Fundamental Structural Analysis.8765.1408344506

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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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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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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

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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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

143

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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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

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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

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365 365 367 376

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Exercises

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Chapter 9 APPLICATIONS OF CASTIGLIANO’S 2nd THEOREM nd

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

Chapter 10 METHOD OF CONSIST DEFORMATION 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

407 407 409 422 432

437 437 441 443 475

479 479 485 501 516 524 536 563 594

REFERENCE

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INDEX

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ACKNOWLEDGEMENT

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Copyright © 2011 J. Rungamornrat

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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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Introduction to Structural Analysis

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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Introduction to Structural Analysis

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

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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

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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

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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

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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

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(a)

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(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

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     

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.

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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

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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

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      

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

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Introduction to Structural Analysis

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

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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.

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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

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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.

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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

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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

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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

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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

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N

i=1

i=1

N

N

i=1

i=1

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Analysis of Determinate Structures

(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

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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

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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

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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

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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

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Analysis of Determinate Structures

[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

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Analysis of Determinate Structures

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).

Copyright © 2011 J. Rungamornrat

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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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

0

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Analysis of Determinate Structures

 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

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Analysis of Determinate Structures

 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

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Analysis of Determinate Structures

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

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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

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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

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

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

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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

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Analysis of Determinate Structures

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

[FY = 0]  

:

69

Analysis of Determinate Structures

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)

Copyright © 2011 J. Rungamornrat

F13

45o 3

F35

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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

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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}

Copyright © 2011 J. Rungamornrat

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Analysis of Determinate Structures

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

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Analysis of Determinate Structures

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

Copyright © 2011 J. Rungamornrat

Y X

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Analysis of Determinate Structures

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

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Analysis of Determinate Structures

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

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Analysis of Determinate Structures

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

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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

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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

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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

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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

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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

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 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

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 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

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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

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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

Copyright © 2011 J. Rungamornrat

L+

L

RAY – VCR – qL = 0

qL

RAM

qL2 A

RAY

B

VCR MCR CR

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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

Copyright © 2011 J. Rungamornrat

E L

X

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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

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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

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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)



Copyright © 2011 J. Rungamornrat

dM(x) dx



V(x)

(2.16)

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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

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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

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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

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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

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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

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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

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 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

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 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

Copyright © 2011 J. Rungamornrat

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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

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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

Copyright © 2011 J. Rungamornrat

L

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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

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Analysis of Determinate Structures

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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Analysis of Determinate Structures

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

(2.27) Copyright © 2011 J. Rungamornrat

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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

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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

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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

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-

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

Copyright © 2011 J. Rungamornrat

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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

Copyright © 2011 J. Rungamornrat

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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

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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

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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

(2.46) Copyright © 2011 J. Rungamornrat

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(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

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 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

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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

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 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

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-

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

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 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

Copyright © 2011 J. Rungamornrat

uD C

D

D

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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

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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

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 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

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Analysis of Determinate Structures

 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

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Analysis of Determinate Structures

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

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Analysis of Determinate Structures

 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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

140

Analysis of Determinate Structures

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

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141

Analysis of Determinate Structures

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

L Copyright © 2011 J. Rungamornrat

2L

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142

Analysis of Determinate Structures

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

2L Copyright © 2011 J. Rungamornrat

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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

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Direct Integration Method

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

(3.4) Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

cos   1 

145

Direct Integration Method

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 : (x) 

d2v dx 2

(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)

(3.10) Copyright © 2011 J. Rungamornrat

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Direct Integration Method

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 

Copyright © 2011 J. Rungamornrat

(3.15)

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Direct Integration Method

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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Direct Integration Method

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  d2v    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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149

Direct Integration Method

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  d2v   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

Copyright © 2011 J. Rungamornrat

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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)

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

v(L)  0 and

v (L)  0

151

Direct Integration Method

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

d2 v (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  d2 v   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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Direct Integration Method

