consequence, the moment equilibrium equation (5.5) can be replaced, without loss, by equation (5.7). The flexibility to choose an arbitrary reference point to form the moment equilibrium equation is useful in various situations; in general, a reference point is commonly chosen in such a way that it reduces the number of unknowns as many as possible. Another important issue is the treatment of terms associated with the curvature diagram. As evident from (5.1) and (5.2), the curvature at any point within the segment appears in terms of its Copyright © 2011 J. Rungamornrat


264


area and its moment of area about the end B. To facilitate the calculation of such terms, it is common to decompose the curvature diagram into several sub-curvature diagrams whose area and centroid can readily be determined and the total area and its moment of area about a particular point can then be obtained from superposition: A M/EI,AB =

A

(5.8)

i M/EI,AB

i

A M/EI,AB x B   (A iM/EI,AB x Bi )

(5.9)

i

where A iM/EI,AB is the area of the ith sub-curvature diagram and x Bi is the corresponding distance from the end B to the centroid of the ith sub-curvature diagram. In the conjugate structure, A iM/EI,AB is treated as a force acting at the centroid of the ith sub-curvature diagram in the positive Z-direction and, thus, terms on the right hand side of (5.8) and (5.9) represent the sum of those forces in the positive Z-direction and the sum of moments produced by those forces in the Y-direction about the end B. As a consequence, the decomposition of the curvature diagram does not change the analogy established above except a slight modification of the free body diagram of the conjugate structure to incorporate equivalent series of forces A iM/EI,AB instead of a single force AM/EI,AB (see Figure 5.3). Y

v AB B

v AB A

u

AB A

θ AB A

1 2 A3M/EI,AB A M/EI,AB A M/EI,AB AB B

θ B

A

u AB B

X

x B1

x B2 x B3

LAB Figure 5.3: Schematic of conjugate structure according to decomposition of total curvature diagram into three sub-curvature diagrams It is important to emphasize that the conjugate structure analogy established here alter neither the meaning nor nature of the original curvature area equations and the length constraint equation. It only offers a simpler interpretation in terms of equilibrium equations of a conjugate system and we believe that forming equilibrium equations requires less memorization and intuition than attempting to write the curvature area equations directly. This argument will be more apparent when we deal with more complex segments. The conjugate structure analogy developed above for a horizontal straight segment has a direct application to deflection and rotation analysis of beams. For each segment chosen (e.g. a Copyright © 2011 J. Rungamornrat


265


AB AB AB AB AB segment AB), it involves six kinematical quantities { AB A ,  B , u A , u B , v A , v B } while only three equilibrium equations (i.e., ΣFZ = 0 , ΣM YB = 0 , ΣM XB = 0 ) can be set up on the corresponding conjugate segment. However, for statically stable beams, there must exist a point within the beam that the horizontal movement is prevented (e.g. point A in beams shown in Figure 5.4); therefore, by using the length constraint equation along with the continuity of the longitudinal displacement at the node between two segments, it can readily be shown that the longitudinal displacement vanishes at every point within the beam. As a result, a set of kinematical quantities for the segment AB AB AB AB simply reduces to { AB A ,  B , v A , v B } and, at the same time, only two equilibrium equations (i.e., ΣFZ = 0 , ΣM YB = 0 ) need be set up on the corresponding conjugate segment. Similar to the method of curvature area, a particular segment (say a segment AB) that contains only two kinematical AB AB AB unknowns from a set { AB A ,  B , v A , v B } must be chosen first in the analysis since these unknowns can be solved directly from the two equilibrium equations of its conjugate segment. Once the transverse displacement and the rotation at certain points are calculated, subsequent segments connecting to one of those points can be chosen and the same procedure is applied until the quantities of interest are obtained.

A

A

A

Figure 5.4: Schematic of statically stable beams and point A where horizontal movement is fully prevented To demonstrate how the conjugate structure analogy works and, more importantly, to adjust ourselves to be familiar with this new technique and ready for further generalization, let us consider following two examples. Example 5.1 Consider a cantilever beam of length 2L and subjected to a concentrated force 2P at the mid span and a concentrated moment PL at the tip as shown below. Determine the tip deflection and the tip rotation of the beam using the conjugate structure analogy. The flexural rigidity of segments AB and BC are 2EI and EI, respectively.

2P A

2EI L

EI B

L

PL C

Solution Since the beam is statically determinate, sub-bending moment diagrams can readily be constructed using superposition technique shown in section 4.4.1 of chapter 4. The corresponding sub-curvature diagram can simply be obtained by dividing the sub-bending moment diagrams by flexural rigidity of each segment. Obtained results are given below. Copyright © 2011 J. Rungamornrat


266


2P PL C

A B

L

L PL

PL

Sub-BMD –2PL PL/EI PL/2EI A2

A3

Sub-curvature diagram

A1 –PL/EI

L/2

L/2

L/3

Since the beam is fully fixed at a point A, the deflection and rotation at that point vanish (i.e., A  v A  0 ). The tip deflection and the tip rotation can then be obtained by considering the segment AC (there are only two unknowns on this segment, i.e., CAC and vCAC ). A conjugate structure of the segment AC is given below Y

v CAC

v AC A 0 A1 A 2

θ AC A 0 u AC A 0

θ CAC

A3

AC C uC  0

A L/3

L/6

L

X

L/2

where areas of the sub-curvature diagrams A1, A2, A3 are given by A1 =

1  PL  PL2 PL2 PL2  PL   PL  L A = L A = L      ; ; 2 3       2  EI  2EI 2EI EI  2EI   EI 

and these three forces are applied to the conjugate segment at X = L/3, L/2 and 3L/2, respectively. From equilibrium of this conjugate structure, we then obtain ΣFZ = 0 :

AC  A1  A 2  A 3  CAC = 0 A Copyright © 2011 J. Rungamornrat


0 



267


PL2 PL2 PL2    CAC = 0 2EI 2EI EI



CAC 

PL2 CCW EI

AC AC v AC A   A (2L)  A1 (5L/3)  A 2 (3L/2)  A 3 (L/2)  v C  0

ΣM YC = 0 :



 PL2   5L   PL2   3L   PL2 0  0(2L)         2EI   3   2EI   2   EI



vCAC 

5PL3 12EI

  L  AC     vC  0  2 

Upward

Example 5.2 Consider a prismatic beam of constant flexural rigidity EI and subjected to loads as shown below. Determine the deflection and rotation at point C.

qL

2q A

B

EI

EI

2L

C

L

Solution All support reactions, sub-bending moment diagrams and sub-curvature diagrams for the given structure are shown below.

qL

2q A

B

EI 3qL/2

EI

C

7qL/2 2L

L 3qL2

Sub-BMD

–qL2 –4qL2 3qL2/EI

A1

A3 A2

2

–qL /EI


–4qL2/EI L/2

L/3

2L/3

From the boundary conditions at points A and B, the deflections at those two points vanish (i.e., v A  v B  0 ). Thus, a segment AB will be considered first since it is the only segment that contains Copyright © 2011 J. Rungamornrat


268


two kinematical unknowns at its ends (i.e., AB and AB A B ). A conjugate structure of this segment is shown below Y

v AB B 0

v AB A 0 θ AB A u AC A 0

θ AB B

A2

A1

X

AB B uB  0

A L/2

L/6

4L/3

where the areas of the sub-curvature diagrams A1 and A2 are given by A1 =

1  3qL2  2  EI

 3qL3 1  4qL2 ; (2L)  A =   2 EI 3  EI 

 8qL3 (2L)    3EI 

and these two forces are applied to the conjugate segment AB at X = 4L/3 and 3L/2, respectively. From equilibrium of this conjugate structure, we then obtain AB A (2L)  A1 (2L/3)  A 2 (L/2)  0

ΣM YB = 0 :



 3qL3   2L   8qL3   L  AB (2L)       0 A    EI   3   3EI   2 



AB A  

qL3 3EI

CW

AB  A1  A 2  BAB = 0 A

ΣFZ = 0 :





qL3 3qL3 8qL3    AB B = 0 3EI EI 3EI



AB B  0

Now, the deflection and rotation at points A and B are completely known. To compute the displacement and rotation at point C, we consider next a segment BC. The corresponding conjugate segment is shown below. Y

v CBC

v BC B θ BC B u BC B 0

θ CBC

A3

C u CBC  0

B L/3

2L/3


X


269


where the area of the sub-curvature diagram A3 is given by A3 =

1  qL2  2  EI

 qL3 (L)    2EI 

and the end displacement v BC and the end rotation BC are known from the continuity of B B BC AB BC displacement and rotation at point B, i.e., v B  v B  0 and B  AB B  0 . Enforcing equilibrium of this conjugate structure leads to BC BC B + A 3  C = 0

ΣFZ = 0 :

0 



qL3  CBC = 0 2EI



CBC  

qL3 CW 2EI

BC BC v BC B   B (L)  A 3 (2L/3)  v C  0

ΣM YC = 0 :

 qL3   2L  BC 0  0(L)       vC  0  2EI   3 



vCBC  



qL4 Downward 3EI

It is remarked that the segment AC can also be chosen to calculate the displacement and rotation at point C; however, the segment BC was selected in this calculation since it involves less computation. Example 5.3 Consider a statically determinate beam of length 3L and subjected to external applied load as shown below. The flexural rigidity of segments AB and BC are denoted by 2EI and EI, respectively. Determine the deflection at point B, the relative hinge rotation, and the rotation at point C.

q A

3EI

qL2 B

2L

EI L

C

Solution All support reactions, sub-bending moment diagrams and sub-curvature diagrams for the given structure are shown below. From boundary conditions at points A and C, we obtain v A  v C  A  0 . Therefore, a segment AB will be considered first since it is the only segment and AB that contains two kinematical unknowns at its ends (i.e., v AB B B ). It is important to emphasize that the segment AC is not allowed to be used in the analysis since the analogy developed above still applies only to a straight segment containing no interior hinge; this restriction will be later removed. A conjugate structure of the segment AB is shown below where areas of the subcurvature diagrams A1 and A2 are given by A1 =

1  4qL2  8qL3 1  2qL2  2qL3 ; A2 =    (2L)    (2L)  3  3EI  9EI 2  3EI  3EI

and these two forces are applied to the conjugate segment AB at X = L/2 and 2L/3, respectively Copyright © 2011 J. Rungamornrat


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2q

qL2

2

2qL

A 3qL

3EI

EI

B

C qL

L

2L 2qL2

Sub-BMD –qL

2

–4qL2 2qL2/3EI

A2 A3

A1


–qL2/EI

–4qL2/3EI

L/3

4L/3

L/2 L/6

Y

v AB B

v AB A 0 θ AB A 0 u AB A 0

θ AB B

A1 A 2

AB B uB  0

A L/2 L/6

X

4L/3

From equilibrium of the conjugate structure, we then obtain AB  A1  A 2  BAB = 0 A

ΣFZ = 0 :

8qL3 2qL3   AB B = 0 9EI 3EI



0 



AB B 

2qL3 CW 9EI

AB AB v AB A   A (2L)  A1 (3L/2)  A 2 (4L/3)  v B  0

ΣM YB = 0 : 



 8qL3   3L   2qL3   4L  AB 0  0(2L)         vB  0  9EI   2   3EI   3  4qL4 Downward v AB B  9EI Copyright © 2011 J. Rungamornrat


271


AB 4 From continuity of the displacement at point B, we can deduce that v BC B  v B  4qL / 9EI . is in general not equal to AB due to the presence of a hinge at point B. However, the rotation BC B B To compute the rotation at point just to the right of a hinge and the rotation at point C, let us consider a segment BC. This segment contains only two unknowns (i.e., BC and CBC ) and its B conjugate structure is shown below

Y

v BC B  

4qL4 9EI

v CBC  0

θ BC B u BC B 0

θ CBC

A3

X

C u CBC  0

B 2L/3

L/3

where the area of the sub-curvature diagram A3 is given by A3 =

1  qL2  2  EI

 qL3 (L)    2EI 

and this force is applied to the conjugate segment at X = 2L/3. Enforcing equilibrium of this conjugate structure yields BC BC v BC B   B (L)  A 3 (L/3)  v C  0

ΣM YC = 0 :





 qL3   L  4qL4 (L)  BC      0  0 B 9EI  2EI   3 



BC B 

qL3 9EI

CCW

11qL3 18EI

CCW

BC BC B + A 3  C = 0

ΣFZ = 0 :



11qL3 qL3   CBC = 0 18EI 2EI



CBC 

The relative hinge rotation at point B, denoted by Δθ B , now becomes AB Δθ B  BC B  B 

11qL3  2qL3  5qL3 CCW   18EI  9EI  6EI

5.2 Conjugate Structure Analogy for Horizontal Segment with Hinges In this section, we generalize the conjugate structure analogy to be capable of to treat a horizontal straight segment containing interior hinges. This generalization will enhance the flexibility of choosing the segments in the analysis and, for various situations, it significantly reduces the Copyright © 2011 J. Rungamornrat


272


computational effort. For instance, it will allow the segment AC in Example 5.3 be chosen in the analysis. Consider a horizontal, straight segment AB containing an interior hinge at point C as shown schematically in Figure 5.5. To construct an analogy for the segment AB, we first divide it into a sub-segment AC and a sub-segment CB. Since both sub-segments contain a hinge only at their end, analogy established in section 5.1 applies and the conjugate structures of these two sub-segments are shown in Figure 5.6. B Y

θ

θ C

C

CB B

θ CAC θ CB C

θ AC A

A

v

v AC A

AC C

v

v CB B

CB C

C

A u AC A

u CAC  u CB C

B

u CB B

X

M/EI xC

xB

LAC

LCB

Figure 5.5: Schematic of deformed and undeformed configurations of horizontal straight segment AB containing hinge at point C and its curvature diagram By applying the conjugate structure analogy to the segment AC, we then obtain following three relations among the end displacements and end rotations ΣFZ = 0



AC  A M/EI,AC  CAC = 0 A

(5.10)

ΣM YB = 0



AC AC AC v AC A   A L AB  A M/EI,AC (x C  L CB )  C L CB  v C  0

(5.11)

ΣM XB = 0



u AC  u CAC = 0 A

(5.12)

Note that the reference point used for taking moment is chosen to be point B (which is out of the segment AC) for convenience in further development. Similarly, the conjugate structure analogy of the segment CB yields ΣFZ = 0



CB  A M/EI,CB  CB C B = 0

(5.13)

ΣM YB = 0



CB CB vCB C  C L CB  A M/EI,CB x B  v B  0

(5.14)

ΣM XB = 0



u CB  u CB C B = 0

(5.15)

By invoking the continuity of the deflection at point C and denoting the relative hinge rotation at this point by C , we obtain the following three relations: Copyright © 2011 J. Rungamornrat


273


v AC A

v CAC A M/EI,AC

θ AC A u

AC A

u CAC

AC C

θ C

A xC LAC

v CAC

u CAC

v CB C

θ CAC θ C θ C

CB

Y

u CB C

v CB C X θ CB C θ

CB C

v CB B A M/EI,CB CB B

θ B

C

u CB B

xB LCB Figure 5.6: Schematic of conjugate structures of segment AC, segment CB and isolated point C AC CB C  C = C

(5.16)

AC v CB C  vC

(5.17)

AC u CB C  uC

(5.18)

To construct an analogy for the relations (5.16)-(5.18), let us imagine that a point C (which is a point connecting a conjugate segment AC and a conjugate segment CB) is subjected to a concentrated force of value C in the positive Z-direction. A free body diagram of this point is shown in Figure 5.6. It is apparent that the relations (5.16), (5.17) and (5.18) are identical to equilibrium equations ΣFZ = 0 , ΣM YC = 0 , ΣM XC = 0 of the isolated point C, respectively. The conjugate segment AC and the conjugate segment CB can then be combined into a single conjugate segment AB at point C via the use of Newton’s third law. The resulting conjugate segment AB is shown in Figure 5.7. Without loss, we replace the superscript C of quantities { v AC A , AC AC CB CB CB u A , A } by the superscript B and quantities { v B , u B , B } by the superscript A to be consistent with the name of the segment. By combining equations (5.10), (5.13) and (5.16), it yields an equation identical to the equilibrium equation of forces in the Z-direction of the conjugate segment AB, i.e., ΣFZ = 0



AB  A M/EI,AC  A M/EI,CB  C  AB A B = 0 Copyright © 2011 J. Rungamornrat

(5.19)


274


Y v AB A θ AB A u AB A

v AB B A M/EI,AC

θ C

A

A M/EI,CB θ AB B B

C xC LAC

u AB B

X

xB LCB

Figure 5.7: Schematic of conjugate structure of segment AB Similarly, by combining equations (5.11), (5.14) and (5.17), it yields an equation identical to the equilibrium equation of moments in the Y-direction about point B of the conjugate segment AB, i.e. ΣM YB = 0



AB AB v AB A   A L AB  A M/EI,AC (x C  L CB )  A M/EI,CB x B  C L CB  v B  0

(5.20)

Finally, by combining equations (5.12), (5.15) and (5.18), it yields an equation identical to the equilibrium equation of moments in the X-direction about point B of the conjugate segment AB, i.e. ΣM XB = 0



u AB  u AB A B = 0

(5.21)

Note that the explicit equations appearing in (5.19) and (5.20) can be considered as the first and second curvature area equations for a horizontal, straight segment containing a hinge at point C whereas the explicit equation appearing in (5.21) represents the length constraint equation. The established correspondence indicates that an analogy between a set of curvature area equations and length constraint equation and a set of equilibrium equations set up on the conjugate structure still exists for a straight segment containing a hinge. The conjugate structure for this particular case is constructed exactly in the same way as that for a segment containing no hinge except at the hinge location where the unknown force equal to the relative hinge rotation is applied in the positive Zdirection. It is worth noting that a reference point chosen to form the moment equilibrium equations (5.20) and (5.21) is arbitrary; other points besides the point B can also be chosen to form a set of equations equivalent to (5.19)-(5.21). Although the analogy is established only for a segment containing only one hinge, it also applies to a segment containing multiple hinges. The only slight modification that must be made is to include unknown forces equal to relative hinge rotations at all locations in the conjugate structure that coincide with the hinge locations. Finally, we remark that the correspondence or conjugate structure analogy established here offers two attractive features: one associated with the enhancement of flexibility to choose a segment containing hinges in the analysis and the other corresponding to the simplicity of forming the first and second curvature area equations (e.g. equations (5.19) and (5.20)), which can possibly be very complex for a segment containing multiple hinges, by simply writing down equilibrium equations of the conjugate segment. Example 5.4 Repeat the problem in Example 5.3 Solution In Example 5.3, the segments AB and BC were considered separately due to the limitation of the analogy established in the section 5.1. Let us investigate this problem again by using the enhanced feature of the analogy established in the section 5.2. Copyright © 2011 J. Rungamornrat


275


Consider the entire beam as a single segment that contains a hinge at point B. A conjugate structure of this segment is shown below. Clearly, this segment contains only two unknowns, i.e., CAC and B , and they can be solved from two equilibrium equations of the conjugate structure as follow: Y

v CAC  0

v AC A 0 θ AC A 0 u AC A 0

θ B

A1 A 2 A

C u CAC  0

B 4L/3

L/2 L/6

θ CAC

A3

2L/3

X

L/3

AC AC v AC A   A (3L)  A1 (5L/2)  A 2 (7L/3)  A 3 (L/3)   B (L)  v C  0

ΣM YC = 0 :



 8qL3   5L   2qL3   7L   qL3   L  0  0(3L)          B (L)  0  0      9EI   2   3EI   3   2EI   3 



B 

5qL3 CCW 6EI

AC  A1  A 2  A 3  B  CAC = 0 A

ΣFZ = 0 :

0

8qL3 2qL3 qL3 5qL3     CAC  0 9EI 3EI 2EI 6EI



CAC 

qL3 CCW 9EI

The deflection at point B can further be computed from either a conjugate segment AB or a conjugate segment BC. It is obvious that if the relative hinge rotation at point B is of interest, consideration of a single segment AC directly yields this unknown quantity. Example 5.5 Consider a continuous beam subjected to external loads as shown below. Determine the relative hinge rotation at point C.

qL A

q

2EI L/2

L/2

B

2EI C L/3

EI 2L/3

D

Solution Since a given structure is statically determinate, all support reactions, sub-bending moment diagrams and sub-curvature diagrams can readily be obtained from static equilibrium with the results shown below. Copyright © 2011 J. Rungamornrat


276


2qL

3q B

A 2qL/3

D

EI

2EI C

2EI

qL

7qL/3

L/2

L/3

L/2

2L/3

2qL2/3 qL2/6

Sub-BMD 2

–qL /3 –qL2 qL2/3EI

A1

qL2/6EI

A3

A4

A2 –qL2/6EI


–qL2/2EI L/2

L/6 L/6 L/6 L/9 2L/9

L/3

L/3

From boundary conditions at points A, B and C, we obtain v A  v B  v D  0 . Therefore, a segment AB AB will be considered first in the analysis since it contains only two unknowns (i.e., AB A and  B ). A conjugate structure of this segment is shown below. Areas of the sub-curvature diagrams A1 and A2 are given by A1 =

1  qL2  qL3 1  qL2   L  qL3 ; A2 =     (L)      2  3EI  6EI 2  2EI   2  8EI

where these two forces are applied to the conjugate structure at X = 2L/3 and 5L/6, respectively. Y

v AB B 0

v AB A 0 θ AB A u AB A 0

θ AB B

A1 A 2

AB B uB  0

A 2L/3

L/6 L/6


X


277


Enforcing equilibrium of this conjugate structure yields AB AB v AB A   A (L)  A1 (L/3)  A 2 (L/6)  v B  0

ΣM YB = 0 :

 qL3   L   qL3   L  0  AB         0  0 A (L)    6EI   3   8EI   6 



 AB A  

5qL3 CW 144EI

AB AB A + A1 + A 2   B = 0

ΣFZ = 0 :





5qL3 qL3 qL3 +   AB B = 0 144EI 6EI 8EI



AB B 

qL3 CCW 144EI

AB 3 With use of continuity of the rotation at point B (i.e., BD B   B  qL / 144EI ), the relative hinge rotation at point C can now be computed from the segment BD since it contains only two unknowns (i.e., C and BD D ). The conjugate structure of the segment BD is shown below and areas of the sub-curvature diagrams A3 and A4 are given by

A3 =

1  qL2   L  qL3 2  qL2   2L  2qL3 ; A4 =         2  6EI   3  36EI 3  6EI   3  27EI

where these two forces are applied to the conjugate structure at X = L/9 and 2L/3, respectively. Y

v BD D 0

v BD B 0 θ BD B u BD B 0

A3

B

θ C

θ BD D

A4

L/9 2L/9

X

BD D uD  0

C L/3

L/3

Enforcing equilibrium of this conjugate structure yields BD BD v BD B   B (L)  A 3 (8L/9)  A 4 (L/3)  C (2L/3)  v D  0

ΣM YD = 0 :



 qL3   qL3   8L   2qL3   L  0  (L)            C (2L/3)  0  0  144EI   36EI   9   27EI   3 



C  

qL3 CW 96EI

BD BD B + A 3 + A 4  C   D = 0

ΣFZ = 0 :



qL3 qL3 2qL3 qL3     BD D = 0 144EI 36EI 27EI 96EI Copyright © 2011 J. Rungamornrat



BD D 

37qL3 CCW 864EI


278


5.3 Conjugate Structure Analogy for Inclined Segment The conjugate structure analogy developed in above two sections has a direct application to analysis of deflection and rotation of beams. The end displacements appearing in the conjugate structure (or, equivalently, in the curvature area equations and the length constraint equation) are in terms of the longitudinal and transverse components or, equivalently, based on a local coordinate system of a member. This therefore poses some difficulty when the analogy is applied to analyze rigid frames that generally contain members with various orientations. More precisely, the difficulty is associated with the enforcement of continuity of the displacement at the joint connecting two segments of different orientation; such continuity relation may not be easily obtained via visual inspection but required a law of coordinate transformation. In this section, we generalize the conjugate structure analogy well-suited for treatment of a straight segment of arbitrary orientation. The main task is to add another feature of the analogy by changing a local reference coordinate system to a global reference coordinate system. Once all displacement components are expressed based on a single global coordinate system, the continuity at the inter-boundary of members can easily be handled. Let us consider a straight segment AB of length LAB and oriented with an angle  to the global X-axis as shown schematically in Figure 5.8 along with its curvature diagram (M/EI diagram). The end displacements and rotations referring to the member local coordinate system and AB AB AB AB AB AB AB AB the global coordinate system are denoted by { u AB A , v A ,  A , u B , v B ,  B } and { U A , VA ,  A , AB AB U AB B , VB ,  B }, respectively. B AB  AB B  θB

v AB B

y

VBAB B

U AB A A VAAB A

v

AB A



AB A

θ

U

AB A

AB B

x

u AB B Y



u AB A X

I M/E

xB xA

LAB

Figure 5.8: Schematic of deformed and undeformed configurations of inclined straight segment AB and its curvature diagram Copyright © 2011 J. Rungamornrat


279


Based on a local coordinate system, the conjugate structure analogy established in section 5.1 still applies, i.e., ΣFz = 0



AB  AB  A M/EI,AB  0 B A

(1st curvature area equation)

(5.22)

ΣM yB = 0



AB v AB  v AB B A   A L AB  A M/EI,AB x B  0

(2nd curvature area equation)

(5.23)

ΣM xB = 0



u AB  u AB B A = 0

(Length constraint equation)

(5.24)

AB AB From a law of coordinate transformation, the end displacements and end rotations { u AB A , v A , A , AB AB AB AB AB AB AB AB u AB B , v B ,  B } and { U A , VA ,  A , U B , VB ,  B } are related by AB  AB A  A

(5.25a)

AB AB U AB A  u A cos   v A sin 

(5.25b)

AB VAAB  v AB A cos   u A sin 

(5.25c)

AB  AB B  B

(5.26a)

AB AB U AB B  u B cos   v B sin 

(5.26b)

AB VBAB  v AB B cos   u B sin 

(5.26c)

Note that the end rotations based on the local and global coordinate systems are identical since both the Z-axis and z-axis are coincident. By applying the coordinate transformation (5.25)-(5.26) to the first and second curvature area equations and the length constraint equation, it leads to an equivalent set of three equations  AB  AB  A M/EI,AB  0 B A

(5.27)

AB U AB  U AB B A   A L AB sin   A M/EI,AB x B sin   0

(5.28)

VBAB  VAAB  AAB L AB cos   A M/EI,AB x B cos   0

(5.29)

More precisely, equation (5.27) is simply obtained by applying (5.25a) and (5.26a) into the first curvature area equation (5.22); equation (5.28) is obtained by first calculating the linear combination (5.24)cos(5.23)sinand then applying (5.25b) and (5.26b); and equation (5.29) is obtained by first calculating the linear combination (5.24)sin(5.23)cosand then applying (5.25c) and (5.26c). Equations (5.27)-(5.29) can be termed the first, second and third curvature area equations for inextensible segment in global coordinate system, respectively. As is evident from (5.27)-(5.29), the curvature area equations consisting of several terms and their form is more complex than that appearing in (5.1)-(5.3). To avoid memorization of (5.27)-(5.29), a conjugate structure analogy can also be established in a similar fashion to the case of a horizontal straight segment. Based on a local coordinate system, a conjugate segment for the segment AB can be obtained using the analogy developed in section 5.1 and such conjugate segment is shown in Figure 5.9. It is worth noting that all forces applied to this conjugate structure are in the Z-direction while moments are either in the xdirection or the y-direction. To establish an equivalent conjugate structure based on the global coordinate system, all local components of forces and moments applied to the conjugate structure shown in Figure 5.9 are transformed to global components using the same law of transformation Copyright © 2011 J. Rungamornrat


280


(5.25)-(5.26) and such global conjugate structure is shown in Figure 5.10. In this global conjugate represents a force acting at the end A in the positive Z-direction,  AB represents a structure,  AB A B force acting at the end B in the negative Z-direction, A M/EI,AB represents a force acting at the centroid of M/EI diagram in the positive Z-direction, U AB A represents a moment acting at the end A AB in the positive X-direction, VA represents a moment acting at the end A in the positive Y-direction, U AB represents a moment acting at the end B in the negative X-direction, and VBAB represents a B moment acting at the end B in the negative Y-direction. It is important to emphasize that forces and moments acting at points A and B possess opposite directions. Clearly, forces and moments are applied, in the positive coordinate directions, to the end of the segment located at the origin of the local coordinate system and this end is termed the start point of the segment; in the contrary, forces and moments are applied, in the negative coordinate directions, to the other end of the segment and, later on, it is termed the end point of the segment. v AB B

x

y

u AB B AB

v

B θB

AB A

A M/EI,AB 

A u

AB A

Y

xB

θ AB A xA

X

LAB

Figure 5.9: Schematic of conjugate structure for inclined straight segment AB shown in Figure 5.8 based on local coordinate system VBAB

B  AB B

VAAB

U

AB A

A M/EI,AB 

A

U AB B

xB

 AB A

xA

LAB

Y

X

Figure 5.10: Schematic of conjugate structure for inclined straight segment AB shown in Figure 5.8 based on global coordinate system Copyright © 2011 J. Rungamornrat


281


Now, it is readily to construct the correspondences between a set of three curvature area equations (5.27)-(5.29) and a set of equilibrium equations of the global conjugate structure shown in Figure 5.10. From the direct correspondences (5.22)-(5.24) established in the local coordinate system and the fact that the law of transformation (5.25) and (5.26) applies equally to the quantities AB AB AB AB AB AB AB AB AB AB AB { u AB A , v A ,  A , u B , v B ,  B } and quantities { U A , VA ,  A , U B , VB ,  B } when they are viewed as either the end displacements and rotations or a set of fictitious forces and moments, it can be deduced that equations (5.27), (5.28) and (5.29) are in fact equilibrium of forces in the Zdirection, equilibrium of moments in the X-direction, and equilibrium of moments in the Ydirection, i.e., ΣFZ = 0



 AB  AB  A M/EI,AB  0 B A

(5.30)

ΣM XB = 0



AB U AB  U AB B A   A L AB sin   A M/EI,AB x B sin   0

(5.31)

ΣM YB = 0



VBAB  VAAB  AB A L AB cos   A M/EI,AB x B cos   0

(5.32)

These correspondences can also be constructed directly by forming equilibrium equations ΣFZ = 0 , ΣM XB = 0 and ΣM YB = 0 of the conjugated structure shown in Figure 5.10. This obtained analogy offers two advantageous features: (i) curvature area equations (5.27)-(5.29) can be formed in terms of equilibrium equations of the conjugate structure and (ii) components of the end displacements and rotations are based on the global coordinate system. The former feature helps to save the memorization while the latter feature allows the continuity of the displacement and rotation at the member inter-boundary be handled in a simple fashion. While the above analogy is established based on the global coordinate system, need of a local coordinate system for each segment is inherent in both the construction of the curvature diagram (or, more directly, the bending moment diagram) and defining the start and end points of the segment. It is clear that the local coordinate system for any segment can be uniquely defined once the start point and end point of that segment are already chosen. To prevent any confusion, a term segment AB is now used to mean a segment containing the start point and end point at A and B, respectively. The local coordinate system can then automatically be generated as follow: the origin is located at the start point A; the positive x-direction is chosen to point from A to B; the positive z-direction is chosen to point outward; and the positive y-direction follows automatically from the right hand rule. Example 5.6 Use conjugate structure analogy established above to determine the displacement and rotation at the free end C of a rigid frame under applied loads shown below. The flexural rigidity EI is assumed to be constant throughout the structure.

P 2P L

B

EI

EI A L

L


C


282


Solution Since a given structure is statically determinate, all support reactions, sub-bending moment diagrams and sub-curvature diagrams for the segment AB and segment BC can readily be obtained from static equilibrium and method of superposition with results shown below. It is emphasized here again that the start point and end point of the two segments are already chosen and clear from their name and, as a result, the corresponding local coordinate system for each segment is also automatically defined.

P 2P

B

C

EI

–PL/EI

–PL

EI A

2P 4PL

–PL

–PL/EI

P –3PL/EI

–3PL

Since the displacement and rotation are fully prescribed at the fixed support, the segment AB is AB AB considered first in the analysis. This segment contains only three unknowns { U AB B , VB ,  B } and its corresponding conjugate structure is shown below. Y

VBAB  AB B

B VAAB  0

U

AB A

0

 AB A 0

U AB B L/2

A2 A1

L/6 L/3

A L/3

X L/6

L/2

Areas of the sub-curvature diagrams A1 and A2 are given by A1 =

1  3PL   2  EI 





2L  

3 2PL2  PL  ; A2 =    2EI  EI 





2L  

2PL2 EI

where these two forces are applied to the conjugate structure as shown above. By considering equilibrium of the conjugate structure AB, it leads to ΣFZ = 0 :

AB  AB A + A1 + A 2   B = 0



0



283


3 2PL2 2PL2   AB B 0 2EI EI

 AB B  



5 2PL2 2EI

CW

AB AB U AB A   A (L)  A1 (2L/3)  A 2 (L/2)  U B  0

ΣM XB = 0 : 

 3 2PL2   2L   2PL2   L  AB 0   0  L             U B  0   2EI 3 EI 2      



U AB B 

3 2PL3 2EI

Rightward

AB VAAB  AB A (L)  A1 (2L/3)  A 2 (L/2)  VB  0

ΣM YB = 0 :



 3 2PL2   2L   2PL2   L  AB 0   0  L              VB  0  2EI 3 EI 2      



VBAB  

3 2PL3 2EI

Downward

By enforcing continuity of the displacement and rotation at point B, we simply obtain the AB 3 displacement and rotation at the start point of the segment BC, i.e., U BC B  U B  3 2PL /2EI , AB 2 VBBC  VBAB  3 2PL3 /2EI and  BC B   B  5 2PL /2EI . Now, the segment BC contains only three unknowns and can therefore be used to compute the displacement and rotation at point C. The conjugate structure of the segment BC is shown below and area of the sub-curvature diagram A3 is given by A3 =

1  PL  PL2 L      2  EI  2EI

VBBC

U

 BC B

BC B

VCBC CBC

A3

L/3

Y

U CBC

X

2L/3

By considering equilibrium of the conjugate structure BC, it leads to  BC  A 3  CBC = 0 B

ΣFZ = 0 :



5 2PL2 PL2   CBC  0 2EI 2EI



CBC  

U CBC 

3 2PL3 2EI

(5 2 +1)PL2 2EI

BC U BC B  UC  0

ΣM XC = 0 :

 ΣM YC = 0 :



3 2PL3  U CBC  0 2EI



BC VBBC  BC B (L)  A 3 (2L/3)  VC  0


Rightward

CW


284


 PL2   2L  3 2PL3  5 2PL2      VCBC  0   L       2EI 2EI   2EI   3  







VCBC  

(12 2 +1)PL3 3EI

Downward

Example 5.7 Use conjugate structure analogy to determine the displacement and rotation at the roller support and point C of a rigid frame due to applied loads shown below. The flexural rigidity of columns and beam are given by 2EI and EI, respectively.

P B

EI

2EI

2EI

L

3P

D

C

E

A L

L

Solution Since a given frame is statically determinate, all support reactions, sub-bending moment diagrams and sub-curvature diagrams for the segment AB, the segment BD and the segment DE can be obtained from static equilibrium and method of superposition with results shown below.

P B

EI

2EI

2EI A

E

3P P

2P

4PL

3PL

4PL/EI

3PL/2EI B

B –PL

A

3P

D

C

C

D

C

D

–PL/EI

E

A


E


285


From boundary conditions at pinned and roller supports, we obtain U A  VA  VE  0 . Due to the limitation that a straight segment must be chosen in the analysis, there is no straight segment within this particular structure that contains exactly three unknowns (i.e., the segment AB contains four AB AB AB BD BD BD BD unknowns {  AB A ,  B , U B , VB }, the segment BD contains six unknowns {  B , U B , VB ,  D , BD DE DE DE DE DE U BD D , VD }, and the segment DE contains five unknowns {  D , U D , VD ,  E , U E }). To solve for these unknowns, all three segments must be considered simultaneously. First, we construct conjugate segments for the three segments AB, BD and DE as shown below.

L/3 VBAB U BD B

U AB B

 AB B

L/3 D

B A2

 BD B



A3

VBBD

B

BD D

U BD VDDE D

VDBD

A1

D

 DE D

U DE D

Y 2L/3

AB U AB A  0 A

A

X

VAAB  0

E

 DE E

U DE E

VEDE  0

The areas A1, A2 and A3 are given by A1 =

1  3PL  3PL2 1  PL  PL2 1  4PL  4PL2 L A = L A = 2L ; ;            2 3       2  2EI  4EI 2  EI  2EI 2  EI  EI

Next, by considering equilibrium of moments in the Y-direction of the conjugate segment AB and the conjugate segment DE, we then obtain ΣM YA = 0 :

VAAB  VBAB  0



VBAB  VAAB  0

ΣM YE = 0 :

VDDE  VEDE  0



VDDE  VEDE  0

From continuity of the displacement at points B and D, we can readily obtain VBBD  VBAB  0 and VDBD  VDDE  0 . Now, the number of unknowns of the segment BD reduces from six to four and this BD allows the end rotations {  BD B ,  D } be computed from equilibrium of moments in the Y-direction and equilibrium of forces in the Z-direction as follow: ΣM YD = 0 :

BD VBBD   BD B (2L)  A 2 (5L/3)  A 3 (4L/3)  VD  0



286




 PL2   5L   4PL2   4L  0  BD   B (2L)     00  2EI   3   EI   3 



 BD B 

9PL2 4EI

CW

 BD  A 2  A3   BD B D = 0

ΣFZ = 0 :





9PL2 PL2 4PL2    BD D 0 4EI 2EI EI



 BD D 

5PL2 4EI

CCW

BD 2 From continuity of the rotation at point B, we simply obtain  AB B   B  9PL /4EI . Next, we move to the conjugate segment AB. By considering equilibrium of forces in the Z-direction and AB equilibrium of moments in the X-direction, it yields {  AB A , U B }:

 AB  A1  AB A B = 0

ΣFZ = 0 :



 AB  A

3PL2  9PL2  4EI  4EI

 0 



 AB A 

3PL2 EI

CW

AB AB U AB A   A (L)  A1 (L/3)  U B  0

ΣM XB = 0 : 

 3PL2 0  EI



U AB B 

  3PL2   L  AB  (L)       UB  0   4EI   3 

11PL3 4EI

Rightward

From continuity of the displacement in the horizontal direction at point B, we obtain AB 3 BD U BD B  U B  11PL /4EI . Now, we return to the conjugate segment BD. The remaining unknown U D can readily be obtained from equilibrium of moments in the X-direction of this conjugate segment: BD U BD B  UD  0

ΣM XB = 0 :



BD U BD D  UB 

11PL3 4EI

Rightward

To obtain the displacement and rotation at the roller support, we first enforce continuity of the BD 3 DE BD 2 displacement and rotation at point D (i.e., U DE D  U D  11PL /4EI and  D   D  5PL /4EI ) and then consider equilibrium of forces in the Z-direction and equilibrium of moments in the Xdirection as shown below. ΣFZ = 0 :

DE  DE D  E = 0

ΣM XE = 0 :

DE DE U DE D   D (L)  U E  0





DE  DE E  D 

11PL3  5PL2  DE   (L)  U E  0 4EI  4EI 



5PL2 4EI

U DE E 

CCW

4PL3 EI

Rightward

The displacement and rotation at points A, B, D and E are now completely known. The displacement and rotation at other points within the structure can therefore be obtained by considering a straight segment containing one of these points as its end point. Copyright © 2011 J. Rungamornrat


287


For example, the displacement and rotation of point C can be obtained by considering either a segment BC or a segment CD. Here, we choose to consider a segment CD since it involves less computational effort. The conjugate segment CD is shown below: Y

L/3

U CD C

C



D A4

CD C



VCCD

BD D

U BD D

X

VDBD

where A4 is the area of curvature diagram over the segment CD which is equal to A4 =

1  2PL  PL2 L    2  EI  EI

By considering equilibrium of the conjugate segment CD, we then obtain the displacement and rotation at point C as follow: CD  A 4  CD C D = 0

ΣFZ = 0 :



CD C 

PL2 5PL2  0 EI 4EI



CD C 

ΣM XD = 0 :

CD U CD C  UD  0

ΣM YD = 0 :

CD VCCD  CD C (L)  A 4 (2L/3)  VD  0



CD U CD C  UD 

PL2 4EI 11PL3 4EI



 PL2   PL2   2L  VCCD    (L)    00  4EI   EI   3 



VCCD  

11PL2 12EI

CCW Rightward

Downward

5.4 Conjugate Structure Analogy for General Segment In this section, we generalize the conjugate structure analogy established above for rigid frames to enhance its capabilities to treat more general segments besides the straight segment. As apparent from the Example 5.7, the restriction of the analogy to consider only straight segment poses some difficulty in the analysis procedure. More precisely for this example, it is impossible to choose a straight segment that contains only three kinematical unknowns at its end and this, therefore, requires consideration of several conjugate segments simultaneously in order to solve for such unknowns. Here, the analogy is extended for a segment that consists of multiple straight segments whose axes may or may not be aligned and may contain interior hinges. This enhanced feature will increase flexibility of the method to treat a broader class of conjugate segments. In particular, it may increase the chance to find a starting segment that contains exactly three unknowns and this, as a result, can avoid substantial calculations such as those required in the Example 5.7. Copyright © 2011 J. Rungamornrat


288


Consider a non-straight segment P1P4 that consists of three straight sub-segments P1P2, P2P3 and P3P4 and contains a hinge at point A2 as shown in Figure 5.11. The orientation of the segments P1P2, P2P3 and P3P4 with respect to the global coordinate system and their lengths are denoted by angles {1, 2, 3} and {L12, L23, L34}, respectively. Note that the orientation angle i is positive if the segment PiPi+1 is oriented in the counter clockwise direction with respect to the X-direction; otherwise it is negative. By following the same convention defined in section 5.3, points P1 and P4 of the non-straight segment P1P4 are defined as its start point and end point, respectively. Similarly, the start point and end point of the straight sub-segments P1P2, P2P3 and P3P4 are also implied automatically from their name; for instance, the start point and end point of the sub-segment P1P2 are P1 and P2, respectively. It is worth noting again that the start point and end point of any segment play a vital role in defining the sign convention of the curvature diagram and the direction of fictitious forces and moments at the ends of the corresponding conjugate structure. 3

P3 2

L23

L34 P4

P2 Y L12 1

X

P1 Figure 5.11: Schematic of a non-straight segment P1P4 consisting of three straight segments P1P2, P2P3 and P3P4 and containing hinge at point P2 The curvature diagrams (or the M/EI-diagrams) for the entire segment P1P4 are shown in Figure 5.12. The areas of the curvature diagram over the sub-segments P1P2, P2P3 and P3P4 are denoted by A12, A23 and A34, respectively, and the distances from the start point and end point of 12 23 23 each segment to the centroid of curvature diagram are denoted by { x12 1 , x 2 }, { x 2 , x 3 } and 34 { x 34 3 , x 4 }, respectively.

x

x 323

23 2

A23

A12

A34

P3

P4

x 34 3

P2

x 34 4 x12 2

P1

x12 1

Figure 5.12: Schematic of curvature diagram for non-straight segment shown in Figure 5.11 Copyright © 2011 J. Rungamornrat


289


The elastic or deformed curve of the segment P1P4 is shown in Figure 5.13 where Uiji , Viij and iji denote the horizontal displacement, vertical displacement and rotation at the start point Pi of the segment PiPj, respectively, and Uijj , Vjij and ijj denote the horizontal displacement, vertical displacement and rotation at the end point Pj of the segment PiPj, respectively.

V414  V434 V323  V334 V212  V223

P2



23 2

P3 323

P3



12 2

34 14 4  4

34 3

U323  U334

23 U12 2  U2

P2

P4

P4

34 U14 4  U4

Y

V114  V112

P1

  14 1

12 1

12 U14 1  U1

P1

X

Figure 5.13: Schematic of deformed and undeformed configurations of non-straight segment P1P4 containing hinge at point P2 Since the sub-segments P1P2, P2P3 and P3P4 are straight and contain no interior hinge, analogy established in section 5.3 can be employed. The conjugate structures for these three subsegments are shown in Figure 5.14. By applying the conjugate structure analogy to the segment P1P2, we then obtain following three relations among the end displacements and end rotations ΣFZ = 0



12  A12  12 1 2 = 0

(5.33)

V112  12 1  L12 cos 1  L 23 cos  2  L 34 cos 3 

ΣM YA4 = 0



 A12 (x12 2 cos 1  L 23 cos  2  L 34 cos 3 ) 

12 2

 L23 cos 2  L34 cos 3   V

12 2

(5.34) 0

12 U12 1  1  L12 sin 1  L 23 sin  2  L 34 sin 3 

ΣM XA4 = 0



 A12 (x12 2 sin 1  L 23 sin  2  L 34 sin 3 ) 

12 2

 L23 sin 2  L34 sin 3   U

12 2

(5.35) 0

Note that the reference point used for taking moment is chosen to be point P4 (which is out of the segment P1P2) for convenience in further manipulation. Similarly, the conjugate structure analogy for the segment P2P3 yields ΣFZ = 0



 23  A 23  323 = 0 2 Copyright © 2011 J. Rungamornrat

(5.36)


290


V223

V223

U323 

P2

23 2

U 23 2

2

U12 2 12 2

P2 

U 23 2

23 2

A 23

V323 V334

V212

U323



12 2

P2 V

P1 

12 1

P3

V323

V212

12 1

323

P3 

23 3

P3

U12 2



34 3



34 3

V434 U34 3

A34

U34 4

Y

A12

U34 3

P4

V334

34 4

X

U12 1

Figure 5.14: Schematic of conjugate segments P1P2, P2P3 and P3P4 and isolated point P2 and P3 ΣM YA4 = 0

ΣM XA4 = 0





V223   223  L 23 cos 2  L34 cos 3   A 23 (x 323 cos 2  L34 cos 3 )  323  L34 cos 3   V323  0 23 23 U 23 2   2  L 23 sin  2  L 34 sin 3   A 23 (x 3 sin  2  L 34 sin 3 )

 323  L34 sin 3   U 323  0

(5.37)

(5.38)

Again, point P4 is chosen as the reference point for taking moments. Finally, the conjugate structure analogy for the segment P3P4 leads to ΣFZ = 0



34  A 34  34 3 4 = 0

(5.39)

ΣM YA4 = 0



34 V334  334  L34 cos 3   A 34 (x 34 4 cos 3 )  V4  0

(5.40)

ΣM XA4 = 0



34 34 34 U34 3  3  L 34 sin 3   A 34 (x 4 sin 3 )  U 4  0

(5.41)

The relations (5.33)-(5.35), (5.36)-(5.38) and (5.39)-(5.41) are in fact the three curvature area equations (in global coordinate system) for the segments P1P2, P2P3 and P3P4, respectively. Next, by invoking the continuity of the displacement at point P2 and denoting the relative hinge rotation at this point by 2 , we obtain the following three relations: 12  23 2   2 =  2

(5.42)

V212  V223



291


23 U12 2  U2

(5.44)

Similarly, by invoking the continuity of the displacement and rotation at point P3, it leads to following three relations: 323  34 3

(5.45)

V323  V334

(5.46)

U323  U 34 3

(5.47)

To construct an analogy for the relations (5.42)-(5.44), let us imagine that a point P2 (which is a point connecting a conjugate segment P1P2 and a conjugate segment P2P3) is subjected to an external concentrated force of value 2 in the positive Z-direction. A free body diagram of this point (including the applied force 2 and those exerted by the end of the two conjugate segments P1P2 and P2P3) is shown in Figure 5.14. It is obvious that the relations (5.42), (5.43) and (5.44) are identical to equilibrium equations ΣFZ = 0 , ΣM YP = 0 , ΣM XP = 0 of the isolated point P2, respectively. 2

2

Similarly, the continuity relations (5.45)-(5.46) can also be viewed as equilibrium equations ΣFZ = 0 , ΣM YP = 0 , ΣM XP = 0 of a conjugate point P3 whose free body diagram is shown in Figure 5.14. 3

3

Since the rotation is continuous at this particular point in the real structure, no external concentrated force is applied to the conjugate point P3. Only forces exerted by the end of the two conjugate segments P2P3 and P3P4 appear in its free body diagram. Now, the conjugate segments P1P2, P2P3 and P3P4 can then be combined into a single conjugate segment P1P4 via the use of Newton’s third law at the connecting points P2 and P3. The resulting conjugate segment P1P4 is shown in Figure 5.15. Without loss, we replace the superscript 2 12 34 34 34 of quantities { V112 , U12 1 , 1 } by the superscript 1 and superscript 3 of quantities { V4 , U 4 ,  4 } by the superscript 1 to be consistent with the name of the segment P1P4. The correspondence between the curvature area equations of the real non-straight segment P1P4 shown in Figure 5.13 and the equilibrium equations of the conjugate segment shown in Figure 5.15 can be demonstrated as follow. By combining equations (5.33), (5.36), (5.39), (5.42) and (5.45), it yields the first curvature area equation of the segment P1P4: 14  A12  A 23  A 34   2  14 1 4 = 0

(5.48) V414 A 23

2

V A12

2

L23

3 A 34

P3

L34

P4

P2

14 1

14 4

U14 4

Y

L12 14 1

1

P1

U14 1

X

Figure 5.15: Schematic of conjugate structure of non-straight segment P1P4 containing a hinge at P2 Copyright © 2011 J. Rungamornrat


292


12 14 34 where 14 have been used. It can readily be verified that (5.48) is identical to 1  1 and  4   4 the force equilibrium equation ΣFZ = 0 of the conjugate segment P1P4. Next, by combining equations (5.34), (5.37), (5.40), (5.43) and (5.46), it leads to the second curvature area equation: 12 V114  14 1  L12 cos 1  L 23 cos  2  L 34 cos 3   A12 (x 2 cos 1  L 23 cos  2  L 34 cos 3 ) 14  A 23 (x 323 cos 2  L34 cos 3 )  A 34 (x 34 4 cos 3 )   2  L 23 cos  2  L 34 cos 3   V4  0

(5.49)

Again, this equation is identical to the moment equilibrium equation ΣM YP = 0 of the conjugate 4

segment P1P4. Finally, by combining equations (5.35), (5.38), (5.41), (5.44) and (5.47), it leads to the third curvature area equation: 14 12 U14 1  1  L12 sin 1  L 23 sin  2  L 34 sin 3   A12 (x 2 sin 1  L 23 sin  2  L 34 sin 3 ) 14  A 23 (x 323 sin 2  L34 sin 3 )  A 34 (x 34 4 sin 3 )   2  L 23 sin  2  L 34 sin 3   U 4  0

(5.50)

Above three correspondences between the curvature area equations of the real segment and the equilibrium equations of the conjugate segment forms the conjugate structure analogy similar to those established in sections 5.1, 5.2 and 5.3. It is important to emphasize that the form of curvature area equations becomes more complex when the considered segment consists of multiple straight segments and contains interior hinges. Memorization of such form is impractical, somewhat cumbersome and, in general, segment-dependent. Existence of the above correspondences allows all curvature area equations for a non-straight segment be formed in a simple manner. The key task is to establish the correct conjugate segment and the rest only involves setting the equilibrium equations. The construction of a conjugate structure for a non-straight segment is similar to the straight segment and can be summarized again as follow:  At the start point of the conjugate segment, a force of value equal to the rotation at this point of the real segment is applied in the positive Z-direction, a moment of value equal to the X-component of the displacement at this point of the real segment is applied in the positive X-direction, and a moment of value equal to the Y-component of the displacement at this point of the real segment is applied in the positive Y-direction.  At the end point of the conjugate segment, a force of value equal to the rotation at this point of the real segment is applied in the negative Z-direction, a moment of value equal to the X-component of the displacement at this point of the real segment is applied in the negative X-direction, and a moment of value equal to the Y-component of the displacement at this point of the real segment is applied in the negative Y-direction.  Known external forces of values equal to areas of sub-curvature diagrams are applied to the conjugate segment at the location of their centroid in the positive Z-direction.  Unknown External force of value equal to the unknown relative rotation of the interior hinge present in the real segment is applied to the conjugate segment at the same point as the hinge location in the positive Z-direction.

Although the analogy is established only for a non-straight segment containing only three straight sub-segments and one interior hinge, it also applies to segments containing multiple straight subsegment and multiple hinges. The only slight modification that must be made is to include unknown external forces of values equal to unknown relative hinge rotations at points in the conjugate structure that coincide with the hinge locations and known external forces of values equal to areas of sub-curvature diagrams at all sub-segments. Finally, we remark that while the established analogy applies to general non-straight segments that may contain interior hinges, the segment must Copyright © 2011 J. Rungamornrat


293


contain exactly one start point and one end point or, equivalently, the segment must be a single chain. For instance, segments such as segments ACB, ACD, ACDEG and GEF of a structure shown in Figure 5.16 are valid for above analogy while segments such as ABCD, BCDEFG, ABCDEG and ABCDEFG are not since they are not single chains. E

C B

F

D

G A

Figure 5.16: Schematic of rigid frame containing certain segments that are not single chains Example 5.8 Use conjugate structure analogy established above to determine the displacement and rotation at point E of a rigid frame due to applied loads shown below. The flexural rigidity EI is constant throughout the structure.

2P A

EI

B

E

EI

EI

C

P L

D

EI L

L

Solution Since a given structure is statically determinate, all support reactions, sub-bending moment diagrams and sub-curvature diagrams for segments AB, BC, CD and DE can readily be obtained from static equilibrium and method of superposition with results shown below.

2P P A 2PL 2P

EI

B

E

EI

EI

C

EI

P –2PL/EI

D

–PL/EI

–PL/EI –PL/EI



294


It is emphasized that the start point and end point of all segments are already chosen and clear from their name and, as a result, the corresponding local coordinate system for each segment is also automatically defined. Since the displacement and rotation are fully prescribed at the fixed support, the displacement and rotation at point E can readily be obtained by considering a single segment AE (containing four sub-segments AB, BC, CD, and DE) since it contains only three unknowns { U AE A , AE AE VA ,  A }. A corresponding conjugate segment of the segment AE is shown below VAAE  0

U

AE A

0

VEAE  AE E

AE A 0

E

A1

A

U AE E 2L/3

Y A2

L/3

A3

X L/3

A4

2L/3

L/2

L/2

where areas of the sub-curvature diagrams A1, A2, A3 and A4 are given by A1 =

1  2PL  PL2 1  PL  PL2 PL2  PL  ; A2  A4 =     L   ; A3      L     L   2  EI  EI 2  EI  2EI EI  EI 

By considering equilibrium of the conjugate structure AE, it leads to AE  AE A + A1 + A 2 + A 3 + A 4   E = 0

ΣFZ = 0 :



0

PL2 PL2 PL2 PL2     AE E 0 EI 2EI EI 2EI



 AE E  

3PL2 EI

CW

AE U AE A  A 2 (2L/3)  A 3 (L)  A 4 (2L/3)  U E  0

ΣM XE = 0 :



 PL2   2L   PL2   PL2   2L  AE  0  L        UE  0 2EI 3 EI 2EI 3        



U AE E 

5PL3 3EI

Rightward

VAAE  AAE (2L)  A1 (5L/3)  A 2 (L)  A3 (L/2)  VEAE  0

ΣM YE = 0 :



 PL2 0   0  2L      EI



VEAE  

8PL3 3EI

  5L   PL2   PL2     L          3   2EI   EI

Downward


 L  AE     VE  0  2 


295


Example 5.9 Use conjugate structure analogy to determine the displacement and rotation at point E of a rigid frame due to applied loads shown below. The flexural rigidity for each segments are clearly indicated in the figure.

qL D

EI

E

L

EI 2q A B

2EI

C

L

L

C

2EI

Solution All support reactions and sub-curvature diagrams for segments AB, BC, BD and DE can readily be obtained from static equilibrium and method of superposition with results shown below.

qL E

D

EI

qL2/EI

EI 2q 0 A B

2EI

2EI

qL

qL2/EI qL2/EI

C –qL2/2EI

2qL

From boundary conditions at the pinned support and roller support, we obtain U A  VA  VC  0 . We start the analysis by first considering a straight segment AC since it is the only segment that AC AC contains three unknowns {  AC A , U C ,  C }. A corresponding conjugate segment of this segment is shown below VAAC  0

U

AC A

0

VCAC  0

 AC A

CAC

4L/3

C

A 2 A3

A1

A

11L/12

L/12

2L/3

where areas of the sub-curvature diagrams A1, A2 and A3 are given by A1 =

1  qL2  qL3 1  qL2  qL3 1  qL2  qL3 ; A2 =   ; A3 =     2L    L   L  2  EI  EI 3  2EI  6EI 2  EI  2EI Copyright © 2011 J. Rungamornrat

Y

U CAC

X


296


By considering equilibrium of the conjugate structure AC, we obtain AC VAAC  AC 0 A (3L)  A1 (5L/3)  A 2 (3L/4)  A 3 (2L/3)  VC

ΣM YC = 0 :



 qL3   5L   qL3   3L   qL3   2L  0  AC    A  2L       00  EI   3   6EI   4   2EI   3 



 AC A 

15qL3 16EI

CW

AC  AC A + A1 + A 2 + A 3   C = 0

ΣFZ = 0 :





15qL3 qL3 qL3 qL3     CAC  0 16EI EI 6EI 2EI

AC U AC A  UC  0

ΣM XC = 0 :



CAC 

19qL3 48EI

CCW

U CAC  U AC A 0



To obtain the displacement and rotation at point E, we consider a segment AE (containing straight sub-segments AB, BD and DE) since the displacement and rotation at point A are completely known. A corresponding conjugate segment of this segment is shown below VEAE L/3 U AE A

V

U AE A 0

 AE E

AE E

E

A5

L/2

0

Y

A4 L/2

 AE A

X

A1

A 4L/3

2L/3

where areas of the sub-curvature diagrams A4 and A5 are given by  qL2  qL3 1  qL2  qL3 ; A4 =  L  A = L         5 EI 2  EI  2EI  EI 

By considering equilibrium of the conjugate structure AE, we obtain AE  AE A + A1 + A 4 + A 5   E = 0

ΣFZ = 0 :

 ΣM YE = 0 :



15qL3 qL3 qL3 qL3     AE E 0 16EI EI EI 2EI



 AE E 

25qL3 16EI

VAAE   AAE (L)  A1 (L/3)  A 4 (L)  A 5 (2L/3)  VEAE  0


CCW


297




 15qL3   qL3   L   qL3   qL3   2L  AE 0   L       L      VE  0 16EI EI 3 EI 2EI 3          



VEAE  

125qL4 48EI

Downward

AE AE U AE A   A (L)  A1 (L)  A 4 (L/2)  U E  0

ΣM XE = 0 :



 15qL3   qL3   qL3   L  AE 0   L     L       UE  0  16EI   EI   EI   2 



U AE E 

9qL4 16EI

Leftward

Example 5.10 Use conjugate structure analogy to determine the displacement and rotation at point E and the relative rotation at hinge B of a rigid frame shown below. The flexural rigidity for each segments are clearly indicated in the figure.

qL

2q EI

B

C EI 2EI

2EI A

E L

D

L

L

L/3

Solution All support reactions and sub-curvature diagrams for segments AB, BC, CD and CE can readily be obtained from static equilibrium and method of superposition with results shown below.

qL

2q B

EI

C EI

qL/3

A

qL/3

E –qL2/6EI

2EI

2EI D

qL2/3EI

–qL2/3EI –qL2/EI

qL/3

8qL/3

From boundary conditions at the pinned supports, we obtain U A  VA  U D  VD  0 . We start by first considering a non-straight segment AD (containing straight sub-segments AB, BC and CD) Copyright © 2011 J. Rungamornrat


298


AD since it is the only segment that contains three unknowns {  AD A ,  D ,  B }. A corresponding conjugate segment of this segment is shown below

L/3 L/4  B

A 2 A3

L/3 A1

Y

2L/3 U AD A 0

 AD A

A

X

D

VAAD  0



AD D

U AD D 0

VDAD  0 L

L

where areas of the sub-curvature diagrams A1, A2 and A3 are given by A1 =

1  qL2  qL3 1  qL2  qL3 1  qL2  qL3 ; ;  L   A = L  A =  L               2 3 2  6EI  12EI 2  3EI  6EI 3  EI  3EI

By considering equilibrium of the conjugate structure AD, we obtain AD U AD A  A1 (2L/3)  A 2 (L)  A 3 (L)   B (L)  U D  0

ΣM XD = 0 :



 qL3   2L   qL3   qL3  0  L      L    B (L)  0  0   12EI   3   6EI   3EI 



 B 

2qL3 9EI

CCW

AD VAAD   AD 0 A (2L)  A 2 (L/3)  A 3 (L/4)   B (L)  VD

ΣM YD = 0 :



 qL3   L   qL3   L   2qL3  0   AD          L  0  0 A (2L)    6EI   3   3EI   4   9EI 



 AD A 

7qL3 72EI

CW

AD  AD A + A1 + A 2 + A 3   B   D = 0

ΣFZ = 0 :

7qL3 qL3 qL3 qL3 2qL3       AD D 0 72EI 12EI 6EI 3EI 9EI







 AD D 

qL3 CW 8EI



299


To obtain the displacement and rotation at point E, we consider a segment AE (containing straight sub-segments AB and BE) since the displacement and rotation at point A and the relative hinge rotation  B are already known. A corresponding conjugate segment of this segment is shown below 2L/3 7L/12 2L/9 E AE A

V

 B

0



A 2 A3 A 4

AE E

U AE E

Y VEAE

U AE A 0

 AE A

X

A

where area of the sub-curvature diagram A4 is given by A4 =

1  qL2   L  qL3      2  3EI   3  18EI

By considering equilibrium of the conjugate structure AE, we obtain AE  AE A + A 2 + A 3 + A 4   B   E = 0

ΣFZ = 0 :

7qL3 qL3 qL3 qL3 2qL3       AE E 0 72EI 6EI 3EI 18EI 9EI







 AE E 

7qL3 CW 72EI

AE AE U AE A   A (L)  U E  0

ΣM XE = 0 :



 7qL3  AE 0   L   UE  0  72EI 



U AE E 

7qL4 72EI

Rightward

VAAE   AAE (2L)  A 2 (2L/3)  A 3 (7L/12)  A 4 (2L/9)   B (L)  VEAE  0

ΣM YE = 0 :



 7qL3   qL3   2L   qL3   7L   qL3   2L   2qL3  AE 0   (2L)        L   VE  0     72EI   6EI   3   3EI   12   18EI   9   9EI 



U AE E 

11qL4 162EI

Downward



300


Exercises 1. Use the method of conjugate structure analogy to determine the relative hinge rotation (if exists) and the deflection and rotation at certain points of beams indicated in the figure below. In the analysis, only bending deformation is considered. Young's modulus and moment of inertia of the cross section are clearly indicated in the figures. 2qL q A

3EI

EI

L/2

B

L/2

q A

EI

qL C

2EI

B

L

EI

2L

L

2P

P B A

D

C

EI

EI

EI

L

2L

L

2qL

q

D

B

A

2EI

2EI

3L

L

L

A

EI

C

2L

P EI

D

P EI L

B

D EI L


C

EI L

PL


301


qL

EI

q

2qL

EI

A

L

2EI

C

B

L

L

L

2. Use the method of conjugate structure analogy to determine the relative hinge rotation (if exists) and displacement and rotation at certain points of rigid frames indicated in the figure below. In the analysis, only bending deformation is considered. Young's modulus and moment of inertia of the cross section are clearly indicated in the figures.

EI

q

2qL

q D

A 2EI

B

A

C

EI

B

EI

L

2EI

EI

qL

C

2EI

qL

L

L

E

EI

L

L/2

L

q A EI

PL B

B L

EI C 2P

D

A

EI

A

L

L q

q EI

D

EI

2EI

L

L

C

EI

B

EI

L/2

B

EI ~ constant

A

2EI

D

C

L/2 C L

E

qL L


L

qL


302


q

qL

P

PL EI

C

EI

B

EI

EI

D

B

L/2 A

3EI

3EI

C

qL

2P

EI ~ constant

L/2 A 2L/3

L/2

L/2

2L/3


4L/3

2L/3

303


Introduction to Work and Energy Theorems

CHAPTER 6 INTRODUCTION TO WORK AND ENERGY THEOREMS The key objective of this chapter is to provide an introduction and fundamental background of work and energy theorems commonly employed in structural mechanics. Various methods (which will be presented in following chapters) derived from such work and energy theorems have gained significant popularity and become crucial tools for analysis of structures. It is worth noting, at the beginning, that the application of work and energy theorems in structural mechanics does not aim to alter the meaning of three basic equations (i.e. equilibrium equations, kinematics, and constitutive law) but aim only to express such equations in other different but equivalent forms that are wellsuited for use in various circumstances. As will be apparent later, formulation of such basic equations in terms of relations involving quantities related to work and energy can significantly ease the analysis procedure. In this chapter, we first introduce the concept of work and complementary work, virtual work and complementary virtual work, strain energy and complimentary strain energy, and virtual strain energy and complementary virtual strain energy. Next, we summarize without a formal proof of work and energy principles that play a crucial role in structural analysis. These include the conservation of work and energy, the principle of virtual work, the principle of complementary virtual work, the principle of stationary total potential energy, the principle of stationary total complementary potential energy, the reciprocal theorem, Castigliano’s 1st theorem, and Castigliano’s 2nd theorem. Applications of these principles to the analysis of internal forces and displacements will be presented in following chapters.

