Structural Analysis 8th Edition R C Hibbeler - Tabelas de Deflexões e Inclinações de Vigas

STRUCTURAL ANALYSIS EIGHTH EDITION R C HIBBELER Table for Evaluating ₀ᴸ m m dx ₀ᴸ m m dx m m m₁ m₂ parabola L L L L L m mmL 12 mmL 12 mm₁ m₂L 23 mmL L L 12 mmL 13 mmL 16 mm₁ 2m₂L 512 mmL L L 12 m m₁ m₂L 16 m m₁ 2m₂L 16 m₁2m₁ m₂ m₂ m₁ 2m₂L 112 m 3m₁ 5m₂L L m 12 mmL 16 mm L a 16 mm L b m₂ L a 112 mm 3 3aL a²L² L m 12 mmL 16 mmL 16 m2m₁ m₂L 14 mmL Beam Deflections and Slopes Loading v θ Equation v P vmax PL³ 3EI at x L θmax PL² 2EI at x L v P 6EI x³ 3Lx² x L Mo vmax MoL² 2EI at x L θmax MoL EI at x L v Mo 2EI x² Beam Deflections and Slopes continued v w wL⁴ 8EI at x L θmax wL³ 6EI at x L v w 24EI x⁴ 4Lx³ 6L²x² x L 2 P vmax PL³ 48EI at x L2 θmax PL² 16EI at x 0 or x L v P 48EI 4x³ 3L²x 0 x L2 P θL PabL b 6LEI θ PabL a 6LEI v Pbx 6LEI L² b² x² 0 x a w vmax 5wL⁴ 384EI at x L2 θmax wL³ 24EI v w x 24EI x³ 2Lx² L³ w θL 3wL³ 128EI θR 7wL³ 384EI v w x 384EI 16x³ 24Lx² 9L³ 0 x L2 v wL 384EI 8x³ 24Lx² 17L²x L³ L2 x L Mo vmax MoL² 93EI θL MoL 6EI θR MoL 3EI v Mo x 6EIL L² x² STRUCTURAL ANALYSIS This page intentionally left blank STRUCTURAL ANALYSIS EIGHTH EDITION R C HIBBELER PRENTICE HALL Boston Columbus Indianapolis New York San Francisco Upper Saddle River Amsterdam Cape Town Dubai London Madrid Milan Munich Paris Montreal Toronto Delhi Mexico City Sao Paulo Sydney Hong Kong Seoul Singapore Taipei Tokyo Library of Congress CataloginginPublication Data on File Vice President and Editorial Director ECS Marcia J Horton Acquisitions Editor Tacy QuinnNorrin Dais Managing Editor Scott Disanno Production Editor Rose Kernan Art Director Kenny Beck Managing Editor AV Management and Production Patricia Burns Art Editor Gregory Dulles Media Editor David Alick Director Image Resource Center Melinda Reo Manager Rights and Permissions Zina Arabia Manager Visual Research Beth Brenzel Manager Cover Visual Research and Permissions Karen Sanatar Manufacturing Manager Alexis HeydtLong Manufacturing Buyer Lisa McDowell Senior Marketing Manager Tim Galligan Cover Designer Kenny Beck About the Cover Background Image Orange Steel girderszimmytwsShutterstock Inset image Building under constructionVladittoShutterstock 2012 by R C Hibbeler Published by Pearson Prentice Hall Pearson Education Inc Upper Saddle River New Jersey 07458 All rights reserved No part of this book may be reproduced in any form or by any means without permission in writing from the publisher Pearson Prentice Hall is a trademark of Pearson Education Inc The author and publisher of this book have used their best efforts in preparing this bookThese efforts include the development research and testing of the theories and programs to determine their effectivenessThe author and publisher make no warranty of any kind expressed or implied with regard to these programs or the documentation contained in this bookThe author and publisher shall not be liable in any event for incidental or consequential damages in connection with or arising out of the furnishing performance or use of these programs Previous editions copyright 2009 2006 2002 1999 1995 1990 1985 by R C Hibbeler Pearson Education Ltd London Pearson Education Australia Pty Ltd Sydney Pearson Education Singapore Pte Ltd Pearson Education North Asia Ltd Hong Kong Pearson Education Canada Inc Toronto Pearson Educación de Mexico SA de CV Pearson EducationJapan Tokyo Pearson Education Malaysia Pte Ltd Pearson Education Upper Saddle River New Jersey Printed in the United States of America 10 9 8 7 6 5 4 3 2 1 ISBN10 013257053X ISBN13 9780132570534 To The Student With the hope that this work will stimulate an interest in Structural Analysis and provide an acceptable guide to its understanding This page intentionally left blank This book is intended to provide the student with a clear and thorough presentation of the theory and application of structural analysis as it applies to trusses beams and frames Emphasis is placed on developing the students ability to both model and analyze a structure and to provide realistic applications encountered in professional practice For many years now engineers have been using matrix methods to analyze structures Although these methods are most efficient for a structural analysis it is the authors opinion that students taking a first course in this subject should also be well versed in some of the more important classicial methods Practice in applying these methods will develop a deeper understanding of the basic engineering sciences of statics and mechanics of materials Also problemsolving skills are further developed when the various techniques are thought out and applied in a clear and orderly way By solving problems in this way one can better grasp the way loads are transmitted through a structure and obtain a more complete understanding of the way the structure deforms under load Finally the classicial methods provide a means of checking computer results rather than simply relying on the generated output New to This Edition Fundamental Problems These problem sets are selectively located just after the example problems They offer students simple applications of the concepts and therefore provide them with the chance to develop their problemsolving skills before attempting to solve any of the standard problems that follow You may consider these problems as extended examples since they all have solutions and answers that are given in the back of the book Additionally the fundamental problems offer students an excellent means of studying for exams and they can be used at a later time to prepare for the exam necessary to obtain a professional engineering license Content Revision Each section of the text was carefully reviewed to enhance clarity This has included incorporating the new ASCE SEI 0710 standards on loading in Chapter 1 an improved explanation of how to draw shear and moment diagrams and the deflection curve of a structure consolidating the material on structures having a variable moment of inertia providing further discussion for analyzing structures having internal hinges using matrix analysis and adding a new Appendix B that discusses some of the common features used for running current structural analysis computer software PREFACE Example Changes In order to further illustrate practical applications of the theory throughout the text some of the examples have been changed and with the aid of photos feature modeling and analysis of loadings applied to actual structures Additional Photos The relevance of knowing the subject matter is reflected by the realistic applications depicted in many new and updated photos along with captions that are placed throughout the book New Problems There are approximately 70 new problems in this editionThey retain a balance between easy medium and difficult applications In addition to the author the problems have been reviewed and checked by four other parties Scott Hendricks Karim Nohra Kurt Norlin and Kai Beng Yap Problem Arrangement For convenience in assigning homework the problems are now placed throughout the text This way each chapter is organized into welldefined sections that contain an explanation of specific topics illustrative example problems and a set of homework problems that are arranged in approximate order of increasing difficulty Organization and Approach The contents of each chapter are arranged into sections with specific topics categorized by title headings Discussions relevant to a particular theory are succinct yet thorough In most cases this is followed by a procedure for analysisguidewhich provides the student with a summary of the important concepts and a systematic approach for applying the theory The example problems are solved using this outlined method in order to clarify its numerical application Problems are given at the end of each group of sections and are arranged to cover the material in sequential order Moreover for any topic they are arranged in approximate order of increasing difficulty Hallmark Elements Photographs Many photographs are used throughout the book to explain how the principles of structural analysis apply to realworld situations Problems Most of the problems in the book depict realistic situations encountered in practice It is hoped that this realism will both stimulate the students interest in structural analysis and develop the skill to reduce any such problem from its physical description to a model or symbolic representation to which the appropriate theory can be applied Throughout the book there is an approximate balance of problems using either SI or FPS units The intent has been to develop VIII PREFACE problems that test the students ability to apply the theory keeping in mind that those problems requiring tedious calculations can be relegated to computer analysis Answers to Selected Problems The answers to selected problems are listed in the back of the book Extra care has been taken in the presentation and solution of the problems and all the problem sets have been reviewed and the solutions checked and rechecked to ensure both their clarity and numerical accuracy Example Problems All the example problems are presented in a concise manner and in a style that is easy to understand Illustrations Throughout the book an increase in twocolor art has been added including many photorealistic illustrations that provide a strong connection to the 3D nature of structural engineering Triple Accuracy Checking The edition has undergone rigorous accuracy checking and proofing of pages Besides the authors review of all art pieces and pages Scott Hendricks of Virginia Polytechnic Institute Karim Nohra of the University of South Florida and Kurt Norlin of Laurel Technical Services rechecked the page proofs and together reviewed the entire Solutions Manual Contents This book is divided into three parts The first part consists of seven chapters that cover the classical methods of analysis for statically determinate structures Chapter 1 provides a discussion of the various types of structural forms and loads Chapter 2 discusses the determination of forces at the supports and connections of statically determinate beams and framesThe analysis of various types of statically determinate trusses is given in Chapter 3 and shear and bendingmoment functions and diagrams for beams and frames are presented in Chapter 4 In Chapter 5 the analysis of simple cable and arch systems is presented and in Chapter 6 influence lines for beams girders and trusses are discussed Finally in Chapter 7 several common techniques for the approximate analysis of statically indeterminate structures are considered In the second part of the book the analysis of statically indeterminate structures is covered in six chapters Geometrical methods for calculating deflections are discussed in Chapter 8 Energy methods for finding deflections are covered in Chapter 9 Chapter 10 covers the analysis of statically indeterminate structures using the force method of analysis in addition to a discussion of influence lines for beams Then the displacement methods consisting of the slopedeflection method in Chapter 11 and moment distribution in Chapter 12 are discussed Finally beams and frames having nonprismatic members are considered in Chapter 13 PREFACE IX The third part of the book treats the matrix analysis of structures using the stiffness methodTrusses are discussed in Chapter 14 beams in Chap ter 15 and frames in Chapter 16 A review of matrix algebra is given in Appendix A and Appendix B provides a general guide for using available software for solving problem in structural analysis Resources for Instructors Instructors Solutions Manual An instructors solutions manual was prepared by the author The manual was also checked as part of the Triple Accuracy Checking program Presentation Resources All art from the text is available in PowerPoint slide and JPEG format These files are available for download from the Instructor Resource Center at wwwpearsonhigheredcom If you are in need of a login and password for this site please contact your local Pearson Prentice Hall representative Video Solutions Located on the Companion Website Video Solutions offer stepbystep solution walkthroughs of representative homework problems from each chapter of the textMake efficient use of class time and office hours by showing students the complete and concise problem solving approaches that they can access anytime and view at their own paceThe videos are designed to be a flexible resource to be used however each instructor and student prefers A valuable tutorial resource the videos are also helpful for student selfevaluation as students can pause the videos to check their understanding and work alongside the video Access the videos at wwwpearsonhigheredcom hibbeler and follow the links for the Structural Analysis text STRAN Developed by the author and Barry Nolan a practicing engineer STRAN is a downloadable program for use with Structural Analysis problems Access STRAN on the Companion Website www pearsonhigheredcomhibbeler and follow the links for the Structural Analysis text Complete instructions for how to use the software are included on the Companion Website Resources for Students Companion Website The Companion Website provides practice and review materials including Video SolutionsComplete stepbystep solution walkthroughs of representative homework problems from each chapterVideos offer Fully worked SolutionsShowing every step of representative homework problems to help students make vital connections between concepts X PREFACE Selfpaced InstructionStudents can navigate each problem and select play rewind fastforward stop and jumpto sections within each problems solution 247AccessHelpwheneverstudentsneeditwithover20hours of helpful review STRANA program you can use to solve two and three dimensional trusses and beams and two dimensional frames Instructions for downloading and how to use the program are available on the Companion Website An access code for the Structural Analysis Eighth Edition Companion Website is included with this text To redeem the code and gain access to the site go to wwwprenhallcomhibbeler and follow the directions on the access code card Access can also be purchased directly from the site Acknowledgments Over one hundred of my colleagues in the teaching profession and many of my students have made valuable suggestions that have helped in the development of this book and I would like to hereby acknowledge all of their comments I personally would like to thank the reviewers contracted by my editor for this new edition namely Thomas H Miller Oregon State University Hayder A Rasheed Kansas State University Jeffrey A Laman Penn State University Jerry R Bayless University of MissouriRolla Paolo Gardoni Texas AM University Timothy Ross University of New Mexico FWayne Klaiber Iowa State University Husam S Najm Rutgers University Also the constructive comments from Kai Beng Yap and Barry Nolan both practicing engineers are greatly appreciated Finally I would like to acknowledge the support I received from my wife Conny who has always been very helpful in preparing the manuscript for publication I would greatly appreciate hearing from you if at any time you have any comments or suggestions regarding the contents of this edition Russell Charles Hibbeler hibbelerbellsouthnet PREFACE XI This page intentionally left blank CREDITS Chapter 1 opener CJ GuntherepaCorbis Figure 16 a Page 7 Mark HarrisPhotodiscGetty Images Chapter 2 opener Joe GoughShutterstock Chapter 3 opener Robert ShantzAlamy Chapter 4 opener Ralf Broskvar123rf Chapter 5 opener Greg Balfour EvansAlamy Chapter 6 opener Accent AlaskacomAlamy Chapter 7 opener David R Frazier Photolibrary IncAlamy Chapter 8 opener PhotographerStoneGetty Images Chapter 9 opener Alamy Images Chapter 10 opener Shutterstock Chapter 11 opener 2011 Photoscom a division of Getty Images All rights reserved Chapter 12 opener FotosearchSuperStock Chapter 13 opener iStockphotocom Chapter 14 opener Corbis RFAlamy Chapter 15 opener Paul A SoudersCORBIS Chapter 16 opener Alan ScheinCorbis Cover 1 zimmytwsShutterstock Cover 2 VladittoShutterstock Other photos provided by the author R C Hibbeler This page intentionally left blank 1 Types of Structures and Loads 3 11 Introduction 3 12 Classification of Structures 4 13 Loads 9 14 Structural Design 26 Problems 27 Chapter Review 31 4 Internal Loadings Developed in Structural Members 133 41 Internal Loadings at a Specified Point 133 42 Shear and Moment Functions 139 43 Shear and Moment Diagrams for a Beam 150 44 Shear and Moment Diagrams for a Frame 163 45 Moment Diagrams Constructed by the Method of Superposition 168 Problems 173 Chapter Review 178 3 Analysis of Statically Determinate Trusses 79 31 Common Types of Trusses 79 32 Classification of Coplanar Trusses 85 33 The Method of Joints 94 34 ZeroForce Members 98 2 Analysis of Statically Determinate Structures 33 21 Idealized Structure 33 22 Principle of Superposition 46 23 Equations of Equilibrium 47 24 Determinacy and Stability 48 25 Application of the Equations of Equilibrium 59 Chapter Review 68 Fundamental Problems 70 Problems 72 Project Problem 77 CONTENTS 5 Cables and Arches 181 51 Cables 181 52 Cable Subjected to Concentrated Loads 182 53 Cable Subjected to a Uniform Distributed Load 184 54 Arches 194 55 ThreeHinged Arch 195 Problems 201 Chapter Review 203 35 The Method of Sections 104 36 Compound Trusses 110 37 Complex Trusses 116 38 Space Trusses 120 Problems 127 Chapter Review 130 XVI CONTENTS 7 Approximate Analysis of Statically Indeterminate Structures 263 71 Use of Approximate Methods 263 72 Trusses 264 73 Vertical Loads on Building Frames 270 74 Portal Frames and Trusses 273 75 Lateral Loads on Building Frames Portal Method 282 76 Lateral Loads on Building Frames Cantilever Method 288 Problems 294 Chapter Review 296 9 Deflections Using Energy Methods 341 91 External Work and Strain Energy 341 92 Principle of Work and Energy 345 93 Principle of Virtual Work 346 94 Method of Virtual Work Trusses 348 95 Castiglianos Theorem 355 96 Castiglianos Theorem for Trusses 356 97 Method of Virtual Work Beams and Frames 364 98 Virtual Strain Energy Caused by Axial Load Shear Torsion and Temperature 375 99 Castiglianos Theorem for Beams and Frames 381 Problems 388 Chapter Review 392 8 Deflections 299 81 Deflection Diagrams and the Elastic Curve 299 82 ElasticBeam Theory 305 83 The Double Integration Method 307 84 MomentArea Theorems 316 85 ConjugateBeam Method 326 Problems 335 Chapter Review 338 6 Influence Lines for Statically Determinate Structures 205 61 Influence Lines 205 62 Influence Lines for Beams 213 63 Qualitative Influence Lines 216 64 Influence Lines for Floor Girders 228 65 Influence Lines for Trusses 232 66 Maximum Influence at a Point due to a Series of Concentrated Loads 240 67 Absolute Maximum Shear and Moment 250 Problems 255 Chapter Review 260 CONTENTS XVII 10 Analysis of Statically Indeterminate Structures by the Force Method 395 101 Statically Indeterminate Structures 395 102 Force Method of Analysis General Procedure 398 103 Maxwells Theorem of Reciprocal Displacements Bettis Law 402 104 Force Method of Analysis Beams 403 105 Force Method of Analysis Frames 411 106 Force Method of Analysis Trusses 422 107 Composite Structures 425 108 Additional Remarks on the Force Method of Analysis 428 109 Symmetric Structures 429 1010 Influence Lines for Statically Indeterminate Beams 435 1011 Qualitative Influence Lines for Frames 439 Problems 446 Chapter Review 448 11 Displacement Method of Analysis SlopeDeflection Equations 451 111 Displacement Method of Analysis General Procedures 451 112 SlopeDeflection Equations 453 113 Analysis of Beams 459 114 Analysis of Frames No Sidesway 469 115 Analysis of Frames Sidesway 474 Problems 482 Chapter Review 485 12 Displacement Method of Analysis Moment Distribution 487 121 General Principles and Definitions 487 122 Moment Distribution for Beams 491 123 StiffnessFactor Modifications 500 124 Moment Distribution for Frames No Sidesway 508 125 Moment Distribution for Frames Sidesway 510 Problems 518 Chapter Review 521 13 Beams and Frames Having Nonprismatic Members 523 131 Loading Properties of Nonprismatic Members 523 132 Moment Distribution for Structures Having Nonprismatic Members 528 133 SlopeDeflection Equations for Nonprismatic Members 534 Problems 536 Chapter Review 537 16 Plane Frame Analysis Using the Stiffness Method 595 161 FrameMember Stiffness Matrix 595 162 Displacement and Force Transformation Matrices 597 163 FrameMember Global Stiffness Matrix 599 164 Application of the Stiffness Method for Frame Analysis 600 Problems 609 Appendices A Matrix Algebra for Structural Analysis 612 B General Procedure for Using Structural Analysis Software 625 Fundamental Problems Partial Solutions and Answers 628 Answers to Selected Problems 665 Index 685 15 Beam Analysis Using the Stiffness Method 575 151 Preliminary Remarks 575 152 BeamMember Stiffness Matrix 577 153 BeamStructure Stiffness Matrix 579 XVIII CONTENTS 14 Truss Analysis Using the Stiffness Method 539 141 Fundamentals of the Stiffness Method 539 142 Member Stiffness Matrix 542 143 Displacement and Force Transformation Matrices 543 144 Member Global Stiffness Matrix 546 145 Truss Stiffness Matrix 547 146 Application of the Stiffness Method for Truss Analysis 552 147 Nodal Coordinates 560 148 Trusses Having Thermal Changes and Fabrication Errors 564 149 SpaceTruss Analysis 570 Chapter Review 571 Problems 572 154 Application of the Stiffness Method for Beam Analysis 579 Problems 592 STRUCTURAL ANALYSIS The diamond pattern framework cross bracing of these highrise buildings is used to resist loadings due to wind 1 3 This chapter provides a discussion of some of the preliminary aspects of structural analysis The phases of activity necessary to produce a structure are presented first followed by an introduction to the basic types of structures their components and supports Finally a brief explanation is given of the various types of loads that must be considered for an appropriate analysis and design 11 Introduction A structure refers to a system of connected parts used to support a load Important examples related to civil engineering include buildings bridges and towers and in other branches of engineering ship and aircraft frames tanks pressure vessels mechanical systems and electrical supporting structures are important When designing a structure to serve a specified function for public use the engineer must account for its safety esthetics and serviceability while taking into consideration economic and environmental constraints Often this requires several independent studies of different solutions before final judgment can be made as to which structural form is most appropriateThis design process is both creative and technical and requires a fundamental knowledge of material properties and the laws of mechanics which govern material responseOnce a preliminary design of a structure is proposed the structure must then be analyzed to ensure that it has its required stiffness and strength To analyze a structure properly certain idealizations must be made as to how the members are supported and connected together The loadings are determined from codes and local specifications and the forces in the members and their displacements are found using the theory of structural analysis which is the subject matter of this text The results of this analysis then can be used to Types of Structures and Loads 4 CHAPTER 1 TYPES OF STRUCTURES AND LOADS 1 redesign the structure accounting for a more accurate determination of the weight of the members and their size Structural design therefore follows a series of successive approximations in which every cycle requires a structural analysis In this book the structural analysis is applied to civil engineering structures however the method of analysis described can also be used for structures related to other fields of engineering 12 Classification of Structures It is important for a structural engineer to recognize the various types of elements composing a structure and to be able to classify structures as to their form and function We will introduce some of these aspects now and expand on them at appropriate points throughout the text Structural Elements Some of the more common elements from which structures are composed are as follows Tie Rods Structural members subjected to a tensile force are often referred to as tie rods or bracing struts Due to the nature of this load these members are rather slender and are often chosen from rods bars angles or channels Fig 11 Beams Beams are usually straight horizontal members used primarily to carry vertical