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(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) (L)  k s v(L)  0 v(L)  0

(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

Copyright © 2011 J. Rungamornrat

ks

(EIv) (L)  k s v(L)  0

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Direct Integration Method

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

d2 v (0)   M 0 dx 2

EI

d2v (L)  M 0 dx 2

and and

d  d2 v   EI  (0)  P0 dx  dx 2  d  d2 v   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  d2v   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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Direct Integration Method

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 d2v 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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Direct Integration Method

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

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Direct Integration Method

EI

d2v q(L  x) 2   dx 2 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) 2    q(x)dx  C1 dx  dx 

(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  C 2 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)

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Direct Integration Method

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

d2 v  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  C1 x  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)

Copyright © 2011 J. Rungamornrat

(3.49)

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Direct Integration Method

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  

EIv(x)     V(x)dxdxdx 

(3.51b)

1 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 d4 v x EI 4  q 0   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

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Direct Integration Method

Performing direction integrations of (3.1.1) yields n

n+1

EI

q L x  d3 v 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 

 C1

(e3.1.6)

n+1

n+2

 C1x  C 2

n+2

q 0 L2 q 0 L3 dv 1 x x 2 EI    dx  C1 x  C 2 x  C3    dx (n+1)(n+2)  L  2 (n+1)(n+2)(n+3)  L  EIv 

q 0 L4 x   (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  C1 x 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) 

2  x  n+3 1 q 0 L3 dv 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

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

160

Direct Integration Method

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

2

[MA = 0]  

:

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

[Mx = 0]  

:

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

n

x

 M(x) 

 x  q 0 L2  (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:  x  n+2  q 0 L2 d2v x EI 2     (n+2)  n +1 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  

 x  q 0 L3   (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

   C1x  C 2 

(e3.1.14)

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

161

Direct Integration Method

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

x R AY   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: EI

n+1  d 3 v q 0 L  x      1 3 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 n+1   x  n+2 q 0 L  x  q 0 L2 d2v x  EI 2      1dx  C1     (n+2)   C1 dx n+1  L  (n+1)(n+2)  L  L  

EI

(e3.1.19)

 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

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

n+4

2

2

   C1 x  C 2 

(e3.1.20)

 1 2 dx  C1 x  C 2 x  C3 2 

3 1  x   1  (n+2)(n+3)(n+4)     C1 x 2  C 2 x  C3 6  L   2

Copyright © 2011 J. Rungamornrat

(e3.1.21)

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

162

Direct Integration Method

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

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

EI(x) 

X

V(x) 

q = q0(x/L)2 2

EI, L

6 3 2 q 0 L4  x  x  x      20    45    360  L  L  L  

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

EI(x) 

X

q 0 L3 24

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

EIv(x) 

Y

5 3 2 q 0 L4  x  x  x      10    20    120  L  L  L  

V(x) 

q 0 L3 60

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

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FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

163

Direct Integration Method

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

Copyright © 2011 J. Rungamornrat

(e3.2.7) n+3

1  C1x 2  C 2 x  C3 (e3.2.8) 2

(e3.2.9)

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

164

Direct Integration Method

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)

By substituting {C1, C2, C3, C4} into (e3.2.6)-(e3.2.9), 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  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  

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

165

Direct Integration Method

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

166

Direct Integration Method

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

Copyright © 2011 J. Rungamornrat

(e3.2.21)

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

167

Direct Integration Method

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 

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

168

Direct Integration Method

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)

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

169

Direct Integration Method

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

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: V(x)  EI

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

170

Direct Integration Method

[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

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

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)

Two constants of integration {C1, C2} can be obtained from boundary conditions (e3.3.11) and (e3.3.12) as follows: v(0)  0



C2  0

v(L)  0



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

By substituting {C1, C2} into (e3.3.13)-(e3.3.14), 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]  

:

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

(e3.3.18) Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

171

Direct Integration Method

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

172

Direct Integration Method

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

V(x)  EI

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

173

Direct Integration Method

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

174

Direct Integration Method

[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 

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

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)