6.1 Work and Complimentary Work Consider a body undergoing a motion from a state 1 to a state 2 as shown in Figure 6.1. The body is subjected to a force vector P at point A and it may or may not be constant along the moving path of point A. From the basic definition in fundamental physics, the incremental work-done, denoted by dW, by the force P during an (infinitesimal) incremental motion can be computed from dW  P  du

(6.1)

Moving path of point A A P State 1

A

du AA

State 2 Figure 6.1: Schematic of a body undergoing motion and subjected to a force P at point A Copyright © 2011 J. Rungamornrat

304



where du denotes an infinitesimal displacement of point A during the incremental motion. It is apparent from (6.1) that only the component of du in the direction of the force P contributes to the incremental work-done dW. Note also that the work-done is a scalar quantity and has a unit of (force)x(length). The total work-done (from now to what follows, the termed work-done will be replaced by work for convenience) of the force P from state 1 to state 2 can readily be computed from the following integral state2



W

state2

dW 

state1



P  du

(6.2)

state1

For a special case that a force vector P is constant during the motion from state 1 to state 2, the total work of the force P is equal to W  P

(6.3)

where P denote the magnitude of the force P and  is the total displacement of point A in the direction of the force P measured from state 1 to state 2; the total work is considered positive if both P and  have the same direction, otherwise it is negative. Now, let focus attention to the case that the force P has the same direction as the displacement u. A formula (6.2), when specialized to this particular case, becomes u0

W   Pdu

(6.4)

0

where P denotes the magnitude of force P, u is the displacement in the direction of the force P, and u0 is the total displacement in the direction of the force P measured from state 1 to the final state 2. It is important to emphasize that the force P is, in general, not constant but is a function of the displacement u, i.e. P = P(u). In addition, the incremental work Pdu is considered positive when P and du have the same direction, otherwise, it is negative. The graphical interpretation of the total work (6.4) can readily be demonstrated by a graph between the force P and the displacement u shown in Figure 6.2(a)-(b). The P-u relations shown in Figure 6.2(a) and Figure 6.2(b) are nonlinear with non-zero and zero force P at the initial state, respectively, while the P-u relation shown in Figure 6.2(c) is linear with zero force P at the initial state. P

P

P

P0

P0

P0

P(u)

P(u)

P(u)

Pi du

(a)

u0

u

u0

du

(b)

u

du u0

u

(c)

Figure 6.2: Relation between the force P and the corresponding displacement u for (a) P = P(u) is nonlinear and P(0) ≠ 0, (b) P = P(u) is nonlinear and P(0) = 0, and (c) P = P(u) is linear and P(0) = 0 Copyright © 2011 J. Rungamornrat

305



It is evident from the expression (6.4) that the total work W due to a force P is simply the area between a graph P = P(u) and the u-axis from u = 0 to u = u0. Note that if we define Pi = P(ui) and P0 = P(u0) as the values of the force P at the initial and final states, respectively, we can equivalently say that W is the total work required for increasing the force P from its initial value Pi to its final value P0. For the common case when Pi = 0 (as shown in Figure 6.2(b)-(c)), W is therefore the total work required in the process of applying the force P until it reaches the final value P0. In many situations, ones just simply say that W is the work done by the force P0 and left behind the fact that the calculation of the work W involves the entire continuous process since the first state when P = 0 to the final state when P = P0. From now to what follows, the work W due to the force P simply means the total work done due to this force since the beginning of its application (P = 0) until it reaches its final value; we restrict attention only to cases (b) and (c) in Figure 6.2. For the special case that P is a linear function in u, the expression for the work W due the force P simply reduces from (6.4) to

W

1 P0 u 0 2

(6.5)

where P0 and u0 are values of the force P and corresponding displacement at the state where the work W is to be computed. Now, let us consider a more general case when a body or a structure is subjected to a set of external loads (forces or moments) {P1, P2, P3, … Pn} and let {u1, u2, u3, … un} be the corresponding movements (displacements or rotations) at locations and in the directions of applied loads. The total work due to this set of external loads can readily be computed by summing the work due to each load as follow: W

uo

n

  P du 0 i 1

i

(6.6)

i

where {u1o, u2o, u3o,… uno} are movements at the state where the work is to be evaluated. Similarly, for the special case that the external loads {P1, P2, P3,… Pn} are proportional to the movements {u1, u2, u3,… un}, the total work W is simply given by W

1 n  Pio u io 2 i 1

(6.7)

where Pio is the value of the force Pi at the state where the work W is to be evaluated. Next, let us introduce another quantity that comes along with the work-done W called the complimentary work-done and denoted by WC. This quantity is defined in an incremental form as the complement part of the incremental work-done dW to complete the differential d(P  u) , i.e. dWC  d(P  u)  dW  u  dP

(6.8)

where dP denotes an infinitesimal increment of the force P during the incremental motion. Again, it is evident from the last term of (6.6) that only the component of dP in the direction of the displacement u contributes to the incremental complementary work dWC. The total complementary work of the displacement u from state 1 to state 2 can be computed from the following integral state2

WC 



state1

state2

dWC 



u  dP

(6.9)

state1


306



For a special case that the displacement u is constant during the motion from state 1 to state 2, the total complementary work of the displacement u is equal to WC  (P)u

(6.10)

where u denote the magnitude of the displacement u and P is the total increment of the force P in the direction of the displacement u. Now, let us focus on the case that the displacement u has the same direction as the force P. A formula (6.9) simply reduces to P0

WC   udP

(6.11)

Pi

where P and u are defined in the same as those appearing in (6.4) and, again, Pi and P0 are initial and final values of the force P. The incremental complementary work udP is considered positive when both u and dP have the same directions, otherwise, it is negative. Similar to the work W, the complementary work WC admits the graphical interpretation as the area between a graph P = P(u) and the P-axis from P = Pi to P = P0 (see Figure 6.3). Based on this interpretation, the work W and the complementary WC are related by a simple relation WC  P0 u 0  W

(6.12)

For the special case where the relation between P and u is linear and Pi = 0, both the work W and the complementary work WC are identical and are equal to WC  W 

1 Po u o 2

(6.13)

Similar to the case of work, the complementary work due to a set of external loads {P1, P2, P3, … Pn} and the corresponding movements {u1, u2, u3, … un} is computed by the complementary work due to each movement as follow: Po

n

0

i 1

  u dP

WC 

i

(6.14)

i

P

P

P

P0 dP

P0 dP

P0 dP

Pi u(P)

(a)

u0

u

u(P)

u0

(b)

u

u(P)

u0

u

(c)

Figure 6.3: Graphical interpretation of the complementary work for (a) u = u(P) is nonlinear and P(0) ≠ 0, (b) u = u(P) is nonlinear and P(0) = 0, and (c) u = u(P) is linear and P(0) = 0 Copyright © 2011 J. Rungamornrat

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where, again, {P1o, P2o, P3o,… Pno} are applied loads at the state where the complementary work is to be evaluated. Similarly, for the special case that the external loads {P1, P2, P3,… Pn} are proportional to the movements {u1, u2, u3,… un}, the total complementary work WC becomes WC 

1 n  u io Pio 2 i 1

(6.15)

and is, clearly, identical to the total work W given by (6.7).

6.2 Virtual Work and Complimentary Virtual Work In this section, we introduce two quantities that are fundamental to work and energy theorems: virtual work and complimentary virtual work. Unlike work and complimentary work, these two quantities, while possessing sound physical meaning as further clearly demonstrated, are not real but result purely from the thought (or virtual) process. Ones who are new in this area may feel somewhat strange at the first glance of these terms but, once getting familiar with them, they gradually absorb their definition in the same fashion as the first study of work and complimentary work. The adjective virtual is in fact used to emphasize the nature of those quantities and remind us about their nature.

Actual configuration

P A

A

u

Virtual configuration

Figure 6.4: Schematic of a body at a particular instant undergoing the virtual displacement Consider a body or a structure where its configuration at a particular instant is shown schematically in Figure 6.4; at this instant, the applied force achieves its value P and its direction as indicated. Now, let us consider the following thought process. As the body occupies its particular configuration shown in Figure 6.4, we imagine that the clock (or time) is stopped in our imagination and the body is completely freezed. Next, it is assumed that the body undergoes a fictitious movement from the freezed configuration. This fictitious movement occurs only in our thought and is commonly termed the virtual displacement. The new configuration of the body in our thought is indicated by the dash line in Figure 6.4. The work done by the force P through the virtual displacement in our imagination, denoted by W, is simply given by W  Pδu

(6.16)

in which P is the value of the applied force at the instant where the time is stopped and u is the virtual displacement at point A in the direction of the applied force P. The work-done W is termed the virtual work of the force P due to the virtual displacement u. The symbol “” is utilized only to Copyright © 2011 J. Rungamornrat

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emphasize that this work-done and the virtual displacement are not real quantities but only occurs in our imagination. It is important to remark that in our thought process, the virtual displacement can be chosen arbitrary and the real force acting on the body does not change their magnitude and direction during the body undergoing the virtual displacement. The latter feature clearly explains why the virtual work is simply computed from the product of the real force at a particular instant and the corresponding virtual displacement. Note that while the virtual work does not really occur, it has the same physical interpretation as the real work. For instance, the virtual work has the same unit as the real work and it is positive if both P and u have the same direction and negative if they are opposite. For a particular case that the body is subjected to a moment M instead of the force P, the virtual work of this applied moment due to the virtual movement can be computed from W  Mδ

(6.17)

where  denotes the rotation of the point A in the direction of the applied moment M resulting from the virtual movement in our though process, concisely termed the virtual rotation. Now, let us generalize this concept to a more general case that the body is subjected to a set of applied loads (forces or moments) {P1, P2, P3, …, Pn}. The virtual work of these loads at any instant due to the corresponding virtual displacements or rotations {u1, u2, u3, …, un} is given by n

W  P1δu1  P2 δu 2  P3δu 3  ...  Pn δu n   Pi δu i

(6.18)

i=1

Remark that the adjective “corresponding” simply means ui is the virtual displacement or rotation in the direction of the real force Pi and at the location where it is applied.

Reference configuration

A

u

P A

Current configuration P

Figure 6.5: Schematic of a body at a particular instant subjected to the virtual force P Now, let us introduce another different thought process. As the body occupies its particular configuration shown in Figure 6.5, we again imagine that the clock (or time) is stopped in our imagination and the body is completely freezed. Next, we assume that the body is subjected to a fictitious force P at point A in addition to the real force P. This force is not real but only applied to the body in our thought and termed the virtual force. Let u be the real displacement of the point A in the direction of the virtual force P at the instant when the time is stopped. The complimentary work due to the displacement u (measured from any selected reference configuration) through the virtual force P in our imagination is termed the complimentary virtual work and it is equal to WC  uδP


309



If, in our thought, the virtual moment M is applied to the point A instead of P, the complimentary virtual work can be computed from WC  δM

(6.20)

where  is the rotation at point A (measured from the reference configuration) at the instant when the time is stopped. Note that the complimentary virtual work can be either positive or negative depending on the direction of a pair {u, P} or a pair {, M}; more specifically, if they have the same directions, the computed WC is positive, otherwise, it is negative. For a more general case that the body is subjected to a set of virtual loads (virtual forces or virtual moments) {P1, P2, P3, …, Pn} at any particular instant, the total complimentary work of the real movement acting through these virtual loads is equal to n

WC  u1δP1  P2 δP2  P3δP3  ...  u n δPn   u i δPi

(6.21)

i=1

where ui is the real displacement or the real rotation at the location and in the direction of the virtual load Pi.

6.3 Strain Energy and Complimentary Strain Energy In this section, the concept of the strain energy and its compliment known as the complimentary strain energy is introduced for treatment of deformable bodies (bodies that can undergo the change in shape, dilatation, and distortion). These two quantities are single parameters providing a measure of the extent of deformation of a body as a whole and, more importantly, they involve in several work and energy theorems state further below. The strain energy is defined as the potential energy stored in the body that undergoes the deformation (e.g. change in shape, distortion, dilatation, etc.). As evident from this definition, the strain energy is a relative quantity and can be computed once the reference or datum state is chosen (this reference state is in fact used to compute the deformation of the body). Within the context of structural analysis, the stress-free state (or, equivalently, the deformation-free state) is generally chosen as the reference state. For simplicity in deriving the general expression of the strain energy, this quantity can be viewed as the work-done of the internal forces (or stress) through the deformation (or strain). First, let us consider an infinitesimal prism element dxdydz at point x = (x, y, z) within the body as shown in Figure 6.6 and also define  = {xx, yy,zz,xy,yz,xz}T and  = {xx, yy, zz, xy, yz, xz}T as the state of stress and state of strain at point x at any instance. Now, consider a typical face of the element, say the face dydz, with the normal and shear components of stresses and strains shown in Figure 6.7. The incremental strain energy dUyz of the element dxdydz due to the incremental of strain components dxx and dxy is equal to the incremental work-done of the forces Fxx = xxdydz and Fxy = xydydz due to the incremental elongation exx = dxxdx and the incremental shear angle xy = dxydx, i.e.

dU yz  Fxx de xx  Fxy dxy   σ xx dε xx  σ xy dγ xy  dxdydz

(6.22)

The incremental strain energy dUxy of the element dxdydz due to the incremental of strain components dzz and dxz and the incremental strain energy dUxz of the element dxdydz due to the incremental of strain components dyy and dyz can readily be computed in the same fashion and this leads to Copyright © 2011 J. Rungamornrat

310



z zz yz yz

xz xz x

xy

dz xx

yy

xy

y

dx

dy x Figure 6.6: Schematic of an infinitesimal prism element dxdydz at point x within the body

dxydx xydx

xy dy

Fxxxxdydz

Fxx

Fxyxydydz

Fxy xy

dx

xxdx dxxdx

xydx

dx

Figure 6.7: Schematic of deformation of a typical face dydz of the infinitesimal element dxdydz dU xy  Fzz de zz  Fxz dxz   σ zz dε zz  σ xz dγ xz  dxdydz

(6.23)

dU xz  Fyy de yy  Fyz d yz   σ yy dε yy  σ yz dγ yz  dxdydz

(6.24)

The total incremental strain energy dU of the element dxdydz due to the incremental strain d = {dxx, dyy, dzz, dxy, dyz, dxz}T is given by dU  σ xx dε xx  σ yydε yy  σ zz dε zz  σ xy dγ xy  σ yzdγ yz  σ zx dγ zx  dxdydz

(6.25)

The strain energy U stored in the body from the reference state to the state that it achieves the final strain o = {xxo, yyo, zzo, xyo, yzo, xzo} T is obtained by first integrating (6.25) from  = 0 to  = o and then integrating the result over the entire volume of the body. The final expression is given by εo

εo

U     σ xx dε xx  σ yy dε yy  σ zz dε zz  σ xy dγ xy  σ yz dγ yz  σ zx dγ zx  dV    σ T dε dV V 0

V 0


(6.26)

311



where T denotes the transpose of . Note that the strain energy of the entire body can also be expressed in another concise form: U   U dV

(6.27)

V

where U , commonly termed the strain energy density function (also known as the modulus of resilience or the strain energy per unit volume), is defined by εo

εo

0

0

U    σ xx dε xx  σ yy dε yy  σ zz dε zz  σ xy dγ xy  σ yz dγ yz  σ zx dγ zx    σ T dε

(6.28)

For the special case of linearly elastic materials (i.e the relationship between the stress  and strain  is linear), the above integral can be directly integrated and the strain energy density function U takes a simple form

U

1 σ xxoε xxo  σ yyoε yyo  σ zzoε zzo  σ xyoγ xyo  σ yzoγ yzo  σ zxo γ zxo  2

(6.29)

where o = {xxo, yyo, zzo, xyo, yzo, xzo}T is the final stress. The strain energy for this particular case is then given by U

1 σ xxoε xxo  σ yyoε yyo  σ zzoε zzo  σ xyo γ xyo  σ yzoγ yzo  σ zxo γ zxo  dV 2 V

(6.30)

It is worth noting that for a body undergoing rigid body displacement (i.e. combination of rigid translations and rigid rotations), the strain field within the body and therefore the corresponding strain energy vanish. Similar to the complimentary work, the complimentary strain energy of a body can be introduced in a similar fashion as follows. First, the complimentary strain energy density function, denoted by U C , is defined by the following differential relation dU C  d(σ T ε)  dU  ε T dσ  ε xx dσ xx  ε yy dσ yy  ε zz dσ zz  γ xy dσ xy  γ yz dσ yz  γ zx dσ zx

(6.31)

By integrating (6.31) from the reference or stress-free state to the final state, we then obtain the explicit expression for the complimentary strain energy density function: σ0

U C   ε dσ  T

0

σ0

 ε

xx

dσ xx  ε yy dσ yy  ε zz dσ zz  γ xy dσ xy  γ yz dσ yz  γ zx dσ zx 

(6.32)

0

Now, the complimentary strain energy of the entire body is simply defined in terms of the integral of the complimentary strain energy density function over the entire volume of the body, and this leads to σ0

U C   U C dV     ε xx dσ xx  ε yy dσ yy  ε zz dσ zz  γ xy dσ xy  γ yz dσ yz  γ zx dσ zx  dV V

V 0


(6.33)

312



For the special case of linear elasticity, the complimentary strain energy density function and the complimentary strain energy of the body are obviously identical to the strain energy density function and the strain energy of the body, i.e. UC  U 

1  σ xxoε xxo  σ yyoε yyo  σzzoε zzo  σ xyo γ xyo  σ yzo γ yzo  σzxo γzxo  2

(6.34)

UC  U 

1  σ xxoε xxo  σ yyoε yyo  σzzoε zzo  σ xyo γ xyo  σ yzo γ yzo  σzxo γ zxo  dV 2 V

(6.35)

where o = {xxo, yyo, zzo, xyo, yzo, xzo}T and o = {xxo, yyo, zzo, xyo, yzo, xzo}T represent the stress and strain at the final state, respectively.

6.4 Virtual Strain Energy and Complimentary Virtual Strain Energy In this section, we generalize the concept of strain energy and complimentary strain energy to introduce another two quantities: the virtual strain energy and the complimentary virtual strain energy. Similar to the virtual work and complimentary virtual work, these two quantities are not associated with the real process but instead corresponds to the thought (or virtual) process. As will become more apparent later, the virtual strain energy and complimentary strain energy are essential ingredients in the development of two fundamental theorems: the principle of virtual work and the principle of complimentary virtual work.


, 

, 


Virtual movement Figure 6.8: Schematic of body at particular instant that undergoes virtual movement Consider a particular instant of a body as shown schematically in Figure 6.8. The stress field and the strain field within the body at this state are denoted by  = {xx, yy,zz,xy,yz,xz}T and  = {xx, yy, zz, xy, yz, xz}T, respectively. Now, let us imagine that the following process takes place in our imagination. The process begins with the body being freezed at that particular instant (the clock or time is virtually stopped) and this therefore renders the real stress state  and real strain state  unchanged throughout the subsequent thought process. Next, we assume that the body undergoes a virtual movement and it achieves a new configuration in our imagination. This movement is accompanied by an additional strain field  = {xx, yy, zz, xy, yz, xz}T termed the virtual strain. The symbol ‘’ and the adjective ‘virtual’ are used only to emphasize the virtual nature of those quantities. Now, the virtual strain energy, denoted by U, is defined as the additional strain energy stored in the body due to such virtual movement or, equivalently, the workdone of the real stress through the virtual strain , i.e. Copyright © 2011 J. Rungamornrat

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U    σ xx ε xx  σ yy ε yy  σ zz ε zz  σ xy γ xy  σ yz γ yz  σ zx γ zx  dV   σ T ε dV V

(6.36)

V

It is worth noting that the virtual strain energy involves only a single integral over the entire volume of a body. This is due primarily to the fact that during the virtual process, the real stress state  remains unchanged and the virtual strain energy density function simply reduces to T. For a special case that the virtual movement in our imagination is a rigid body displacement, the accompanied virtual strain  identically vanishes and this therefore produces zero virtual strain energy, i.e. U = 0. Another crucial remark is associated with the arbitrary nature of the virtual strain; this feature results directly from the arbitrariness of the virtual movement.

, 

Actual state

, 

Virtual state associated with virtual loadings

Figure 6.9: Schematic of body at particular instant that is subjected to virtual loadings Next, let us consider another though process. Once the body is freezed at a particular instant of interest (the actual stress state  and the actual strain state  are held fixed), we assume that the body is subjected to additional virtual loadings in our imagination and this produces an additional stress field  called the virtual stress (see Figure 6.9). The new state of the body is commonly termed the virtual state. Now, the complimentary virtual strain energy UC is defined as the additional complimentary strain energy stored in the body due to the introduction of the virtual stress or, equivalently, the complimentary work of the real strain  through the virtual stress , i.e. U C    ε xx σ xx  ε yy σ yy  ε zz σ zz  γ xy σ xy  γ yz σ yz  γ zx σ zx  dV   ε T σ dV V

(6.37)

V

It is emphasized that during this virtual process, the real strain  remains unchanged and the complimentary virtual strain energy density simply reduces to T. This is significantly different from the expression of the complimentary strain energy density that involves the integration over a continuous process (see the relation (6.32)). Similar to the virtual strain, the virtual stress field  is also arbitrary in nature in that a set of virtual loadings producing  can be chosen arbitrary. As a final remark, the virtual strain energy and the complimentary virtual strain energy involve either the actual stress state or the actual strain state only at the particular instant of interest (an instant when the time is stopped) while the strain energy and the complimentary strain energy involve either the actual stress state or actual strain state throughout the continuous process.

6.5 Conservation of Work and Energy While a law of conservation of work and energy is fundamental and well-known in classical physics, its applications in the field of structural mechanics have commonly been found. This law, Copyright © 2011 J. Rungamornrat

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when specialized to a mechanical system (i.e. a system involving only mechanical energies such as the potential and kinetic energies), states that the work-done to the system due to the applied forces simply interchanges to the kinetic energy and potential energy of the system or, mathematically, U1  K1  W  U 2  K 2

(6.38)

where U1 and K1 are the potential energy and the kinetic energy of the system at a state 1, U2 and K2 are the potential energy and the kinetic energy of the system at a state 2, and W is the total mechanical work-done to the system from the state 1 to the state 2. Let us focus attention further to a system of a deformable body (i.e. a body that can undergo deformation such as dilatation and distortion) with all involved mechanical processes occurring in a very slow or time-independent fashion (i.e. the velocity and acceleration of the body resulting from the processes are infinitesimally small and can be neglected). For this particular system (sometimes called a quasi-static system), the kinetic energy is relatively small and commonly discarded in the calculation, and the potential energy stored in the body resulting from the deformation is given in terms of the strain energy. By choosing a reference state or undeformed state of the body where it is free of deformation and internal force as the state 1, the conservation of work and energy (6.38) now becomes WU

(6.39)

where U is the strain energy of the body at a new state or deformed state and W is the total workdone due to applied forces from the reference state to the deformed state. To clearly demonstrate the meaning of (6.39), let us consider a deformable body shown in Figure 6.10. A set of external loads {P1, P2,…, Pn} is applied continuously to the body in a quasi-static manner and, during this continuous process, it produces a stress field  = {xx, yy,zz,xy,yz,xz}T. Let u0 = {u10, u20,…, un0} and o = {xxo, yyo, zzo, xyo, yzo, xzo} T be the corresponding movements and the strain field at the final deformed state (of interest) of the body. The conservation of work and energy (6.39) leads to the following relation W

uo

εo

n

 Pi du i    σ T dε dV  U



i 1

0

(6.40)

V 0

Equation (6.39) or (6.40) is sometimes called the real work equation. Applications of this equation to the deformation and displacement analysis of structures are extensively discussed in the next chapter. P2 = 0

P3 = 0 Pn = 0

  u

Undeformed state

P2 = P20

P2

P3 P1 = 0 Pn

  u

State during process

P3 = P30 P1 Pn = Pn0

0 0 uu0

Deformed state

Figure 6.10: Schematic of deformable body subjected to a set of applied loads Copyright © 2011 J. Rungamornrat

P1 = P10

315



6.6 Principle of Virtual Work (PVW) The principle of virtual work (PVW), also known as the principle of virtual displacement (PVD), has a very long history; its applications were found in the study of rigid bodies since the era of Jean Bernoulli (1667-1748) and Daniel Bernoulli (1700-1782). Nowadays, the principle of virtual work has been broadened its applications to various fields of mechanics and approximation theories (e.g. structural mechanics, solid mechanics, finite element methods, etc.). When applied to deformable bodies such as those commonly used to model engineering structures (e.g. trusses, beams, frames, etc.), this principle provides a means to set up an alternative but equivalent set of equilibrium equations (relating external forces and induced internal forces) in a form attractive for structural modeling. Here, we only outline the principle, without a formal proof, for structural problems with the key objective to get familiar with the principle and ready for uses in structural analysis.

P2


P3




un 

P1

Pi

Pn

Figure 6.11: Schematic of a body at particular instant subjected to virtual movement Consider a deformable body subjected to a set of external loads P = {P1, P2, P3, …, Pn}T as shown in Figure 6.11. Let u and  = {xx, yy,zz,xy,yz,zx}T be the displacement and stress at any point within the structure caused by the applied loads P and the configuration of the body surrounded by the solid line be the current or deformed configuration at the instant that P achieves the value {P1, P2, P3, …, Pn}T. Now, let imagine that the body undergoes arbitrary virtual displacement field u from such actual configuration and this takes the body to a new configuration, called the virtual configuration, in our imagination. This virtual displacement u produces a virtual strain field  = {xx, yy,zz,xy,yz,zx}T within the body. During this thought process, the virtual work W of the external loads P = {P1, P2, P3, …, T Pn} due to the virtual displacement field u is given by n

δW  P1δu1  P2 δu 2  ...  Pn δu n   Pi δu i

(6.41)

i 1

where {u1, u2, u3, …, un} is a set of virtual displacements at the location and in the direction of the applied loads {P1, P2, P3, …, Pn}. The virtual work W is usually termed the external virtual work to emphasize that it is associated with the external loads. It is important to emphasize that the expression (6.41) is valid with no regard to the actual relationship between the external loads {P1, P2, P3, …, Pn} and the corresponding actual displacements {u1, u2, u3, …, un}. The virtual strain energy of the actual stress field  = yy,yy,zz,xy,yz,zx}T due to the virtual strain field  = {xx, yy,zz,xy,yz,zx}T is given by δU    σ xx δε xx  σ yy δε yy  σ zz δε zz  σ xy δγ xy  σ yz δγ yz  σ zx δγ zx  dV   σ T δεdV V

V


(6.42)

316



The virtual strain energy is sometimes called the internal virtual work. Similar to the case of the external virtual work, the expression (6.42) is valid with no regard to the behavior of materials constituting the body. The principle of virtual work states that the configuration of the body is an equilibrium configuration (i.e. the external applied loads P = {P1, P2, P3, …, Pn}T are in equilibrium with the stress field ) if and only if the external virtual work is equal to the virtual strain energy for any admissible choice of the virtual displacement u, i.e. Body is in equilibrium



n

δW   Pi δu i  δU   σ T δεdV i 1

δu

(6.43)

V

The principle ensures the implication in both directions; the implication  indicates the necessary conditions for the body to be in equilibrium while the implication  provides the sufficient conditions to confirm the equilibrium of the body. It is also apparent from this principle that the virtual work equation (i.e. W = U) is in fact an alternative form of the equilibrium equation expressed in terms of the virtual work and the virtual strain energy. Equilibrium of applied loads and the stress field within the body is guaranteed by the satisfaction of the virtual work equation for arbitrary choices of the virtual displacement. The arbitrariness of the virtual displacement is the essential component of the virtual work principle; satisfaction of the virtual work equation only for certain choices of the virtual displacement is insufficient to deduce equilibrium of the body. On the contrary, if it is known a priori that the body is in equilibrium, the virtual work equation is always satisfied no matter what the choice of virtual displacement being chosen. For a special case that the virtual displacement is arbitrary chosen from a space of rigid body motions, the virtual strain energy U vanishes and the virtual work equation simply reduces to δW  0

 rigid body displacement δu

(6.44)

This equation is simply the statement of equilibrium of all external applied loads or, equivalently, the necessary condition that all external applied loads must be satisfied for a body to be in equilibrium. One useful application of Eq. (6.44) is the determination of the support reactions and internal forces of statically determinate structures as demonstrated by following examples. Example 6.1 Consider a statically determinate beam subjected to external loads shown below. Use the principle of virtual work to determine all support reactions and the shear force and bending moment at point B.

q

4qL qL2

B

A L/2

D

C L/2

L/2

L/2

E

Solution Since the body is in equilibrium with all applied loads, the virtual work equation is satisfied for any choice of the virtual displacement. First, let us choose a special virtual displacement associated with the rigid body motion of the beam resulting from removing the roller support and then introducing a unit vertical displacement at point E (as indicated by the virtual displacement 1 in the figure below). It is worth noting that any statically determinate structure become unstable when either external release or internal release is introduced in the structure. Copyright © 2011 J. Rungamornrat

317



q MA

B

A

4qL C

qL2 E

D

RAY

REY 1/L 1/2

B

A

1

C

1

B

A

C

D

1

E

1/2 1/L

Virtual displacement 1


D

E

1

L

B

C

L/2

D

1

A

Virtual displacement 3 E

For this particular choice of the virtual displacement, the virtual strain energy vanishes (i.e. U = 0) while the external virtual work associated with the applied loads and all support reactions is given by 1 1 δW  (4qL)    (qL2 )    R EY (1) 2 L

By applying the virtual work equation, it yields the support reaction at the roller support as follow 1 1 δW  (4qL)    (qL2 )    R EY (1)  0  R EY  qL Upward 2 L

To determine the support reaction RAY at the fixed support, we choose a virtual displacement associated with a rigid body motion of the beam resulting from removing the vertical constraint from the fixed support and then introducing a unit vertical displacement at point A as indicated by the virtual displacement 2 in above figure. Again, by applying the virtual work equation for this choice of the virtual displacement, we obtain 1 1 δW  (q)(L) 1  (4qL)    (qL2 )    R AY (1)  0  R AY  4qL Upward 2 L

Similarly, the moment reaction MA at the fixed support can be obtained by choosing a virtual displacement associated with a rigid body motion of the beam resulting from removing the Copyright © 2011 J. Rungamornrat

318



rotational constraint from the fixed support and then introducing a unit rotation at point A as indicated by the virtual displacement 3. The virtual work equation then results in L δW  (q)(L)  L/2   (4qL)    (qL2 ) 1  M A (1)  0  M A  7qL2 /2 CCW 2

To determine the shear force and bending moment at point B of the beam, we first introduce a fictitious cut at that cross section and this allows us to see those two internal forces (see figure below). In this sense, the shear force VB and the bending moment MB can be viewed as external loads applied to the end of the segments resulting from the cut. To obtain the shear force VB, we choose a special virtual displacement associated with the rigid body motion of the beam resulting from removing the shear constraint at point B and then introducing a unit separation as indicated by the virtual displacement 4 below. Note that removal of the internal shear constraint is readily achieved by putting the shear release at that point. By writing the virtual work equation for this particular virtual rigid body motion, we then obtain 1 1 δW  (q)(L/2) 1  (4qL)    (qL2 )    VB (1)  0  VB  7qL/2 2 L

Similarly, by choosing a virtual displacement to be the rigid body motion of the beam resulting from removing the moment constraint at point B and then introducing a unit relative rotation as indicated by the virtual displacement 5 below, we can readily obtain the bending moment MB from the virtual work equation as follows: L 1 δW  (q)(L/2)(L/4)  (4qL)    (qL2 )    M B (1)  0  M B  13qL2 /8 4 2

q MA

A RAY

MB B VB

4qL

MB B

C

VB

qL2 E

D

REY

A

1

1

B

C

1/2 1/L


D

E A

B

1

C L/2

D L/4


1/2

E Copyright © 2011 J. Rungamornrat

319



Example 6.2 Consider a statically determinate frame subjected to external loads shown below. Use the principle of virtual work to determine all support reactions and the internal forces at point D.

4P C

B

D

P

E

L F L A L

L

Solution To determine all support reactions of the given frame (i.e. RAX, RAY, and RFY), three choices of the virtual displacement are chosen as shown in the figure below. Each special virtual displacement is simply obtained by removing the movement constraint in the direction that the support reaction to be computed and then introducing a unit movement in that direction. It should be

4P P

E

D

C

B

C

B

D

F

RAX

F 1

RFY

A

E

A 1

RAY

B

C

Virtual displacement 1 1

1 D

1/2

E

C

B

D

E

1/2 F 1

F A 1

1 A Virtual displacement 3

Virtual displacement 2 Copyright © 2011 J. Rungamornrat

320



noted that these virtual displacements are purely rigid body movements since the original statically determinate structure becomes statically unstable once one constraint was removed. As a result, the virtual strain energy associated with such virtual displacement identically vanishes (i.e. U = 0). By applying the virtual displacement 1 to set up the virtual work equation, it yields the reaction RAX as follows: Virtual displacement 1: δW  P 1  (4P)  0   R AX (1)  δU  0  R AX  P Leftward Similarly, by considering the virtual displacement 2 and the virtual displacement 3, it leads to the reactions RAY, and RFY, respectively, as follows: Virtual displacement 2: δW  P 1  (4P) 1/ 2   R AY (1)  δU  0  R AY  P Upward Virtual displacement 3: δW  P 1  (4P) 1/ 2   R FY (1)  δU  0  R FY  3P Upward

C

B

1

MD MD

4P

D VD

FD

D VD

E

P

B

C

D

D

F

F

RAX

1

RFY

A

E

A


RAY

1

1 1/2 B

C

L/2

D 1

B

E

L/4 C 1

D

E D F

F

A

A


Virtual displacement 5 Copyright © 2011 J. Rungamornrat

321



To determine the internal forces at point D (i.e. the axial force FD, the shear force VD, and the bending moment MD), we first introduce a fictitious cut at that cross section and this allows us to access to those internal forces (see figure below). In the other word, the axial force FD, the shear force VD, and the bending moment MD can now be regarded as external loads applied to the end of the segments resulting from the cut. By using the same idea as demonstrated in Example 6.1, the virtual displacements used in the calculation of the internal forces are simply obtained by removing the internal constraint associated with the internal force to be computed and then introducing a unit movement along the released direction (see above figures). For instance, the virtual displacement resulting from removing the axial constraint at point D and introducing a unit longitudinal separation, as indicated by the virtual displacement 4, is utilized for computing FD. Again, these chosen virtual displacements are purely rigid body movements and, therefore, produces zero virtual strain energy. By applying the virtual displacement 4, the virtual displacement 5, and the virtual displacement 6 to set up three virtual work equations, the internal forces FD, VD, and MD can readily be solved as follows: Virtual displacement 4: δW  P 1  (4P)  0   FD (1)  δU  0  FD  P Virtual displacement 5: δW  P 1  (4P) 1/ 2   VD (1)  δU  0  VD  3P Virtual displacement 6: δW  P  L / 2   (4P)  L / 4   M D (1)  δU  0  M D  3PL/2

6.7 Principle of Complimentary Virtual Work (PCVW) The principle of complementary virtual work (PCVW) is a fundamental principle that comes along with the principle of virtual work and its vast applications have been found in the deformation and displacement analysis of structures. Due to its fundamental concept which will become more apparent later, this principle is also known as the principle of virtual force (PVF). Unlike the principle of virtual work, PCVW aims to establish the correspondence between the compatibility condition of the displacement and deformation of a structure and a balance of the complimentary virtual work and the complimentary virtual strain energy of the system. This principle mainly provides an alternative to derive the compatibility condition without the direct geometric consideration. The direct application of PCVW to the displacement/deformation analysis of structures and analysis of statically indeterminate structures can be found in Chapter 8 and Chapter 10, respectively. Here, the PCVW is only outlined without the formal proof and involving terms are clearly stated. P2

P3  u

P3

P2

Pi

P1

 P1 Pn

Pn (a)

(b)

Figure 6.12: (a) Schematic of actual structure and (b) Schematic of virtual structure Copyright © 2011 J. Rungamornrat

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Consider a deformable body subjected to a set of applied loads P = {P1, P2, P3,…, Pn}T. Let u = {u1, u2, u3}T and  = {xx, yy,zz,xy,yz,zx}T be the displacement and strain at any point within the structure resulting from the applied loads P. This given structure, as indicated in Figure 6.12(a), is termed as the “actual system” or the “actual structure” and the displacement u and the strain  associated with this actual structure are termed as the “actual displacement” and the “actual strain”, respectively. Now, let us imagine (in our thought process) that there is another structure possessing an identical geometry to the actual system but subjected to a set of applied loads P = {P1, P2, P3, …, Pm}T as shown schematically in Figure 6.12(b). This fictitious system which exists only in our thought process is termed as the “virtual system” or the “virtual structure”. The applied loads P = {P1, P2, P3, …, Pm}T is also termed as the “virtual applied loads” and the corresponding stress field  = {xx, yy,zz,xy,yz,zx}T that is in equilibrium with the virtual applied loads is termed as the “virtual stress”. It is important to emphasize that the virtual applied loads are arbitrary in the sense that their magnitude and direction and locations of their application can be arbitrary and are not necessarily identical to the actual applied loads P. In addition, we denote {u1, u2, u3,…, um}T as the actual displacement of the actual structure at the same locations and in the same directions of the virtual applied loads. Now, let’s imagine that the deformed actual structure is subjected further to a set of virtual applied loads P and this introduces an additional stress equal to the virtual stress . This fictitious process occurs only in our imagination and is generally termed as the “virtual process”. Apparently, the virtual system represents, in fact, the actual structure undergoing the virtual process in our imagination. It should also be addressed that throughout the virtual process, the displacement field u and the strain field  of the actual structure remain fixed. The complementary work done by the actual displacement {u1, u2, u3, …, um}T during the virtual process is denoted by WC and commonly termed as the “external complementary virtual work”. Since the actual displacement of the actual structure remains fixed during the virtual process, the external complementary virtual work WC can readily be computed from δWC



u1δP1  u 2δP2  ...  u m δPm



m

 u δP i 1

i

(6.45)

i

It is important to emphasize that validity of the expression (6.45) is independent of the relationship between the actual displacement {u1, u2, u3, …, um}T and the actual applied loads P = {P1, P2, P3, …, Pn}T. The complementary strain energy introduced during the virtual process is equal to the complementary work done by the actual strain  = {xx, yy, zz, xy, yz, zx}T in the actual structure through the virtual stress  = {xx, yy,zz,xy,yz,zx}T developed in the virtual structure and is commonly termed as the “complementary virtual strain energy” or the “internal complementary virtual work”. Since the actual strain in the actual structure does not alter during the virtual process, the complementary virtual strain energy, denoted by UC, can simply be computed from δU C



 ε

xx

δσ xx  ε yyδσ yy  ε zz δσ zz  γ xy δσ xy  γ yzδσ yz  γ zx δσ zx  dV

V

  ε T δσdV

(6.46)

V

Similar to the case of the external complementary virtual work, validity of the expression (6.46) is independent of the constitutive model. The principle of complementary virtual work states that if the actual displacement field u and the strain field  are compatible (i.e. there is no gap, overlapping, or undesirable discontinuity


323



within the structure), the external complementary virtual work is equal to the complementary virtual strain energy for any choice of virtual applied loads P, i.e. Displacement u and strain  are compatible



m

WC   u i δPi  δU C   ε T δσdV P (6.47) i 1

V

Equation (6.47) is known as the “complementary virtual work equation”. Conversely, if the complementary virtual work equation is satisfied for any choice of the virtual loads (i.e., δWC  δU C δP ), the displacement u and the strain field  are compatible. The complementary virtual work equation is in fact an alternative form of the compatibility condition expressed in terms of complementary virtual work and complementary virtual strain energy. Compatibility between the displacement and the strain within the actual structure is guaranteed by the satisfaction of the complementary virtual work equation for arbitrary choices of the virtual applied loads. Such arbitrariness feature of the virtual applied loads is a necessary ingredient; satisfaction of the complementary virtual work equation only for certain choices of the virtual applied loads does not imply compatibility between the displacement and the strain. However, if the displacement and strain for a given structure are compatible, the complementary virtual work equation is always satisfied for any particular choice of virtual applied loads.

6.8 Principle of Stationary Total Potential Energy (PSTPE) The principle of stationary total potential energy is a fundamental variational principle well-known in the area of structural mechanics. This principle provides an alternative to formulate a boundary value problem (i.e., the governing equation and a set of boundary conditions) in terms of work- and energy-related quantities. The positive feature of such formulation stems primarily from its form well-suited for constructing approximate solutions. Furthermore, the principle can also be used to derive a set of governing differential equations and boundary conditions of a problem in which the total potential energy is clearly defined. Here, we first state the principle and then provide a formal proof by the principle of virtual work. P2

P3 Admissible displacement uˆ 

P1

Pi

Pn Figure 6.13: Schematic of a body subjected to a set of applied loads Let us consider a deformable body that is subjected to a set of applied loads P = {P1, P2, P3,…, Pn}T and properly constrained against the rigid body motion by supports at certain locations as shown schematically in Figure 6.13. The material constituting the body is assumed to be elastic with the strain energy density function being denoted by U  U(ε) (i.e. the stress-strain relation is given by σ  U(ε )/ε ) and the displacement u and the strain  are related through the kinematics. Any displacement uˆ is said to be kinematically admissible if it is sufficiently smooth (e.g. no gap, no overlapping, and no undesirable discontinuity) and also satisfies the essential boundary conditions at all supports (i.e., satisfies the prescribed displacement-like boundary conditions). The Copyright © 2011 J. Rungamornrat

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kinematically admissible displacement uˆ is said to be the solution or the correct displacement of the body if the deformation generated by this displacement also produces the internal force that is in equilibrium with the applied loads P. Next, let us define the total potential energy functional of the body, denoted by , for any kinematically admissible displacement uˆ by  (uˆ )  U(uˆ )  W (uˆ )

(6.48)

where U and W are strain energy functional and load potential functional defined in terms of any kinematically admissible displacement uˆ by U(uˆ )   U (ε(uˆ ))dV

(6.49)

V

n

W (uˆ )   Pi uˆ i (uˆ )

(6.50)

i 1

where the function form ε(uˆ ) depends primarily on the kinematics and uˆ i (uˆ ) is the displacement at the location and in the direction of the applied load Pi resulting from the kinematically admissible displacement uˆ . The principle of stationary total potential energy states that a kinematically admissible displacement uˆ is the correct displacement if and only if the total potential energy is stationary at uˆ , i.e.,  (uˆ )  0 . Proof (  ) Assume uˆ is the correct displacement of a body. Let uˆ be arbitrary but sufficiently smooth displacement that satisfies the homogeneous essential boundary conditions at all supports. It is obvious that the displacement uˆ  uˆ is arbitrary and also kinematically admissible (i.e., satisfies the prescribed essential boundary conditions at all supports). The total potential energy associated with uˆ  uˆ is given by  (uˆ  uˆ )  U(uˆ  uˆ )  W (uˆ  uˆ )

(6.51)

By expanding the strain energy functional U about the correct displacement uˆ , we obtain U (uˆ  uˆ )  U(uˆ )  U (uˆ ; uˆ )

(6.52)

where U (uˆ ; uˆ ) is the difference between the strain energy associated with uˆ  uˆ and uˆ and it can be expressed in terms of the first variation U and its higher order variations  2 U , 3 U , … as U (uˆ ; uˆ )  U (uˆ ; uˆ )   2 U (uˆ ; uˆ )  3 U (uˆ ; uˆ )  ...

(6.53)

From the definition of the strain energy U, the first variation U(uˆ ; uˆ ) is obtained as U (ε(uˆ )) ε(uˆ )dV   σ (ε(uˆ ))ε(uˆ )dV ε  V V

U (uˆ ; uˆ )  

(6.54)

Similarly, by expanding the load potential functional W about the correct displacement uˆ , we obtain W (uˆ  uˆ )  W (uˆ )  W (uˆ ; uˆ )

(6.55)

where W (uˆ ; uˆ )  W (uˆ ; uˆ )   2W (uˆ ; uˆ )  3W (uˆ ; uˆ )  ... Copyright © 2011 J. Rungamornrat

(6.56)

325



n

W (uˆ ; uˆ )   Pi uˆ i (uˆ )

(6.57)

i 1

Upon utilizing (6.51), (6.52) and (6.55), it leads to the expansion of the total potential energy about the solution uˆ :  (uˆ  uˆ )  U (uˆ  uˆ )  W (uˆ  uˆ )

(6.58)

where U (uˆ  uˆ )  U(uˆ )  U (uˆ ; uˆ ) ( n ) (uˆ ; uˆ )  ( n ) U (uˆ ; uˆ )  ( n ) W (uˆ ; uˆ )

(6.59) (6.60)

n  1

Since uˆ is the correct displacement, the stress generated by uˆ , i.e. σ (ε(uˆ )) , must be in equilibrium with the applied loads P. From the principle of virtual work, it can then be concluded that the first variation  vanishes for every uˆ , i.e. n

 (uˆ ; uˆ )   σ (ε (uˆ ))ε(uˆ )dV   Pi uˆ i (uˆ )  0 V

uˆ

(6.61)

i 1

The condition (6.61) implies the stationary of  at uˆ . (  ) Assume uˆ is a kinematically admissible displacement that makes the total potential energy functional stationary, i.e. the condition (6.61) is satisfied. From the principle of virtual work, the condition (6.61) implied that the stress generated by the displacement uˆ , i.e. σ (ε(uˆ )) , is in equilibrium with the applied loads P. Therefore, uˆ must be the correct displacement of the body. Various applications of the PSTPE have been found in the area of structural and solid mechanics. The principle along with the calculus of variation provides an alternative to derive the governing differential equation of a boundary value problem. In addition, it can also be used directly to obtain the formulation well-suited for constructing its approximate solution. Applications of the PSTPE in the buckling analysis can be found in the work of Watcharakorn et al. (2009), Watcharakorn (2010), and Watcharakorn and Rungamornrat (2011). Within the context of linear structural analysis (i.e. all governing equations are in a linear form), the principle of stationary total potential energy can further be replaced by a stronger theorem known as the principle of minimum total potential energy (proof is left as an exercise). This theorem precisely states that for all kinematically admissible displacements, one that minimizes the total potential energy functional (6.48) is the correct displacement of a body. To clearly demonstrate applications of the PSTPE, two examples, one associated with the calculation of the displacement of a single degree of freedom structure and the other corresponding to finding an approximate solution of a beam under uniformly distributed load, are presented. Example 6.3 Consider a three-member truss subjected to a vertical load P shown below. The crosssectional area A and Young modulus E are assumed constant. Use the principle of stationary total potential energy to obtain the relation between the load P and the vertical displacement at point A.

L

B

L C

D

o 45o 45

A P Copyright © 2011 J. Rungamornrat

L

326



Solution From the symmetry of geometry and loading condition, the horizontal component of the displacement at point A vanishes. Since all members (i.e., members AB, AC, and AD) are axial members, the displacement at any point of the member can therefore be fully described by the vertical displacement at point A, denoted by . Elongation, axial strain, and stress of the members AB, AC, and AD can be computed in terms of the displacement  as follows. e AB  e AD   cos 45o   / 2

L

 AB   AD  ( / 2 ) / 2L   / 2L

B

L C

 AB   AD  E / 2L

D

o 45o 45

e AC  

eAD

L

A 

 AC   / L   AC  E / L

By using the formula (6.29) and (6.30) along with the fact that the stress and strain states within the axial member are uniform, we then obtain the strain energy for each member as

 

1  E  Δ  U AB  U AD    A  2L  2  2L  2L 

2EAΔ 2 8L

1  E  Δ  EAΔ 2 ; U AC    A L   2  L  L  2L

The strain energy of the entire structure is then given by U  U AB  U AC  U AD 

(2  2 )EAΔ 2 4L

The load potential can readily be obtained by W  PΔ

The total potential energy of the system now becomes (2  2 )EAΔ 2  UW   PΔ 4L

From the PSTPE,  is the correct displacement of this structure if and only if  

 (2  2 )EAΔ  (2  2 )EAΔΔ  PΔ    P Δ  0 2L 2L  

Due to the arbitrariness of , it finally leads to the relationship between the displacement  and the load P: P

(2  2 )EAΔ 2L Copyright © 2011 J. Rungamornrat

327



Example 6.4 Consider a cantilever beam of length L and constant flexural rigidity EI and subjected to a uniformly distributed load q0 as shown in the figure. Apply the PSTPE to approximate the deflection at any point of this particular beam due to bending effect by assuming the solution form in terms of (i) a cubic polynomial function and (ii) a complete polynomial function of the fourth degree.

Y q0 X

EI L

Solution For any given deflection v = v(x), the curvature , the flexural strain  and the flexural stress  within the beam can readily be obtained as follows:   v(x)

,

   y   yv(x) ,   E  Eyv(x)

where y is the distance from the neutral axis. From the formula (6.29) and (6.30), the strain energy stored in the beam due to the deflection v = v(x) is given by L

L

L

1 1 1 2 U    σεdAdx    ( Eyv(x))( yv(x))dAdx    Ey 2 v(x) dAdx 2 2 20A 0A 0A 

L 1  2  2 E  y dA v(x) dx  2 0 A 



1 2 EIv(x) dx  20

L

The load potential for uniformly distributed load is given by L

W    q 0 v( x) dx 0

The total potential energy now becomes L

 UW 

L

1 2 EIv(x) dx   q 0 v( x) dx  20 0

(e6.4.1)

From the PSTPE, the deflection v = v(x) is a solution of the problem if an only if it is kinematically admissible and renders the total potential energy stationary, i.e.   0 . To obtain the exact solution for v = v(x), it requires the knowledge of calculus of variation and this is beyond the scope of this text. To construct an approximate solution, we may search the best solution in a smaller set of kinematically admissible deflections by assuming the function form of v. This particular strategy changes the problem of finding the unknown function to a much simpler problem of finding a finite number of unknown scalars. Copyright © 2011 J. Rungamornrat

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(i) Let the deflection v assume a form v(x)  c 0  c1x  c 2 x 2  c3 x 3

where c0, c1, c2 and c3 are constants. Since the deflection v must be kinematically admissible, it must satisfy the essential boundary conditions, i.e. v(0) = v′(0) = 0. These conditions provide the constraint on the four constants {c0, c1, c2, c3}; only two of them are independent and can be varied arbitrarily. By enforcing such two constraints, we then obtain v(0)  0  c0  c1 (0)  c 2 (0) 2  c3 (0)3



v(0)  0  c1  2c 2 (0)  3c3 (0) 2

c1  0



c0  0

Thus, the most general form of kinematically admissible deflections containing polynomials of degree less than or equal to three (for this particular beam) is v(x)  c 2 x 2  c3 x 3

(e6.4.2)

By substituting (e6.4.2) into (e6.4.1), we then obtain the total potential energy of the beam as L



L





1 q 0 L3 q 0 L4 2 2 3 2 2 3 2   EI 2c  6 c x dx  q c x  c x dx  2 EILc  6 EIL c c  6 EIL c  c  c3 2 2 3 3 2 2 3 3 0 0 2 2 0 3 4

The best choice of c2 and c3 is one that renders the total potential energy stationary, i.e.

 

Π Π c 2  c3  0 c 2 c3

Since c2 and c3 are arbitrary, it implies that

Π q L3  4EILc2  6EIL2c3  0  0 c 2 3

(e6.4.3)

Π q L4  6EIL2c 2  12EIL3c3  0  0 c3 4

(e6.4.4)

By solving (e6.4.3) and (e6.4.4) simultaneously, it leads to

5q 0 L2 c2  24EI

and

c3  

q0L 12EI

The best approximate deflection of the beam based on a family of cubic polynomial functions is finally obtained by substituting the value of c2 and c3 (e6.4.2) and this yields q 0 L4   x  x 5    2   24EI   L  L 2

v(x) 

3

   Copyright © 2011 J. Rungamornrat

329



(ii) Let the deflection v assume the form v(x)  c0  c1x  c 2 x 2  c3 x 3  c 4 x 4

where c0, c1, c2, c3 and c4 are constants. Again, by enforcing the essential boundary conditions (i.e. v(0) = v′(0) = 0), it leads to c0 = c1 = 0 and the kinematically admissible deflection becomes v(x)  c 2 x 2  c3 x 3  c 4 x 4

(e6.4.5)

By substituting the deflection (e6.4.5) into (e6.4.1), we obtain the total potential energy as L



L







2 1    EI 2c2  6c3 x  12c 4 x 2 dx   q 0 c 2 x 2  c3 x 3  c 4 x 4 dx 20 0

The best choice of c2, c3 and c4 is one that renders the total potential energy stationary, i.e.

 

Π Π Π c 2  c3  c 4  0 c 2 c3 c 4

Since c2, c3 and c4 are arbitrary, the above condition implies that L

L

Π q L3   EI 2c2  6c3 x  12c 4 x 2 (2)dx   q 0 x 2 dx  4EILc2  6EIL2c3  8EIL3c 4  0  0 c2 0 3 0 L





L

Π q L4   EI 2c2  6c3 x  12c4 x 2 (6x)dx   q 0 x 3 dx  6EIL2c 2  12EIL3c3  18EIL4c4  0  0 c3 0 4 0 L



(e6.4.6)



(e6.4.7)

L

Π 144EIL5 q L5   EI 2c2  6c3 x  12c 4 x 2 (12x 2 )dx   q 0 x 4 dx  8EIL3c 2  18EIL4c3  c4  0  0 (e6.4.8) c 4 0 5 5 0





By solving (e6.4.6)-(e6.4.8) simultaneously, we obtain

c2 

q 0 L2 qL , c3   0 4EI 6EI

, c4 

q0 24EIL

The best approximate deflection of the beam based on a family of polynomial functions of the fourth degree is then given by q L4   x  x x v(x)  0 6    4      24EI   L  L L 2

3

4

  

It is evident that the deflection v = v(x) obtained in the case (i) is only an approximate solution whereas that obtained in the case (ii) is identical to the exact solution shown in Example 3.1. This is due to that the space or family of kinematically admissible deflections used in the solution search for the case (ii) contains the exact deflection and it is therefore the best approximate solution. Quality or accuracy of an approximate solution obtained using this particular strategy clearly depends on the assumed solution form. Copyright © 2011 J. Rungamornrat

330



6.9 Principle of Stationary Total Complementary Potential Energy Another important variational principle that comes along with the PSTPE is the principle of stationary total complementary potential energy (PSTCPE). Although this principle is not as wellknown as the PSTPE within the context of modern structural analysis and approximation theories but its significance in this area should be recognized. It is worth noting that the relation between the PSTPE and PSTCPE and that for the PVW and PCVW possesses some similarities. Clearly, the PVW establishes the correspondence between the equilibrium state of a body and the balance of the virtual work and virtual strain energy whereas the PCVW states the equivalence between the compatibility of the displacement and deformation and the balance of the complementary virtual work and the complementary virtual strain energy. As already described in the previous section, the PSTPE arises directly from the PVW and the stationary condition is obviously equivalent to a set of equilibrium equations. As will be apparent below, the PSTCPE states the compatibility condition of the displacement and deformation in terms of the variational formulation involving the complementary work and complementary strain energy. A proof of this principle can be established directly by employing the PCVW. P2

P3 Actual configuration u

P1

Pi

Pn Figure 6.14: Schematic of a body subjected to a set of applied loads Let us consider a deformable body that is subjected to a set of applied loads P = {P1, P2, P3,…, Pn}T and properly constrained against the rigid body motion by supports at certain locations as shown schematically in Figure 6.14. The material constituting the body is assumed to be elastic with the complimentary strain energy density function being denoted by U C  U C (σ ) and the stress-strain relation is therefore governed by the constitutive law ε  U C (σ )/σ . Any stress filed σˆ is said to be statically admissible if its functional form is sufficiently smooth (for mathematical operations such as differentiation can be performed) and it is in equilibrium with the applied loads P. The statically admissible stress filed σˆ is said to be the solution or the correct stress field of the body if the deformation resulting from the strain ε obtained from σˆ through the constitutive relation is compatible with the displacement and essential boundary conditions of the body. Next, let us define the total complimentary potential energy functional of the body, denoted by C, for any statically admissible stress filed σˆ by  C (σˆ )  U C (σˆ )  WC (σˆ )

(6.62)

where U C and WC are complimentary strain energy functional and complimentary load potential functional defined in terms of any statically admissible stress filed σˆ by U C (σˆ )   U C (σˆ )dV

(6.63)

V


331



m

WC (σˆ )   u ri Pˆri (σˆ )

(6.64)

i 1

where Pˆri (σˆ ) are loads (in fact the support reactions) generated by the statically admissible stress filed σˆ at the location and in the direction of the prescribed essential boundary conditions u ri . For the special case that all essential boundary conditions are homogeneous (i.e. no support settlements), the complimentary load potential vanishes, i.e. WC (σˆ )  0 . The principle of stationary total complimentary potential energy states that a statically admissible stress filed σˆ is the correct stress filed if and only if the total complimentary potential energy Cis stationary at σˆ , i.e.,  C (σˆ )  0 . Proof (  ) Assume σˆ is the correct stress field. Let σˆ be arbitrary but sufficiently smooth stress field that are in equilibrium with zero applied loads. It should be emphasized that σˆ can generate nonzero reactions at the supports. With this arbitrary choice of σˆ , the stress field σˆ  σˆ is therefore arbitrary and also statically admissible. The total complimentary potential energy associated with σˆ  σˆ is given by  C (σˆ  σˆ )  U C (σˆ  σˆ )  WC (σˆ  σˆ )

(6.65)

By expanding the complimentary strain energy functional UC about the correct stress filed σˆ , we obtain U C (σˆ  σˆ )  U C (σˆ )  U C (σˆ ; σˆ )

(6.66)

where U C (σˆ ; σˆ ) is the difference between the complimentary strain energy associated with σˆ  σˆ and σˆ and it can be expressed in terms of the first variation U C and its higher order variations  2 U C , 3 U C , … as U C (σˆ ; σˆ )  U C (σˆ ; σˆ )   2 U C (σˆ ; σˆ )  3 U C (σˆ ; σˆ )  ...

(6.67)

From the definition of the complimentary strain energy functional UC, the first variation U C (σˆ ; σˆ ) can be obtained in an explicit form as U C (σˆ ) σˆ dV   ε(σˆ )σˆ dV ˆ σ  V V

U C (σˆ ; σˆ )  

(6.68)

Similarly, by expanding the complimentary load potential functional WC about the correct stress filed σˆ , we obtain WC (σˆ  σˆ )  WC (σˆ )  WC (σˆ ; σˆ )

(6.69)

where WC (σˆ ; σˆ )  WC (σˆ ; σˆ )   2WC (σˆ ; σˆ )  3WC (σˆ ; σˆ )  ... m

WC (σˆ ; σˆ )   u riPˆri (σˆ )

(6.70) (6.71)

i 1


332



Upon utilizing (6.65), (6.66) and (6.69), it leads to the expansion of the total complimentary potential energy functional about the solution σˆ :  C (σˆ  σˆ )   C (σˆ )   C (σˆ ; σˆ )   2 C (σˆ ; σˆ )  3 C (σˆ ; σˆ )  ...

(6.72)

where  C (σˆ )  U C (σˆ )  WC (σˆ )

(6.73)

( n ) C (σˆ ; σˆ )  ( n ) U C (σˆ ; σˆ )  ( n ) WC (σˆ ; σˆ )

n  1

(6.74)

Since σˆ is the correct stress filed, the deformation resulting from the strain ε obtained from σˆ through the constitutive relation (i.e. ε  ε(σˆ ) ) is compatible with the displacement and essential boundary conditions of the body. From the principle of complementary virtual work, it can then be concluded that the first variation  C vanishes for every σˆ , i.e. m

 C (σˆ ; σˆ )   ε(σˆ )σˆ dV   u riPˆri (σˆ )  0 V

σˆ

(6.75)

i 1

The condition (6.75) implies the stationary of  C at σˆ . (  ) Assume σˆ is a statically admissible stress field that makes the total complimentary potential energy functional stationary, i.e. the condition (6.75) is satisfied. From the principle of complimentary virtual work, the condition (6.75) implied that the strain ε obtained from σˆ through the constitutive relation (i.e. ε  ε(σˆ ) ) is compatible with the displacement and essential boundary conditions of the body. Therefore, σˆ must be the correct stress field associated with the applied loads P. Similar to the PSTPE, for a linear system, the principle of stationary total potential energy the PSTCPE reduces to a stronger theorem known as the principle of minimum total complimentary potential energy (proof is left as an exercise). More specifically, the theorem states that for all statically admissible stress, one that minimizes the total complimentary potential energy functional (6.62) is the correct stress field. To clearly demonstrate applications of the PSTCPE, two examples are presented below. Example 6.5 Consider a two-member truss subjected to the horizontal load P as shown below. The cross-sectional area of the members AB and AC are 2A and A, respectively, and two models for the uniaxial stress-strain relationship, (i)   E and (ii)    3 , are considered. Use the principle of stationary total complimentary potential energy to determine the vertical displacement at point A.

L

L C

B o 45o 45

A

L P

Solution Let  be the value of the vertical displacement at point A to be determined. To enable the use of the PSTCPE, let us imagine that the given structure is subjected to the prescribed vertical displacement  in additional to the applied load P. Now, we choose the statically admissible internal Copyright © 2011 J. Rungamornrat

333



forces FAB and FAC in the members AB and AC, respectively, and let Pˆ be the load at point A in the direction of the prescribed displacement  generated by (or in equilibrium with) these internal forces. Since the given structure is statically determinate (i.e. DI = 4 + 2 – 2(3) = 0), the axial forces FAB and FAC can readily be obtained in terms of P and Pˆ by enforcing equilibrium at a joint A and final results are given by FAB  (P  Pˆ) / 2

,

FAC  (Pˆ  P) / 2

o 45o 45

The axial stresses in both members now become AB  FAB /A AB

ˆ / 2 2A  (P  P)

,

 AC  FAC /A AC

C

B

 (Pˆ  P) / 2A

P

A

Pˆ

(i) By using the linear stress-strain relationship    / E and the formula (6.32), the complimentary strain energy density function is equal to σ0

U C   dσ  0

σ0

  02 dσ  0 E 2E

Since the stress and strain states within the axial member are uniform, we then obtain the complementary strain energy for any axial member of length L and cross-sectional area A as  2   2 AL U Cmember   0 AL  0 2E  2E 

The complementary strain energy of the entire structure is then given by U C  U CAB  U CAC 

ˆ 2L 2(P  P) 2(Pˆ  P) 2 L  4EA 2EA

The complimentary load potential can readily be obtained as WC  ΔPˆ

The total complementary potential energy of the system now becomes  C  U C  WC 

2 (P  Pˆ) 2 L 2 (Pˆ  P) 2 L   ΔPˆ 4EA 2EA

From the PSTCPE, FAB and FAC are the correct internal forces of this structure if  C 

  2 2PˆL 2 (P  Pˆ)LPˆ 2 (Pˆ  P)LPˆ 2PL   ΔPˆ     Δ Pˆ  0 2EA EA 2EA   EA

Due to the arbitrariness of Pˆ , it finally leads to the relationship between the displacement  and the load P and Pˆ : Copyright © 2011 J. Rungamornrat

334



Δ

2 2PˆL 2PL  EA 2EA

Since there is no vertical applied load at point A of the given structure, we choose Pˆ  0 and finally obtain the upward vertical displacement  in terms of the load P as 2PL  2EA

Δ

(ii) By using the nonlinear stress-strain relationship    3 and the formula (6.32), the complimentary strain energy density function is equal to σ0

σ0

0

0

U C   dσ    3dσ 

1 4  0 4

Since the stress and strain states within the axial member are uniform, we then obtain the complementary strain energy for any axial member of length L and cross-sectional area A is 1 1  U Cmember    04 AL   04 AL 4 4 

The complementary strain energy of the given structure is then given by 4

UC  U

AB C

U

4

2  P  Pˆ  2  Pˆ  P      AL   AL 2  2 2A  4  2A 

AC C

The complimentary load potential can readily be obtained as WC  ΔPˆ

The total complementary potential energy of the system now becomes 4

4

2  Pˆ  P  2  P  Pˆ   AL  ΔPˆ  AL    C  U C  WC   2  2 2A  4  2A 

From the PSTCPE, FAB and FAC are the correct internal forces of this structure if 3 3 3   P  Pˆ 3  ˆ P  P  Pˆ  ˆ  Pˆ  P  ˆ  P  P  L  P  ΔPˆ  L    L   Δ  Pˆ  0  C  L   2A   2A    2 2A    2 2A       

Due to the arbitrariness of Pˆ , it finally leads to the relationship between the displacement  and the load P and Pˆ : 3

 P  Pˆ   Pˆ  P    L  Δ  L   2A  2 2 A    

3


335



Since there is no vertical applied load at point A of the given structure, we choose again Pˆ  0 and this leads to Δ

7LP3 LP3 LP3     16 2A 3 2 2A 3 16 2A 3

Example 6.6 Consider a statically indeterminate beam of length L and constant flexural rigidity EI and subjected to a uniformly distributed load q0 as shown in the figure. Apply the PSTCPE to obtain the bending moment of this beam by assuming a quadratic function form.