loads Quite often they are classified according to the way they are supported as indicated in Fig 12 In particular when the cross section varies the beam is referred to as tapered or haunched Beam cross sections may also be built up by adding plates to their top and bottom Beams are primarily designed to resist bending moment however if they are short and carry large loads the internal shear force may become quite large and this force may govern their design When the material used for a beam is a metal such as steel or aluminum the cross section is most efficient when it is shaped as shown in Fig 13 Here the forces developed in the top and bottom flanges of the beam form the necessary couple used to resist the applied moment M whereas the web is effective in resisting the applied shear V This cross section is commonly referred to as a wide flange and it is normally formed as a single unit in a rolling mill in lengths up to 75 ft 23 m If shorter lengths are needed a cross section having tapered flanges is sometimes selected When the beam is required to have a very large span and the loads applied are rather large the cross section may take the form of a plate girder This member is fabricated by using a large plate for the web and welding or bolting plates to its ends for flangesThe girder is often transported to the field in segments and the segments are designed to be spliced or joined together rod tie rod bar angle channel typical cross sections Fig 11 simply supported beam cantilevered beam fixedsupported beam continuous beam Fig 12 12 CLASSIFICATION OF STRUCTURES 5 1 flange flange web V M Fig 13 at points where the girder carries a small internal moment See the photo below Concrete beams generally have rectangular cross sections since it is easy to construct this form directly in the field Because concrete is rather weak in resisting tension steel reinforcing rods are cast into the beam within regions of the cross section subjected to tension Precast concrete beams or girders are fabricated at a shop or yard in the same manner and then transported to the job site Beams made from timber may be sawn from a solid piece of wood or laminated Laminated beams are constructed from solid sections of wood which are fastened together using highstrength glues The prestressed concrete girders are simply supported and are used for this highway bridge Shown are typical splice plate joints used to connect the steel girders of a highway bridge The steel reinforcement cage shown on the right and left is used to resist any tension that may develop in the concrete beams which will be formed around it 6 CHAPTER 1 TYPES OF STRUCTURES AND LOADS 1 Columns Members that are generally vertical and resist axial compressive loads are referred to as columns Fig 14 Tubes and wideflange cross sections are often used for metal columns and circular and square cross sections with reinforcing rods are used for those made of concrete Occasionally columns are subjected to both an axial load and a bending moment as shown in the figure These members are referred to as beam columns Types of Structures The combination of structural elements and the materials from which they are composed is referred to as a structural system Each system is constructed of one or more of four basic types of structures Ranked in order of complexity of their force analysis they are as follows Trusses When the span of a structure is required to be large and its depth is not an important criterion for design a truss may be selected Trusses consist of slender elementsusually arranged in triangular fashion Planar trusses are composed of members that lie in the same plane and are frequently used for bridge and roof support whereas space trusses have members extending in three dimensions and are suitable for derricks and towers Due to the geometric arrangement of its members loads that cause the entire truss to bend are converted into tensile or compressive forces in the members Because of this one of the primary advantages of a truss compared to a beam is that it uses less material to support a given load Fig 15 Also a truss is constructed from long and slender elements which can be arranged in various ways to support a load Most often it is Wideflange members are often used for columns Here is an example of a beam column beam column column Fig 14 12 CLASSIFICATION OF STRUCTURES 7 1 economically feasible to use a truss to cover spans ranging from 30 ft 9 m to 400 ft 122 m although trusses have been used on occasion for spans of greater lengths Cables and Arches Two other forms of structures used to span long distances are the cable and the arch Cables are usually flexible and carry their loads in tension They are commonly used to support bridges Fig 16a and building roofs When used for these purposes the cable has an advantage over the beam and the truss especially for spans that are greater than 150 ft 46 m Because they are always in tension cables will not become unstable and suddenly collapse as may happen with beams or trusses Furthermore the truss will require added costs for construction and increased depth as the span increases Use of cables on the other hand is limited only by their sag weight and methods of anchorage The arch achieves its strength in compression since it has a reverse curvature to that of the cable The arch must be rigid however in order to maintain its shape and this results in secondary loadings involving shear and moment which must be considered in its design Arches are frequently used in bridge structures Fig 16b dome roofs and for openings in masonry walls Fig 15 Fig 16 Loading causes bending of truss which develops compression in top members tension in bottom members Cables support their loads in tension a Arches support their loads in compression b 8 CHAPTER 1 TYPES OF STRUCTURES AND LOADS 1 Frames Frames are often used in buildings and are composed of beams and columns that are either pin or fixed connected Fig 17 Like trusses frames extend in two or three dimensionsThe loading on a frame causes bending of its members and if it has rigid joint connections this structure is generally indeterminate from a standpoint of analysisThe strength of such a frame is derived from the moment interactions between the beams and the columns at the rigid joints Surface Structures A surface structure is made from a material having a very small thickness compared to its other dimensions Sometimes this material is very flexible and can take the form of a tent or airinflated structure In both cases the material acts as a membrane that is subjected to pure tension Surface structures may also be made of rigid material such as reinforced concrete As such they may be shaped as folded plates cylinders or hyperbolic paraboloids and are referred to as thin plates or shells These structures act like cables or arches since they support loads primarily in tension or compression with very little bending In spite of this plate or shell structures are generally very difficult to analyze due to the threedimensional geometry of their surface Such an analysis is beyond the scope of this text and is instead covered in texts devoted entirely to this subject pinned rigid rigid pinned Frame members are subjected to internal axial shear and moment loadings Fig 17 Here is an example of a steel frame that is used to support a crane rail The frame can be assumed fixed connected at its top joints and pinned at the supports The roof of the Georgia Dome in Atlanta Georgia can be considered as a thin membrane 13 LOADS 9 1 13 Loads Once the dimensional requirements for a structure have been defined it becomes necessary to determine the loads the structure must support Often it is the anticipation of the various loads that will be imposed on the structure that provides the basic type of structure that will be chosen for design For example highrise structures must endure large lateral loadings caused by wind and so shear walls and tubular frame systems are selected whereas buildings located in areas prone to earthquakes must be designed having ductile frames and connections Once the structural form has been determined the actual design begins with those elements that are subjected to the primary loads the structure is intended to carry and proceeds in sequence to the various supporting members until the foundation is reached Thus a building floor slab would be designed first followed by the supporting beams columns and last the foundation footings In order to design a structure it is therefore necessary to first specify the loads that act on it The design loading for a structure is often specified in codes In general the structural engineer works with two types of codes general building codes and design codesGeneral building codes specify the requirements of governmental bodies for minimum design loads on structures and minimum standards for construction Design codes provide detailed technical standards and are used to establish the requirements for the actual structural designTable 11 lists some of the important codes used in practice It should be realized however that codes provide only a general guide for design The ultimate responsibility for the design lies with the structural engineer TABLE 11 Codes General Building Codes Minimum Design Loads for Buildings and Other Structures ASCESEI 710American Society of Civil Engineers International Building Code Design Codes Building Code Requirements for Reinforced Concrete Am Conc Inst ACI Manual of Steel Construction American Institute of Steel Construction AISC Standard Specifications for Highway Bridges American Association of State Highway and Transportation Officials AASHTO National Design Specification for Wood ConstructionAmerican Forest and Paper Association AFPA Manual for Railway Engineering American Railway Engineering Association AREA Since a structure is generally subjected to several types of loads a brief discussion of these loadings will now be presented to illustrate how one must consider their effects in practice Dead Loads Dead loads consist of the weights of the various structural members and the weights of any objects that are permanently attached to the structure Hence for a building the dead loads include the weights of the columns beams and girders the floor slab roofing walls windows plumbing electrical fixtures and other miscellaneous attachments In some cases a structural dead load can be estimated satisfactorily from simple formulas based on the weights and sizes of similar structures Through experience one can also derive a feeling for the magnitude of these loadings For example the average weight for timber buildings is 4050 lbft2 1924 kNm2 for steel framed buildings it is 6075 lbft2 2936 kNm2 and for reinforced concrete buildings it is 110130 lbft2 5362 kNm2 Ordinarily though once the materials and sizes of the various components of the structure are determined their weights can be found from tables that list their densities The densities of typical materials used in construction are listed in Table 12 and a portion of a table listing the weights of typical building TABLE 12 Minimum Densities for Design Loads from Materials lbft3 kNm3 Aluminum 170 267 Concrete plain cinder 108 170 Concrete plain stone 144 226 Concrete reinforced cinder 111 174 Concrete reinforced stone 150 236 Clay dry 63 99 Clay damp 110 173 Sand and gravel dry loose 100 157 Sand and gravel wet 120 189 Masonry lightweight solid concrete 105 165 Masonry normal weight 135 212 Plywood 36 57 Steel colddrawn 492 773 Wood Douglas Fir 34 53 Wood Southern Pine 37 58 Wood spruce 29 45 Reproduced with permission from American Society of Civil Engineers Minimum Design Loads for Buildings and Other Structures ASCESEI 710 Copies of this standard may be purchased from ASCE at wwwpubsasceorg components is given in Table 13 Although calculation of dead loads based on the use of tabulated data is rather straightforward it should be realized that in many respects these loads will have to be estimated in the initial phase of design These estimates include nonstructural materials such as prefabricated facade panels electrical and plumbing systems etc Furthermore even if the material is specified the unit weights of elements reported in codes may vary from those given by manufacturers and later use of the building may include some changes in dead loading As a result estimates of dead loadings can be in error by 15 to 20 or more Normally the dead load is not large compared to the design load for simple structures such as a beam or a singlestory frame however for multistory buildings it is important to have an accurate accounting of all the dead loads in order to properly design the columns especially for the lower floors TABLE 13 Minimum Design Dead Loads Walls psf kNm2 4in 102 mm clay brick 39 187 8in 203 mm clay brick 79 378 12in 305 mm clay brick 115 551 Frame Partitions and Walls Exterior stud walls with brick veneer 48 230 Windows glass frame and sash 8 038 Wood studs 2 4 in 51 102 mm unplastered 4 019 Wood studs 2 4 in 51 102 mm plastered one side 12 057 Wood studs 2 4 in 51 102 mm plastered two sides 20 096 Floor Fill Cinder concrete per inch mm 9 0017 Lightweight concrete plain per inch mm 8 0015 Stone concrete per inch mm 12 0023 Ceilings Acoustical fiberboard 1 005 Plaster on tile or concrete 5 024 Suspended metal lath and gypsum plaster 10 048 Asphalt shingles 2 010 Fiberboard in 13 mm 075 004 Reproduced with permission from American Society of Civil Engineers Minimum Design Loads for Buildings and Other Structures ASCESEI 710 12 CHAPTER 1 TYPES OF STRUCTURES AND LOADS 1 The live floor loading in this classroom consists of desks chairs and laboratory equipment For design the ASCE 710 Standard specifies a loading of 40 psf or 192 kNm2 The floor beam in Fig 18 is used to support the 6ft width of a lightweight plain concrete slab having a thickness of 4 in The slab serves as a portion of the ceiling for the floor below and therefore its bottom is coated with plaster Furthermore an 8fthigh 12inthick lightweight solid concrete block wall is directly over the top flange of the beam Determine the loading on the beam measured per foot of length of the beam SOLUTION Using the data in Tables 12 and 13 we have Ans Here the unit k stands for kip which symbolizes kilopounds Hence 1 k 1000 lb Concrete slab Plaster ceiling Block wall Total load 8 lb1ft2 in214 in216 ft2 192 lbft 15 lbft2216 ft2 30 lbft 1105 lbft3218 ft211 ft2 840 lbft 1062 lbft 106 kft EXAMPLE 11 3 ft 3 ft 8 ft 4 in 12 in Fig 18 Live Loads Live Loads can vary both in their magnitude and location They may be caused by the weights of objects temporarily placed on a structure moving vehicles or natural forces The minimum live loads specified in codes are determined from studying the history of their effects on existing structures Usually these loads include additional protection against excessive deflection or sudden overload In Chapter 6 we will develop techniques for specifying the proper location of live loads on the structure so that they cause the greatest stress or deflection of the members Various types of live loads will now be discussed Building Loads The floors of buildings are assumed to be subjected to uniform live loads which depend on the purpose for which the building is designed These loadings are generally tabulated in local state or national codes A representative sample of such minimum live loadings taken from the ASCE 710 Standard is shown in Table 14The values are determined from a history of loading various buildings They include some protection against the possibility of overload due to emergency situations construction loads and serviceability requirements due to vibration In addition to uniform loads some codes specify minimum concentrated live loads caused by hand carts automobiles etc which must also be applied anywhere to the floor system For example both uniform and concentrated live loads must be considered in the design of an automobile parking deck TABLE 14 Minimum Live Loads Live Load Live Load Occupancy or Use psf kNm2 Occupancy or Use psf kNm2 Assembly areas and theaters Residential Fixed seats 60 287 Dwellings one and twofamily 40 192 Movable seats 100 479 Hotels and multifamily houses Garages passenger cars only 50 240 Private rooms and corridors 40 192 Office buildings Public rooms and corridors 100 479 Lobbies 100 479 Schools Offices 50 240 Classrooms 40 192 Storage warehouse Corridors above first floor 80 383 Light 125 600 Heavy 250 1197 Reproduced with permission from Minimum Design Loads for Buildings and Other Structures ASCESEI 710 For some types of buildings having very large floor areas many codes will allow a reduction in the uniform live load for a floor since it is unlikely that the prescribed live load will occur simultaneously throughout the entire structure at any one time For example ASCE 710 allows a reduction of live load on a member having an influence area KLL AT of 400 ft2 372 m2 or more This reduced live load is calculated using the following equation L Lo 025 15 sqrtKLL AT FPS units L Lo 025 457 sqrtKLL AT SI units where L reduced design live load per square foot or square meter of area supported by the member Lo unreduced design live load per square foot or square meter of area supported by the member see Table 14 KLL live load element factor For interior columns KLL 4 AT tributary area in square feet or square meters The reduced live load defined by Eq 11 is limited to not less than 50 of Lo for members supporting one floor or not less than 40 of Lo for members supporting more than one floor No reduction is allowed for loads exceeding 100 lbft2 479 kNm2 or for structures used for public assembly garages or roofs Example 12 illustrates Eq 11s application Specific examples of the determination of tributary areas for beams and columns are given in Sec 21 A twostory office building shown in the photo has interior columns that are spaced 22 ft apart in two perpendicular directions If the flat roof loading is determine the reduced live load supported by a typical interior column located at ground level 20 lbft2 EXAMPLE 12 SOLUTION As shown in Fig 19 each interior column has a tributary area or effective loaded area of A groundfloor column therefore supports a roof live load of This load cannot be reduced since it is not a floor load For the second floor the live load is taken from Table 14 Since then and the live load can be reduced using Eq 11Thus The load reduction here is OK Therefore The total live load supported by the groundfloor column is thus Ans F FR FF 968 k 143 k 240 k FF 12955 lbft221484 ft22 14 300 lb 143 k 12955502100 591 7 50 L 50a025 15 21936 b 2955 lbft2 1936 ft2 7 400 ft2 4AT 41484 ft22 1936 ft2 KLL 4 Lo 50 lbft2 FR 120 lbft221484 ft22 9680 lb 968 k AT 122 ft2122 ft2 484 ft2 14 CHAPTER 1 TYPES OF STRUCTURES AND LOADS 1 AT 22 ft 22 ft 22 ft 22 ft Fig 19 Highway Bridge Loads The primary live loads on bridge spans are those due to traffic and the heaviest vehicle loading encountered is that caused by a series of trucks Specifications for truck loadings on highway bridges are reported in the LRFD Bridge Design Specifications of the American Association of State and Highway Transportation Officials AASHTO For twoaxle trucks these loads are designated with an H followed by the weight of the truck in tons and another number which gives the year of the specifications in which the load was reported Hseries truck weights vary from 10 to 20 tons However bridges located on major highways which carry a great deal of traffic are often designed for twoaxle trucks plus a oneaxle semitrailer as in Fig 110 These are designated as HS loadings In general a truck loading selected for design depends upon the type of bridge its location and the type of traffic anticipated The size of the standard truck and the distribution of its weight is also reported in the specificationsAlthough trucks are assumed to be on the road all lanes on the bridge need not be fully loaded with a row of trucks to obtain the critical load since such a loading would be highly improbableThe details are discussed in Chapter 6 Railroad Bridge Loads The loadings on railroad bridges as in Fig 111 are specified in the Specifications for Steel Railway Bridges published by the American Railroad Engineers Association AREA Normally E loads as originally devised by Theodore Cooper in 1894 were used for design B Steinmann has since updated Coopers load distribution and has devised a series of M loadings which are currently acceptable for design Since train loadings involve a complicated series of concentrated forces to simplify hand calculations tables and graphs are sometimes used in conjunction with influence lines to obtain the critical loadAlso computer programs are used for this purpose 13 LOADS 15 1 Fig 110 Fig 111 Impact Loads Moving vehicles may bounce or sidesway as they move over a bridge and therefore they impart an impact to the deck The percentage increase of the live loads due to impact is called the impact factor I This factor is generally obtained from formulas developed from experimental evidence For example for highway bridges the AASHTO specifications require that I 50L 125 but not larger than 03 where L is the length of the span in feet that is subjected to the live load In some cases provisions for impact loading on the structure of a building must also be taken into account For example the ASCE 710 Standard requires the weight of elevator machinery to be increased by 100 and the loads on any hangers used to support floors and balconies to be increased by 33 Wind Loads When structures block the flow of wind the winds kinetic energy is converted into potential energy of pressure which causes a wind loading The effect of wind on a structure depends upon the density and velocity of the air the angle of incidence of the wind the shape and stiffness of the structure and the roughness of its surface For design purposes wind loadings can be treated using either a static or a dynamic approach For the static approach the fluctuating pressure caused by a constantly blowing wind is approximated by a mean velocity pressure that acts on the structure This pressure q is defined by its kinetic energy q 12ρV2 where ρ is the density of the air and V is its velocity According to the ASCE 710 Standard this equation is modified to account for the importance of the structure its height and the terrain in which it is located It is represented as qz 000256KzKztKdV2 lbft2 qz 0613KzKztKdV2 Nm2 13 LOADS 17 1 example the interior portion of the continental United States reports a wind speed of 105 mih 47 ms if the structure is an agricultural or storage building since it is of low risk to human life in the event of a failure The wind speed is 120 mih 54 ms for cases where the structure is a hospital since its failure would cause substantial loss of human life the velocity pressure exposure coefficient which is a function of height and depends upon the ground terrainTable 15 lists values for a structure which is located in open terrain with scattered lowlying obstructions a factor that accounts for wind speed increases due to hills and escarpments For flat ground a factor that accounts for the direction of the wind It is used only when the structure is subjected to combinations of loads see Sec 14 For wind acting alone Kd 10 Kd Kzt 10 Kzt Kz Hurricane winds caused this damage to a condominium in Miami Florida TABLE 15 Velocity Pressure Exposure Coefficient for Terrain with LowLying Obstructions z ft m Kz 015 046 085 20 61 090 25 76 094 30 91 098 40 122 104 50 152 109 TABLE 21 Supports for Coplanar Structures Type of Connection Idealized Symbol Reaction Number of Unknowns 1 light cable weightless link One unknown The reaction is a force that acts in the direction of the cable or link 2 rollers rocker One unknown The reaction is a force that acts perpendicular to the surface at the point of contact 3 smooth contacting surface One unknown The reaction is a force that acts perpendicular to the surface at the point of contact 4 smooth pinconnected collar One unknown The reaction is a force that acts perpendicular to the surface at the point of contact 5 smooth pin or hinge Two unknowns The reactions are two force components 6 slider fixedconnected collar Two unknowns The reactions are a force and a moment 7 fixed support Three unknowns The reactions are the moment and the two force components 18 CHAPTER 1 TYPES OF STRUCTURES AND LOADS 1 Design Wind Pressure for Enclosed Buildings Once the value for is obtained the design pressure can be determined from a list of relevant equations listed in the ASCE 710 Standard The choice depends upon the