Two constants of integration {C1, C2} can be obtained from boundary conditions (e3.4.11) and (e3.4.12) as follows: v(0)  0



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 

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FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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Direct Integration Method

By substituting {C1, C2} into (e3.4.13)-(e3.4.14), 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]  

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 

Copyright © 2011 J. Rungamornrat

(e3.4.21)

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176

Direct Integration Method

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

177

Direct Integration Method

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

Copyright © 2011 J. Rungamornrat

C

X

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

178

Direct Integration Method

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

Copyright © 2011 J. Rungamornrat

(e3.5.5a)

(e3.5.5b)

(e3.5.5c)

(e3.5.5d)

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179

Direct Integration Method

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         

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

180

Direct Integration Method

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)

Boundary conditions: v(0)  0

(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)

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

2EIv  

q 0 L4 24

q L4 EIv  0 3

181

Direct Integration Method

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)

Boundary conditions: v(0)  0

(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)

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

182

Direct Integration Method

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

183

Direct Integration Method

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

184

Direct Integration Method

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)

Boundary conditions: v(0)  0

(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





Copyright © 2011 J. Rungamornrat

(e3.6.8b) (e3.6.8c) (e3.6.8d)

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

v(3L/4 )  0



v(L/2 )  v(L/2 )

185

Direct Integration Method

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

186

Direct Integration Method

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)

Copyright © 2011 J. Rungamornrat

(e3.7.1) (e3.7.2)

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

187

Direct Integration Method

Boundary conditions: v(0)  0

(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)

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

188

Direct Integration Method

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 

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FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

189

Direct Integration Method

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

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

190

Direct Integration Method

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 )

(3.66) Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

191

Direct Integration Method

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 )

Copyright © 2011 J. Rungamornrat

(3.72)

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

192

Direct Integration Method

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

EI

193

Direct Integration Method

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)

Performing direction integrations of (3.8.2) yields EI

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

194

Direct Integration Method

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)

(e3.8.12) Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

195

Direct Integration Method

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)

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

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)

Performing direction integrations of (3.8.15) yields EI

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)

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



C2  0 Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

196

Direct Integration Method



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)  





Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

197

Direct Integration Method

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)

Performing direction integrations of (3.9.1) yields EI

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)

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



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

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198

Direct Integration Method

 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

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199

Direct Integration Method

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

200

Direct Integration Method

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)

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

v(L)  0





201

Direct Integration Method

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

202

Direct Integration Method

Performing direction integrations of (e3.10.11) yields EI

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

203

Direct Integration Method

Performing direction integrations of (e3.10.19) yields EI

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)

(e3.11.1) Copyright © 2011 J. Rungamornrat

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204

Direct Integration Method

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

205

Direct Integration Method

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

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206

Direct Integration Method

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  

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

207

Direct Integration Method

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

208

Direct Integration Method

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

Copyright © 2011 J. Rungamornrat

a

L–a

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

209

Direct Integration Method

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 Copyright © 2011 J. Rungamornrat

L–a

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

210

Direct Integration Method

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

Copyright © 2011 J. Rungamornrat

a

L–a

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

212

Method of Curvature (Moment) Area

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|, || 0; otherwise, it is in the clockwise direction. Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

467

Method of Consistent Deformation

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

0

M

–3qL2/2

C

Structure 0 A 0

V

qL 2

qL

3qL /2 Copyright © 2011 J. Rungamornrat

qL

D

qL

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

468

Method of Consistent Deformation

L

MI VI

–1

C 0

B

0

D

1 L

1

0

Structure I A

I

I

M

V

0 L

1

2L

MII VII

–1

D 0

B

C

0 1

2L

1

0

Structure II A

II

M

II

V

0 2L

1

u1r  u1o   f11 f12  R 1         u 2 r  u 2 o  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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