Y q0 X

EI L

Solution For any given bending moment M = M(x), the flexural stress  and the flexural strain  can readily be obtained as follows: 

M(x)y I

,



σ M(x)y  E EI

where y is the distance from the neutral axis. From the formula (6.34) and (6.35), the complimentary strain energy functional of the entire beam associated with the given bending moment M = M(x) is given by L

L

L

1 1  M(x)y  M(x)y  1 M 2 (x)y 2 U C    εσdAdx     dAdx  dAdx    2 2  EI  I  2 0 A EI 2 0A 0A 

L 1 M 2 (x)  2   y dA dx  2 0 EI 2  A  L

1 M 2 (x)   dx 2 0 EI

The complementary load potential functional for this particular case vanishes (i.e. WC  0 ) since all the prescribed essential boundary conditions are homogeneous (i.e. no support settlement). The total complementary potential energy now becomes L

 C  U C  WC 

1 M 2 (x) dx 2 0 EI

(e6.6.1)

From the PSTCPE, the bending moment M = M(x) is a solution of the problem if an only if it is statically admissible and renders the total complementary potential energy stationary, i.e.  C  0 . To obtain the exact solution for M = M(x), it requires the knowledge of calculus of variation and Copyright © 2011 J. Rungamornrat

336



this is beyond the scope of this text. To construct an approximate solution, we may search the best solution in a smaller set of statically admissible bending moment by assuming the function form of M. This particular strategy changes the problem of finding the unknown function to a much simpler problem of finding a finite number of unknown scalars. Let assume the bending moment M = M(x) in a quadratic form M(x)  C1  C 2 x  C3 x 2 where C1, C2 and C3 are unknown constants. For the bending moment M(x) to be statically admissible, it is required that M(x) must be in equilibrium with the uniformly distributed load q0 and M(L) = 0, i.e. dM(x)  V(x)  C 2  2C3 x dx

and

dV(x)  q 0  2C 3  C 3  q 0 / 2 dx

M (L)  C1  C 2 L  C3 L2  0  C 2  C1/L  q 0 L/ 2

Thus, only C1 is arbitrary whereas C 2  C1  q 0 L / 2 and C3 = q0/2. Therefore, the assumed bending moment M(x) becomes  x 1  x  x  M(x)  C1 1    q 0 L2    1  L 2  L  L 

(e6.6.2)

By substituting the bending moment (e6.6.2) into (e6.6.1), we obtain the total complementary potential energy as 2

L

1   x 1  x  x   C   C1 1    q 0 L2    1  dx  2EI 0   L  2  L  L   The best choice of C1 is one that renders the total complementary potential energy stationary, i.e.

 C 

 C C1  0 C1

Since C1 is arbitrary, the above condition implies that L

q L3 Π C 1   x 1 L  x  x   x  C1  0  0    C1 1    q 0 L2    1 1  dx  3EI 24EI C1 EI 0   L  2  L  L   L 

(e6.6.3)

By solving (e6.6.3), we obtain C1  q 0 L2 / 8 . The bending moment becomes M(x) 

q 0 L2  x x 1  5   4   8  L L

2

   

It can readily be verified that this approximate bending moment is identical to the exact M(x). This is due to that the assumed form of the bending moment is the correct form of the exact solution.


337



6.10 Reciprocal Theorem The reciprocal theorem can be viewed as a product of the PVW or the PCVW in the sense that the theorem itself can readily be derived from either one of those two important principles. Although this theorem is applicable only to linearly elastic materials, its vast applications have been found in structural analysis; for instance, the determination of influence lines and proof of certain crucial properties of flexibility and stiffness matrices in the analysis by matrix methods. Here, we only outline the theorem and give the formal proof by using the PVW while leaving the proof by using the PCVW to readers as an exercise.

P2II

I 3

P

I 2

P

I

I

u ,ε ,σ

P1II

I

PiI

I 1

P

PnI

P3II

u II , ε II , σ II

PiII

PmII

(a)

(b)

Figure 6.15: Schematic of a body subjected to two sets of applied loads Let us consider a deformable body that is made of a linearly elastic material and properly constrained against the rigid body motion by supports at certain locations as shown schematically in Figure 6.15. Assume that this body is subjected to two sets of applied loads, one associated with P I  {P1I , P2I , P3I ,..., PnI } as shown schematically in Figure 6.15(a) and the other associated with

P II  {P1II , P2II , P3II ,..., PmII } as shown in Figure 6.15(b). Let {u I , ε I , σ I } and {u II , ε II , σ II } be two elastic states indicating the displacement, strain and stress of the body due to the applied loads P I and P II , respectively. To be the elastic state, the displacement, strain and stress must satisfy all three basic governing equations (i.e. equilibrium equations, constitutive laws and kinematics) and all boundary conditions. The reciprocal theorem states that the external work done by the applied loads P I in the first state acting through the displacement in the second state (at the same location and in the same direction of P I ) is equal to the external work done by the applied loads P II in the second state acting through the displacement in the first state (at the same location and in the same direction of P II ), i.e. n

m

i 1

i 1

 PiI u iII   PiIIu iI

(6.75)

where u iI is the displacement of the first state at the same location and in the same direction as the applied load PiII and u iII is the displacement of the second state at the same location and in the same direction as the applied load PiI . Proof Since the stress field σ II in the second state is in equilibrium with the applied loads P II and the displacement u I in the first state can be chosen as a virtual displacement (i.e. it is sufficiently smooth and compatible to the deformation or strain ε I ) , the principle of virtual work yields Copyright © 2011 J. Rungamornrat

338



m

P i 1

II i

u iI   (σ II ) T ε I dV

(6.76)

V

From the constitutive relation for linear elastic materials σ  Cε where C denotes the compliance matrix along with the symmetry of C (i.e. CT  C ), equation (6.76) becomes m

P i 1

II i

u iI   (σ II ) T ε I dV   (Cε II ) T ε I dV   (ε II ) T CT ε I dV   (CT ε I ) T ε IIdV   (Cε I ) T ε IIdV V

V

V

V

(6.77)

V

Again, by noting the relation σ I  Cε I , we then get m

P i 1

II i

u iI   (σ I ) T ε IIdV

(6.78)

V

By following the same reasoning, the stress field σ I in the first state is essentially in equilibrium with the applied loads P I and the displacement u II in the second state can be chosen as a virtual displacement (i.e. it is sufficiently smooth and compatible to the deformation or strain ε II ) , the principle of virtual work leads to n

I T II I II  (σ ) ε dV   Pi u i

V

(6.79)

i 1

Combining (6.78) and (6.79) will complete the proof.

6.11 Castigliano’s 1st and 2nd Theorems In this final section, two classical theorems that are fundamental and well-known in the context of structural analysis by energy principles (i.e. Castigliano’s 1st and 2nd theorems) are presented. The first theorem has a very close connection to the PVW and PSTPE and it can be employed as a basis for the development of displacement methods (i.e. methods where the displacements are treated as primary unknowns) for analysis of statically indeterminate structures, while the second theorem can be derived from the PCVW and it has direct applications to the deformation/displacement analysis of structure and also to forming a force method (a method where the static quantities such as reactions and internal forces are treated as primary unknowns) in the analysis of statically indeterminate structures.

6.11.1 Castigliano’s 1st Theorems P2 , u 2

P3 ,u 3 P1 , u1

Undeformed state Deformed state

σ ,ε Pi , u i Pn , u n

Figure 6.16: Schematic of deformed and undeformed state of an elastic structure Copyright © 2011 J. Rungamornrat

339



Consider a statically stable structure that is made from an elastic material. Assume that the deformed state of the structure (the displacement and strain at any point) can be completely characterized by a set of n components of displacements and/or rotations at certain points called the “degrees of freedom of the structure”. Now, let the structure be subjected to a set of forces and/or moments {P1 , P2 , P3 ,..., Pn } applied at each degree of freedom and let {u1 , u 2 , u 3 ,..., u n } be the displacements and/or rotations (corresponding to the same degrees of freedom) produced by this set of loads as shown schematically in Figure 6.16. All components of the strain ε and stress σ at any point within the structure can readily be expressed as functions of {u1 , u 2 , u 3 ,..., u n } as ε  ε(u1 , u 2 , u 3 ,..., u n )

(6.80)

σ  σ (u1 , u 2 , u 3 ,..., u n )

(6.81)

Note that the stress σ can simply be obtained from the constitutive relation when the strain ε is known. Similarly, the total strain energy stored in the body due to the external applied loads can also be written as a function of {u1 , u 2 , u 3 ,..., u n } as U    σ T dεdV  U(u1 , u 2 , u 3 ,..., u n )

(6.82)

V

Now, let us imagine that the given structure is subjected to infinitesimally small and arbitrary displacements {u1 , u 2 , u 3 ,..., u n } in addition to {u1 , u 2 , u 3 ,..., u n } resulting in the final displacements {u1  u1 , u 2  u 2 , u 3  u 3 ,..., u n  u n } . Due to their arbitrariness, {u1, u2, u3,…, un} can be considered as the virtual displacements that are fictitiously introduced to the structure in addition to the actual displacements u1 , u 2 ,..., u i ,..., u n  . The strain associated with this new deformed state can be approximated, to the first order, by ε new  ε  ε

(6.83)

where ε is the strain resulting from the virtual displacement {u1, u2, u3,…, un} and, similarly, it can be considered as the virtual strain associated with the virtual displacements {u1, u2, u3,…, un}. Now, the total strain energy associated with this new deformed state, denoted by Unew, can be approximated, to the first order, by U new    σ T dεdV   σ T εdV  U  U V

(6.84)

V

where the first variation U is defined by U   σ T εdV

(6.85)

V

It is evident from (6.85) that U is, in fact, the internal virtual work (or the virtual strain energy) associated with the virtual strain ε . Since U  U(u1 , u 2 , u 3 ,..., u n ) , the first variation U can therefore be expressed as U 

U U U U u1  u 2  u 3  ... u n u1 u 2 u 3 u n


(6.86)

340



Now, let imagine that a set of external loads {P1 , P2 , P3 ,..., Pn } acts through the virtual displacements {u1 , u 2 , u 3 ,..., u n } . The corresponding external virtual work associated with such virtual displacements is then given by W  P1u1  P2 u 2  P3u 3  ...Pn u n

(6.87)

Since the structure is in equilibrium, from the principle of virtual work it implies that the external virtual work is equal to the virtual strain energy for any admissible virtual displacement and this therefore leads to     U  U U  U  u 3  ... Pn  u 2   P3  u1   P2   P1  u 3  u n u 2  u1     

 u n  0 

(6.88)

Since {u1 , u 2 , u 3 ,..., u n } are arbitrary, it can be concluded from (6.88) that Pi 

U u i

for i  1, 2, 3,..., n

(6.89)

The relation (6.89) is known as Castigliano’s first theorem. More precisely, this theorem states that “for a given elastic structure that is in equilibrium with external applied loads, the partial derivative of the total strain energy with respect to any degree of freedom is equal to the force or the moment acting to that degree of freedom”. To demonstrate the use of Castigliano’s first theorem, let us consider the following example.

Example 6.7 Consider a three-member truss subjected to a vertical load P1 and a horizontal load P2 at point A as shown below. The Young modulus E for all members are assumed constant whereas the cross-sectional area for members AB, AC and AD are given by 2A, A and A, respectively. Use Castigliano’s 1st theorem to find the relationship between the loads P1 and P2 and the vertical and horizontal displacements at point A. L

L

B

D

C o 45o 45

A

L P2

P1

Solution Let the vertical and horizontal components of the displacement at point A be denoted by 1 and 2, respectively. The deformation of all three members, represented by the elongations eAB, eAC, and eAD, can be fully described by 1 and 2. Such relations can be obtained via the geometric consideration of the deformed configuration of the structure as follows. e AB  1 cos 45o   2 sin 45o  (1   2 ) / 2 e AC  1  0  1 e AD  1 cos 45o   2 sin 45o  (1   2 ) / 2 Copyright © 2011 J. Rungamornrat

341


B


D

C

o 45o 45  A

o 45o 45

eAD

A



D

C

B

eAB

eAB

Elongations due to 1

eAD

Elongations due to 2

The axial strains of the three members now become  AB  e AB / 2L  (1   2 ) /2L

(e6.7.1)

 AC  e AC /L  1/L

(e6.7.2)

 AD  e AD / 2L  (1   2 ) /2L

(e6.7.3)

For linear elastic materials, the strain energy density function U due to the axial strain is given by 0

0

0

0

U   σd   Ed 

 02 E 2

where  0 is the final axial strain. Since the stress and strain states within the axial member are uniform, we then obtain the strain energy for any axial member of length L and cross-sectional area A as  2E  1 U member  UV   0 AL  EAL 02 2  2 

(e6.7.4)

By using (e6.7.1)-(e6.7.3) and (e6.7.4), we obtain the total strain energy of the structure in terms of the displacements 1 and  2 as U  U AB  U AC  U AD  

1 1 1 E(2A)( 2L) 2AB  E(A)(L) 2AC  E(A)( 2L) 2AD 2 2 2 2EA EA 2 2EA (1   2 ) 2  1  (1   2 ) 2 4L 2L 8L

Now, by applying Castigliano’s 1st theorem, it leads to the expressions of P1 and P2 in terms of 1 and 2: P1 

U  1

2EA EA 2EA (3 2  4)EA 2EA (1   2 )  1  (1   2 )  1  2 2L L 4L 4L 4L

P2 

U   2

2EA 2EA (1   2 )  0  (1   2 )(1)  2L 4L

2EA 3 2EA 1  2 4L 4L


342



6.11.2 Castigliano’s 2nd Theorems Let us consider a structure that is made from elastic material and is subjected to a set of independent forces and/or moments {P1, P2, P3,…, Pn} as shown in Figure 6.16. Let {u1, u2, u3,…, un} be the displacements and/or rotations introduced at the same location and in the same direction of the applied loads {P1, P2, P3,…, Pn}. Let {R1, R2, R3,…, Rm} and {us1, us2, us3,…, usm} be a set of components of all support reactions and a set of their corresponding prescribed movements, respectively. More precisely, usi is the prescribed movement of the support in the direction of the support reaction Ri. Any component of the support reactions Ri, the stress field  and the strain field  of the structure can be expressed as a function of {P1, P2, P3,…, Pn}, i.e. R i  R i (P1 , P2 , P3 ,..., Pn )

(6.90)

σ  σ (P1 , P2 , P3 ,..., Pn )

(6.91)

ε  ε(P1 , P2 , P3 ,..., Pn )

(6.92)

Now, the complementary strain energy of the entire structure can also be written as a function of {P1, P2, P3,…, Pn} as U C    ε T dσdV  U C (P1 , P2 , P3 ,..., Pn )

(6.93)

V

Next, let the structure be subjected to a set of infinitesimally small and arbitrary loads {P1, P2, P3,…, Pn} in addition to {P1, P2, P3,…, Pn}. It should be noted that, due to their arbitrariness, {P1, P2, P3,…, Pn} can also be viewed as the virtual loads applied to the virtual structure that is identical to the actual structure. All support reactions and the stress field of the actual structure produced by a new set of applied loads {P1 +P1, P2 +P2, P3 +P3,…, Pn +Pn}, denoted by Ri,new and new, can be approximated to the first order by R i,new  R i  R i

(6.94)

σ new  σ  σ

(6.95)

where  is the additional stress field induced by the additional loads {P1, P2, P3,…, Pn} and n

R i   j1

R i δPj Pj

(6.96)

It worth noting that can also be viewed as the virtual stress in the virtual system subjected to the virtual forces {P1, P2, P3,…, Pn}. The complementary strain energy associated with the new set of applied loads {P1 +P1, P2 +P2, P3 +P3,…, Pn +Pn}, denoted by UC,new, can be approximated to the first order by U C,new    ε T dσdV   ε T σdV  U C  U C V

(6.97)

V

where the first variation UC is defined by U C   ε T σdV

(6.98)

V


343



It is evident from (6.98) that UC is in fact the complementary virtual strain energy associated with the virtual stress  acting through the actual strain . Since U C  U C (P1 , P2 , P3 ,..., Pn ) , UC can also be expressed as n

U C   i 1

U C δPi Pi

(6.99)

The complementary virtual work of a set of virtual loads {P1, P2, P3,…, Pn} and the corresponding support reactions {R1, R2, R3,…, Rm} in the virtual structure acting through the actual displacements {u1, u2, u3,…, un} and the actual prescribed movements of the supports {us1, us2, us3,…, usm} is given by n

m

i 1

j1

WC   u i δPi   u sjδR j n m  n R j    u i δPi   u sj   δPi  i 1 j1  i 1 Pi  n  m n  R j    WCs   δPi    u i  δPi    u i    u sj  Pi  Pi   i 1  i 1  j1 

(6.100)

where WCs is the complementary work of the support reactions acting through their prescribed movements by m

WCs   u si R i

(6.101)

i 1

Since the displacement and strain of the actual structure at the same state are essentially compatible, the principle of complementary virtual work implies that the complementary virtual work is equal to the complimentary virtual strain energy, i.e. WC = UC, and this results in  U C WCs  δPi  0    u i   Pi  Pi i 1  n

(6.102)

Since {P1, P2, P3,…, Pn} are arbitrary, it can be concluded from (6.102) that ui 

U C WCs  Pi Pi

i  1,2,3,..., n

(6.103)

The relation (8.103) is known as the Castigliano’s second theorem. More precisely, this theorem states that “for a given elastic structure that is compatible, the partial derivative of the complementary strain energy less the complementary work due to the support movements with respect to the applied concentrated load is equal to the displacement or rotation at the location and in the direction of that load”. It is important to emphasize that the second term on the right hand side of (6.103) represents the displacement due to the support settlements. For the special case when there is no movement of all supports, the complementary work WCs vanishes (i.e. WCs = 0) and the relation (6.103) simply reduces to original Castigliano’s second theorem, i.e. Copyright © 2011 J. Rungamornrat

344



ui 

U C Pi

i  1,2,3,..., n

(6.104)

Applications of Castigliano’s 2nd theorem to the deformation/displacement analysis of statically determinate structures and the analysis of statically indeterminate structures are described extensively in Chapter 9.

Exercises 1. Prove the principle of minimum total potential energy 2. Prove the principle of minimum total complementary energy 3. Prove the reciprocal theorem by the principle of complementary virtual work 4. Apply the principle of virtual work to determine all support reactions of following statically determinate structures 2qL q A

B

L/2

L/2 2qL q

A

B L/4

L/4

L/4

P

L/4 P C

B A L

L

PL

L

L P

2P

L 2P

L

L 3P L

L

L

L


L

L

345



q 2qL L

L

qL qL2

L

qL

qL2

2qL L

q

L

L

2L qL

2qL L 3L

qL2

qL

L

2qL2 q

L

L

L

2L

2L

5. Apply the principle of stationary total potential energy along with the given assumed solution forms to estimate the deflection of linearly elastic beams shown below. The flexural rigidity EI is constant throughout. 4q0x(L – x)/L2 v(x) = c0 + c1x + c2x2 v(x) = c0 + c1x + c2x2 + c3x3 v(x) = c0 sin(x/L)

L

P

L/2

L/2

v(x) = c0 + c1x + c2x2 + c3x3 v(x) = c0 + c1x + c2x2 + c3x3+ c4x4 v(x) = c0 + c1x + c2x2 + c3x3+ c4x4+ c5x5


346



q v(x) = c0 + c1x + c2x2 + c3x3+ c4x4 v(x) = c0 + c1cos(2x/L) L

6. Apply either (i) the principle of stationary total complimentary potential energy or (ii) Castigliano’s 1st theorem to determine both the vertical and horizontal displacements at point A of two trusses shown below. 2L

L

L

B

B

C

D

C

L A

L

P

A

P

7. Consider a statically stable structure shown below. Use the reciprocal theorem to prove that the displacement at point A (in the direction of the applied load at point B) due to the applied unit load at point B is equal to the displacement at point B (in the direction of the applied load at point A) due to the applied unit load at point A. 1

A

A B

B

1

Structure under unit load at point A

Structure under unit load at point B


347


Deformation/Displacement Analysis by PRW

CHAPTER 7 DEFORMATION/DISPLACEMENT ANALYSIS BY PRW This chapter devotes mainly to the application of conservation of work and energy (see section 6.5 of chapter 6) to the deformation and displacement analysis of statically determinate, linearly elastic structures such as trusses, beams and frames. Since quantities involved in such conservation principle are work and energy associated with the system undergoing a real process, it is called, here, the principle of real work (PRW) in order to differentiate from the principle of virtual work (PVW) in which work and energy are associated with the fictitious or virtual process existing only in our thought. First, the conservation equation (6.40) is specialized to structures subjected to a single concentrated load (either a force or a moment). Next, explicit expressions for computing the strain energy in a member due to various effects (e.g. axial effect, shear effect, bending effect, and torsion effect) are presented. Several example problems are considered in order to clearly demonstrate applications of the PRW. Finally, limitations of the principle are listed.

7.1 Real Work Equation P A u

Figure 7.1: Schematic of structure subjected to concentrated load P at point A Consider a statically stable, linearly elastic structure shown schematically in Figure 7.1. This structure is subjected only to a concentrated load P (either a force or a moment) at point A such that its direction remains fixed throughout its application and its final magnitude is denoted by P0. The external work done W by the load P is given by 0

W   P(u)du

(7.1)

0

where u is the displacement at point A in the direction of load P and  is the value of the displacement u at P = P0. In general, the function form of the load-displacement relation P = P(u) can be either linear or nonlinear depends primarily on the behavior of a given structure. For a given linear structure, the integral appearing in (7.1) can be integrated directly to obtain W

1 P0 Δ 0 2

(7.2)


348



Since the constituting material is assumed to be linearly elastic, the strain energy of the structure resulting from the applied load P depends only on either the stress field  or the strain field  at the final state associated with P = P0 and takes a form 1 U   σ T0ε 0 dV 2 V

(7.3)

Conservation of work and energy (stated in section 6.5 of chapter 6) implies that WU 

1 1 P0 Δ 0   σ T0 ε 0 dV 2 2 V

(7.4)

Equation (7.4) is known as the real work equation and it involves following four quantities: (i) the prescribed final concentrated load P0, (ii) the unknown displacement 0 at the location and in the direction of load P0, (iii) the stress field at the final state  and (iv) the strain field at the final state . For statically determinate skeleton structures, the internal forces at any cross section of all members can be obtained from static equilibrium and, when supplied with simplified kinematics of the cross section, it yields the stress field . The strain field  can directly be determined from the constitutive law. As a result, for statically determinate structures, the real work equation (7.4) can be solved for the unknown displacement 0. It is important to emphasize that the displacement 0 obtained from (7.4) is the displacement at the location and in the direction of the applied load P0.

7.2 Strain Energy for Various Effects In this section, the general expression of the strain energy (7.3) is specialized to treat certain basic one-dimensional structural members such as axial member, flexural member and frame member. The final formulae are given in terms of the internal forces (i.e. resultants of the stress over the cross section) instead of the stress and strain fields and this should significantly reduce the computational effort in the analysis. Let’s consider a generic straight, one-dimensional, structural element of length L and made of a linearly elastic material with Young’s modulus E and shear modulus G as shown in Figure 7.2(a). The area, moment of inertia and torsional constant of the cross section are denoted by A, I and J, respectively. Note that in the development follows, E, G, A, I and J are allowed to be a function of position along the axis of the member, i.e. E = E(x), G = G(x), A = A(x), I = I(x) and J = J(x) where x is the local coordinate of the member defined in a usual fashion as shown in Figure 7.2(a). Let’s further define F = F(x), V = V(x), M = M(x) and Mt = Mt(x) as the axial force, the shear force, the bending moment and the torque at any cross section (located at the coordinate x) of the member as shown in Figure 7.2(b). y V(x) E, G, A, I, J L

x x

(a)

Mt(x)

F(x) M(x) (b)

Figure 7.2: (a) Schematic of generic straight structural member and (b) diagram indicating internal forces at any cross section Copyright © 2011 J. Rungamornrat

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7.2.1 Strain energy due to axial effect The axial force F(x) produces the uni-axial state of stress for the entire member (i.e. the stress field simply reduces to  = {}T where is the stress component normal to the cross section) and, in addition, the normal stressis uniform across the section. The value of the normal stress and the corresponding normal strain  can be obtained in terms of the internal force F(x) as follows. 

F A



,

σ F  E EA

(7.6)

The strain energy density is equal to Ua 

1 1  F  F  F2 σ      2 2  A  EA  2EA 2

(7.7)

The subscript “a” is added to emphasize the contribution from the axial force. It is evident from (7.7) that both U a is independent of the position within the cross section. By using the relations (7.3, it leads to the strain energy of the entire member: L

L

L

F2 F2 1 F2 U a   U a dV    dA dx  dA dx  dx 0 2EA 2 A 2EA 2 2 0 EA V 0A

(7.8)

For the special case that F, E and A are constant for the entire member, the expression (7.8) reduces directly to Ua 

F2 L 2EA

(7.9)

7.2.2 Strain energy due to bending effect From the conventional beam theory, the bending moment M(x) also produces the uni-axial state of stress  = {}T where denotes the flexural stress or the stress component normal to the cross section. Unlike the axial force case, both the flexural stress  and the corresponding strain vary linearly across the section and their values are given, in terms of the bending moment M(x), by 

My I

,



σ My  E EI

(7.10)

where y is the distance from the neutral axis. The strain energy density due to bending effect, denoted by U b , is given by Ub 

1 σ  2

1  My  My  M 2 y 2    2  I  EI  2EI 2

(7.11)


350



For this particular case, the strain energy density U b is a function of position across the section, i.e. the distance y from the neutral axis. By inserting (7.10) into (7.3), we then obtain the strain energy of the entire member as L

L

V

L

M 2 y2 M2 1 M2 2 dA dx  y dA dx  dx 2 2    2EI 2EI 2 EI 0A 0 A 0

U b   U b dV   

(7.12)

where the definition of the moment of inertia, i.e. I   y 2dA , has been utilized. For the special case A

that M, E and I are constant for the entire member, the expression (7.12) can be integrated directly and this yields M 2L Ub  2EI

(7.13)

7.2.3 Strain energy due to torsional effect From a member with a circular cross section, the internal torque Mt(x) produces a state of stress in pure shear with a non-zero shear stress  given by the torsional formula 

Mtr J

(7.14)

where r is the distance from the center of the cross section. The shear strain  can readily be obtained from the linear constitutive law, i.e. 

τ Mtr  G GJ

(7.15)

The strain energy density due to torsional effect, denoted by U t , becomes 1 1  M t r  M t r  M 2t r 2 U t       2 2  J  GJ  2GJ 2

(7.16)

Similar to the bending case, the strain energy density U t varies across the section (i.e. it is a function of the distance r from the center of the cross section). By substituting (7.16) into (7.3), we then obtain the strain energy of the entire member due to torsional effect as L

V

L

L

M 2t r 2 M 2t 1 M 2t 2 dA dx r dA dx   dx 0 2GJ 2 A 2GJ 2 2 0 GJ 0A

U t   U t dV   

(7.17)

where the definition of the torsional constant, i.e. J   r 2dA , has been employed. For the special A

case that Mt, G and J are constant throughout the member, the expression (7.17) reduces to Ut 

M 2t L 2GJ


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7.2.4 Strain energy due to shear effect From the conventional beam theory, the non-zero shear stress, denoted by, and the corresponding shear strain, denoted by, due to the shear force V(x) can be computed from 

VQ(y) Ib(y)



,

 VQ(y)  G GIb(y)

(7.19)

where y is the distance from the neutral axis, b = b(y) is the width of the cross section at the distance y from the neutral axis, and Q = Q(y) is defined by h

Q(y)   ξb(ξ)dξ

(7.20)

y

where h is the distance from the neutral axis to the extreme fiber. The strain energy density due to the presence of the shear force, denoted by U s , is then given by Us 

1 1  VQ  VQ  V 2 Q 2      2 2  Ib  GIb  2GI 2 b 2

(7.21)

It is evident from (7.21) that the strain energy density U s varies across the section. Again, by inserting (7.21) into (7.3), it leads to the expression of the strain energy of the entire member due to the shear effect: L L  V 2Q 2 V2  A Q2 1 V 2  dA dx dA dx dx    2 2 2 2   2GI b 2GA I b 2 GA 0A 0 0   A

L

U s   U s dV    V

(7.22)

where  is the form factor of the cross section defined by A Q2   2  2 dA I Ab

(7.23)

This parameter depends only on the geometry of the cross section. For rectangular and circular cross sections, the form factors are equal to 1.2 and 10/9, respectively. For the special case that V, G,  and A are constant for the entire member, the strain energy U s is obtained explicitly from V 2 L Us  2GA

(7.24)

7.2.5 Strain energy of member due to all possible effects The total strain energy of a structural member due to all possible effects depends primarily on the member type and the internal forces present in that member. Once the internal forces within the member are determined, the strain energy due to each effect can be computed directly from (7.8), (7.12), (7.17) or (7.22) and the total strain energy is obtained by superposing contributions from all possible effects. For a prismatic truss member, only the effect of axial deformation exists and the total strain energy simply becomes Copyright © 2011 J. Rungamornrat

352



L

U  Ua 

1 F2 F2 L dx  2 0 EA 2EA

(7.25)

For a beam member, only the shear force and bending moment exist and, as a result, the total strain energy becomes L

U  U b  Us 

2

L

1 M 1 V 2 dx dx  2 0 EI 2 0 GA

(7.26)

For a member in a plane frame subjected only to in-plane loadings, the axial force, the bending moment and the shear force generally exist. The total strain energy for this particular case is 2

1 F2 1 M 1 V 2 U  Ua  Ub  Us   dx   dx   dx 2 0 EA 2 0 EI 2 0 GA L

L

L

(7.27)

For a member in a plane frame subjected only to out-of-plane loadings, only the bending moment, the shear force and internal torque may exist. The total strain energy becomes L

2

L

L

1 M 1 V 2 1 M2 U  U b  Us  U t   dx   dx   t dx 2 0 EI 2 0 GA 2 0 GJ

(7.28)

7.3 Applications of Real Work Equation In this section, applications of the real work equation (7.4) to determine the displacement/rotation of statically determinate structures subjected to a single concentrated load are demonstrated via several example problems including trusses, beams and frames. To form the real work equation, the work done by the applied load can readily be computed by (7.2) but the computation of the strain energy of the system is generally non-trivial since the internal forces of all members must be determined a priori. For statically determinate structures, this task can be achieved via the enforcement of static equilibriums as extensively described in Chapter 2. Once the real work equation is established, it can readily be solved to obtain the displacement/rotation at the location and in the direction of the applied concentrated load. Example 7.1 Determine the horizontal displacement at joint D of a linearly elastic truss subjected to a concentrated load P shown below. The axial rigidities of members AB, CD, AC, BC, and BD are given by 2EA, 2EA, EA, EA, and EA, respectively.

C

D

P

L

B

A L


353



Solution The work done by the horizontal applied load P is equal to W

1 PΔ D 2

where D is the horizontal displacement the joint D. To compute the strain energy, we first perform the static analysis for the support reactions and the internal forces in all members. Since the given truss is statically determinate, those static quantities can readily be obtained via static equilibrium and either the method of joints or the method of sections. The member forces for all members are reported in a table below. The strain energy for each member is also computed using the expression (7.25) and shown in the same table. Member

Ei Ai

AB

2EA

L

P

CD

2EA

L

P

AC

EA

L

P

BC

EA

2L

BD

EA

Ui =

Fi

Li

L

Fi2 Li 2E i A i

C

P2L 4EA P2L 4EA P2L 2EA 2P 2 L EA

 2P

0

D

A

P

P

B P

P

0

The strain energy of the entire structure is therefore equal to U

P2L P2L P2L 2P 2 L P2L     0  (1  2 ) 4EA 4EA 2EA EA EA

By applying the real work equation (7.4), we then obtain the displacement D as follows: 1 P2L PΔ D  (1  2 ) 2 EA



Δ D  2(1  2 )

PL Rightward EA

Example 7.2 Determine the vertical displacement at joint F of a linearly elastic truss subjected to a vertical concentrated load P shown below. The axial rigidity EA of all members is assumed constant.

A

C

B

L F E

D L

L

P


G L

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Solution The work done by the vertical concentrated load P is equal to W

1 PΔ F 2

where F is the vertical displacement the joint F. Again, to compute the strain energy, we first perform the analysis for the support reactions and the internal forces in all members. Specifically, the reactions are obtained from equilibrium of the entire structure (see the free body diagram below) and all member forces are determined by either the method of joints or the method of sections. Results including the strain energy for each member are reported in the table below. P/3 0

A

C

B

F

G

E

D

2P/3

P

Member

Ei Ai

Li

Fi

AB

EA

L

–P/3

BC

EA

L

–2P/3

DE

EA

L

0

EF

EA

L

P/3

FG

EA

L

2P/3

AD

EA

L

0

Ui =

Fi2 Li 2E i Ai

P 2L 18EA 2P 2 L 9EA

0 P 2L 18EA 2P 2 L 9EA

Member

Ei Ai

Li

Fi

AE

EA

2L

2P/3

BE

EA

L

BF

EA

2L

CF

EA

L

CG

EA

2L

–P/3 2P/3

2P/3  2 2P/3

Ui =

Fi2 Li 2E i Ai

2P 2 L 9EA P 2L 18EA 2P 2 L 9EA 2P 2 L 9EA 4 2P 2 L 9EA

0

The strain energy of the entire structure is therefore equal to  2P 2 L   P 2 L   2P 2 L  4 2P 2 L (5  4 2)P 2 L U  3  3   2  2(0)       9EA 6EA  18EA   9EA   9EA 

By applying the real work equation (7.4), we then obtain the vertical displacement F as follows: 1 (5  4 2)P 2 L PΔ F  2 6EA



ΔF 

(5  4 2)PL Downward 3EA


355



Example 7.3 Determine the deflection at the mid-span of a prismatic, linearly elastic, simplysupported beam of length L by taking both bending and shear effects into account. Given that E, G, I, A and  are constant for the entire beam.

P

L/2

L/2

Solution Since the given beam is statically determinate, support reactions, the shear force and bending moment diagrams can readily be obtained from static equilibrium with results given below.

y P 0

x P/2

P/2 L/2

L/2 P/2

SFD –P/2 PL/4 BMD The work done by the vertical concentrated load P is equal to W

1 PΔ 2

where  is the downward deflection at the mid-span of the beam. By using the equation (7.26), the strain energy of the entire beam due to both shear and bending effects is given by 2 L L  1 L/2(Px/2) 2   1 L/2(P/2) 2  1 M 1 V 2 U  U b  Us   dx   dx  2  dx   2  dx  2 0 EI 2 0 GA 2 EI 2 GA  0   0 



P 2 L P 2 L3  8GA 96EI

By applying the real work equation (7.4), we then obtain Copyright © 2011 J. Rungamornrat

356



WU



P 2 L P 2 L3 1  PΔ  2 8GA 96EI PL PL3 Downward Δ  4GA 48EI

Note that the direction of the deflection follows that of the external load P. Example 7.4 Determine the rotation at the tip of a linearly elastic beam under a concentrated moment M by taking both bending and shear effects into account. E, G, I, A and  for each segment are shown in the figure.

M

L/2

L/2

L/2

Solution Since the given beam is statically determinate, support reactions, the shear force and bending moment diagrams can readily be obtained from static analysis and results are shown below.

M 0

E, G, 2I, 2A, 

M

E, G, I, A, 

2M/L

2M/L

L/2

L/2

2M/L

2M/L

L/2

SFD M

M BMD

–M

The work done by the concentrated moment M is equal to W

1 M 2 Copyright © 2011 J. Rungamornrat

357



where  is the rotation at the tip of the beam. By using the equation (7.26), the strain energy of the entire beam due to both shear and bending effects is given by 1 U  U b  Us  2



L/2

1 (2Mx/L ) 2 0 2EI dx  2

L/2

1 (2Mx/L ) 2 0 EI dx  2

L/2

1 M2 0 EI dx  2

L/2

 (2M/L ) 2 1 0 2GA dx  2

L/2

(2M/L ) 2 0 GA dx

M 2 L M 2 L M 2 L M 2 2M 2 3M 2 L 3M 2       24EI 12EI 4EI EIL2 GAL2 8EI GAL

By applying the real work equation (7.4), we then obtain WU

1 3M 2 L 3M 2 M   2 8EI GAL





3ML 6M CCW  4EI GAL

Note that the direction of the rotation follows the direction of the applied moment M. Example 7.5 Determine the vertical displacement at point C of a linearly elastic frame subjected to a vertical concentrated load P shown below. Given that E, G, I,  and A are constant for the entire frame.

x

P B

P

B C

C L

x A

A P

L

0 PL

Solution The work done due to a vertical concentrated load P can readily be obtained as W

1 P C 2

where C is the vertical displacement at point C. To compute the strain energy of the entire frame, we first perform the static analysis to obtain the support reactions and the internal forces (i.e. the axial force, the shear force and the bending moment). The strain energy for each member is calculated as follows. Member AB: The axial force F, the shear force V and the bending moment M of this member are constant and given by F  P

;

M  PL

;

V0 Copyright © 2011 J. Rungamornrat

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The strain energy of this member due to all effects, denoted by U AB , can be computed from (7.27) as follows. L

AB U AB  U aAB  U AB  b  Us

L

1 (  P) 2 1 (PL) 2 dx  dx  0 2 0 EA 2 0 EI

P 2 L P 2 L3   2EA 2EI

Member BC: The axial force F, the shear force V and the bending moment M of this member are given as function of x by F0

M  P(L  x)

;

VP

;

The strain energy of this member due to all effects, denoted by U BC , can be computed again by using the formula (7.27): L

U

BC

U

BC a

U

BC b

U

BC s

L

1 (P(L  x)) 2 1 P 2 dx   dx 0  20 EI 2 0 GA 

P 2 L3 P 2 L  6EI 2GA

Thus, the strain energy of the entire frame is equal to U  U AB  U BC 



P 2 L P 2 L3 P 2 L3 P 2 L    2EA 2EI 6EI 2GA P 2 L 2P 2 L3 P 2 L   2EA 3EI 2GA




1 P 2 L 2P 2 L3 P 2 L P C    2 2EA 3EI 2GA C 

PL 4PL3 PL Downward   EA 3EI GA

Now, let further explore the contribution of the axial force, bending moment and shear force on the total displacement  C . Assume that the members AB and BC have a rectangular cross section of width b and depth h (A = bh and I = bh3/12). The ratios between the displacement due to axial effect (a) and the displacement due to the shear effect (s) and the displacement due to the bending effect (b) are given by  a  PL   4PL3  1  h       /  b  EA   3EI  16  L 

2


359


 a  PL   4PL3   E  h     /    b  GA   3EI  16 G  L 


2

It is evident that effects of the axial force and the shear force on the deformation of the structure depend primarily on the ratio between the depth and length of the member. A member with its length much larger than its depth, the deformation due to both axial and shear effects can be negligible. Example 7.6 Determine the rotation at point C of a linearly elastic frame subjected to a concentrated moment M shown below. Given that E, G, I,  and A are constant for the entire frame.

x

M

B

B

C

L/2

D L/2

C x D

x A

A

M

0

M/L

M/L

L

Solution The work done due to a concentrated moment M can readily be obtained as 1 M C 2

W

where C is the rotation at point C. To compute the strain energy of the entire frame, we first perform the static analysis to obtain the support reactions and the internal forces (i.e. the axial force, the shear force and the bending moment). The strain energy for each member is calculated as follows. Member AB: The axial force F, the shear force V and the bending moment M of this member are constant and given by F  M/L

;

M0

V0

;

The strain energy of this member due to all effects, denoted by U AB , can be computed from (7.27) as follows. L

U

AB

U

AB a

U

AB b

U

AB s

1 (M/L) 2 M2   dx  0  0  2 0 EA 2EAL

Member BC: The axial force F, the shear force V and the bending moment M of this member are given as function of x by

F0

;

M  Mx/L

;

V  M/L Copyright © 2011 J. Rungamornrat

360



The strain energy of this member due to all effects, denoted by U BC , can also be computed by using the formula (7.27): L

BC U BC  U aBC  U BC 0 b  Us

L

1 (Mx/L) 2 1 (M/L) 2 M 2 L M 2   dx dx  2 0 EI 2 0 GA 6EI 2GAL

Member CD: The axial force F, the shear force V and the bending moment M of this member are constant and given by F  M/L

;

M0

;

V0

The strain energy of this member due to all effects, denoted by U CD , is given by U

CD

U

CD a

U

CD b

U

CD s

1  2

L/2

(M/L) 2 M2    dx 0 0 0 EA 4EAL

Thus, the strain energy of the entire frame is equal to U  U AB  U BC  U CD 



M2 M 2 L M 2 M2    2EAL 6EI 2GAL 4EAL

3M 2 M 2 L M 2   4EAL 6EI 2GAL




3M 2 M 2 L M 2 1 M C    2 4EAL 6EI 2GAL C 

3M ML M CW   2EAL 3EI GAL

Again, let further explore the contribution of the axial force, bending moment and shear force on the total rotation C . Assume that the members AB, BC and CD have a rectangular cross section of width b and depth h (A = bh and I = bh3/12). The ratios between the rotation due to axial effect (a) and the rotation due to the shear effect (s) and the rotation due to the bending effect (b) are given by θ a  3M   ML  9  h    /    θ b  2EAL   3EI  2  L 

2

θs  λM   ML  Eh  /   3λ   θ b  GAL   3EI  GL

2

It is evident that effects of the axial force and the shear force on the deformation of the given frame depend primarily on the ratio between the depth and length of the member. A member with its length much larger than its depth, the deformation due to both axial and shear effects can be negligible. Copyright © 2011 J. Rungamornrat

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7.4 Limitations of PRW While the principle of real work is simple and can be employed in the deformation/displacement analysis of statically determinate structures, the method itself possesses various limitations. For instance, the displacement (or rotation) can only be calculated at the location and in the direction of the applied concentrated force (or applied concentrated moment) and, more importantly, there must be only to a single concentrated force (or a single concentrated moment) applied to the structure. Those limitations exclude various situations to be treated; for examples, structures subjected to distributed loads or a set of many concentrated loads and structures whose of the displacements and rotations at points or in directions different from the applied load are of interest. q

P1 v

(a)

P2

1  2

P3

P1

3

(b)

2

1

(c)

Figure 7.3: Examples of (a) structure subjected to uniformly distributed load, (b) structure subjected to multiple concentrated loads, and (c) structure whose the displacement of interest is not at the point and in the direction of applied load Since the principle of real work provides only one independent equation, the work done by the applied loads must contain only one unknown displacement (or one unknown rotation) in order to be able to solve that existing unknown. The work done due to the applied load of a linear elastic beam of length L shown in Figure 7.3(a) is given by L

1 W   qv(x)dx 20

(7.29)

where v(x) is the deflection at any point x. It is apparent that the external work done given by (7.29) involve an unknown function v = v(x) and, as a result, it cannot be determined from the real work equation. Similarly, for a linear elastic beam subjected to a set of three concentrated loads shown in Figure 7.3(b), the work done by these applied loads is given by W

1 1 1 P1Δ1  P2 Δ 2  P3Δ 3 2 2 2

(7.30)

Apparently, all three unknown deflections {1,2,3} appearing in the work done (7.30) cannot be determined from the (single) real work equation. Finally, let consider a linear elastic cantilever beam subjected only to a concentrated load at the tip as shown in Figure 7.3(c). For this particular case, the tip deflection 1 can readily be obtained from the real work equation since the work done consists only of one unknown deflection, i.e. W = P11/2. However, the deflection at other points such as 2 cannot be determined since such unknown quantity does not contribute to the external work done.


362



Exercises 1. Determine the deflection (or rotation) at a point where the concentrated load (or concentrated moment) is applied of following beams using the principle of real work. Include both the bending and shear effects in the calculation. E, G and  are assumed to be constant throughout while the cross sectional area and the moment of inertia are clearly indicated in the figure. A, I

A, I

L

L

A, I

M

L

L P 2A, 2I

A, I

L

A, I

L

L

L P

A, I

A, 2I

L

L

A, I

L

L

P A, I

A, I

A, I

L

2L

L

2A, 2I

A, I

2L

L

A, I

L

M

2. Determine the deflection (or rotation) at a point where the concentrated load (or concentrated moment) is applied of following beams using the principle of real work. Include all possible effects in the calculation. E, G, A, I, and  are assumed to be constant throughout. Copyright © 2011 J. Rungamornrat

363



L P

L

L

M

L L

P

L

L

L

L

M

P

L

L

L

L


364



3. Determine the displacement at a point of load application of following trusses using the principle of real work. The axial rigidity of each member is given in the figure.

2EA

L

P

EA EA

EA

2EA EA

EA

2EA

EA

EA

L

EA

EA

1

P 1

L

L

2EA

2EA

L

L

P EA

L

EA

EA

EA

EA

EA

2EA

EA

2EA

EA

EA

EA

L

EA

EA

EA

EA

L

L

L

P L

L



365

Deformation/Displacement Analysis by PCVW

CHAPTER 8 DEFORMATION/DISPLACEMENT ANALYSIS BY PCVW In this chapter, applications of the principle of complementary virtual work (PCVW) to the deformation/displacement analysis of statically determinate structures such as trusses, beams and frames are presented. As clearly described in the Chapter 6, the PCVW is applicable not only to linear elastic structures but also to structures that are made of nonlinear materials. In addition, limitations posed by the PRW can readily be handled by the PCVW. Before it is specialized to each type of structures, use of the PCVW to derive an equation necessary and sufficient for determining the displacement or rotation at any point and in any direction of a structure is presented.

8.1 PCVW with Single Virtual Concentrated Load Consider a statically stable structure that is subjected to a system of forces and moments as shown in Figure 8.1. The displacement and strain at any point of the structure produced by this set of applied loads are denoted by u and , respectively. This structure under consideration is generally termed the actual structure or the actual system. Let A be a particular point within the actual structure where the displacement A in a certain direction is of interest and to be determined. q P2 u

Deformed state θA ε

P1

A

ΔA

M2

Pi

M1

Undeformed state Pn Figure 8.1: Schematic of actual structure subjected to a system of applied loads δP

δσ

A

Figure 8.2: Schematic of virtual structure subjected to a concentrated virtual force P Copyright © 2011 J. Rungamornrat


366


Now, let’s introduce another statically stable structure called the virtual structure or the virtual system as shown in Figure 8.2. This virtual structure is chosen to have an identical geometry to the actual structure, but it is subjected only to a virtual force P at the point A in the same direction as that for the displacement A. The virtual stress introduced within the virtual structure due to the action of the virtual force P is denoted by . Note that the resultants of this virtual stress can be represented in terms of the virtual internal forces such as the axial force, the shear force, the bending moment, and the internal torque depending on the member type. Since the displacement u and the strain  of the actual structure are geometrically compatible, from the principle of complementary virtual work along with a particular choice of the virtual forces and virtual stress (which are in equilibrium) associated with the virtual system, it leads to the relation WC  U C



P   A  WCs   δσ TεdV

(8.1)

V

where WCs is the complementary virtual work by the support reactions in the virtual structure acting through the movement of the supports in the actual structure (i.e. support settlements). It is important to emphasize that validity of the relation (8.1) is independent of the magnitude of the virtual force P and, by choosing particularly P = 1, it yields 1   A   A   δσ TεdV  WCs

(8.2)

V

where  is now the stress field in the virtual structure due to the unit virtual force P. The method corresponding to this particular choice of unit virtual force is also known as the unit load method. It should be noted that for statically determinate structures, the support reactions and internal forces in both actual and virtual structures can readily be computed via static equilibrium and this implies that both the actual and virtual stress fields are also known. In addition, the actual strain field  can also be obtained from the actual stress field  via the constitutive relation. As a result, for statically determinate structures, the equation (8.1) or the equation (8.2) is sufficient for determining a single unknown displacement A. In a similar fashion, if a rotation at point A of the actual structure (denoted by A) is to be computed, the virtual system shown below is employed. This particular virtual structure is chosen such that it is subjected only to a virtual moment M at the point A and in the same direction as that for A as shown in Figure 8.3.

δσ

A

M

Figure 8.3: Schematic of virtual structure subjected to a concentrated virtual moment M Copyright © 2011 J. Rungamornrat


367


Since the displacement, rotation and strain of the actual structure are geometrically compatible, from the principle of complementary virtual work along with the particular choice of the virtual force and the virtual stress associated with the virtual system shown in Figure 8.3, it leads to WC  U C



M  A  WCs   δσ TεdV

(8.4)

V

Again, the relation (8.4) is valid for any arbitrary virtual moment M and, by choosing M = 1, it yields 1  A  A   δσ TεdV  WCs

(8.5)

V

Note again that for statically determinate structures, the virtual reactions, the virtual internal forces and the actual internal forces can be obtained from static equilibrium. Thus, the virtual stress  and the actual strain  are known and this renders the equation (8.4) or the equation (8.5) containing only one unknown, i.e. A .

8.2 Applications to Trusses Consider a statically determinate truss structure that consists of m straight, prismatic axial members. The (actual) structure is subjected to a set of nodal forces as indicated in Figure 8.4. Let Li and Ai be the length and the cross sectional area of the ith member, respectively, Fi , σ i , ε i  be the internal axial force, the normal stress, and the normal strain of the ith member, respectively. To determine the displacement at a particular nodal point and in a particular direction, we select a virtual structure (with its geometry identical to the actual structure) such that it is subjected to a unit virtual force P = 1 at the same point and in the same direction as that for the displacement to be computed (see Figure 8.4). Let define δFi , δσ i  as the virtual internal axial force and the virtual normal stress of the ith member of the virtual structure, respectively. P2

P1 P3

vA

A

A P = 1

(a)

(b)

Figure 8.4: (a) Schematic of actual truss and (b) schematic of virtual truss Copyright © 2011 J. Rungamornrat


368


By applying the principle of complementary virtual work to this particular case, we obtain m

m

1  v A  v A   δU Ci  WCs    ε i δi dV  WCs i 1

(8.6)

i 1 V

where UCi is the complementary virtual strain energy of the ith member. By recalling that the internal force of a truss member consists only of the axial member, the complementary virtual strain energy UCi can be simplified further as follows δU Ci   ε i δi dV  V

Li

Li

Li

  ε δσ dAdx   ε δσ A dx   ε δF dx  ε L δF i

i

i

0A

0

i

i

i

i

i

i

i

(8.7)

0

where we have used the fact that the member is prismatic and made of a homogeneous material. Ai Fi Li

ε i  ε fi  ε ni σ i  Fi /A i

Fi

Actual ith member

Ai δFi  A i δσ i

δFi

Virtual ith member

Li

Figure 8.5: Schematic of actual and virtual truss members It is important to remark that the equation (8.7) is valid for both linear and nonlinear materials. The actual strain i of the ith member can be decomposed into two parts, the first part fi termed the mechanical strain and the other ni termed the non-mechanical strain, i.e. ε i  ε fi  ε ni

(8.8)

The mechanical strain fi is the strain produced by the internal force Fi and, for a linear elastic material, this strain is given by ε fi 

Fi EiAi

(8.9)

where E i is the Young modulus of the ith member. For nonlinear materials, the mechanical strainfi can also be computed from the constitutive law, fi = fi(Fi/Ai). The non-mechanical strain fi is the strain produced by non-mechanical effects such as the fabrication errors and temperature change. By assuming that the ith member is subjected to the temperature change ΔTi and the lack of fit ei , the non-mechanical strain ε ni is then given by ε ni  α i ΔTi 

ei Li



369


where α i is the coefficient of thermal expansion of the ith member. As a result, the total actual strain ε i for both linear and nonlinear materials becomes e Fi  α i ΔTi  i Li Ei Ai e ε i  ε fi  α i ΔTi  i Li

εi 

for linear elastic materials

(8.11)

for nonlinear materials

(8.12)

By substituting (8.11) or (8.12) into (8.7), we obtain the expression of the complementary virtual strain energy of the ith member UCi as U Ci 

Fi Li Fi  α i Li ΔTi Fi  ei Fi Ei Ai

U Ci  ε fi Li Fi  α i Li ΔTi Fi  ei Fi


(8.13)


(8.14)

Finally, the displacement vA in the equation (8.6) is given by m

vA   i 1

m Fi L i Fi m   α i L i ΔTi Fi   e i Fi  WCs EiAi i 1 i 1

m

m

m

i 1

i 1

i 1

v A   ε fi Li Fi   α i Li ΔTi Fi   ei Fi  WCs


(8.15)


(8.16)

To clearly demonstrate the application of the PCVW (i.e. the equation (8.15) or equation (8.16)) to the deformation/displacement analysis of trusses, several example problems are presented below. Example 8.1 Consider a statically determinate, linearly elastic truss shown below. Properties of all members are given in a table below. Determine the horizontal and vertical displacements at a joint C due to the following four effects: (i) a horizontal force P is applied to a joint B, (ii) a member AC is heated resulting in the temperature change ΔT , (iii) a member CD is subjected to the fabrication error with its length eo shorter than its original length L, and (iv) a roller support (at a joint A) is subjected to the upward settlement  s . B

C

B

P

C

L

L

D

A

D

A

L



370


Member

Area, A i

Young modulus, E i

Coefficient of thermal expansion, i

AB BC AC AD CD

2A A 3A A 2A

E 2E E 2E E

    

The actual system and the virtual systems I and II for determining the horizontal and vertical displacements at a joint C are shown below. Since all three systems are statically determinate, the support reactions and all member forces due to the applied loads can readily be obtained from static equilibrium via either the method of joints or the method of sections. P = 1

–P

P B

2P

0

0

1

1

Actual System

0

–1

D

1 A

–1

C

B

–1

D

P A

P

P

2

0

–P

0

C

B

–P

D

A

P = 1

0

C

0

0 1

0

Virtual System I

Virtual System II

The complimentary virtual strain energy associated with the actual system and the virtual system I is given in the table below. Member

Ai

Ei

αi

Li

Fi

ΔTi

ei

δFi

Fi δFi Li Ai Ei

α i Li ΔTi δFi

ei δFi

AB

2A

E



L

0

0

0

0

0

0

0

BC

A

2E



L

–P

0

0

0

0

0

0

AC

3A

E



2L

ΔT

0

2

2α L ΔT

0

AD

A

2E



L

–P

0

0

–1

0

0

CD

2A

E



L

–P

0

 eo

–1

0

eo

2α L ΔT

eo

2P

 Copyright © 2011 J. Rungamornrat

2 2PL 3AE PL 2AE PL 2AE  2 2  PL  1   3  AE 


371


By applying the principle of complementary virtual work, we then obtain the horizontal displacement at the joint C, uC, as follows: WC  P  u C  WCs  1 u C  1  (  s )  u C   s m

U C   i 1

m m  2 2  PL Fi L i Fi    α i L i ΔTi Fi   e i Fi  1   AE  2 α LT  e 0 3 EiAi i 1 i 1  

WC  U C

 2 2  PL   u C  1   AE  2 α LT  e 0   s Rightward 3  

The direction of the displacement uC is dictated by the direction of the virtual force P and the sign of uC. For this particular case, uC is positive and P is assumed to be rightward; as a result, the displacement uC is rightward. The complimentary virtual strain energy associated with the actual system and the virtual system II is given in the table below.

Member

Ai

Ei

αi

Li

Fi

ΔTi

ei

δFi

Fi δFi Li Ai Ei

α i Li ΔTi δFi

ei δFi

AB

2A

E



L

0

0

0

0

0

0

0

BC

A

2E



L

–P

0

0

0

0

0

0

AC

3A

E



2L

ΔT

0

0

0

0

0

AD

A

2E



L

–P

0

0

0

0

0

0

CD

2A

E



L

–P

0

 eo

–1

PL 2AE

0

eo



PL 2AE

0

eo

2P

By applying the principle of complementary virtual work, we then obtain the vertical displacement at the joint C, vC, as follows: WC  P  v C  WCs  1  v C  0  (  s )  v C m

U C   i 1

m m Fi L i Fi PL   α i L i ΔTi Fi   e i Fi   0  e0 2AE EiAi i 1 i 1

WC  U C

 vC 

PL  e 0 Downward 2AE

The direction of the displacement vC is again dictated by the direction of the virtual force P and the sign of vC. For this case, vC is positive and P is assumed to be downward; as a result, the displacement vC is downward. Copyright © 2011 J. Rungamornrat


372


Example 8.2 Consider a statically determinate, linearly elastic truss shown below. Properties of all members are given in a table below. Determine the displacement at a joint E due to the following four effects: (i) a vertical force P is applied to a joint D, (ii) a member AB is heated resulting in the temperature change ΔT , (iii) a member BC is subjected to the fabrication error with its length eo longer than its original length L, and (iv) a pinned support (at a joint A) is subjected to the upward settlement  s . A

B

L E D

C

1

P

1

L

L

Member

Area, A i

Young modulus, E i

Coefficient of thermal expansion, α i

AB, AC, BD, CD, DE

A

E



BC, BE

2A

E



The actual system and the virtual system for determining the displacement at the joint E are shown below. Since both structures are statically determinate, all support reactions and member forces can be obtained from static equilibrium and the method of joints and the method of sections. Results are reported in the table below. 2

0 B

–P

P

A 0

0

C

A  2

P

0

E

P

2

0

 2

C

D

B –2

 2P

0

2 2

2 2

2P

Actual System

 2

E

D P = 1

3

Virtual System

The complimentary virtual strain energy associated with the above actual and virtual systems is computed and shown in the same table. Copyright © 2011 J. Rungamornrat


373


Member

Ai

Ei

αi

Li

Fi

ΔTi

ei

δFi

Fi δFi Li Ai Ei

α i Li ΔTi δFi

ei δFi

AB

A

E



L

-P

ΔT

0

2 2

2 2PL AE

 2 2αLΔT

0

AC

A

E



L

0

0

0

 2

0

0

0

BC

2A

E



2L

0

0

eo

2

0

0

2eo

BD

A

E



L

P

0

0

0

0

0

0

BE

2A

E



2L

 2P

0

0

-2

2PL AE

0

0

CD

A

E



L

0

0

0

 2

0

0

0

DE

A

E



L

0

0

0

 2

0

0

0

PL 2  2 2  AE

 2 2αLΔT

2eo



By applying the principle of complementary virtual work, we then obtain the displacement at the joint E, E, as follows: WC  P   E  WCs  1   E  2  (  s )   E  2 s m

U C   i 1

m Fi Li Fi m PL   α i Li ΔTi Fi   ei Fi  (2  2 2 )  2 2αLΔT  2e 0 EiAi AE i 1 i 1

WC  U C

  E  (2  2 2 )

PL  2 2αLΔT  2e 0  2 s AE

The displacement E has the same direction as that for the virtual force P if E is positive; otherwise, its direction reverses.

Example 8.3 Consider a statically determinate truss as shown below. Determine the horizontal and vertical displacements at a joint B due to a horizontal force P acting at a joint B for the following two cases: (i) all members are made of a linear elastic material with the uni-axial stress-strain relation governed by   E and (ii) all members are made of nonlinear elastic material with the uni-axial stress-strain relation governed by   1 / 3 . P

2A

A

B

A 60o

30o

3L

L


C


374


The actual system and the virtual systems I and II for determining the horizontal and vertical displacements are shown below. Since the structure in each system is statically determinate, the internal force for both members can be obtained directly from equilibrium of the joint B and results are indicated in the figure below.

P

B –P/2

3P/2

60o

30o

A

C

L

3L Actual System

P = 1 P = 1

B  1/2

–1/2

3/2

60o

30o

A

B

3L Virtual System I

A

C

60o

30o

3L

L

 3/2

C

L

Virtual System II

Case (i): The complementary strain energy of each member is reported in the table below for both the virtual system I and the virtual system II.

Member

Ai

Ei

Li

AB

2A

E

2 3L

BC

A

E

2L

Fi

δFiI

3P/2

3/2

 P/2

1/ 2

δFiII

1/ 2  3/2



Fi Li δFiI EiAi

Fi Li δFiII Ei Ai

3 3PL 4EA PL 2EA



 3 3 1  PL    4  2  EA  

3PL 4EA

3PL 2EA  3 3  PL    2  4  EA  

By employing the principle of complementary virtual work with the virtual system I, we obtain the horizontal displacement u B as Copyright © 2011 J. Rungamornrat


WC  1  u B  u B ;

m

375


U C   i 1

Fi Li δFiI  3 3 1  PL     Ei Ai 2  EA  4

 3 3 1  PL  u B     Rightward 2  EA  4

WC  U C

By employing the principle of complementary virtual work with the virtual system II, we obtain the vertical displacement v B as WC  1  v B  v B ;

m

U C   i 1

Fi Li δFiII  3 3  PL     Ei Ai  2 4  EA

 3 3  PL  v B     Downward 2 4 EA  

WC  U C

Case (ii): The complementary strain energy of each member for this particular case is shown in the table below for both the virtual system I and the virtual system II.

Member

Ai

Li

AB

2A

2 3L

BC

A

2L

Fi 3P 2



P 2

i 

Fi Ai

3P 4A P  2A

   fi   i  

3

3 3P 3 64A 3 α 3



P3 8A 3α 3

δFiI

δFiII

 fi Li δFiI

 fi Li δFiII 

3 2



1 2

9 3P 3 L 64A 3 α 3

1 2



3 2

P3L 8A 3α 3

3P 3 L 8A 3α 3

 9 3 1  P3L    64  8  A 3α 3  

 3 9  P3L    8  64  A 3 α 3  





9P 3 L 64A 3α 3

By employing the principle of complementary virtual work with the virtual system I, we obtain the horizontal displacement u B as WC  1  u B  u B ;

WC  U C

m  9 3 1  P 3L U C    fi Li δFiI     3 3 i 1  64 8  A α

 9 3 1  P 3L  u B     3 3 Rightward  64 8  A α

By employing the principle of complementary virtual work with the virtual system II, we obtain the vertical displacement v B as WC  1  v B  v B ;

WC  U C

m  3 9  P3L U C    fi Li δFiII     3 3 i 1  8 64  A α

 3 9  P 3L  v B     3 3 Downward  8 64  A α Copyright © 2011 J. Rungamornrat


376


8.3 Applications to Flexure-Dominating Structures Consider a statically stable, flexure-dominating structure (beam or rigid frame) that consists of m beam or frame members. This (actual) structure is subjected to a set of forces (concentrated or distributed forces) and moments as shown in Figure 8.6. Let {Li, Ai, Ii, Ji, i} be the length, crosssectional area, moment of inertia, torsional constant, and form factor of the ith member, respectively, {Fi, Mi, Vi, Mti} be the axial force, bending moment, shear force, and internal torque of the ith member introduced by the applied loads, respectively, and σ i and ε i be the stress and strain field within the ith member, respectively. To determine the displacement and rotation at a particular point A and in a particular direction as indicated in Figure 8.6, two virtual structures (with their geometry identical to that of the actual structure), one subjected to a unit virtual force at the point Aand in the direction of the displacement to be computed and the other subjected a unit virtual moment at the point A, are chosen. Let {Fi, Mi, Vi, Mti} be the virtual axial force, virtual bending moment, virtual shear force, and virtual internal torque of the ith member of the virtual structure due to the unit virtual load, respectively, and let σ i be the corresponding virtual stress field induced within the ith member. θA

A

A

(a) δM  1

P = 1 A

A

(b)

(c)

Figure 8.6: Schematic of (a) actual system subjected to applied loads, (b) virtual system I subjected to unit concentrated force at point A and in the direction of A, and (c) virtual system II subjected to unit moment at point A Copyright © 2011 J. Rungamornrat


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Since the displacement and strain of the actual structure is compatible, from the principle of complementary virtual work along with a special choice of the virtual system I, it leads to the complementary virtual work equation governed the displacement A: WC  U C



m

m

1   A  WCs   δU Ci    ε iT δσ i dV i 1

(8.17)

i 1 V

where UCi is the complementary virtual strain energy of the ith member due to the stress in the actual structure acting through the strain in the virtual structure I and WCs is the complementary virtual work due to the support reactions in the virtual structure I acting through the prescribed settlement of the supports in the actual structure. Similarly, by applying the principle of complementary virtual work along with a special choice of the virtual system II, it leads to the complementary virtual work equation governed the rotation A: WC  U C



m

m

1  A  WCs   δU Ci    ε iT δσ i dV i 1

(8.18)

i 1 V

where UCi is again the complementary virtual strain energy of the ith member due to the stress in the actual structure acting through the strain in the virtual structure II and WCs is the complementary virtual work due to the support reactions in the virtual structure II acting through the prescribed settlement of the supports in the actual structure. It is important to emphasize that the unit force and the unit moment applied to the virtual structure I and the virtual structure II have been chosen only for convenience; other magnitudes can also be selected in the analysis and they essentially yield the same solution. It is worth noting that for statically determinate structures, virtual support reactions, σ i and ε i can be determined from static equilibrium and the constitutive relation and this therefore renders either the equation (8.17) or the equation (8.18) containing only one unknown (i.e. either uA or A). As a result, it is sufficient to solve for such unknown displacement or unknown rotation. Note however that the term on the right hand side of (8.17) or (8.18) is still in terms of the stress and strain field and this is not well-suited within the context of structural analysis in which the internal forces are usually represented by the resultants of stress components across the cross section of any member, e.g. the axial force, the shear force, the bending moment and the internal torque. In the following subsection, more straightforward formulae for the complementary strain energy due to axial, shear, bending and torsional effects are derived.