flexibility and height of the structure and whether the design is for the main windforce resisting system or for the buildings components and cladding For example using a directional procedure the windpressure on an enclosed building of any height is determined using a twotermed equation resulting from both external and internal pressures namely 13 Here for the windward wall at height z above the ground Eq 12 and for the leeward walls side walls and roof where the mean height of the roof a windgust effect factor which depends upon the exposure For example for a rigid structure a wall or roof pressure coefficient determined from a table These tabular values for the walls and a roof pitch of are given in Fig 112 Note in the elevation view that the pressure will vary with height on the windward side of the building whereas on the remaining sides and on the roof the pressure is assumed to be constant Negative values indicate pressures acting away from the surface the internal pressure coefficient which depends upon the type of openings in the building For fully enclosed buildings Here the signs indicate that either positive or negative suction pressure can occur within the building Application of Eq 13 will involve calculations of wind pressures from each side of the building with due considerations for the possibility of either positive or negative pressures acting on the buildings interior 1GCpi2 018 1GCpi2 u 10 Cp G 085 G z h q qh qz q p qGCp qh1GCpi2 qz Wind blowing on a wall will tend to tip a building or cause it to sidesway To prevent this engineers often use cross bracing to provide stabilityAlso see p 46 Surface LB Cp Use with Windward wall All values 08 qz Leeward wall 01 2 4 05 03 02 qh Side walls All values 07 qh Wall pressure coefficients Cp Wind direction Windward angle θ Leeward angle hL 10 θ 10 Normal to ridge 025 05 10 07 09 13 03 05 07 Maximum negative roof pressure coefficients Cp for use with qh Fig 112 For highrise buildings or those having a shape or location that makes them wind sensitive it is recommended that a dynamic approach be used to determine the wind loadings The methodology for doing this is also outlined in the ASCE 710 Standard It requires windtunnel tests to be performed on a scale model of the building and those surrounding it in order to simulate the natural environment The pressure effects of the wind on the building can be determined from pressure transducers attached to the model Also if the model has stiffness characteristics that are in proper scale to the building then the dynamic deflections of the building can be determined EXAMPLE 13 The enclosed building shown in the photo and in Fig 113a is used for storage purposes and is located outside of Chicago Illinois on open flat terrain When the wind is directed as shown determine the design wind pressure acting on the roof and sides of the building using the ASCE 710 Specifications SOLUTION First the wind pressure will be determined using Eq 12 The basic wind speed is V 105 mih since the building is used for storage Also for flat terrain Kzt 10 Since only wind loading is being considered Kd 10 Therefore qz 000256 KzKztKdV2 000256 Kz10101052 2822 Kz From Fig 113a h 75 tan 10 1322 ft so that the mean height of the roof is h 25 13222 316 ft Using the values of Kz in Table 15 calculated values of the pressure profile are listed in the table in Fig 113b Note the value of Kz was determined by linear interpolation for z h ie 104 09840 30 104 Kz40 316 Kz 0990 and so qh 28220990 279 psf In order to apply Eq 13 the gust factor is G 085 and GCpi 018 Thus p qGCp qhGCpi q085Cp 279018 085qCp 503 1 The pressure loadings are obtained from this equation using the calculated values for qz listed in Fig 113b in accordance with the windpressure profile in Fig 112 z ft Kz qz psf 015 085 240 20 090 254 25 094 265 h 316 0990 279 13 LOADS 21 1 Windward Wall Here the pressure varies with height z since must be used For all values of so that from Eq 1 Leeward Wall Here so that Also and so from Eq 1 Side Walls For all values of and therefore since we must use in Eq 1 we have Windward Roof Here so that and Thus Leeward Roof In this case therefore with we get These two sets of loadings are shown on the elevation of the building representing either positive or negative suction internal building pressure Fig 113c The main framing structure of the building must resist these loadings as well as for separate loadings calculated from wind blowing on the front or rear of the building p 122 psf or 209 psf q qh Cp 03 p 216 psf or 116 psf q qh Cp 07 hL 31621752 0211 6 025 p 216 psf or 116 psf q qh Cp 07 LB p 169 psf or 684 psf q qh Cp 05 LB 21752150 1 p25 130 psf or 231 psf p20 122 psf or 223 psf p015 113 psf or 213 psf LB Cp 08 qzGCp 113 psf 122 psf 130 psf 216 psf 122 psf 169 psf c 213 psf 223 psf 231 psf 116 psf 209 psf 684 psf Design Wind Pressure for Signs If the structure represents a sign the wind will produce a resultant force acting on the face of the sign which is determined from F qhGCfAs Here qh the wind pressure evaluated at the height h measured from the ground to the top of the sign G the windgust coefficient factor defined previously Cf a force coefficient which depends upon the aspect ratio width B of the sign to height s of the sign and the clear area ratio sign height s to the elevation h measured from the ground to the top of the sign For cases of wind directed normal to the sign and through its center for Bs 4 values are listed in Table 16 As the area of the face of the sign in ft² m² To allow for normal and oblique wind directions the calculated resultant force is assumed to act either through the geometric center of the face of the sign or at other specified locations on the face of the sign which depend upon the ratios sh and Bs Hurricane winds acting on the face of this sign were strong enough to noticeably bend the two supporting arms causing the material to yield Proper design would have prevented this 13 LOADS 23 1 Snow Loads In some parts of the country roof loading due to snow can be quite severe and therefore protection against possible failure is of primary concern Design loadings typically depend on the buildings general shape and roof geometry wind exposure location its importance and whether or not it is heated Like wind snow loads in the ASCE 710 Standard are generally determined from a zone map reporting 50year recurrence intervals of an extreme snow depth For example on the relatively flat elevation throughout the midsection of Illinois and Indiana the ground snow loading is However for areas of Montana specific case studies of ground snow loadings are needed due to the variable elevations throughout the state Specifications for snow loads are covered in the ASCE 710 Standard although no single code can cover all the implications of this type of loading If a roof is flat defined as having a slope of less than 5 then the pressure loading on the roof can be obtained by modifying the ground snow loading by the following empirical formula 15 Here an exposure factor which depends upon the terrain For example for a fully exposed roof in an unobstructed area whereas if the roof is sheltered and located in the center of a large city then a thermal factor which refers to the average temperature within the building For unheated structures kept below freezing whereas if the roof is supporting a normally heated structure then the importance factor as it relates to occupancy For example for agriculture and storage facilities and for schools and hospitals If then use the largest value for either computed from the above equation or from If 1096 kNm22 then use pf Is120 lbft22 pg 7 20 lbft2 pf Ispg pf pg 20 lbft2 1096 kNm22 Is 120 Is 080 Is Ct 10 Ct 12 Ct Ce 12 Ce 08 Ce pf 07Ce Ct Ispg pg 20 lbft2 1096 kNm22 Excessive snow and ice loadings act on this roof 24 CHAPTER 1 TYPES OF STRUCTURES AND LOADS 1 Earthquake Loads Earthquakes produce loadings on a structure through its interaction with the ground and its response characteristics These loadings result from the structures distortion caused by the grounds motion and the lateral resistance of the structure Their magnitude depends on the amount and type of ground accelerations and the mass and stiffness of the structure In order to provide some insight as to the nature of earthquake loads consider the simple structural model shown in Fig 115 This model may represent a singlestory building where the top block is the lumped mass of the roof and the middle block is the lumped stiffness of all the buildings columns During an earthquake the ground vibrates both horizontally and vertically The horizontal accelerations create shear forces in the column that put the block in sequential motion with the ground If the column is stiff and the block has a small mass the period of vibration of the block will be short and the block will accelerate with the same motion as the ground and undergo only slight relative displacements For an actual structure which is designed to have large amounts of bracing and stiff connections this can be beneficial since less stress is developed in the members On the other hand if the column in Fig 115 is very flexible and the block has a large mass then earthquakeinduced motion will cause small accelerations of the block and large relative displacements In practice the effects of a structures acceleration velocity and displacement can be determined and represented as an earthquake The unheated storage facility shown in Fig 114 is located on flat open terrain in southern Illinois where the specified ground snow load is Determine the design snow load on the roof which has a slope of 4 15 lbft2 EXAMPLE 14 Fig 114 SOLUTION Since the roof slope is 5 we will use Eq 15 Here due to the open area and Thus Since then also By comparison choose Ans pf 18 lbft2 pf Ipg 12115 lbft22 18 lbft2 pg 15 lbft2 6 20 lbft2 07108211221082115 lbft22 806 lbft2 pf 07Ce Ct Ispg Is 08 Ct 12 Ce 08 lumped mass of columns lumped mass of roof Fig 115 13 LOADS 25 1 response spectrum Once this graph is established the earthquake loadings can be calculated using a dynamic analysis based on the theory of structural dynamics This type of analysis is gaining popularity although it is often quite elaborate and requires the use of a computer Even so such an analysis becomes mandatory if the structure is large Some codes require that specific attention be given to earthquake design especially in areas of the country where strong earthquakes predominate Also these loads should be seriously considered when designing highrise buildings or nuclear power plants In order to assess the importance of earthquake design consideration one can check the seismic groundacceleration maps published in the ASCE 710 Standard These maps provide the peak ground accelerations caused by an earthquake along with risk coefficients Regions vary from low risk such as parts of Texas to very high risk such as along the west coast of California For small structures a static analysis for earthquake design may be satisfactory This case approximates the dynamic loads by a set of externally applied static forces that are applied laterally to the structure One such method for doing this is reported in the ASCE 710 Standard It is based upon finding a seismic response coefficient determined from the soil properties the ground accelerations and the vibrational response of the structure For most structures this coefficient is then multiplied by the structures total dead load W to obtain the base shear in the structureThe value of is actually determined from where the spectral response acceleration for short periods of vibration a response modification factor that depends upon the ductility of the structure Steel frame members which are highly ductile can have a value as high as 8 whereas reinforced concrete frames can have a value as low as 3 the importance factor that depends upon the use of the building For example for agriculture and storage facilities and for hospitals and other essential facilities With each new publication of the Standard values of these coefficients are updated as more accurate data about earthquake response become available Hydrostatic and Soil Pressure When structures are used to retain water soil or granular materials the pressure developed by these loadings becomes an important criterion for their design Examples of such types of structures include tanks dams ships bulkheads and retaining walls Here the laws of hydrostatics and soil mechanics are applied to define the intensity of the loadings on the structure Ie 15 Ie 1 Ie R SDS Cs SDS RIe Cs Cs The design of this retaining wall requires estimating the soil pressure acting on it Also the gate of the lock will be subjected to hydrostatic pressure that must be considered for its design 26 CHAPTER 1 TYPES OF STRUCTURES AND LOADS 1 Other Natural Loads Several other types of live loads may also have to be considered in the design of a structure depending on its location or use These include the effect of blast temperature changes and differential settlement of the foundation 14 Structural Design Whenever a structure is designed it is important to give consideration to both material and load uncertainties These uncertainties include a possible variability in material properties residual stress in materials intended measurements being different from fabricated sizes loadings due to vibration or impact and material corrosion or decay ASD Allowablestress design ASD methods include both the material and load uncertainties into a single factor of safety The many types of loads discussed previously can occur simultaneously on a structure but it is very unlikely that the maximum of all these loads will occur at the same time For example both maximum wind and earthquake loads normally do not act simultaneously on a structure For allowablestress design the computed elastic stress in the material must not exceed the allowable stress for each of various load combinations Typical load combinations as specified by the ASCE 710 Standard include dead load LRFD Since uncertainty can be considered using probability theory there has been an increasing trend to separate material uncertainty from load uncertainty This method is called strength design or LRFD load and resistance factor design For example to account for the uncertainty of loads this method uses load factors applied to the loads or combinations of loads According to the ASCE 710 Standard some of the load factors and combinations are 14 dead load 09 dead load 10 wind load 09 dead load 10 earthquake load In all these cases the combination of loads is thought to provide a maximum yet realistic loading on the structure 12 1dead load2 16 1live load2 05 1snow load2 06 1dead load2 07 1earthquake load2 06 1dead load2 06 wind load 14 STRUCTURAL DESIGN 27 1 11 The floor of a heavy storage warehouse building is made of 6inthick stone concrete If the floor is a slab having a length of 15 ft and width of 10 ft determine the resultant force caused by the dead load and the live load 12 The floor of the office building is made of 4inthick lightweight concrete If the office floor is a slab having a length of 20 ft and width of 15 ft determine the resultant force caused by the dead load and the live load 14 The New Jersey barrier is commonly used during highway construction Determine its weight per foot of length if it is made from plain stone concrete PROBLEMS Prob 12 26 in 40 in 8 in 10 in Prob 13 12 in 4 in 24 in 6 in 55 75 Prob 14 13 The Tbeam is made from concrete having a specific weight of 150 lbft3 Determine the dead load per foot length of beam Neglect the weight of the steel reinforcement 15 The floor of a light storage warehouse is made of 150mmthick lightweight plain concrete If the floor is a slab having a length of 7 m and width of 3 m determine the resultant force caused by the dead load and the live load 16 The prestressed concrete girder is made from plain stone concrete and four in coldform steel reinforcing rods Determine the dead weight of the girder per foot of its length 3 4 8 in 8 in 4 in 4 in 6 in 6 in 6 in 20 in Prob 16 28 CHAPTER 1 TYPES OF STRUCTURES AND LOADS 1 25 m Prob 17 17 The wall is 25 m high and consists of 51 mm 102 mm studs plastered on one side On the other side is 13 mm fiberboard and 102 mm clay brick Determine the average load in kNm of length of wall that the wall exerts on the floor 111 A fourstory office building has interior columns spaced 30 ft apart in two perpendicular directions If the flatroof live loading is estimated to be 30 lbft2 determine the reduced live load supported by a typical interior column located at ground level 112 A twostory light storage warehouse has interior columns that are spaced 12 ft apart in two perpendicular directions If the live loading on the roof is estimated to be 25 lbft2 determine the reduced live load supported by a typical interior column at a the groundfloor level and b the secondfloor level 113 The office building has interior columns spaced 5 m apart in perpendicular directions Determine the reduced live load supported by a typical interior column located on the first floor under the offices 18 A building wall consists of exterior stud walls with brick veneer and 13 mm fiberboard on one side If the wall is 4 m high determine the load in kNm that it exerts on the floor 19 The interior wall of a building is made from 2 4 wood studs plastered on two sides If the wall is 12 ft high determine the load in lbft of length of wall that it exerts on the floor 110 The second floor of a light manufacturing building is constructed from a 5inthick stone concrete slab with an added 4in cinder concrete fill as shown If the suspended ceiling of the first floor consists of metal lath and gypsum plaster determine the dead load for design in pounds per square foot of floor area 114 A twostory hotel has interior columns for the rooms that are spaced 6 m apart in two perpendicular directions Determine the reduced live load supported by a typical interior column on the first floor under the public rooms 4 in cinder fill 5 in concrete slab ceiling Prob 110 Prob 113 Prob 116 Prob 115 14 STRUCTURAL DESIGN 29 1 115 Wind blows on the side of a fully enclosed hospital located on open flat terrain in Arizona Determine the external pressure acting over the windward wall which has a height of 30 ftThe roof is flat 116 Wind blows on the side of the fully enclosed hospital located on open flat terrain in Arizona Determine the external pressure acting on the leeward wall which has a length of 200 ft and a height of 30 ft 118 The light metal storage building is on open flat terrain in central Oklahoma If the side wall of the building is 14 ft high what are the two values of the external wind pressure acting on this wall when the wind blows on the back of the building The roof is essentially flat and the building is fully enclosed 117 A closed storage building is located on open flat terrain in central Ohio If the side wall of the building is 20 ft high determine the external wind pressure acting on the windward and leeward walls Each wall is 60 ft long Assume the roof is essentially flat Prob 117 Prob 118 30 CHAPTER 1 TYPES OF STRUCTURES AND LOADS 1 120 A hospital located in central Illinois has a flat roof Determine the snow load in kNm2 that is required to design the roof 119 Determine the resultant force acting perpendicular to the face of the billboard and through its center if it is located in Michigan on open flat terrain The sign is rigid and has a width of 12 m and a height of 3 m Its top side is 15 m from the ground Prob 119 121 The school building has a flat roof It is located in an open area where the ground snow load is 068 kNm2 Determine the snow load that is required to design the roof 122 The hospital is located in an open area and has a flat roof and the ground snow load is 30 lbft2 Determine the design snow load for the roof Prob 121 Prob 122 CHAPTER REVIEW 31 1 The basic structural elements are Tie RodsSlender members subjected to tension Often used for bracing BeamsMembers designed to resist bending moment They are often fixed or pin supported and can be in the form of a steel plate girder reinforced concrete or laminated wood ColumnsMembers that resist axial compressive force If the column also resists bending it is called a beam column tie rod beam column column simply supported beam cantilevered beam Loads are specified in codes such as the ASCE 710 code Dead loads are fixed and refer to the weights of members and materials Live loads are movable and consist of uniform building floor loads traffic and train loads on bridges impact loads due to vehicles and machines wind loads snow loads earthquake loads and hydrostatic and soil pressure The types of structures considered in this book consist of trusses made from slender pinconnected members forming a series of triangles cables and arches which carry tensile and compressive loads respectively and frames composed of pin or fixedconnected beams and columns CHAPTER REVIEW Oftentimes the elements of a structure like the beams and girders of this building frame are connected together in a manner whereby the analysis can be considered statically determinate 2 33 In this chapter we will direct our attention to the most common form of structure that the engineer will have to analyze and that is one that lies in a plane and is subjected to a force system that lies in the same plane We begin by discussing the importance of choosing an appropriate analytical model for a structure so that the forces in the structure may be determined with reasonable accuracy Then the criteria necessary for structural stability are discussed Finally the analysis of statically determinate planar pinconnected structures is presented 21 Idealized Structure An exact analysis of a structure can never be carried out since estimates always have to be made of the loadings and the strength of the materials composing the structure Furthermore points of application for the loadings must also be estimated It is important therefore that the structural engineer develop the ability to model or idealize a structure so that he or she can perform a practical force analysis of the members In this section we will develop the basic techniques necessary to do this Analysis of Statically Determinate Structures 34 CHAPTER 2 ANALYSIS OF STATICALLY DETERMINATE STRUCTURES Fig 22 2 Notice that the deck of this concrete bridge is made so that one section can be considered roller supported on the other section Support Connections Structural members are joined together in various ways depending on the intent of the designerThe three types of joints most often specified are the pin connection the roller support and the fixed joint A pinconnected joint and a roller support allow some freedom for slight rotationwhereas a fixed joint allows no relative rotation between the connected members and is consequently more expensive to fabricate Examples of these joints fashioned in metal and concrete are shown in Figs 21 and 22 respectively For most timber structures the members are assumed to be pin connected since bolting or nailing them will not sufficiently restrain them from rotating with respect to each other Idealized models used in structural analysis that represent pinned and fixed supports and pinconnected and fixedconnected joints are shown in Figs 23a and 23b In reality however all connections exhibit some stiffness toward joint rotations owing to friction and material behavior In this case a more appropriate model for a support or joint might be that shown in Fig 23c If the torsional spring constant the joint is a pin and if k q the joint is fixed k 0 Fig 21 weld weld stiffeners b typical fixedsupported connection concrete a typical pinsupported connection metal a typical rollersupported connection concrete b typical fixedsupported connection metal When selecting a particular model for each support or joint the engineer must be aware of how the assumptions will affect the actual performance of the member and whether the assumptions are reasonable for the structural design For example consider the beam shown in Fig 24a which is used to support a concentrated load P The angle connection at support A is like that in Fig 21a and can therefore be idealized as a typical pin support Furthermore the support at B provides an approximate point of smooth contact and so it can be idealized as a roller The beams thickness can be neglected since it is small in comparison to the beams length and therefore the idealized model of the beam is as shown in Fig 24b The analysis of the loadings in this beam should give results that closely approximate