469

Method of Consistent Deformation

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

M 0M I dx  EI

1  2qL    2  EI

2

AB BC δU Ca  δU Ca  δU Ca  δU CD Ca 

L AB

 0

L BC

 0

M0M I dx  EI

L CD

 0

M0M 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  AB BC δU Cs  δU Cs  δU Cs  δU CD Cs 

L AB

 0

V 0 V I dx  GA

L BC

 0

V 0 V I dx  GA

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 AB BC δU Cb  δU Cb  δU Cb  δU CD Cb 

L AB

 0

M 0 M II 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 0 M II dx  EI

L CD

 0

M 0 M II dx EI

 1  qL   7 L  55qL4 L 2L     L    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  AB BC δU Cs  δU Cs  δU Cs  δU CD Cs 

L AB

 0

V 0 V I dx  GA

L BC

 0

V 0 V I dx  GA

Copyright © 2011 J. Rungamornrat

L CD

 0

V 0 V I dx GA

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

470

Method of Consistent Deformation

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

MIMI dx  EI

L CD

 0

MIMI dx EI

1  L   2L  4L3 L   0   L L    L   2  EI   3  3EI  EI  AB BC δU Ca  δU Ca  δU 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 

δU Cs  δU

AB Cs

 δU

BC 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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

AB BC δU Ca  δU Ca  δU Ca  δU CD Ca 

L AB

 0

471

Method of Consistent Deformation

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   AB BC δU Cs  δU Cs  δU 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 AB BC δU Cb  δU Cb  δU Cb  δU CD Cb 

L AB

 0

M IIM 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

FIIFII 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 IIV 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   4 2 2   o  qL 27  qL 1 qL 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 

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

472

Method of Consistent Deformation

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

2

qL /8

B –L

–L

C

C

B

–L

0 A

0

M0 0

0

MI D

A

1

1

Structure I

Structure 0 Copyright © 2011 J. Rungamornrat

D

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

473

Method of Consistent Deformation

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

M0M I dx  EI

L CD

 0

M0M I F0 FI L AD dx  EI EA

2  qL  qL4  0  L  L  0  0       3  8EI  12EI 2

δWC  δU C



u1o  

qL4 Gap 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 AB BC AD δU C  δU Cb  δU Cb  δU CD Cb  δU Ca 

L AB

 0

M IM I dx  EI

L BC

 0

MIMI dx  EI

L CD

 0

MIMI FI FI L AD dx  EI EA

L 5L3 L 1  L   2L   L       L          L L 2    EA 3EI EA 2  EI   3   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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

474

Method of Consistent Deformation

The shear force and bending moment diagrams of the original structure are given below. q

qL2/12

qL/2

C

B

C

B

(qL/24)

A

D

2 –qL2/24 –qL /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 δU C  δU

AB Cb

 δU

BC Cb

 δU

CD Cb

 δU

AD 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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

475

Method of Consistent Deformation

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

476

Method of Consistent Deformation

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

Copyright © 2011 J. Rungamornrat

q EI, 2L

q

M EI, L

EI, L2

EI, L

EI, L

b

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

477

Method of Consistent Deformation

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

478

Method of Consistent Deformation

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

Copyright © 2011 J. Rungamornrat

L

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

481

Influence Lines

l2

l2

l2

l1

l1

l1

R A  R A ( q )   R A ( qdx )   R A (qdx  U (x))   R A (U (x))qdx

(11.3)

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)

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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

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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

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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

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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

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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

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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

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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  

L

C   0

1 5L2 x  9x 3 81EI





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 dx  EI 

1  L  x   L   4L  1  L  3x   L   7 x  2L  1  2x   2L   4L           x         2  3EI   3   27  2  3EI   3   6 2L  9  2  3EI   3   27  1  x  L  L2  18Lx  9x 2 162EI





L

MδM II 1  L  x   L   2  1  L  3x   L    7 x  2  1  2x   2L   4  dx   C          x             EI 2  3EI   3   9  2  3EI   3    6 2L  3  2  3EI   3   9  0 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