8.3.1 Complementary virtual strain energy due to bending effect Let Mi and Mi be the actual bending moment and the virtual bending moment of the generic ith member in the actual and virtual structures, respectively, as shown in Figure 8.7. Note that both Mi and Mi can be a function of position along the member, i.e. Mi = Mi(x) and Mi = Mi(x). The complementary virtual strain energy of the ith member due to the bending effect, denoted by UCbi, can be obtained as follows: i δi dV  Ei V

δU Cbi   i δi dV   V

Li

L

L

i i  M i y  δM i y  M i δM i 2 M δM     dAdx  y dAdx  0 A  Ei Ii  Ii  0 Ei Ii2 A 0 Ei i Ii i dx (8.19)

where y is the distance from the neutral axis, Ei denotes the Young modulus of the ith member and the flexural formula has been employed to compute the actual and virtual stress fields in terms of the actual and virtual bending moments, i.e. i  M i y/Ii and i  M i y/Ii . Copyright © 2011 J. Rungamornrat


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y

y Mi

δM i

δM i

Mi E , I , L i i i

x

x dx

dx

(a)

(b)

Figure 8.7: Schematic of the generic ith member under the bending moment in: (a) actual structure and (b) virtual structures

8.3.2 Complementary virtual strain energy due to shear effect Normally, the contribution of shear effect on the deformation of a flexure-dominating structure is neglected since the member is practically slender. However, for members whose depth to span ratio becomes large (e.g. deep beams), the shear effect can be significant and needs to be included in the analysis. Let’s consider the generic ith member and let Vi and Vi be the shear forces of this member in the actual structure and in the virtual structure, respectively, as shown in Figure 8.8. Similar to the bending moment, both the Vi and Vi can be a function of position along the member axis, i.e. Vi = Vi(x) and Vi = Vi(x). The complementary virtual strain energy of the ith member due to the shear effect, denoted by UCsi, can be obtained as follows: δU Csi

    i δi dV   i δi dV  Gi V V

Li

 Vi Q i  δVi Q i   dAdx I b i i i i i    0A

   G I b

Li

V δV  i i G i Ai 0

Li  A i Qi2   i Vi δVi dx  2  2 dA dx   G i Ai 0   Ii A bi

(8.20)

where bi is the width of the cross section at the distance y from the neutral axis, Gi denotes the elastic shear modulus of the ith member, i is the form factor of the cross section, and the shear formula has been employed to compute the actual and virtual stress fields in terms of the actual and virtual shear forces, i.e. i  Vi Qi /Ii bi and i  Vi Qi /Ii bi . Note that the form factor  i of any cross section shown in Figure 8.9(a) can be computed from the following formula A Q2 A  i  2i  2i dA  2i Ii A b i Ii

Ct

Qi2 (y)  bi (y) dy C b

(8.21) y

y Vi

Vi

λ i , G i , A i , Li

Vi

Vi

x

x

dx

dx

(a)

(b)

Figure 8.8: Schematic of the generic ith member under the shear force in: (a) actual structure and (b) virtual structures Copyright © 2011 J. Rungamornrat


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b

ct

d/2

A y 

NA

y dy

y

NA

yg c b

A y 

yg

d/2

by 

(a)

(b)

Figure 8.9: Schematic of (a) general cross section and (b) rectangular cross section where Qi(y) is the moment of an area of a region either above or below the distance y from the neutral axis (as indicated by the shaded region in Figure 8.9) about the neutral axis and it is given mathematically by y

Q i (y)  A(y)yg   ξb i ()dy

(8.22)

Cb

For a rectangular cross section of width b and depth d as shown in Figure 8.9(b), the form factor can readily be computed as follows: d  y d  Q(y)  A(y)yg  b  y    2  2 4 

d/2

d/2

2

2





A Q 2 (y) bd d  y d dy  b  y     dy 2  3  I  d / 2 b(y) (bd / 12)  d / 2  2   2 4





15d 5  10d 5  3d 5  2b 2d    1.2 4(bd 3 / 12) 2  480 

The form factor for other types of cross sections can also be calculated in the same fashion using the expressions (8.21) and (8.22) and values for certain cross sections are reported in Table 8.1. Table 8.1: Values of form factor for certain cross sections Section

Form factor 1.2 10/9 2

d

tw

A (Approximate) dt w



380


8.3.3 Complementary virtual strain energy due to axial effect Next, consider the generic ith member due to the axial effect. Let Fi and Fi denote the axial force and the virtual axial force of this particular member in the actual structure and in the virtual structure, respectively, as shown in Figure 8.10. In general, both Fi and Fi are allowed to be a function of position along the member, i.e. Fi = Fi(x) and Fi = Fi(x). The complementary virtual strain energy of the ith member due to the axial effect, denoted by UCai, can be obtained as follows: i δi dV  E V i

δU Cai   i δi dV   V

Li

L

L

i i  Fi  δFi  Fi δFi F δF     dAdx dAdx   0 A  Ei Ai  Ai  0 Ei Ai2 A 0 Eii Aii dx

(8.23)

where the actual stress i and the virtual stress i were obtained in terms of the actual and virtual internal forces Fi and Fi as i  Fi /A i and i  Fi /A i , respectively. For the special case that Fi and Fi are independent of the position along the member axis (e.g. truss members) and the member is prismatic and made of a homogeneous material (i.e. cross sectional area and the Young modulus are constant), the relation (8.23) reduces to δU Cai 

Fi Li δFi  ei δFi EiAi

(8.24)

where ei  Fi Li / E i A i is the elongation of the ith member. y

y Fi

Fi

Fi E i , A i , L i

x

Fi

x

dx

dx (a)

(b)

Figure 8.10: Schematic of the generic ith member under the axial force in: (a) actual structure and (b) virtual structures

8.3.4 Complementary virtual strain energy due to torsional effect Finally, consider the generic ith member with a circular cross section and under torsion. Let Mti and Mti be the internal torque of this member in the actual and virtual structures, respectively, as shown in Figure 8.11. The internal torques Mti and Mti are generally a function of position along the member axis, i.e. Mti = Mti(x) and Mti = Mti(x). The complementary virtual strain energy of the ith member due to the torsional effect, denoted byUCti, can be obtained as follows:

V

L

L

L

i i i  M r  M ti r  M δM M δM i dAdx   ti 2 ti  r 2dAdx   ti ti dx (8.25) δi dV     ti  G G J J G J GiJi i i i i i  i  0 0 A 0 A V

δU Cai    i δi dV  

where the actual shear stress i and the virtual shear stress i were obtained from the torsional formula as i  M ti r/G i J i and i  M ti r/G i J i , respectively, and the property of the torsional constant has been used. For the special case that Mti and Mti are independent of the position along Copyright © 2011 J. Rungamornrat


381


the member axis and the member is prismatic and made of a homogeneous material (i.e. torsional constant and the Shear modulus are constant), the relation (8.25) reduces to δU Cti 

M ti L i δM ti  i δM ti GiJi

(8.26)

where i  M ti L i / G i J i is the relative angle of twist between two ends of the ith member. y

y M ti G i , J i , L i

M ti

M ti

x

M ti

x

dx

dx (a)

(b)

Figure 8.11: Schematic of the generic ith member under the torsion in: (a) actual structure and (b) virtual structures

8.3.5 Complementary virtual strain energy for certain basic members The complementary virtual strain energy δU Ci of the generic ith member due to all possible effects (i.e. the axial, shear, bending and torsional effects) is simply the direct sum of the complementary virtual strain energy due to each effect. The general expression of δU Ci is therefore given by δU Ci 

Li

L

L

L

i i i  i Vi δVi M δM M i δM i Fi δFi    dx dx dx 0 E i A i 0 G i A i 0 E i Ii 0 Gti i J i ti dx

(8.27)

For a prismatic truss member made from a homogeneous material, only the axial force is non-zero and, as a result, δU Ci is simply equal to δU Ci  δU Cai 

Fi L i δFi EiAi

(8.28)

For a beam member, only the axial force and shear force are non-zero and, therefore, δU Ci with consideration of all possible effects is given by δU Ci  δU Csi  δU Cbi

L

Li

i  V δV M δM   i i i dx   i i dx G i Ai E i Ii 0 0

(8.29)

For a two-dimensional frame member subjected only to in-plane loadings, the axial force, shear force and bending moment are present and, as a result, δU Ci with all possible effects is given by δU Ci  δU Cai  δU Csi  δU Cbi

Li

L

L

i i  V δV M δM F δF   i i dx   i i i dx   i i dx E i Ii G i Ai Ei Ai 0 0 0

(8.30)

For a two-dimensional frame member subjected only to out-of-plane loadings (i.e. loads normal to the plane of a structure), the shear force, the bending moment and the internal torque are present and, as a result, δU Ci with all possible effects is given by Copyright © 2011 J. Rungamornrat


382


δU Ci  δU Csi  δU Cbi  δU Cti

Li

L

L

i i  V δV M δM M δM   i i i dx   i i dx   ti ti dx GiJi E i Ii G i Ai 0 0 0

(8.31)

8.3.6 Special integration formula As evident from the expression of the complementary virtual strain energy, it is required, for any member, to compute integrals whose integrand is a product of two functions, one related to quantities from the actual structure and the other corresponding to the virtual internal force of the virtual structure. It is worth noting that all internal forces in the virtual structure are at most piecewise linear (since it is subjected only to either a unit concentrated force in a particular direction or a unit concentrated moment) while the internal forces in the actual structure and other related quantities (e.g. Young’s modulus, shear modulus, cross sectional area, moment of inertia, and torsional constant) can be spatially nonlinear. The piecewise linear function form of the virtual internal force renders the calculation of those involved integrals in a simple fashion as demonstrated below. Let’s consider a definite integral whose integrand is a product of a linear function g = g(x) (see Figure 8.12(b)) and any function f = f(x) (see Figure 8.12(a)) and range of integration is from x = a to x = b: b

I   f(x)g(x)dx

(8.32)

a

Since g is a linear function, it can be written by g(x)  mx  c

(8.33)

where m and c are constants. By inserting (8.32) into (8.33) and then simplify the arithmetic, it leads to following results b

b

b

a

a

a

b

b

a

a

I   f(x)(mx  c)dx  m  f(x)xdx  c  f(x)dx  mx f  f(x)dx  c  f(x)dx

     f(x)dx mx f  c   A f,ab g(x f ) a  b

(8.34)

where Af,ab is the area under the graph of f = f(x) over the interval [a,b] and gx f  is the value of the function g(x) evaluated at the centroid of the area Af,ab as illustrated in Figure 8.12. f

g f = f(x) g = g(x) A f,ab a

g x f 

C.G. xf

b

x

a

xf

b

Figure 8.12: Schematic of any function f = f(x) and a linear function g = g(x) Copyright © 2011 J. Rungamornrat

x


383


The area under a graph and its centroid for certain functions (e.g. constant function, linear function, and polynomial functions), commonly found in the computation of the complementary virtual strain energy, are shown in the table below. It is important to remark that for the last case shown in the table, the function must attain its maximum at the middle of the interval.

Function

Constant x

Figure

Cx n

Linear x

x

x h

h

h

Parabola

L

h

L

L

Area

Lh

Lh 2

Lh n 1

2Lh 3

x

L 2

L 3

L n2

L 2

x

L

Example 8.4 Consider a statically determinate beam subjected to external loads as shown in the figure below. The moment of inertia I, the cross sectional area A, the form actor , the Young modulus E, and the shear modulus G are assumed to be constant throughout the structure. Determine the deflection at a point C, denoted by v C , and the rotation at point A, denoted by θ A , due to all possible effects.

2qL

q

C A

B 4L

L

Solution First, let’s determine the deflection at a point C. The actual system and the virtual system I (associated with a structure subjected to a unit vertical force at a point C) are given below. Since the structures in both systems are statically determinate, all support reactions, the bending moment and the shear force for the entire beam can readily be obtained from static equilibrium. The bending moment diagram, the M/EI diagram, the shear force diagram, and V/GA diagram are shown below. Copyright © 2011 J. Rungamornrat


384


P = 1

2qL

q A 3qL/2

A

C

B

1/4

9qL/2 4L

B

L

5/4 4L

L 1

2qL

3qL/2

C

V

δV 1/ 4

–4qL 2qL/GA

3qL/2GA

λV GA

δM

–4qL/GA 6qL

L

2

M –2qL

2

–8qL2 6qL2/EI

–2qL2/EI

M EI

–8qL2/EI

Actual System

Virtual System I

Since there is no support settlement in the actual structure, the complementary virtual work associated with all reactions in the virtual structure I vanishes, i.e. M Cs  0 . The total complementary virtual work produced by the actual displacement through the virtual forces in the virtual structure I is therefore given by WC  P  v C  WCs  1  v C  0  v C

The complementary virtual strain energy due to bending effect is given by U Cb  U

AB cB

 U

BC cB



L AB

 0

M δM dx  EI

L BC

 0

M δM dx EI



385


2 2 1  6qL2   3L  1  2qL   2L   2L  1  8qL  L   4L   4L              2  EI   4  2  EI   3   3  3  EI  2qL4  3EI

The complementary virtual strain energy due to bending effect is given by δU Cs  δU

AB Cs

 δU

BC Cs



L AB

 0

V δV λ dx  GA

L BC

 0

λ

V δV dx GA

 3λ qL   1  1  4λ qL   1   2λ qL   4L       4L      L 1  2GA   4  2  GA   4   GA  

5λ qL2 2GA

The total complementary virtual strain energy due to all possible effects is equal to 2qL4 5λ qL2 δU C  δU Cb  δU Cs   3EI 2GA

Since the strain field and the displacement field of the actual structure is compatible, from the principle of complementary virtual work along with a particular choice of the virtual system I, it leads to the displacement v C δWC  δU C  v C  v Cb  v Cs 

2qL4 5λ qL2  Downward 3EI 2GA

For a beam with a rectangular cross section of depth h and Poisson ratio v  0.25 , the ratio between the displacement due to the shear effect ( v Cs ) and the displacement due to the bending effect ( v Cb ) is given by v Cs 5λ qL2 3EI 15  E  I  15  I  15  h     λ  2   λ 1  ν  2     4 v Cb 2GA 2qL 4  G  AL  2  AL  4  L 

2

It is obvious that the contribution of the shear effect to the deflection of the beam can become significant for large h/L. Next, let’s determine the rotation at a point A. The actual system remains unchanged while the virtual system II subjected to a unit concentrated moment at the point A is chosen in the analysis. Since this virtual structure is also statically determinate, all support reactions, the bending moment and the shear force can be obtained, again, from static equilibrium and results are indicated in the figure below. Since there is no support settlement in the actual structure, the complementary virtual work associated with the virtual reactions vanishes, i.e. δWCs  0 . The complementary virtual work produced by the actual displacement via the virtual forces in the virtual structure II is therefore given by WC  M   A  WCs  1   A  0   A

The complementary virtual strain energy due to the bending effect is equal to Copyright © 2011 J. Rungamornrat


386


M = 1 A

B

1/4L

1/4L 4L

C

L 1/4L δV

δM

–1 Virtual System II

AB BC  δU Cb  δU Cb  δU Cb

L AB

 0

M δM dx  EI

1  6qL   2  EI

2

L BC

 0

M δM dx EI

2  4qL3  1  1  1  8qL  4L      4L       3EI  4  3  3  EI  

The complementary virtual strain energy due to the shear effect is equal to AB BC  δU Cs  δU Cs  δU Cs

L AB

 0

λ

V δV dx  GA

L BC

λ 0

V δV dx GA

λqL  3λ qL   1  1  4λ qL   1   4L      4L     2GA  4L   2GA   4L  2  GA 

The total complementary virtual strain energy due to all possible effects is equal to δU C  δU Cb  δU Cs  

4qL3 λqL  3EI 2GA

Since the strain field and the displacement of the actual structure is compatible, from the principle of complementary virtual work, the complimentary virtual work and the complementary virtual strain energy must also be balanced for a particular choice of the virtual system II and this yields the rotation at the point A as follows: δWC  δU C



θ A  θ Ab  θ As  

4qL3 λqL  CW 3EI 2GA

The negative sign of A indicates that its direction is opposite to that of the unit virtual moment M. Copyright © 2011 J. Rungamornrat


387


Example 8.5 Consider a simply-supported beam subjected to uniformly distributed load q as shown below. The moment of inertia I, the cross sectional area A, the form factor , the Young modulus E, and the shear modulus G are assumed to be constant. Determine both the deflection and rotation at point B due to all possible effects.

q A

C

B L

2L

Solution The actual structure and two virtual structures (one for determining the deflection at the point B and the other for determining the rotation at the point B) are given below. Since all three systems are statically determinate, the support reactions, the bending moment and the shear force for the entire beam can be readily obtained from static equilibrium. The SFD and BMD diagrams for all structures and M/EI and V/GA diagrams for the actual structure are given below

P = 1

q A

A

C

B

3qL/2

2/ 3

3qL/2 2qL

3qL/2

C

B

1/ 3 2/ 3

δV

V –qL 3qL/2GA

 1/ 3

–3qL/2 2qL/GA

2L/3 λV GA

–qL/GA

δM

–3qL/2GA 3qL

2

3qL2/2

M –qL2/2 –2qL2 3qL2/EI 2

3qL /2EI

M EI

–qL2/2EI –2qL2/EI

Virtual System I

Actual System Copyright © 2011 J. Rungamornrat


388


M = 1

A

C

B

1/3L

1/3L δV –1/3L 2/3

δM –1/3

Virtual System II By applying the principle of complementary virtual work along with the virtual system I, we can obtain the deflection at the point B as follows: δWC  δP  v B  δWCs  1  v B  0  v B δU Cb  δU

AB Cb

 δU

BC Cb

L AB





0

M δM dx  EI



1  3qL  2  2EI



11qL4 12EI

2

L AB



AB BC  δU Cs  δU Cs  δU Cs

0

λ

L BC

 0

M δM dx EI

2   4L  1  qL2   L  1  3qL2   4L  1  2qL  L L  L     2L  2L           9 3 2EI 2 2 EI 9 3 EI    2          

V δV dx  GA

L BC

λ 0

V δV dx GA

 3 λqL   2  1  λqL   2  1  2 λqL   1   3 λqL   1  L      L     2L       2L    2GA 3 2 GA 3 2 GA 3 2GA                3



λqL2 GA

δU C  δU Cb  δU Cs 

11qL4 λqL2  12EI GA

δWC  δU C

v B  v Bb  v Bs 



11qL4 λqL2  12EI GA

Downward

To further investigate the contribution of the shear effect to the total deflection, let’s assume that the cross section of the beam is of rectangular shape with depth h and Poisson ratio is equal to 0.25. Copyright © 2011 J. Rungamornrat


389


The ratio between the deflection due to the shear effect (vBs) and the deflection due to the bending effect (vBb) is therefore given by v Bs λqL2 12EI 12  E  I  6  I  3 h    λ  2   λ1  ν  2     4 v Bb GA 11qL 11  G  AL  11  AL  44  L 

2

This clearly indicates that the contribution of the shear effect on the total deformation of the beam depends primarily on the depth to span ratio; in particular, the shear effect becomes significant and needs to be considered when h/L is relatively large. By employing the principle of complementary virtual work along with the virtual system II, we obtain the rotation at the point B as follows: δWC  δM  θ A  δWCs  1  θ A  0  θ A L AB



AB BC  δU Cb  δU Cb  δU Cb

0

δU Cs  δU

AB Cs

 δU

BC Cs

M δM dx  EI



1  3qL  2  2EI



13qL3 24EI



2

L AB

 0

L BC

 0

M δM dx EI

2   2  1  qL2   1  1  3qL2   4  1  2qL  1 L       L      2L      2L    9  3  EI  2   9  3  2EI   4  2  EI 

V δV λ dx  GA

L BC

 0

λ

V δV dx GA

 3 λqL   1  1  λqL   1  1  2 λqL   1   3 λqL   1   L      L     2L      2L     2GA   3L  2  GA   3L  2  GA   3L   2GA   3L 

0

δU C  δU Cb  δU Cs 

13qL3 0 24EI

δWC  δU C

θ B  θ Bb  θ Bs 



13qL3 CW 24EI

Note again that the direction of the deflection and rotation at the point B follows directly that of the virtual load if their computed value is positive; otherwise (i.e. the computed value is negative), their direction is opposite to that of the virtual load. Example 8.6 Consider a statically determinate beam subjected to a concentrated moment as shown below. The moment of inertia, the form factor, the cross sectional area, the Young modulus, and the shear modulus of the segment AB and segment BC are given by 2I, λ,2A, E, G and I, λ, A, E, G , respectively. Assume that during load application, the fixed support moves downward with an amount s. Determine the relative rotation of a hinge B, Δθ B , due to bending and shear effects and the support settlement.

M A

B C L

2L



390


Solution The actual structure and the virtual structure for determining the relative rotation at the hinge B are shown below. Since the relative rotation at the hinge B is defined by Δθ B  θ BR  θ BL , two unit virtual moments of opposite directions are then applied to the hinge B in the virtual system such that one is applied just to the left of the point B and the other is applied just to the right of the point B as shown in the figure below. All support reactions, the SFD and BMD for both actual and virtual structures, and M/EI and V/GA diagrams for the actual structure are given below M/2

A

A

C

B

M/2L

C

B

1/2L

M/2L

L

δM  1 δM  1

3/2

M

1/2L

L

2L

2L 1/2L

M/2L

V

δV

λV GA

δM

M/2GAL

M/4GAL

–1

M

–3/2

M –M/2

M/EI

M EI

–M/4EI

Actual System

Virtual System

By applying the principle of complementary virtual work along with the chosen virtual system, the relative rotation at the hinge B can be obtained as follows: Δ  1  WC  1  BR  1  BL   Δ s  ΔB  s 2L  2L  AB BC  δU Cb  δU Cb  δU Cb

L AB

 0



M δM dx  EI

L BC

 0

M δM dx EI

ML 1 1 M  1 M    1  L   1     2L      6EI 3  2  EI  2  4EI    3 Copyright © 2011 J. Rungamornrat


AB BC  δU Cs  δU Cs  δU Cs

L AB

 0

λ

391


V δV dx  GA

L BC

 0

λ

V δV dx GA

 λM   1   λM   1  5 λM    L     2L     4LGA   2L   2LGA   2L  8LGA

δU C  δU Cb  δU Cs  

δWC  δU C



ML 5λ M  6EI 8LGA

ΔB  ΔBb  ΔBs  ΔB,settlement  

ML 5M Δ s   6EI 8LGA 2L

The relative rotation at the hinge B is in a counter clockwise direction if Δθ B is positive; otherwise, it is in a clockwise direction. Example 8.7 Consider a statically determinate frame subjected to external loads as shown in the figure below. The moment of inertia, the form factor, the cross sectional area, the Young modulus, and the shear modulus of the segment AB, segment BC, and segment CD are given by 2I, λ,2A, E, G, 2I, λ, A, E, G , and I, λ,2A, E, G, respectively. Assume that the fixed support at a point A rotates in a counter clockwise direction with an amount of θ o . Determine both the vertical displacement and the rotation of a point D due to all possible effects. 2q

B

A

L qL

2qL2

D

C

L

L

Solution The actual structure and the virtual structure I used to determine the vertical displacement at the point D are shown in the figure below. Since frames in both systems are statically determinate, all support reactions, the axial force, the bending moment, and the shear force for the entire structure can readily be obtained from static equilibrium. The F/AE, M/EI and V/GA diagrams for the actual structure and AFD, SFD and BMD for the virtual structure are shown below. By employing the principle of complementary virtual work along with the chosen virtual system I, the vertical displacement at the point D due to all possible effects can be obtained as follows: Copyright © 2011 J. Rungamornrat


392


Actual System 2q

qL2 0

Virtual System I 2L 0

B

A

B

A 1

3qL qL

2qL2

P =1

D

C

0

0

F/EA

D

C

F

qL/EA

1

0

0

qL/2GA 

qL/2GA

V/GA

V





qL/2GA



qL2/2EI

–L –2L

2

–qL /2EI

/EI

qL2/2EI

–L

M

–L

–qL2/EI Copyright © 2011 J. Rungamornrat


393


WC  1  v D  WCs  v D  2Lo AB BC U Ca  U Ca  U Ca  U CD Ca



L AB

 0

FF dx  EA

L BC

FF dx  EA

 0

L CD

 0

FF dx EA

2

qL  qL   0 L 1  0  EA  EA  AB BC U Cb  U Cb  U Cb  U CD Cb



L AB

 0



MM dx  EI

L BC

 0

MM dx  EI

LCD

 0

MM dx EI

1  qL   L  1  qL2   3L   qL2  1  qL2   2L   L   L      L   L  L  L     L       2  2EI   3  3  2EI   4   2EI  2  EI   3  2



5qL4 24EI

AB BC U Cs  U Cs  U Cs  U CD Cs



LAB





0

VV dx  GA

LBC

 0



VV dx  GA

LCD

 0

VV dx GA

1  3qL  qL   3qL2  qL    L 1  L 1   2GA 2 2GA  2GA  

U C  U Ca δWC  δU C

qL2 5qL4 3qL2  U Cb  U Cs    EA 24EI 2GA



v D  v Da  v Db  v Ds  v Dss

qL2 5qL4 3qL2     2Lo EA 24EI 2GA

The vertical displacement at point D directs downward if v D is positive; otherwise it directs upward. To compute the rotation at the point D, the virtual structure II shown in the figure below is chosen in the analysis. All support reactions, the axial force, the bending moment and the shear force diagrams for this virtual structure are given below. 1 0

A

B

0

0

F

C

M =1 D

Virtual System II Copyright © 2011 J. Rungamornrat

0 0


394


–1 

V

M



–1

 –1

It is evident that the virtual axial force F and the virtual shear force V identically vanish for the entire frame in the virtual system II. This implies that there is no contribution of the axial and shear effects to the rotation at the point D. By employing the principle of complementary virtual work along with the chosen virtual structure II, we can obtain the rotation at the point D as follows: WC  1  D  WCs  D  o

U Ca  U

AB Ca

 U

BC Ca

 U

CD Ca



L AB

FF dx  EA

 0

AB BC U Cs  U Cs  U Cs  U CD Cs 

LAB





0

L BC

FF dx  EA

 0

VV dx  GA

L BC

 0



L CD

 0

FF dx  0 EA

VV dx  GA

LCD

 0

VV dx  0 GA

AB BC U Cb  U Cb  U Cb  U CD Cb



LAB

 0

MM dx  EI

LBC

 0

MM dx  EI

LCD

 0

MM dx EI

 qL2  1  qL2  1  qL  1  qL  qL3 L  1   L  1    L  1   L  1      2  EI  3  2EI  2  2EI  12EI  2EI  2

2

U C  U Ca  U Cb  U Cs  0  δWC  δU C



qL3 qL3 0   12EI 12EI

D  Da  Db  Ds  Dss  

qL3  o CCW 12EI

Since the computed rotation D is negative, the direction of such rotation is opposite to that of the unit virtual moment M. Example 8.8 Consider a statically determinate frame subjected to external loads as shown in the figure below. The moment of inertia, the form factor, the cross sectional area, the Young modulus, and the shear modulus of the segments AB and BC are given by 2I, ,2A, E, G and I, , A, E, G, respectively. Assume that the support at a point C settles downward with an amount of  o . Determine the both the displacement and rotation at a point C due to all possible effects. Copyright © 2011 J. Rungamornrat


395


2P P

C

B

L

A

3L/4

3L/4

Solution The actual structure and the virtual structure I for determining the horizontal displacement at the point C are shown below. Again, the frames in both systems are statically determinate, all support reactions the axial force, the bending moment, and the shear force can be obtained from static analysis. The F/AE, M/EI and V/GA diagrams for the actual structure and AFD, SFD and BMD for the virtual structure I are given below. By employing the principle of complementary virtual work along with the virtual structure I, we then obtain the horizontal displacement at the point C due to all possible effects, denoted by uC, as follows:

2 o 3

WC  1  u C  WCs  u C  AB BC  U Ca  U Ca  U Ca

L AB

 0

FF dx  EA

L BC

 0

FF dx EA

PL  P  2  L    0  9EA  6EA   3 

U Cb  U

AB Cb

 U

BC Cb



LAB

 0

MM dx  EI

LBC

 0

MM dx EI 3

1  PL   2L  1  5PL  3L  2L  1  3PL  3L  5L  91PL              L  2  2EI   3  2  2EI  2  3  2  2EI  4  6  96EI AB BC  U Cs  U Cs  U Cs

L AB

 0



VV dx  GA

L BC

 0

VV dx GA

 P  3L  2   5P  3L  2  7PL  P               L 1    3GA  4  3   3GA  4  3  6GA  2GA 

U C  U Ca  U Cb  U Cs 

PL 91PL3 7PL   9EA 96EI 6GA

δWC  δU C  u C  u Ca  u Cb  u Cs  u Css 

PL 91PL3 7PL 2 o Rightward    9EA 96EI 6GA 3



396


Actual System

Virtual System I

Virtual System II

2P P

B

B

B

C

C

C

A

A

A

0

1

F/EA

–P/6EA

P/2GA

 = 1

P = 1

2/3

F

0

2L/3

F

P/3GA 

P/3GA

V/GA



V

L 

V

5PL/2EI L PL/2EI

L –3PL/2EI

/EI

M


0

M

–1


397


To compute the rotation at the point C, the virtual structure II subjected to the unit moment at the point C is chosen in the analysis. The AFD, SFD and BMD for this virtual structure are given in the figure above. Again, by employing the principle of complementary virtual work along with the virtual structure II, the rotation at the point C can readily be computed as follows: WC  1  C  WCs  C  U Ca  U

AB Ca

 U

BC Ca



L AB

 0

AB BC  U Cb  U Cb  U Cb

L AB

 0

2 o 3L FF dx  EA

L BC

FF P  P   2  dx   L    0  EA 9EA  6EA   3L 

 0

MM dx  EI

L BC

 0

MM dx EI

1  5PL  3L  1  1  3PL  3L  1  61PL2 0               2  2EI  2  3  2  2EI  4  6  96EI U Cs  U

AB Cs

 U

BC Cs



L AB

 0

VV  dx  GA

L BC

 0

VV dx GA

 P  3L  2   5P  3L  2  2P  0              3GA  4  3L   3GA  4  3L  3GA

U C  U Ca  U Cb  U Cs  δWC  δU C



P 51PL2 2P   9EA 96EI 3GA

C  Ca  Cb  Cs  Css 

P 51PL2 2P 2 o    9EA 96EI 3GA 3L

The rotation at the point C directs clockwise if C is positive (the same direction as the virtual moment M); otherwise its direction reverses. Example 8.9 Consider a statically determinate structure subjected to a load P as shown below. The cross sectional area, the Young modulus and the coefficient of thermal expansion of the axial member BD is given by A, E,  , respectively, and the moment of inertia, the form factor, the cross sectional area, the Young modulus, and the shear modulus of the frame member ABC is given by I, ,2A,2E, G, respectively. Determine the vertical displacement and the rotation at a point C due to all possible effects. Assume that the support at a point D moves downward with an amount  o and the temperature of the member BD increases with an amount of To .

D L P A

C

B L Copyright © 2011 J. Rungamornrat

L


398


Solution The actual structure and the virtual structure I for computing the vertical displacement at the point C are shown in the figure below. All support reactions, the axial force, the bending moment and the shear force for both structures are obtained from static equilibrium and the F/AE, M/EI and V/GA diagrams for the actual structure and AFD, SFD and BMD for the virtual structure I are shown below.

D

2P

2P

L

2 2 P/EA

P A 2P

P

B

–P/2EA

0

L

C

L P/2GA

V/GA P/2GA

M/EI –PL/2EI

Actual System

D

2 2

L

2 2

P = 1 A 2

P

–2

B

L

0

C

L 

δV 1

δM –L

Virtual System I Copyright © 2011 J. Rungamornrat


399


By employing the principle of complementary virtual work along with the virtual structure I, we then obtain the vertical displacement at the point C due to all possible effects as follows: WC  1  vC  WCs  vC  2 o AB BC BD U Ca  U Ca  U Ca  U Ca



L AB

 0

FF dx  EA

L BC

  F L F FF dx   BD BD BD   BD TBD L BDFBD  EA   E BD A BD

 0

 2 2P  P        L  2   0    2EA   EA 







2L 2 2  αΔTo



 



2L 2 2  1  8 2

PL  EA  4 T L o

AB BC BD U Cb  U Cb  U Cb  U Cb



L AB

 0

MM dx  EI

L BC

 0

MM dx  EI

L BD

 0

MM dx EI

PL3 1  PL   2L  1  PL   2L    L      L   0 3EI 2  2EI   3  2  2EI   3  AB BC BD U Cs  U Cs  U Cs  U Cs



L AB

 0

VV  dx  GA

L BC





0

VV dx  GA

L BD

 0

VV dx GA

PL  P   P    L  1    L  1  0  GA  2GA   2GA 



U C  U Ca  U Cb  U Cs  1  8 2 δWC  δU C

PL PL PL  EA  4T L   3EI GA 3

o

v C  v Ca  v Cb  v Cs  v Css  v CT





 1 8 2

PL PL PL  EA    2 3EI GA 3

o

 4To L Downward

To compute the rotation at the point C, the virtual structure II subjected to a unit moment at the point C is chosen in the analysis. All support reactions, the AFD, SFD and BMD for this particular virtual structure are shown in the below. By applying the principle of complementary virtual work along with the virtual structure II, the rotation at the point C due to all possible effects can be obtained as follows: WC  1  C  WCs  C 

o L

AB BC BD U Ca  U Ca  U Ca  U Ca



L AB

 0

FF dx  EA

L BC

 0

 F L F  FF dx   BD BD BD   BD TBD L BDFBD  EA  E BD A BD 

 

 2 2P   2  2P   1   2L     L     0     L   To  4EA   L   EA   

 2L L2    12  4








 P 2  2To  EA


400


D

1/L

1/L

L

2 /L

M = 1

A 2/L

B

–2/L

1/L

L

0

C

L 

δV

L

δM –1

–1

Virtual System II

AB BC BD U Cb  U Cb  U Cb  U Cb



L AB

 0

MM dx  EI

L BC

 0

MM dx  EI

L BD

 0

MM dx EI

5PL2 1  PL   2  1  PL    L  1  0  L       12EI 2  2EI   3  2  2EI  AB BC BD U Cs  U Cs  U Cs  U Cs



L AB





0

VV dx  GA

L BC

 0



VV dx  GA

L BD

 0

VV dx GA

P  P    1    L    0  0  2GA  2GA   L 

P 5PL2 1  P U C  U Ca  U Cb  U Cs    4 2   2To   12EI 2GA 2  EA

δWC  δU C



C  Ca  Cb  Cs  Css  CT 5PL2 P  1  P    4 2    o  2To CW 2  EA 12EI 2GA L

Since the obtained rotation at the point C is positive, its direction therefore follows directly that of the virtual moment M. Copyright © 2011 J. Rungamornrat


401


Example 8.10 Consider a statically determinate structure fully fixed against the displacement and rotation at a point C and subjected to a concentrated load in the negative Z-direction as shown in the figure below. Note that the structure is contained entirely in the X-Y plane. The moment of inertia I, the form factor , the cross sectional area A, the torsional constant J, the Young modulus E, and the shear modulus G are assumed constant throughout. Determine the vertical displacement at a point D due to all effects.

A

L Z

P

B D

Y

X

L L/2

C

Solution The actual structure and the virtual structure for computing the vertical displacement at the point D are shown below. Since both structures are statically determinate and the applied load is normal to the plane of a structure, only the bending moment, shear force, and internal torque exist and they can be obtained from static equilibrium. The M/EI, V/GA and Mt/GJ diagrams for the actual structure and SFD, BMD and ITD (internal torque diagram) for the virtual structure are given below. By applying the principle of complementary virtual work along with this choice of the virtual structure, we then obtain the vertical displacement at the point D as follows:

WC  1  v D  v D AB BC U Cb  U Cb  U Cb  U CD Cb



L AB

 0

MM dx  EI

L BC

MM dx  EI

 0

LCD

MM dx EI

 0

1  PL  L  L  1  PL   2L   PL  1  PL   L  11PL3   L     L 0     L          2  2EI  2  3  2  EI   3   2EI  2  EI   6  24EI AB BC U Cs  U Cs  U Cs  U CD Cs



L AB

 0



VV dx  GA

L BC





0

VV dx  GA

LCD

 0

VV dx GA

5PL  P  L   P   P      1    L  1    L  1  2GA  GA  2   GA   GA  AB BC U Ct  U Ct  U Ct  U CD Ct



L AB

 0

M t M t dx  GJ

L BC

 0

M t M t dx  GJ

LCD

 0

M t M t dx GJ

5PL3  PL   L   PL  0 L     L L   4GJ  2GJ   2   GJ  Copyright © 2011 J. Rungamornrat


402


A

A

P

Z

B D

Y

P = 1

B

D

X

C

C

Actual System

D

L/2

V/GA /EI t/GJ C

L

V/GA

Virtual System

C

D

P/GA

V

–PL/2EI

M

0

t

B

/EI

–PL/EI

t/GJ

–PL/2GJ

B

L

 L/2 0 L

V

B

1

M

L

t

A

C

1

C

P/GA

L/2

L/2

B

L

A

V/GA

P/GA

V

1

/EI

–PL/2EI –PL/EI

M

L/2 L

t/GJ

–PL/GJ

t

L



U C  U Cb  U Cs  U Ct 

δWC  δU C

403


11PL3 5PL 5PL3   24EI 2GA 4GJ

v D  v Cb  v Cs  v Ct 



11PL3 5PL 5PL3   Downward 24EI 2GA 4GJ

Since the computed displacement vD is positive, the direction of such rotation is identical to that of the unit virtual force P.

Exercises 1. Use the principle of complementary virtual work (or the unit load method) to determine the horizontal and vertical displacement at a joint Q of statically determinate trusses shown below. Assume that all trusses are made from linear elastic materials. Change in temperature, lack of fit, and support settlement are indicated in each figure (if exists). 2EA

2EA

EA

P

P

EA, eo

L

EA is constant

EA EA, –T,

EA

Q

EA

EA

Q

o

2P L

L

4L

Q

Q 3L

2EA, –T,

EA

L

EA is constant

L

EA, –eo

P P

L

o 4L

2.25L

L

L

2. Use the principle of complementary virtual work (or the unit load method) to determine the deflection and rotation at a point Q and the hinge rotation (if exists) of statically determinate beams shown below. Assume that all beams are made from linear elastic materials with constant Young modulus E and shear modulus G and that the form factor  is constant throughout. The cross sectional area, the moment of inertia and support settlements (if exists) are indicated in each figure. Consider all possible effects. Copyright © 2011 J. Rungamornrat


404


qL

q Q

A, I

A, I

2A, 2I

L

o

2L

L

M A, I

2A, 2I

Q

L

2A, 2I 2L

L

qL

q o

2A, 2I 2L

A, I

L

L

Q

qL

q o

A, I

2A, 3I

A, I

L

L

Q

2A, 2I L

L P

2P A, I

Q

L

A, I L

A, I

A, I

L

L

3. Use the principle of complementary virtual work (or the unit load method) to determine the horizontal and vertical deflection and rotation at a point Q and the hinge rotation (if exists) of statically determinate frames shown below. Assume that all frames are made from linear elastic materials with constant Young modulus E and shear modulus G and that the form factor  is constant throughout. The cross sectional area, the moment of inertia, length of the segment, and support settlements (if exists) are indicated in each figure. Consider all possible effects.

Q

A, I, 2L

qL o

2A, I, L/2

A, I, L q A, I, 2L

P 2A, I, L/2

o


Q


405


q Q

Q

2qL

2A, 2I, L

A, I, L

A, I, L

PL

A, I, 1.5L

A, 2I, L

A, 3I, L

2A, 2I, L qL2

P

o PL Q

A, I, L 45o

45o

45o

2qL

A, I, L

Q

2A, 2I, L

A, I, L

q

q

P A, I, L

A, I, 2L

A, I, L

2P

Q

Q

A, I, L A, I, L

A, I, 2L 2A, 2I, L

2A, 2I, 2L

4. Use the principle of complementary virtual work (or the unit load method) to determine the distance between the point A and B after a pair of point forces or point moments is applied. Consider only the bending effect and assume that all structures are made from linear elastic materials and the flexural rigidity EI is constant throughout. P

A

B

L

L

P

AB

P

L

P L

L

L

L

L PL

L

A

PL L/2

L/2

P

P P

P

A

AB

A

B P

2L

B P

2L

60o

60o

L

L

L

L


L

B


406


5. Use the principle of complementary virtual work (or the unit load method) to determine the displacement and rotation at point B of a structure shown below. The axial rigidity EA of all members of a truss is assumed to be constant and the flexural rigidity, axial rigidity and the shear rigidity of the segment AB are given by EI, EA and GA, respectively. Consider all possible effects. P

–T,

T, 

–eo

P

L

B

A

o

2P L

L

L


L

407


Applications of Castigliano’s 2nd Theorem

CHAPTER 9 APPLICATIONS OF CASTIGLINO’S 2nd THEOREM In this chapter, we demonstrate the application of Castigliano’s 2nd theorem outlined in Chapter 6 to the analysis of two-dimensional, linearly elastic skeleton structures (i.e. structures consisting of one-dimensional members). In the first part, the theorem is applied to perform the deformation/displacement analysis of statically determinate structures such as trusses, beams and frames. Since Castigliano’s 2nd theorem derives directly from the principle of complementary virtual work (PCVW), it provides an alternative analysis tool of similar capability (at least within the context of structural analysis focused in this book) but offers a different analysis procedure that may attract those searching for their own best approach. The second part of this chapter devotes to the application of the theorem to form a force method (i.e. a method where static quantities such as support reactions and internal forces are treated as primary unknowns) for the analysis of statically indeterminate structures.

9.1 Castigliano’s 2nd Theorem for Linearly Elastic Structures P2 , u 2

P3 ,u 3 P1 , u1

Undeformed state Deformed state

σ ,ε Pi , u i Pn , u n

Figure 9.1: Schematic of linearly elastic structure subjected to applied loads {P1, P2, P3,…, Pn} Let’s consider a statically stable structure that is made from linearly elastic material and is subjected to a set of forces and/or moments {P1, P2, P3,…, Pn} as shown in Figure 9.1. It is worth noting that the complementary strain energy UC of a structure made from a linearly elastic material is identical to the strain energy U, i.e. UC = U (see section 6.3 in Chapter 6 for more details). As a result, Castigliano’ 2nd theorem (i.e. equation (6.103) in Chapter 6), when specialized to this particular linearly elastic structure, takes the form ui 

U WCs  Pi Pi

i  1,2,3,..., n

(9.1)

where ui is the displacement (or rotation) at the location and in the direction of the applied load Pi. It is emphasized that both the strain energy U and the complimentary work due to support settlement WCs must be expressed as a function of the applied loads {P1, P2, P3,…, Pn}, i.e. U = U(P1, P2, P3,…, Pn) and WCs = WCs(P1, P2, P3,…, Pn). In addition, the applied loads P1, P2, P3,…, Pn must be all independent. For a general two-dimensional skeleton structure, the strain energy U due to all possible effects is given by U  Ua  U b  Us  U t


408



where U a denotes the strain energy due to the axial effect, U b denotes the strain energy due to the bending effect, U s denotes the strain energy due to shear effect, and U t denotes the strain energy due to the torsional effect. For a structure consisting of m members, {U a , U b , U s , U t } are given in explicit forms by

 1 L Fi2  U a     dx  i 1  2 0 E i A i 

(9.3)

 1 L M i2  U b     dx  i 1  2 0 E i I i 

(9.4)

L m  1  V2  U s     i i dx  i 1  2 0 G i A i 

(9.5)

L m  1 M2  U t     ti dx  i 1  2 0 G i J i 

(9.6)

m

m

where {Fi , M i , Vi , M ti } denote the axial force, the bending moment, the shear force, and the internal torque of the ith member expressed as a function of the applied loads {P1, P2, P3,…, Pn}, i.e. Fi  Fi (P1 , P2 , P3 ,..., Pn ) , M i  M i (P1 , P2 , P3 ,..., Pn ) , Vi  Vi (P1 , P2 , P3 ,..., Pn ) and M ti  M ti (P1 , P2 , P3 ,..., Pn ) . As a result, the total displacement (or total rotation) ui due to all possible effects can be decomposed as u i  u i, a  u i, b  u i,s  u i, t  u i, support

(9.7)

where the displacement due to the axial effect u i, a , the bending effect u i, b , the shear effect u i,s , the torsional effect u i, t and the support settlement u i,support are given by

u i, a

L m  m  L Fi Fi  U a   1 Fi2     dx      dx   Pi i 1 Pi  2 0 E i A i  i 1  0 E i A i Pi 

(9.8)

u i, b 

L L m U b   1 M i2  m  M i M i          dx   i 1  E I P dx  Pi i 1 Pi  2 0 E i Ii i  0 i i 

u i,s 

L L m U s   1  i Vi2  m   i Vi Vi          dx   i1  G A P dx  Pi i 1 Pi  2 0 G i A i  0 i i i 

(9.10)

u i, t 

L L m U t   1 M 2ti  m  M ti M ti         dx     G J P dx  Pi i 1 Pi  2 0 G i J i i  1 i i i  0 

(9.11)

u i,support  

N WCs R i   u si Pi Pi i 1

(9.9)

(9.12)

where usi is the prescribed displacement (or rotation) at the support, R i  R i (P1 , P2 , P3 ,..., Pn ) is the support reaction at the location and in the direction of the prescribed usi, and N is the number of prescribed movements of all supports. Copyright © 2011 J. Rungamornrat

409



It should be noted that for a truss structure consisting of prismatic members made of homogeneous materials, there is no contribution of the bending, shear, and torsional effects to the displacement (i.e. u i ,b  u i ,s  u i ,t  0 ) and, in addition, integrals appearing in (9.7) can directly be integrated. The relation (9.7) can finally be simplified to m  F L F  N R i u i  u i, a  u i,support    i i i    u si Pi i 1 i 1  E i A i Pi 

(9.13)

For a beam, there is no contribution of the axial and torsional effects to the displacement or rotation (i.e. u i ,a  u i ,t  0 ). The relation (9.7), when applied to this particular case, reduces to L m L R i M M i  m   i Vi Vi  N u i  u i, b  u i,s  u i, support     i dx      dx    u si Pi i 1  0 E i I i Pi  i 1  0 G i A i Pi  i 1

(9.14)

For a two-dimensional frame subjected to in-plane loadings, the contribution of the torsional effect to the displacement or rotation vanishes (i.e. u i ,t  0 ) and, as a result, the relation (9.7) becomes L L m L F Fi  m  M i M i  m   i Vi Vi  N R i (9.15) u i  u i, a  u i, b  u i, s  u i, support     i dx      dx      dx    u si Pi i 1  0 E i A i Pi  i 1  0 E i Ii Pi  i 1  0 G i A i Pi  i 1

For a two-dimensional frame subjected to out-of-plane loadings, the contribution of the axial effect to the displacement or rotation disappears (i.e. u i, a  0 ). The relation (9.7) reduces, for this case, to L L m L M M i  m   i Vi Vi  m  M ti M ti  N R i (9.16) u i  u i,a  u i,b  u i,s  u i,support     i dx      dx      dx    u si E I P G A P G J P  Pi   i 1  0 i i i i  i1  i1  0 i i i  i1  0 i i

It is important to point out that the partial derivative with respect to Pi is taken inside the integrals to reduce the computational cost associated with the evaluation of all involved integrals.

9.2 Applications to Statically Determinate Structures In this section, we present the direct application of Castigliano’s 2nd theorem to the deformation and displacement analysis of statically determinate structures. It is known that for statically determinate structures, all support reactions Ri and the internal forces at any point within the structure {Fi , M i , Vi , M ti } can readily be obtained in terms of applied loads via static equilibrium. This therefore enables the equation (9.7) to compute the displacement (or rotation) at a particular point of interest. In such analysis, following steps are recommended: Step 1 Identifying the location and the direction where the displacement (or the rotation) to be determined Step 2 Applying a fictitious concentrated force (or a fictitious concentrated moment) at the location and in the direction of the displacement (or the rotation) to be computed and then labeling it independently from other applied loads acting to that structure. If the given structure is subjected, a priori, to a concentrated force (or a concentrated moment) at the location and in the direction of the displacement (or the rotation) to be computed, this concentrated load can be treated as a fictitious load but must be labeled independently from other applied loads. Copyright © 2011 J. Rungamornrat

410



Step 3 Applying static equilibrium to determine all support reactions {R1, R2, R3,…, RN} and the internal forces {Fi , M i , Vi , M ti } for all members in terms of the fictitious concentrated load and other applied loads Step 4 Appling equations (9.7)-(9.12) to determine the displacement (or the rotation) of interest Step 5 Replacing the fictitious concentrated load by the actual load acting at that location and in the same direction. If there is no applied load at that location and in that direction in the original structure, it is simply replaced by zero. If the calculated displacement (or rotation) is positive, its direction follows directly that of the fictitious; otherwise, its direction is opposite.

3P P A

B 2PL

Figure 9.2: Schematic of a model structure subjected to external loads

3P P PA

3P P

B

B

A

A 2PL

MA

(a)

(b) 3P

PB

P B

A

A 2PL

3P

MB B

2PL

(c)

(d)

Figure 9.3: Structures used for computing of (a) vertical displacement at point A, (b) rotation at point A, (c) horizontal displacement at point B, and (d) rotation at point B Copyright © 2011 J. Rungamornrat

411



To clearly demonstrate the above procedure, let’s consider a model structure subjected to external loads as shown in Figure 9.2 and assume that the vertical displacement and the rotation at a point A and the horizontal displacement and the rotation at a point B are to be determined. By following the step 2, the fictitious vertical concentrated force PA is applied at the point A in addition to other loads as shown in Figure 9.3(a) to obtain a structure suitable for determining the vertical displacement at the point A. Similarly, the applied moment 2PL is changed to MA and used as a fictitious concentrated moment as shown in Figure 9.3(b) and this structure is now suitable for determining the rotation at the point A. The structures suitable for computing the horizontal displacement and the rotation at the point B can also be obtained in the same manner and are shown in Figures 9.3(c) and 9.3(d), respectively. The steps 3 should be straightforward since the given structure is statically determinate whereas the step 4 involves mainly the integration. It is important to emphasize that the displacement or the rotation obtained in the step 4 is still not the solution of the original problem since the fictitious concentrated load still appears. By following the step 5, the fictitious force PA is replaced by 0 since there is no vertical applied force at the point A; the fictitious moment MA is replaced by the actual applied moment 2PL at the point A; the fictitious force PB is replaced by the actual horizontal applied load P at the point B; and the fictitious moment MB is replaced by 0 since there is no applied moment at the point B. Following example problems demonstrate the use of Castigliano’s 2nd theorem, or more specifically the equation (9.7)-(9.12), to determine the displacements/rotations of statically determinate truss, beam and frame. Example 9.1 Consider a statically determinate truss subjected to nodal loads as shown below. The cross sectional area and the Young modulus of each member are indicated in the figure. Determine the vertical displacement at a point C and the horizontal displacement at a point A for the following two cases: (i) no support settlement and (2) the pinned support subjected to the downward settlement  o .

2P B

P

EA

A

EA

2EA

C

EA

EA

L

D

L Solution To determine the vertical displacement at a point C, we first re-labeled the joint load from 2P to PC to make it independent of the other joint load P acting to the joint B and the resulting structure with the fictitious force PC is shown below. Since this structure is statically determinate, all support reactions can be obtained from equilibrium of the entire structure whereas all member forces can readily be determined from either the method of joints or the method of sections. A table shown below summarizes the axial rigidity, the length, the member force in terms of the applied load P and the fictitious force PC, and terms involved in the calculation of derivative of the complementary strain energy of all members. Copyright © 2011 J. Rungamornrat

412



Member

EA

L

F

F PC

FL F AE PC

AB

EA

L

0

0

0

BC

EA

L

–P

0

0

AC

2EA

2L

0

0

AD

EA

L

–P

0

0

CD

EA

L

–P– PC

–1

2P



P  PC L EA

PC B

P

0

C

–P–PC

2P

D

A

–P

P  PC L EA

–P

P

P

P+PC

Case (i): Since there is no support settlement, the complementary work WCs vanishes (i.e. WCs = 0) and the vertical displacement at the point C can be obtained from equation (9.13) as follows vC 

vC 

5 F L F P  PC L U a WCs j   j j 0 PC PC EA j 1 E jA j PC

P  2P L  3PL EA

EA

Downward

Note that the fictitious force PC is replaced by 2P in the last step in order to represent the original structure. Since the calculated displacement is positive, its direction is identical to that of the fictitious force PC. Case (ii): For this particular case, the complimentary work WCs associated with the support settlement is given by WCs =   o P  PC  The strain energy is still the same as the previous case and, therefore, the vertical displacement at the point C becomes vC 

vC 

5 F L F P  PC L   U a WCs j   j j  o  o PC PC EA j 1 E j A j PC

P  2P L   EA

o



3PL   o Downward EA

Similarly, to determine the horizontal displacement at the point A, a fictitious horizontal force PA is applied to the joint A as shown in the figure below. Since, this structure is statically determinate, all support reactions and member forces can readily be obtained from static equilibrium. Results are summarized in the table below. Copyright © 2011 J. Rungamornrat

413


Member

EA

L

F


F PA

FL F AE PA

AB

EA

L

0

0

0

BC

EA

L

–P

0

0

AC

2EA

2L

0

0

AD

EA

L

–P– PA

–1

CD

EA

L

–3P

0

2P



2P B

P

–P

0

C

–3P

2P

P  PA L EA 0

D

PA

A

P  PA L EA

P

–P – PA

P+PA 3P

Case (i): Since there is no support settlement, the complementary work WCs vanishes (i.e. WCs = 0) and the horizontal displacement at the point A can be obtained as follows uA 

uA 

5 FL Fj P  PA L U a WCs  j j 0  EA PA PA j 1 E jA j PA

P  0L  EA

PL Rightward EA

Note again that the fictitious force PA is replaced by 0 in the last step since there is no horizontal force applied to the joint A in the original structure. Case (ii): For this particular case, the complimentary work WCs associated with the support settlement is given by WCs =   o 3P  The strain energy is still the same as that for the case (i) and, as a result, the horizontal displacement at the point A is equal to uA  uA 

5 FL Fj P  PA L U a WCs   j j 0 PA PA EA j 1 E jA j PA

P  0L  EA

PL EA

Rightward

It should be noted that the downward settlement at the pinned support does not affect the horizontal displacement at the point A. This is also clear from that the complimentary work due to the support settlement WCs is independent of the fictitious force PA and, as a result, the derivative of WCs with respect to PA vanishes. Copyright © 2011 J. Rungamornrat

414



Example 9.2 Consider a cantilever beam subjected to concentrated and uniformly distributed loads as shown below. The cross sectional area, the moment of inertia, the form factor, the Young modulus, and the shear modulus of segments AB and BC are given by 2A,4I, , E, G and A, I, , E, G , respectively. Determine the deflection at a point B and the rotation at a point C due to both bending and shear effects.

qL q C

B

A

L

L

Solution To determine the deflection at a point B, we first re-labeled the joint load from qL to PB to make it independent of the uniformly distributed load q and the resulting structure with the fictitious force PB is shown below. Since this structure is statically determinate, the shear force and bending moment for the segments AB and BC can be obtained from static equilibrium as shown below.

PB

PB

q M(x) C

B

A

V(x) M(x)

q

C

B

x

x

q

V(x)

x

C x

L

FBD2

FBD1 Member AB (0 < x < L): By considering equilibrium of the FBD1, it yields

FY  0



:

Vx   PB  qL

M  0 +

:

L  Mx   PB x  qL  x  2 

Member BC (0 < x < L): By considering equilibrium of the FBD2, it leads to

FY  0   M  0 +

Vx   qx 1 Mx    qx 2 2

: :

A table shown below summarizes the flexural and shear rigidities, the form factor, the origin of reference axis, the shear force and bending moment and their derivatives with respect to PB. Member

EI

GA



x=0

Vx 

M x 

V/PB

M/PB

AB

4EI

2GA



B

PB  qL

L   PB x  qL  x  2 

1

–x

BC

EI

GA



C

qx

1  qx 2 2

0

0


415



By applying Castigliano’s 2nd theorem, the deflection at the point B due to both the bending and shear effects can be obtained as follows vB 

U WCs U b U s    PB PB PB PB

(No support settlement  WCs  0 )

v B ,b

  PB x  qL(L/2  x ) U b M M PB L3 7qL4    dx    ( x )dx  0  12EI 48EI 4EI PB EI PB 0

v B,s

V V PB  qL L PB  qL U s   (1)dx  0   dx   2GA 2GA PB GA PB 0

L

L

 v B  v B, b  v B, s 

PB L3 7qL4 PB  qL L   12EI 48EI 2GA

Finally, the fictitious force PB is replaced by the actual load qL and this leads to vB 

11qL4 qL2  48EI GA

Downward

Next, to determine the rotation at the point C, a fictitious concentrated moment MC is applied to that point in addition to other applied loads as shown in the figure below. By applying static equilibrium, the shear force and the bending moment at any cross section of the segments AB and BC can be obtained and results are shown below qL

qL q M(x) C MC

B

A

x

x

q

V(x)

V(x) M(x) C MC

B x

L FBD3

Member AB (0 < x < L): By considering equilibrium of the FBD3, it yields

FY  0



:

Vx   2qL

M  0 +

:

L  Mx   qLx  qL  x   M C 2 


FY  0



:

Vx   qx

M  0 +

:

1 Mx    qx 2  M C 2

q


C MC x FBD4

416



Member

EI

GA



x=0

V(x)

AB

4EI

2GA



B

2qL

BC

EI

GA



C

qx

M x 

L   qLx  qL  x   M C 2  1  qx 2  M C 2

V/M C

M/M C

0

–1

0

–1

By applying Castigliano’s 2nd theorem, the rotation at the point C due to both the bending and shear effects can be obtained as follows C 

U WCs U b U s    M C M C M C M C

C , b 

U b M M   dx M C EI M C L



 qLx  qL(L/2  x )  M C  (1)dx  L  qx 2 / 2  M C (1)dx  5qL3  M C L

0

C,s 

(No support settlement  WCs  0 )



4EI

0

4EI

12EI

2EI

U s V V   dx  0  0  0 M C GA M C

 C  C ,b  C , s 

5qL3 M C L  12EI 2EI

Finally, the fictitious concentrated moment MC is replaced by the actual moment, i.e. M C  0 , and we then obtain C 

5qL3 CW 12EI

Example 9.3 Consider a statically determinate frame subjected to external loads shown below. The cross sectional area, the moment of inertia, the form factor, the Young modulus, and the shear modulus of the segments AB, BC, and CD are given by 2A,2I, , E, G , 2A,2I, , E, G and A, I, , E, G , respectively. Determine the displacement and rotation at the roller support due to all possible effects. Given that the roller support is subjected to a downward settlement 0.

q D

C

L/2

2qL

B L/2

A L Copyright © 2011 J. Rungamornrat

417



Solution To determine the horizontal displacement at the point D, a fictitious concentrated load HD is applied to the joint D in addition to other applied loads. Since the given structure is statically determinate, the axial force, the shear force and the bending moment within the segments AB, BC, and CD can readily be obtained using static equilibrium as shown below.

q D C

2qL

x

B

x

HD

3 H D  qL 2

F(x) M(x) V(x)

H D  2qL

x

x

A

H D  2qL

A

1 H D  qL 2

1 H D  qL 2

FBD1

F(x) M(x) V(x) 2qL

B

x

H D  2qL

q

M(x)

L/2

D

F(x)

A

HD 3 H D  qL 2

V(x)

1 H D  qL 2

x

FBD2

FBD3

Member AB (0 < x < L/2): By considering equilibrium of the FBD1, it yields

FY  0



FX  0   M  0 +

:

1 Fx   H D  qL 2

:

Vx   H D  2qL

:

M x   H D  2qL x

Member BC (0 < x < L/2): By considering equilibrium of the FBD2, it yields

FY  0



:

1 Fx   H D  qL 2 Copyright © 2011 J. Rungamornrat

418



FX  0



M  0 +

:

Vx   H D

:

L L   Mx   H D  2qL  x    2qLx  H D  x    qL2 2 2  

Member CD (0 < x < L): By considering equilibrium of the FBD3, it yields

FY  0



:

Fx   H D

FX  0



:

3 Vx   qx  H D  qL 2

:

3  1  M x    H D  qL  x  qx 2 2  2 

M  0 + Member

EA

EI

GA



x=0

Fx 

Vx 

M x 

AB

2EA

2EI

2GA



A

1 H D  qL 2

H D  2qL

H D  2qL x

BC

2EA

2EI

2GA



B

1 H D  qL 2

HD

L  H D  x    qL2 2 

CD

EA

EI

GA



D

HD

3 qx  H D  qL 2

3  1 2   H D  qL  x  qx 2  2 

Member

F H D

V H D

M H D

AB

1

1

x

BC

1

1

x

CD

1

–1

x

L 2

By applying Castigliano’s 2nd theorem, the horizontal displacement at the point D due to all possible effects can be obtained as follows 3   WCs   H D  qL Δ 0 2   uD 

U WCs U a U b U s      Δ0 H D H D H D H D H D

u D ,a 

F F U a dx    EA H D H D 

H D  qL / 2 (1)dx  L/2H D  qL / 2 (1)dx  L H D (1)dx

L/2

 0

2EA

 0

2

3H D L qL  2EA 4EA Copyright © 2011 J. Rungamornrat

2EA

 EA 0

419


u D,b 

M M U b dx   EI H D H D L/2





H D  2qLx (x )dx  L/2 H D (x  L/2)  qL2 (x  L/2)dx  L (H D  3qL/2)x  qx 2 /2(x)dx

0

 u D ,s 




2EI



2EI

0

EI

0

25H D L3 2qL4  48EI 3EI

U s V V dx   GA H D H D 

H D  2qL  (1)dx  0 2GA



3H D L 3qL2  2GA 2GA

L/2

 u D  u D,a  u D,b  u D,s   0 

L/2

L

H (qx  H D  3qL/2) (1)dx 0 2GAD (1)dx  0 GA

3H D L qL2 25H D L3 2qL4 3H D L 3qL2       0 2EA 4EA 48EI 3EI 2GA 2GA

Finally, the fictitious force HD is replaced by the actual load, i.e. H D  0 , and we finally obtain uD 

qL2 2qL4 3qL2     0 Rightward 4EA 3EI 2GA

Next, to determine the rotation at the point D, a fictitious concentrated moment MD is applied to the joint D in addition to other applied loads. All support reactions and the axial force, the shear force and the bending moment within the segments AB, BC, and CD can readily be obtained from static equilibrium and results are provided below. q

D MD

C

2qL

B

x

x



MD 3  qL L 2

F(x) M(x) V(x)

2qL

A

x

x

2qL 



MD 1  qL L 2

A

MD 1  qL L 2 FBD4


420



F(x) M(x) V(x) 2qL

x

B



MD

F(x)

A

2qL

q

M(x)

L/2

D

V(x)



MD 1  qL L 2

MD 3  qL L 2

x FBD6

FBD5

Member AB (0 < x < L/2): By considering equilibrium of the FBD4, it yields

FY  0



FX  0   M  0 +

MD 1  qL L 2

:

Fx   

:

Vx   2qL

:

M x   2qLx

Member BC (0 < x < L/2): By considering equilibrium of the FBD5, it yields MD 1  qL L 2

FY  0



:

Fx   

FX  0



:

V x   0

:

L  Mx   2qL  x    2qLx  qL2 2 

M  0 +

Member CD (0 < x < L): By considering equilibrium of the FBD6, it yields

FY  0



:

Fx   0

FX  0



:

Vx   qx 

:

3  1  M M x     D  qL  x  qx 2  M D 2  2  L

M  0 +

MD 3  qL L 2

Member

EA

EI

GA



x=0

Fx 

Vx 

M x 

AB

2EA

2EI

2GA



A

M 1 qL  D 2 L

2qL

2qLx

BC

2EA

2EI

2GA



B

1 M qL  D 2 L

0

qL2

CD

EA

EI

GA



D

0

qx 


MD 3  qL L 2

M  1 3  qL  D  x  qx 2  M D L 2 2  

421



Member

F M D

V M D

M M D

AB

1/ L

0

0

BC

1/ L

0

0

CD

0

1/ L

1

x L

By applying Castigliano’s 2nd theorem, the rotation at the point D due to all possible effects can be obtained as follows 3   M WCs    D  qL Δ 0 2   L D 

U a U WCs U b U s Δ 0      M D M D M D M D M D L

D ,a 

F F U a dx   EA M D M D L/2





 M D /L  qL / 2   1 dx  L/2 M D /L  qL / 2   1 dx  0  2EA

0

D, b 

M M U b dx   EI M D M D L

00 0

D ,s 

   L

 M

D

 0

2EA

   L

MD qL  2LEA 4EA



M L 5qL3 /L  3qL / 2 x  qx 2 /2  M D  x  1  dx  D  3EI 24EI EI  L

U s V V dx   EI M D M D

qx  M D /L  3qL / 2   1  M D qL   dx  LGA GA GA L 0

L

00

 D  D,a

Δ0 MD qL M D L 5qL3 M D qL Δ 0  D,b   D,s         L 2LEA 4EA 3EI 24EI LGA GA L

Finally, the fictitious concentrated moment MD is replaced by the actual moment, i.e. M D  0 , and it finally leads to

qL 5qL3 qL Δ 0 D      4EA 24EI GA L The computed rotation D follows the direction of the fictitious moment MD if it is positive; otherwise its direction is opposite. Copyright © 2011 J. Rungamornrat

422



9.3 Applications to Statically Indeterminate Structures In this section, applications of Castigliano’s second theorem to the analysis of statically indeterminate structures are presented. The theorem is mainly employed to establish additional equations, termed the compatibility conditions, sufficient for determining extra static unknowns that cannot be solved from equilibrium equations alone. To clearly demonstrate such procedure and provide some useful remarks, let’s consider the following model problem, a statically indeterminate beam with two degrees of static indeterminacy (i.e. DI = 2 or, equivalently, there are two static unknowns (or two redundants) exceeding the number of independent equilibrium equations) as shown in Figure 9.4(a).