the loadings in the actual beam To show that the model is appropriate consider a specific case of a beam made of steel with P 8 k 8000 lb and L 20 ft One of the major simplifications made here was assuming the support at A to be a pin Design of the beam using standard code procedures indicates that a W10 19 would be adequate for supporting the load Using one of the deflection methods of Chapter 8 the rotation at the pin support can be calculated as θ 00103 rad 059 From Fig 24c such a rotation only moves the top or bottom flange a distance of Δ θr 00103 rad512 in 00528 in This small amount would certainly be accommodated by the connection fabricated as shown in Fig 21a and therefore the pin serves as an appropriate model Codes such as the Manual of Steel Construction American Institute of Steel Construction 36 CHAPTER 2 ANALYSIS OF STATICALLY DETERMINATE STRUCTURES 2 Other types of connections most commonly encountered on coplanar structures are given in Table 21 It is important to be able to recognize the symbols for these connections and the kinds of reactions they exert on their attached members This can easily be done by noting how the connection prevents any degree of freedom or displacement of the member In particular the support will develop a force on the member if it prevents translation of the member and it will develop a moment if it prevents rotation of the member For example a member in contact with a smooth surface 3 is prevented from translating only in one direction which is perpendicular or normal to the surface Hence the surface exerts only a normal force F on the member in this direction The magnitude of this force represents one unknownAlso note that the member is free to rotate on the surface so that a moment cannot be developed by the surface on the memberAs another example the fixed support 7 prevents both translation and rotation of a member at the point of connection Therefore this type of support exerts two force components and a moment on the member The curl of the moment lies in the plane of the page since rotation is prevented in that plane Hence there are three unknowns at a fixed support In reality all supports actually exert distributed surface loads on their contacting members The concentrated forces and moments shown in Table 21 represent the resultants of these load distributions This representation is of course an idealization however it is used here since the surface area over which the distributed load acts is considerably smaller than the total surface area of the connecting members A typical rocker support used for a bridge girder Rollers and associated bearing pads are used to support the prestressed concrete girders of a highway bridge The short link is used to connect the two girders of the highway bridge and allow for thermal expansion of the deck Typical pin used to support the steel girder of a railroad bridge Fig 26 a D A C B joist slab column girder idealized framing plan b Fig 25 Idealized Structure Having stated the various ways in which the connections on a structure can be idealized we are now ready to discuss some of the techniques used to represent various structural systems by idealized models As a first example consider the jib crane and trolley in Fig 25a For the structural analysis we can neglect the thickness of the two main members and will assume that the joint at B is fabricated to be rigid Furthermore the support connection at A can be modeled as a fixed support and the details of the trolley excludedThus the members of the idealized structure are represented by two connected lines and the load on the hook is represented by a single concentrated force F Fig 25b This idealized structure shown here as a line drawing can now be used for applying the principles of structural analysis which will eventually lead to the design of its two main members Beams and girders are often used to support building floors In particular a girder is the main loadcarrying element of the floor whereas the smaller elements having a shorter span and connected to the girders are called beams Often the loads that are applied to a beam or girder are transmitted to it by the floor that is supported by the beam or girder Again it is important to be able to appropriately idealize the system as a series of models which can be used to determine to a close approxi mation the forces acting in the members Consider for example the framing used to support a typical floor slab in a building Fig 26a Here the slab is supported by floor joists located at even intervals and these in turn are supported by the two side girders AB and CD For analysis it is reasonable to assume that the joints are pin andor roller connected to the girders and that the girders are pin andor roller connected to the columns The top view of the structural framing plan for this system is shown in Fig 26b In this graphic scheme notice that the lines representing the joists do not touch the girders and the lines for the girders do not touch the columns This symbolizes pin and or rollersupported connections On the other hand if the framing plan is intended to represent fixedconnected members such as those that are welded 38 CHAPTER 2 ANALYSIS OF STATICALLY DETERMINATE STRUCTURES 2 F 3 m A B actual structure a 4 m 4 m B A 3 m F idealized structure b Fig 27 21 IDEALIZED STRUCTURE 39 2 instead of simple bolted connections then the lines for the beams or girders would touch the columns as in Fig 27 Similarly a fixed connected overhanging beam would be represented in top view as shown in Fig 28 If reinforced concrete construction is used the beams and girders are represented by double lines These systems are generally all fixed connected and therefore the members are drawn to the supports For example the structural graphic for the castinplace reinforced concrete system in Fig 29a is shown in top view in Fig 29b The lines for the beams are dashed because they are below the slab Structural graphics and idealizations for timber structures are similar to those made of metal For example the structural system shown in Fig 210a represents beamwall construction whereby the roof deck is supported by wood joists which deliver the load to a masonry wall The joists can be assumed to be simply supported on the wall so that the idealized framing plan would be like that shown in Fig 210b fixedconnected beam idealized beam fixedconnected overhanging beam idealized beam a idealized framing plan b a idealized framing plan b Fig 28 Fig 29 Fig 210 Fig 211 Tributary Loadings When flat surfaces such as wallsfloorsor roofs are supported by a structural frame it is necessary to determine how the load on these surfaces is transmitted to the various structural elements used for their support There are generally two ways in which this can be done The choice depends on the geometry of the structural system the material from which it is made and the method of its construction OneWay System A slab or deck that is supported such that it delivers its load to the supporting members by oneway action is often referred to as a oneway slab To illustrate the method of load transmission consider the framing system shown in Fig 211a where the beams AB CD and EF rest on the girders AE and BF If a uniform load of is placed on the slab then the center beam CD is assumed to support the load acting on the tributary area shown dark shaded on the structural framing plan in Fig 211b Member CD is therefore subjected to a linear distribution of load of shown on the idealized beam in Fig 211c The reactions on this beam 2500 lb would then be applied to the center of the girders AE and BF shown idealized in Fig 211d Using this same conceptdo you see how the remaining portion of the slab loading is transmitted to the ends of the girder as 1250 lb 1100 lbft2215 ft2 500 lbft 100 lbft2 40 CHAPTER 2 ANALYSIS OF STATICALLY DETERMINATE STRUCTURES 2 a 10 ft E F D B 100 lbft2 A C 5 ft 5 ft idealized framing plan b A B C D E F 25 ft 25 ft 25 ft 25 ft A C D B 500 lbft 2500 lb idealized beam c 2500 lb 10 ft A E 1250 lb 2500 lb 1250 lb 5 ft 5 ft idealized girder d The structural framework of this building consists of concrete floor joists which were formed on site using metal pansThese joists are simply supported on the girders which in turn are simply supported on the columns An example of oneway slab construction of a steel frame building having a poured concrete floor on a corrugated metal deck The load on the floor is considered to be transmitted to the beams not the girders For some floor systems the beams and girders are connected to the columns at the same elevation as in Fig 212a If this is the case the slab can in some cases also be considered a oneway slab For example if the slab is reinforced concrete with reinforcement in only one direction or the concrete is poured on a corrugated metal deck as in the above photo then oneway action of load transmission can be assumed On the other hand if the slab is flat on top and bottom and is reinforced in two directions then consideration must be given to the possibility of the load being transmitted to the supporting members from either one or two directions For example consider the slab and framing plan in Fig 212b According to the American Concrete Institute ACI 318 code if L2 L1 and if the span ratio L2L1 2 the slab will behave as a oneway slab since as L1 becomes smaller the beams AB CD and EF provide the greater stiffness to carry the load TwoWay System If according to the ACI 318 concrete code the support ratio in Fig 212b is L2L1 2 the load is assumed to be delivered to the supporting beams and girders in two directions When this is the case the slab is referred to as a twoway slab To show one method of treating this case consider the square reinforced concrete slab in Fig 213a which is supported by four 10ftlong edge beams AB BD DC and CA Here L2L1 1 Due to twoway slab action the assumed tributary area for beam AB is shown dark shaded in Fig 213b This area is determined by constructing diagonal 45 lines as shown Hence if a uniform load of 100 lbft² is applied to the slab a peak intensity of 100 lbft²5 ft 500 lbft will be applied to the center of beam AB resulting in a triangular load distribution shown in Fig 213c For other geometries that cause twoway action a similar procedure can be used For example if L2L1 15 it is then necessary to construct 45 lines that intersect as shown in Fig 214a A 100lbft² loading placed on the slab will then produce trapezoidal and triangular distributed loads on members AB and AC Fig 214b and 214c respectively 21 IDEALIZED STRUCTURE 43 2 The ability to reduce an actual structure to an idealized form as shown by these examples can only be gained by experienceTo provide practice at doing this the example problems and the problems for solution throughout this book are presented in somewhat realistic form and the associated problem statements aid in explaining how the connections and supports can be modeled by those listed in Table 21 In engineering practice if it becomes doubtful as to how to model a structure or transfer the loads to the members it is best to consider several idealized structures and loadings and then design the actual structure so that it can resist the loadings in all the idealized models EXAMPLE 21 The floor of a classroom is to be supported by the bar joists shown in Fig 215a Each joist is 15 ft long and they are spaced 25 ft on centers The floor itself is to be made from lightweight concrete that is 4 in thick Neglect the weight of the joists and the corrugated metal deck and determine the load that acts along each joist SOLUTION The dead load on the floor is due to the weight of the concrete slabFrom Table 13 for 4 in of lightweight concrete it is From Table 14 the live load for a classroom is Thus the total floor load is For the floor system and Since the concrete slab is treated as a oneway slabThe tributary area for each joist is shown in Fig215b Therefore the uniform load along its length is This loading and the end reactions on each joist are shown in Fig215c w 72 lbft2125 ft2 180 lbft L2L1 7 2 L2 15 ft L1 25 ft 32 lbft2 40 lbft2 72 lbft2 40 lbft2 14218 lbft22 32 lbft2 15 ft 25 ft b 1350 lb c 1350 lb 180 lbft Fig 215 a 44 CHAPTER 2 ANALYSIS OF STATICALLY DETERMINATE STRUCTURES 2 The flat roof of the steelframe building shown in the photo is intended to support a total load of over its surface Determine the roof load within region ABCD that is transmitted to beam BCThe dimensions are shown in Fig 216a 2 kNm2 EXAMPLE 22 2 m a B A C D 15 m 15 m 2 m 2 m 4 m B C 2 m b 4 kNm 15 m 15 m A B C D Fig 216 SOLUTION In this case and Since we have twoway slab actionThe tributary loading along each edge beam is shown in Fig 216a where the lighter shaded trapezoidal area of loading is transmitted to member BCThe peak intensity of this loading is As a result the distribution of load along BC is shown in Fig 216b 12 kNm2212 m2 4 kNm L2L1 125 6 2 L1 4 m L2 5 m This process of tributary load transmission should also be calculated for the region to the right of BC shown in the photo and this load should also be placed on BC See the next example EXAMPLE 23 The concrete girders shown in the photo of the passenger car parking garage span 30 ft and are 15 ft on center If the floor slab is 5 in thick and made of reinforced stone concrete and the specified live load is 50 lbft² see Table 14 determine the distributed load the floor system transmits to each interior girder SOLUTION Here L2 30 ft and L1 15 ft so that L2L1 2 We have a twoway slab From Table 12 for reinforced stone concrete the specific weight of the concrete is 150 lbft³ Thus the design floor loading is p 150 lbft³ 512 ft 50 lbft² 1125 lbft² A trapezoidal distributed loading is transmitted to each interior girder AB from each of its sides The maximum intensity of each of these distributed loadings is 1125 lbft²75 ft 84375 lbft so that on the girder this intensity becomes 284375 lbft 16875 lbft Fig 217b Note For design consideration should also be given to the weight of the girder 46 CHAPTER 2 ANALYSIS OF STATICALLY DETERMINATE STRUCTURES 2 22 Principle of Superposition The principle of superposition forms the basis for much of the theory of structural analysis It may be stated as follows The total displacement or internal loadings stress at a point in a structure subjected to several external loadings can be determined by adding together the displacements or internal loadings stress caused by each of the external loads acting separately For this statement to be valid it is necessary that a linear relationship exist among the loads stresses and displacements Two requirements must be imposed for the principle of superposition to apply 1 The material must behave in a linearelastic manner so that Hookes law is valid and therefore the load will be proportional to displacement 2 The geometry of the structure must not undergo significant change when the loads are applied ie small displacement theory applies Large displacements will significantly change the position and orientation of the loads An example would be a cantilevered thin rod subjected to a force at its end Throughout this text these two requirements will be satisfied Here only linearelastic material behavior occurs and the displacements produced by the loads will not significantly change the directions of applied loadings nor the dimensions used to compute the moments of forces wind The walls on the sides of this building are used to strengthen its structure when the building is subjected to large hurricane wind loadings applied to its front or back These walls are called shear walls 23 EQUATIONS OF EQUILIBRIUM 47 2 M M V V N N internal loadings Fig 218 23 Equations of Equilibrium It may be recalled from statics that a structure or one of its members is in equilibrium when it maintains a balance of force and moment In general this requires that the force and moment equations of equilibrium be satisfied along three independent axes namely 21 The principal loadcarrying portions of most structures however lie in a single plane and since the loads are also coplanar the above requirements for equilibrium reduce to 22 Here and represent respectively the algebraic sums of the x and y components of all the forces acting on the structure or one of its members and represents the algebraic sum of the moments of these force components about an axis perpendicular to the xy plane the z axis and passing through point O Whenever these equations are applied it is first necessary to draw a freebody diagram of the structure or its members If a member is selected it must be isolated from its supports and surroundings and its outlined shape drawn All the forces and couple moments must be shown that act on the member In this regard the types of reactions at the supports can be determined using Table 21 Also recall that forces common to two members act with equal magnitudes but opposite directions on the respective freebody diagrams of the members If the internal loadings at a specified point in a member are to be determined the method of sections must be used This requires that a cut or section be made perpendicular to the axis of the member at the point where the internal loading is to be determined A freebody diagram of either segment of the cut member is isolated and the internal loads are then determined from the equations of equilibrium applied to the segment In general the internal loadings acting at the section will consist of a normal force N shear force V and bending moment M as shown in Fig 218 We will cover the principles of statics that are used to determine the external reactions on structures in Sec 25 Internal loadings in structural members will be discussed in Chapter 4 MO Fy Fx MO 0 Fy 0 Fx 0 Fx 0 Fy 0 Fz 0 Mx 0 My 0 Mz 0 48 CHAPTER 2 ANALYSIS OF STATICALLY DETERMINATE STRUCTURES 2 24 Determinacy and Stability Before starting the force analysis of a structure it is necessary to establish the determinacy and stability of the structure Determinacy The equilibrium equations provide both the necessary and sufficient conditions for equilibriumWhen all the forces in a structure can be determined strictly from these equations the structure is referred to as statically determinate Structures having more unknown forces than available equilibrium equations are called statically indeterminate As a general rule a structure can be identified as being either statically determinate or statically indeterminate by drawing freebody diagrams of all its members or selective parts of its members and then comparing the total number of unknown reactive force and moment components with the total number of available equilibrium equations For a coplanar structure there are at most three equilibrium equations for each part so that if there is a total of n parts and r force and moment reaction components we have 23 In particular if a structure is statically indeterminate the additional equations needed to solve for the unknown reactions are obtained by relating the applied loads and reactions to the displacement or slope at different points on the structure These equations which are referred to as compatibility equations must be equal in number to the degree of indeterminacy of the structure Compatibility equations involve the geometric and physical properties of the structure and will be discussed further in Chapter 10 We will now consider some examples to show how to classify the determinacy of a structureThe first example considers beams the second example pinconnected structures and in the third we will discuss frame structures Classification of trusses will be considered in Chapter 3 r 7 3n statically indeterminate r 3n statically determinate Drawing the freebody diagrams is not strictly necessary since a mental count of the number of unknowns can also be made and compared with the number of equilibrium equations 24 DETERMINACY AND STABILITY 49 2 EXAMPLE 24 Classify each of the beams shown in Fig 219a through 219d as statically determinate or statically indeterminate If statically indeterminate report the number of degrees of indeterminacy The beams are subjected to external loadings that are assumed to be known and can act anywhere on the beams SOLUTION Compound beams ie those in Fig 219c and 219d which are composed of pinconnected members must be disassembled Note that in these cases the unknown reactive forces acting between each member must be shown in equal but opposite pairs The freebody diagrams of each member are shownApplying or the resulting classifications are indicated r 7 3n r 3n a c b d Fig 219 Statically determinate Ans n 1 3 3112 r 3 Statically indeterminate to the second degree Ans n 1 5 7 3112 r 5 Statically determinate Ans n 2 6 3122 r 6 Statically indeterminate to the first degree Ans r 10 n 3 10 7 3132 50 CHAPTER 2 ANALYSIS OF STATICALLY DETERMINATE STRUCTURES 2 Classify each of the pinconnected structures shown in Fig 220a through 220d as statically determinate or statically indeterminate If statically indeterminate report the number of degrees of indeterminacy The structures are subjected to arbitrary external loadings that are assumed to be known and can act anywhere on the structures SOLUTION Classification of pinconnected structures is similar to that of beams The freebody diagrams of the members are shown Applying or r 7 3n the resulting classifications are indicated r 3n EXAMPLE 25 a b c Statically indeterminate to the fourth degree Ans n 2 10 7 6 r 10 Statically determinate Ans n 3 9 9 r 9 d Statically determinate Ans n 3 9 9 r 9 Statically indeterminate to the first degree Ans n 2 7 7 6 r 7 Fig 220 24 DETERMINACY AND STABILITY 51 2 EXAMPLE 26 Classify each of the frames shown in Fig 221a and 221b as statically determinate or statically indeterminate If statically indeterminate report the number of degrees of indeterminacy The frames are subjected to external loadings that are assumed to be known and can act anywhere on the frames SOLUTION Unlike the beams and pinconnected structures of the previous examples frame structures consist of members that are connected together by rigid joints Sometimes the members form internal loops as in Fig 221a Here ABCD forms a closed loop In order to classify these structures it is necessary to use the method of sections and cut the loop apart The freebody diagrams of the sectioned parts are drawn and the frame can then be classified Notice that only one section through the loop is required since once the unknowns at the section are determined the internal forces at any point in the members can then be found using the method of sections and the equations of equilibrium A second example of this is shown in Fig 221b Although the frame in Fig 221c has no closed loops we can use this same method using vertical sections to classify it For this case we can also just draw its complete freebody diagram The resulting classifications are indicated in each figure c B A C D a Statically indeterminate to the third degree Ans n 2 9 7 6 r 9 Statically indeterminate to the sixth degree Ans n 1 9 7 3 r 9 This frame has no closed loops Fig 221 Statically indeterminate to the ninth degree Ans n 3 18 7 9 r 18 b Statically indeterminate to the sixth degree Ans n 4 18 7 12 r 18 a c 52 CHAPTER 2 ANALYSIS OF STATICALLY DETERMINATE STRUCTURES 2 Fig 222 Fig 223 Stability To ensure the equilibrium of a structure or its members it is not only necessary to satisfy the equations of equilibrium but the members must also be properly held or constrained by their supports Two situations may occur where the conditions for proper constraint have not been met Partial Constraints In some cases a structure or one of its members may have fewer reactive forces than equations of equilibrium that must be satisfied The structure then becomes only partially constrained For example consider the member shown in Fig 222 with its corresponding freebody diagram Here the equation will not be satisfied for the loading conditions and therefore the member will be unstable Improper Constraints In some cases there may be as many unknown forces as there are equations of equilibrium however instability or movement of a structure or its members can develop because of improper constraining by the supports This can occur if all the support reactions are concurrent at a point An example of this is shown in Fig 223 From the freebody diagram of the beam it is seen that the summation of moments about point O will not be equal to zero thus rotation about point O will take place Another way in which improper constraining leads to instability occurs when the reactive forces are all parallelAn example of this case is shown in Fig 224 Here when an