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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

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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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

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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

L

B   0

MδM I 1 x  1 x L dx      x      3Lx 2  2x 3   EI 2  EI   3 2  12EI MδM II 1 x  1 2 dx      x 1   x EI 2  EI  2EI

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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

MδM I 1  L  2x   L   L  1  L   L   L  dx   6L2 x  L3             EI  2EI   2   4  2  2EI   2   3  48EI



MδM II 1  L  L  1  L  2x   L  dx   L2  4Lx    1       1  EI 2  2EI   2  8EI  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]

Copyright © 2011 J. Rungamornrat

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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

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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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)

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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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. x IL(MA)

U=1 C

A IL(RA)

B

L/2

IL(MB)

D IL(RD) L

L/2 B IL(VB)

C

L D IL(RD)

Copyright © 2011 J. Rungamornrat

Case I

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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)

Copyright © 2011 J. Rungamornrat

Case I

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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)

Copyright © 2011 J. Rungamornrat

Case I

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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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

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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

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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

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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

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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

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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

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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 h 1 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

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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

Copyright © 2011 J. Rungamornrat

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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

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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

Copyright © 2011 J. Rungamornrat

F

L/2

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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

Copyright © 2011 J. Rungamornrat

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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  ; h4  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  h2    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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

A

C

B

L/4

511

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

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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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

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

A

C

B

L/4

513

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

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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)

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

A

C

B

L/4

515

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

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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)

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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*)

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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)

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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)

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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

C Girder

D

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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

Copyright © 2011 J. Rungamornrat

–2/6

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

U=1

x A

B

D

C

h

1

529

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.

D

C

h

G

P

B

A

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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

Copyright © 2011 J. Rungamornrat

–7h/6

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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

531

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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

Copyright © 2011 J. Rungamornrat

h/2

h/2

I

H

G h

h

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

542

Influence Lines

U=1

x B

A

C

E

D

F

G

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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)

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

G

545

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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

A

h

(a)

B 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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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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)

D MD VCD

C

h (b)

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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 

(11.31) (11.32) (11.33)

M C = MC   FIJ a = MC  IL(FIJ )   IL(MC ) / a M J = MD 

FCD a = MD  IL(FCD )  IL(MD ) / a

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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

E

F 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

Copyright © 2011 J. Rungamornrat

h

F h

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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

E

h

F 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

Copyright © 2011 J. Rungamornrat

C

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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

C

h

E

D

h

h

h

F 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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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

h

F

E h

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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

o

60 60o

A

B

h/2

60o

D

C

h

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 h/2

J

I

H 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 o

A h/2

B h

60

C h

H

J

I

FCJ o

60

G

60o o

60o

60

FCD A

1

h/2 Copyright © 2011 J. Rungamornrat

B h

C h

2

FJK FDJ

FCD D h

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

557

Influence Lines

U=1

x

G

H

J

I

K

L

M

o

60 o

60o

60

A

B

h/2

D

C

h

h

h

I

J

F

E h

h

h/2

U=1

x

F′

A′ G

H

1.1

h

h/2

h/2 1

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

x

G

559

Influence Lines

U=1

H

J

I

K

L

M

o

60 o

60o

60

h/2

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

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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

D

C h

h

O

U=1

x

J

N

F

E h

h

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

D

C h

h

F

E h

h

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

Copyright © 2011 J. Rungamornrat

h

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

561

Influence Lines

M  1 2

K

L

FMN

U=1

x

FEM

J RAX

A

B

D

C

RAY h

h

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

h

E

D

C

F

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

Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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

h

E

D

C

F

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

Copyright © 2011 J. Rungamornrat

3 5 /8

 5/4

 5/8

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

563

Influence Lines

M 1 2

L

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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon 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

L Copyright © 2011 J. Rungamornrat

FUNDAMENTAL STRUCTURAL ANALYSIS Jaroon Rungamornrat

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