A

A B

B

C

(a)

(b)

C

A

B RB

C RC

(c)

Figure 9.4: Schematics of (a) statically indeterminate beam with DI = 2, (b) primary structure resulting from moving roller supports at B and C, and (c) admissible structure Let’s introduce a statically determinate and stable structure resulting from removing two roller supports at points B and C from the original structure as shown in Figure 9.4(b) and then term this structure as the primary structure. Next, let’s introduce a class of admissible structures such that a structure belonging to this class is obtained by taking the primary structure and then applying the redundants RB and RC to the location and in the direction of the released reactions. It is worth noting that the magnitude of the redundants RB and RC can be chosen arbitrarily and, as a direct consequence, this leads to an infinite number of admissible structures. The adjective admissible is used only to emphasize that an admissible structure can become the original structure if the redundants RB and RC are properly chosen. To obtain a correct choice of the redundants RB and RC that renders an admissible structure identical to the original structure, kinematical conditions at the location of the releases must be recovered. More precisely, the redundants RB and RC must be chosen, for this particular case, such that the deflections at points B and C of an admissible structure vanish in order to mimic the constraint provided by the roller supports at points B and C in the original structure, i.e. structure structure v admissible  voriginal 0 B B

(9.17)

structure structure v admissible  v original 0 C C

(9.18)

Since the two redundants RB and RC are independent of all other applied loads, the deflections at points B and C of the admissible structure can readily be obtained from Castigliano’s 2nd theorem as follows structure v admissible  B

U C WCs  R B R B

(9.19)

structure v admissible  C

U C WCs  R C R C


423



where UC is the complementary strain energy and WCs is the complementary work associated with the support settlement of the admissible structure. It should be noted that both UC and WCs can readily be expressed in terms of the redundants RB, RC and other applied loads since the admissible structure is statically determinate; all support reactions and the internal forces within any member can be obtained directly from static equilibrium. By inserting (9.19) and (9.20) into (9.17) and (9.18), it leads to the following compatibility equations

U C WCs  0 R B R B

(9.21)

U C WCs  0 R C R C

(9.22)

These two compatibility equations are necessary and sufficient for solving the two redundants RB and RC. Such solutions are in fact the reactions of the roller supports at points B and C of the original structure. Once the redundants RB and RC are known, the original structure becomes statically determinate. All remaining support reactions and the internal forces can subsequently be obtained from static equilibrium. For the special case that there is no support settlement and the structure is made from a linearly elastic material, the complimentary work WCs vanishes and the complementary strain energy can be replaced by the strain energy. The compatibility equations (9.21) and (9.22) simply reduce to U 0 R B

(9.23)

U 0 R C

(9.24)

These two equations are also known as the least work equations. It is apparent for this special case that the correct value of the redundants RB and RC produces the minimum value of the strain energy of the structure (or, equivalently, the minimum value of the work done by all external loads). Now, let’s consider a more general case that the roller supports at the points B and C in the original structure are subjected to prescribed support settlements  B and  C , respectively. The only modification required is to change the compatibility equations from (9.17) and (9.18) to structure structure v admissible  v original  ΔB B B

(9.25)

structure structure v admissible  v original  ΔC C C

(9.26)

With use of Castigliano’s 2nd theorem (9.19) and (9.20) to determine the deflections vB and vC of the admissible structure in terms of external applied loads and the redundants RB and RC along with the compatibility conditions (9.25) and (9.26), it leads to two equations sufficient for determining the redundants RB and RC: U C WCs   ΔB R B R B

(9.27)

U C WCs   ΔC R C R C


424



Another crucial remark is that a primary structure chosen to represent the original structure and its corresponding set of redundants are not unique. In fact, any statically determinate and stable structure resulting from releasing external or internal constraints within the original structure can be employed as a primary structure. Once the primary structure (or the choice of redundants) was chosen, the admissible structure can readily be obtained by adding the redundants back to the primary structure in terms of applied loads at the location and in the direction of all releases. Structures shown below are valid candidates of the admissible structure and the final choice chosen in the analysis depends primarily on the matter of taste or preference.

MA

MA

B A

RB

A RA

C

(a)

A

B

RA

RB

(b)

MA C

MB A

(c)

MA B

MB B

C

(d)

C A

C

B

RC

C

A RA

(e)

B

RC

(f)

Figure 9.5: Valid candidates of admissible structure for original structure shown in Figure 9.4(a) If the moment reaction of the fixed support at point A (MA) and the reaction of the roller support at point B (RB) are chosen as the redundants with the corresponding admissible structure shown in Figure 9.5(a), the general compatibility equations are given by structure structure admissible  original  0 A A

(9.29)

structure structure v admissible  v original  ΔB B B

(9.30)

where o is the prescribed rotation of the fixed support at point A and B is the prescribed deflection at the roller support at point B. If both the moment and force reactions of the fixed support at point A (MA and RA) are chosen as the redundants with the corresponding admissible structure shown in Figure 9.5(b), the general compatibility equations are given by structure structure admissible  original  0 A A

(9.31)

structure structure v admissible  v original  ΔA A A


425



where A is the prescribed deflection of the fixed support at point A. If the reaction of the fixed support at point A (RA) and the reaction of the roller support at point B (RB) are chosen as the redundants with the corresponding admissible structure shown in Figure 9.5(c), the compatibility equations are given by structure structure v admissible  v original  ΔA A A

(9.33)

structure structure v admissible  v original  ΔB B B

(9.34)

If the moment reaction of the fixed support at point A (RA) and the bending moment at the point B (MB) are chosen as the redundants with the corresponding admissible structure shown in Figure 9.5(d), the compatibility equations are given by structure structure admissible  original  0 A A

(9.35)

structure structure admissible  original 0 B B

(9.36)

where B is the relative rotation at the point B. If the moment reaction of the fixed support at point A (MA) and the reaction of the roller support at point C (RC) are chosen as the redundants with the corresponding admissible structure shown in Figure 9.5(e), the two compatibility equations are given by structure structure admissible  original  0 A A

(9.37)


(9.38)

where C is the prescribed deflection of the roller support at point C. Finally, if the force reaction of the fixed support at point A (RA) and the reaction of the roller support at point C (RB) are chosen as the redundants with the corresponding admissible structure shown in Figure 9.5(f), the general form of compatibility equations is given by structure structure v admissible  v original  ΔA A A

(9.39)


(9.40)

Example 9.4 Consider a statically indeterminate truss subjected to a horizontal load P at a joint B as shown below. The cross sectional area and the Young modulus of all members are clearly indicated in the figure. Determine all support reactions and all member forces in terms of P.

B

P

EA

2EA

EA

A

EA

C

EA

D


L

426



Solution The given structure is statically indeterminate with DI = 4 + 5(1) – 4(2) – 0 = 1; in particular, there is only one redundant. Structures shown below are examples of admissible structures and the corresponding choice of redundant that can be employed in the analysis. The compatibility condition for each admissible structure is indicated below the figure.

B

P

C

B

P

RC

D

A

C

D

A

B

P

C

D

A

RD

RA Compatibility: u C  0

Compatibility: v A  0

Compatibility: u D  0

Let choose the first admissible structure in the following analysis. All support reactions and member forces of this admissible structure are obtained in terms of the applied load P and the redundant RC using static equilibrium and the method of joints. All member forces and their derivatives with respect to the redundant RC is summarized in a table below.

C

–RC

B

P

RC

2 R C  P 

P – RC

0

D

A

RC – P

0 P – RC

RC – P

Member

EA

L

F

F R C

FL F EA R C

AB

EA

L

P  RC

–1

 R C  P  L/EA

BC

EA

L

 RC

–1

R C L/EA

BD

2EA

2L

AD

EA

L

0

0

0

CD

EA

L

0

0

0



PL 2  2  REAL  1  2  EA

2 R C  P 

2


2  R C  P  L/EA

C

427



By applying Castigliano’s second theorem, the horizontal displacement at the point C in terms of the applied load P and the redundant RC can be obtained as uC 









5 FL Fj U a WCs R L PL   j j  0  2  2 C  1 2 R C R C EA EA j1 E j A j R C

The redundant R C can then be obtained by solving the compatibility equation u C  0 :



uC  2  2

PL  REAL  1  2  EA 0

 1 2  P R C    2 2 



C

Leftward

Other support reactions and all member forces can subsequently be obtained via the direct substitution of R C  (1  2)P/(2  2) and final results are given by 1 P Downward 2 2 1 R DX  R C  P   P Leftward 2 2 1 R DY  P  R C  P Upward 2 2 1 P (Tension) FAB  P  R C  2 2 1 2 P (Compression) FBC   R C   2 2 2 P (Compression) FBD  2 R C  P    2 2 FCD  FAD  0 RA  RC  P  

Example 9.5 Consider a statically indeterminate beam subjected to a concentrated load P at its mid-span as shown below. The cross sectional area, the moment of inertia, the form factor, the Young modulus, and the shear modulus of the segments AB and BC are given by 2A,2I, , E, G and A, I, , E, G , respectively. Determine all support reactions and sketch the SFD and BMD. Let’s assume that during the application of the load P, the roller support is subjected to the downward settlement 0.

P C B

A L

L

Solution Since the given structure is statically indeterminate with DI = 3 + 2(2) – 3(2) = 1, only one redundant must be released to construct a primary structure. Figures shown below are examples of valid admissible structures and the corresponding choice of redundant that can be used in the analysis. The compatibility condition associated with each admissible structure is indicated below. Copyright © 2011 J. Rungamornrat

428



P C

P C

A

B

A

P

MA

C

B

A

RC

B

RA Compatibility: A  0

Compatibility: v C   0

Compatibility: v A  0

Let’s choose the first admissible structure in the following analysis. All support reactions, the shear force and bending moment of such admissible structure can readily be obtained in terms of the applied load P and the redundant RC from static equilibrium. The shear force and bending moment within the segments AB and BC and their derivatives with respect to the redundant RC are summarized below. P M(x)

C B

A

P

V(x)

C

V(x) M(x)

C

B RC

RC

x

x

x

RC x

L

FBD2

FBD1


FY  0   M  0 +

Vx   P  R C M x    Px  R C L  x 

: :


FY  0   M  0 +

V x    R C M x   R C x

: :

Member

EI

GA



x=0

V(x)

M(x)

V R C

M R C

AB

2EI

2GA



B

P  RC

 Px  R C L  x 

–1

L+x

BC

EI

GA



C

 RC

R Cx

–1

x

By applying Castigliano’s second theorem, the deflection at the point C in terms of the applied load P and the redundant RC can be obtained as v C  v C,b  v C ,s 

U b U s  R C R C

(No support settlement in the admissible structure  WCs  0 ) Copyright © 2011 J. Rungamornrat

429



v C ,b 

 Px  R C (L  x ) (L  x )dx  R C x (x )dx  3R C L3  5PL3 U b M M   dx   0 EI 2EI R C EI R C 2EI 12EI 0

v C,s 

P  R C   R C  U s V V 3R C L PL   dx    (1)dx   (1)dx  R C GA R C 2GA 2GA 2GA GA 0 0

L

L

L

v C  v C,b  v C,s

L

3R C L3 5PL3 3R C L PL     2EI 12EI 2GA 2GA

The redundant R C can then be obtained by solving the compatibility equation v C   0 : λEI  5   2 3R C L 5PL 3R C L PL vC       0  R C   18 3GAL 2EI 12EI 2GA 2GA  1  λEI  GAL2  3

3

 2EI    3 P   3L  1  λEI    GAL2  

   0 Downward   

Once the redundant R C is determined, other support reactions can readily be obtained from static equilibrium and the SFD and BMD are given below. P C  13 2 EI   2EI  B A     2  3  0 Upward R A  P  R C   18 3GAL P   3L  1  λEI    GAL2  

λEI 4   2 M A  PL  2R C L   9 3GAL  1  λEI  GAL2 

 1  λEI    GAL2  

 4EI    2 PL   3L  1  λEI    GAL2  

L

L RA    0 CCW   

SFD RCL

–RC BMD

–MA If the shear effect is neglected and  0  0 , all support reactions and the SFD and BMD simply reduce to P C B 5P A RC  Downward 18 L

13P Upward RA  18

L

13P/18

SFD

4PL MA  CCW 9

5PL/18

–5P/18

BMD –4PL/9 Copyright © 2011 J. Rungamornrat

430



Example 9.6 Consider a statically indeterminate frame subjected to a concentrated moment M at a point B as shown below. The cross sectional area A, the moment of inertia I, the form factor , the Young modulus E, and the shear modulus G are assumed constant throughout the structure. Determine all support reactions and sketch the SFD and BMD by considering only the bending effect. Let’s assume that during the application of the moment M, the fixed support moves downward with an amount of 0.

M

C

B

L

A L Solution Since the given structure is statically indeterminate with DI = 5 + 2(3) – 3(3) = 2, two redundants must be released to obtain the primary structure. In this analysis, reactions of a pinned support at a point C, RCX and RCY, are chosen as the redundants and the corresponding admissible structure is shown below. The compatibility conditions required to make the admissible structure identical to the original structure are

uC  0

vC  0

and

First, all support reactions and the bending moment of the selected admissible structure can readily be obtained in terms of the applied moment M and the two redundants RCX and RCY from static equilibrium as follows.

M

C B

M(x) V(x)

RCY F(x)

x V(x)

R CX

A

M  R CY L  R CX L R CY

x R CX

RCY

x M(x)

x

C

F(x)

RCX

A

M  R CY L  R CX L

R CY FBD1


FBD2

RCX

431




M  0 +

Mx   R CX L  x   R CY L  M

:

Member BC (0 < x < L): By considering equilibrium of the FBD2, it yields

M  0 +

Mx   R CY x

:

Member

EI

x=0

M x 

M R CX

M R CY

AB

EI

A

R CX L  x   R CY L  M

L–x

L

BC

EI

C

R CY x

0

x

By applying Castigliano’s second theorem, the horizontal and vertical displacements at the point C in terms of the applied moment M and the redundants RCX and RCY can be obtained as WCs  R CY Δ 0 u C  u C,b 

U b WCs M M   dx  0  R CX R CX EI R CX L



R CX (L  x )  R CY L  M (L  x )dx  0 EI

0

R CX L3 R CY L3 ML2    3EI 2EI 2EI v C  v C,b 

U b WCs M M    dx   0 R CY R CY EI R CY L



R CX (L  x )  R CY L  M (L)dx  L R CY x (x )dx   EI

0



 0

EI

0

R CX L3 4R CY L3 ML2    0 2EI 3EI EI

The two compatibility equations now become uC 

R CX L3 R CY L3 ML2   0 3EI 2EI 2EI

vC 

R CX L3 4R CY L3 ML2    0  0 2EI 3EI EI

By solving above two linear equations simultaneously, it yields the unknown redundants R CX and R CY Copyright © 2011 J. Rungamornrat

432



R CX 

6M 18EI 0  7L 7L3

;

R CY 

3M 12EI 0  7L 7L3

The remaining support reactions can then be computed from equilibrium of the entire structure and results are given below R AX 

6M 18EI 0  7L 7L3

; R AY 

3M 12EI 0  7L 7L3

;

MA  

2M 6EI 0  7 7L2

The SFD and BMD for 0 = 0 are given below 3M/7L

3M / 7 C

B

B

C

–4M/7

–6M/7L BMD

SFD 2M / 7

A

A

Exercises 1. Use Castiglino’s second theorem to determine the deflection at a point a and the rotation at a point b of following statically determinate, linearly elastic beams due to both bending and shear effects. Assume that the Young modulus E, the shear modulus G and the form factor  are constant throughout and the support settlements are indicated in the figure. 2P 2I, 2A, L

a

q

P I, A, L

3I, 2A, L

b

b

I, A, 2L

a

o

q M

a

b 2I, 2A, L

I, A, L

2qL2 I, A, L

I, A, L

a

I, A, L

b

o

2. Use Castiglino’s second theorem to determine the horizontal displacement at a point a and the vertical displacement at a point b of following statically determinate, linearly elastic trusses. The axial rigidity of each member and support settlements are given in the figure.


433



2P a

P

L

2EA

L a

EA

L

b

2EA P EA is constant

2EA

2EA

EA

b

EA

EA

2P

L

EA

o

P

L

L

L

EA is constant a

P

EA

L

b

3EA

L

L

L

a

3EA

EA

EA

EA

L

2EA b

o

2P

L

3P

P

L

L

L

L

L

3. Use Castiglino’s second theorem to determine the horizontal displacement at a point a, the vertical displacement at a point b, and the rotation at a point c of following statically determinate, linearly elastic frames due to all possible effects. Assume that the Young modulus E, the shear modulus G, and the form factor  are constant throughout and support settlements are clearly indicated in the figure. 2qL c I, A, L

q

q a I, A, L b

qL

2I, 2A, 2L

b

I, A, L

c

2I, 2A, L

a

o Copyright © 2011 J. Rungamornrat

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2q qL

b

2PL

a

I, A, 2L

a

2I, 2A, 2L

I, A, L

I, A, L

3P

I, A, 2L

2qL2

c

I, A, L

P

o

I, A, L b

c

4. Use Castiglino’s second theorem to determine a gap (between the two points where the loads are applied) produced by a pair of concentrated loads as shown in the figure below. Consider only the bending effect. M

P

M EI, L/2

EI, L/2

P

P EI, L/2

EI, L/2

EI, L/2

EI, L/2

P 2EI, L

2EI, L

2EI, L

2EI, L

EI, L

2EI, L

2EI, L

EI, L

EI, L

5. Analyze following statically indeterminate structures (for all support reactions and the internal forces such as the axial force, the bending moment, and the shear force) by Castiglino’s second theorem. Consider only the bending effect for flexure-dominating members. 2P P

L

L

2EA

2EA EA

P

EA is constant

L

L

2EA

2EA EA

EA

EA

EA 2P

L

L

L

L


P

L

L

435


2P

P


2P

2P

2P

2P

P

P h

EA constant h

h

h

h

h

h

q M EI, 2L

3EI, L

2EI, L

EI, L

q EI, L

EI, L

EI, L

P EI, 2L 2EI, 3L 2qL EI, L

2PL

qL

EI, L

2EI, 2L EI, L

q

3P

2EI, 2L

EI, L q

EI, L

EI, L

2qL EI, L/2

EI, L/2

2EI, L


qL

436



2q qL

EI, 2L

P

2EI, 2L

2EI, 2L

EA, 2 L

P EA, 2L

EA, L

EA, L EI, L

EA, L

EI, L

2P


EA, 2 L

437


Method of Consistent Deformation

CHAPTER 10 METHOD OF CONSISTENT DEFORMATION A method of consistent deformation, also known as a flexibility method or a force method, is a wellknown technique widely used in the analysis of statically indeterminate structures. In this method, a set of independent static quantities (e.g. support reactions and internal forces at certain locations) that cannot be determined from static equilibrium equations is chosen as a set of primary unknowns called “redundants”. An additional set of equations called “compatibility equations” is formed to solve all such unknown redundants. After all redundants are resolved, the structure becomes statically determinate and, as a consequence, all remaining static quantities can readily be determined from static equilibrium equations and the displacement and rotation at any point within the structure can be computed using several techniques discussed previously (e.g. method curvature area, conjugate structure analogy, energy methods).

10.1 Basic Concept To clearly demonstrate the basic concept underlying the method of consistent deformation, let consider a statically indeterminate frame with the degree of static indeterminacy equal to 2, i.e. DI = 2, and subjected to external loads as shown schematically in Figure 10.1. Y q RCX

C

B

RCY

P A

X

Figure 10.1: Schematic of a statically indeterminate frame with DI = 2 For this particular structure, there are two extra static unknowns or two redundants that cannot be determined by static equilibrium. Let RCX and RCY be support reactions at a pinned support at point C. If both the support reactions RCX and RCY are known, the given structure now becomes statically determinate and all other static quantities (remaining support reactions and all internal forces) can readily be calculated from static equilibrium. Now, let consider the following thought process. First, we remove the pinned support at the point C from the original structure or, equivalently, release both the support reactions RCX and RCY. The resulting structure, shown in Figure 10.2(a), is therefore statically determinate and is generally termed as the primary structure. Next, we apply two forces R1 and R2, in addition to existing external loads acting to the original structure, to the primary structure at point C in the direction of the support reactions RCX and RCY, respectively. The resulting structure, shown in Figure 10.2(b), is still statically determinate and is termed as the admissible structure. It is important to note that the two forces R1 and R2 up to this state can take any value. Since both the primary structure and the Copyright © 2011 J. Rungamornrat

438



admissible structure are statically determinate, all support reactions and the internal forces can be determined from static equilibrium and the displacement and rotation at any point within both structures can readily be calculated from various techniques such as the method of curvature area, the conjugate structure analogy, the principle of complementary virtual work, etc. Y

Y

q

q u1o C

B

u2o

R1

C

B

R2

P A

A

X

X (b)

(a)

Figure 10.2: (a) Schematic of primary structure and (b) admissible structure of the original structure shown in Figure 10.1 If the structure under consideration is linear, the method of superposition can be employed to simplify the analysis procedure. For instance, any response of the admissible structure can be obtained by superposing the response of the following structures: (i) the primary structure shown in Figure 10.2(a), (ii) the structure subjected to the force R1 as shown in Figure 10.3(a), and (iii) the structure subjected to the force R2 as shown in Figure 10.4(a). Let u1 and u2 be the displacement at the point C in the direction of R1 and R2 of the admissible structure. From the method of superposition, we then obtain u1  u1o  u11  u12

(10.1)

u 2  u 2 o  u 21  u 22

(10.2) Y

Y

f11

u11 B

A

C

R1 u21

A

X

C

B

1 f21

X (b)

(a)

Figure 10.3: (a) Schematic of the released structure subjected to the force R1 and (b) schematic of the released structure subjected to a unit force in the direction of R1. Copyright © 2011 J. Rungamornrat

439



f12

Y

u12

Y

f22

u22 C

B

C

B

1

R2

A

A

X

X (b)

(a)

Figure 10.4: (a) Schematic of the released structure subjected to the force R2 and (b) schematic of the released structure subjected to a unit force in the direction of R2. where u1o and u2o are the displacements at the point C of the primary structure shown in Figure 10.2(a) in the direction of the force R1 and the force R2, respectively; u11 and u12 are the displacements at the point C of the structure shown in Figure 10.3(a) in the direction of the force R1 and the force R2, respectively; and u21 and u22 are the displacements at the point C of the structure shown in Figure 10.4(a) in the direction of the force R1 and the force R2, respectively. It is worth noting that calculation of uij (i, j = 1, 2) involves analyses of a statically determinate structure (a statically stable released structure of the original structure shown in Figure 10.1) subjected either to a force R1 or a force R2 one at a time. To further simplify such analysis, we exploit the linearity of the structure and then analyze the structures subjected to a unit force shown in Figure 10.3(b) and 10.4(b) instead of the structures in Figure 10.3(a) and 10.4(a), respectively. The displacements uij at the point C can then be obtained, in terms of response of the structure subjected to the unit force, by a linearity rule as u11  f11R 1

;

u 21  f 21R 1

(10.3)

u12  f12 R 2

;

u 22  f 22 R 2

(10.4)

where f11 and f21 are the displacements at the point C of the released structure subjected to a unit load in the direction of the force R1 (as shown in Figure 10.3(b)) in the direction of the force R1 and the force R2, respectively; and f12 and f22 are the displacements at the point C of the released structure subjected to a unit load in the direction of the force R2 (as shown in Figure 10.4(b)) in the direction of the force R1 and the force R2, respectively. Again, fij (i, j = 1, 2) can be calculated from any convenient method such as the method of curvature area, conjugate structure analogy, the principle of complementary virtual work, etc. By substituting (10.3) and (10.4) into the relations (10.1) and (10.2), we obtain the displacements u1 and u2 at the point C of the admissible structure in the direction of the force R1 and the force R2, respectively, as u1  u1o  f11R1  f12 R 2

(10.5)

u 2  u 2 o  f 21R1  f 22 R 2

(10.6)

These two equations can, alternatively, be expressed in a matrix form as Copyright © 2011 J. Rungamornrat

440



u1   u1o  f11 f12  R1            uo  f R u 2   u 2o  f 21 f 22  R 2 

(10.7)

where uo is a vector containing components of the displacement at the point C of the primary structure given by u1o  uo    u 2o 

(10.8)

R is a vector containing arbitrary loads applied to the admissible structure at the locations and directions of releases given by R 1  R  R 2 

(10.9)

and f is termed as a flexibility matrix of the released structure defined by f11 f12  f   f 21 f 22 

(10.10)

with fij (i, j = 1, 2) denoting the flexibility coefficients. It is evident from equation (10.7) that the displacement at the point C of the admissible structure shown in Figure 10.2(b) can equivalently be obtained by performing analysis of a set of simpler structures (structures shown in Figure 10.2(a), Figure 10.3(b), and Figure 10.4(b)) and following by employing linearity of the structure via the method of superposition. Equivalence between the admissible structure and the set of simpler structures is clearly illustrated by Figure 10.5. It is important to remark that other quantities or responses of the admissible structure can also be determined using the method of superposition in the same fashion. As a final step in our thought process, we aim to make a connection between the statically determinate admissible structure (shown in Figure 10.2(b)) and the statically indeterminate original structure (shown in Figure 10.1). The key difference between the admissible structure and the original structure is that the kinematical conditions at the released locations of the admissible structure (the point C for this particular case) are, in general, not the same as those of the original structure. This results from that the additional applied loads R1 and R2 in the admissible structure can vary arbitrarily. For instance, the vertical and horizontal components of the displacement at the point C of the admissible structure are, in general, not equal to zero. For the admissible structure to become the original structure, the applied loads R1 and R2 must be chosen to be identical to the support reactions RCX and RCY of the original structure. Since both RCX and RCY are unknown a priori, R1 and R2 must be chosen such that the kinematical conditions at the released locations of the admissible structure are identical to those of the original structure. Let ur be a vector containing the (prescribed) displacement components of the original structure at the released point. The kinematical conditions or compatibility conditions for determining R1 and R2 are ur  uo  f R



u1r  u1o  f11 f12  R 1        u 2 r  u 2o  f 21 f 22  R 2 


(10.11)

441



Y q R1

C

B

R2 P A

Y

X

Y q C

u2o

B

C

+ R1

P A

1 f21

C

B

1

+ R2 A

X

f22

f11

u1o B

f12

Y

X

A

X

Figure 10.5: Equivalence between the admissible structure and a set of simpler structures Equation (10.11) is known as the “compatibility equation” or the “continuity equation”. For the special case when there is no support settlement taking place at the point C, i.e. ur = 0, and the compatibility equation (10.11) reduces to uo  f R  0



u1o  f11 f12   R1  0         u 2o  f 21 f 22   R 2  0 

(10.12)

The forces R1 and R2 obtained by solving the compatibility equation (10.11) are therefore the support reactions RCX and RCY of the original structure. The admissible structure with the known or solved R1 and R2 is now identical to the original structure and, as a consequence, all quantities of interest of the original structure can equivalently be obtained from the admissible structure or from superposition of responses of a set of simpler structures as indicated by the correspondence shown in Figure 10.5.

10.2 Choice of Released Structures While the basic concept of the method of consistent deformation has already been demonstrated in the previous section, two very important issues remain to be addressed. These issues are related to the following questions: “how many redundants to be removed from the original structure to obtain a statically determinate released structure?” and “what is the necessary condition for the released structure?” Copyright © 2011 J. Rungamornrat

442



To answer the first question, the degree of static indeterminacy of the given structure must be determined and this number is in fact the number of all unknown redundants to be removed from the original structure to render the resulting released structure statically determinate. By recalling discussion in Chapter 1, the degree of static indeterminacy of a structure, denoted by DI, can readily be computed from DI  ra  n m  n j  n c

(10.13)

where ra is the number of all components of the support reactions, nm is the number of components of the internal force of all members (e.g. the number of components of the internal force for an individual truss, beam, and frame members is 1, 2, and 3, respectively), nj is the number of independent equilibrium equations at all nodes (e.g. the number of independent equilibrium equations for an individual node of truss, beam, and frame structures is 2, 2, and 3, respectively), and nc is the number of static conditions associated with all internal releases present within the structure. For instance, the degree of static indeterminacy of the frame structure shown in Figure 10.1 is DI = (3 + 2) + (3  2) – (3  3) – (0) = 2. q

Y

q

Y

C

C B

B

P

R1

P A

R2

X q

Y

A

X q

Y

C

C B

B

P

P A

X

A R2

R1

q

Y

q

Y B

C B

X

C R2

P

P A

A R1

X (a)

X (b)

Figure 10.6: Examples of (a) primary structure and (b) admissible structures of original structure shown in Figure 10.1 Copyright © 2011 J. Rungamornrat

443



Response to the second question is somewhat more difficult. The technique requires only that the original structure must be released to obtain a primary structure that is statically determinate and, of course, must also be statically stable. Note that while the degree of static indeterminacy or the number of redundants of a given structure can readily be determined, the process of constructing the primary structure is nontrivial since a choice of redundants is not unique and, at the same time, is not fully arbitrary. In fact they must be chosen such that the released structure is statically stable; i.e. there is no development of rigid body motion of the entire structure or any portion of the structure. Examples of valid sets of redundants, primary structures and admissible structures of the original structure shown in Figure 10.1 are given, in addition to those indicated in Figure 10.2, in Figure 10.6. It is noted that a valid set of redundants is not necessary consisting of only components of support reactions, the internal force such as bending moment, shear force, and axial force at certain locations can also be chosen as redundants. For instance, the last primary structure shown in Figure 10.6 is obtained by removing the moment reaction at the point A and the bending moment at the point B. Examples of invalid sets of redundants that produce statically unstable, released structures are shown in Figure 10.7; these released structures cannot be used as the primary structures in the analysis by the method of consistent deformation. q

Y

C B P

q

Y B

C

P A

A

X

X

Figure 10.7: Examples of statically unstable released structures

10.3 Compatibility Equations for General Case Consider a statically indeterminate structure with the degree of static indeterminacy DI = N (i.e. the number of redundants is equal to N). Let R = {R1, R2, R3, …, RN} be a proper set of redundants chosen to construct the primary structure and let ur = {u1r, u2r, u3r, …, uNr} be a vector of kinematical conditions of the original structure where uir (i = 1, 2, …, N) denotes the prescribed displacement (or prescribed rotation or prescribed internal discontinuity depending on the type of the redundant) of the original structure at a location and in the direction of the released redundant Ri. For the constructed primary structure (a structure resulting from releasing all the redundants R and subjected to the same set of external applied loads as that for the original structure), let define uo = {u1o, u2o, u3o, …, uNo} as a vector of kinematical quantities of the primary structure where uio (i = 1, 2, …, N) denotes the displacement (or the rotation or the discontinuity condition) of the primary structure at a location and in the direction associated with the released redundant Ri. Next, consider a released structure subjected only to a unit value of the redundant Ri and let define the flexibility coefficient fij (j = 1, 2, …, N) as the displacement (or rotation or the discontinuity condition) of this released structure at the location and in the direction associated with the released redundant Rj. A set of compatibility equations necessary and sufficient for solving all the unknown redundants R is given by Copyright © 2011 J. Rungamornrat

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 u1r   u1o   f11 f12 u  u   f  2 r   2o   21 f 22               u Nr  u No  f N1 f N 2

 f1N   R1   f 2 N   R 2           f NN  R N 

(10.14)

or, in a matrix form, by ur  uo  f R

(10.15)

where f is the flexibility matrix of the released structure given by  f11 f12 f f 22 f   21     f N1 f N 2

 f1N   f 2 N       f NN 

(10.16)

10.3.1 Vector of kinematical conditions ur A vector of kinematical conditions ur for a given statically indeterminate structure is known a priori. Determination of this vector requires only the knowledge of a choice of redundants, support conditions at locations and in directions where support reactions are chosen as redundants, and continuity conditions at locations where the internal forces are chosen as redundants. If the support reaction Ri is chosen as one of redundants, the corresponding kinematical condition uir vanishes if there is no support settlement in the direction of Ri and uir =  if the support settlement takes place in the direction of Ri with an amount of . If the internal force Ri is chosen as one of redundants, the corresponding kinematical condition uir vanishes if there is no discontinuity (e.g. relative rotation, gap, overlapping) taking place in the direction of Ri and uir =  if the discontinuity takes place in the direction of Ri with an amount of . Note that such discontinuity occurs only at a point where the internal constraint cannot completely be developed, e.g. a flexible rotational point and an extensible point.

10.3.2 Vector of kinematical quantities uo As evident from the definition described above, a vector uo contains the displacement components, rotations, or internal discontinuity (e.g. relative rotation, gap, overlapping) of the primary structure at the locations and in the directions of released redundants. Since the primary structure is statically determinate, computation of the vector uo can readily be achieved by using various techniques such as the method of curvature area, the conjugate structure analogy, and the principle of complementary virtual work or the unit load method once the static analysis for the internal forces of such structure is completed.

10.3.3 Flexibility matrix f The flexibility coefficient fij can readily be computed as follow: (i) obtain the released structure by releasing all the redundants and removing all external applied loads applied from the original structure, (ii) apply a unit value of the redundant Ri to the released structure, (iii) compute the displacement (or rotation or internal discontinuity) at the location and in the direction of the Copyright © 2011 J. Rungamornrat

445



released redundant Rj by using various techniques similar to those used to compute uo, and (iv) set fij equal to the quantity computed from (iii). This process is repeated until all coefficients fij are obtained. The flexibility matrix f of the released structure possesses following two important properties: symmetry and positive definiteness. The former significantly reduces the computational effort and storage while the latter implies that the flexibility matrix f is nonsingular or invertible and, as a consequence, the system of linear algebraic equations (10.14) possesses a unique solution. To prove these two properties, let consider a released structure obtained by releasing all the redundants and removing all external applied loads from the original structure. For convenience, let define the location and the direction of the released redundant Ri of the released structure as the ith degree of freedom. Next, let the released structure be subjected to a set of loads P = {P1, P2, P3, …, PN} where Pi is a concentrated load acting at the ith degree of freedom. This process is assumed to be continuous in the sense that the loads P increases continuously from zero to their final values {P1, P2, P3, …, PN}. The corresponding kinematical quantity (e.g. displacement, rotation, internal discontinuity) at the ith degree of freedom due to the loads P is denoted by ui and it can readily be computed by a method of superposition, i.e. uf P

(10.17)

where f is the flexibility matrix of the released structure given by (10.16) and u = {u1, u2, u3, …, uN}. Note that the flexibility coefficient fij can be viewed as the value of kinematical quantity at the ith degree of freedom due to a unit load acting only at the jth degree of freedom. The total work done W due to the loads P for the entire process is given, for a linearly elastic structure, by

W

1 T 1 P u  P Tf P 2 2

(10.18)

Since the work done W is a scalar quantity, taking the transpose of (10.18) yields the identical result, i.e.

W

1 T 1 u P  P Tf T P 2 2

(10.19)

Combining (10.18) and (10.19) leads to P T f  f T  P  0

(10.20)

Since (10.20) is valid for an arbitrary set of loads P, it can be deduced that f  fT

 f ij  f ji

(10.21)

or the flexibility matrix is essentially symmetric. The property (10.21) is also known as the Maxwell’s reciprocal theorem. This theorem simply states that the value of kinematic quantity at the ith degree of freedom due to a unit load acting at the jth degree of freedom is equal to the value of kinematic quantity at the jth degree of freedom due to a unit load acting at the ith degree of freedom. To clearly demonstrate the theorem, let consider two identical simply-supported beams as shown in Figure 9.8. One of the beams is subjected to a unit (vertical) load at a point 1 and the other is subjected to a unit (vertical) load at a point 2. From Maxwell’s reciprocal theorem, it can be deduced that the (vertical) deflection at the point 2 due to the unit (vertical) load applied at the Copyright © 2011 J. Rungamornrat

446



point 1, f21, is equal to the (vertical) deflection at the point 1 due to the unit (vertical) load applied at the point 2, f12. P=1 1

P=1 2

1

f21

f12

2

Figure 10.8: Schematic of two identical beams subjected to a unit load applied at different locations Next, let’s consider two identical rigid frames as shown in Figure 10.9. One of the frames is subjected to a unit horizontal load at a point 1 and the other is subjected to a unit vertical load at a point 2. From Maxwell’s reciprocal theorem, it can be deduced that the vertical displacement at the point 2 due to the unit horizontal load applied at the point 1, f21, is equal to the horizontal deflection at the point 1 due to the unit vertical load applied at the point 2, f12. P=1 2

2 f21

P=1

1

1

f12

Figure 10.9: Schematic of two identical rigid frames subjected to a unit load applied at different locations Similarly, let’s consider the same frames as shown in Figure 10.10; one of them is subjected to a unit horizontal load at a point 1 and the other is subjected to a unit moment at a point 2. From Maxwell’s reciprocal theorem, it implies that the rotation at the point 2 due to the unit horizontal load applied at the point 1, f21, is equal to the horizontal deflection at the point 1 due to the unit moment applied at the point 2, f12. M=1

f21 2 P=1

1

1

2 f12

Figure 10.10: Schematic of two identical identical frames subjected to a unit load applied at different locations Copyright © 2011 J. Rungamornrat

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Since the primary structure is statically stable, any set of non-zero loads P produces a set of non-zero displacements u such that the external work done is always greater than zero, i.e. W

1 T P u0 2

(10.22)

By employing the relation (10.17) along with (10.22), we then obtain P Tf P  0

(10.23)

for every set of non-zero loads P. This completes the proof of positive definiteness of the flexibility matrix. The positive definiteness of f implies that all its eigen values are positive. Now, by choosing a particular set of applied loads P = {P1 = 0, P2 = 0, … Pi-1 = 0, Pi = 1, Pi+1 = 0, …, PN = 0}, the condition (10.23) implies that P Tf P  f ii  0

(10.24)

That is all diagonal entries must be positive. Example 10.1 Let’s consider a statically indeterminate truss shown below. Properties of each member (e.g. cross sectional area, Young modulus, and coefficient of thermal expansion) are given in a table below. Analyze this structure for all support reactions and member forces when it is subjected to the following four effects: (i) applied loads P and 3P at joints 2 and 4, respectively, (ii) increase in temperature T in a member 35, (iii) the length of a member 13 is eo longer than L due to fabrication error, and (iv) a pinned support at a joint 1 is subjected to the leftward settlement s. In addition, determine the vertical displacement at a joint 4. 3P P

2

6 4

L

5

1 3 L

L Members

Area, A i

Young modulus, E i

Coefficient of thermal expansion, i

13, 35

2A

E



24, 46

2A

E



12, 34, 56

A

E



23, 36

A

E




448



Solution First, we determine the degree of static indeterminacy or the number of redundants, DI = (2 + 2) + (9  1) – (6  2) = 13 – 12 = 1. Thus, the structure is statically indeterminate with degree of static indeterminacy equal to 1. Next, we choose a horizontal reaction at a joint 5, R1, as a redundant and the corresponding primary structure, denoted by “structure 0”, is given in the figure below. It is worth noting that the primary structure is subjected to the same loading conditions, temperature change, error from fabrication, and support settlement as those for the original structure except that the redundant is released. Consider also the other system used for computing the flexibility matrix f, a released structure subjected only to a unit load at the location and in the direction of the redundant denoted by “structure I”, as shown in the figure below.

3P P

2

(-2P)

(-2P) 4

( 2 P) (-P) P

6

(0)

(-2P) 5

1

P

(0)

(0)

6

(0)

(+1)

(+1)

3

0

2P Structure 0 Primary structure

(0) 5

1

1

(0)

3

(0) 4

( 2 2 P)

(-3P)

(+P)

(0)

2

1

0

Structure I Released structure loaded by unit redundant

Since both the structure 0 and the structure I are statically determinate, all support reactions and member forces can readily be calculated from static equilibrium and the method of joints or the method of sections. Results are reported in above figures. To determine the redundant R1, it is required to set up the following compatibility equation

u1r   u1o   f11 R1

(e10.1.1)

Since there is no settlement in the horizontal direction of the support 5 of the original structure, then u1r which is the horizontal displacement at the joint 5 of the original structure vanishes, i.e. u1r = 0. The horizontal displacement u1o at the joint 5 of the primary structure can readily be computed by using the principle of complementary virtual work. By choosing the structure I as the virtual structure, u1o can be obtained as follows: Member

Ai

Ei

αi

Li

Fi0

ΔTi

ei

FiI

Fi0 FiI Li Ai Ei

α i Li ΔTi FiI

ei FiI

13

2A

E



L

P

0

eo

1

0

eo

35

2A

E



L

0

ΔT

0

1

PL 2AE 0

αLΔT

0

24

2A

E



L

-2P

0

0

0

0

0

0

46

2A

E



L

-2P

0

0

0

0

0

0


449



12

A

E



L

-P

0

0

0

0

0

0

23

A

E



2L

2P

0

0

0

0

0

0

34

A

E



L

-3P

0

0

0

0

0

0

36

A

E



2L

2 2P

0

0

0

0

0

0

56

A

E



L

-2P

0

0

0

0

0

0

PL 2AE

αLΔT

eo

 δWC  1  u1o  1  Δ s   u1o  Δ s m

δU C   i 1

m PL Fi δFi Li m   α i Li ΔTi δFi   ei δFi  + α L ΔT  e o Ai Ei 2AE i 1 i 1

δWC  δU C



u1o 

PL + α L ΔT  eo   s 2AE

The flexibility coefficient f11 or the horizontal displacement at the joint 5 of the structure I can also be computed from the principle of complementary virtual work by choosing the structure I itself as the virtual structure. Details of calculations are given below. Member

Ai

Ei

αi

Li

FiI

FiI

FiI FiI Li AiEi

13

2A

E



L

1

1

35

2A

E



L

1

1

24

2A

E



L

0

0

L 2AE L 2AE 0

46

2A

E



L

0

0

0

12

A

E



L

0

0

0

23

A

E



0

0

0

34

A

E



2L L

0

0

0

36

A

E



0

0

0

56

A

E



2L L

0

0

0



L AE

δWC  1  f11  f11 m

L Fi δFi Li  AE i 1 A i E i

δU C  

δWC  δU C  f11 

L AE Copyright © 2011 J. Rungamornrat

450



By substituting u1r, u1o, and f11 into the compatibility equation (e10.1.1), the redundant R1 can be solved as follows:

PL   L  R1  LT  eo   s     2AE   AE 

0  

 R1   P  AET  AEeo  AE s 2

L

L

After the redundant R1 was solved, the response of the original structure can be obtained by superposing the response of the primary structure and the response of the structure I multiplied by the computed redundant R1. Thus, the remaining support reactions of the original structure, in addition to R1, are given by R 1x  P  (1)R1  

P AEeo AE s  AET   2 L L

; R1y  P  (0)R 1  P ; R 5 y  2P  (0)R1  2P

All member forces are given by Member

Member forces Primary structure

Member forces Structure I

Member forces Original structure

13

P

1

P AEeo AE s  AET   2 L L

35

0

1

24

-2P

0

-2P

46

-2P

0

-2P

12

-P

0

-P

23

2P

0

2P

34

-3P

0

-3P

36

2 2P

0

2 2P

56

-2P

0

-2P



P AEeo AE s  AET   2 L L

Similarly, the vertical displacement at the joint 4 of the original structure, u4y, is equal to the sum of the vertical displacement at the joint 4 of the primary structure, u4yo, and the vertical displacement at the joint 4 of the structure I, u4yI, multiplied by the redundant R1. To proceed, we again employ the principle of complementary virtual work (or the unit load method) with a virtual structure shown below. 1 2

(-1/2) 4 ( 2/2) (-1)

(-1/2) 0

(-1/2)

6

( 2 /2) (-1/2) 5

1 (0)

3

(0)

1/2

1/2 Copyright © 2011 J. Rungamornrat

451



Member

Ai

Ei

αi

Li

Fi0

ΔTi

ei

δFi

Fi0δFi Li Ai Ei

α i Li ΔTi δFi

ei δFi

13

2A

E



L

P

0

eo

0

0

0

0

35

2A

E



L

0

ΔT

0

0

0

0

0

24

2A

E



L

-2P

0

0

-1/2

0

0

46

2A

E



L

-2P

0

0

-1/2

0

0

12

A

E



L

-P

0

0

-1/2

0

0

23

A

E



2L

2P

0

0

2 /2

0

0

34

A

E



L

-3P

0

0

-1

0

0

36

A

E



2L

2 2P

0

0

2 /2

0

0

56

A

E



L

-2P

0

0

-1/2

0

0

0

0



PL 2AE PL 2AE PL 2AE 2PL AE 3PL AE 2 2PL AE PL AE  11  PL  3 2 2  AE

The displacement u4yo is obtained as follow

δWC  1  u 4yo  0  u 4yo m

δU C   i 1

m Fi δFi Li m  11  PL   α i Li ΔTi δFi   ei δFi    3 2  Ai Ei 2  AE i 1 i 1

δWC  δU C



 11  PL u 4yo    3 2  2  AE

The displacement u4yI is obtained is obtained in a similar fashion as follows: δWC  1  u 4yI  u 4yI m

δU C   i 1

Fi δFi Li 0 AiEi

δWC  δU C



u 4yI  0


452



Member

Ai

Ei

αi

Li

FiI

δFi

FiI δFi Li Ai Ei

13

2A

E



L

1

0

0

35

2A

E



L

1

0

0

24

2A

E



L

0

-1/2

0

46

2A

E



L

0

-1/2

0

12

A

E



L

0

-1/2

0

23

A

E



2L

0

2 /2

0

34

A

E



L

0

-1

0

36

A

E



2L

0

2 /2

0

56

A

E



L

0

-1/2

0



0

Therefore, the vertical displacement at the joint 4 of the original structure, u4y, is equal to

 11  PL  11  PL u 4y  u 4yo  u 4yI  R 1    3 2  0  3 2 Downward 2  AE 2  AE It is important to emphasize that the virtual structure chose for computing the displacement of a statically indeterminate truss is not necessary to be of identical geometry to the original truss but can be its (statically determinate and statically stable) released structure. Use of this statically determinate structure in the analysis significantly reduces the computational effort and avoids solving another statically indeterminate structure.

Example 10.2 Determine all support reactions and member forces of a statically indeterminate truss subjected to external applied loads as shown in the figure below. Given that the axial rigidity EA is constant throughout. 3P 2

P

6

4

L

1

5

3 L

L


453



Solution First, we determine the degree of static indeterminacy or the number of redundants, DI = (2 + 1) + (11  1) – (6  2) = 14 – 12 = 2. Thus, the structure is statically indeterminate with degree of static indeterminacy equal to 2. Next, we choose the internal force of the member 23, denoted by F1, and the internal force of the member 36, denoted by F2, as two redundants and the corresponding primary structure obtained by releasing the two redundants via cutting the two members 23 and 36 and denoted by a “structure 0”, is given in the figure below. Consider also the other two systems used for computing the flexibility matrix f: one associated with a released structure subjected to a pair of unit and opposite forces at the cut of the member 23, denoted by a “structure I”, and the other associated with a released structure subjected to a pair of unit and opposite forces at the cut of the member 36, denoted by a “structure II”. 3P 2

P

(-P)

6

(0)

4

(0)

(0) (0)

P

(0)

(0)

1

(  2 P)

(  2 2 P)

(2P)

3

5

(2P)

P

2P

Structure 0 Primary structure

Since all three structures, the structure 0, the structure I and the structure II, are statically determinate, all support reactions and member forces can readily be calculated from static equilibrium and the method of joints or the method of sections. Results are reported in the figures. To determine the redundants F1 and F2, we need to set up the following compatibility equations u1r  u1o   f11 f12  F1        u 2 r  u 2 o  f 21 f 22  F2 

2

(e10.2.1)

( 1/ 2 )

(0)

4

6 (0)

(1) ( 1/ 2 )

( 1/ 2 ) (1) 0

1

(0)

1

(0)

1 ( 1/ 2 ) 0

3

5

(0)

Structure I Copyright © 2011 J. Rungamornrat

0

454



2

(0)

( 1/ 2 )

4

(0)

(1) ( 1/ 2 )

(0)

1

(0)

1

0

6

( 1/ 2 ) 0

3

( 1/ 2 )

1

(1) 5

(0) 0

Structure II

Since there is no gap and overlapping of the members 23 and 36 in the original structure, then u1r = u2r = 0. The overlapping generated at the member 23, u1o, and the overlapping generated at the member 36, u2o, of the primary structure (the structure 0) can be computed using the principle of complementary virtual work with the structure I and structure II be chosen as the virtual structures, respectively. Member

Ai

Ei

Li

Fi0

FiI

FiII

13

A

E

L

2P

1/ 2

0

35

A

E

L

2P

0

1/ 2

0

24

A

E

L

-P

1/ 2

0

2PL 2AE

0

46

A

E

L

0

0

1/ 2

0

0

12

A

E

L

0

1/ 2

0

0

0

14

A

E

2L

 2P

1

0

2PL AE

0

23

A

E

2L

0

1

0

0

0

34

A

E

L

0

1/ 2

1/ 2

0

0

45

A

E

2L

 2P

0

1

0

36

A

E

2L

0

0

1

0

0

56

A

E

L

0

0

1/ 2

0

0

 Copyright © 2011 J. Rungamornrat

Fi0 FiI Li Ai Ei

Fi0 FiII Li AiEi

2PL AE





 2  PL    2   AE 2  

0







2PL AE

2PL AE

 2 2

PL  AE

455



δWC  δU C



 Fi0 FiI Li 2  PL    2  u1o    AE 2 i 1 A i E i  

δWC  δU C



u 2o  

11





Fi0 FiII Li PL   2 2 AE i 1 A i E i 11

Note that the minus sign of u1o and u2o indicates that a gap occurs for both the member 23 and the member 36 in the original structure. The flexibility coefficient fij can also be computed from the principle of complementary virtual work by properly choosing a pair of actual and virtual structure from the structure I and the structure II as shown below. By using the symmetric property of the flexibility matrix, only three flexibility coefficients f11, f12 and f22 needs to be computed. In particular, the entry f11 is obtained by choosing the structure I as the actual and virtual structures, the entry f22 is obtained by choosing the structure II as the actual and virtual structures, and the entry f12 is obtained by choosing the structure I as the actual structure and the structure II as the virtual structure. Details of calculation are shown below. Member

Ai

Ei

Li

FiI

FiII

13

A

E

L

1/ 2

35

A

E

L

24

A

E

46

A

12

FiI FiI Li AiEi

FiI FiII Li Ai Ei

FiII FiII Li Ai Ei

0

L 2AE

0

0

0

1/ 2

0

0

L 2AE

L

1/ 2

0

L 2AE

0

0

E

L

0

1/ 2

0

0

L 2AE

A

E

L

1/ 2

0

L 2AE

0

0

14

A

E

2L

1

0

2L AE

0

0

23

A

E

2L

1

0

34

A

E

L

1/ 2

1/ 2

45

A

E

2L

0

36

A

E

2L

56

A

E

L

2L AE L 2AE

0

0

L 2AE

L 2AE

1

0

0

2L AE

0

1

0

0

0

1/ 2

0

0

L 2  2 2  AE

L 2AE




2L AE L 2AE

L 2  2 2  AE

456



δWC  δU C



δWC  δU C



δWC  δU C

11

f11   i 1 11

f12   i 1





FiI FiI Li L  22 2 Ai Ei AE FiI FiII Li L   f 21 Ai Ei 2AE





FiII FiII Li L  22 2 f 22   AE i 1 A i E i 11



By substituting u1r, u2r, u1o, u2o, and f into the compatibility equations (e10.2.1), the two redundants F1 and F2 can be solved as follows: 0 PL 2  2 / 2 L 2  2 2      AE  2  2  AE  1 / 2 0

1/ 2 22

  F1    2  F2 

 F1  P  20  18 2  0.4927       P F2  47  32 2  28  31 2  0.7787  Once the redundants F1 and F2 were solved, the responses of the original structure can be obtained by superposing the responses of the primary structure, the responses of the structure I multiplied by F1 and the responses of the structure I multiplied by F2. All support reactions of the original structure are given by R 1x  P  (0)F1  (0)F2  P

R 1y  P  (0)F1  (0)F2  P

;

;

R 5 y  2P  (0)F1  (0)F2  2P

Similarly, all member forces are given in the table below. Member

Fi0

FiI

FiII

Fi  Fi0  FiI (0.4927 P)  FiII (0.7787 P)

13

2P

1/ 2

0

1.6516P

35

2P

0

1/ 2

1.4494P

24

-P

1/ 2

0

-1.3484P

46

0

0

1/ 2

-0.5506P

12

0

1/ 2

0

-0.3484P

14

 2P

1

0

-0.9215P

23

0

1

0

0.4927P

34

0

1/ 2

1/ 2

-0.8990P

45

 2P

0

1

-0.6355P

36

0

0

1

0.7787P

56

0

0

1/ 2

-0.5506P


457



Example 10.3 Consider a statically indeterminate beam as shown in the figure below. The flexural rigidity EI is assumed to be constant throughout. The beam is subjected to a uniformly distributed load 2q on the segment AB and a concentrated load qL at a point C and, during load application, the support at B settles downward with an amount of o. Determine all support reactions, sketch the SFD and BMD for the case that o = 0, and compute the (vertical) deflection at point C. Consider only the bending effect in the analysis. 2q

qL B

A

C 2L

L

Solution First, we determine the degree of static indeterminacy or the number of redundants, DI = (2 + 1) + (2  2) – (3  2) = 7 – 6 = 1. Thus, the structure is statically indeterminate with degree of static indeterminacy 1. Next, we choose a moment reaction at point A, M1, as a redundant and the corresponding primary structure, denoted by “structure 0”, is given in the figure below. Consider also a released structure (used for computing the flexibility matrix f) subjected only to a unit moment at point A, denoted by “structure I”, as shown in the figure below. 2q

qL 1

B A

B C

C

3qL/2

1/2L

7qL/2

1/2L

3qL2 M0 -qL2 -4qL2

MI -1

Structure 0

Structure I

Since both the structure 0 and the structure I are statically determinate, all support reactions and the bending moment diagram can readily be obtained from static equilibrium with results given in above figures. To determine the redundant M1, we need to set up the following compatibility equation

u1r   u1o  f11 M1

(e10.3.1)

Since there is no support rotation at the fixed support, then u1r = 0. The rotation at the point A, u1o, of the primary structure (structure 0) can readily be computed using the principle of complementary virtual work with the structure I being chosen as the virtual structure as follows: Copyright © 2011 J. Rungamornrat

458



δWC  1  u1o  (1/2L) o  u1o   o / 2L

δU C  δU

AB C

 δU

BC C



L AB

 0

M0M I dx  EI

L BC

 0

M 0M I dx EI

1  3qL   2  EI

2  qL3  1  1  4qL   1 2L       2L     0   3EI  4  3  3  EI   qL3  o u1o    CW 3EI 2L 2



δWC  δU C

The flexibility coefficient f11 can also be computed from the principle of complementary virtual work by choosing the structure I as both the actual and virtual structures as shown below. δWC  1  f11  f11 δU C  δU

AB C

 δU

BC C



L AB

 0



δWC  δU C

M IM I dx  EI

f11 

L BC

 0

M IM I 1 1   2  2L dx    2L     2  EI  EI  3  3EI

2L CCW 3EI

By substituting u1r, u1o, and f11 into the compatibility equation (e10.3.1), the redundant M1 can be solved as follows: 

3

0   qL

 3EI



 o   2L  M1  2L   3EI 

 M1  qL  3EI2 o CCW 2

2

4L

After the redundant M1 was solved, the response of the original structure can be obtained by superposing the responses of the primary structure and the responses of the structure I multiplied by M1. Thus, the remaining support reactions of the original structure are given by R Ay 

3qL  1  qL2 3EI o  7qL 3EI o 7qL  1  qL2 3EI o  13qL 3EI o ;     R           By 2 4L2  4 8L3 2 4L2  4 8L3  2L  2  2L  2

The shear force diagram and bending moment diagram of the entire beam is shown below. 2q M1 = qL2/2

qL

B

A

C

RAy = 7qL/4 RBy = 13qL/4 qL

7qL/4

SFD 17qL2/64

–9qL/4

BMD –qL2/2 –qL2 Copyright © 2011 J. Rungamornrat

459



The deflection at the point C of the original structure, vC, is equal to the sum of the deflection at the point C of the primary structure, vCo, and the deflection at the point C of the structure I, vCI, multiplied by the redundant M1. To compute such displacements, the principle of complementary virtual work is employed with a special choice of the virtual structure shown below. P = 1

B A

C

1/2

3/2 M

–L Virtual Structure The deflection vCo is obtained as follows δWC  1  vCo  (3/2) o  vCo  3 o / 2 δU C  δU

AB C

 δU

BC C



L AB

 0

M 0M dx  EI

L BC

 0

M 0M dx EI

2 2 4 1  3qL   2L  1  4qL   3L  1  qL   2L  qL 2L   2L   L              2  EI   3  3  EI   4  2  EI   3  3EI qL4 3 o vCo   Downward 3EI 2 2

δWC  δU C



The deflection vCI is obtained as follows δWC  1  vCI  vCI δU C  δU

AB C

 δU

BC C



L AB

 0

δWC  δU C



M I M dx  EI

v CI 

L BC

 0

2 M I M 1 1   L L dx    2L     2  EI  EI  3  3EI

2

L 3EI

Downward

Therefore, the deflection at the point C of the original structure, vC, is equal to v C  v Co  v CI  M1 

qL4 3 o  L2   qL2 3EIΔ o     3EI 2  3EI   2 4L2

 qL4 7 o  Downward  4  2EI

It is important to emphasize again that the virtual structure chosen for computing the deflection or rotation of a statically indeterminate beam is not necessary to be of identical geometry to the original beam but can be its (statically determinate and statically stable) released structure. Use of this statically determinate structure in the analysis significantly reduces the computational effort. Copyright © 2011 J. Rungamornrat

460



Example 10.4 Determine all support reactions and sketch the SFD and BMD of a statically indeterminate beam subjected to uniformly distributed load q shown below. The flexural rigidity of the segments AB and BC is given by EI and 2EI, respectively. In the analysis, let’s consider only the bending effect.

q A C

B L

L

Solution First, we determine the degree of static indeterminacy of the beam: DI = (2 + 1 + 1) + (2  2) – (3  2) = 8 – 6 = 2. Thus, the structure is statically indeterminate with degree of static indeterminacy equal to 2. Next, we choose a moment reaction M1 at point A and the bending moment M2 at point B as redundants and the corresponding primary structure, denoted by a “structure 0”, is given in the figure below. Consider also the other two systems used for computing the flexibility matrix f: one associated with a released structure subjected to a unit moment at a point A, denoted by a “structure I”, and the other associated with a released structure subjected to a pair of unit and opposite moments at a point B, denoted by a “structure II”.

q 1 A qL/2

B

A

C

qL

B

1/L

qL/2

1/L

C 0

qL2/2 MI

M0 –qL2/2

–1

Structure 0

Structure I 1

B

1

A 1/L

C 1/L

0 MII

–1 Structure II Copyright © 2011 J. Rungamornrat

461



Since all three structures (i.e. the structure 0, the structure I and the structure II) are statically determinate, all support reactions and bending moment diagram can readily be obtained from static equilibrium with results reported in above figures. To determine the redundants M1 and M2, we need to set up the following compatibility equations u1r  u1o   f11 f12  M1          u 2 r  u 2o  f 21 f 22  M 2 

(e10.4.1)

Since there is no support rotation at the point A and no rotational discontinuity at point B in the original structure, then u1r = u2r = 0. The rotation at the point A, u1o, and the relative rotation at the hinge point B, u2o, of the primary structure (structure 0) can be computed using the principle of complementary virtual work with the structure I and structure II be chosen as the virtual structure, respectively. Details of calculations are given below. Computation of u1o: the structure 0 is treated as the actual structure whereas the structure I is chosen as the virtual structure δWC  1  u1o  u1o

δU C  δU

AB C

 δU

BC C



L AB

 0

M0M I dx  EI

L BC

 0

M0M I dx EI

qL3 1  qL   1  1  qL2   1  L     0   L         24EI 2  2EI   3  3  2EI   4  2

δWC  δU C



u1o  

qL3 CW 24EI

Computation of u2o: the structure 0 is treated as the actual structure whereas the structure II is chosen as the virtual structure δWC  1  u 2o  u 2o δU C  δU

AB C

 δU

BC C



L AB

 0

M 0 M II dx  EI

L BC

 0

M 0 M II dx EI

qL3 1  qL2   3  1  qL   2  L   (2)   L   (2)      12EI 3  2EI   4  2  2EI   3  2

δWC  δU C



u 2o  

qL3 CW 12EI

Note that the minus sign of u1o and u2o indicates that the point A of the primary structure rotates in a clockwise direction and the relative rotation at the point B of the primary structure is in THE clockwise direction. The flexibility coefficient fij can also be computed from the principle of complementary virtual work by properly choosing a pair of actual and virtual structures from the structure I and the structure II as shown below. By using the symmetric property of the flexibility matrix, only three flexibility coefficients f11, f12 and f22 needs to be computed. Copyright © 2011 J. Rungamornrat

462



Computation of f11: the structure I is treated as the actual and virtual structures δWC  1  f11  f11 δU C  δU

AB C

 δU

BC C



L AB

 0

δWC  δU C



M IM I dx  EI

f11 

L BC

 0

1 1   2 L MIMI dx    L     2  EI   3  3EI EI

L 3EI

Computation of f12: the structure I is treated as the actual structure whereas the structure II is chosen as the virtual structure δWC  1  f12  f12 δU C  δU

AB C

 δU

BC C



L AB

 0

δWC  δU C



M I M II dx  EI

f12 

L BC

 0

M I M II 1 1   1 L dx    L     2  EI   3  6EI EI

L  f 21 6EI

Computation of f22: the structure II is treated as the actual and virtual structures δWC  1  f 22  f 22 δU C  δU

AB C

 δU

BC C



L AB

 0

δWC  δU C



M II M II dx  EI

f 22 

L BC

 0

M II M II 1 1   2 2L dx    L   (2)  EI 2  EI   3  3EI

2L 3EI

By substituting u1r, u2r, u1o, u2o, and f into the compatibility equations (e10.4.1), the two redundants M1 and M2 can be solved as follow: 0 qL3 1  L 2 1   M1           0 24 EI   2 6EI 1 4 M 2 

 M1  qL2 2     M 2  28 3



Once the redundants M1 and M2 were solved, the responses of the original structure can be obtained by superposing the responses of the primary structure and the responses of the structure I multiplied by M1 and the responses of the structure II multiplied by M2. Thus, all support reactions of the original structure are given by R Ay 

qL  1  qL2   1  3qL2  13qL          2  L  14   L  28  28

2 2  1  qL   2  3qL  8qL       R By  qL     7  L  14   L  28 

;

;

M A  M1 

R Cy 

qL2 14

 qL2   1  3qL2  11qL qL         0  2 28  14   L  28 


463



The total shear force diagram (SFD) and the total bending moment diagram (BMD) of the original structure are given below. q MA = qL2/14

C

B

A RAy = 13qL/28

RBy = 8qL/7

RCy = 11qL/28

L

L 17qL/28

13qL/28

SFD –11qL/28

–15qL/28 121qL2/1568 2

57qL /1568

BMD –qL2/14

–3qL2/28

Example 10.5 Consider a statically indeterminate rigid frame as shown in the figure below. Given that EI for the vertical and horizontal segments are 2EI and EI, respectively. The frame is subjected to external loads as shown in the figure and, during load applications, the roller support at a point C settles downward with an amount of o. Determine all support reactions, sketch the SFD and BMD for the case that o = 0, and compute the rotation at point C. Consider only deformation due to bending.

qL C B L

q

A L/2

L/2

Solution First, we determine the degree of static indeterminacy of the given rigid frame: DI = (3 + 1) + (2 3) – (3 3) = 10 – 9 = 1. Thus, the structure is statically indeterminate with degree of static indeterminacy equal to 1. Next, we choose the support reaction of the roller support at the point C, denoted by R1, as a redundant and the corresponding primary structure, denoted by a “structure 0”, Copyright © 2011 J. Rungamornrat

464



is given in the figure below. Consider also a released structure (used for computing the flexibility matrix f) subjected only to a unit load at the point C, denoted by a “structure I”, as shown in the figure below. Structure 0

Structure I

qL C

C

B

B 1

q

A

A

qL qL2

0 L

qL

1 L 2

B

–qL /2

C

B

C

2

–qL /2

M0

2

A

MI

L

A

–qL

Since both structure 0 and structure I are statically determinate, all support reactions and the bending moment diagrams can readily be obtained from static equilibrium with results reported in above figures. To determine the redundant R1, we need to set up the following compatibility equation

u1r   u1o   f11 R1

(e10.5.1)

Since there is downward settlement at the roller support equal to o, then u1r = –o. The vertical displacement at the point C, u1o, of the primary structure (structure 0) can readily be computed using the principle of complementary virtual with the structure I be chosen as the virtual structure as follows: δWC  1  u1o  0  u1o δU C  δU CAB  δU CBC 

L AB

 0

M 0M I dx  EI

L BC

 0

M 0M I dx EI

1  qL  1  qL2  1  qL2  L  5L  5qL4 L L     L L             2  4EI  3  4EI  2  2EI  2  6  16EI 2


465




δWC  δU C

u1o  


5qL4 16EI

Downward

The flexibility coefficient f11 can also be computed from the principle of complementary virtual work by choosing the structure I as both the actual and virtual structures as shown below. δWC  1  f11  f11 δU C  δU