inclined force P is applied the summation of forces in the horizontal direction will not equal zero 1Pd Z 02 Fx 0 A C B P O O FB d FA FC P d concurrent reactions P A B C P FA FB FC parallel reactions P A A P FA MA partial constraints Fig 224 24 DETERMINACY AND STABILITY 53 2 In general then a structure will be geometrically unstablethat is it will move slightly or collapseif there are fewer reactive forces than equations of equilibrium or if there are enough reactions instability will occur if the lines of action of the reactive forces intersect at a common point or are parallel to one another If the structure consists of several members or components local instability of one or several of these members can generally be determined by inspection If the members form a collapsible mechanism the structure will be unstable We will now formalize these statements for a coplanar structure having n members or components with r unknown reactions Since three equilibrium equations are available for each member or component we have 24 If the structure is unstable it does not matter if it is statically determinate or indeterminate In all cases such types of structures must be avoided in practice The following examples illustrate how structures or their members can be classified as stable or unstable Structures in the form of a truss will be discussed in Chapter 3 r 6 3n unstable r Ú 3n unstable if member reactions are concurrent or parallel or some of the components form a collapsible mechanism The Kbracing on this frame provides lateral support from wind and vertical support of the floor girders Notice the use of concrete grout which is applied to insulate the steel to keep it from losing its stiffness in the event of a fire Classify each of the structures in Fig 225a through 225d as stable or unstable The structures are subjected to arbitrary external loads that are assumed to be known SOLUTION The structures are classified as indicated EXAMPLE 27 54 CHAPTER 2 ANALYSIS OF STATICALLY DETERMINATE STRUCTURES 2 Fig 225 B a A A B b B A B C c A B C D d The member is stable since the reactions are nonconcurrent and nonparallel It is also statically determinate Ans The member is unstable since the three reactions are concurrent at B Ans The beam is unstable since the three reactions are all parallel Ans The structure is unstable since so that by Eq 24 Also this can be seen by inspection since AB can move horizontally without restraint Ans r 6 3n 7 6 9 r 7 n 3 24 DETERMINACY AND STABILITY 55 2 Probs 2122 Probs 232425 21 The steel framework is used to support the rein forced stone concrete slab that is used for an office The slab is 200 mm thick Sketch the loading that acts along members BE and FED Take Hint See Tables 12 and 14 22 Solve Prob 21 with b 4 m a 3 m b 5 m a 2 m 26 The frame is used to support a 2inthick plywood floor of a residential dwelling Sketch the loading that acts along members BG and ABCD Set Hint See Tables 12 and 14 27 Solve Prob 26 with 28 Solve Prob 26 with b 15 ft a 9 ft b 8 ft a 8 ft b 15 ft a 5 ft PROBLEMS 23 The floor system used in a school classroom consists of a 4in reinforced stone concrete slab Sketch the loading that acts along the joist BF and side girder ABCDE Set Hint See Tables 12 and 14 24 Solve Prob 23 with 25 Solve Prob 23 with b 20 ft a 75 ft b 15 ft a 10 ft b 30 ft a 10 ft 29 The steel framework is used to support the 4in reinforced stone concrete slab that carries a uniform live loading of Sketch the loading that acts along members BE and FED Set Hint See Table 12 210 Solve Prob 29 with a 4 ft b 12 ft b 10 ft a 75 ft 500 lbft2 A B C D E F b a a A E b a a a a B C D F F H G E a a a b C A B D Probs 262728 A B C D E F b a a Probs 29210 56 CHAPTER 2 ANALYSIS OF STATICALLY DETERMINATE STRUCTURES 2 a b c d e Prob 211 211 Classify each of the structures as statically determinate statically indeterminate or unstable If indeterminate specify the degree of indeterminacy The supports or connections are to be assumed as stated 212 Classify each of the frames as statically determinate or indeterminate If indeterminate specify the degree of indeterminacyAll internal joints are fixed connected a b c d Prob 212 24 DETERMINACY AND STABILITY 57 2 213 Classify each of the structures as statically determinate statically indeterminate stable or unstable If indeterminate specify the degree of indeterminacy The supports or connections are to be assumed as stated 214 Classify each of the structures as statically determinate statically indeterminate stable or unstable If indeterminate specify the degree of indeterminacy The supports or connections are to be assumed as stated 215 Classify each of the structures as statically determinate statically indeterminate or unstable If indeterminate specify the degree of indeterminacy a b c Prob 215 roller fixed pin a fixed fixed b pin pin c pin pin Prob 213 rocker fixed a pin pin b fixed roller roller pin pin fixed c fixed fixed pin Prob 214 58 CHAPTER 2 ANALYSIS OF STATICALLY DETERMINATE STRUCTURES 2 a b c d Prob 216 a b c d Prob 217 216 Classify each of the structures as statically determi nate statically indeterminate or unstable If indeterminate specify the degree of indeterminacy 217 Classify each of the structures as statically determinate statically indeterminate stable or unstable If indeterminate specify the degree of indeterminacy 25 APPLICATION OF THE EQUATIONS OF EQUILIBRIUM 59 2 25 Application of the Equations of Equilibrium Occasionally the members of a structure are connected together in such a way that the joints can be assumed as pins Building frames and trusses are typical examples that are often constructed in this manner Provided a pinconnected coplanar structure is properly constrained and contains no more supports or members than are necessary to prevent collapse the forces acting at the joints and supports can be determined by applying the three equations of equilibrium to each member Understandably once the forces at the joints are obtained the size of the members connections and supports can then be determined on the basis of design code specifications To illustrate the method of force analysis consider the threemember frame shown in Fig 226a which is subjected to loads and The freebody diagrams of each member are shown in Fig 226b In total there are nine unknowns however nine equations of equilibrium can be written three for each member so the problem is statically determinate For the actual solution it is also possible and sometimes convenient to consider a portion of the frame or its entirety when applying some of these nine equations For example a freebody diagram of the entire frame is shown in Fig 226c One could determine the three reactions and on this rigid pinconnected system then analyze any two of its members Fig 226b to obtain the other six unknowns Furthermore the answers can be checked in part by applying the three equations of equilibrium to the remaining thirdmemberTo summarize this problem can be solved by writing at most nine equilibrium equations using freebody diagrams of any members andor combinations of connected members Any more than nine equations written would not be unique from the original nine and would only serve to check the results Cx Ay Ax P2 P1 1Fx 0 Fy 0 MO 02 Bx By Dy Ay Dx Ax Dy Dx P1 Ex Ey Bx By P2 Ey Ex Cx b P1 Cx Ax Ay c P2 Fig 226 B D E C P2 P1 a A 60 CHAPTER 2 ANALYSIS OF STATICALLY DETERMINATE STRUCTURES 2 Consider now the twomember frame shown in Fig 227a Here the freebody diagrams of the members reveal six unknowns Fig 227b however six equilibrium equations three for each member can be written so again the problem is statically determinateAs in the previous case a freebody diagram of the entire frame can also be used for part of the analysis Fig 227c Although as shown the frame has a tendency to collapse without its supports by rotating about the pin at B this will not happen since the force system acting on it must still hold it in equilibrium Hence if so desired all six unknowns can be determined by applying the three equilibrium equations to the entire frame Fig 227c and also to either one of its members The above two examples illustrate that if a structure is properly supported and contains no more supports or members than are necessary to prevent collapse the frame becomes statically determinate and so the unknown forces at the supports and connections can be determined from the equations of equilibrium applied to each member Also if the structure remains rigid noncollapsible when the supports are removed Fig 226c all three support reactions can be determined by applying the three equilibrium equations to the entire structure However if the structure appears to be nonrigid collapsible after removing the supports Fig 227c it must be dismembered and equilibrium of the individual members must be considered in order to obtain enough equations to determine all the support reactions B A D C P2 P1 a P2 P1 b Ay Ax Bx By Bx By Cx Cy B P1 c Ay Ax Cx Cy P2 Fig 227 25 APPLICATION OF THE EQUATIONS OF EQUILIBRIUM 61 2 Procedure for Analysis The following procedure provides a method for determining the joint reactions for structures composed of pinconnected members FreeBody Diagrams Disassemble the structure and draw a freebody diagram of each member Also it may be convenient to supplement a member freebody diagram with a freebody diagram of the entire structure Some or all of the support reactions can then be determined using this diagram Recall that reactive forces common to two members act with equal magnitudes but opposite directions on the respective free body diagrams of the members All twoforce members should be identified These members regardless of their shape have no external loads on them and therefore their freebody diagrams are represented with equal but opposite collinear forces acting on their ends In many cases it is possible to tell by inspection the proper arrowhead sense of direction of an unknown force or couple moment however if this seems difficult the directional sense can be assumed Equations of Equilibrium Count the total number of unknowns to make sure that an equivalent number of equilibrium equations can be written for solution Except for twoforce members recall that in general three equilibrium equations can be written for each member Many times the solution for the unknowns will be straightforward if the moment equation is applied about a point O that lies at the intersection of the lines of action of as many unknown forces as possible When applying the force equations and orient the x and y axes along lines that will provide the simplest reduction of the forces into their x and y components If the solution of the equilibrium equations yields a negative magnitude for an unknown force or couple moment it indicates that its arrowhead sense of direction is opposite to that which was assumed on the freebody diagram Fy 0 Fx 0 MO 0 EXAMPLE 28 Determine the reactions on the beam shown in Fig 228a Fig 228 SOLUTION FreeBody Diagram As shown in Fig 228b the 60k force is resolved into x and y components Furthermore the 7ft dimension line is not needed since a couple moment is a free vector and can therefore act anywhere on the beam for the purpose of computing the external reactions Equations of Equilibrium Applying Eqs 22 in a sequence using previously calculated results we have ΣFx 0 Ax 60 cos 60 0 Ax 300 k Ans ΣMA 0 60 sin 6010 60 cos 601 By14 50 0 By 385 k Ans ΣFy 0 60 sin 60 385 Ay 0 Ay 134 k Ans EXAMPLE 29 Determine the reactions on the beam in Fig 229a SOLUTION FreeBody Diagram As shown in Fig 229b the trapezoidal distributed loading is segmented into a triangular and a uniform load The areas under the triangle and rectangle represent the resultant forces These forces act through the centroid of their corresponding areas Equations of Equilibrium ΣFx 0 Ax 0 Ans ΣFy 0 Ay 60 60 0 Ay 120 kN Ans ΣMA 0 604 606 MA 0 MA 600 kN m Ans 25 APPLICATION OF THE EQUATIONS OF EQUILIBRIUM 63 2 EXAMPLE 210 Determine the reactions on the beam in Fig 230aAssume A is a pin and the support at B is a roller smooth surface SOLUTION FreeBody Diagram As shown in Fig 230b the support roller at B exerts a normal force on the beam at its point of contactThe line of action of this force is defined by the 345 triangle Equations of Equilibrium Resolving into x and y components and summing moments about A yields a direct solution for Why Using this result we can then obtain and Ans Ans Ans Ay 270 k Ay 3500 3 51133152 0 cFy 0 Ax 107 k Ax 4 51133152 0 Fx 0 NB 13315 lb 133 k 35001352 A4 5BNB142 A3 5BNB1102 0 dMA 0 Ay Ax NB NB 35 ft 65 ft 3500 lb A b Ax Ay 4 ft 3 4 5 NB 7 ft 3 ft 500 lbft A B 4 ft a Fig 230 EXAMPLE 211 The compound beam in Fig 231a is fixed at A Determine the reactions at A B and C Assume that the connection at B is a pin and C is a roller Fig 231 SOLUTION FreeBody Diagrams The freebody diagram of each segment is shown in Fig 231b Why is this problem statically determinate Equations of Equilibrium There are six unknowns Applying the six equations of equilibrium using previously calculated results we have Segment BC ΣMC 0 6000 By15 0 By 400 lb Ans ΣFy 0 400 Cy 0 Cy 400 lb Ans ΣFx 0 Bx 0 Ans Segment AB ΣMA 0 MA 800010 40020 0 MA 720 k ft Ans ΣFy 0 Ay 8000 400 0 Ay 760 k Ans ΣFx 0 Ax 0 0 Ax 0 Ans 25 APPLICATION OF THE EQUATIONS OF EQUILIBRIUM 65 2 EXAMPLE 212 Determine the horizontal and vertical components of reaction at the pins A B and C of the twomember frame shown in Fig 232a 3 kNm a 2 m 4 3 5 8 kN 2 m C 15 m 2 m A B 2 m 4 3 5 A 2 m 8 kN By Bx Ax Ay 15 m C 6 kN 1 m Bx By 1 m Cx Cy b Fig 232 SOLUTION FreeBody Diagrams The freebody diagram of each member is shown in Fig 232b Equations of Equilibrium Applying the six equations of equilibrium in the following sequence allows a direct solution for each of the six unknowns Member BC Ans Member AB Ans Ans Ans Member BC Ans Ans cFy 0 3 6 Cy 0 Cy 3 kN Fx 0 147 Cx 0 Cx 147 kN cFy 0 Ay 4 5182 3 0 Ay 940 kN Fx 0 Ax 3 5182 147 0 Ax 987 kN dMA 0 8122 3122 Bx1152 0 Bx 147 kN dMC 0 By122 6112 0 By 3 kN 66 CHAPTER 2 ANALYSIS OF STATICALLY DETERMINATE STRUCTURES 2 The side of the building in Fig 233a is subjected to a wind loading that creates a uniform normal pressure of 15 kPa on the windward side and a suction pressure of 5 kPa on the leeward side Determine the horizontal and vertical components of reaction at the pin connections A B and C of the supporting gable arch EXAMPLE 213 60 kNm 20 kNm 60 kNm 20 kNm B 3 m 3 m 3 m 3 m b A C 3 m 3 m 4 m 3 m 2 m 2 m 3 m 4 m wind a A C B Fig 233 SOLUTION Since the loading is evenly distributed the central gable arch supports a loading acting on the walls and roof of the darkshaded tributary area This represents a uniform distributed load of on the windward side and 20 kNm on the leeward side Fig 233b 15 kNm2214 m2 14 m2 60 kNm 115 kNm22 FreeBody Diagrams Simplifying the distributed loadings the freebody diagrams of the entire frame and each of its parts are shown in Fig 233c Equations of Equilibrium Simultaneous solution of equations is avoided by applying the equilibrium equations in the following sequence using previously computed results Entire Frame ΣMA 0 180 6015 2546 849 cos 4545 2546 sin 4515 849 sin 4545 Cy6 0 Cy 2400 kN Ans ΣFy 0 Ay 2546 sin 45 849 sin 45 2400 0 Ay 1200 kN Ans Member AB ΣMB 0 Ax6 12003 18045 2546212 0 Ax 2850 kN Ans ΣFx 0 2850 180 2546 cos 45 Bx 0 Bx 750 kN Ans ΣFy 0 1200 2546 sin 45 By 0 By 3000 kN Ans Member CB ΣFx 0 Cx 60 849 cos 45 750 0 Cx 1950 kN Ans The problem can also be solved by applying the six equations of equilibrium only to the two members If this is done it is best to first sum moments about point A on member AB then point C on member CB By doing this one obtains two equations to be solved simultaneously for Bx and By CHAPTER REVIEW SupportsStructural members are often assumed to be pin connected if slight relative rotation can occur between them and fixed connected if no rotation is possible stiffeners weld weld typical pinsupported connection metal typical fixedsupported connection metal Idealized StructuresBy making assumptions about the supports and connections as being either roller supported pinned or fixed the members can then be represented as lines so that we can establish an idealized model that can be used for analysis actual beam idealized beam The tributary loadings on slabs can be determined by first classifying the slab as a oneway or twoway slab As a general rule if L2 is the largest dimension and L2L1 2 the slab will behave as a oneway slab If L2L1 2 the slab will behave as a twoway slab oneway slab action requires L2L1 2 twoway slab action requires L2L1 2 CHAPTER REVIEW 69 2 P partial constraint P concurrent reactions P parallel reactions Principle of SuperpositionEither the loads or displacements can be added together provided the material is linear elastic and only small displacements of the structure occur EquilibriumStatically determinate structures can be analyzed by disassembling them and applying the equations of equilibrium to each memberThe analysis of a statically determinate structure requires first drawing the freebody diagrams of all the members and then applying the equations of equilibrium to each member The number of equations of equilibrium for all n members of a structure is 3n If the structure has r reactions then the structure is statically determinate if and statically indeterminate if The additional number of equations required for the solution refers to the degree of indeterminacy StabilityIf there are fewer reactions than equations of equilibrium then the structure will be unstable because it is partially constrained Instability due to improper constraints can also occur if the lines of action of the reactions are concurrent at a point or parallel to one another r 7 3n r 3n g MO 0 g Fy 0 g Fx 0 FUNDAMENTAL PROBLEMS F21 Determine the horizontal and vertical components of reaction at the pins A B and C F24 Determine the horizontal and vertical components of reaction at the roller support A and fixed support B F22 Determine the horizontal and vertical components of reaction at the pins A B and C F25 Determine the horizontal and vertical components of reaction at pins A B and C of the twomember frame F23 Determine the horizontal and vertical components of reaction at the pins A B and C F26 Determine the components of reaction at the roller support A and pin support C Joint B is fixed connected FUNDAMENTAL PROBLEMS 71 2 A B C D 4 ft 2 ft 8 ft 2 kft 05 kft F29 A B C D 3 kNm 8 kN 8 kN 2 m 2 m 2 m 3 m 4 m F27 A B C D 2 m 3 m 6 kN 4 kN 6 kN 3 m 2 m 2 m F28 F27 Determine the horizontal and vertical components of reaction at the pins A B and D of the threemember frameThe joint at C is fixed connected F29 Determine the components of reaction at the fixed support D and the pins A B and C of the threemember frame Neglect the thickness of the members F28 Determine the components of reaction at the fixed support D and the pins A B and C of the threemember frame Neglect the thickness of the members F210 Determine the components of reaction at the fixed support D and the pins A B and C of the threemember frame Neglect the thickness of the members 15 kNm 6 m 2 m 2 m 2 m 6 kN 8 kN 8 kN 6 kN A B C D F210 PROBLEMS 218 Determine the reactions on the beam Neglect the thickness of the beam 221 Determine the reactions at the supports A and B of the compound beam Assume there is a pin at C 219 Determine the reactions on the beam 222 Determine the reactions at the supports A B D and F 220 Determine the reactions on the beam 223 The compound beam is pin supported at C and supported by a roller at A and B There is a hinge pin at D Determine the reactions at the supports Neglect the thickness of the beam 224 Determine the reactions on the beam The support at B can be assumed to be a roller 225 Determine the reactions at the smooth support C and pinned support A Assume the connection at B is fixed connected 226 Determine the reactions at the truss supports A and B The distributed loading is caused by wind 227 The compound beam is fixed at A and supported by a rocker at B and C There are hinges pins at D and E Determine the reactions at the supports 228 Determine the reactions at the supports A and B The floor decks CD DE EF and FG transmit their loads to the girder on smooth supports Assume A is a roller and B is a pin 229 Determine the reactions at the supports A and B of the compound beam There is a pin at C 235 Determine the reactions at the supports A and B 236 Determine the horizontal and vertical components of reaction at the supports A and B Assume the joints at C and D are fixed connections 237 Determine the horizontal and vertical components force at pins A and C of the twomember frame 238 The wall crane supports a load of 700 lb Determine the horizontal and vertical components of reaction at the pins A and D Also what is the force in the cable at the winch W 230 Determine the reactions at the supports A and B of the compound beam There is a pin at C 231 The beam is subjected to the two concentrated loads as shown Assuming that the foundation exerts a linearly varying load distribution on its bottom determine the load intensities w1 and w2 for equilibrium a in terms of the parameters shown b set P 500 lb L 12 ft 232 The cantilever footing is used to support a wall near its edge A so that it causes a uniform soil pressure under the footing Determine the uniform distribution loads wA and wB measured in lbft at pads A and B necessary to support the wall forces of 8000 lb and 20000 lb 233 Determine the horizontal and vertical components of reaction acting at the supports A and C 234 Determine the reactions at the smooth support A and the pin support B The joint at C is fixed connected 76 CHAPTER 2 ANALYSIS OF STATICALLY DETERMINATE STRUCTURES 2 239 Determine the resultant forces at pins B and C on member ABC of the fourmember frame 241 Determine the horizontal and vertical reactions at the connections A and C of the gable frameAssume that A B and C are pin connections The purlin loads such as D and E are applied perpendicular to the center line of each girder 240 Determine the reactions at the supports A and D Assume A is fixed and B and C and D are pins 242 Determine the horizontal and vertical components of reaction at ACand DAssume the frame is pin connected at A C and D and there is a fixedconnected joint at B 2 ft 150 lbft 4 ft 5 ft 5 ft 2 ft A F E D B C Prob 239 A B D C w w L 15L Prob 240 800 lb 600 lb 600 lb 400 lb 400 lb D G E C F A B 120 lbft 800 lb 6 ft 6 ft 6 ft 6 ft 10 ft 5 ft Prob 241 A C D B 50 kN 40 kN 4 m 6 m 15 m 15 m 15 kNm 2 m Prob 242 243 Determine the horizontal and vertical components at A B and C Assume the frame is pin connected at these points The joints at D and E are fixed connected 244 Determine the reactions at the supports A and B The joints C and D are fixed connected PROJECT PROBLEM 21P The railroad trestle bridge shown in the photo is supported by reinforced concrete bents Assume the two simply supported side girders track bed and two rails have a weight of 05 kft and the load imposed by a train is 72 kft see Fig 111 Each girder is 20 ft long Apply the load over the entire bridge and determine the compressive force in the columns of each bent For the analysis assume all joints are pin connected and neglect the weight of the bent Are these realistic assumptions The forces in the members of this bridge can be analyzed using either the method of joints or the method of sections 3 79 In this chapter we will develop the procedures for analyzing statically determinate trusses using the method of joints and the method of sections First however the determinacy and stability of a truss will be discussed Then the analysis of three forms of planar trusses will be considered simple compound and complex Finally at the end of the chapter we will consider the analysis of a space truss 31 Common Types of Trusses A truss is a structure composed of slender members joined together at their end points The members commonly used in construction consist of wooden struts metal bars angles or channels The joint connections are usually formed by bolting or welding the ends of the members to a common plate called a gusset plate as shown in Fig 31 or by simply passing a large bolt or pin through each of the members Planar trusses lie in a single plane and are often used to support roofs and bridges Analysis of Statically Determinate Trusses The gusset plate is used to connect eight members of the truss supporting structure for a water tank gusset plate Fig 31 80 CHAPTER 3 ANALYSIS OF STATICALLY DETERMINATE TRUSSES Roof Trusses Roof trusses are often used as part of an industrial building frame such as the one shown in Fig 32 Here the roof load is transmitted to the truss at the joints by means of a series of purlins The roof truss along with its supporting columns is termed a bent Ordinarily roof trusses are supported either by columns of wood steel or reinforced concrete or by masonry wallsTo keep the bent rigid and thereby capable of resisting horizontal wind forces knee braces are sometimes used at the supporting columns The space between adjacent bents is called a bay Bays are economically spaced at about 15 ft 46 m for spans around 60 ft 18 m and about 20 ft 61 m for spans of 100 ft 30 m Bays are often tied together using diagonal bracing in order to maintain rigidity of the buildings structure Trusses used to support roofs are selected on the basis of the span the slope and the roof material Some of the more common types of trusses used are shown in Fig 33 In particular the scissors truss Fig 33a can be used for short spans that require overhead clearanceThe Howe and Pratt trusses Figs 33b and 33c are used for roofs of moderate span about 60 ft 18 m to 100 ft 30 mIf larger spans are required to support the roof the fan truss or Fink truss may be used Figs 33d and 33eThese trusses may be built with a cambered bottom cord such as that shown in Fig 33f If a flat roof or nearly flat roof is to be selected the Warren truss Fig 33g is often used Also the Howe and Pratt trusses may be modified for flat roofs Sawtooth trusses Fig 33h are often used where column spacing is not objectionable and uniform lighting is important A textile mill would be an example The bowstring truss Fig 33i is sometimes selected for garages and small airplane hangars and the arched truss Fig 33j although relatively expensive can be used for high rises and long spans such as field houses gymnasiums and so on 3 span bottom cord knee brace gusset plates top cord roof bay purlins Fig 32 Although more decorative than structural these simple Pratt trusses are used for the entrance of a building 31 COMMON TYPES OF TRUSSES 81 3 Fig 33 Fink e Howe b Pratt c fan d cambered Fink f Warren g sawtooth h roof window roof window bowstring i threehinged arch j scissors a 82 CHAPTER 3 ANALYSIS OF STATICALLY DETERMINATE TRUSSES 3 bottom cord panel floor beam portal end post stringers portal bracing top lateral bracing sway bracing top cord deck Fig 34 Bridge Trusses The main structural elements of a typical bridge truss are shown in Fig 34 Here it is seen that a load on the deck is first transmitted to stringers then to floor beams and finally to the joints of the two supporting side trussesThe top and bottom cords of these side trusses are connected by top and bottom lateral bracing which serves to resist the lateral forces caused by wind and the sidesway caused by moving vehicles on the bridge Additional stability is provided by the portal and sway bracing As in the case of many longspan trusses a roller is provided at one end of a bridge truss to allow for thermal expansion A few of the typical forms of bridge trusses currently used for single spans are shown in Fig 35 In particular the Pratt Howe and Warren trusses are normally used for spans up to 200 ft 61 m in length The most common form is the Warren truss with verticals Fig 35c For larger spans a truss with a polygonal upper cord such as the Parker truss Fig 35d is used for some savings in material The Warren truss with verticals can also be fabricated in this manner for spans up to 300 ft 91 m The greatest economy of material is obtained if the diagonals have a slope between 45 and 60 with the horizontal If this rule is maintained then for spans greater than 300 ft 91 m the depth of the truss must increase and consequently the panels will get longer This results in a heavy deck system and to keep the weight of the deck within tolerable limits subdivided trusses have been developedTypical examples include the Baltimore and subdivided Warren trusses Figs 35e and 35f Finally the Ktruss shown in Fig 35g can also be used in place of a subdivided truss since it accomplishes the same purpose Parker trusses are used to support this bridge 31 COMMON TYPES OF TRUSSES 83 3 Fig 35 Pratt a Howe b Warren with verticals c Parker d Baltimore e subdivided Warren f Ktruss g Assumptions for Design To design both the members and the connections of a truss it is first necessary to determine the force developed in each member when the truss is subjected to a given loading In this regard two important assumptions will be made in order to idealize the truss 1 The members are joined together by smooth pins In cases where bolted or welded joint connections are used this assumption is generally satisfactory provided the center lines of the joining members are concurrent at a point as in Fig 31 It should be realized however that the actual connections do give some rigidity to the joint and this in turn introduces bending of the connected members when the truss is subjected to a load The bending stress developed in the members is called secondary stress whereas the stress in the members of the idealized truss having pinconnected joints is called primary stressA secondary stress analysis of a truss can be performed using a computer as discussed in Chapter 16 For some types of truss geometries these stresses may be large 2 All loadings are applied at the joints In most situations such as for bridge and roof trusses this assumption is true Frequently in the force analysis the weight of the members is neglected since the force supported by the members is large in comparison with their weight If the weight is to be included in the analysis it is generally satisfactory to apply it as a vertical force half of its magnitude applied at each end of the member Because of these two assumptions each truss member acts as an axial force member and therefore the forces acting at the ends of the member must be directed along the axis of the member If the force tends to elongate the member it is a tensile force T Fig 36a whereas if the force tends to shorten the member it is a compressive force C Fig 36b In the actual design of a truss it is important to state whether the force is tensile or compressive Most often compression members must be made thicker than tension members because of the buckling or sudden instability that may occur in compression members 84 CHAPTER 3 ANALYSIS OF STATICALLY DETERMINATE TRUSSES 3 Fig 36 C C b a T T 32 CLASSIFICATION OF COPLANAR TRUSSES 85 3 32 Classification of Coplanar Trusses Before beginning the force analysis of a truss it is important to classify the truss as simple compound or complex and then to be able to specify its determinacy and stability Simple Truss To prevent collapse the framework of a truss must be rigid Obviously the fourbar frame ABCD in Fig 37 will collapse unless a diagonal such as AC is added for supportThe simplest framework that is rigid or stable is a triangle Consequently a simple truss is constructed by starting with a basic triangular element such as ABC in Fig 38 and connecting two members AD and BD to form an additional element Thus it is seen that as each additional element of two members is placed on the truss the number of joints is increased by one B C D A P D B C A B C E A D F simple truss B A C E D simple truss Fig 37 Fig 38 Fig 39 Fig 310 An example of a simple truss is shown in Fig39where the basic stable triangular element is ABC from which the remainder of the joints D E and F are established in alphabetical sequence For this method of construction however it is important to realize that simple trusses do not have to consist entirely of triangles An example is shown in Fig 310 where starting with triangle ABC bars CD and AD are added to form joint D Finally bars BE and DE are added to form joint E 86 CHAPTER 3 ANALYSIS OF STATICALLY DETERMINATE TRUSSES 3 Compound Truss A compound truss is formed by connecting two or more simple trusses together Quite often this type of truss is used to support loads acting over a large span since it is cheaper to construct a somewhat lighter compound truss than to use a heavier single simple truss There are three ways in which simple trusses are joined together to form a compound truss The trusses may be connected by a common joint and bar An example is given in Fig 311a where the shaded truss ABC is connected to the shaded truss CDE in this manner The trusses may be joined by three bars as in the case of the shaded truss ABC connected to the larger truss DEF Fig 311bAnd finally the trusses may be joined where bars of a large simple truss called the main truss have been substituted by simple trusses called secondary trusses An example is shown in Fig 311c where dashed members of the main truss ABCDE have been replaced by the secondary shaded trusses If this truss carried roof loads the use of the secondary trusses might be more economical since the dashed members may be subjected to excessive bending whereas the secondary trusses can better transfer the load Complex Truss A complex truss is one that cannot be classified as being either simple or compoundThe truss in Fig 312 is an example Fig 311 a A B E D C simple trusses b A B C E D F simple trusses c C B A D E secondary simple truss secondary simple truss secondary simple truss main simple truss Various types of compound trusses Complex truss Fig 312 32 CLASSIFICATION OF COPLANAR TRUSSES 87 3 Determinacy For any problem in truss analysis it should be realized that the total number of unknowns includes the forces in b number of bars of the truss and the total number of external support reactions r Since the truss members are all straight axial force members lying in the same planethe force system acting at each joint is coplanar and concurrent Consequently rotational or moment equilibrium is automatically satisfied at the joint or pin and it is only necessary to satisfy and to ensure translational or force equilibrium Therefore only two equations of equilibrium can be written for each joint and if there are j number of joints the total number of equations available for solution is 2j By simply comparing the total number of unknowns with the total number of available equilibrium equations it is therefore possible to specify the determinacy for either a simple compound or complex truss We have 31 In particular the degree of indeterminacy is specified by the difference in the numbers Stability If a truss will be unstable that is it will collapse since there will be an insufficient number of bars or reactions to constrain all the joints Also a truss can be unstable if it is statically determinate or statically indeterminate In this case the stability will have to be determined either by inspection or by a force analysis External Stability As stated in Sec 24 a structure or truss is externally unstable if all of its reactions are concurrent or parallel For example the two trusses in Fig 313 are externally unstable since the support reactions have lines of action that are either concurrent or parallel b r 6 2j 1b r2 2j b r 7 2j statically indeterminate b r 2j statically determinate 1b r2 Fy 0 Fx 0 unstable concurrent reactions unstable parallel reactions Fig 313 88 CHAPTER 3 ANALYSIS OF STATICALLY DETERMINATE TRUSSES 3 Internal Stability The internal stability of a truss can often be checked by careful inspection of the arrangement of its members If it can be determined that each joint is held fixed so that it cannot move in a rigid body sense with respect to the other joints then the truss will be stable Notice that a simple truss will always be internally stable since by the nature of its construction it requires starting from a basic triangular element and adding successive rigid elements each containing two additional members and a joint The truss in Fig 314 exemplifies this construction where starting with the shaded triangle element ABC the successive joints D E F G H have been added If a truss is constructed so that it does not hold its joints in a fixed position it will be unstable or have a critical formAn obvious example of this is shown in Fig 315 where it can be seen that no restraint or fixity is provided between joints C and F or B and E and so the truss will collapse under load Fig 314 Fig 315 D C E A B F H G D C E A B F H G O A B D E F C Fig 316 To determine the internal stability of a compound truss it is necessary to identify the way in which the simple trusses are connected together For example the compound truss in Fig 316 is unstable since the inner simple truss ABC is connected to the outer simple truss DEF using three barsAD BE and CF which are concurrent at point OThus an external load can be applied to joint A B or C and cause the truss ABC to rotate slightly 32 CLASSIFICATION OF COPLANAR TRUSSES 89 3 If a truss is identified as complex it may not be possible to tell by inspection if it is stable For example it can be shown by the analysis discussed in Sec 37 that the complex truss in Fig 317 is unstable or has a critical form only if the dimension If it is stable The instability of any form of truss be it simple compound or complex may also be noticed by using a computer to solve the 2j simultaneous equations written for all the joints of the truss If inconsistent results are obtained the truss will be unstable or have a critical form If a computer analysis is not performed the methods discussed previously can be used to check the stability of the trussTo summarize if the truss has b bars r external reactions and j joints then if 32 Bear in mind however that if a truss is unstable it does not matter whether it is statically determinate or indeterminate Obviously the use of an unstable truss is to be avoided in practice b r 6 2j unstable b r Ú 2j unstable if truss support reactions are concurrent or parallel or if some of the components of the truss form a collapsible mechanism d Z d d d d s d Fig 317 90 CHAPTER 3 ANALYSIS OF STATICALLY DETERMINATE TRUSSES 3 Fig 318 Classify each of the trusses in Fig 318 as stable unstable statically determinate or statically indeterminate The trusses are subjected to arbitrary external loadings that are assumed to be known and can act anywhere on the trusses SOLUTION Fig 318a Externally stable since the reactions are not concurrent or parallel Since then or Therefore the truss is statically determinate By inspection the truss is internally stable 22 22 b r 2j r 3 j 11 b 19 EXAMPLE 31 Fig 318b Externally stable Since then or The truss is statically indeterminate to the first degree By inspection the truss is internally stable 19 7 18 b r 7 2j r 4 j 9 b 15 a b 32 CLASSIFICATION OF COPLANAR TRUSSES 91 3 Fig 318c Externally stable Since then or The truss is statically determinate By inspection the truss is internally stable 12 12 b r 2j r 3 j 6 b 9 Fig 318d Externally stable Since then or 15 6 16 The truss is internally unstable b r 6 2j r 3 j 8 b 12 c d 92 CHAPTER 3 ANALYSIS OF STATICALLY DETERMINATE TRUSSES 3 Prob 31 Prob 32 31 Classify each of the following trusses as statically determinate statically indeterminate or unstable If indeterminate state its degree 32 Classify each of the following trusses as stable unstable statically determinate or statically indeterminate If indeterminate state its degree PROBLEMS a b c d a b c 32 CLASSIFICATION OF COPLANAR TRUSSES 93 3 Prob 33 Prob 34 33 Classify each of the following trusses as statically determinate indeterminate or unstable If indeterminate state its degree 34 Classify each of the following trusses as statically determinate statically indeterminate or unstable If indeterminate state its degree a b c a b c d 33 The Method of Joints If a truss is in equilibrium then each of its joints must also be in equilibrium Hence the method of joints consists of satisfying the equilibrium conditions ΣFx 0 and ΣFy 0 for the forces exerted on the pin at each joint of the truss When using the method of joints it is necessary to draw each joints freebody diagram before applying the equilibrium equations Recall that the line of action of each member force acting on the joint is specified from the geometry of the truss since the force in a member passes along the axis of the member As an example consider joint B of the truss in Fig 319a From the freebody diagram Fig 319b the only unknowns are the magnitudes of the forces in members BA and BC As shown FBA is pulling on the pin which indicates that member BA is in tension whereas FBC is pushing on the pin and consequently member BC is in compression These effects are clearly demonstrated by using the method of sections and isolating the joint with small segments of the member connected to the pin Fig 319c Notice that pushing or pulling on these small segments indicates the effect of the member being either in compression or tension In all cases the joint analysis should start at a joint having at least one known force and at most two unknown forces as in Fig 319b In this way application of ΣFx 0 and ΣFy 0 yields two algebraic equations that can be solved for the two unknowns When applying these equations the correct sense of an unknown member force can be determined using one of two possible methods 1 Always assume the unknown member forces acting on the joints freebody diagram to be in tension iepulling on the pin If this is done then numerical solution of the equilibrium equations will yield positive scalars for members in tension and negative scalars for members in compression Once an unknown member force is found use its correct magnitude and sense T or C on subsequent joint freebody diagrams 2 The correct sense of direction of an unknown member force can in many cases be determined by inspection For example in Fig 319b must push on the pin compression since its horizontal component sin 45 must balance the 500N force Likewise is a tensile force since it balances the vertical component cos 45 In more complicated cases the sense of an unknown member force can be assumed then after applying the equilibrium equations the assumed sense can be veri fied from the numerical results A positive answer indicates that the sense is correct whereas a negative answer indicates that the sense shown on the freebody diagram must be reversed This is the method we will use in the example problems which follow 1Fy 02 FBC FBA 1Fx 02 FBC FBC 33 THE METHOD OF JOINTS 95 3 Procedure for Analysis The following procedure provides a means for analyzing a truss using the method of joints Draw the freebody diagram of a joint having at least one known force and at most two unknown forces If this joint is at one of the supports it may be necessary to calculate the external reactions at the supports by drawing a freebody diagram of the entire truss Use one of the two methods previously described for establishing the sense of an unknown force The x and y axes should be oriented such that the forces on the freebody diagram can be easily resolved into their x and y componentsApply the two force equilibrium equations and solve for the two unknown member forces and verify their correct directional sense Continue to analyze each of the other joints where again it is necessary to choose a joint having at most two unknowns and at least one known force Once the force in a member is found from the analysis of a joint at one of its ends the result can be used to analyze the forces acting on the joint at its other end Remember a member in compression pushes on the joint and a member in tension pulls on the joint Fy 0 Fx 0 EXAMPLE 32 Determine the force in each member of the roof truss shown in the photo The dimensions and loadings are shown in Fig 320a State whether the members are in tension or compression SOLUTION Only the forces in half the members have to be determined since the truss is symmetric with respect to both loading and geometry Joint A Fig 320b We can start the analysis at joint A Why The freebody diagram is shown in Fig 320b ΣFy 0 4 FAG sin 30 0 FAG 8 kN C Ans ΣFx 0 FAB 8 cos 30 0 FAB 6928 kN T Ans Joint G Fig 320c In this case note how the orientation of the x y axes avoids simultaneous solution of equations ΣFy 0 FGB sin 60 3 cos 30 0 FGB 300 kN C Ans ΣFx 0 8 3 sin 30 300 cos 60 FGF 0 FGF 500 kN C Ans Joint B Fig 320d ΣFy 0 FBF sin 60 300 sin 30 0 FBF 173 kN T Ans ΣFx 0 FBC 173 cos 60 300 cos 30 6928 0 FBC 346 kN T Ans EXAMPLE 33 Determine the force in each member of the scissors truss shown in Fig 321a State whether the members are in tension or compression The reactions at the supports are given SOLUTION The truss will be analyzed in the following sequence Joint E Fig 321b Note that simultaneous solution of equations is avoided by the x y axes orientation ΣFy 0 1910 cos 30 FED sin 15 0 FED 6391 lb C Ans ιΣFx 0 6391 cos 15 FEF 1910 sin 30 0 FEF 5218 lb T Ans Joint D Fig 321c ιΣFx 0 FDF sin 75 0 FDF 0 Ans tΣFy 0 FDC 6391 0 FDC 6391 lb C Ans Joint C Fig 321d ΣFx 0 FCB sin 45 6391 sin 45 0 FCB 6391 lb C Ans ΣFy 0 FCF 175 26391 cos 45 0 FCF 7288 lb T Ans Joint B Fig 321e ΣFy 0 FBF sin 75 200 0 FBF 2071 lb C Ans ΣFx 0 6391 2071 cos 75 FBA 0 FBA 6927 lb C Ans Joint A Fig 321f ΣFx 0 FAF cos 30 6927 cos 45 1414 0 FAF 7289 lb T Ans ΣFy 0 1254 6927 sin 45 7289 sin 30 0 check Notice that since the reactions have been calculated a further check of the calculations can be made by analyzing the last joint F Try it and find out Fig 321 34 ZeroForce Members Truss analysis using the method of joints is greatly simplified if one is able to first determine those members that support no loading These zeroforce members may be necessary for the stability of the truss during construction and to provide support if the applied loading is changed The zeroforce members of a truss can generally be determined by inspection of the joints and they occur in two cases Case 1 Consider the truss in Fig 322a The two members at joint C are connected together at a right angle and there is no external load on the joint The freebody diagram of joint C Fig 322b indicates that the force in each member must be zero in order to maintain equilibrium Furthermore as in the case of joint A Fig 322c this must be true regardless of the angle say θ between the members Case 2 Zeroforce members also occur at joints having a geometry as joint D in Fig 323a Here no external load acts on the joint so that a force summation in the y direction Fig 323b which is perpendicular to the two collinear members requires that FDF 0 Using this result FC is also a zeroforce member as indicated by the force analysis of joint F Fig 323c In summary then if only two noncollinear members form a truss joint and no external load or support reaction is applied to the joint the members must be zeroforce members Case 1 Also if three members form a truss joint for which two of the members are collinear the third member is a zeroforce member provided no external force or support reaction is applied to the joint Case 2 Particular attention should be directed to these conditions of joint geometry and loading since the analysis of a truss can be considerably simplified by first spotting the zeroforce members Fig 322 Fig 323 34 ZEROFORCE MEMBERS 99 3 EXAMPLE 34 Using the method of joints indicate all the members of the truss shown in Fig 324a that have zero force SOLUTION Looking for joints similar to those discussed in Figs 322 and 323 we have Joint D Fig 324b Ans Ans Joint E Fig 324c Ans Note that and an analysis of joint C would yield a force in member CF Joint H Fig 324d Ans Joint G Fig 324e The rocker support at G can only exert an x component of force on the joint ie Hence Ans FGA 0 cFy 0 Gx FHB 0 QFy 0 FEC P FEF 0 Fx 0 FDE 0 FDE 0 0 Fx 0 FDC 0 FDC sin u 0 cFy 0 A B C D E F G H a P y x D FDC FDE b u y x E FEC FEF 0 P c y x H FHB FHF d FHA y x G FGA Gx e FGF Fig 324 FUNDAMENTAL PROBLEMS F31 Determine the force in each member of the truss and state whether it is in tension or compression F31 F32 Determine the force in each member of the truss and state whether it is in tension or compression F32 F33 Determine the force in each member of the truss and state whether it is in tension or compression F33 F34 Determine the force in each member of the truss and state whether it is in tension or compression F34 F35 Determine the force in each member of the truss and state whether it is in tension or compression F35 F36 Determine the force in each member of the truss and state whether it is in tension or compression F36 PROBLEMS 35 A sign is subjected to a wind loading that exerts horizontal forces of 300 lb on joints B and C of one of the side supporting trusses Determine the force in each member of the truss and state if the members are in tension or compression 36 Determine the force in each member of the truss Indicate if the members are in tension or compression Assume all members are pin connected 