AB C

 δU

BC C



L AB

 0



δWC  δU C

M IM I dx  EI

5L3 f11  6EI

L BC

 0

M IM I 1  L   2L  5L3  L  dx   L L    L    2  EI   3  6EI EI  2EI 

Upward

By substituting u1r, u1o, and f11 into the compatibility equation (e10.5.1), the redundant R1 can be solved as follows:  5qL4   5L3    o      R 1  16EI   6EI 



R1 

3qL 6EI o  8 5L3

After the redundant R1 was solved, the response of the original structure can be obtained by superposing the response of the primary structure and the response of the structure I multiplied by R1. Thus, all support reactions of the original structure are given by  3qL 6EIΔ o  R Ax  qL   0     qL Leftward 5L3   8  3qL 6EIΔ o  5qL 6EIΔ o R Ay  qL   1     Upward 5L3  8 5L3  8 2  3qL 6EI o  5qL 6EI o  M A  qL2   L    CCW  5L3  8 5L2  8 3qL 6EIΔ o R Cy  R1   8 5L3

The total shear force diagram (SFD) and the total bending moment diagram (BMD) of the original structure for the case that o = 0 are given below. 3qL2/16

5qL/8

qL

C

C B

B –3qL/8

q

3qL/8

A

qL

B –qL2/8

SFD

qL

A

2

qL qL


C

–qL2/8

BMD

A

–5qL2/8

466



Similarly, the rotation at the point C of the original structure, C, is equal to the sum of the rotation at the point C of the primary structure, Co, and the rotation at the point C of the structure I, CI, multiplied by the redundant R1. To proceed, we again employ the principle of complementary virtual work with a virtual structure shown below. 1 B

C

B

C

1 1

A

A

0 1

0 M

Virtual Structure The rotation Co is obtained as follows: δWC  1  Co  Co δU C  δU CAB  δU CBC 

L AB

 0

δWC  δU C



M 0M dx  EI

L BC

 0

M 0M dx EI

1  qL2  1  qL2  1  qL2  L  qL3 L 1    L 1     1      2  4EI  3  4EI  2  2EI  2  3EI qL3 Co   3EI

The rotation CI is obtained as follows: δWC  1  CI  CI δU C  δU

AB C

 δU

BC C



L AB

 0

δWC  δU C



M IM dx  EI

CI 

L BC

 0

1 L  L2 M I M  L  dx   L 1   L 1  2  EI  EI EI  2EI 

2

L EI

Therefore, the rotation at the point C of the original structure, C, is equal to C  C0  CI  R1  

qL3  L2   3qL 6EI 0  qL3 6 0       3EI  EI   8 5L3  24EI 5L

The rotation at the point C is in the counter clockwise direction if C > 0; otherwise, it is in the clockwise direction. Copyright © 2011 J. Rungamornrat

467



Example 10.6 Consider a statically indeterminate rigid frame as shown in the figure below. Given that E, G, A, I,  are throughout. The frame is subjected to external loads as shown in the figure below and, during load applications, the roller support at a point C settles downward with an amount of o. Choose a proper set of redundants and then set up the corresponding set of compatibility equations to solve for those redundants. In the analysis, let’s consider all possible effects. q

qL

B C

L

D

A L

L

Solution First, we determine the degree of static indeterminacy of the given frame as follow: DI = (3 + 1 + 1) + (3 3) – (4 3) = 14 – 12 = 2. Thus, the structure is statically indeterminate with degree of static indeterminacy equal to 2. Next, we choose a reaction R1 at the point C and a reaction R2 at the point D as redundants and the corresponding primary structure, denoted by a “structure 0”, is given in the figure below. Consider also the other two systems used for computing the flexibility matrix f: one associated with a released structure subjected to a unit vertical load at the point C, denoted by a “structure I”, and the other associated with a released structure subjected to a unit vertical load at the point D, denoted by a “structure II”. Since all three structures (i.e. the structure 0, the structure I and the structure II) are statically determinate, all support reactions, the axial force diagram, the shear force diagram and the bending moment diagrams can readily be obtained from static equilibrium with results reported in the figures below. To determine the redundants R1 and R2, we need to set up the following compatibility equations

M0 –qL2/2 qL

V0 –qL2/2 qL

B –qL

qL

M

0

–3qL2/2

C

Structure 0 A 0

V

qL

qL

3qL2/2 Copyright © 2011 J. Rungamornrat

qL

D

qL

468



L

MI VI

–1

C 0

B

0

D

1 L

1

0

Structure I A

M

I

I

V

0 L

1

2L

MII VII

–1

D 0

B

C

0 1

2L

1

0

Structure II A

M

II

II

V

0 2L

1

u1r  u1o   f11 f12  R 1        u 2 r  u 2o  f 21 f 22  R 2 

(e10.6.1)

Since the roller support at the point C settles downward with an amount of o and there is no support settlement of the roller support at the point D in the original structure, then u1r = –o and u2r = 0. The vertical displacement at the point C, u1o, and the vertical displacement at the point D, u2o, of the primary structure (structure 0) can be computed using the principle of complementary virtual work with the structure I and structure II be chosen as the virtual structure, respectively. Details of calculations are given below. Copyright © 2011 J. Rungamornrat

469



Computation of u1o: the structure 0 is treated as the actual structure whereas the structure I is chosen as the virtual structure δWC  1  u1o  u1o δU C  δU Cb  δU Ca  δU Cs δU Cb  δU

AB Cb

 δU

BC Cb

 δU

CD Cb



L AB

 0

M0M I dx  EI

1  2qL    2  EI

2

δU Ca  δU

AB Ca

 δU

BC Ca

 δU

CD Ca



L AB

 0

L BC

 0

M 0M I dx  EI

L CD

 0

M 0M I dx EI

 1  qL   3L  9qL4 L L     L    0   3  2EI   4  8EI  2

F0 F I dx  EA

L BC

 0

F0 F I dx  EA

L CD

 0

F0 F I dx EA

2

qL  qL    L 1  0  0   EA  EA 

δU Cs  δU

AB Cs

 δU

BC Cs

 δU

CD Cs



L AB

 0

V 0 V I dx  GA

L BC

 0


L CD

 0

V 0 V I dx GA

qL 1  qL  0  L  1  0   2  GA  2GA 2

δWC  δU C



u1o  

9qL4 qL2 qL2   8EI EA 2GA

Computation of u2o: the structure 0 is treated as the actual structure whereas the structure II is chosen as the virtual structure δWC  1  u 2o  u 2o δU C  δU Cb  δU Ca  δU Cs δU Cb  δU

AB Cb

 δU

BC Cb

 δU

CD Cb



L AB

 0

M 0 M II dx  EI

1  2qL    2  EI

2

AB BC δU Ca  δU Ca  δU Ca  δU CD Ca 

L AB

 0

L BC

 0

M 0 M II dx  EI

L CD

 0

M 0 M II dx EI

 1  qL   7 L  55qL4 L  L 2L     0     3  2EI   4  24EI 

F0 F I dx  EA

2

L BC

 0

F0 F I dx  EA

L CD

 0

F0 F I dx EA

2

qL  qL    L 1  0  0   EA  EA 

δU Cs  δU

AB Cs

 δU

BC Cs

 δU

CD Cs



L AB

 0


L BC

 0



L CD

 0

V 0 V I dx GA

470



qL2 1  qL  0  L  1  0   2  GA  2GA

δWC  δU C



u 2o  

55qL4 qL2 qL2   24EI EA 2GA

The flexibility coefficient fij can also be computed from the principle of complementary virtual work by properly choosing a pair of actual and virtual structures from the structure I and the structure II as shown below. By using the symmetric property of the flexibility matrix, only three flexibility coefficients f11, f12 and f22 needs to be computed. Computation of f11: the structure I is treated as the actual and virtual structures δWC  1  f11  f11 δU C  δU Cb  δU Ca  δU Cs δU Cb  δU

AB Cb

 δU

BC Cb

 δU

CD Cb



L AB

 0

M IM I dx  EI

L BC

 0

M IM I dx  EI

L CD

 0

M IM I dx EI

1  L   2L  4L3 L 0   L L    L     2  EI   3  3EI  EI 

δU Ca  δU

AB Ca

 δU

BC Ca

 δU

CD Ca



L AB

 0

FI FI dx  EA

L BC

 0

FI FI dx  EA

L CD

 0

FI FI dx EA

L  1   L 1  0  0  EA  EA  AB BC δU Cs  δU Cs  δU Cs  δU CD Cs 

L AB

 0

V I V I dx  GA

L BC

 0

V I V I dx  GA

L CD

 0

V I V I dx GA

L     0   L  1  0  GA  GA  δWC  δU C



f11 

L 4L3 L   3EI EA GA

Computation of f12: the structure I is treated as the actual structure whereas the structure II is chosen as the virtual structure δWC  1  f12  f12 δU C  δU Cb  δU Ca  δU Cs AB BC δU Cb  δU Cb  δU Cb  δU CD Cb 

L AB

 0

M I M II dx  EI

L BC

 0

M I M II dx  EI

L CD

 0

M I M II dx EI

1  L   5L  17L3 L   L 2L    L    0  2  EI   3  6EI  EI  Copyright © 2011 J. Rungamornrat

471


δU Ca  δU

AB Ca

 δU

BC Ca

 δU

CD Ca



L AB

 0


FI FII dx  EA

L BC

FI FII dx  EA

 0

L CD

 0

FI FII dx EA

L  1   L 1  0  0  EA  EA  δU Cs  δU

AB Cs

 δU

BC Cs

 δU



CD Cs

L AB

 0

V I V II dx  GA

L BC

 0

V I V II dx  GA

L CD

 0

V I V II dx GA

L     0   L  1  0  GA  GA  δWC  δU C



f12 

17L3 L L    f 21 6EI EA GA

Computation of f22: the structure II is treated as the actual and virtual structures δWC  1  f 22  f 22 δU C  δU Cb  δU Ca  δU Cs δU Cb  δU

AB Cb

 δU

BC Cb

 δU

CD Cb



L AB

 0

M II M II dx  EI

L BC

 0

M II M II dx  EI

L CD

 0

M II M II dx EI

3 1  2L   2L   4L  20L   L 2L    2L   2  EI   EI   3  3EI

δU Ca  δU

AB Ca

 δU

BC Ca

 δU

CD Ca



L AB

 0

FII FII dx  EA

L BC

 0

FII FII dx  EA

L CD

 0

FII FII dx EA

L  1   L 1  0  0  EA  EA  δU Cs  δU

AB Cs

 δU

BC Cs

 δU

CD Cs



L AB

 0

V II V II dx  GA

L BC

 0

V II V II dx  GA

L CD

 0

V IIV II dx GA

2L     0   2L  1  0  GA  GA  δWC  δU C



f 22 

20L3 L 2L   3EI EA GA

By substituting u1r, u2r, u1o, u2o, and f into (e10.6.1), we obtain a set of two compatibility equations governing the two unknown redundants R1 and R2:  4L3 L L    o  qL4 27  qL2 1 qL2 1  3EI EA GA          24EI 55  EA 1 2GA 1 17L3 L L  0   6EI  EA  GA Copyright © 2011 J. Rungamornrat

L  17L3 L    6EI EA GA   R1    20L3 L 2L   R2    3EI EA GA 

472



Example 10.7 Consider a statically indeterminate rigid frame as shown in the figure below. The flexural rigidity EI is assumed to be constant throughout and the axial member AD has constant axial rigidity EA. The frame is subjected to a uniformly distributed load q on the top segment BC. Determine all support reactions, the member force in the axial member, the shear force diagram and the bending moment diagram for the entire frame, and then compute the displacement at a point D. Consider only deformation due to bending effect for the segments AB, BC, and CD and assume that I = AL2/3. q

C

B

L

D

A L

Solution First, we determine the degree of static indeterminacy of the given structure as follow: DI = (2 + 1) + (3 3+1) – (3 4) = 13 – 12 = 1. Thus, the structure is statically indeterminate with degree of static indeterminacy equal to 1. Next, we choose the axial force in member AD, denoted by F1, as a redundant and the corresponding primary structure, denoted by a “structure 0”, is given in the figure below. Consider also a released structure (used for computing the flexibility matrix f) subjected only to a pair of unit and opposite forces at the point of release, denoted by a “structure I”, as shown in the figure below.

q C

B

0

A

C

B

D

0

1

A

D 1

qL/2

qL/2

0

0

qL2/8 –L C

B –L C

B

–L

0 A

0

M0 0

0

MI D

A

1

1

Structure I

Structure 0 Copyright © 2011 J. Rungamornrat

D

473



Since both the structure 0 and the structure I are statically determinate, all support reactions and the bending moment diagram can readily be obtained from static equilibrium with results given in above figures. To determine the redundant F1, we need to set up the following compatibility equation

u1r   u1o   f11 F1

(e10.7.1)

Since there is no internal discontinuity in the member AD, then u1r = 0. The overlapping generated in the member AD of the primary structure (structure 0) , denoted by u1o, can readily be computed using the principle of complementary virtual work with the structure I being chosen as the virtual structure as follows: δWC  1  u1o  0  u1o δU C  δU

AB Cb

 δU

BC Cb

 δU

CD Cb

 δU

AD Ca



L AB

 0

M0M I dx  EI

L BC

 0

M 0M I dx  EI

L CD

 0

M 0M I F0 FI L AD dx  EI EA

2  qL  qL4  0       L L 0 0     3  8EI  12EI 2

δWC  δU C

qL4 Gap u1o   12EI



The flexibility coefficient f11 can also be computed from the principle of complementary virtual work by choosing the structure I as the actual and virtual structures as shown below. δWC  1  f11  f11 δU C  δU

AB Cb

 δU

BC Cb

 δU

CD Cb

 δU

AD Ca



L AB

 0

M IM I dx  EI

L BC

 0

M IM I dx  EI

L CD

 0

M IM I FI FI L AD dx  EI EA

1  L   2L  L 5L3 L  L       L          2 L L    2  EI   3  EA 3EI EA  EI 

δWC  δU C



f11 

5L3 L 5L3 L3 2L3     3EI EA 3EI 3EI EI

Overlapping

By substituting u1r, u1o, and f11 into the compatibility equation (e10.7.1), the redundant R1 can be solved as follow: 

qL4   2L3    F1  12EI   EI 

0  

 F1  qL 24

Once the redundant F1 was solved, the response of the original structure can be obtained by superposing the responses of the primary structure and the responses of the structure I multiplied by R1. Thus, all support reactions of the original structure are given by  qL  R Ax  0  0   0  24 

;

R Ay 

qL  qL  qL  0    ; 2  24  2 Copyright © 2011 J. Rungamornrat

R Dy 

qL  qL  qL  0    2  24  2

474



The shear force and bending moment diagrams of the original structure are given below. q

C

B

C

B

qL2/12

qL/2

(qL/24)

A

D

–qL2/24 –qL2/24

–qL /24

–qL/2 –qL/24

0

B 2

–qL2/24

qL/24

A

D

A

D BMD

SFD qL/2

C

qL/2

Similarly, the displacement at the point D of the original structure, denoted by uD, is equal to the sum of the displacement at the point D of the primary structure, denoted by uDo, and the displacement at the point D of the structure I, denoted by uDI, multiplied by the redundant F1. To proceed, we again employ the principle of complementary virtual work with a virtual structure shown below. L L

C

B

B

C

L

M

1

A

D

0

A

1

0

0

D

0

Virtual Structure

The displacement uDo is obtained as follows δWC  1  u Do  u Do AB BC AD δU C  δU Cb  δU Cb  δU CD Cb  δU Ca 

L AB

 0

M 0M dx  EI

L BC

 0

M 0M dx  EI

L CD

 0

2  qL  qL  0    L  L   0  0  3  8EI  12EI 2

δWC  δU C



u Do 

qL4 Rightward 12EI Copyright © 2011 J. Rungamornrat

4

M 0M F0FL AD dx  EI EA


475


The displacement uDI is obtained as follows δWC  1  u DI  u DI

δU C  δU

AB Cb

 δU

BC Cb

 δU

CD Cb

 δU



AD Ca

L AB

 0

M IM dx  EI

L BC

 0

M I M dx  EI

L CD

 0

M IM FIFL AD dx  EI EA

1  L   2L  5L3  L    L    2    L L    2  EI   3  3EI  EI 

δWC  δU C



u DI  

5L3 Leftward 3EI

Therefore, the displacement at the point D of the original structure (uD) is equal to u D  u Do  u DI  F1 

qL4  5L3   qL  qL4  Rightward   12EI  3EI   24  72EI

Exercises 1. Analyze following statically indeterminate trusses for support reactions and member forces by the method of consistent deformation. The axial rigidity of each member, the support settlement (if exists), the temperature change, and error from fabrication are shown in the figure. 2P 2EA

P

2P

L

EA

L

2EA

2EA EA

EA

2EA

2EA

L

P 2EA

2EA

EA

EA EA

EA

o

L

L EA is constant

, T

eo

L

L

EA is constant

L

P

P

L eo

L

L L Copyright © 2011 J. Rungamornrat

, T

, –T

L


476


P

P EA

EA is constant

L L

2EA

2EA

L

eo

, T

o L

L

L

L

L

2. Analyze following statically indeterminate beams for all support reactions, SFD and BMD, the deflection at a point a and the rotation at a point b by the method of consistent deformation. In the analysis, let’s consider only bending effect. The flexural rigidity and length of each member and the support settlement are clearly indicated in the figure.

2q

qL a, b

EI, L

2EI, 3L

q EI, L

EI, L

o

o

P

2EI, L

a

P EI, L

EI, L1

a, b

qL M 2EI, L

a

EI, L

b

EI, L a, b EI, L

b


q EI, 2L

q

M EI, L

EI, L2

EI, L

EI, L

b


477


3. Analyze following statically indeterminate frames for all support reactions, AFD, SFD, BMD, the deflection at a point a and the rotation at a point b by the method of consistent deformation. In the analysis, let’s consider only the bending effect except for axial members. The flexural rigidity and length of each segment and the support settlement are clearly indicated in the figure. b

b

2EI, L

q

EI, L

q

EI, L

2qL a 2EI, L

o

a

2EI, L

q a EI, 2L

a, b

P

EI, 2L EI, L/2 P

2EI, L

EI, L

EI, L

EI, L/2 EI, L

b

EI, L

, T

a, b

EA, L

EA, L EI, L

EI, L

q 2EI, 2L a, b

EI, 2L q P EI, L EI, L

a, b EA, L

EI, L 2PL

o Copyright © 2011 J. Rungamornrat

a, b

2EI, L


478


a

EA,

2P

2L

EA,

2L

EA, L

EI, L

EA, L

b

2EI, L

EA, L P 2P

2P

P

2P

P

3P

L

EA constant 2P

3L

EI

EI

L

L

L


L

479


Influence Lines

CHAPTER 11 INFLUENCE LINES Structural modeling and analysis focused on in previous chapters are aimed primarily to determine responses of structures under an action of applied loads whose magnitude, direction and point of application being entirely fixed; i.e. only a single load pattern is considered in the analysis. However, in practices, there are various types of structures that are purposely designed and built to be functioned under an action of non-stationary excitations generally termed moving loads, for instance, bridges, railway and subway structures, cranes, etc. Analysis of such structures to identify the critical loading patterns and to predict the corresponding maximum structural responses is essential and obligatory to ensure both the safe and economical designs. In this chapter, the concept of influence lines and how to construct them is first presented and its connection and applications to the analysis of statically determinate structures (e.g. beams, floor systems, trusses) under moving loads are subsequently demonstrated. Finally, the concept of influence lines is extended to statically indeterminate structures.

11.1 Introduction to Concept of Influence Lines

P1

P2

P3 P4

P5

Loading path Figure 11.1: Schematic of truss structure subjected to series of axle loads along loading path on bottom chords To clearly demonstrate the definition and basic concept of influence lines and their applications to the analysis of structures under moving loads, let’s introduce a simple model problem as shown schematically in Figure 11.1. A given truss structure is subjected to a series of axle loads P1, P2, …, P5 resulting from two vehicles moving on the structure along what is termed the loading path. The magnitude of these moving loads remains unaltered whereas the point of application varies along the loading path. It is worth noting that the loading path of different structures can be different depending primarily on the structural configuration and their designed functions; the loading path of a truss shown in Figure 11.1 is on bottom chords whereas that of a truss shown in Figure 11.2 is on top chords. The moving load Pi is conveniently and commonly represented in terms of a unit moving load by Pi  Pi  U (x)| x  x i

(11.1)

where Pi denotes the magnitude of Pi, U(x) is the unit moving load acting to the structure at any point x along the loading path, and x = xi indicates the location of the moving load Pi at any instant. Copyright © 2011 J. Rungamornrat

480


Influence Lines

P1

P2

P3 P4

P5

Loading path

Figure 11.2: Schematic of truss structure subjected to series of axle loads along loading path on top chords Now, by focusing only on linearly elastic structures (i.e. their responses is linear with respect to input excitations or loadings), any response of interest at any particular point A of the truss shown in Figure 11.1, denoted by RA, due to the action of all moving loads {P1, P2, …, P5} at a particular instant can be obtained using superposition as 5

5

5

i 1

i 1

i 1

R A  R A (P1 , P2 , P3 , P4 , P5 )   R A (Pi )   R A (Pi  U (x)|x  x i )   Pi R A (U (x)| x  x i )

(11.2)

It is evident that to compute the response RA due to all moving loads, it is required to know only the response at the same point due to a unit moving load U(x), i.e. R A (U (x)) . More specifically, the response due to an individual moving load Pi located at any point xi (at a particular instant) is simply equal to the product of the magnitude of the moving load Pi and the response due to the unit moving load acting to the structure at x = xi, R A (U (x)|x  xi ) . The response R A (U (x)) due to the

unit moving load U(x) is termed the influence function of a particular response at point A and a graph of this influence function plotted along the entire loading path is termed the influence line. In particular, R A (U (x)|x  xi ) represents, in fact, the value of the influence function or the height of the

influence line at a point x = xi. While the above model problem involves only the moving concentrated loads, treatment of moving distributed loads follows the same procedure. To demonstrate such the idea, let q be a moving distributed load acting to the structure along the loading path as shown in see Figure 11.3. A particular response of a point A of the structure RA due to the action of the moving distributed load at a particular instant can readily be expressed in termed of the influence function using the method of superposition as follows. q Loading path

Figure 11.3: Schematic of truss structure subjected to moving distributed load q Copyright © 2011 J. Rungamornrat

481

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat l2

Influence Lines

l2

l2

R A  R A ( q )   R A ( qdx )   R A (qdx  U (x))   R A (U (x))qdx l1

l1

(11.3)

l1

where l1 and l2 are coordinates indicating the location of the moving distributed load at a particular instant. Clearly, a problem of finding a particular response at a particular point due to a set of moving loads can be reduced to an equivalent problem of finding the influence function or influence line of that particular response. Once the influence function or the influence line is constructed, a particular response of interest can readily be obtained using the relations such as (11.2) and (11.3). For convenience and brevity in following presentations, a symbol IL(RA) will be employed to represent the influence function of the response at a point A, i.e. IL(R A )  R A (U (x)) , and the term influence line will be used to mean the influence function. Once the concept of the influence lines is sufficiently motivated, we are now ready to see their vast applications. First, consider a beam subjected to three different load patterns as shown in Figure 11.4. The objective here is to compare a particular response, say the support reaction at a point A, due to those three different load patterns. Since the given beam is statically determinate, the reaction at the point A can readily be obtained from static equilibrium. However, three analyses, one for each load pattern, are required. Now, let’s attack the problem using an alternative approach based on the influence line concept. In this approach, the influence line of the support reaction at the point A, IL(RA), is first constructed as indicated in the figure below (how to construct this influence line will be clearly demonstrated in following sections). The support reaction at the point A due to the three load patterns, RA,LOAD1, RA,LOAD2 and RA,LOAD3, can then be obtained by using the relations similar to (11.2) and (11.3) as follows: x

U=1

A

B

IL(RA)

L/4 2qL

qL

1

3qL

L/4 1

L/4

L/4

1 1/2

A

IL(RA)

B

RA,LOAD1

L/4

L/4

L/4

L/4

1

q

A

1 1/2 IL(RA)

B

A RA,LOAD2

1

qL

1

q

1

1 1/2

B

RA,LOAD3

Figure 11.4: Schematic of beam subjected to three different load patterns Copyright © 2011 J. Rungamornrat

IL(RA)

482


Influence Lines

Load pattern 1: R A, LOAD1  qL  IL(R A )| x  L/4  2qL  IL(R A )| x  L/2  3qL  IL(R A )| x  3L/4  qL(1)  2qL(1)  3qL(1/2)  9qL/2

Load pattern 2: L

L

0

0

R A, LOAD2   q  IL(R A )dx  q   IL(R A )dx  q(1/2)(1)(L  L/2)  3qL/4

Load pattern 3: L

L

L/2

L/2

R A,LOAD3  qL  IL(R A )|x L/4   q  IL(R A )dx  qL(1)  q   IL(R A )dx  qL  q(1/2)(1)(L/2)  5qL/4

It should be noted that the support reaction at the point A due to other load patterns can also be obtained in the same fashion once the influence line IL(RA) is constructed. Another obvious application of the influence lines is found in the design of continuous beams under the action of dead loads (loads that are permanent or relatively constant over time, e.g. self-weight of structural and nonstructural components) and live loads (loads that are temporary or can vary with time, e.g. occupants, movable objects). One crucial step in the design procedure is to determine the maximum bending moment and shear force within the beam due to the design load. For relatively low live load in comparison with the dead load, it may be convenient (as suggested by design specifications) to determine such maximum responses by applying the live load in addition to the dead load over all spans of the beam. Nevertheless, this strategy can considerably underestimate the maximum responses when the magnitude of the live load becomes significant. The location to which the live load must be applied to produce the maximum effects is not necessary the entire beam and it can readily be identified using the influence lines. To clearly demonstrate the idea, let’s consider a three-span continuous beam shown in Figure 11.5 and let the maximum negative moment at the support A, the positive moment at the mid-span B and C, and the negative shear force at the left of the support A due to the uniformly distributed dead load DL and uniformly distributed live load LL be of interest. To identify the correct load pattern for each maximum response, we first construct the influence lines for the negative moment at the support A, IL(MA), the positive moment at the point B and C, IL(MB) and IL(MC), and the negative shear force at the left of the support A, IL(VAL), as shown in Figure 11.6. DL LL B

A

C

Figure 11.5: Three-span continuous beam subjected to dead load DL and live load LL Since the dead load is permanent, it is obligatory to place it over the entire beam. In the contrary, the live load can be chosen to be placed on a portion of the beam that provides only the positive contribution to the response of interest and this, as a result, yields its maximum value. It is evident from the influence line IL(MA) that the live load must be applied only to the first two spans of the Copyright © 2011 J. Rungamornrat

483


Influence Lines

beam in order to produce the maximum negative moment at the support A; clearly, applying the live load to the third span clearly reduces the value of the such negative moment. Similarly, from the influence lines IL(MB), IL(MC) and IL(VAL), the placement of the live load to attain the maximum positive moment at point B and C and the maximum negative shear force at left of the support A is shown in Figure 11.6. B

A

C

LL

IL(MA) LL

LL

IL(MB) LL

IL(MC) LL IL(VAL)

Figure 11.6: Influence lines IL(MA), IL(MB), IL(MC) and IL(VAL) and the placement of live load to produce the maximum MA, MB, MC and VAL The final example is associated with the determination of maximum responses of a structure (e.g. support reactions, internal forces and displacements) due to the action of a series of moving concentrated loads such as axles of vehicles, trains, cranes, etc. Let’s consider a statically determinate beam subjected to a series of three concentrated loads moving along the loading path from the left to the right of the beam as shown in Figure 11.7. The interest here is to determine the location of the moving loads that produces the maximum support reaction at the support A, the maximum positive shear at the right of the support A, and the maximum positive bending moment at the mid-span B and to compute the corresponding maximum responses. Copyright © 2011 J. Rungamornrat

484


Influence Lines

P P L

2L

2P B

A

4L

6L

5L

5L P P L

2L

2P Pattern 1

P P L

2L

2P Pattern 2

P P L

2L

2P Pattern 3

1

1.2 1.3 1.4

1.5

1.2

0.9

0.6

IL(RA) P P L

2L

2P Pattern 4

1

1

IL(VAR) P P Pattern 5

L

2L

2P

P P Pattern 6

L

2L

P P L

Pattern 7 2.4L 1.8L 1.2L 0.6L

2L

2P

2L

2P

1.6L 1.2L

IL(MB) -2.0L

Figure 11.7: Statically determinate beam under a series of moving loads, influence lines IL(RA), IL(VAR) and IL(MB) and load patterns for obtaining the maximum RA, VAR and MB Copyright © 2011 J. Rungamornrat

485


Influence Lines

To determine the maximum reaction at the support A, we first construct the influence line IL(RA) as shown in Figure 11.7. Apparently, from this influence line, there are only three possible locations of the moving loads (i.e. Pattern 1, Pattern 2 and Pattern 3 as indicated in Figure 11.7) that can produce the maximum RA. The support reaction RA for each load pattern can be computed as follows: Load pattern 1: Load pattern 2: Load pattern 3:

R A  2P(1.5)  P(1.3)  P(1.2)  5.5P R A  2P(0.9)  P(1.5)  P(1.4)  4.7P R A  2P(0.6)  P(1.2)  P(1.5)  3.9P

As a result, the load pattern 1 produces the maximum reaction at the support A and its corresponding value is 5.5P. Similarly, the maximum shear force at the right of the support A can also be obtained by using the influence line IL(VAR). For this particular case, it is clear that any position of the moving loads within a portion of the beam that IL(VAR) is constant (i.e. Pattern 4 as indicated in Figure 11.7) yields the maximum VAR = 2P(1) + P(1) + P(1) = 4P. Finally, the maximum positive bending moment at the mid-span B can be obtained from the influence line IL(MB) in the same fashion. There are three possible load patterns (i.e. Pattern 5, Pattern 6 and Pattern 7 as indicated in Figure 11.7) that can produce the maximum MB and its value for each case is given as follows: Load pattern 5: Load pattern 6: Load pattern 7:

M B  2P(2.4L)  P(1.2L)  P(0.6L)  6.6PL M B  2P(1.6L)  P(2.4L)  P(1.8L)  7.4PL M B  2P(1.2L)  P(2L)  P(2.4L)  6.8PL

Therefore, the maximum positive bending moment at the mid-span B is 7.4PL due to the load pattern 6. It is evident from above three examples that the influence lines constitutes a basis and provides useful information for determining structural responses of interest due to various types of loading such as static loads, temporary loads, and moving loads. However, it still remains to show how to construct influence lines for various structures and following sections serve that purpose.

11.2 Influence Lines for Determinate Beams by Direct Method In this section, we present a basic technique for constructing the influence lines of various responses, e.g. support reactions, internal forces, deflections and rotations, for statically determinate beams. The technique, also termed the direct method, is based primarily on the definition of the influence line along with other essential methods introduced in all previous chapters (e.g. static equilibrium for determining support reactions, method of sections for determining the bending moment and shear forces, and energy methods for computing the deflections and rotations). While the direct method is presented in this section only for statically determinate beams, the technique itself is fundamental and applies equally to both statically determinate and indeterminate structures such as trusses, beams and frames. From its definition, an influence line for any response of interest of a given beam can be constructed directly by performing the analysis of that beam under the action of a unit concentrated load moving along its loading path as shown in Figure 11.8(a). While this applied load is not stationary and its position changes continuously along the structure, a particular response of the beam when the unit load moves to the point x (i.e. the value of the corresponding influence line at the point x) is equal to the response of the same beam subjected to a stationary unit load acting to the point x provided that the dynamic effect is negligible. With such analogy, finding the influence


486


Influence Lines

line of a particular response of a given structure is equivalent to obtain that particular response of the same structure subjected to a stationary unit load acting to any point x (see Figure 11.8(b)): U=1

x

(a) P=1

x

(b) Figure 11.8: (a) beam subjected to a moving unit concentrated load and (b) the same beam subjected to a stationary unit concentrated load acting at any point x The procedure for constructing the influence line of a statically determinate beam by the direct method can be summarized as follows: (1) Identify the loading path and then choose a reference coordinate x along the loading path of a beam; (2) Place a stationary unit concentrated load at any point x along the loading path; (3) Identify the response of interest in which the influence is to be constructed; (4) Construct the influence function (or the influence line) using appropriate methods. For instance, support reactions can be determined from static equilibrium; bending moment and shear force can be obtained from the method of section and static equilibrium; and deflection and rotation can be obtained from moment area method, conjugate structure analogy, or energy methods. It should be noted that to determine the influence function, it may requires to dividing the loading location along the loading path into several cases depending on the beam configuration and response of interest. (5) Summarize the influence function for all cases and then draw the corresponding influence line To clearly demonstrate the above procedure, influence lines for support reactions, bending moments, shear forces, deflections and rotations for several statically determinate beams are constructed in following examples. Example 11.1 Construct influence functions for support reactions at A and B, bending moment and shear force at C, deflection and rotation at C due to bending effect, and then draw corresponding influence lines for a simply-supported beam shown below A

C

B EI

L/3

2L/3


487


Influence Lines

Solution The beam used to construct the influence functions is shown below along with the chosen coordinate x. For this particular beam, the loading path is defined by x [0, L]. U=1

x A

B EI

C L/3

2L/3

First, let’s construct the influence functions for the support reactions at A and B, IL(RA) and IL(RB). Since the beam is statically determinate, the support reactions can readily be obtained from equilibrium of the entire structure as follows. x

U=1

A

B IL(RB)

IL(RA) L [MA = 0]  

:

IL(RB) ×(L) – (1)(x) = 0



IL(RB) = x/L

[FY = 0]  

:

IL(RA) + IL(RB) – 1 = 0



IL(RA) = 1 – IL(RB) = 1 – x/L

Next, let’s construct the influence functions for the bending moment and shear force at the point C, IL(MC) and IL(VC). Since all support reactions were already computed, the bending moment and shear force can be obtained from the method of sections. To obtain such influence functions, two separate cases are considered: case I associated with the unit load on the left of the point C (0 ≤ x < L/3) and case II associated with the unit load on the right of the point C (L/3 < x ≤ L). x

U=1

A

B IL(RA)

C

B IL(RA)

C

IL(RB)

L/3

2L/3

2L/3

U=1

A IL(RA)

A

IL(RB)

L/3 x

U=1

x

C

IL(MC) IL(VC)

A IL(RA)

C

IL(MC) IL(VC)

L/3

L/3 Case I

Case II Copyright © 2011 J. Rungamornrat

488


Influence Lines

For case I (0 ≤ x < L/3), equilibrium of the left portion of the beam yields [MC = 0]  

:

IL(MC) + (1)(L/3 – x) – IL(RA)×(L/3) = 0

[FY = 0]  

:

IL(RA) – IL(VC) – 1 = 0





IL(MC) = 2x/3

IL(VC) = –x/L

For case II (L/3 < x ≤ L), equilibrium of the left portion of the beam yields [MC = 0]  

:

IL(MC) – IL(RA)×(L/3) = 0

[FY = 0]  

:

IL(RA) – IL(VC) = 0





IL(MC) = (L – x)/3

IL(VC) = 1 – x/L

Thus, the influence functions IL(MC) and IL(VC) for the entire loading path are given by , x  [0, L/3) 2x/3 IL(M C )   (L  x)/3 , x  (L/3, L]

 x/L IL(VC )   1  x/L

, x  [0, L/3) , x  (L/3, L]

Finally, let’s construct the influence functions for the deflection and rotation at the point C, IL(vC) and IL(C), by using principle of complementary virtual work. Similar to the previous task, following two cases are considered: case I associated with the unit load on the left of the point C (0 ≤ x < L/3) and case II associated with the unit load on the right of the point C (L/3 < x ≤ L). The bending moment diagrams for each case can readily be obtained as shown below. The virtual structures for computing the deflection and rotation at the point C and the corresponding bending moment diagrams are shown below.

x

U=1

A

A

B IL(RA)

C

L/3

U=1

x IL(RB)

B C

IL(RA) L/3

2L/3

IL(RB) 2L/3

2x/3

(L – x)/3 2x/3

(L – x)/3

M

M

x – L/3

Case I: Actual structure

L/3 – x

Case II: Actual structure


489


Influence Lines

P = 1

A C

2/3

1/3

L/3

M = 1

A

B

C

1/L L/3

2L/3

B 1/L 2L/3

2L/9 1/3

MI

MII –2/3

Virtual structure II

Virtual structure I

For case I (0 ≤ x < L/3), by applying the principle of complementary virtual work along with the actual structure for case I and the virtual structures I and II, we obtain L

vC   0

MδM I 1  L  x   L   4L  1  3x  L  L   2 x  2L  1  2x   2L   4L  dx         x           EI 2  3EI   3   27  2  3EI  3   3 L  9  2  3EI   3   27  1  5L2 x  9x 3 81EI



L

C   0



MδM II 1  L  x  L   2  1  3x  L   L   2 x  1  1  2x   2L   4  dx            x            EI 2  3EI  3   9  2  3EI   3   3 L  3  2  3EI   3   9  1  L2 x  3x 3 18EIL





For case II (L/3 < x ≤ L), by applying the principle of complementary virtual work along with the actual structure for case II and the virtual structures I and II, we obtain L

vC   0

MδM I 1  L  x   L   4L  1  L  3x  L   7 x  2L  1  2x   2L   4L    dx        x         EI 2  3EI   3   27  2  3EI  3   6 2L  9  2  3EI   3   27  

L

C   0

1  x  L  L2  18Lx  9x 2 162EI





MδM II 1  L  x  L   2  1  L  3x   L    7 x  2  1  2x   2L   4  dx           x             EI 2  3EI  3   9  2  3EI   3    6 2L  3  2  3EI   3   9  1   L  x  L2  6Lx  3x 2 18EIL





Thus, the influence functions IL(vC) and IL(C) for the entire loading path are given by Copyright © 2011 J. Rungamornrat

490


Influence Lines

 1 2 3  81EI 5L x  9x IL(vC )    1  x  L  L2  18Lx  9x 2 162EI







1  2 3  18EIL L x  3x IL(C )    1  L  x  L2  6Lx  3x 2 18EIL



, x  [0, L/3)



, x  (L/3, L]





, x  [0, L/3)



, x  (L/3, L]

Influence lines for all influence functions obtained above can be drawn as shown below. U=1

x A

B C

EI

L/3

2L/3

1 IL(RA) 1 IL(RB) 2L/9

IL(MC) 2/3

IL(VC) -1/3 4L3/243EI

IL(vC) IL(C) -4L2/162EI


491


Influence Lines

Example 11.2 Construct influence functions for support reactions at A, bending moment and shear force at B, deflection and rotation at B due to bending effect, and then draw corresponding influence lines for a cantilever beam shown below B

A

EI L/2

L/2

Solution The beam used to construct the influence functions is shown below along with the chosen coordinate x. For this particular beam, the loading path is defined by x [0, L]. U=1

x

B

A

EI L/2

L/2

First, let’s construct the influence functions for the support reactions at A, IL(RA) and IL(MA). Since the beam is statically determinate, the support reactions can be obtained from equilibrium of the entire structure as follows. x

U=1

A

IL(MA)

IL(RA) L [MA = 0]  

:

IL(MA) – (1)(x) = 0

[FY = 0]  

:

IL(RA) – 1 = 0





IL(MA) = x

IL(RA) = 1

Next, let’s construct the influence functions for the bending moment and shear force at the point B, IL(MB) and IL(VB). Such bending moment and shear force can readily be obtained from the method of sections. To obtain such influence functions, following two cases are considered: case I associated with the unit load on the left of the point B (0 ≤ x < L/2) and case II associated with the unit load on the right of the point B (L/2 < x ≤ L). For case I (0 ≤ x < L/2), equilibrium of the right portion of the beam yields [MB = 0]  

:

IL(MB) = 0

[FY = 0]  

:

IL(VB) = 0


492


Influence Lines

x

IL(MA) A

U=1 B

IL(RA)

U=1

x

IL(MA) A

B

IL(RA)

L/2

L/2

L/2

L/2 U=1

IL(MB)

IL(MB)

B IL(VB) Case I

B IL(VB) Case II

For case II (L/2 < x ≤ L), equilibrium of the right portion of the beam yields [MB = 0]  

:

[FY = 0]  

:



–IL(MB) – 1×(x – L/2) = 0 

IL(VB) – 1 = 0

IL(MB) = L/2 – x

IL(VB) = 1

Thus, the influence functions IL(MB) and IL(VB) for the entire loading path are given by , x  [0, L/2) 0 IL(M B )   L/2  x , x  (L/2, L] 0 IL(VB )   1

, x  [0, L/2) , x  (L/2, L]

Finally, let’s construct the influence functions for the deflection and rotation at the point B, IL(vB) and IL(B), by using principle of complementary virtual work. Again, following two cases are considered: case I associated with the unit load on the left of the point B (0 ≤ x < L/2) and case II associated with the unit load on the right of the point B (L/2 < x ≤ L). The bending moment diagrams for each case can readily be obtained as shown below. The virtual structures for computing the deflection and rotation at the point B and the corresponding bending moment diagrams are shown below. For case I (0 ≤ x < L/2), by applying the principle of complementary virtual work along with the actual structure for case I and the virtual structures I and II, we obtain L

vB   0

MδM I dx  EI

1 1 x  x L 3Lx 2  2x 3   x    2  EI   3 2  12EI



L

1 2 MδM II 1 x  x B   dx      x 1   2EI EI 2  EI  0




493


Influence Lines

U=1

x

U=1

x A

A

B

B L/2

L/2

L/2

L/2

M –x

M –x

Case II: Actual structure

Case I: Actual structure P = 1

M = 1

A

A

B

B L/2

L/2

L/2

L/2

1

MII

MI –L/2

Virtual structure II

Virtual structure I

For case II (L/2 < x ≤ L), by applying the principle of complementary virtual work along with the actual structure for case II and the virtual structures I and II, we obtain L

vB   0

L

B   0

1 MδM I  L  2x   L   L  1  L  L   L  6L2 x  L3 dx                EI  2EI   2   4  2  2EI  2   3  48EI



1 MδM II 1  L  L   L  2x  L  L2  4Lx dx     1      1  8EI EI 2  2EI  2   2EI  2 







Thus, the influence functions IL(vB) and IL(B) for the entire loading path are given by  1 2 3 12EI 3Lx  2x IL(v B )    1 6L2 x  L3  48EI









, x  [0, L/2) , x  (L/2, L]


494


Influence Lines

 1 2  2EI x IL(B )    1 L2  4Lx  8EI



, x  [0, L/2)



, x  (L/2, L]

Influence lines for all influence functions obtained above can be drawn as shown below. U=1

x

B

A

EI L/2

L/2

1 IL(RA) L IL(MA)

IL(MB) –L/2 1 IL(VB) 5L3/48EI L3/24EI

IL(vB)

IL(B) 2

-L /8EI -3L2/8EI


495


Influence Lines

Example 11.3 Construct the influence lines for the support reactions, i.e. IL(RA), IL(MA) and IL(RD), and then use them to obtain the influence lines of bending moments and shear forces IL(MB), IL(VB), IL(VC), IL(MDL), IL(VDL), IL(MDR) and IL(VDR) for a statically determinate beam shown below A

C

B

L/2

D

L

L/2

L

Solution The beam used to construct the influence functions is shown below along with the chosen coordinate x. The loading path for this case is defined by x [0, 3L]. U=1

x A

C

B

L/2

D

L

L/2

L

First, let’s construct the influence lines for all support reactions. This can be achieved by considering equilibrium of the beam and the condition at the hinge at a point C. Two cases are considered in the analysis: case I associated with the unit load on the left of the hinge (0 ≤ x < L) and case II associated with the unit load on the right of the hinge (L < x ≤ 3L). Results obtained for each case are shown below. x IL(MA)

U=1 C

A

D

IL(RA) L/2

IL(RD) L

L/2 C

D IL(RD)

C

D

IL(RA) L/2

FBD2

U=1

x A

Case I

L

IL(VC)

IL(MA)

FBD1

FBD1 IL(RD)

L

L/2

L

U=1 D

C IL(VC)

Case II FBD2

IL(RD)


496


Influence Lines

Case I (0 ≤ x < L):

[FBD2; MC = 0]  

:

IL(RD) ×L = 0

[FBD2;FY = 0]  

:

IL(VC) + IL(RD) = 0

[FBD1; MA = 0]  

:

IL(MA) + IL(RD) ×2L – x = 0  IL(MA) = x – IL(RD) ×2L

[FBD1;FY = 0]  

:

IL(RA) + IL(RD) – 1 = 0

[FBD2; MC = 0]  

:

IL(RD) ×L – 1×(x – L) = 0

[FBD2;FY = 0]  

:

IL(VC) + IL(RD) – 1 = 0

[FBD1; MA = 0]  

:

IL(MA) + IL(RD) ×2L – x = 0  IL(MA) = x – IL(RD) ×2L

[FBD1;FY = 0]  

:

IL(RA) + IL(RD) – 1 = 0



IL(RD) = 0 

IL(VC) = –IL(RD) 

IL(RA) = 1 – IL(RD)

Case II (L < x ≤ 3L):  



IL(RD) = x/L – 1 IL(VC) = 1 – IL(RD) IL(RA) = 1 – IL(RD)

It is evident that the influence function IL(RD) is completely obtained whereas the influence functions IL(MA), IL(RD) and IL(VC) are fully expressed in terms of the known IL(RD) for the entire loading path. Results are summarized again below. 0  IL(R D )   x  L  1

, x  [0, L) , x  (L, 3L]

IL(M A )  x  IL(R D )  (2L) , x  [0, 3L] IL(R A )  1  IL(R D ) , x  [0, 3L]   IL(R D ) IL(VC )   1  IL(R D )

, x  [0, L) , x  (L, 3L]

It is evident that the influence lines IL(MA), IL(RA) and IL(VC) can readily be constructed from the influence line for the reaction at point D. The influence lines for the bending moment and shear force at the point B can be obtained in terms of the influence line IL(RD) using the method of section. Two cases are considered as follows: case I associated with the unit load on the left of the point B (0 ≤ x < L/2) and case II associated with the unit load on the right of the point B (L/2 < x ≤ 3L). Results obtained for each case are given below.

IL(MA)

x A

U=1 C

IL(RA)

B

L/2

IL(MB)

D IL(RD) L

L/2 B IL(VB)

C

L D IL(RD)


Case I

497


Influence Lines

U=1

x C

A

IL(MA)

D

IL(RA)

IL(RD)

L/2

L

L/2

L

Case II

U=1 C

B

IL(MB)

D IL(RD)

IL(VB) Case I (0 ≤ x < L/2):

[MB = 0]  

:

[FY = 0]  

:



–IL(MB) + IL(RD) ×(3L/2) = 0 IL(VB) + IL(RD) = 0



IL(MB) = IL(RD) ×(3L/2)

IL(VB) = –IL(RD)

Case II (L/2 < x ≤ 3L):

[MB = 0]  

:

–IL(MB) – (x – L/2) + IL(RD) ×(3L/2) = 0 

[FY = 0]  

:

IL(MB) = IL(RD) ×(3L/2) + L/2 – x 

IL(VB) + IL(RD) – 1 = 0

IL(VB) = 1 – IL(RD)

Thus, the influence functions IL(MB) and IL(VB) for the entire loading path are given by , x  [0, L/2) IL(R D )  (3L/2) IL(M B )   L/2  x + IL(R D )  (3L/2) , x  (L/2, 3L]

IL(R D ) IL(VB )   1  IL(R D )

, x  [0, L/2) , x  (L/2, 3L]

The influence lines for the bending moment and shear force at the left of the roller-support D can be obtained in terms of the influence line IL(RD) in the same manner. Two cases, case I associated with the unit load on the left of the support D (0 ≤ x < 2L) and case II associated with the unit load on the right of the support D (2L < x ≤ 3L), are considered and obtained results are given below. x IL(MA)

U=1 C

A

D

IL(RA) L/2

IL(RD) L/2

L IL(MDL) IL(VDL)

L D IL(RD)


Case I

498


Influence Lines

U=1

x IL(MA)

C

A

D

IL(RA) L/2

IL(RD) L

L/2

L

Case II

U=1 D

IL(MDL)

IL(RD)

IL(VDL) Case I (0 ≤ x < 2L):

[MD = 0]  

:

[FY = 0]  

:

IL(VDL) + IL(RD) = 0

[MD = 0]  

:

–IL(MDL) – (x – 2L) = 0

[FY = 0]  

:



–IL(MDL) = 0

IL(MDL) = 0 

IL(VDL) = –IL(RD)

Case II (2L < x ≤ 3L): 

IL(MD) = 2L – x 

IL(VDL) + IL(RD) – 1 = 0

IL(VD) = 1 – IL(RD)

Thus, the influence lines IL(MDL) and IL(VDL) for the entire loading path are given by 0 IL(M DL )   2L  x

, x  [0, 2L) , x  (2L, 3L]

IL(R D ) IL(VDL )   1  IL(R D )

, x  [0, 2L) , x  (2L, 3L]

Similarly, the influence lines for the bending moment and shear force at the right of the roller-support D can be obtained in terms of the influence line IL(RD) using the method of section. Two cases, case I associated with the unit load on the left of the support D (0 ≤ x < 2L) and case II associated with the unit load on the right of the support D (2L < x ≤ 3L), are considered and obtained results are given below. x IL(MA)

U=1 C

A

D

IL(RA) L/2

IL(RD) L/2

L

L

IL(MDR)

D

IL(VDR)


Case I

499


Influence Lines

U=1

x IL(MA)

C

A

D

IL(RA)

IL(RD)

L/2

L

L/2

L

Case II

U=1 D

IL(MDR) IL(VDR) Case I (0 ≤ x < 2L):

[MD = 0]  

:

–IL(MDL) = 0

[FY = 0]  

:

IL(VDL) = 0



IL(MDL) = 0

Case II (2L < x ≤ 3L):

[MD = 0]  

:

[FY = 0]  

:



–IL(MDL) – (x – 2L) = 0 

IL(VDL) – 1 = 0

IL(MD) = 2L – x

IL(VD) = 1

Thus, the influence lines IL(MDR) and IL(VDR) for the entire loading path are given by 0 IL(M DR )   2L  x 0 IL(VDR )   1

, x  [0, 2L) , x  (2L, 3L]

, x  [0, 2L) , x  (2L, 3L]

Influence lines obtained above are shown below. U=1

x A

C

B

L/2

D

L

L/2

L 2 1

IL(RD) L

IL(MA) –L Copyright © 2011 J. Rungamornrat

500


Influence Lines

U=1

x A

C

B

L/2

D

L

L/2

L 2 1

IL(RD) 1

1

IL(RA) –1 L/2

IL(MB) –L/2 1

1

IL(VB) –1

1

IL(VC) –1

IL(MDL) –L

IL(VDL) –1

–1

IL(MDR) 1

–L 1

IL(VDR) Copyright © 2011 J. Rungamornrat

501


Influence Lines

11.3 Influence Lines by Müller-Breslau Principle It is evident from the previous section that construction of the influence lines by a direct approach requires an extensive analysis and, in addition, it does not provide any qualitative information such as their shape until the influence functions are obtained and plotted. In this section, we present an alternative and convenient method, known as Müller-Breslau principle, to construct the influence lines of support reactions, bending moments and shear forces for statically determinate beams. The principle simply states the one-to-one correspondence between the influence line and a quantity that can readily be established and interpreted. This correspondence allows the influence lines to be constructed qualitatively and quantitatively without the direct analysis for the explicit influence functions. For simplicity in the demonstration of Müller-Breslau principle, let’s consider its applications to a statically determinate beam subjected to a moving unit concentrated load as shown in Figure 11.9.

x

U=1

B

A L

L

Figure 11.9: Statically determinate beam subjected to moving unit concentrated load

11.3.1 Influence Lines of Support Reactions The key component for establishing Müller-Breslau principle is the principle of virtual work (outlined in Chapter 6) along with a special choice of the virtual displacement v. To construct the influence line of any support reaction, the virtual displacement v is simply chosen from the rigid body motion of the given beam resulting from releasing that particular support reaction. The procedure to obtain such virtual displacement can be summarized in details as follows: (i) the support reaction whose influence line is to be constructed is first removed from the given beam and this renders the beam statically unstable to the first degree of kinematical indeterminacy, (ii) the rigid body displacement is then introduced to this unstable beam by imposing a unit displacement (or rotation if the moment reaction is considered) in the direction of the released reaction, (iii) the displacement at any point within the beam is uniquely determined using the similar triangle rule (the uniqueness results directly from that the unstable beam possesses only one degree of kinematical indeterminacy and the unit displacement is imposed in the direction where the support reaction is released), and (iv) such rigid body displacement is finally chosen as the virtual displacement v. Examples of the virtual displacement resulting from releasing the support reactions RA, MA and RB are shown in Figure 11.10(b), (c) and (d), respectively. Since the beam is in equilibrium under the action of a moving unit load and all support reactions, from the principle of virtual work, it implies that the virtual work equation (6.43) is satisfied for any choice of the virtual displacement v. By choosing the virtual displacementv1, we then obtain W  IL(R A ) 1  1 δv1  IL(R A )  δv1 ; U  0

W  U  IL(R A )  δv1 Copyright © 2011 J. Rungamornrat

502


Influence Lines

U=1

x IL(MA)

B

A IL(RA)

IL(RB) L

L (a)

1

v1(x)

B

A (b)

v2(x)

1

B

A (c)

v3(x)

1

A (d) Figure 11.10: (a) Free body diagram of a beam shown in Figure 11.9, (b) rigid body displacement resulting from removing the reaction RA and then introducing a unit displacement at point A in the direction of RA, (c) rigid body displacement resulting from removing the reaction MA and then introducing a unit rotation at point A in the direction of MA, and (d) rigid body displacement resulting from removing the reaction RB and then introducing a unit displacement at point B in the direction of RB Note that the virtual strain energy vanishes since the rigid body displacement produces no deformation. Similarly, by choosing the virtual displacementv2, it leads to W  IL(M A ) 1  1 δv 2  IL(M A )  δv 2 ; U  0 W  U  IL(M A )  δv 2

Finally, by choosing the virtual displacementv3, it results in W  IL(R B ) 1  1 δv3  IL(R B )  δv3 ; U  0 W  U  IL(R B )  δv3 Copyright © 2011 J. Rungamornrat

503


Influence Lines

It is evident from above results that the influence line of any support reaction possesses an identical shape and magnitude to the rigid body displacement resulting from removing that particular support reaction from the beam and then imposing a unit displacement (or unit rotation if the moment reaction is to be determined) at the location and in the direction of the released reaction. This is commonly known as Müller-Breslau principle for support reactions. As a result of this principle, the influence lines for all support reactions of the beam shown in Figure 10.9, i.e. IL(RA), IL(MA) and IL(RB), can be summarized in Figure 11.11. U=1

x

B

A L 1

L 1 IL(RA) L IL(MA) 1 IL(RB)

Figure 11.11: Influence lines for all support reactions of beam shown in Figure 10.9

11.3.2 Influence Lines of Bending Moments To construct the influence line of the bending moment at any cross section of a beam, the virtual displacement v must be chosen from the rigid body motion of the given beam resulting from the release of the moment constraint at that particular cross section. The procedure to obtain such virtual displacement can be summarized as follows: (i) the moment constraint at the cross section of the beam where the influence line is to be constructed is released (by adding the moment release or hinge) and the beam now becomes statically unstable and possesses one degree of kinematical indeterminacy, (ii) the rigid body displacement is then introduced by imposing a unit relative rotation at the hinge following the direction of the released moment, (iii) the displacement at any point within the beam is uniquely determined using the similar triangle rule (the uniqueness results directly from that the unstable beam possesses only one degree of kinematical indeterminacy and the unit relative rotation is imposed at the location and in the direction where the bending moment is released), and (iv) such rigid body displacement is finally chosen as the virtual displacement v. Copyright © 2011 J. Rungamornrat

504


Influence Lines

Examples of the virtual displacement resulting from releasing the bending moment at points C and D are shown in Figures 11.12(b) and 11.13(b). Since the beam is in equilibrium under the action of a moving unit load, from the principle of virtual work, it implies that the virtual work equation (6.43) is satisfied for any choice of the virtual displacement v. By choosing the virtual displacementv1 shown in Figure 11.12(b), we then obtain W  IL(M C )  1  IL(M C )  2  1 δv1  IL(M C )  (1  2 )  δv1  IL(M C )  δv1 ; U  0 W  U  IL(M C )  δv1

Similarly, by choosing the virtual displacementv2 shown in Figure 11.13(b), we then obtain W  IL(M D )  1  1 δv 2  IL(M D )  δv 2 ; U  0 W  U  IL(M D )  δv 2

U=1

x

B

A IL(MC) L

IL(VC) L

(a)

v1(x)

1 1 A (b)

2

B

C

Figure 11.12: (a) Beam shown in Figure 11.9 with the sectioning at point C, (b) rigid body displacement resulting from removing the bending moment MC and then introducing a unit relative rotation at point C in the direction of MC U=1

x

B

A IL(MD)

IL(VD) L

L (a) B

A

D

1

v2(x) (b)

Figure 11.13: (a) Beam shown in Figure 11.9 with the sectioning at point D, (b) rigid body displacement resulting from removing the bending moment MD and then introducing a unit relative rotation at point D in the direction of MD Copyright © 2011 J. Rungamornrat

505


Influence Lines

It is evident from above results that the influence line of bending moment at any cross section possesses an identical shape and magnitude to the rigid body displacement resulting from removing that particular bending moment from the beam and then imposing a unit relative rotation at the location and in the direction of the released bending moment. This is also known as Müller-Breslau principle for bending moment. The influence lines for bending moments at points C and D of the beam shown in Figure 10.9, i.e. IL(MC) and IL(MD), are summarized in Figure 11.14. The value of h1 and h2 can readily be computed from the geometry as follows: h1 h1 ab   1  h1  a b ab

;

h2  1  h2  c c

U=1

x D

C

B

A L

L 1 h1 a

b

IL(MC)

c

IL(MD)

1 –h2

Figure 11.14: Influence lines for bending moments at points C and D of beam shown in Figure 10.9

11.3.3 Influence Lines of Shear Forces To construct the influence line of the shear force at any cross section of a beam, the virtual displacement v must be chosen from the rigid body motion of the given beam resulting from the release of the shear constraint at that particular cross section. The procedure to obtain such special virtual displacement can be summarized as follows: (i) the shear constraint at the cross section of the beam where the influence line is to be constructed is released (by introducing the shear release) and the beam now becomes statically unstable and possesses one degree of kinematical indeterminacy, (ii) the rigid body displacement is then introduced by imposing a unit relative displacement at the shear release following the direction of the released shear force, (iii) the displacement at any point within the beam is uniquely determined using the similar triangle rule (the uniqueness results directly from that the unstable beam possesses only one degree of kinematical indeterminacy and the unit relative displacement is imposed at the location and in the direction where the shear force is released), and (iv) such rigid body displacement is finally chosen as the virtual displacement v. Examples of the virtual displacement resulting from releasing the shear force at points C and D are shown in Figures 11.15(b) and 11.16(b). It should be noted that at Copyright © 2011 J. Rungamornrat

506


Influence Lines

the shear release, the moment constraint still exists (except at the hinge) and, as a result, the two segments connecting to the shear release (e.g. segments EC and CB in Figure 11.15(b) and segments AD and DE in Figure 11.16(b)) must maintain their slope or, equivalently, be parallel. If the shear release is introduced at the pre-existing hinge (e.g. point E), the two segments connecting to that shear release need not to maintain their slope (since there is no moment constraint a priori). For instance, the segments AE and EB due to the shear release at the point E in Figure 11.17(b) are not necessarily parallel. Since the beam is in equilibrium under the action of a moving unit load, from the principle of virtual work, the virtual work equation (6.43) is satisfied for any choice of the virtual displacement v. By choosing the virtual displacementv1 shown in Figure 11.15(b), we then obtain W  IL(VC )  h1  IL(VC )  h 2  1 δv1  IL(VD )  (h1  h 2 )  δv1  IL(VC )  δv1 ; U  0 W  U  IL(VC )  δv1

Similarly, by choosing the virtual displacementv2 shown in Figure 11.16(b), we then obtain W  IL(VD ) 1  1 δv 2  IL(VD )  δv 2 ; U  0 W  U  IL(VD )  δv 2

U=1

x E

A

B

IL(MC) L

IL(VC) L

(a) C h2

1

E A

v1(x)

B

h1

(b)

C

Figure 11.15: (a) Beam shown in Figure 11.9 with the sectioning at point C, (b) rigid body displacement resulting from removing the shear force VC and then introducing a unit relative displacement at point C in the direction of VC It is evident from above results that the influence line of shear force at any cross section possesses an identical shape and magnitude to the rigid body displacement resulting from removing that particular shear force from the beam and then imposing a unit relative displacement at the location and in the direction of the released shear force. This is also known as Müller-Breslau principle for shear force. The influence lines for shear forces at points C, D and E, i.e. IL(VC), IL(VD) and


507


Influence Lines

IL(VE), are shown in Figure 11.18. The height h1 and h2 of the influence line IL(VC) can readily be computed in terms of a and b as follows: h1 h 2  a b

and

h1  h 2  1  h1 

a b , h2  ab ab

U=1

x

B

A IL(MD)

IL(VD) L

E

D 1

A

L

(a)

v2(x)

B

D (b)

Figure 11.16: (a) Beam shown in Figure 11.9 with the sectioning at point D, (b) rigid body displacement resulting from removing the shear force VD and then introducing a unit relative displacement at point D in the direction of VD U=1

x A

B

E IL(VE) L

L

(a) E 1

A

v2(x)

B

E (b)

Figure 11.17: (a) Beam shown in Figure 11.9 with the sectioning at the hinge E, (b) rigid body displacement resulting from removing the shear force VE and then introducing a unit relative displacement at point E in the direction of VE From above results, it can be concluded in general that the influence line of either the support reaction or the bending moment and shear force at any cross section of the statically determinate beam possesses an identical shape and magnitude to the rigid body displacement resulting from removing the constraint associated with that quantity and then imposing a unit movement (e.g. unit Copyright © 2011 J. Rungamornrat

508


Influence Lines

displacement for force reactions, unit rotation for moment reactions, unit relative rotation for bending moments, unit relative displacement for shear forces) at the location and in the direction of the released constraint. Since the rigid body displacement involves only rigid translation and rigid rotation, the influence lines of support reactions, bending moments and shear forces for statically determinate beams are always piecewise linear. From Müller-Breslau principle, the statement of finding the influence line for any static quantities (i.e. support reactions, bending moments and shear forces at certain cross section) of a statically determinate beam simply reduces to finding the rigid body displacement of an unstable beam resulting from the release of a proper constraint. In the sketch of rigid body displacement, it is worth noting that the type and location of release must be chosen correctly; each rigid segment can undergo only the rigid translation and the rigid rotation without kinking; and conditions at all supports must be maintained except the one being released. U=1

x E

D

C

B

A L

L

1

h2

IL(VC)

–h1 a 1

b

1

IL(VD) 1

IL(VE) Figure 11.18: Influence lines for shear forces at points C, D and E of beam shown in Figure 10.9 Example 11.4 Use Muller-Breslau to construct the influence lines of all support reactions IL(RA), IL(RD) and IL(RF), shear forces IL(VB), IL(VC), IL(VDL), IL(VDR) and IL(VE), and bending moments IL(MB), IL(MD) and IL(ME) of a statically determinate beam shown below.

A

C

B

L/4

L/4

E

D

L/2

L/2


F

L/2

509


Influence Lines

Solution The influence line of the support reaction IL(RD) is obtained as follows: 1) release the roller support at the point D, 2) introduce the rigid body displacement, 3) impose the unit upward displacement at the point D (the same direction as the support reaction RD), and 4) the resulting rigid displacement is the influence line IL(RD) as shown below.

Released RD 1

A

B L/4

D

C L/4

h2

L/2

h1

F

E L/2

1

L/2

h3

IL(RD)

Values of the influence line at other points can be readily determined from the geometry and property of similar triangles, for instance, h1  1 

3L/2 3 L/4 3 L/2 1 ; h 2  h1  ; h 3  1    L 2 L/2 4 L 2

The influence line of the shear force at the point E, IL(VE), is obtained as follows: 1) release the shear constraint at the point E, 2) introduce the rigid body displacement, 3) impose the unit relative displacement at the point E (in the same direction as the shear force VE), and 4) the resulting rigid displacement is the influence line IL(VE) as shown below. Released VE D

A

B

E

F

C 1

L/4

L/4

h4

L/2

L/2

h3

L/2

h1

IL(VE) –h2


510


Influence Lines

Values of the influence line at the point E and other points can be readily determined from the geometry and property of similar triangles as follows: h1 h  2 L/2 L/2

h1  h 2  1  h1  h 2 

and

1 L/2 1 L/4 1 ; h3  h 2  ; h 4  h3    2 L/2 2 L/2 4

The influence line of the bending moment at the point E, IL(ME), is obtained as follows: 1) release the moment constraint at the point E, 2) introduce the rigid body displacement, 3) impose the unit relative rotation at the point E (in the same direction as the bending moment ME), and 4) the resulting rigid displacement is the influence line IL(ME) as shown below.