37 Determine the force in each member of the truss State whether the members are in tension or compression Set P 8 kN 38 If the maximum force that any member can support is 8 kN in tension and 6 kN in compression determine the maximum force P that can be supported at joint D 39 Determine the force in each member of the truss State if the members are in tension or compression 102 CHAPTER 3 ANALYSIS OF STATICALLY DETERMINATE TRUSSES 3 312 Determine the force in each member of the truss State if the members are in tension or compressionAssume all members are pin connected FE ED AG GF 310 Determine the force in each member of the truss State if the members are in tension or compression 313 Determine the force in each member of the truss and state if the members are in tension or compression Prob 310 Prob 311 Prob 312 Prob 313 B C E F G H D A 10 ft 10 ft 10 ft 10 ft 12 ft 3 k 2 k 3 k 8 kN 8 kN 4 kN 4 kN 8 kN A B C F G E D 4 m 4 m 2 m E D C B F A 5 m 3 m 5 kN 4 kN 3 m 3 m 3 m A G F E B C D 3 m 2 m 2 m 2 m 5 kN 5 kN 5 kN 311 Determine the force in each member of the truss State if the members are in tension or compressionAssume all members are pin connected 314 Determine the force in each member of the roof russ State if the members are in tension or compression 315 Determine the force in each member of the roof truss State if the members are in tension or compression Assume all members are pin connected 316 Determine the force in each member of the truss State if the members are in tension or compression 317 Determine the force in each member of the roof truss State if the members are in tension or compression Assume B is a pin and C is a roller support 35 The Method of Sections If the forces in only a few members of a truss are to be found the method of sections generally provides the most direct means of obtaining these forces The method of sections consists of passing an imaginary section through the truss thus cutting it into two parts Provided the entire truss is in equilibrium each of the two parts must also be in equilibrium and as a result the three equations of equilibrium may be applied to either one of these two parts to determine the member forces at the cut section When the method of sections is used to determine the force in a particular member a decision must be made as to how to cut or section the truss Since only three independent equilibrium equations ΣFx 0 ΣFy 0 ΣMo 0 can be applied to the isolated portion of the truss try to select a section that in general passes through not more than three members in which the forces are unknown For example consider the truss in Fig 325a If the force in member GC is to be determined section aa will be appropriate The freebody diagrams of the two parts are shown in Figs 325b and 325c In particular note that the line of action of each force in a sectioned member is specified from the geometry of the truss since the force in a member passes along the axis of the member Also the member forces acting on one part of the truss are equal but opposite to those acting on the other partNewtons third law As shown members assumed to be in tension BC and GC are subjected to a pull whereas the member in compression GF is subjected to a push 35 THE METHOD OF SECTIONS 105 3 The three unknown member forces and can be obtained by applying the three equilibrium equations to the freebody diagram in Fig 325b If however the freebody diagram in Fig 325c is considered the three support reactions and will have to be determined first Why This of course is done in the usual manner by considering a freebody diagram of the entire truss When applying the equilibrium equations consider ways of writing the equations so as to yield a direct solution for each of the unknownsrather than having to solve simultaneous equations For example summing moments about C in Fig 325b would yield a direct solution for since and create zero moment about C Likewise can be obtained directly by summing moments about G Finally can be found directly from a force summation in the vertical direction since and have no vertical components As in the method of joints there are two ways in which one can determine the correct sense of an unknown member force 1 Always assume that the unknown member forces at the cut section are in tension ie pulling on the member By doing this the numerical solution of the equilibrium equations will yield positive scalars for members in tension and negative scalars for members in compression 2 The correct sense of an unknown member force can in many cases be determined by inspection For example is a tensile force as represented in Fig 325b since moment equilibrium about G requires that create a moment opposite to that of the 1000N force Also is tensile since its vertical component must balance the 1000N force In more complicated cases the sense of an unknown member force may be assumed If the solution yields a negative scalar it indicates that the forces sense is opposite to that shown on the freebody diagram This is the method we will use in the example problems which follow FGC FBC FBC FBC FGF FGC FBC FGC FBC FGF Ex Dy Dx FGF FGC FBC A truss bridge being constructed over Lake Shasta in northern California 106 CHAPTER 3 ANALYSIS OF STATICALLY DETERMINATE TRUSSES 3 Procedure for Analysis The following procedure provides a means for applying the method of sections to determine the forces in the members of a truss FreeBody Diagram Make a decision as to how to cut or section the truss through the members where forces are to be determined Before isolating the appropriate section it may first be necessary to determine the trusss external reactions so that the three equilibrium equations are used only to solve for member forces at the cut section Draw the freebody diagram of that part of the sectioned truss which has the least number of forces on it Use one of the two methods described above for establishing the sense of an unknown force Equations of Equilibrium Moments should be summed about a point that lies at the intersection of the lines of action of two unknown forces in this way the third unknown force is determined directly from the equation If two of the unknown forces are parallel forces may be summed perpendicular to the direction of these unknowns to determine directly the third unknown force An example of a Warren truss with verticals EXAMPLE 35 Determine the force in members GJ and CO of the roof truss shown in the photo The dimensions and loadings are shown in Fig 326a State whether the members are in tension or compression The reactions at the supports have been calculated Fig 326 SOLUTION Member CF FreeBody Diagram The force in member GJ can be obtained by considering the section aa in Fig 326a The freebody diagram of the right part of this section is shown in Fig 326b Equations of Equilibrium A direct solution for FGJ can be obtained by applying ΣMI 0 Why For simplicity slide FGJ to point G principle of transmissibility Fig 326b Thus ΣMI 0 FGJ sin 306 3003464 0 FGJ 346 lb C Ans Member GC FreeBody Diagram The force in CO can be obtained by using section bb in Fig 326a The freebody diagram of the left portion of the section is shown in Fig 326c Equations of Equilibrium Moments will be summed about point A in order to eliminate the unknowns FOP and FCD ΣMA 0 3003464 FCO6 0 FCO 173 lb T Ans EXAMPLE 36 Determine the force in members GF and GD of the truss shown in Fig 327a State whether the members are in tension or compression The reactions at the supports have been calculated Fig 327 SOLUTION FreeBody Diagram Section aa in Fig 327a will be considered Why The freebody diagram to the right of this section is shown in Fig 327b The distance EO can be determined by proportional triangles or realizing that member GF drops vertically 45 3 15 m in 3 m Fig 327a Hence to drop 45 m from G the distance from C to O must be 9 m Also the angles that FGD and FGF make with the horizontal are tan¹453 563 and tan¹459 266 respectively Equations of Equilibrium The force in GF can be determined directly by applying ΣMD 0 Why For the calculation use the principle of transmissibility and slide FGF to point O Thus ΣMD 0 FGF sin 2666 73 0 FGF 783 kN C Ans The force in GD is determined directly by applying ΣMO 0 For simplicity use the principle of transmissibility and slide FGD to D Hence ΣMO 0 73 26 FGD sin 5636 0 FGD 180 kN C Ans EXAMPLE 37 Determine the force in members BC and MC of the Ktruss shown in Fig 328a State whether the members are in tension or compression The reactions at the supports have been calculated SOLUTION FreeBody Diagram Although section aa shown in Fig 328a cuts through four members it is possible to solve for the force in member BC using this section The freebody diagram of the left portion of the truss is shown in Fig 328b Equations of Equilibrium Summing moments about point L eliminates three of the unknowns so that ΣML 0 290015 FBC20 0 FBC 2175 lb T Ans FreeBody Diagrams The force in MC can be obtained indirectly by first obtaining the force in MB from vertical force equilibrium of joint B Fig 328c ie FMB 1200 lb T Then from the freebody diagram in Fig 328b ΣFy 0 2900 1200 1200 FML 0 FML 2900 lb T Using these results the freebody diagram of joint M is shown in Fig 328d Equations of Equilibrium ΣFx 0 313 FMC 313 FMK 0 ΣFy 0 2900 1200 213 FMC 213 FMK 0 FMK 1532 lb C FMC 1532 lb T Ans Sometimes as in this example application of both the method of sections and the method of joints leads to the most direct solution to the problem It is also possible to solve for the force in MC by using the result for FBC In this case pass a vertical section through LK MK MC and BC Fig 328a Isolate the left section and apply ΣMK 0 36 Compound Trusses In Sec 32 it was stated that compound trusses are formed by connecting two or more simple trusses together either by bars or by joints Occasionally this type of truss is best analyzed by applying both the method of joints and the method of sections It is often convenient to first recognize the type of construction as listed in Sec 32 and then perform the analysis using the following procedure EXAMPLE 38 Indicate how to analyze the compound truss shown in Fig 329a The reactions at the supports have been calculated SOLUTION The truss is a compound truss since the simple trusses ACH and CEG are connected by the pin at C and the bar HG Section aa in Fig 329a cuts through bar HG and two other members having unknown forces A freebody diagram for the left part is shown in Fig 329b The force in HG is determined as follows ΣMC 0 54 42 FHG4 sin 60 0 FHG 346 kN C We can now proceed to determine the force in each member of the simple trusses using the method of joints For example the freebody diagram of ACH is shown in Fig 329c The joints of this truss can be analyzed in the following sequence Joint A Determine the force in AB and AI Joint H Determine the force in HI and HJ Joint I Determine the force in IJ and IB Joint B Determine the force in BC and BJ Joint J Determine the force in JC Fig 329 EXAMPLE 39 Compound roof trusses are used in a garden center as shown in the photo They have the dimensions and loading shown in Fig 330a Indicate how to analyze this truss SOLUTION We can obtain the force in EF by using section aa in Fig 330a The freebody diagram of the right segment is shown in Fig 330b ΣMO 0 11 12 13 14 15 056 66 FEF6 tan 30 0 FEF 520 kN T Ans By inspection notice that BT EO and HJ are zeroforce members since ΣFy 0 at joints B E and H respectively Also by applying ΣFy 0 perpendicular to AO at joints P Q S and T we can directly determine the force in members PU QU SC and TC respectively Fig 330 EXAMPLE 310 Indicate how to analyze the compound truss shown in Fig 331a The reactions at the supports have been calculated SOLUTION The truss may be classified as a type 2 compound truss since the simple trusses ABCD and FEHG are connected by three nonparallel or nonconcurrent bars namely CE BH and DG Using section aa in Fig 331a we can determine the force in each connecting bar The freebody diagram of the left part of this section is shown in Fig 331b Hence ΣMB 0 36 FDG6 sin 45 FCE cos 4512 FCE sin 456 0 1 ΣFy 0 3 3 FBH sin 45 FCE sin 45 0 2 ΣFx 0 FBH cos 45 FDG FCE cos 45 0 3 From Eq 2 FBH FCE then solving Eqs 1 and 3 simultaneously yields FBH FCE 268 k C FDG 378 k T Analysis of each connected simple truss can now be performed using the method of joints For example from Fig 331c this can be done in the following sequence Joint A Determine the force in AB and AD Joint D Determine the force in DC and DB Joint C Determine the force in CB Fig 331 36 COMPOUND TRUSSES 113 3 F38 FUNDAMENTAL PROBLEMS F37 Determine the force in members HG BG and BC and state whether they are in tension or compression F310 Determine the force in members GF CF and CD and state whether they are in tension or compression F38 Determine the force in members HG HC and BC and state whether they are in tension or compression F311 Determine the force in members FE FC and BC and state whether they are in tension or compression F39 Determine the force in members ED BD and BC and state whether they are in tension or compression F312 Determine the force in members GF CF and CD and state whether they are in tension or compression F37 F39 F310 F311 F312 A H G F E B C D 2 k 2 k 2 k 5 ft 5 ft 5 ft 5 ft 5 ft A B C D F E G H 400 lb 400 lb 400 lb 400 lb 400 lb 8 ft 8 ft 8 ft 8 ft 6 ft 6 ft A B C D F E G 15 m 15 m 15 m 3 m 15 m 2 kN 2 kN 4 kN 3 m 3 m A B C D F E G H 500 lb 500 lb 500 lb 4 ft 4 ft 4 ft 4 ft 3 ft 1 ft A B C D F E G H I J 4 ft 4 ft 4 ft 4 ft 3 ft 600 lb 600 lb 600 lb 600 lb 600 lb 2 m 8 kN 6 kN A B C D E 2 m 2 m 114 CHAPTER 3 ANALYSIS OF STATICALLY DETERMINATE TRUSSES 3 319 Determine the force in members JK JN and CD State if the members are in tension or compression Identify all the zeroforce members 320 Determine the force in members GF FC and CD of the cantilever truss State if the members are in tension or compressionAssume all members are pin connected 322 Determine the force in members BG HG and BC of the truss and state if the members are in tension or compression 318 Determine the force in members GF FC and CD of the bridge truss State if the members are in tension or compressionAssume all members are pin connected 321 The Howe truss is subjected to the loading shown Determine the forces in members GF CD and GC State if the members are in tension or compression Assume all members are pin connected PROBLEMS Prob 318 Prob 319 Prob 320 Prob 321 Prob 322 10 k B 15 k C D E A F H G 30 ft 15 ft 40 ft 40 ft 40 ft 40 ft 5 kN G H F A E B C D 2 m 2 m 2 m 2 m 3 m 5 kN 5 kN 2 kN 2 kN 7 kN B 6 kN C 4 kN D E A F H G 45 m 3 m 12 m 4 3 m B C N O E F G H I J K L M D A 2 k 20 ft 20 ft 20 ft 30 ft 2 k 2 m 2 m 2 m 3 m A B C D E F G 12 kN 12 kN 12 kN 323 Determine the force in members GF CF and CD of the roof truss and indicate if the members are in tension or compression 325 Determine the force in members IH ID and CD of the truss State if the members are in tension or compression Assume all members are pin connected 326 Determine the force in members JI IC and CD of the truss State if the members are in tension or compression Assume all members are pin connected 324 Determine the force in members GF FB and BC of the Fink truss and state if the members are in tension or compression 327 Determine the forces in members KJ CD and CJ of the truss State if the members are in tension or compression 116 CHAPTER 3 ANALYSIS OF STATICALLY DETERMINATE TRUSSES 3 37 Complex Trusses The member forces in a complex truss can be determined using the method of joints however the solution will require writing the two equilibrium equations for each of the j joints of the truss and then solving the complete set of 2j equations simultaneously This approach may be impractical for hand calculations especially in the case of large trusses Therefore a more direct method for analyzing a complex truss referred to as the method of substitute members will be presented here Procedure for Analysis With reference to the truss in Fig 332a the following steps are necessary to solve for the member forces using the substitute member method P E D F A C B a P E D F A C B b Si forces E D F A C B c 1 1 si forces Fig 332 This can be readily accomplished using a computer as will be shown in Chapter 14 37 COMPLEX TRUSSES 117 3 Reduction to Stable Simple Truss Determine the reactions at the supports and begin by imagining how to analyze the truss by the method of joints ie progressing from joint to joint and solving for each member force If a joint is reached where there are three unknowns remove one of the members at the joint and replace it by an imaginary member elsewhere in the truss By doing this reconstruct the truss to be a stable simple truss For example in Fig 332a it is observed that each joint will have three unknown member forces acting on it Hence we will remove member AD and replace it with the imaginary member EC Fig 332b This truss can now be analyzed by the method of joints for the two types of loading that follow External Loading on Simple Truss Load the simple truss with the actual loading P then determine the force in each member i In Fig 332b provided the reactions have been determined one could start at joint A to determine the forces in AB and AF then joint F to determine the forces in FE and FC then joint D to determine the forces in DE and DC both of which are zero then joint E to determine EB and EC and finally joint B to determine the force in BC Remove External Loading from Simple Truss Consider the simple truss without the external load P Place equal but opposite collinear unit loads on the truss at the two joints from which the member was removed If these forces develop a force in the ith truss member then by proportion an unknown force x in the removed member would exert a force in the ith member From Fig 332c the equal but opposite unit loads will create no reactions at A and C when the equations of equilibrium are applied to the entire truss The forces can be determined by analyzing the joints in the same sequence as before namely joint A then joints F D E and finally B Superposition If the effects of the above two loadings are combined the force in the ith member of the truss will be 1 In particular for the substituted member EC in Fig 332b the force Since member EC does not actually exist on the original truss we will choose x to have a magnitude such that it yields zero force in EC Hence 2 or Once the value of x has been determined the force in the other members i of the complex truss can be determined from Eq 1 x SEC œ sEC SEC œ xsEC 0 SEC SEC œ xsEC Si Si œ xsi si xsi si Si œ EXAMPLE 311 Determine the force in each member of the complex truss shown in Fig 333a Assume joints B F and D are on the same horizontal line State whether the members are in tension or compression SOLUTION Reduction to Stable Simple Truss By inspection each joint has three unknown member forces A joint analysis can be performed by hand if for example member CF is removed and member DB substituted Fig 333b The resulting truss is stable and will not collapse External Loading on Simple Truss As shown in Fig 333b the support reactions on the truss have been determined Using the method of joints we can first analyze joint C to find the forces in members CB and CD then joint F where it is seen that FA and FE are zeroforce members then joint E to determine the forces in members EB and ED then joint D to determine the forces in DA and DB then finally joint B to determine the force in BA Considering tension as positive and compression as negative these Si forces are recorded in column 2 of Table 1 Remove External Loading from Simple Truss The unit load acting on the truss is shown in Fig 333c These equal but opposite forces create no external reactions on the truss The joint analysis follows the same sequence as discussed previously namely joints C F E D and BThe results of the force analysis are recorded in column 3 of Table 1 Superposition We require Substituting the data for and where is negative since the force is compressive we have The values of are recorded in column 4 of Table 1 and the actual member forces are listed in column 5 Si Si œ xsi xsi 250 x111672 0 x 2143 SDB œ sDB SDB œ SDB SDB œ xsDB 0 si 37 COMPLEX TRUSSES 119 3 C D B F A E 1 k c 1 k TABLE 1 Member CB 354 202 T CD 505 C FA 0 0833 179 T FE 0 0833 179 T EB 0 153 C ED 491 C DA 534 381 T DB 1167 0 BA 250 196 T CB 214 T 0536 0250 250 250 153 0712 0536 0250 438 153 0712 179 179 152 0707 354 152 0707 Si xsi si Si œ 120 CHAPTER 3 ANALYSIS OF STATICALLY DETERMINATE TRUSSES 3 38 Space Trusses A space truss consists of members joined together at their ends to form a stable threedimensional structure In Sec 32 it was shown that the simplest form of a stable twodimensional truss consists of the members arranged in the form of a triangle We then built up the simple plane truss from this basic triangular element by adding two members at a time to form further elements In a similar manner the simplest element of a stable space truss is a tetrahedron formed by connecting six members together with four joints as shown in Fig 334 Any additional members added to this basic element would be redundant in supporting the force P A simple space truss can be built from this basic tetrahedral element by adding three additional members and another joint forming multiconnected tetrahedrons Determinacy and Stability Realizing that in three dimensions there are three equations of equilibrium available for each joint then for a space truss with j number of joints 3j equations are available If the truss has b number of bars and r number of reactionsthen like the case of a planar truss Eqs31 and 32 we can write 33 The external stability of the space truss requires that the support reactions keep the truss in force and moment equilibrium about any and all axesThis can sometimes be checked by inspection although if the truss is unstable a solution of the equilibrium equations will give inconsistent results Internal stability can sometimes be checked by careful inspection of the member arrangement Provided each joint is held fixed by its supports or connecting members so that it cannot move with respect to the other joints the truss can be classified as internally stableAlso if we do a force analysis of the truss and obtain inconsistent results then the truss configuration will be unstable or have a critical form Assumptions for Design The members of a space truss may be treated as axialforce members provided the external loading is applied at the joints and the joints consist of ballandsocket connections This assumption is justified provided the joined members at a connection intersect at a common point and the weight of the members can be neglected In cases where the weight of a member is to be included in the analysis it is generally satisfactory to apply it as a vertical force half of its magnitude applied to each end of the member For the force analysis the supports of a space truss are generally modeled as a short link plane roller joint slotted roller joint or a ballandsocket joint Each of these supports and their reactive force components are shown in Table 31 b r 7 3j statically indeterminatecheck stability b r 3j statically determinatecheck stability b r 6 3j unstable truss Fy 0 Fz 02 1Fx 0 Fig 334 P The roof of this pavilion is supported using a system of space trusses 38 SPACE TRUSSES 121 3 TABLE 31 Supports and Their Reactive Force Components z y x z y x Fy short link z y x 2 1 roller z y x Fz 3 z y x z y x Fz Fx slotted roller constrained in a cylinder 4 z y x ballandsocket z y x Fz Fx Fy A FA z x y D C B FD FC FB Fig 336 122 CHAPTER 3 ANALYSIS OF STATICALLY DETERMINATE TRUSSES 3 x y z Force Components Since the analysis of a space truss is threedimensional it will often be necessary to resolve the force F in a member into components acting along the x y z axes For example in Fig 335 member AB has a length l and known projections x y z along the coordinate axes These projections can be related to the members length by the equation 34 Since the force F acts along the axis of the member then the components of F can be determined by proportion as follows 35 Notice that this requires 36 Use of these equations will be illustrated in Example 312 ZeroForce Members In some cases the joint analysis of a truss can be simplified if one is able to spot the zeroforce members by recognizing two common cases of joint geometry Case 1 If all but one of the members connected to a joint lie in the same plane and provided no external load acts on the joint then the member not lying in the plane of the other members must be subjected to zero force The proof of