1

Released ME D

A

B

L/4

E

F

C

L/4

L/2

L/2

L/2

h1

IL(VE) –h3

–h2

Values of the influence line at the point E and other points can be readily determined from the geometry and property of similar triangles as follows: h1 h L  1  1  h1  L/2 L/2 4

; h 2  h1 

L/2 L L/4 L ; h3  h 2    L/2 4 L/2 8

Other influence lines can be constructed in the same manner and results are given below A

C

B

L/4

L/4

E

D

L/2

L/2

F

L/2

1 1/2

IL(RA) Copyright © 2011 J. Rungamornrat

511


A

C

B

L/4

Influence Lines

E

D

L/4

L/2

L/2

F

L/2 1 1/2

IL(RF) –1/4

–1/2

1/2

IL(VB) –1/2

IL(VC) –1/2 –1

IL(VDL) –1/2 –1

–1 1

1/2

1/2

1/4

IL(VDR)

L/8

IL(MB)

IL(MD) –L/4 –L/2


512


Influence Lines

Example 11.5 Use Muller-Breslau to construct the influence lines of all support reactions IL(RA), IL(MA) and IL(RE), shear forces IL(VB), IL(VC), IL(VD), IL(VEL) and IL(VER), and bending moments IL(MB), IL(MD) and IL(ME) of a statically determinate beam shown below.

A

C

B

L/4

E

D

L/4

L/2

L/2

F

L/2

Solution By applying the Muller-Breslau principle, we obtain the influence lines for any quantities as follows: 1) release the constraint associated with the quantity whose influence line is to be constructed, 2) introduce a rigid body displacement, 3) impose unit movement in the direction of the released constraint and 4) the resulting rigid body displacement is the influence line to be determined. It is noted that values of the influence line can be readily and completely obtained from the geometry and properties of similar triangles. Results are given below.

A

C

B

L/4 1

L/4 1

E

D

L/2

L/2

F

L/2

1 1/2

IL(RA) –1/2 L/2 L/4

L/4

IL(MA) –L/4 3/2 1 1/2

IL(RE)

1

1 1/2

IL(VB) –1/2


513


A

C

B

L/4

Influence Lines

E

D

L/4

L/2

L/2

F

L/2

1 1/2

IL(VC) –1/2 1/2

IL(VD) –1/2

–1/2

IL(VEL) –1/2

–1/2 –1 1

1

IL(VER) L/8

IL(MB) –L/8 –L/4 L/4

IL(MD) –L/4

IL(ME) –L/2


514


Influence Lines

Example 11.6 Use Muller-Breslau to construct the influence lines of all support reactions IL(RA), IL(MA), IL(RD) and IL(RF), shear forces IL(VB), IL(VC), IL(VDL), IL(VDR) and IL(VE), and bending moments IL(MB) and IL(MD) of a statically determinate beam shown below.

A

C

B

L/4

E

D

L/4

L/2

L/2

F

L/2

Solution By applying the Muller-Breslau principle, we obtain the influence lines for any quantities as follows: 1) release the constraint associated with the quantity whose influence line is to be constructed, 2) introduce a rigid body displacement, 3) impose unit movement in the direction of the released constraint and 4) the resulting rigid body displacement is the influence line to be determined. It is noted that values of the influence line can be readily and completely obtained from the geometry and properties of similar triangles. Results are given below.

A

C

B

L/4 1

L/4 1

E

D

L/2

L/2

F

L/2

1

IL(RA) –1

L/2 L/4

IL(MA) –L/2 2 1

IL(RD) 1

IL(RF)


515


A

C

B

L/4

Influence Lines

L/4 1

E

D

L/2

L/2

F

L/2

1

IL(VB) –1

1

IL(VC) –1

IL(VDL) –1

–1 1

1

IL(VDR) 1

IL(VE)

L/4

IL(MB) –L/4

IL(MD) –L/2


516


Influence Lines

11.4 Influence Lines for Beams with Loading Panels In this section, we demonstrate how to construct the influence lines for any response of a beam containing a loading panel along its loading path as shown in Figure 11.19. A loading panel is a load transferring system where any load acting on it transfers to a beam through its panel points. Here, we focus on a simple loading panel containing only two panel points and possessing simple load transferring mechanism, for instance, a loading panel BC shown in Figure 11.19 containing two panel points at B and C. x

U=1

A

D

E

C

B

Figure 11.19: Beam containing a loading panel BC The load transferring mechanism of a loading panel due to a moving unit concentrated load can readily be obtained from the simply-supported beam action. From equilibrium of the loading panel BC shown in Figure 11.20, the support reactions RB and RC transferred to the beam at the panel points B and C are given by RB  1

ξ h

;

RC 

ξ h

(11.4)

It is evident from (11.4) that a moving unit concentrated load acting at the panel points is transferred to the beam as if it acts directly to the beam at the panel points, i.e. for  = 0, RB = 1 and RC = 0 and for  = h, RB = 0 and RC = 1. This implies that any response of the beam with a loading panel (e.g. support reactions, bending moments, shear forces, displacements, rotations, etc.) due to a unit concentrated load acting to the panel point is identical to that of the beam without the loading panel due to the same unit load acting directly to the beam at the panel points. U=1

A

D

E

D

E

C

B  U = 1

RB

RC

A C

B h

Figure 11.20: Load transferring mechanism due to moving unit concentrated load traveling along loading panel Copyright © 2011 J. Rungamornrat


517

Influence Lines

A moving unit concentrated load acting within the loading panel (between two panel points 0 < < h) is transferred to the beam at two panel points with their magnitude varying linearly with the location of the moving unit concentrated load. This implies that any response of the beam due to the action of the unit concentrated load acting within the loading panel is equal to the response of the beam without a loading panel under the action of two concentrated loads RB and RC acting at the panel points B and C. For instance, any response f of the beam due to a moving unit concentrated load acting within the panel BC (0 <  < h) or, equivalently, the value of the influence line at , denoted by IL(f), can be calculated from  ξ ξ IL(f )  R B  IL(f ) B  R C  IL(f )C  1   IL(f ) B    IL(f )C  h h

(11.5)

where IL(f)B and IL(f)C are values at points B and C of the influence line of the response f. It is evident from (11.5) that once values of the influence line at two panel points, i.e. IL(f)B and IL(f)C, are known, value of the influence line at any point within the loading panel BC is also known and, clearly, the influence line over a loading panel is simply a straight line connecting those two known points. As a result, to construct a portion of the influence line over the loading panel, it only requires to finding values of the influence line at the two panel points. To find IL(f)B and IL(f)C, we simply use the fact that there is no difference between the response of a beam containing a loading panel and that of the same beam with the loading panel being removed if the moving unit load is applied directly to the panel point (see Figure 11.21). More specifically, IL(f)B and IL(f)C are identical to values of the influence line at points B and C of the beam with the loading panel being removed. From this strategy, it is obvious that construction of the influence line of any response of a beam containing a loading panel requires only the information of the influence line of the same response of the beam with the loading panel being removed. U=1 A B

D

E

D

E

D

E

D

E

C

U=1 A B

C U=1

A B

C U=1

A B

C

Figure 11.20: Schematic indicating the similarity of two beams, one containing loading panel subjected to a unit load at the panel point and the other with loading panel being removed and subjected to a unit load at the same point Copyright © 2011 J. Rungamornrat


518

Influence Lines

Procedures for constructing the influence line of any response f of a beam containing loading panels, denoted by IL(f), can be summarized as follows: (i) the influence line of the same response f of a beam with the loading panel being fictitiously removed, denoted by IL(f *), is first constructed by using either the direct method or Müller-Breslau principle as shown in Figure 11.21; (ii) the influence line IL(f) for any portion of the beam except those containing the loading panel (e.g. portion AB and portion CDE shown in Figure 11.22) is taken from IL(f *) since they are identical; (iii) values of the influence line IL(f) at all panel points are taken from values of IL(f *) at the same point (e.g. IL(f) at the panel points B and C are identical to IL(f *) at points B and C, respectively); and (iv) the influence line IL(f) for any loading panel can be obtained by simply connecting the two panel points by a straight line. It is important to remark that the above strategy applies equally to both statically determinate and statically indeterminate beams and beams containing more than one loading panels. Following two examples are considered to clearly demonstrate the procedures.

x

U=1

A

D B

hB

E

C

hC

IL(f *)

Figure 11.21: Influence line of any response f of a beam shown in Figure 11.19 with loading panel BC being removed

x

U=1

A

D B

hB

E

C

hC

IL(f *)

Figure 11.22: Influence line of any response f of a beam shown in Figure 11.19 Copyright © 2011 J. Rungamornrat


519

Influence Lines

Example 11.7 Construct the influence lines of all support reactions IL(RA), IL(MA) and IL(RG), shear forces IL(VB), IL(VC) and IL(VE), and bending moments IL(MB) and IL(ME) of a statically determinate beam containing a loading panel DF shown below. A

C

B

G D

L/2

L/2

L/2

E

F L/2

L/2

L/2

Solution To construct the influence lines of any response for this beam, we first introduce a fictitious structure, called the structure I, by removing the loading panel DF from the original structure as shown below. Structure I

A

C

B

G D

L/2

L/2

L/2

F

E L/2

L/2

L/2

The influence line of the support reaction for the original structure, IL(MA), can be constructed as follows. The influence line of the support reaction IL(MA*) for the structure I is constructed first by using Müller-Breslau principle and result is given in the figure below. Values of the influence line IL(MA*) at the panel points D and F are 3L/4 and L/4, respectively. Structure I

A

C

B

G D

L/2

L/2

L/2

F

E L/2

L/2

L/2

L 3L/4 L/2

L/2

L/4 IL(MA*)

The influence line of the support reaction IL(MA) for the original structure is identical to the influence line IL(MA*) for the portion ABCD and the portion FG. The influence line IL(MA) for the portion DF (the loading panel) is obtained by connecting two known points on the influence line, i.e. D and F, by a straight line. The final influence line IL(MA) is shown below Original Structure

A

C

B

G D

L/2

L/2

L/2

F

E L/2

L/2

L/2

L 3L/4 L/2

L/2

L/4 IL(MA)



520

Influence Lines

The influence line of the shear force at the point E for the original structure, IL(VE), can be constructed as follows. The influence line of the shear force at the point E for the structure I, IL(VE*), is constructed by using Müller-Breslau principle and result is given in the figure below. Values of the influence line IL(VE*) at the panel points D and F are –1/4 and 1/4, respectively. Structure I

A

C

B

G D

L/2

L/2

L/2

F

E L/2

L/2

L/2

1/2 1/4 IL(VE*) –1/4 –1/2

The influence line of the shear force at the point E for the original structure, IL(VE), is identical to the influence line IL(VE*) for the portion ABCD and the portion FG. The influence line IL(VE) for the loading panel DF is obtained by connecting two known points on the influence line, i.e. D and F, by a straight line. The final influence line IL(VE) is shown below Original Structure

A

C

B

G D

L/2

L/2

L/2

F

E L/2

L/2

L/2

1/2 1/4 IL(VE) –1/4 –1/2

The influence line of the bending moment at the point E for the original structure, IL(ME), can be constructed as follows. The influence line of the bending moment at the point E for the structure I, IL(ME*), is constructed by using Müller-Breslau principle and result is given in the figure below. Values of the influence line IL(ME*) at the panel points D and F are L/4 and L/4, respectively. Structure I

A

C

B

G D

L/2

L/2

L/2

F

E L/2

L/2

L/2

L/2 L/4

L/4 IL(ME*)



521

Influence Lines

The influence line of the bending moment at the point E for the original structure, IL(ME), is identical to the influence line IL(ME*) for the portion ABCD and the portion FG. The influence line IL(ME) for the loading panel DF is obtained by connecting two known points on the influence line, i.e. D and F, by a straight line. The final influence line IL(ME) is shown below

Original Structure

A

C

B

G D

L/2

L/2

E

L/2

F L/2

L/2

L/2

L/2 L/4

L/4 IL(ME)

The influence lines of other responses can be constructed in the same fashion and results are indicated below.

Original Structure

A

C

B

G D

L/2

L/2

E

L/2

1

L/2

1

F L/2

L/2

3/4 1/4 IL(RA) 1

3/4 1/4

IL(RG) 1

1

3/4 1/4 IL(VB)

1

3/4 1/4 IL(VC) 3/4

1/4 IL(MB) –L/8

–L/2

–3L/8



522

Influence Lines

Example 11.8 Construct the influence lines of the shear force and bending moment at point C, IL(VC) and IL(MC), for a statically determinate beam containing two loading panels BD and CE shown below. A

F B L/2

D

C L/2

E L/4

L/2

L/4

Solution To construct the influence lines for the given beam, we first introduce two fictitious structure, called the structure I and structure II. The structure I is obtained by removing the loading panel CE from the original structure whereas the structure II results from removing both loading panels from the original structure as shown below.

Structure I

A

F B L/2

Structure II

D

C L/2

E L/4

L/2

L/4

A

F B L/2

D

C L/2

E L/4

L/2

L/4

First, the influence lines of the shear force and bending moment at the point C are constructed using Müller-Breslau principle and results are given below. Values of both influence lines at the panel points B and D are also indicated in the figure. Structure II

A

F B L/2

D

C L/2

L/2

E L/4

L/4

1/2 1/4 IL(VC) –1/4 –1/2 L/2 L/4

L/4

IL(MC)



523

Influence Lines

The influence lines of the shear force and bending moment at point C of the Structure I can now be obtained as follows: (i) the influence lines over the portions AB and DF of the structure I (including the panel points B and D) are identical to those of the structure II and (ii) the influence lines along the loading panel BD of the structure I can be obtained by connecting two known points on the influence lines (i.e. points B and D) by a straight line. Results are shown below.

Structure I

A

F B L/2

D

C L/2

E L/4

L/2

L/4

1/2 1/4 1/8 IL(VC) –1/4 –1/2 L/2 L/4

L/4

L/4

L/8 IL(MC)

Finally, the influence lines of the shear force and bending moment at point C of the original beam can be obtained in a similar fashion to those for the Structure I: (i) portions of the influence lines along the loading path AB, BC and EF of the original structure are identical to those for the structure I and (ii) a portion of the influence lines along the loading panel CE of the original structure can be obtained by connecting two known points on the influence lines (i.e. points C and E) by a straight line. The final influence lines are shown schematically in the figure below. Original Structure

A

F B L/2

D

C L/2

L/2

E L/4

1/4

L/4

1/8 IL(VC)

–1/4

L/4

L/4

L/4

L/8 IL(MC)



524

Influence Lines

11.5 Influence Lines for Determinate Floor System In this section, we generalize results established in the previous section to construct the influence lines of any response of a girder in a statically determinate floor system. A real floor system consists of several levels of structural components and possesses very complex load transferring mechanism. For instance, a floor system shown in Figure 11.23 consists of four basic components including the pavement surface, stringers, floor beams and girders. The load transferring mechanism for this particular floor system can be described as follows: (i) applied loads acting on the pavement are transferred directly to stringers; (ii) loads distributed to the stringers are transferred to the floor beams; and (iii) reactions from the floor beams are finally transferred to the girders. It is apparent that applied loads acting on the pavement surface are not transferred directly to the girder but through a complex sequence of load transferring mechanism.

Stringers Pavement surface

Girder

Floor beam

Figure 11.23: Example of a floor system consisting of several levels of structural components

Loading panel

Stringers Floor beams

Girder Panel points Figure 11.24: Idealized girder in the floor system Copyright © 2011 J. Rungamornrat

Pavement surface

525


Influence Lines

To simplify the load transferring mechanism, a girder in the above floor system can be idealized as shown in Figure 11.24. By assuming that the pavement surface has no load resistance, all applied loads acting on the pavement surface are therefore transferred directly to the stringers; each stringer is termed a loading panel. Loads from each stringer are transferred to the girder through the floor beams by the simply-supported beam action; each floor beam is termed the panel point. It is evident from this idealized girder that the load transferring mechanism for each loading panel (see Figure 11.25) is exactly the same as that described in the previous section. More specifically, the reactions at the two panel points due to the moving unit concentrated load vary linearly with its position along the loading panel, i.e. RD  1

ξ h

;

RE 

ξ h

(11.6)

where h is the width of the loading panel and  is the distance of the moving unit load from the panel point D. It is clear from (11.6) that RD = 1, RE = 0 for  = 0 and RD = 0, RE = 1 for  = h. This implies that a girder with loading panels and a girder with all loading panels being removed have exactly the same behavior if the moving unit concentrated load act to both girders at the panel point (see Figure 11.26 for the equivalence between the two girders when the moving unit load is at the panel points). Due to such similar load transferring mechanism, the influence lines of any response for a girder in the floor system can be constructed using the same strategy as that demonstrated in the previous section. The only difference is that a girder considered in this section may contain several loading panels over its entire length. U=1

A

B

C Girder

D

E

F

G

H

 U = 1

RD Girder

RE E

D

h Figure 11.25: Load transferring mechanism for each loading panel To describe such procedure again, let’s consider a statically determinate floor system subjected to a moving unit concentrated load along the loading path as shown in Figure 11.27. For this particular structure, the loading path AH consists of seven loading panels denoted by AB, BC, CD, DE, EF, FG and GH with eight panel points denoted by A, B, C, D, E, F, G, and H. Copyright © 2011 J. Rungamornrat

526


Influence Lines

U=1

B

A

C Girder

D

E

F

G

H

E

F

G

H

F

G

H

F

G

H

U=1 B

A

D

C Girder

U=1

B

A

C Girder

D

E

U=1 B

A

D

C Girder

E

Figure 11.26: Schematic indicating the similarity between two girders, one containing loading panels and subjected to a moving unit concentrated load at the panel points D and E and the other with all loading panels being removed and subjected to a moving unit concentrated load at the same points D and E U=1

x

G A

B

C

D

E

F

H

Figure 11.27: Statically determinate floor system consisting of seven loading panels and eight panel points Copyright © 2011 J. Rungamornrat

527


Influence Lines

Let IL(f) denote the influence line of a particular response of the girder due to the unit concentrated load traveling along the girder through the loading panels. By employing results established in the section 11.4, the influence line IL(f) can readily be constructed as follows. First, an intermediate structure resulting from removing all loading panels along the loading path of the girder is introduced as shown in Figure 11.28. Next, the influence line of the same response for this intermediate structure, denoted by IL(f *), is constructed using either a direct approach or the Müller-Breslau principle with the result given in Figure 11.28. Values of the influence line IL(f *) at all panel points (i.e. points A, B, C, D, E, F, G and H) are determined and marked. By using the similarity between the two structures (i.e. structures with and without loading panels) when the moving unit load is at the panel points, the influence line IL(f) is therefore identical to IL(f *) at all panel points A, B, C, D, E, F, G and H and the influence line IL(f) within each loading panel can readily be obtained by connecting points of the influence line at its panel points by straight lines. The complete influence line IL(f) is shown in Figure 11.29. U=1

x

G A

B

D

C

E

H

F

IL(f *)

Figure 11.28: Influence line IL(f *) of a girder with all loading panel being removed and values at all panel points being marked U=1

x

G A

B

C

D

E

F

H

IL(f )

Figure 11.29: Influence line IL(f ) of a floor system; dash line represents the influence line IL(f *) Copyright © 2011 J. Rungamornrat

528


Influence Lines

It is important to remark that the construction of a complete influence line for a floor system simply reduces to determination of a set of points on the influence line associated with all panel points of the loading panels; the process of connecting those points by a straight line is trivial. The key task of finding such a set of points can be readily achieved by considering the intermediate structure resulting from removing all loading panels. Example 11.9 Construct the influence lines of all support reactions IL(RA) and IL(RG), the shear force within the panels DE, FG and GH, IL(VDE), IL(VFG), and IL(VGH), and the bending moment at points D, P and Q, IL(MD), IL(MP) and IL(MQ) of the floor system shown below. P and Q are the mid-points of the loading panels DE and GH, respectively.

A

B h

D

C h

h

P

E

G Q

F

h

h

h

I

H

h

h

Solution An intermediate structure of the above floor system is obtained by removing all loading panels as shown below.

A

B h

D

C h

h

P

E

G Q

F

h

h

h

I

H

h

h

The influence line of the support reaction IL(RA) is obtained as follows: (1) the influence line of the support reaction of the intermediate structure IL(RA*) is obtained by Müller-Breslau principle; (2) values of IL(RA*) at all panel points are determined and marked; and (3) the complete influence line IL(RA) is obtained by connecting points obtained in the step (2) by straight lines.

1

G

A

B h 1

D

C h

5/6

h 4/6

P

E

H I

F

h

3/6

Q

h

2/6

h

h

h

1/6

IL(RA*) –1/6


–2/6

529


U=1

x A

B

D

C

h

1

Influence Lines

h

P

h

5/6

E

h

4/6

G Q

F h

3/6

h

2/6

I

H

h

h

1/6

IL(RA) –1/6

–2/6

Since the moving unit concentrated load is transferred to the girder only at the panel points, the shear forces at any cross section within the loading panel are the same. As a result, the influence lines for the shear force at any points within the loading panel are identical. The influence line of the shear force within the panel DE, IL(VDE), is then obtained as follows: (1) the influence line of the shear force at any point  within the loading panel DE of the intermediate structure IL(V*) is obtained by Müller-Breslau principle; (2) values of IL(V*) at all panel points are determined and marked; and (3) the complete influence line IL(VDE) is obtained by connecting points obtained in the step (2) by straight lines. B

A

D

C

h

G

P

h

1

h

E

–1/6

h

h

2/6

h

h

1/6

IL(VP*)

–3/6

–2/6

–7/12

U=1

B h

I

–1/6

–2/6

x A

H

F

h 5/12

Q

D

C h

h

P

E

G Q

F

h

h 2/6

h

h

I

H h

1/6

IL(VDE) –1/6

–2/6

–1/6 –3/6 Copyright © 2011 J. Rungamornrat

–2/6

530


Influence Lines

It is remarked that the point  can be chosen arbitrary within the panel DE. The result shown above corresponds to a particular  at the mid-point of the panel DE. If a point just to the right of the point D or a point just to the left of the point E is chosen, we obtain the same final influence line IL(VDE) as shown below. U=1

x A

B h

D

C h

h

P

E

h

h

h

2/6

–1/6

G Q

F

h

h

1/6

IL(VDE)

IL(VE*)

–2/6

I

H

–1/6

–2/6

–3/6

IL(VD*) 2/6

1/6

IL(VDE) –1/6

–2/6

–1/6

–2/6

–3/6

The influence line of the bending moment at the point P, IL(MP), is obtained as follows: (1) the influence line of the bending moment at the point P of the intermediate structure IL(MP*) is obtained by Müller-Breslau principle; (2) values of IL(MP*) at all panel points are determined and marked; and (3) the complete influence line IL(MP) is obtained by connecting points obtained in the step (2) by straight lines. 1

G

A

h

5h/12

D

C

B

h

h

5h/6

5h/4

P

E

H I

F

h 35h/24

Q

h

7h/6

h

h

h

7h/12

IL(MP*) –7h/12


–7h/6

531


U=1

x A

B h

D

C h

h

E

P

G Q

F

h

h

35h/24

5h/4

5h/6

5h/12

Influence Lines

h

7h/6

I

H

h

h

7h/12

IL(MP) –7h/12

–7h/6

Other influence lines can also be constructed in the same fashion. First, the influence lines for the intermediate structure are obtained and shown with their values at all panel points marked below. U=1

x A

B h

h

1/6

D

C

3/6

G Q

F

h

h

2/6

E

P

h

h 5/6

4/6

h

h 1

I

H

8/6

7/6

IL(RG*) IL(VGL*)

–1/6

–2/6

–1/6

–2/6

–3/6 –4/6

–5/6

–1

1

1

IL(VQ*)

h/2

h

3h/2

h

h/2

IL(MD*) –h/2

–h

IL(MQ*) –h/2 –3h/2 Copyright © 2011 J. Rungamornrat

532


Influence Lines

Influence lines of the same quantities for the floor system can subsequently be obtained as shown below. U=1

x A

B

D

C

h

h

h

G Q

F h

h

h

h 8/6

7/6

1

I

H

h

5/6

4/6

3/6

2/6

1/6

E

P

IL(RG) IL(VFG)

–1/6

–1/6

–2/6

–2/6

–3/6 –4/6

–5/6

1

1

IL(VGH) 3h/2

h

h/2

h

h/2

IL(MD) –h/2

–h

IL(MQ) –h/2 –3h/2

Example 11.10 Construct the influence lines of all support reactions IL(RC), IL(RI) and IL(MI), the shear force IL(VBC), IL(VCD), IL(VEF) and IL(VG), and the bending moment IL(MC), IL(ME) and IL(MP) of the floor system shown below.

C A h

E

D

B h

h

h

P

F h


h/2

h/2

I

H

G h

h

533


Influence Lines

Solution An intermediate structure of the above floor system is obtained by removing all loading panels (i.e. panels AB, BD, DE, EF, FH and HI) as shown below. C A

E

D

B h

h

h

P

F

h

h/2

h

I

H

G h/2

h

h

The influence lines for the floor system are obtained as follows: (1) the influence line of any response of the intermediate structure is constructed first by Müller-Breslau principle; (2) values of the influence line at all panel points are determined and marked; and (3) the complete influence line for the floor system is obtained by connecting points obtained in the step (2) by straight lines. Influence lines for the intermediate structure are shown below with their values at all panel points marked. C A h 3/2

E

D

B h 5/4

h

h

1

1/4

h/2

h

3/4

1/2

1/2

P

F

h/2

I

H

G h

h

1/4

3/4

IL(RC*) 1

1

IL(RI*) –1/2

h

–1/4

h/2

IL(MI*) –h/2

–h

–h –3h/2

IL(VCL*) –1

–1


534


Influence Lines

C A h

1/2

E

D

B h

h

h

h/2

h

3/4

1/2

1/4

P

F

h/2

I

H

G h

h

1/4

IL(VCR*)

1/2

1/4

1/4

IL(VER*) –1/4

1/2

–1/2

1/4

IL(VG*) –1/4

–1/2

–3/4

IL(MC*) –h –2h

h h/2

h/2

IL(ME*) –h/2 –h

h/8 –h/4

h/4

3h/8

IL(MP*)

–h/8


535


Influence Lines

All influence lines of support reactions, shear forces and bending moments for the given floor system are shown below. C A

3/2

E

D

B h

h 5/4

h

h

1

h/2

h

3/4

P

F

1/2

h/2

I

H

G h

h

1/4

IL(RC) 1/2

1/4

1

3/4

1

IL(RI) –1/2 h

–1/4

h/2

IL(MI) –h/2

–h

–h

–3h/2

IL(VBC) –1

1/2

–1 3/4

1/2

1/4

1/4

IL(VCD)

1/2

1/4

1/4

IL(VEF) –1/4

1/2

–1/2

1/4

IL(VG) –1/4

–1/2

–3/4


536


Influence Lines

C A

E

D

B h

h

h

h

P

F h/2

h

h/2

I

H

G h

h IL(MC)

–h –2h h h/2

h/2

IL(ME) –h/2 –h h/8 –h/4

h/4

3h/8

IL(MP)

–h/8

11.6 Influence Lines for Determinate Trusses In this section, we demonstrate how to construct influence lines of any response of two-dimensional trusses (with a primary focus on a statically determinate case) using both the direct method and the analogy between the floor system and the truss. It should be noted that a loading path of any given truss must is considered prescribed a priori in the construction of the influence lines and such path generally depends on geometry and designed function of that particular truss. For instance, a truss shown in Figure 11.1 possesses a loading path along its lower chords whereas a truss shown in Figure 11.2 has a different loading path along its upper chords. In following subsections, the loading transferring mechanism along the loading path, essential characteristics of the influence lines of any responses, determination of the influence lines by a direct method and its pros and cons, and finally a more advanced and efficient method based on the floor-system-truss analogy are presented.

11.6.1 Load Transferring Mechanism In the modeling, moving loads travelling along the loading path of a given truss are assumed to transfer strictly to its joints through the simple loading-panel mechanism similar to that of the floor system considered in the section 11.5. Figure 11.30 illustrates a moving unit concentrated load on the loading path along the lower chord of a truss. Panel points of all loading panels (i.e. points where a moving load travelling along the loading path is transferred to the truss) are located only at joints of the truss. A unit load is transferred to the truss at both panel points of a loading panel via the simply-supported beam action. The reactions R1 and R2 at the two panel points vary linearly as a function of the loading location  as follows Copyright © 2011 J. Rungamornrat

537


Influence Lines

U=1

Loading path

 U = 1

Panel point

h Truss joint Loading panel

 U = 1

R1

R2

R1

R2 h

Figure 11.30: Schematic indicating load transferring mechanism of given truss under moving unit load along its loading path R1  1 

ξ h

;

R2 

ξ h

(11.7)

Clearly, R1 = P and R2 = 0 for  = 0 whereas R1 = 0 and R2 = P for  = h. This indicates that when a moving concentrated load travels to the panel point, it is transferred directly and entirely to the truss joint coincident with that panel point. Due to such load transferring mechanism, any linear response function of the truss due to a moving unit concentrated load acting within the loading panel, denoted by f, is always a linear function of the location of the moving unit concentrated load, i.e., Copyright © 2011 J. Rungamornrat

538


Influence Lines

 ξ ξ f    1   f1    f 2  h h

(11.8)

where f1 and f2 are responses due to the moving unit concentrated load acting to the left panel point ( = 0) and the right panel point ( = h), respectively. It is evident from (11.8) that any response of a given truss due to the moving unit concentrated load traveling to any interior point of the loading panel can completely be obtained once the response is known for the moving unit concentrated load traveling to the left and right panel points. In addition, both the responses f1 and f2 can directly be obtained by ignoring the presence of the loading panels but simply placing the unit concentrated load at the truss joints coincident with the left and right panel points. Schematic shown in Figure 10.31 clearly indicates the meaning of the relation (11.8).

 U = 1

(a)

=  ξ  1     h

1

(b)

+

1

ξ    h

(c) Figure 11.31: (a) Truss subjected to moving unit concentrated load within loading panel, (b) truss subjected to unit concentrated load at truss joint coincident with the left panel point, and (c) truss subjected to unit concentrated load at truss joint coincident with the right panel point Copyright © 2011 J. Rungamornrat

539


Influence Lines

11.6.2 Characteristics of Influence Lines From the load transferring mechanism described above, an influence line (or influence function) of any response of a given truss possesses two common characteristics as follows. (1) The influence line is piecewise linear; i.e., a segment of the influence line for any loading panel is always a straight line. This results directly from the fact that the influence function f is only a linear function of the position  of the moving unit concentrated load as indicated by the relation (11.8). For instance, segments of the influence line IL(r) for all loading panels AB, BC, CD, DE, EF and FG of a truss shown in Figure 11.32 must always be straight lines. (2) The influence line is fully known if and only if its values at all panel points are known. This statement should be very clear from the relation (11.8). In particular, values of the influence line at any interior point of any loading panel can simply be obtained by linear interpolating values of the influence line at both panel points. For instance, a value of any influence line at any interior point of a particular loading panel BC, denoted by h, can readily be computed by carrying out the linear interpolation of values of the influence line at the panel points B and C, denoted by h and hC, respectively. This strategy can be applied to all loading panels and, as a result, the entire influence line IL(r) is known once {hA, hB, hC, hD, hE, hF, hG} are determined.

U=1

x

A

hA

C

B

hB

h

D

hC

hD

E

hE

F

G

hF hG

IL(r)

Figure 11.32: Influence line of a given truss containing 6 loading panels AB, BC, CD, DE, EF and FG

11.6.3 Construction of Influence Lines by Direct Procedure The influence line of any response of a given truss can be constructed directly by exploiting its special features described above. The procedure is straightforward and involves the analysis of a series of truss under the action of a unit concentrated force at each panel point. Details of such procedure can be outlined as follows: Copyright © 2011 J. Rungamornrat

540


Influence Lines

U=1

x B

A

C

D

E

F

G

1 Figure 11.33: Schematic of a truss possessing a loading panel along its top chords with 6 loading panels and 7 panel points. An influence line is to be constructed for a particular response, i.e., the member force of a member 1. (1) Identify the response of interest of a given truss (e.g. support reactions, member forces) whose influence line is to be determined. For instance, let’s construct the influence line of a member force of a member 1 of a truss shown in Figure 11.33, denoted by IL(F1). (2) Identify the loading path, loading panels and all panel points along the loading path. For the truss shown in Figure 11.33, it has a loading path along its top chord and contains 6 loading panels (i.e., AB, BC, CD, DE, EF and FG) and 7 panel points (i.e., A, B, C, D, E, F and G). (3) Analyze a series of truss under the action of a single unit concentrated force acting to each panel point for the response of interest identify in step (1). For instance, the member force of the member 1 due to a unit concentrated load applied to the panel points A, B, C,…, F and G is obtained by performing the analysis of a truss shown in Figure 11.34(a)-11.34(g), respectively. Let F1A, F1B,…, F1F and F1G be results from such series of analysis. (4) Use results from step (3) to obtain values of the influence line at all panel points. For instance, values of the influence line IL(F1) at the panel points A, B, C, …, F and G are equal to F1A, F1B, …, F1F and F1G, respectively. (5) Complete the influence line by connecting all points obtained in step (4) by straight segments. The complete influence line IL(F1) is shown in Figure 11.35. 1 A

(a)

F1A = 0 1 B

(b)

F1B Copyright © 2011 J. Rungamornrat

541


Influence Lines

1 C

(c)

F1C 1 D

(d)

F1D 1 E

(e)

F1E 1 F

(f)

F1F 1 G (g)

F1G = 0

Figure 11.34: A series of trusses subjected to unit concentrated force at (a) panel point A, (b) panel point B, (c) panel point C, (d) panel point D, (e) panel point E, (f) panel point F and (g) panel point G. Member force of the member 1 obtained for each case is indicated in each figure. Copyright © 2011 J. Rungamornrat

542


Influence Lines

U=1

x B

A

C

E

D

G

F

1

F1C F1D F1E

F1B

F1F F1A

F1G

IL(F1)

Figure 11.35: Influence line of a member force of the member 1, IL(F1). Values of the influence line at all panel points are clearly indicated by solid dots. It is worth noting that the direct method outlined above can equally be applied to both statically determinate and statically indeterminate trusses. While the method is very straightforward, a number of analyses must be performed in order to obtain values of the influence line at all panel points. This renders the method inefficient and cumbersome when a truss under consideration is relatively complex and consists of several loading panels. Example 11.11 Use the direct method to construct the influence lines of support reactions IL(RAY) and IL(RE) and the member forces IL(FCD), IL(FJK) and IL(FDJ) of a truss shown below. G

J

H U=1

x

A

h

C

B h

L

K

h

E

D h

h

Solution The loading path for a given truss is along its bottom chords and consists of 4 loading panels (i.e., AB, BC, CD and DE) and 5 panel points (i.e., A, B, C, D, and E). First, we perform a series of analyses of the truss under the action of a unit concentrated load acting to each panel point to obtain values the influence lines IL(RAY), IL(RE), IL(FCD), IL(FJK) and IL(FDJ) at all panel points Copyright © 2011 J. Rungamornrat

543


Influence Lines

A, B, C, D, and E. Since the structure is statically determinate, all support reactions and member forces can be determined using only static equilibrium; in particular, the member force of the members CD, JK, and DJ can readily be computed from a method of section (with a particular section chosen passing through those three members). Results from the analyses are given below.

G

H

J

(0)

K

L

(0) 1

0

A

(0) B

C

E

D

1

0 Unit load applied to panel point A

G

H

J

(-1/4) K

L

(-2/4) 1 0

A

B

(1/2) C

E

D

3/4

1/4 Unit load applied to panel point B

G

H

J

(-1/2) K

L

(-2/2) 1 0

A

B

(1)

C

D

1/2

E 1/2

Unit load applied to panel point C


544


Influence Lines

G

J

H

L

(-3/4) K (2/4) (1/2)

0

A

C

B

1 E

D

1/4

3/4 Unit load applied to panel point D

G

J

H

(0)

L

K

(0) 1

(0)

0

A

C

B

E

D

0

1 Unit load applied to panel point E

Using results from above analyses, the influence lines IL(RAY), IL(RE), IL(FCD), IL(FJK) and IL(FDJ) are plotted at all panel points as indicated in the figure below. The complete influence lines are readily obtained by simply connected these points by straight segments. G

A

L

K

U=1

x

h

C

B h

1

J

H

h 3/4

E

D h

1/2

h

1/4 IL(RAY)


545


G

Influence Lines

J

H

L

K

U=1

x

h

A

C

B h

h

1/4

h

1/2

E

D h 1

3/4

IL(RE) 1 1/2

1/2

IL(FCD) 2/4

IL(FDJ) -2/4 -2/2 IL(FJK) -1/4 -1/2 -3/4

11.6.4 Construction of Influence Lines by Truss-Floor System Analogy For certain statically determinate trusses with a simple configuration and a loading path such as those shown in Figure 11.36, the influence lines of the support reactions and the member forces can alternatively be obtained by using a method called the truss-floor system analogy. In this method, a given truss is first represented by an equivalent floor system, and the influence lines of certain quantities of this analogous structure are employed as a basis for constructing the influence lines of support reactions and member forces for the original truss. This analogy follows directly from the equivalence of the form of static equilibrium equations of the two structures. To demonstrate and clarify such analogy and also provide a step-by-step analysis procedure, let’s consider an example of a statically determinate truss with a loading path located along its bottom chords as shown in Figure 11.37(a) and let’s focus attention to the construction of the influence lines of all support reactions IL(RAX), IL(RAY) and IL(RFY) and the member force of certain members IL(FCD), IL(FCJ), IL(FDJ) and IL(FIJ). Copyright © 2011 J. Rungamornrat

546


Influence Lines

U=1

U=1

U=1

U=1

U=1

Figure 11.36: Schematic of statically determinate trusses with simple geometry and straight loading path H

a

C

B h

L

K

U=1

x

A

J

I

h

h

F

E

D h

h

G h

(a) Y

H

I

A RAY

B

K

D

E

L

U=1

x

RAX

J

C

(b)

F

G

X

RFY

Figure 11.37: (a) Example of statically determinate truss used to demonstrate the truss-floor system analogy and (b) free body diagram of the entire structure Copyright © 2011 J. Rungamornrat

547


Influence Lines

All support reactions RAX, RAY and RFY of this truss due to the moving unit concentrated load can readily be computed by enforcing overall equilibrium of the entire structure (see a free body diagram of the truss in Figure 11.37(b)): ΣFX = 0  R AX  0

(11.9)

ΣFY = 0  R AY  R FY  1

(11.10)

ΣM A = 0  R FY (5h)  1  x  0

(11.11)

To construct the analogy, we first form an equivalent floor system, i.e. a system with the same span, the same loading panels, and the same pattern of boundary conditions. For the above particular truss, the equivalent floor system can be obtained as shown in Figure 11.38(a). U=1

x

A

B

D

C

h

h

h

E h

F h

G h

(a) Y U=1

x RAX

A

B

RAY

C

D (b)

E

F

G

X

RFY

Figure 11.38: (a) Equivalent floor system for truss shown in Figure 11.37(a) and (b) corresponding free body diagram From the free body diagram shown in Figure 11.38(b), all support reactions RAX, RAY and RFY of the equivalent floor system can be computed from following three equilibrium equations (note that the script letter is used here only to distinguish between quantities associated with the original truss and those corresponding to the equivalent floor system): ΣFX = 0  RAX  0

(11.12)

ΣFY = 0  RAY  RFY  1

(11.13)

ΣM A = 0  RFY (5h)  1  x  0

(11.14)

It is obvious from the two systems of equations (11.9)-(11.11) and (11.12)-(11.14) that three static equilibrium equations for both the truss and the equivalent floor system possess the same form for any location of the moving unit concentrated load along the loading path. This implies that the support reactions and their influence lines of the two structures are identical, i.e. Copyright © 2011 J. Rungamornrat

548


Influence Lines

R AX = RAX  IL(R AX )  IL(RAX )

(11.15) (11.16) (11.17)

R AY = RAY  IL(R AY )  IL(RAY )

R FY = RFY  IL(R FY )  IL(RFY )

The analogy (11.15)-(11.17) is useful in the construction of influence lines for support reactions of trusses since such task is simply reduced to constructing the influence lines of support reactions of an equivalent floor system. Note that the influence line of support reactions for the floor system can be readily obtained using Müller-Breslau principle as previously discussed in Section 11.5. Now, let’s turn attention to form an analogy for an influence line of a member force. To determine the member force of members CD, CJ and IJ, a method of sections supplemented by a fictitious section passing through those three members is employed. The free body diagram of a left part of the truss resulting from the sectioning is shown in Figure 11.39(a). Resultants of all member forces appearing on the fictitious section in terms of the shear force at the panel CD, denoted by VCD, and the moments about two selected points C and J, denoted by MC and MJ, are given by VCD =  FCJ sin 

(11.18) (11.19) (11.20)

M C =  FIJ a M J = FCD a

It is worth noting that all loading panels are not considered as a part of the structure but simply a loading system. Next, resultants of all external loads acting to the left part of the truss (i.e. support reactions at point A and forces transferring from the moving unit concentrated load to panel points A, B and C through the load transferring mechanism) in terms of the shear force at the panel CD, denoted by VCD,ext, and moments about points C and J, denoted by MC,ext and MJ,ext, can also be obtained as VCD,ext = R AY  Lleft

(11.21)

M C,ext = R AY (2h)  M C,left

(11.22)

M J,ext = R AY (3h)  M J,left

(11.23)

where Lleft, MC,left and MJ,left are force resultant, moment resultant about the point C and moment resultant about the point J of forces transferring to all panel points on the left part of the truss, respectively. Note that both the internal forces and external applied loads can be equivalently represented by their resultant shear forces and moments about points C and J as shown in Figure 11.39 (b). H

I U=1

x

FIJ

H

J

I



FCJ

MJ,ext J  MJ VCD,ext VCD

RAX RAY

A

C

B h

h

FCD

B

A

h

(a)

h

MC,ext h (b)

C MC h

Figure 11.39: (a) Free body diagram of the left part of a truss resulting from sectioning through members IJ, CJ and CD and (b) resultant shear forces and moments about points C and J of both internal forces and external applied loads Copyright © 2011 J. Rungamornrat

549


Influence Lines

Since the structure is in equilibrium, any of its parts must also be in equilibrium and this finally leads to VCD = VCD,ext

(11.24)

M C = M C,ext

(11.25)

M J = M J,ext

(11.26)

Now, let’s turn attention to the equivalent floor system in Figure 11.38(a). By introducing a fictitious cut within the panel CD at a point just to the right of the point C, it leads to the corresponding free body diagram of its left part as shown in Figure 11.40(a). Similarly, all loading panels are not considered, again, as a part of the structure but simply a loading system. Y

Y U=1

x

RAX A RAY

B

MC

C VCD

h

h

U=1

x

RAX A RAY

B

h

(a)

C

h (b)

D MD VCD h

Figure 11.40: (a) Free body diagram of the left part of an equivalent floor system resulting from sectioning at a point just to the right of the point C and (b) free body diagram of the left part of an equivalent floor system resulting from sectioning at a point just to the left of the point D Equilibrium of the left portion of the equivalent floor system shown in Figure 11.40(a) requires that VCD = VCD,ext  RAY  Lleft

(11.27)

MC = MC,ext  RAY (2h)  MC,left

(11.28)

where VCD and MC represent the shear force in the panel CD and the bending moment at a point C, respectively, and Lleft and MC,left are the force resultant and the moment resultant about the point C of forces transferring to all panel points on the left part of the equivalent floor system shown in Figure 11.40(a). By introducing another fictitious cut within the panel CD at a point just to the left of the point D, we obtain the free body diagram of the left part of the floor system as shown in Figure 11.40(b). Again, equilibrium of the left portion of the equivalent floor system shown in Figure 11.40(b) requires that VCD = VCD,ext  RAY  Lleft

(11.29)

MD = MD,ext  RAY (3h)  MD,left

(11.30)

where MD represents the bending moment at a point D and MD,left is the moment resultant about the point D of forces transferring to all panel points on the left part of the equivalent floor system shown in Figure 11.40(b). Copyright © 2011 J. Rungamornrat

550


Influence Lines

By comparing a set of equilibrium equations for the truss (11.24)-(11.26) supplemented by (11.18)(11.23) and a set of equilibrium equations for the equivalent floor system (11.27)-(11.30) and noting the analogy for the support reactions (11.15)-(11.17) and fact that the load transferring mechanics of both structures are equivalent (i.e., Lleft = Lleft, MC,left = MC,left and MD,left = MJ,left), it leads to VCD = VCD   FCJ sin  = VCD  IL(FCJ )   IL(VCD ) / sin  M C = MC   FIJ a = MC  IL(FIJ )   IL(MC ) / a M J = MD  FCD a = MD  IL(FCD )  IL(MD ) / a

(11.31) (11.32) (11.33)

The analogy (11.31)-(11.33) concludes that (1) A shear force resultant within a particular loading panel (e.g. the loading panel CD) of all member forces appearing on the fictitious section of a truss is identical to the shear force within the same panel of the equivalent floor system; (2) A moment resultant about a particular reference point (e.g. point C or point J) of all member forces appearing on the fictitious section of a truss is identical to the bending moment at an equivalent point of the equivalent floor system. Such equivalent point of the equivalent floor system is simply obtained by projecting the reference point (used to compute the moment resultant of the truss) onto the floor. For instance, an equivalent point of the reference point C is the point C itself whereas an equivalent point of the reference point J is the point D as illustrated in Figure 11.41. Since the shear force and moment resultants are related to the member forces along the fictitious section through the relations (11.18)-(11.20), the above analogy directly yields the relations between the member forces of the truss (and their influence lines) and the shear force and bending moment of the equivalent floor system (and their influence lines). These relations are significantly useful in the construction of the influence lines of the member force of a truss since it is only required (i) to obtain expressions of the shear force and moment resultants in terms of that member force (e.g. the relations (11.18)-(11.20)) and (ii) to construct the influence lines of the shear force and bending moment of the equivalent floor system. The former task can readily be achieved using the method of sections whereas the latter task can be accomplished by using Müller-Breslau principle as discussed previously in section 11.5. H

U=1

x

B h



C

B

A

J

U=1

x

A

I

C h

E

D h

h

F h

G h

Figure 11.41: Schematic indicating equivalent points on the equivalent floor system of the reference points, C and J, used to compute the moment resultant of the truss Copyright © 2011 J. Rungamornrat

551


Influence Lines

By following the same procedure, the influence line of the member force FDJ can readily be obtained as follows. By introducing a fictitious section passing through the members IJ, CJ, DJ, and DE and then considering equilibrium of the left part of the truss shown in Figure 11.42, we first obtain the shear force resultant within the loading panel DE in terms of all member forces along the fictitious section and then form the following analogy VDE   FCJ sin   FDJ  VDE  IL(FDJ )  IL(FCJ ) sin   IL(VDE )

(11.34)

where VDE is the shear force within the loading panel DE of the equivalent floor system. Since the influence line of the member force FCJ, IL(FCJ), was already obtained from (11.31) and the influence line of the shear force VDE, IL(VDE), can readily be constructed using Müller-Breslau principle, the influence line of the member force FDJ, IL(FDJ), can directly be obtained from the relation (11.34). Y

H

I U=1

x

RAX

A

FIJ



FCJ

FDJ

C

B

J

FDE

D

X

RAY Figure 11.42: Free body diagram of the left part of a truss resulting from sectioning through members IJ, CJ, DJ and DE Example 11.12 Construct the influence lines of support reactions IL(RAY) and IL(RDY) and the member forces IL(FBC), IL(FCH), IL(FHI), IL(FCI) and IL(FDK) of a truss shown below

G

H

L

K

U=1

x

A

J

I

h

h

D

C

B h

h

F

E h

h

Solution To use the truss-floor system analogy, we first construct an equivalent floor system for the given truss. This floor system must consist of five loading panels of total span length 5h and the pinned support and the roller support are located at the panel point A and the panel point D, respectively. The schematic of the equivalent floor system is shown in the figure below U=1

x

A

B h

h

E

D

C h


h

F h

552


Influence Lines

The influence lines of the support reactions IL(RAY) and IL(RDY) for the above floor system can readily be constructed by using Müller-Breslau principle with results given below and the influence lines of the support reactions IL(RAY) and IL(RDY) for the truss follow immediately from the trussfloor system analogy, i.e. RAYI = RAYI and RDYI = RDYI. G

H

h

D

C

B h

h

F

E

h

h

h

U=1

x

A 1

L

K

U=1

x

A

J

I

B h

E

D

C h

h

F h

h

2/3 1/3 IL(RAY) = IL(RAY) –1/3

1/3

4/3

1

2/3

–2/3 5/3

IL(RDY) = IL(RDY) To construct influence lines of the member forces FBC, FCH, and FHI, we introduce a fictitious section passing through three members BC, CH, and HI (section 1) and the free body diagram of the left part of the truss is given below. G

H

FHI

U=1

x

FCH

h



RAX A RAY

B h

1

FBC h


C

553


Influence Lines

The shear force resultant within the loading panel BC and the moment resultant about the reference points C and H of all member forces appearing at the fictitious section are obtained as (also see these representative resultants in the figure below) VBC  FCH sin 45o  FCH / 2 ; M C   FHI h ; M H  FBC h

(e11.12.1)

H MJ

G

U=1

x

VBC

h

MC



RAX A RAY

B h

C

1

h

Next, influence lines of the shear force within the loading panel BC, IL(VBC), and the bending moment at points B and C, IL(MB) and IL(MC), for the equivalent floor system are constructed using Müller-Breslau principle and final results are given below. U=1

x

A

B h

E

D

C h

h

F h

h

1/3 IL(VBC) –1/3

–1/3

–2/3

–2/3

2h/3 h/3

IL(MB) –h/3 2h/3

–2h/3

h/3

IL(MC) –2h/3 –4h/3 Copyright © 2011 J. Rungamornrat

554


Influence Lines

By using the relation (e11.12.1) and the truss-floor system analogy, i.e. VBC = VBC, MC = MC and MH = MB, influence lines of the member forces FBC, FCH and FHI can readily be obtained and results are shown below. H

G

J

I

L

K

U=1

x

h

A

D

C

B h

h

E h

h

F h

2 /3

IL(FCH) =  2 /3

 2 /3

2/3

2 IL(VBC)

2 2 /3

1/3 IL(FBC) = IL(MB)/h –1/3 –2/3 4/3 2/3 IL(FHI) = –IL(MC)/h –1/3 –2/3 Influence lines of the member forces FCI and FDK can be obtained in a similar fashion by introducing the following two fictitious sections, one passing through the members HI, CI and CD (section 2), and the other passing through the members JK, DK, and DE (section3). The free body diagrams of parts of the truss resulting from those above two cuts are shown below. G x

RAX A RAY

H

U=1

h

FJK

FCI

C

B

L

K

FHI

FDK FCD 2

 D FDE

3

h

h Copyright © 2011 J. Rungamornrat

h

F

E h

555


Influence Lines

The shear force resultant within the loading panel CD and the shear force resultant within the loading panel DE can be computed in terms of the member force FCI and the member force FHI as follows VCD   FCI ; VDE   FDK sin 45o   FDK / 2

(e11.12.2)

Next, we construct an influence line of the shear force within the panel CD, VCD, and an influence line of the shear force within the panel DE, VDE, of the equivalent floor system by using MüllerBreslau principle and the final results are shown below. By using the relation (e11.12.2) and the truss-floor system analogy, i.e. VCD = VCD and VDE = VDE, the influence lines of the member forces FCI and FHI can directly be obtained and results are indicated further below. U=1

x

A

B h

C h

E

D h

h

F h

IL(VCD) –1/3

–1/3 –2/3

–2/3

–1 1

1

1

IL(VDE) G

H

L

K

U=1

x

A

J

I

h

h

D

C

B h

h

h

h

2/3

2/3 1/3

F

E

1/3 IL(FCI) = –IL(VCD) IL(FDK) = – 2 IL(VDE)

 2 Copyright © 2011 J. Rungamornrat

 2

556


Influence Lines

Example 11.13 Construct the influence lines of support reactions IL(RAY) and IL(RFY) and the member forces IL(FCD), IL(FCJ), IL(FIJ), IL(FDJ) and IL(FJK) of a truss shown below U=1

x

G

H

J

I

K

L

M

60o 60

A

o

B

D

C

h

h/2

60o

h

h

F

E h

h

h/2

Solution An equivalent floor system for the given truss consists of five loading panels (panels GH, HI, IJ, JL and LM) and its total span length is 6h. The system is constrained by a pinned support at the mid-point of the loading panel GH, denoted by a point A´, and a roller support at the mid-point of the loading panel LM, denoted by a point F´, as shown in the figure below. It is important to emphasize that, for this particular structure, the joint K located on the loading path is not a panel point. U=1

x

F′

A′ G

I

H h/2

J

h

h/2

h

M

L

K h

h

h/2

h/2

Influence lines of the support reactions RA′Y and RF′Y for the above floor system can readily be constructed by using Müller-Breslau principle with final results shown below. Influence lines of the support reactions RAY and RFY for the given truss then follow immediately from the truss-floor system analogy of the support reactions, i.e. RAY = RA′Y and RFY = RF′Y. Influence lines of the member forces FCD, FCJ, FIJ, FDJ and FJK can be obtained by first introducing the following two fictitious sections, one passing through the members CD, CJ and IJ (section 1), and the other passing through the members CD, DJ, and JK (section 2). Free body diagrams of the left part of the truss resulting from the sectioning are shown below. U=1

x

U=1

x FIJ

G

H

J

I 60o 60

A h/2

B h

o

h

H

o

J

I

FCJ 60

C

G

60o 60

FCD A

1

h/2 Copyright © 2011 J. Rungamornrat

B h

o

60

C h

2

o

FJK FDJ

FCD D h

557


Influence Lines

U=1

x

G

H

J

I

K

L

M

60o 60

A

o

60o

B

h/2

D

C

h

h

h

F

E h

h

h/2

U=1

x

F′

A′ G

I

H h/2

1.1

h

h/2

1

J h

0.9

h

0.7

M

L

K h

h/2

h/2

0.5 0.1 –0.1 0.9

0.1

0.3

1

IL(RAY) = IL(RA′Y)

1.1

0.5 IL(RFY) = IL(RF′Y)

–0.1

From above free body diagrams, the shear force resultants within the loading panels IJ and JL and the moment resultants about points C, D and J of the member forces along the fictitious sections are obtained as follows VIJ   FCJ sin60o   3FCJ /2 ; VJL  FDJ sin60o  3FDJ /2

(e11.13.1)

M J  FCD hsin60o  3FCD h /2 ; M C  FIJ hsin60o   3FIJ h /2 ; M D  FJK hsin60o   3FJK h /2 (e11.13.2)

Next, we construct influence lines of the shear forces with the loading panels IJ and JL and the bending moments at points C´(a mid-point of the loading panel IJ), D´(a quarter-point of the loading panel JL) and J of the equivalent floor system by using Müller-Breslau principle. Obtained results are given below. Copyright © 2011 J. Rungamornrat

558


Influence Lines

U=1

x

F′

A′ G

I

H h/2

h/2

h

J

C′

D′

h

L

K

h

h

M h/2

h/2

0.5 0.1

0.1 –0.1

–0.3

IL(VIJ) –0.1

–0.5

0.1

0.1 –0.1

IL(VJL) –0.1

–0.3 –0.5 0.9 0.9h

1.2h

h

0.3h

0.2h

IL(MC´) –0.2h

–0.3h

0.2h

0.6h

h

1.2h 0.3h

IL(MD´) –0.3h

–0.2h 1.25h 0.75h 0.25h

0.25h

IL(MJ) –0.25h

–0.25h

By using the relations (e11.13.1)-(e11.13.2) and the truss-floor system analogy, i.e. VIJ = VIJ, VJL = VJL, MC = MC´, MD = MD´ and MJ = MJ, influence lines of the member forces FCD, FCJ, FIJ, FDJ and FJK can directly be obtained and results are given below. Copyright © 2011 J. Rungamornrat

559


x

G

Influence Lines

U=1

H

J

I

K

L

M

60o 60

h/2

o

60o

A

B

C

D

E

h/2

h

h

h

h

F h/2

h/2

0.6/ 3

0.2/ 3

0.2/ 3 0.2/ 3

IL(FCJ)

0.2/ 3 1/ 3

0.2/ 3

0.2/ 3

IL(FDJ)

0.2/ 3

0.2/ 3 0.6/ 3 1/ 3

0.9h 0.6/ 3

0.4/ 3 0.4/ 3

0.6/ 3 1.8/ 3

IL(FIJ)

2/ 3

0.6/ 3

0.4/ 3

IL(FJK) 0.4/ 3

0.6/ 3

1.2/ 3 2/ 3 2.5/ 3 1.5/ 3

0.5/ 3

0.5/ 3

IL(FCD) 0.5/ 3

0.5/ 3


560


Influence Lines

Example 11.14 Construct the influence lines of member forces IL(FDE), IL(FEL), IL(FLM), IL(FEM) and IL(FMN) of a truss shown below

M 1 2

L

K

A

B

P

E

D

C h

h

O

U=1

x

J

N

h

h

F

G h

h

I

H h

h

Solution An equivalent floor system for the given truss consists of eight loading panels (i.e. panels AB, BC, CD, DE, EF, FG, GH and HI) and its total span length is 8h. This floor system is constrained by a pinned support at the panel point A and a roller support at the panel point I as shown in the figure below. U=1

x A

B h

h

E

D

C

h

h

F

G h

h

I

H h

h

Influence lines of the member forces FDE, FEL, FLM, FEM, and FMN can be obtained by considering the following two fictitious sections, one passing through the members DE, EL and LM (i.e., section 1) and the other passing through the members DE, EL, EM and MN (i.e., section 2). The free body diagrams of the left part of the truss resulting from the sectioning are shown in the figure below.

 1 2

K x

J RAX

L

FLM

U=1

FEL FDE

A

B

E

D

C

RAY

1

h

h

h


h

561


Influence Lines

M  1 2

K

RAX

A

B

D

C

RAY h

h

FMN

U=1

x

J

L FEM FEL FDE 2

h

h

From the first free body diagram, the shear force resultant within the loading panel DE and the moment resultants about points E and L of the member forces FDE, FEL and FLM are given by VDE  3FEL / 13  FLM / 5

(e11.14.1)

M E  (2FLM / 5)(2h)  4FLM h / 5 M L  FDE (3h /2)

(e11.14.2) (e11.14.3)

Similarly, from the second free body diagram, the shear force resultant within the loading panel DE and the moment resultant about the point E of the member forces FDE, FEL, FEM and FMN are given by VDE  3FEL / 13  FEM  FMN / 5

(e11.14.4)

M E  (2FMN / 5)(2h)  4FMN h / 5

(e11.14.5)

Next, we construct influence lines of the shear force within the loading panel DE and the bending moments at a point D (a projection of the point L onto the loading path) and a point E of the equivalent floor system by using Müller-Breslau principle. Final results are shown below. U=1

x A

B h

D

C h

F

E h

h

G h

h

1/2

3/8

h

h

1/4

I

H

1/8 IL(VDE)

–1/8

–1/4

–3/8

–1/2


562


Influence Lines

U=1

x A

B h

h

G

2h

h

h

3h/2

I

H

h

h

3h/2

h

F

h

h

h/2

E

D

C

h

h/2

IL(ME) 15h/8

3h/2

5h/4

9h/8

3h/4

5h/8

3h/8 IL(MD)

By using the relations (e11.14.1)-(e11.14.5) and the truss-floor system analogy, i.e. VDE = VDE, ML = MD, and ME = ME, influence lines of the member forces FDE, FEL, FLM, FEM and FMN can readily be obtained and results are shown below. M 1 2

L

N

K

O

U=1

x

P

J A

B h

D

C h

F

E h

h

5/4 5/6

G h

h

1

3/4

5/12

I

H h

h

1/2

1/4

IL(FDE)

IL(FLM)  5/8

 5/4

3 5/8

 5/2


3 5/8

 5/4

 5 /8

563


Influence Lines

M 1 L

2

K

A

B

P

D

C h

h

O

U=1

x

J

N

F

E h

h

G h

h

I

H h

h

IL(FEL)  13/12

 13/6

 13/4

IL(FMN)  5 /8

1/4

 5 /4

1/2

3 5 /8

3/4

 5/2

1

3 5 /8

3/4

 5 /4

1/2

 5/8

1/4 IL(FEM)

11.7 Influence Lines for Statically Indeterminate Structures This entire section is devoted to how to construct qualitatively and quantitatively an influence line of statically indeterminate structures such as those shown in Figure 11.43. While the definition (of the influence function and its corresponding graphical representation, i.e., the influence line) remains unaltered and some basic principles previously introduced still apply, the procedure to construct such influence line is relatively complex (in comparison with that for statically determinate structures), requires significant amount of work due to the static indeterminacy present within the structure, and, as a result, deserves an extra section for its clear explanation and discussion. It is worth noting that, for this particular class of structures, only static equilibrium is insufficient to completely determine all support reactions and internal forces; the constitutive law and kinematics must be properly incorporated to obtain a complete set of governing equations. We first outline a direct approach, based primarily on its definition, for constructing the influence line of various quantities, e.g. support reactions, internal forces, and displacements. Subsequently, a more advanced technique based on Muller-Breslau’s principle is introduced. This technique not only provides a more convenient means to construct the influence line but also allows it to be physically interpreted in a simple fashion. The latter feature is found significantly useful in the qualitative sketch of the influence line without carrying out a comprehensive analysis. Several examples are then presented to clearly demonstrate all procedures and techniques described above. Copyright © 2011 J. Rungamornrat

564


Influence Lines

x

U=1

x

U=1

x

U=1

x

U=1

Figure 11.43: Examples of statically indeterminate structures under moving unit concentrated load

11.7.1 Influence Lines by Direct Procedure Influence lines for statically indeterminate structures can also be constructed in a direct fashion following the same procedure as that previously employed for statically determinate structures. The key difference is that the analysis technique utilized in this particular case must be capable of handling statically indeterminate structures. Presence of the static redundancy of the structure renders all support reactions and internal member forces not be directly computed via static equilibrium. Such direct procedure is summarized here again as follows: (1) Identify a response of interest of a given structure (e.g. support reactions, internal forces, and displacements and rotations at certain points) whose influence line is to be constructed. (2) Identify a loading path and loading panels and all panel points along the loading path (for floor systems, trusses and other structures with the same load transferring mechanism). (3) Construct an influence function by carrying out the analysis of a structure under the action of a moving unit concentrated load along the loading path for the response of Copyright © 2011 J. Rungamornrat

565


Influence Lines

interest identify in step (1). In this step, various methods can be employed in the analysis of statically indeterminate structures and such specific choice is simply a matter of taste. Note in addition that for a structure with a load transferring mechanism similar to that of the floor system and truss described in sections 11.5 and 11.6, an analysis for the influence function is not necessary and can be replaced by an analysis of a series of structures under a unit concentrated load acting at each panel point. (4) Use information obtained in step (3) to sketch the influence line. It is noted again that for the floor system and other structures with a similar load transferring mechanism, values of the influence function obtained at all panel points in step (3) are connected by straight segments to form the complete influence line. To clearly demonstrate applications of above procedure, two examples involving a continuous beam and a simple frame are presented. Example 11.15 Construct influence lines of the support reaction (RBY), the bending moment at a point B (MB), the shear force at a point just to the left of the point B (VBL), and the rotation at the point B (B). The Young modulus (E) and the moment of inertia of the cross section (I) are constant throughout. x

U=1 C

B A L

L

Solution It is clear from the problem statement that the influence line is to be constructed for four different quantities, i.e., RBY, MB, VBL and B, and the loading path is defined along the entire beam, i.e., 0 ≤ x ≤ 2L. To construct the influence functions of all those four quantities, various analysis methods (e.g., Castigliano’ 2nd theorem described in Chapter 9, method of consistent deformation described in Chapter 10, slope-deflection method, etc) can be chosen. We note by passing that the given beam is statically indeterminate to the second degree and, for convenience in the analysis, two load cases, one associated with a unit concentrated load moving along the span AB (0 ≤ x ≤ L) and the other corresponding to a unit concentrated load moving along the span BC (L ≤ x ≤ 2L), may be treated separately as shown in the figure below. 0≤x≤L

U=1 C

B A L

L U=1

L ≤ x ≤ 2L

C

B A L


566


Influence Lines

Since the analysis procedure of statically indeterminate beams under a unit load with a variable location as shown in the above figures is quite standard (readers may consult Chapter 9 and Chapter 10) but rather lengthy, they are not presented here for brevity and to prevent the lost of our main focus. Only final results for the influence functions are provided below.

R BY

 24  x  2 17  x 3       7 L  7 L  2 3  2 15  x  12  x  2  x      7 7  L  7  L  7  L  

; 0
(e11.15.1) ; L < x < 2L

 3L  x  2 3L  x 3       7 L  7 L MB   2 3 12L 22L  x  12L  x  2L  x   7  7  L   7  L   7  L  

VBL

; 0
(e11.15.2) ; L < x < 2L

 15  x 2 8  x 3        7 L 7L  2 3  18 33  x  18  x  3  x   7  7  L   7  L   7  L  

; 0
(e11.15.3) ; L < x < 2L

 L2  x 2 L2  x 3        7EI  L  7EI  L  B   2 3 2 2 2 L2  x   6L 11L  x  6L  x   7EI  7EI  L   7EI  L   7EI  L  

; 0
(e11.15.4) ; L < x < 2L

Finally, the influence lines for RBY, MB, VBL and B are plotted as shown below. x

U=1 C

B A L

L 1

IL(RBY)

1


IL(MB)

567


Influence Lines

U=1

x

C

B A L

L IL(VBL) 1

IL(B)

Example 11.16 Construct influence lines of the support reaction RAY and RDY, the bending moment at a point just to the left of the point B (MBL), the shear force at a point just to the left of the point B (VBL), the horizontal displacement at the point B (uB), and the rotation at the point B (B). The Young modulus (E) and the moment of inertia of the cross section (I) are constant throughout and a loading path is along the segment ABC. U=1

x

B

A

C

L

D L

L

Solution Again, it is clear from the problem statement that the influence line is to be constructed for six different quantities, i.e., RAY, RDY, MBL, VBL, uB and B, and the loading path is defined along the segment ABC, i.e., 0 ≤ x ≤ 2L. To construct the influence functions of all those six quantities, a method of consistent deformation (see Chapter 10) is chosen. We note by passing that the given frame is statically indeterminate to the first degree (the reaction at the roller support C is chosen as the redundant) and, for convenience in the analysis, two load cases, one associated with a unit concentrated load moving along the span AB (0 ≤ x ≤ L) and the other corresponding to a unit concentrated load moving along the span BC (L ≤ x ≤ 2L), are treated separately as shown in the figure below. Copyright © 2011 J. Rungamornrat

568


Influence Lines

U=1

0≤x≤L

B

A

C

L

D L

L U=1

L ≤ x ≤ 2L B

A

C

L

D L

L

Again, the analysis procedure by the method of consistent deformation for above statically indeterminate frames is standard and quite lengthy; as a result, only final results for the influence functions are presented as follows.  5  x  1  x 3 1        4L 4L R AY   3 x  1 x  1  2   2      4  L 4 L   3  x  1  x 3      2  L  2  L  R DY   3 x  1 x 3  2 2        2  L 2 L   L  x  L  x 3        4L 4L M BL   3 x  L x  L  2   2      4  L 4 L 

; 0
(e11.16.1) ; L < x < 2L ; 0
(e11.16.2) ; L < x < 2L ; 0
(e11.16.3) ; L < x < 2L Copyright © 2011 J. Rungamornrat

569


Influence Lines

 5  x  1  x 3        4L 4L VBL   3 x  1 x  1  4  2  L   4  2  L    L3  x  L3  x 3       12EI  L  12EI  L  uB   3 3 x  L3  x  L  12EI  2  L   12EI  2  L    L2  x  L2  x 3      12EI  L  12EI  L  B   3 2 x  L2  x  L   12EI  2  L   12EI  2  L  

; 0
(e11.15.4) ; L < x < 2L ; 0
(e11.15.5) ; L < x < 2L ; 0
(e11.15.6) ; L < x < 2L

Finally, the influence lines for RAY, RDY, MBL, VBL, uB and B are plotted as shown below. U=1

x

B

A

C

L

D L

L 1

IL(RAY)

1 IL(RDY)

1


IL(MBL)

570


Influence Lines

U=1

x

B

A

C

L

D L

L

IL(VBL) 1

IL(uB)

IL(B)

11.7.2 Influence Lines by Muller-Breslau Principle It is evident from the previous section that construction of the influence lines by a direct approach is not convenient and requires a comprehensive analysis of a statically indeterminate structure under several load cases especially when the configuration and loading path of the structure are relatively complex. In addition, such method does not provide an easy way to sketch the qualitative influence lines without carrying out an analysis for the influence functions. To circumvent this difficulty, an alternative approach based on the reciprocal theorem (see more details in section 6.10 of Chapter 6) is introduced and it is commonly known as Muller-Breslau principle for indeterminate structures. First, we provide a brief review of the reciprocal theorem and then demonstrate the application of this theorem to construct the influence lines of various quantities. Let’s consider a linear elastic structure whose displacement and deformation are described by a linearized kinematics. Two elastic (or equilibrium) states of this structure corresponding to different sets of applied loads are shown schematically in Figure 11.44. The reciprocal theorem simply states that the external virtual work of a set of applied loads in the elastic state 1 through the displacements from the elastic state 2 is equal to the external virtual work of a set of applied loads in the state 2 through the displacements from the elastic state 2. This theorem can be written mathematically as Copyright © 2011 J. Rungamornrat

571


Influence Lines

q1

2 1

M

1 1

M

P11 P31

P22

P32

q2

P21

M12

P12

P42

Elastic state 1

Elastic state 2

Figure 11.44: Schematic of two elastic states of a frame subjected to different sets of applied concentrated forces, concentrated moments and distributed forces W1  W 2

(11.35)

where W1 and W2 are defined, for a structure subjected to concentrated forces, concentrated moments and distributed forces as shown above, by W1  Pi1u i2  M1i i2   q1u 2 ds

(11.36)

W 2  Pi2 u1i  M i2 1i   q 2 u1ds

(11.37)

in which u1i , u1 and 1i are displacements and rotation of the structure at the elastic state 1 measured at the same location and in the same direction as the applied loads Pi2 , q 2 and M i2 at the elastic state 2 and u i2 , u 2 and i2 are displacements and rotation of the structure at the elastic state 2 measured at the same location and in the same direction as the applied loads Pi1 , q1 and M1i at the elastic state 1. To deduce Müller-Breslau principle from the reciprocal theorem for constructing influence lines for support reactions and internal forces, let’s consider a following model problem: a statically indeterminate frame with a loading path along the spans AB and BC as shown in Figure 11.45.

F

Y

E x

U=1

A

X D

B

C

E Figure 11.45: Statically indeterminate frame with loading path along the spans AB and BC Copyright © 2011 J. Rungamornrat

572


Influence Lines

11.7.2.1 Influence lines for support reactions Without lost of generality, we first establish the principle for a specific support reaction at point C (RCY) of the model problem for clarity and then conclude the principle in a general context applicable for any component of support reactions. To proceed, following two elastic states of the structure are introduced: State 1 associated with the actual structure under the action of a moving unit concentrated load along its loading path (see Figure 11.45) and State 2 associated with the released structure (obtained by removing the support constraint at the point C) subjected to a concentrated load at the point C whose direction coincides with that of positive RCY and magnitude is chosen to produce a unit displacement in the same direction (as the positive RCY), see Figure 11.46. It is worth noting that the positive sign convention for the support reaction RCY can be defined arbitrarily and, for this particular case, the positive RCY is chosen to direct along the Y-axis of the global coordinate system {X, Y, Z}. F

Y X C

A B

P such that v C2  1

E Figure 11.46: Elastic state 2 for determining influence line of support reaction at point C. The magnitude of the force P applied to the point C is chosen such that it produces a unit displacement in the direction of positive RCY. The external virtual works W1 and W2 associated with above two elastic states can be computed as follows W1  1  v 2 (x)  R CY  vC2   v 2 (x)  R CY

;

W 2  0

From the reciprocal theorem (i.e.,W1 = W2), it can be concluded that R CY  v 2 (x)

(11.38)

This relation indicates that the value of the influence line of the support reaction at point C (RCY) is identical to the value of the vertical displacement (i.e. displacement in the same direction as that of the moving unit concentrated load) of the elastic state 2 at a point coincident with the location of the moving unit concentrated load in the elastic state 1. As a result, the construction of an influence line for RCY simply reduces to determination of the displacement v2(x) along the loading path of a simpler system subjected only to a concentrated load with a fixed location. One crucial conclusion of the relation (11.38) is that the shape of the influence line is identical to the elastic or deformed curve of the loading path of the structure in the elastic state 2; this analogy is significantly useful Copyright © 2011 J. Rungamornrat

573


Influence Lines

for sketching qualitative influence lines of support reactions, see Figure 11.47 for the shape of the influence line RCY and the elastic curve of the portion ABC of the released structure. In summary, an influence line of any support reaction is qualitatively and quantitatively identical to an elastic curve of the loading path of the released structure (resulting from removing the constraint associated with the support reaction to be determined) subjected to an applied load at the released point whose direction coincides with that of the support reaction and magnitude is chosen to produce a unit movement at the same point and in the same direction. F

Y X v C2  1

v 2 (x)

C

A x

P

B

E 1 2

v (x)

IL(RCY)

Figure 11.47: Schematic indicating the shape of the influence line of RCY and the elastic curve of the loading path ABC

11.7.2.2 Influence lines for bending moments Again, for clarity in describing the principle, we first demonstrate how to construct an influence line for the bending moment at a particular point B (MB) of the model problem and its generalization to the bending moment at any location of any structure is established at the end of this section. Two elastic states of the structure are introduced, for this particular case, as follows: State 1 associated with the actual structure under the action of a moving unit concentrated load along its loading path (see Figure 11.45) and State 2 associated with the released structure (obtained by removing the bending moment constraint at the point B or, equivalently, by introducing a hinge at that point) subjected to a pair of equal and opposite moments at the released point whose direction coincides with that of the positive MB and magnitude is chosen to produce a unit relative rotation in the same direction (as the positive MB), see Figure 11.48. Note that standard sign convention for the bending moment is utilized throughout. The external virtual works W1 and W2 associated with above two elastic states can be computed as follows W1  1  v 2 (x)  M B  2BR  M B  2BL   v 2 (x)  M B (2BR  2BL )   v 2 (x)  M B

;

W 2  0

From the reciprocal theorem (i.e.,W1 = W2), it can be concluded that M B  v 2 (x)


574


Influence Lines

F

Y X M such that 2BR  2BL  1

A B

C

E Figure 11.48: Elastic state 2 for determining influence line of bending moment at point B. The magnitude of the two moments M applied to the point B is chosen such that they produce a unit relative rotation at the point B in the direction of positive MB. This relation indicates that the value of the influence line of the bending moment at point B (MB) is identical to the value of the vertical displacement (i.e. displacement in the same direction as that of the moving unit concentrated load) of the elastic state 2 at a point coincident with the location of the moving unit concentrated load in the elastic state 1. As a result, the influence line for MB can readily be obtained by simply determining the displacement v2(x) along the loading path of the corresponding released structure. The relation (11.39) also implies that the shape of the influence line is identical to the elastic curve of the loading path of the structure in the elastic state 2; this useful analogy can be applied in the sketch of qualitative influence lines of the bending moment, see Figure 11.49 for the shape of the influence line MB and the elastic curve of the portion ABC of the released structure. In summary, an influence line of the bending moment at a particular point is qualitatively and quantitatively identical to an elastic curve of the loading path of a released structure (resulting from removing the constraint associated with the bending moment to be determined) subjected to a pair of equal and opposite moments whose direction coincides with the bending moment and magnitude is chosen to produce a unit relative rotation at the same point and in the same direction. F

Y X M

2BR  2BL  1

A x

B

v 2 (x)

C

E IL(MB) v 2 (x)

Figure 11.49: Schematic indicating the shape of the influence line of MB and the elastic curve of the loading path ABC Copyright © 2011 J. Rungamornrat

575


Influence Lines

11.7.2.3 Influence lines for shear forces To construct, as an example, an influence line of the shear force at a particular point D (VD) of the model problem, we introduce following two elastic states: State 1 associated with the actual structure under the action of a moving unit concentrated load along its loading path (see Figure 11.45) and State 2 associated with the released structure (obtained by removing the shear force constraint at the point D or, equivalently, by introducing a shear release at that point) subjected to a pair of equal and opposite forces at the released point whose direction coincides with that of the positive VD and magnitude is chosen to produce a unit relative displacement in the same direction (as the positive VD), see Figure 11.50. Remark that standard sign convention for the shear force is utilized throughout. The external virtual works W1 and W2 associated with above two elastic states can be computed as follows 2 2 2 2 W1  1  v 2 (x)  VD  v DR  VD  v DL   v 2 (x)  VD (v DR  v DL )   v 2 (x)  VD

;

W 2  0

From the reciprocal theorem (i.e.,W1 = W2), it can be concluded that VD  v 2 (x)

(11.40) F

Y X

2 1 V such that v 2DR  v DL

A

D B

V

C

E Figure 11.50: Elastic state 2 for determining influence line of shear force at point D. The magnitude of the two forces V applied to the point D is chosen such that they produce a unit relative displacement at point D in the direction of positive VD. Again, the relation (11.40) indicates that the value of the influence line of the shear force at the point D (VD) is identical to the value of the vertical displacement (i.e. displacement in the same direction as that of the moving unit concentrated load) of the elastic state 2 at a point coincident with the location of the moving unit concentrated load in the elastic state 1. As a result, the influence line for VD can be constructed alternatively by determining the displacement v2(x) along the loading path of the corresponding released structure, also see Figure 11.51 for the similarity between the shape of the influence line VD and the elastic curve of the portion ABC of the released structure. The relation (11.40) also provides an analogy useful for sketching qualitative influence lines of the shear force at any point within the structure. In summary, an influence line of the shear force at a particular point is qualitatively and quantitatively identical to an elastic curve of the loading path of a released structure (resulting from removing the constraint associated with the shear force to be determined) subjected to a pair of equal and opposite forces whose direction Copyright © 2011 J. Rungamornrat

576


Influence Lines

coincides with the shear force and magnitude is chosen to produce a unit relative displacement at the same point and in the same direction.

F

Y X

2 v 2DR  v DL 1

V

2

v (x)

A

D V

B

x

C

E v 2 (x)

IL(VD)

Figure 11.51: Schematic indicating the shape of the influence line of VD and the elastic curve of the loading path ABC

11.7.2.4 Influence lines for displacements Now, let’s establish the Muller-Breslau principle for constructing an influence line of the displacement at any point within the structure. For simplicity in the demonstration, let’s construct an influence line of the vertical displacement at the point D (vD) of the model structure. In this case, following two elastic states are chosen: State 1 associated with the actual structure under the action of a moving unit concentrated load along its loading path (see Figure 11.45) and State 2 associated with the same structure subjected to a unit downward concentrated force at the point D, see Figure 11.52. Note that the positive sign convention for the displacement can be defined arbitrary as is convenient and, here, it simply follows the global coordinate system {X, Y, Z}. Also, the direction of the unit concentrated force in State 2 is chosen to direct along the negative Y-axis (similar to the moving unit concentrated load in State 1) in order to conveniently interpret the final result. F

Y X

A

1 D

B

C

E Figure 11.52: Elastic state 2 for determining influence line of vertical displacement at point D Copyright © 2011 J. Rungamornrat

577


Influence Lines

The external virtual works W1 and W2 associated with above two elastic states can be computed as follows W1  1  v 2 (x)   v 2 (x)

;

W 2  1 v D   v D

From the reciprocal theorem (i.e.,W1 = W2), it can be concluded that v D  v 2 (x)

(11.41)

The relation (11.41) indicates that the value of the influence line of the vertical displacement at the point D (vD) is identical to the value of the vertical displacement (i.e. displacement in the same direction as that of the moving unit concentrated load) of the elastic state 2 at a point coincident with the location of the moving unit concentrated load in the elastic state 1. As a result, the influence line for vD can be constructed instead by determining the displacement v2(x) along the loading path of the structure in State 2. Figure 11.53 shows the similarity between the shape of the influence line of vD and the elastic curve of the portion ABC of the structure in State 2. F

Y X 2

v (x)

A

1 D

B

x

C

E v 2 (x)

IL(vD)

Figure 11.53: Schematic indicating similarity between the shape of the influence line of vD and the elastic curve of the loading path ABC Next, we demonstrate how to construct an influence line of the horizontal displacement at the point G (uG) of the model structure by Muller-Breslau principle. In this particular case, following two elastic states are chosen: State 1 associated with the actual structure under the action of a moving unit concentrated load along its loading path (see Figure 11.45) and State 2 associated with the same structure subjected to a unit leftward concentrated force at the point G, see Figure 11.54. Again, the positive sign convention for the displacement follows the global coordinate system {X, Y, Z}. The direction of the unit concentrated force in State 2 is chosen to direct along the negative X-axis instead of the positive X-axis for convenience in the interpretation of the final result. The external virtual works W1 and W2 associated with above two elastic states can be computed as follows W1  1  v 2 (x)   v 2 (x)

;

W 2  1  u G   u G Copyright © 2011 J. Rungamornrat

578


Influence Lines

F 1

Y

G

X

A B

C

E Figure 11.54: Schematic of elastic state 2 for determining influence line of horizontal displacement at point G From the reciprocal theorem (i.e.,W1 = W2), it can be concluded that u G  v 2 (x)

(11.42)

The relation (11.42) indicates that the value of the influence line of the horizontal displacement at the point G (uG) is identical to the value of the vertical displacement (i.e. displacement in the same direction as that of the moving unit concentrated load) of the elastic state 2 at a point coincident with the location of the moving unit concentrated load in the elastic state 1. As a result, the influence line for uG can be constructed alternatively by determining the displacement v2(x) along the loading path of the structure in State 2. Figure 11.55 shows the similarity between the shape of the influence line of uG and the elastic curve of the portion ABC of the structure in State 2. F 1

Y X

G

2

v (x)

A B

x

C

E v 2 (x)

IL(uG)

Figure 11.55: Schematic indicating similarity between the shape of the influence line of uG and the elastic curve of the loading path ABC Copyright © 2011 J. Rungamornrat

579


Influence Lines

The analogies (11.41) and (11.42) provide a very convenient way to sketch qualitative influence lines of both vertical and horizontal displacements at any point within the structure. In summary, an influence line of the displacement component at a particular point is qualitatively and quantitatively identical to an elastic curve of the loading path of the same structure subjected to a unit concentrated load in the direction opposite to that of the displacement component to be determined.

11.7.2.5 Influence lines for rotations Finally, let’s deduce the Muller-Breslau principle for constructing an influence line of the rotation at any point within the structure. As an example, let’s construct an influence line of the rotation at the point A (A) of the model structure. In this case, following two elastic states are chosen: State 1 associated with the actual structure under the action of a moving unit concentrated load along its loading path (see Figure 11.45) and State 2 associated with the same structure subjected to a unit clockwise concentrated moment at the point A, see Figure 11.56. Note that the positive sign convention for the displacement and rotation follows the global coordinate system {X, Y, Z}. Also, the direction of the unit concentrated moment in State 2 is chosen to direct along the negative Zaxis for convenience in the interpretation of the final result. F

Y X

1 A B

C

E Figure 11.56: Elastic state 2 for determining influence line of rotation at point A The external virtual works W1 and W2 associated with above two elastic states can be computed as follows W1  1  v 2 (x)   v 2 (x)

;

W 2  1 A  A

From the reciprocal theorem (i.e.,W1 = W2), it can be concluded that A  v 2 (x)

(11.43)

The relation (11.43) indicates that the value of the influence line of the rotation at the point A (vA) is identical to the value of the vertical displacement (i.e. displacement in the same direction as that of the moving unit concentrated load) of the elastic state 2 at a point coincident with the location of the moving unit concentrated load in the elastic state 1. As a result, the influence line for A can be constructed conveniently by determining the displacement v2(x) along the loading path of the structure in State 2. Figure 11.57 shows the similarity between the shape of the influence line of A and the elastic curve of the portion ABC of the structure in State 2. Copyright © 2011 J. Rungamornrat

580


Influence Lines

F

Y X

G 1 A

v 2 (x)

x

B

C

E

IL(uG) v 2 (x)

Figure 11.57: Schematic indicating similarity between the shape of the influence line of A and the elastic curve of the loading path ABC The analogy (11.43) can be employed in the sketch of qualitative influence lines of rotation at any point within the structure without carrying out the comprehensive analysis. In summary, an influence line of the displacement at a particular point is qualitatively and quantitatively identical to an elastic curve of the loading path of the same structure subjected to a unit concentrated moment in the direction opposite to that of the rotation to be determined.

11.7.2.6 Final remarks While Muller-Breslau principle deduced above is demonstrated for statically indeterminate structures, it can also be equally applied to statically determinate structures. For influence lines of the displacement and rotation of statically determinate structures, the elastic state 2 only involves a statically determinate structure subjected to either a unit concentrate force or a unit concentrated moment at a location where the displacement or rotation whose influence line is to be constructed, respectively. For influence lines of support reactions and internal forces of statically determinate structures, the released structure in the elastic state 2 (after releasing a proper constraint associated with the quantity whose influence line is to be constructed) becomes statically unstable. The displacement of this released structure due to the load of a finite value (no matter how small it is) applied to the structure at the location and in the direction of a released constraint is therefore infinite. To circumvent this difficulty, we therefore use the displacement control strategy. The finite movement of the unstable structure (actually the rigid body motion) is obtained by first specifying the unit movement in the direction of a released constraint. The displacement at all other points of the structure can readily be determined since the released structure is statically unstable to the first degree of kinematical indeterminacy. It is important to remark that the process of introducing this rigid body movement requires zero external applied loads; as a result, the virtual work W2 vanishes. Note also that the virtual work W1 is the virtual work of a moving unit concentrated load in the elastic state 1 through the finite movement of the unstable structure in the elastic state 2. Copyright © 2011 J. Rungamornrat

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Influence Lines

Since the movement of the released structure in the elastic state 2 is the rigid body displacement, the deformed shape of this structure is piecewise linear. From Müller-Breslau principle, it can therefore be deduced that the influence lines of support reactions and internal forces of the statically determinate structures are always piecewise linear. On the contrary, influence lines of any response of statically indeterminate structures are, in general, not piecewise linear except structures possessing a load transferring mechanism similar to the floor systems (see section 11.5) whose influence lines consist of straight segments connecting points associated with the panel points. Finally, we remark that Müller-Breslau principle is found very useful when the qualitative (or the shape of) influence lines is needed. The principle allows such information be readily obtained via the analogy such as (11.38)-(11.43) without carrying out the comprehensive analysis. The reduction of analysis effort becomes more apparent when the complexity of structures increases. Example 11.17 Construct influence lines of the support reaction MA and REY, the bending moment at a point C (MC), the shear force at a point D (VD), the deflection at a point D (vD), and the rotation at a point D (D). The Young modulus (E) and the moment of inertia of the cross section (I) are constant throughout and a loading path is along the entire beam. U=1

x

E

C A

D

B L

L

L

L

Solution The given structure is statically indeterminate to the first degree (i.e., there exists only one redundant). Before we construct influence lines for various quantities of this beam, basic steps based on Müller-Breslau principle are summarized first as follows: (i) choose two elastic states where the state 1 corresponds to the given structure subjected to a moving unit concentrated load along its loading path (as shown in the above figure) and the state 2 corresponds to either the released structure or the same structure subjected to a special set of load depending on the quantity whose influence line is to be determined; (ii) obtain the analogy by applying the reciprocal theorem along with the two elastic states introduced in step (i). The correspondence between the influence line of the response of interest of the given structure and the deformed shape of the loading path of the structure in the elastic state 2 is deduced. Results obtained from this step yield the qualitative influence line of the response of interest provided that the deformed shape of the loading path of the structure in the elastic state 2 can be sketched qualitatively. (iii) Obtain the quantitative influence line. The deformed shape of the loading path of the structure in the elastic state 2 is computed quantitatively and, along with the analogy deduced in the step (ii), the quantitative influence line can be completely obtained. Influence line of the moment reaction MA: the elastic state 2 is chosen to be associated with the released beam (obtained by removing the moment constraint at the fixed support A) subjected to a concentrated moment at the point A whose direction coincides with that of the positive MA and magnitude is chosen to produce a unit rotation in the same direction as that of the positive MA (see the figure below). Note that the released beam for this particular case is statically determinate. Copyright © 2011 J. Rungamornrat

582


Influence Lines

x v 2 (x)

M

A

E

D

C

1 B L

L

L

L

Elastic state 2 for IL(MA) From the reciprocal theorem (i.e., W1  1 v 2 (x)  1  M A  W 2  0 ), it can be concluded that MA = v2(x). The qualitative influence line of the moment reaction, MA, can be obtained immediately from the above analogy and results are given below U=1

x

E

C A

D

B L

L

L

L

IL(MA)

Finally, the displacement v2(x) of the structure in the elastic state 2 is obtained, in terms of the applied moment M, using the principle of complementary virtual work. Final results are given by  M 3 2 2  6EIL x  3x L  8L x   M 2 v (x)   x 3  3x 2 L  8L2 x  12L3  6EIL M  3 2 2 3  12EIL x  12x L  44L x  48L 







; 0




; L < x < 2L



; 2L < x < 4L

By setting θ A  v 2 (0)  1 , we then obtain M = 3EI/4L. By inserting the value of M into v2(x), it leads to the influence function of MA:  1 3 2 2  8L2 x  3x L  8L x   1 M A (x)   2 x 3  3x 2 L  8L2 x  12L3  8L  1 3 2 2 3  16L2 x  12x L  44L x  48L 







; 0




; L < x < 2L



; 2L < x < 4L

Influence line of the reaction REY: the elastic state 2 is chosen to be associated with the released beam (obtained by removing the constraint at the roller support E) subjected to a concentrated load Copyright © 2011 J. Rungamornrat

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Influence Lines

at the point E whose direction coincides with that of the positive REY and magnitude is chosen to produce a unit displacement in the same direction as that of the positive REY (see the figure below). It is noted that this released beam is statically determinate.

x

1

B

v 2 (x)

C

A

E

D L

L

L

L

P

Elastic state 2 for IL(REY) From the reciprocal theorem (i.e., W1  1 v 2 (x)  1 R EY  W 2  0 ), it can be concluded that REY = v2(x). The qualitative influence line of the support reaction REY can be obtained immediately from the above analogy and results are given below U=1

x

E

C A

D

B L

L

L

L

IL(REY)

Finally, the displacement v2(x) of the structure in the elastic state 2 is obtained, in terms of the applied force P, using the principle of complementary virtual work. Final results are given by  R 3 2  3EI x  3x L   R 2 v (x)   x 3  3x 2 L  4L2 x  4L3  3EI  R 3 2 2 3  6EI x  12x L  28L x  16L 







; 0




; L < x < 2L



; 2L < x < 4L

By setting v E  v 2 (4L)  1 , we then obtain P = 3EI/16L3. By inserting the value of P into v2(x), it leads to the influence function of REY:


584


Influence Lines

 1 3 2 16L3 x  3x L   1 R EY (x)   x 3  3x 2 L  4L2 x  4L3 3 16L   1 3 2 2 3  32L3 x  12x L  28L x  16L 







; 0




; L < x < 2L



; 2L < x < 4L

Influence line of the bending moment MC: the elastic state 2 is chosen to be associated with the released beam (obtained by removing the bending moment constraint at the point C) subjected to a pair of equal and opposite moments at the point C whose direction coincides with that of the positive MC and magnitude is chosen to produce a unit relative rotation in the same direction as that of the positive MC (see the figure below). Again, this released beam is statically determinate.

x M

B

C

M

D 1

A L

v (x)

L

L

L

E

2

Elastic state 2 for IL(MC) From the reciprocal theorem (i.e., W1  1 v 2 (x)  1 M C  W 2  0 ), it can be concluded that MC = v2(x). The qualitative influence line of the bending moment at the point C (MC) can be obtained immediately from the above analogy and results are given below U=1

x

E

C A

D

B L

L

L

L IL(MC)

Finally, the displacement v2(x) of the structure in the elastic state 2 is obtained, in terms of the applied moment M, using the principle of complementary virtual work. Final results are given by  M 3 2  6EIL x  3x L   M 2 v (x)   x 3  3x 2 L  4L2 x  4L3  6EIL M  3 2 2 3  12EIL x  12x L  44L x  48L 







; 0




; L < x < 2L



; 2L < x < 4L


585


Influence Lines

By setting CR  CL  v 2 (2L )  v 2 (2L )  1 , we then obtain M = 3EI/4L. By inserting the value of M into v2(x), it leads to the influence function of MC:  1 3 2  8L2 x  3x L   1 M C (x)   2 x 3  3x 2 L  4L2 x  4L3  8L  1 3 2 2 3   16L2 x  12x L  44L x  48L 







; 0




; L < x < 2L



; 2L < x < 4L

Influence line of the shear force VD: the elastic state 2 is chosen to be associated with the released beam (obtained by removing the shear force constraint at the point D) subjected to a pair of equal and opposite forces at the point D whose direction coincides with that of the positive VD and magnitude is chosen to produce a unit relative displacement in the same direction as that of the positive VD (see the figure below). The released beam is apparently statically determinate.

x B

v 2 (x)

V 1

C

A

E

D V L

L

L

L

Elastic state 2 for IL(VD) From the reciprocal theorem (i.e., W1  1  v 2 (x)  1 VD  W 2  0 ), it can be concluded that VD = v2(x). The qualitative influence line of the shear force at the point D (VD) can be obtained immediately from the above analogy and results are shown below. U=1

x

E

C A

D

B L

L

L

L

IL(VD)

Finally, the displacement v2(x) of the structure in the elastic state 2 is obtained, in terms of the applied force V, using the principle of complementary virtual work. Final results are given by Copyright © 2011 J. Rungamornrat

586


Influence Lines

 V 3 2  3EI x  3x L   V x 3  3x 2 L  4L2 x  4L3  3EI v 2 (x)    V x 3  12x 2 L  28L2 x  16L3  6EI  V  x 3  12x 2 L  28L2 x  16L3  6EI







; 0


; L < x < 2L





; 2L < x < 3L





; 3L < x < 4L

By setting v DR  v DL  v 2 (3L )  v 2 (3L )  1 , we then obtain V = 3EI/16L3. By inserting the value of V into v2(x), it leads to the influence function of VD:  1 3 2  16L3 x  3x L   1 x 3  3x 2 L  4L2 x  4L3  16L3 VD (x)    1 x 3  12x 2 L  28L2 x  16L3  32L3  1  x 3  12x 2 L  28L2 x  16L3  32L3







; 0


; L < x < 2L





; 2L < x < 3L





; 3L < x < 4L

Influence line of the deflection vD: the elastic state 2 is chosen to be associated with the original beam subjected to a unit downward concentrated load (opposite to the direction of the positive deflection) at the point D as shown in the figure below. Note that the beam in the elastic state 2 is statically indeterminate to the first degree.

x B

1

v 2 (x)

C

A

E

D L

L

L

L

Elastic state 2 for IL(vD) From the reciprocal theorem (i.e., W1  1  v 2 (x)  W 2  1  v D ), it can be concluded that vD = v2(x). The qualitative influence line of the deflection at the point D can be obtained immediately from the above analogy with results shown below. U=1

x

E

C A

D

B L

L

L

L IL(vD)


587


Influence Lines

Finally, the deflection v2(x) of the structure in the elastic state 2 can be obtained by first applying the method of consistent deformation to determine the redundant and then employing the principle of complementary virtual work to post-process for the deflection. Next, by applying the above analogy, i.e., vD = v2(x), the influence function of the deflection at the point D, vD(x), can immediately be obtained. Since details of the analysis for the deflection v2(x) are quite standard (consulting Chapter8 and Chapter 10), they are not presented here for brevity but left as a good exercise for readers. Influence line of the rotation C: the elastic state 2 is chosen to be associated with the original beam subjected to a unit clockwise concentrated moment (opposite to the direction of the positive rotation) at the point C as shown in the figure below. Similar to the previous case, the beam in the elastic state 2 is statically indeterminate to the first degree.

x B

v 2 (x)

C 1

A L

E

D L

L

L

Elastic state 2 for IL(C)

From the reciprocal theorem (i.e., W1  1  v 2 (x)  W 2  1  C ), it can be concluded that C = v2(x). The qualitative influence line of the rotation at the point C can be obtained immediately from the above analogy with results shown below. U=1

x

E

C A

D

B L

L

L

L

IL(C)

Finally, the deflection v2(x) of the structure in the elastic state 2 can be obtained by first applying the method of consistent deformation to determine the redundant and then employing the principle of complementary virtual work to post-process for the deflection. Next, by applying the above analogy, i.e., C = v2(x), the influence function of the rotation at the point C, C(x), can immediately be obtained. Since details of the analysis for the deflection v2(x) are quite standard (consulting Chapter 8 and Chapter 10), they are not presented here for brevity but left as a good exercise for readers. Copyright © 2011 J. Rungamornrat

588


Influence Lines

Example 11.18 Sketch qualitative influence lines of the support reactions RAY and MG, the bending moment at a point just to the left of a point C (MCL) and the bending moment at a point D (MD), the shear force at a point D (VD), the vertical displacement at a point D (vD), and the rotation at a point C (C) of a frame shown below. Assume that the loading path is along the segment ABCEF and the axial deformation of all members is negligible.

U=1

x B

C

D

E

A

F

G Solution By applying Müller-Breslau principle, qualitative influence lines of all quantities of the given frame indicated in the problem statement can be obtained as follows (note that the structure under the action of a moving unit concentrated load shown in the problem statement is always chosen as the elastic state 1). Influence line of the reaction RAY: the elastic state 2 is chosen to be associated with the released frame (obtained by removing the vertical constraint at the pinned support A) subjected to a concentrated force P at the point A whose direction coincides with that of the positive RAY and magnitude is chosen to produce a unit displacement in the same direction as that of the positive RAY (see the figure below).

1 A

B

C

D

E F

P

G Elastic state 2 for IL(RAY) From the reciprocal theorem (i.e., W1  1  v 2 (x)  1  R AY  W 2  0 ), it can be deduced that RAY = v2(x). The qualitative influence line of the support reaction RAY can be obtained immediately from the above analogy and the result is given below. It is worth noting that the magnitude of the force P is not necessary to be computed since only the shape of the influence line is needed. Copyright © 2011 J. Rungamornrat

589


Influence Lines

U=1

x B

C

D

E

A

F G

IL(RAY)

Influence line of the moment reaction MG: the elastic state 2 is chosen to be associated with the released frame (obtained by removing the moment constraint at the fixed support G) subjected to a concentrated moment M at the point G whose direction coincides with that of the positive MG and magnitude is chosen to produce a unit rotation in the same direction as that of the positive MG (see the figure below).

D B

E

C

F

A 1 M G Elastic state 2 for IL(MG)

From the reciprocal theorem (i.e., W1  1  v 2 (x)  1  M G  W 2  0 ), it can be deduced that MG = v2(x). The qualitative influence line of the moment reaction at the fixed support G, MG, can be obtained immediately from the above analogy and the result is given below. Again, the magnitude of the applied moment M at the point G is not necessary to be computed since only the shape of the influence line is needed. Copyright © 2011 J. Rungamornrat

590


Influence Lines

U=1

x B

C

D

E

A

F

G

IL(MG)

Influence line of the bending moment MCL: the elastic state 2 is chosen to be associated with the released frame (obtained by fictitiously introducing a hinge at a point just to the left of point C) subjected to a pair of equal and opposite moment M at the hinge whose direction coincides with that of the positive MCL and magnitude is chosen to produce a unit relative rotation at the hinge in the same direction as that of the positive MCL (see the figure below).

M B A

M

1 D

C

E F

G Elastic state 2 for IL(MCL) From the reciprocal theorem (i.e., W1  1  v 2 (x)  1  M CL  W 2  0 ), it can be deduced that MCL = v2(x). The qualitative influence line of the bending moment at a point just to the left of the point C, MCL, can be obtained immediately from the above analogy and the result is given below. Again, the magnitude of the applied moment M at the hinge is not necessary to be computed since only the shape of the influence line is needed. Influence line of the bending moment MD: the elastic state 2 is chosen to be associated with the released frame (obtained by fictitiously introducing a hinge at a point D) subjected to a pair of equal and opposite moments M at the hinge whose direction coincides with that of the positive MD Copyright © 2011 J. Rungamornrat

591


Influence Lines

and magnitude is chosen to produce a unit relative rotation at the hinge in the same direction as that of the positive MD (see the figure below).

1 C

B

M

A

D

M

E F

G Elastic state 2 for IL(MD) From the reciprocal theorem (i.e., W1  1  v 2 (x)  1 M D  W 2  0 ), it can be deduced that MD = v2(x). The qualitative influence line of the bending moment at the point D, MD, can be obtained immediately from the above analogy and the result is given below. Again, the magnitude of the applied moments M at the hinge is not necessary to be computed since only the shape of the influence line is needed.

U=1

x B

C

D

A

E F

G

IL(MCL)

IL(MD)


592


Influence Lines

Influence line of the shear force VD: the elastic state 2 is chosen to be associated with the released frame (obtained by fictitiously introducing a shear release at a point D) subjected to a pair of equal and opposite forces V at the shear release whose direction coincides with that of the positive VD and magnitude is chosen to produce a unit relative displacement at the hinge in the same direction as that of the positive VD (see the figure below).

V D

C

B A

1

E F

V

G Elastic state 2 for IL(VD) From the reciprocal theorem (i.e., W1  1 v 2 (x)  1 VD  W 2  0 ), it can be deduced that VD = v2(x). The qualitative influence line of the shear force at the point D, VD, can be obtained immediately from the above analogy and the result is given below. Again, the magnitude of the applied forces V at the shear release is not necessary to be computed since only the shape of the influence line is required.

U=1

x B

C

D

A

E F

G

IL(VD)


593


Influence Lines

Influence line of the vertical displacement vD: the elastic state 2 is chosen to be associated with the same original frame but subjected to a unit downward concentrated load at the point D as shown in the figure below.

1 C

B

D

E F

A

G Elastic state 2 for IL(vD) From the reciprocal theorem (i.e., W1  1 v 2 (x)  W 2  1  v D ), it can be deduced that vD = v2(x). The qualitative influence line of the vertical displacement at the point D, vD, can be obtained immediately from the above analogy and the result is given below.

U=1

x B

C

D

A

E F

G

IL(vD)

Influence line of the rotation C: the elastic state 2 is chosen to be associated with the same original frame but subjected to a unit clockwise concentrated moment at the point C as shown in the figure below.


594


Influence Lines

1 C

B

D

E

A

F G Elastic state 2 for IL(C)

From the reciprocal theorem (i.e., W1  1  v 2 (x)  W 2  1  C ), it can be deduced that C = v2(x). The qualitative influence line of the rotation at the point C, C, can be obtained immediately from the above analogy as shown below.

U=1

x

C

B

D

E

A

F G

IL(C)

Exercises 1. Use either the direct procedure or Müller-Breslau principle to construct influence lines of support reactions, shear forces, bending moments, deflections and rotations of statically determinate beams shown below. Quantities of interest are clearly indicated in each figure. 1.1

A

B

L

C

L

D

2L

RBY, RDY, VBL, VBR, VC, VDL, VDR, MB, MC, MD, vA, vC, A, C Copyright © 2011 J. Rungamornrat

E

L

595


1.2

A

Influence Lines

B

L

L

D

C

E

L

L

F

L

RAY, RCY, REY, VCL, VCR, VD, VEL, VER, MB, MC, ME, vB, vD, vF, A, F 1.3

A

C

B

L

L

D

E

L

L

F

L

RAY, MA, REY, VB, VC, VD, VEL, VER, MB, MD, ME, vB, vD, vF, B, F 1.4

A

B

L

L

D

C

E

L

L

F

L

RAY, MA, RCY, RFY, VB, VCL, VCR, VD, VE, MC, ME, vB, vD, vE, C, E 1.5

A

B

L

L

D

C

E

L/2

L

G

F

L/2

L

RAY, RCY, RDY, RGY, VB, VCL, VCR, VDL, VDR, VE, VF, MC, MD, MF, vB, vF, C, 1.6

A

B

L

L

D

C

E

L

L

F

L

RAY, MA, RFY, MF, VB, VC, VD, VE, MC, ME, vB, vC, vD, C, E 1.7

A

B

L

L

D

C

E

L

L

F

L

RAY, RCY, MF, VB, VCL, VCR, VD, VE, MB, MC, ME, vB, vD, vE, B, C, E 1.8

A

B

L

D

C

L

L

E

L

F

L

RAY, MA, RCY, RFY, MF, VB, VCL, VCR, VD, VE, MC, vB, vD, vE, C Copyright © 2011 J. Rungamornrat

596


Influence Lines

2. Use information of influence lines obtained in Problem 1 to determine the maximum responses of all indicated quantities due to following live loads (they can be applied to any location of the beam): 2.1 Uniform load of intensity q and a point load of magnitude 3qL 2.2 A pair of axle loads moving from the left to the right of the beam 2P

P

L/4 2.3 A series of axle loads moving from the left to the right of the beam 2P

P

P

L/2

L/4

3. Use Müller-Breslau principle to construct influence lines of support reactions, shear forces, and bending moments of following statically determinate beams containing some loading panels. Quantities of interest are clearly indicated in each figure.

3.1

B

C

E

D

F

A

G L

L

L

L

L

L

RAY, RGY, VB, VCL, VCE, VER, VF, MB, MC, MD, ME, MF

3.2

B

C

D

E

F

A

G L

L

L

L

L

L

RAY, RGY, VBL, VBC, VCE, VEF, VFR, MB, MC, MD, ME, MF

3.3

B

C

D

E

F

A

G L

L

L

L

RAY, RGY, VB, VCL, VCE, VEF, MB, MC, MD, ME, MF Copyright © 2011 J. Rungamornrat

L

L

597


3.4

B

Influence Lines

C

E

A

F

D L

L

L

L

G L

L

RAY, RDY, RGY, VB, VCL, VCD, VDF, VFR, MB, MC, MD, MF 3.5

B

C

E

A

F

D L

L

L

L

G L

L

RAY, RDY, RGY, VB, VCL, VCD, VDF, VFR, MB, MC, MD, MF 3.6

B

C

G

E

A

D L

L

F L

L

L

L

RAY, MA, RDY, RFY, VBC, VCD, VDR, VE, VFL, VFR, MC, MD, MF

3.6

B

C

G

E

A

D L

L

F L

L

L

L

RAY, MA, RDY, RFY, VBL, VBC, VCD, VDE, VER, VFL, VFR, MC, MD, MF 4. Use information of influence lines obtained in Problem 3 to determine the maximum responses of all indicated quantities due to following live loads (they can be applied to any location of the beam): 4.1 Uniform load of intensity q and a point load of magnitude 3qL 4.2 A pair of axle loads moving from the left to the right of the beam 2P

P

L/4 4.3 A series of axle loads moving from the left to the right of the beam 2P

L/4

P

P

L/2 Copyright © 2011 J. Rungamornrat

598


Influence Lines

5. Use Müller-Breslau principle to construct influence lines of support reactions, shear forces, and bending moments of statically determinate floor systems shown below. Quantities of interest are clearly indicated in each figure.

5.1

B

H

C

F

D

A

G

E L/2

L

L

L/2

L

L

L

RAY, REY, VBC, VCD, VDE, VEF, VFG, MB, MC, MD, ME, MF, MH

5.2

B

E

C

A

H

F G

D L

L

L/2

L

L/2

L/2

L/2

L

RAY, RDY, RGY, VBC, VCD, VDE, VEF, VFG, MB, MC, MD, ME, MF, MH

5.3

B

E

H

C

A

F G

D L

L

L/2

L

L/2

L

L

RAY, MA, RDY, RGY, VAB, VBD, VDE, VEF, VFG, MB, MD, ME, MH

5.4

B

D

C

F

H

E

G

A L

L

L/2

L

L

L/2

L

RAY, MA, RGY, MG, VAB, VBC, VCD, VDE, VEF, VFG, MB, MD, MF, MH

5.5

B

C

D

G

F

A

I

H

J

L

E L

L

L

L

M

K L

L

L

L

L

L

L

L

RAY, REY, RKY, VAC, VDE, VEF, VGI, VJK, VKL, MB, MD, ME, MG, MJ, MK


599


5.6

B

C

Influence Lines

D

F

A

G

J

H

E L

L

L

M

I L

L

L

K

L

L

L

L

L

L

L

RAY, REY, RIY, RMY, VBD, VEF, VFH, VHI, VJL, MB, MC, ME, MH, MI, MJ

5.7

B

A

D

C

F

G

J

H

E L

L

L

M

I L

L

L

K

L

L

L

L

L

L

L

RAY, MA, REY, RIY, RMY, VBD, VEF, VFH, VHI, VJL, MB, ME, MH, MI, MJ 6. Use information of influence lines obtained in Problem 5 to determine the maximum responses of all indicated quantities of the floor systems due to following live loads (they can be applied to any location of the structure): 6.1 Uniform load of intensity q and a point load of magnitude 3qL 6.2 A pair of axle loads moving from the left to the right of the floor system 2P

P

L/4 6.3 A series of axle loads moving from the left to the right of the floor system 2P

P

P

L/4

L/2

7. Use direct procedure or truss-floor system analogy to construct influence lines of support reactions and member forces of statically determinate trusses shown below. Quantities of interest are clearly indicated in each figure. H

7.1

h C

B h

L

K

U=1

x

A

J

I

h

h

F

E

D h

RAY, RFY, FIJ, FCJ, FCD, FCI, FJD, FLG, FLF Copyright © 2011 J. Rungamornrat

h

G h

600


Influence Lines

H

7.2

J

I U=1

x

A

h C

B h

L

K

E

D

h

h

h

G

F h

h

RAY, RGY, FIJ, FCJ, FCD, FCI, FAH, FAB, FHI I

H

7.3

K

J

h

B h

N

U=1

x

A

M

L

C h

h

h

G

F

E

D

h

h

RBY, RFY, FBI, FJK, FCK, FCD, FKD, FAH, FFM U=1

x 7.4

F

G

I

H

L

K

J

h B

A h

h

D

C h

h

E h

h

RFY, RLY, FAF, FHI, FBI, FBC, FIC, FAG U=1

x 7.5

E

F

G

K

J

I

H

h

h

C

B

A h

h

h

REY, RKY, FFG, FAG, FAB, FBH, FGH, FBC Copyright © 2011 J. Rungamornrat

D h

h

601


Influence Lines

7.6

J h/3

K

I H U=1

x

A

h/3

L

h C

B h

E

D

h

h

h

G

F h

h

RAY, RGY, FBC, FCH, FHI, FCD, FDI, FIJ, FJD 7.7

I

H

J U=1

x

A

h C

B h

M

L

K

h

h

F

E

D h

h

G h

RAY, RCY, RFY, FBC, FBJ, FIJ, FCJ, FCD, FDJ, FLF, FLG 7.8

J h H

I U=1

x

A

h C

B h

L

K

h

D h

E h

G

F h

h

RAY, RDY, RGY, FHI, FBI, FBC, FIJ, FCD, FDI 8. Use information of influence lines obtained in Problem 7 to determine the maximum responses of all indicated quantities of the trusses due to following live loads (they can be applied to any location of the structure): 8.1 Uniform load of intensity q and a point load of magnitude 3qL 8.2 A pair of axle loads moving from the left to the right of the truss Copyright © 2011 J. Rungamornrat

602


Influence Lines

2P

P

h/4 8.3 A series of axle loads moving from the left to the right of the truss 3P

2P

P

h/2

h/4

9. Use direct procedure or Müller-Breslau principle to construct both qualitative and quantitative influence lines of support reactions, internal forces, and displacements and rotations of statically indeterminate structures shown below. Quantities of interest are clearly indicated in each figure. EI constant

9.1

B

F

D

C

A

G

E L

L

L

L

L

L

RAY, REY, RGY, VC, VEL, VER, VF, MC, ME, MF, vC, vF, A, D EI constant

9.2

B

A

E

C

F

D L

L

L

3L/2

3L/2

RAY, MA, RDY, RFY, VB, VC, VDL, VDR, VE, MB, MD, ME, vC, vE, D, F 9.3

EI constant D

B A

F

C L

L

L

G

E L

L

L

RAY, RCY, REY, RGY, VCL, VCR, VB, VD, MB, MC, MD, vB, vD, A, C 9.4

I

H

N h

D

C B EA constant h

M

L

U=1

x

A

K

J

h

h

E h

RAY, RDY, RGY, FBC, FBJ, FIJ, FCD, FCK, FJK, FJC, FDK Copyright © 2011 J. Rungamornrat

G

F h

h

603


9.5

Influence Lines

I

H

K

J

N

U=1

x

A

M

L

h

C B EA constant h

E

D

h

h

G

F

h

h

h

RAY, RGY, FBC, FBJ, FIJ, FCI, FJC, FCK, FJK, FCD 9.6

B

D

C

L

EI constant, axially rigid Loading path along BCD

L

EI constant, axially rigid Loading path along DEF

A 1.5L

1.5L

RAX, RAY, MA, RDY, VC, MB, MC, uB, vC, B, C, D 9.7

D

A

F

H

E

G

B L/2

L/2

C L/2

L/2

RAX, RAY, RBY, RCY, VDR, VG, VEL, VER, VH, MD, MG, MEL, MER uD, vG, vH, D, E, H 10.Use Müller-Breslau principle to sketch qualitative influence lines of indicated quantities of statically indeterminate structures shown below and then identify the location to place the uniformly distributed live load to produce the maximum responses. 10.1

A

B

C

D

E

F

G

RAY, MA, RCY, REY, RGY, VCL, VCR, VEL, VER, MB, MC, MD, ME, MF, vB, vD, C, E 10.2

A

B

C

D

E

F

RAY, RCY, REY, RGY, VB, VCL, VCR, VD, MB, MC, vB, vD, A, B, C Copyright © 2011 J. Rungamornrat

G

604


Influence Lines

10.3

A

C

B

E

D

G

F

I

H

RHX, RHY, MH, RIX, RIY, MI, VAR, VB, VCL, VCR, VD, MAR, MB MCL, MCR, MD, vB, vD, A, B, C 10.4

A

C

B

E

D

G

F

I

H

RHX, RHY, MH, RIX, RIY, MI, REY, RGY, VAR, VB, VCL, VCR, VEL, VER MAR, MB, MCL, MCR, MD, MEL, MER, MF, vB, vD, vF, A, B, C, E, G 10.5 H

I

K

J

M

L

N h

A

D

C B EA constant h

h

h

E h

RAY, RDY, RGY, FBC, FBJ, FIJ, FCD, FCK, FJK, FJD, FDK


G

F h

h


R-1

References

REFERENCES Argyris J.H., Energy Theorems and Structural Analysis, 1st Edition, Springer, 1995. DaDeppo D.A., Introduction to Structural Mechanics and Analysis, 1st Edition, Prentice-Hall, 1998. Danmongkoltip P. “Large curvature analysis of inextensible nonlinear elastic beams”. Master thesis, Chulalongkorn University, Thailand, 2009. Danmongkoltip P., Rungamornrat J. “Large curvature analysis of nonlinear elastic inextensible, cantilever beam”. Proceedings of the 14th National Convention on Civil Engineering, Nakornrachasima, May 13-15, 2009. Douanevanh J. “Large curvature analysis of two-dimensional linear elastic frames by direct stiffness method”. Master thesis, Chulalongkorn University, Thailand, 2011. Douanevanh J., Rungamornrat J., Thepvongsa K., Wijeyewickrema A.C. “Semi-analytical technique for large displacement and rotation analysis of 2D frames”. Proceedings of The Janpan-Thailand-Laos P.D.R. Joint Friendship International Conference on Applied Electrical and Mechanical Engineering 2011, Nhongkai, Thailand, September 21-23, 2011. Durka F., Structural Mechanics, 7th Edition, Prentice-Hall, 2010. French S.E., Fundamentals of Structural Analysis, 1st Edition, Delmar, 1995. Gere J.M. and Goodno B.J., Mechanics of Materials, 7th Edition, CL-Engineering, 2008. Hibbeler R.C., Mechanics of Materials, 8th Edition, Prentice-Hall, 2010. Hibbeler R.C., Structural Analysis, 7th Edition, Prentice-Hall, 2008. Hjelmstad K.D., Fundamentals of Structural Mechanics, 2nd Edition, Springer, 2010. Kassimali A., Structural Analysis, 4th Edition, CL-Engineering, 2010. Leet K.M., Uang C.M., and Gilbert A., Fundamentals of Structural Analysis, 4th Edition, McGrawHill, 2010. McCormac J.C., Structural Analysis: Using Classical and Matrix Methods, 4th Edition, Wiley, 2006. Megson T.H.G., Structural and Stress Analysis, 2nd Edition, Butterworth-Heinemann, 2005. Nayfeh A.H. and Pai P.F., Linear & Nonlinear Structural Mechanics, 1st Edition, Wiley, 2002. Pinyochotiwong Y., Pinitpanich M., Rungamornrat J. “Analysis of nonlinear elastic beams by direct stiffness method”. Proceedings of the 14th National Convention on Civil Engineering, Nakornrachasima, May 13-15, 2009 Rajan S.D., Introduction to Structural Analysis & Design, 1st Edition, Wiley, 2000. Reddy J.N., Energy Principles and Variational Methods in Applied Mechanics, 2nd Edition, Wiley, 2002. Rungamornrat J., Jamnongpipatkul A., Boon-intra S. “Comparison study of truss analysis with consideration of various nonlinear effects”. Proceedings of the 13th National Convention on Civil Engineering, Chonburi, May 14-16, 2008. Rungamornrat J., Tangnovarad P. “Analysis of linearly elastic inextensible frames undergoing large displacement and rotation”. Mathematical Problems in Engineering, 2011, vol. 2011, Article ID 592958, 37 pages, doi:10.1155/2011/592958. Tangnovarad P., “Large curvature analysis of linear elastic, inextensible structures by a direct stiffness method”. Master thesis, Chulalongkorn University, Thailand, 2008. Tangnovarad P., Rungamornrat J. “Large curvature analysis of linear elastic inextensible structures by a direct stiffness method”. Proceedings of the 13th National Convention on Civil Engineering, Chonburi, May 14-16, 2008.



R-2

References

Tangnovarad P., Rungamornrat J., “Large curvature analysis of linear elastic, inextensible structures by direct stiffness method”. Proceedings of the Twenty-Second KKCNN Symposium on Civil Engineering, Chiangmai, Thailand, October 31-November 2, 2009. Tartaglione L.C., Structural Analysis, McGraw-Hill, 1991. Watcharakorn N. “Determination of buckling load by adaptive shape functions”. Master thesis, Chulalongkorn University, Thailand, 2010. Watcharakorn N., Atthavongpisarn B., Rungamornrat J. “Estimation of elastic buckling load using adaptive basis functions”. Proceedings of the 14th National Convention on Civil Engineering, Nakornrachasima, May 13-15, 2009. Watcharakorn N., Rungamornrat J. “Determination of buckling load by adaptive shape functions”. Proceedings of the 16th National Convention on Civil Engineering, Pattaya, Chonburi, May 18-20, 2011. West H.H. and Geschwindner L.F., Fundamentals of Structural Analysis, 2nd Edition, Wiley, 2002. Williams A., Structural Analysis: In Theory and Practice, 1st Edition, Butterworth-Heinemann, 2008.


I-1


Index

INDEX A admissible structure analogy applied load arch axial deformation axial effect axial force axial force diagram axial member axial release axial rigidity

422, 426, 430 222, 261-264, 272-273, 278, 287, 485, 536, 550 10, 20-21, 49 3, 24 215, 219, 249 347, 349, 380, 408-409 60, 117 121, 251, 254, 467 3, 18, 23, 367 35 353, 411

B beam beam joint beam member bending effect bending moment bending moment diagram boundary conditions boundary value problem

3, 9, 16, 24 33, 37 16, 37 211, 347, 349, 377 19, 80, 116 78, 84, 91, 223-231 143, 149-156, 176-177, 198-199, 211, 220, 330 143, 154, 156-157, 198

C coefficient of thermal expansion compatible compatibility equation complimentary strain energy complimentary virtual strain energy complimentary virtual work complimentary work compound joint concentrated load concurrent forces conjugate segment conjugate structure connection consistent deformation constitutive law continuity coordinate system coordinate transformation cross sectional area curvature curvature diagram

369 322-323, 330, 332, 337-338, 343, 366-367 423-425, 437, 441, 443 309, 311, 330 312-313, 321, 343 307-309, 312, 321, 332 303, 305, 307-309, 311, 313, 407, 423 33-34, 37 81, 92, 190, 220, 343, 361, 409, 445, 480, 501 516, 525, 536, 564, 572 23, 32, 64, 72 265, 273-274, 279, 285, 287, 291 261, 271, 278, 287, 437, 444, 486 3, 7-8, 17, 32, 33 437, 441, 443, 587 20, 146, 303, 330, 337, 368, 563 11, 61, 79, 114, 120, 213, 278, 572 11, 261, 278 249, 367, 380, 382 3, 20, 106, 127, 144, 211, 213,215, 218, 260, 437 211, 213-214, 223-224, 227, 278, 288


I-2


Index

D dead load deflection deformation

displacement method distributed load

482 3, 106-107, 143, 145, 148-151, 177, 220-221, 361, 424 6, 19, 81, 108, 130, 148, 215, 252, 312, 353, 389 440, 475, 505, 570, 591 6, 108, 147, 214, 265, 318, 343 109, 146, 215, 317, 342, 368, 410 17, 32, 224, 328, 342, 448 34, 41, 48, 66, 440, 445, 451, 466, 475 29, 501, 503, 505, 580 31, 52, 101, 151, 201, 253, 319, 369, 398, 414, 451 462, 504, 568, 587 143, 156-157, 176, 198 84, 89, 92-94, 121, 124-127, 176-177, 188, 190 176, 188, 192 13, 17-19, 32 6, 47, 115, 131, 201, 214, 250, 292, 342, 387, 446 504, 566, 582 23, 201, 341 87, 92, 107, 158, 228, 330, 463, 484

E element elongation equilibrium equation error from fabrication excitation external loads external virtual work

13, 23, 83, 125, 148, 312, 351 7, 19, 23, 120, 312, 329, 343, 383 20-24, 31, 50, 63, 71, 89, 123, 143, 263, 281 488 4, 13, 53, 214, 480 25, 35, 245, 278, 308, 343, 386, 394, 419, 470, 551 315-317, 340, 370, 372, 575, 577

deformed configuration deformed state degree of freedom degree of static indeterminacy degree of kinematical indeterminacy determinate direct integration discontinuity discontinuity-function discrete structure displacement

F fictitious section fixed support flexible support flexural member flexural rigidity floor system force method form factor frame frame joint frame member free body diagram

548, 550-551 6, 112 7 18-19 148, 154, 211 479, 524-527, 545-549 20, 21 351, 376, 378-379 3, 9, 19, 25 33, 37 16, 19 24, 49, 120

G geometry girder global coordinate system governing equation guided support

2-3, 9-11, 211, 322, 376 524-527 11, 114, 213, 278, 572, 579 20, 148-149, 337, 563 6, 51 Copyright © 2011 J. Rungamornrat

I-3


Index

H Heaviside function hinge hinged support hinge joint I idealization idealized structure indeterminate inextensibility inflection point influence function influence line internal force internal release internal virtual work

188 34-35 6 7

1-2, 41, 212 1-3, 5, 8-10, 24-25, 27, 60, 78, 113 28-31, 37-42, 54, 149, 198, 218, 321 338, 344, 407, 422, 437, 485, 563, 570 144, 214, 220, 249 106-107, 128 480-481, 486-497, 501, 539, 563, 570 337, 479, 485, 501, 516, 524, 536, 563-564, 579 4, 9, 16, 18-19, 21, 32, 35, 49, 60 78, 113, 315, 348, 435, 564 14, 26, 31, 33, 35-36, 119, 129, 261 316, 339

J joint joint equilibrium

3, 8, 32, 63, 78, 113, 278, 352, 536 32

K kinematics kinematical indeterminacy kinematical quantities kinematically determinate kinematically indeterminate L lack of fit length constraint linearized kinematic live load load loading panel local coordinate system longitudinal loading M mathematical model mechanism member member deformation member load method of joints method of sections

6, 19, 32, 108, 146, 215, 306, 326, 351, 566 27, 32, 504, 506, 583 17, 23, 221, 234, 251, 446 32, 34 32, 34

368 215, 219, 221, 249-250, 261, 263-265, 274, 278 145, 212, 570 482-483 1, 5, 9, 18-20, 23-26, 33, 49-50, 53, 60, 78, 122, 143-144 154, 211, 305-306, 361, 376, 407, 423, 437, 440, 479 516-518, 525, 527, 538-539, 547 11, 114-115, 120, 123, 213, 278 9

1-2 24-26, 43-45, 525, 536-537, 539, 548, 564-565, 581 3, 7, 10, 32, 38, 40, 60, 78, 113 16, 211 18 63, 70 70, 72, 78, 83, 89, 113, 120, 485 Copyright © 2011 J. Rungamornrat

I-4


Index

method of superposition moment moment of inertia moment release moving loads Müller-Breslau principle

223, 438, 440, 445, 480 4, 6, 17, 20, 34, 49, 80, 94, 107, 211, 261, 501 107, 147, 348, 350, 376 34, 78, 107, 503 479, 536 501, 508, 570, 576, 579

N nodal point node nonlinearity

367 7, 10, 31, 40, 261 77, 147

P panel point partially rigid joint pinned support Poisson ratio primary structure principle of complementary virtual work principle of real work principle of virtual work R real work real work equation reciprocal theorem redundancy redundant reference coordinate system relative end rotation released structure rigid body motion rigid joint roller support rotation

516-518, 524-528, 536-540, 548 11 20, 58, 64, 81, 114, 135, 246, 298, 375, 414, 433 450, 554, 563 388, 391 201, 204, 209, 425, 430, 440, 451, 464, 477 306, 324, 335, 346, 368, 380, 389, 402, 442, 464 496, 585, 590 350, 364, 367 306, 315, 318, 324, 340, 350, 504, 509

308, 314, 347-348, 352, 361 314, 347-348, 352, 361 303, 337, 570, 575 31, 564 31, 198, 422, 437, 441 11, 61, 79, 114, 222, 278 16, 215 438-443, 572, 575 16, 316, 443, 501, 503, 505, 580 7, 33, 113, 132 5, 17, 61, 78, 110, 151, 422 3, 11, 14, 19, 34, 60, 127, 143, 177, 211, 261, 278 342, 365, 409, 437, 485

S section shear deformation shear effect shear force shear force diagram shear modulus shear release shear rigidity

20, 38, 71, 73, 212, 349, 351 504, 506-507, 548, 553, 563 105, 127, 143, 212, 250

347, 351, 378, 408 4, 9, 17, 33, 78, 80, 90, 92, 114, 124, 162, 278 316, 351, 378, 474, 486, 501, 505, 548, 575 78, 84, 90, 123, 383, 393, 458, 465 4, 348, 378, 382, 391, 397 35, 38, 54, 107, 129, 176, 213, 318, 505, 575, 592 406


I-5


Index

statically admissible statically determinate

330-332 28, 30, 38, 49, 60, 70, 78, 113, 120, 198, 221, 347 365, 409, 437, 479, 485, 501, 507, 524, 580 28, 30, 39, 54, 149, 198, 218, 321, 338, 407 422, 437, 479, 518, 563, 570 24, 27, 37, 42, 198, 265, 339, 347, 365, 407, 443 25, 42, 320, 443, 501, 503, 580 5, 20-21, 30, 50, 54, 63, 70, 83, 89, 92, 122, 199 212, 218, 236, 261, 352, 409, 437, 485, 545, 563 24, 28, 31, 37, 63, 198, 422, 437, 563 14, 17, 20, 28, 49, 61, 198, 338, 353, 407, 437, 508 24, 27, 41, 43, 45 4, 16, 20, 145, 146, 212, 249, 250, 303 313, 322, 323, 326, 330, 337, 365 303, 309, 312, 347-351 311-313, 323, 330, 333-334, 349-351 4, 16, 18, 49, 80, 116, 146, 212, 309, 311 313, 322, 330, 337, 342, 366, 376, 380 1, 2, 8, 10, 14, 21, 24-26, 28, 29, 38 49, 53, 113, 198, 211, 261, 263, 271 278, 287, 366, 376, 409, 422, 441, 563 264, 438, 440-441 5, 6, 24, 54, 61, 78, 94, 111, 113 150, 151, 224, 343, 422, 556 5, 6, 17, 28, 31, 49, 54, 62, 80, 117, 342 407, 437, 481, 501, 503, 508, 563, 572 331, 343, 366, 384, 407, 423

statically indeterminate statically stable statically unstable static equilibrium static indeterminacy static quantities static stability strain strain energy strain energy density stress structure superposition support support reaction support settlement T temperature change transformation matrix translation transverse displacement transverse load truss truss joint torque torsional constant torsional effect total potential energy total complementary potential energy U undeformed configuration undeformed state unit load unit load method unit moving load unit-step function

1, 368 13 3, 6, 154, 508 127, 144, 212 9, 23, 143 3, 7, 9, 38, 60, 367, 536 32, 537 4, 17, 49, 348, 350 348, 350, 376, 382 350, 377, 380-381, 408-409 303, 323-325 330, 332

3, 10, 14, 20, 144, 211, 213, 249, 262, 272, 278, 289 106, 143, 145, 212, 314, 338, 365, 407 366, 439, 444-446 366, 444, 450 479-480 188


I-6


Index

V virtual force virtual reaction virtual strain virtual strain energy virtual work virtual work equation

308-309, 342, 366, 376 367 303, 312-313 303, 312, 330, 339, 340, 502 303, 307, 308, 312, 315, 330, 501, 570, 580 316, 501

Y Young modulus

128, 368, 377, 380



A-1

Acknowledgement

ACKNOWLEDGEMENT

The author would like to express his greatest gratitude to everyone who advised, supported, and helped him to complete this book. First of all, he would like to gratefully acknowledge the Department of Civil Engineering, Faculty of Engineering, Chulalongkorn University for providing him the great opportunity in teaching of fundamental courses in structural analysis since the first year he was appointed as a faculty member. The author would also like to express his deepest appreciation to Mr. Nithi Nikornpakorn and several of his students (Mr. Panupong Boonphennimit, Mr. Mattana Pinitpanich, Miss Porjan Tuttipongsawat, Mr. Thanachote Techawongsakorn, Mr. Natdanai Sinsamutpadung, Miss Varanich Chinprapinporn, Miss Sitanan Tanyasakulkit, Mr. Suttinan Sakulpinyo, Mr. Jessada Sresagoolchai) for their invaluable help in proofreading and verifying various examples appearing throughout this book. Furthermore, special thanks to his sister-in-law, Wisa A. Kompayak, for her beautiful design of the book cover. Last but not least, the author would like to express his sincere gratitude to all family members, especially his wife and beloved son, for their encouragement and endless love. Without their continuous support, the author would never finish this writing.


Fundamental Structural Analysis (2)

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