this statement is shown in Fig 336 where members A B C lie in the xy plane Since the z component of must be zero to satisfy member D must be a zeroforce member By the same reasoning member D will carry a load that can be determined from if an external force acts on the joint and has a component acting along the z axis Fz 0 Fz 0 FD F 2F2 x F2 y F2 z Fx Fa x l b Fy Fa y l b Fz Fa z l b l 2x2 y2 z2 Fig 335 z Fy F Fz Fx B x x A l z y y Because of their cost effectiveness towers such as these are often used to support multiple electric transmission lines Case 2 If it has been determined that all but two of several members connected at a joint support zero force then the two remaining members must also support zero force provided they do not lie along the same line This situation is illustrated in Fig 337 where it is known that A and C are zeroforce members Since FD is collinear with the y axis then application of Fx 0 or Fz 0 requires the x or z component of FB to be zero Consequently FB 0 This being the case FD 0 since Fy 0 Particular attention should be directed to the foregoing two cases of joint geometry and loading since the analysis of a space truss can be considerably simplified by first spotting the zeroforce members Procedure for Analysis Either the method of sections or the method of joints can be used to determine the forces developed in the members of a space truss Method of Sections If only a few member forces are to be determined the method of sections may be used When an imaginary section is passed through a truss and the truss is separated into two parts the force system acting on either one of the parts must satisfy the six scalar equilibrium equations Fx 0 Fy 0 Fz 0 Mx 0 My 0 Mz 0 By proper choice of the section and axes for summing forces and moments many of the unknown member forces in a space truss can be computed directly using a single equilibrium equation In this regard recall that the moment of a force about an axis is zero provided the force is parallel to the axis or its line of action passes through a point on the axis Method of Joints Generally if the forces in all the members of the truss must be determined the method of joints is most suitable for the analysis When using the method of joints it is necessary to solve the three scalar equilibrium equations Fx 0 Fy 0 Fz 0 at each joint Since it is relatively easy to draw the freebody diagrams and apply the equations of equilibrium the method of joints is very consistent in its application 124 CHAPTER 3 ANALYSIS OF STATICALLY DETERMINATE TRUSSES 3 Determine the force in each member of the space truss shown in Fig 338a The truss is supported by a ballandsocket joint at A a slotted roller joint at B and a cable at C EXAMPLE 312 z 8 ft 8 ft B x A E y 4 ft 4 ft D C Ez 600 lb a z 8 ft 8 ft x y 4 ft 600 lb b Ay Ax By Bx Cy Az SOLUTION The truss is statically determinate since or Fig 338b Support Reactions We can obtain the support reactions from the freebody diagram of the entire truss Fig 338b as follows Az 600 lb Az 600 0 Fz 0 Ay 600 lb Ay 600 0 Fy 0 Ax 300 lb 300 Ax 0 Fx 0 By 600 lb By182 600182 0 Mx 0 Cy 0 Mz 0 Bx 300 lb 600142 Bx182 0 My 0 9 6 3152 b r 3j Fig 338 Joint B We can begin the method of joints at B since there are three unknown member forces at this joint Fig 338c The components of FBE can be determined by proportion to the length of member BE as indicated by Eqs 35 We have ΣFy 0 600 FBE812 0 FBE 900 lb T Ans ΣFx 0 300 FBC 900412 0 FBC 0 Ans ΣFz 0 FBA 900812 0 FBA 600 lb C Ans Joint A Using the result for FBA 600 lb C the freebody diagram of joint A is shown in Fig 338d We have ΣFz 0 600 600 FAC sin 45 0 FAC 0 Ans ΣFy 0 FAE25 600 0 FAE 6708 lb C Ans ΣFx 0 300 FAD 670815 0 FAD 0 Ans Joint D By inspection the members at joint D Fig 338a support zero force since the arrangement of the members is similar to either of the two cases discussed in reference to Figs 336 and 337 Also from Fig 338e ΣFx 0 FDE 0 Ans ΣFz 0 FDC 0 Ans Joint C By observation of the freebody diagram Fig 338f FCE 0 Ans EXAMPLE 313 Determine the zeroforce members of the truss shown in Fig 339a The supports exert components of reaction on the truss as shown SOLUTION The freebody diagram Fig 339a indicates there are eight unknown reactions for which only six equations of equilibrium are available for solution Although this is the case the reactions can be determined since b r 3j or 16 8 38 To spot the zeroforce members we must compare the conditions of joint geometry and loading to those of Figs 336 and 337 Consider joint F Fig 339b Since members FC FD FE lie in the xy plane and FG is not in this plane FG is a zeroforce member ΣFz 0 must be satisfied In the same manner from joint E Fig 339c EF is a zeroforce member since it does not lie in the yz plane ΣFx 0 must be satisfied Returning to joint F Fig 339b it can be seen that FFD FFC 0 since FFE FFG 0 and there are no external forces acting on the joint Use this procedure to show that AB is a zero force member The numerical force analysis of the joints can now proceed by analyzing joint G FGF 0 to determine the forces in GH GB GC Then analyze joint H to determine the forces in HE HB HA joint E to determine the forces in EA ED joint A to determine the forces in AB AD and Az joint B to determine the force in BC and Bx Bz joint D to determine the force in DC and Dy Dz and finally joint C to determine Cx Cy Cz PROBLEMS 328 Determine the forces in all the members of the complex truss State if the members are in tension or compression Hint Substitute member AD with one placed between E and C 329 Determine the forces in all the members of the lattice complex truss State if the members are in tension or compression Hint Substitute member JE by one placed between K and F 330 Determine the force in each member and state if the members are in tension or compression 331 Determine the force in all the members of the complex truss State if the members are in tension or compression 332 Determine the force developed in each member of the space truss and state if the members are in tension or compression The crate has a weight of 150 lb 334 Determine the force in each member of the space truss and state if the members are in tension or compression The truss is supported by balland socket joints at C D E and G Note Although this truss is indeterminate to the first degree a solution is possible due to symmetry of geometry and loading 130 CHAPTER 3 ANALYSIS OF STATICALLY DETERMINATE TRUSSES 3 Trusses are composed of slender members joined together at their end points to form a series of triangles If the number of bars or members of a truss is b and there are r reactions and j joints then if the truss will be statically determinate the truss will be statically indeterminate b r 7 2j b r 2j T T C C For analysis we assume the members are pin connected and the loads are applied at the joints Thus the members will either be in tension or compression Trusses can be classified in three ways Simple trusses are formed by starting with an initial triangular element and connecting to it two other members and a joint to form a second triangle etc Compound trusses are formed by connecting together two or more simple trusses using a common joint andor additional member Complex trusses are those that cannot be classified as either simple or compound CHAPTER REVIEW simple truss compound truss simple trusses complex truss compound truss simple trusses CHAPTER REVIEW 131 3 unstableconcurrent reactions unstableparallel reactions unstable internally The truss will be externally unstable if the reactions are concurrent or parallel Internal stability can be checked by counting the number of bars b reactions r and joints j If the truss is unstable If it may still be unstable so it becomes necessary to inspect the truss and look for bar arrangements that form a parallel mechanism without forming a triangular element b r Ú 2j b r 6 2j Planar trusses can be analyzed by the method of joints This is done by selecting each joint in sequence having at most one known force and at least two unknowns The freebody diagram of each joint is constructed and two force equations of equilibrium are written and solved for the unknown member forces The method of sections requires passing a section through the truss and then drawing a freebody diagram of one of its sectioned parts The member forces cut by the section are then found from the three equations of equilibrium Normally a single unknown can be found if one sums moments about a point that eliminates the two other forces Compound and complex trusses can also be analyzed by the method of joints and the method of sections The method of substitute members can be used to obtain a direct solution for the force in a particular member of a complex truss Fx 0 Fy 0 The simply supported beams and girders of this building frame were designed to resist the internal shear and moment acting throughout their lengths 4 133 Before a structural member can be proportioned it is necessary to determine the force and moment that act within it In this chapter we will develop the methods for finding these loadings at specified points along a members axis and for showing the variation graphically using the shear and moment diagrams Applications are given for both beams and frames 41 Internal Loadings at a Specified Point As discussed in Sec 23 the internal load at a specified point in a member can be determined by using the method of sections In general this loading for a coplanar structure will consist of a normal force N shear force V and bending moment M It should be realized however that these loadings actually represent the resultants of the stress distribution acting over the members crosssectional area at the cut section Once the resultant internal loadings are known the magnitude of the stress can be determined provided an assumed distribution of stress over the crosssectional area is specified Internal Loadings Developed in Structural Members Threedimensional frameworks can also be subjected to a torsional moment which tends to twist the member about its axis 134 CHAPTER 4 INTERNAL LOADINGS DEVELOPED IN STRUCTURAL MEMBERS 4 Sign Convention Before presenting a method for finding the internal normal force shear force and bending moment we will need to establish a sign convention to define their positive and negative values Although the choice is arbitrary the sign convention to be adopted here has been widely accepted in structural engineering practice and is illustrated in Fig 41a On the lefthand face of the cut member the normal force N acts to the right the internal shear force V acts downward and the moment M acts counterclockwise In accordance with Newtons third law an equal but opposite normal force shear force and bending moment must act on the righthand face of the member at the section Perhaps an easy way to remember this sign convention is to isolate a small segment of the member and note that positive normal force tends to elongate the segment Fig 41b positive shear tends to rotate the segment clockwise Fig 41c and positive bending moment tends to bend the segment concave upward so as to hold water Fig 41d Fig 41 This will be convenient later in Secs 42 and 43 where we will express V and M as functions of x and then plot these functions Having a sign convention is similar to assigning coordinate directions x positive to the right and y positive upward when plotting a function y f1x2 M N V N V M a N N b V V c M M d 41 INTERNAL LOADINGS AT A SPECIFIED POINT 135 4 Procedure for Analysis The following procedure provides a means for applying the method of sections to determine the internal normal force shear force and bending moment at a specific location in a structural member Support Reactions Before the member is cut or sectioned it may be necessary to determine the members support reactions so that the equilibrium equations are used only to solve for the internal loadings when the member is sectioned If the member is part of a pinconnected structure the pin reactions can be determined using the methods of Sec 25 FreeBody Diagram Keep all distributed loadings couple moments and forces acting on the member in their exact location then pass an imaginary section through the member perpendicular to its axis at the point where the internal loading is to be determined After the section is made draw a freebody diagram of the segment that has the least number of loads on it At the section indicate the unknown resultants N V and M acting in their positive directions Fig 41a Equations of Equilibrium Moments should be summed at the section about axes that pass through the centroid of the members crosssectional area in order to eliminate the unknowns N and V and thereby obtain a direct solution for M If the solution of the equilibrium equations yields a quantity having a negative magnitude the assumed directional sense of the quantity is opposite to that shown on the freebody diagram 136 CHAPTER 4 INTERNAL LOADINGS DEVELOPED IN STRUCTURAL MEMBERS 4 EXAMPLE 41 1 m a 1 m 1 m 1 m 1 m 1 m 1 m 1 m 1 m 1 m 1 m 1 m 12 m 12 m 12 m 36 kN 36 kN 72 kN girder 432 kN 432 kN C 72 kN 72 kN edge beam girder Fig 42 b 72 kN beam 05 m 05 m 18 kNm 7 m 72 kN girder c MC VC 432 kN 1 m 1 m 04 m 12 m 12 m 36 kN 72 kN 72 kN C The building roof shown in the photo has a weight of and is supported on 8m long simply supported beams that are spaced 1 m apart Each beam shown in Fig 42b transmits its loading to two girders located at the front and back of the building Determine the internal shear and moment in the front girder at point C Fig 42a Neglect the weight of the members 18 kNm2 FreeBody Diagram The freebody diagram of the girder is shown in Fig 42a Notice that each column reaction is The freebody diagram of the left girder segment is shown in Fig 42c Here the internal loadings are assumed to act in their positive directions Equations of Equilibrium 12136 kN2 11172 kN22 432 kN Ans Ans MC 302 kN m MC 721042 721142 361242 4321122 0 dMC 0 VC 252 kN 432 36 21722 VC 0 cFy 0 SOLUTION Support Reactions The roof loading is transmitted to each beam as a oneway slab The tributary loading on each interior beam is therefore The two edge beams support From Fig 42b the reaction of each interior beam on the girder is 118 kNm218 m22 72 kN 09 kNm 118 kNm2211 m2 18 kNm 1L2L1 8 m1 m 8 7 22 41 INTERNAL LOADINGS AT A SPECIFIED POINT 137 4 EXAMPLE 42 Determine the internal shear and moment acting at a section passing through point C in the beam shown in Fig 43a SOLUTION Support Reactions Replacing the distributed load by its resultant force and computing the reactions yields the results shown in Fig 43b FreeBody Diagram Segment AC will be considered since it yields the simplest solution Fig 43c The distributed load intensity at C is computed by proportion that is Equations of Equilibrium Ans Ans This problem illustrates the importance of keeping the distributed loading on the beam until after the beam is sectioned If the beam in Fig 43b were sectioned at C the effect of the distributed load on segment AC would not be recognized and the result and MC 54 k ft would be wrong VC 9 k 9162 3122 MC 0 MC 48 k ft dMC 0 VC 6 k 9 3 VC 0 cFy 0 wC 16 ft18 ft213 kft2 1 kft A 6 ft 18 ft C a 3 kft B 27 k 12 ft b 6 ft 9 k 18 k Fig 43 1 kft NC MC 9 k 6 ft 2 ft 3 k c VC 138 CHAPTER 4 INTERNAL LOADINGS DEVELOPED IN STRUCTURAL MEMBERS 4 EXAMPLE 43 The 9k force in Fig 44a is supported by the floor panel DE which in turn is simply supported at its ends by floor beams These beams transmit their loads to the simply supported girder AB Determine the internal shear and moment acting at point C in the girder 12 ft 6 ft 4 ft 2 ft 9 k 3 k 525 k 24 ft 375 k b 3 k 6 k 6 k C MC VC NC A 12 ft 3 ft 375 k 6 k c Ans Ans 3751152 6132 MC 0 MC 3825 k ft dMC 0 VC 225 k 375 6 VC 0 cFy 0 SOLUTION Support Reactions Equilibrium of the floor panel floor beams and girder is shown in Fig 44b It is advisable to check these results FreeBody Diagram The freebody diagram of segment AC of the girder will be used since it leads to the simplest solution Fig 44c Note that there are no loads on the floor beams supported by AC Equations of Equilibrium a 6 ft 6 ft 6 ft 4 ft 2 ft 9 k A B C D E 15 ft 24 ft Fig 44 42 SHEAR AND MOMENT FUNCTIONS 139 4 42 Shear and Moment Functions The design of a beam requires a detailed knowledge of the variations of the internal shear force V and moment M acting at each point along the axis of the beam The internal normal force is generally not considered for two reasons 1 in most cases the loads applied to a beam act perpendicular to the beams axis and hence produce only an internal shear force and bending moment and 2 for design purposes the beams resistance to shear and particularly to bending is more important than its ability to resist normal force An important exception to this occurs however when beams are subjected to compressive axial forces since the buckling or instability that may occur has to be investigated The variations of V and M as a function of the position x of an arbitrary point along the beams axis can be obtained by using the method of sections discussed in Sec 41 Here however it is necessay to locate the imaginary section or cut at an arbitrary distance x from one end of the beam rather than at a specific point In generalthe internal shear and moment functions will be discontinuous or their slope will be discontinuous at points where the type or magnitude of the distributed load changes or where concentrated forces or couple moments are applied Because of this shear and moment functions must be determined for each region of the beam located between any two discontinuities of loading For example coordinates and will have to be used to describe the variation of V and M throughout the length of the beam in Fig 45a These coordinates will be valid only within regions from A to B for from B to C for and from C to D for Although each of these coordinates has the same origin as noted here this does not have to be the case Indeed it may be easier to develop the shear and moment functions using coordinates having origins at A B and D as shown in Fig 45b Here and are positive to the right and x3 is positive to the left x2 x1 x2 x3 x1 x3 x2 x1 x3 x2 x1 Additional reinforcement provided by ver tical plates called stiffeners is used over the pin and rocker supports of these bridge girders Here the reactions will cause large internal shear in the girders and the stiff eners will prevent localized buckling of the girder flanges or web Also note the tipping of the rocker support caused by the thermal expansion of the bridge deck A B C D P w x1 x2 x3 a A B C D P w x1 x2 x3 b Fig 45 140 CHAPTER 4 INTERNAL LOADINGS DEVELOPED IN STRUCTURAL MEMBERS 4 Procedure for Analysis The following procedure provides a method for determining the variation of shear and moment in a beam as a function of position x Support Reactions Determine the support reactions on the beam and resolve all the external forces into components acting perpendicular and parallel to the beams axis Shear and Moment Functions Specify separate coordinates x and associated originsextending into regions of the beam between concentrated forces andor couple moments or where there is a discontinuity of distributed loading Section the beam perpendicular to its axis at each distance x and from the freebody diagram of one of the segments determine the unknowns V and M at the cut section as functions of x On the free body diagram V and M should be shown acting in their positive directions in accordance with the sign convention given in Fig 41 V is obtained from and M is obtained by summing moments about the point S located at the cut section The results can be checked by noting that and where w is positive when it acts upward away from the beamThese relationships are developed in Sec 43 dVdx w dMdx V MS 0 Fy 0 The joists beams and girders used to support this floor can be designed once the internal shear and moment are known throughout their lengths Determine the shear and moment in the beam shown in Fig 46a as a function of x SOLUTION Support Reactions For the purpose of computing the support reactions the distributed load is replaced by its resultant force of 30 k Fig 46b It is important to remember however that this resultant is not the actual load on the beam Shear and Moment Functions A freebody diagram of the beam segment of length x is shown in Fig 46c Note that the intensity of the triangular load at the section is found by proportion that is wx 230 or w x15 With the load intensity known the resultant of the distributed loading is found in the usual manner as shown in the figure Thus ΣFy 0 30 12 x15 x V 0 V 30 00333 x2 Ans ΣMS 0 600 30x 12 x15 x x3 M 0 M 600 30x 00111 x3 Ans Note that dMdx V and dVdx x15 w which serves as a check of the results EXAMPLE 45 Determine the shear and moment in the beam shown in Fig 47a as a function of x Support Reactions The reactions at the fixed support are V 108 k and M 1588 kft Fig 47b Shear and Moment Functions Since there is a discontinuity of distributed load at x 12 ft two regions of x must be considered in order to describe the shear and moment functions for the entire beam Here x1 is appropriate for the left 12 ft and x2 can be used for the remaining segment 0 x1 12 ft Notice that V and M are shown in the positive directions Fig 47c Fy 0 108 4x1 V 0 V 108 4x1 Ans Ms 0 1588 108x1 4x1x12 M 0 M 1588 108x1 2x1² Ans 12 ft x2 20 ft Fig 47d Fy 0 108 48 V 0 V 60 Ans Ms 0 1588 108x2 48x2 6 M 0 M 60x2 1300 Ans These results can be partially checked by noting that when x2 20 ft then V 60 k and M 100 kft Also note that dMdx V and dVdx w 42 SHEAR AND MOMENT FUNCTIONS 143 4 Fig 48 EXAMPLE 46 Determine the shear and moment in the beam shown in Fig 48a as a function of x 10 kNm 30 kNm 9 m x a 75 kN 20 kNm 9 m b 90 kN 90 kN 6 m 45 m 105 kN 10 kNm x 20 kNm 9 10 kNm 75 kN 10x x2 x2 x3 M V c 1 x 20 x 2 9 SOLUTION Support Reactions To determine the support reactions the distributed load is divided into a triangular and rectangular loading and these loadings are then replaced by their resultant forces These reactions have been computed and are shown on the beams free body diagram Fig 48b Shear and Moment Functions A freebody diagram of the cut section is shown in Fig 48c As above the trapezoidal loading is replaced by rectangular and triangular distributions Note that the intensity of the triangular load at the cut is found by proportion The resultant force of each distributed loading and its location are indicatedApplying the equilibrium equations we have Ans Ans M 75x 5x2 0370x3 75x 110x2ax 2 b c1 2 1202ax 9 bxd x 3 M 0 dMS 0 V 75 10x 111x2 75 10x c1 2 1202a x 9 bxd V 0 cFy 0 FUNDAMENTAL PROBLEMS F41 Determine the internal normal force shear force and bending moment acting at point C in the beam F44 Determine the internal normal force shear force and bending moment acting at point C in the beam F42 Determine the internal normal force shear force and bending moment acting at point C in the beam F45 Determine the internal normal force shear force and bending moment acting at point C in the beam F43 Determine the internal normal force shear force and bending moment acting at point C in the beam F46 Determine the internal normal force shear force and bending moment acting at point C in the beam 42 SHEAR AND MOMENT FUNCTIONS 145 4 F47 Determine the internal shear and moment in the beam as a function of x F410 Determine the internal shear and moment in the beam as a function of x throughout the beam F48 Determine the internal shear and moment in the beam as a function of x F411 Determine the internal shear and moment in the beam as a function of x throughout the beam F49 Determine the internal shear and moment in the beam as a function of x throughout the beam F412 Determine the internal shear and moment in the beam as a function of x throughout the beam B 3 m 6 kN 18 kNm A x F47 A B 12 kNm 6 m x F48 A B 8 kNm 4 m 4 m x F49 5 kNm A B 2 m 2 m x 15 kN m 20 kN m F410 5 kNm A 2 m 2 m 15 kN x F411 A B 2 kft 12 ft 12 ft x 18 k F412 PROBLEMS 41 Determine the internal normal force shear force and bending moment in the beam at points C and D Assume the support at A is a pin and B is a roller 43 The boom DF of the jib crane and the column DE have a uniform weight of 50 lbft If the hoist and load weigh 300 lb determine the internal normal force shear force and bending moment in the crane at points A B and C 42 Determine the internal normal force shear force and bending moment in the beam at points C and D Assume the support at B is a roller Point D is located just to the right of the 10k load 44 Determine the internal normal force shear force and bending moment at point D Take w 150 Nm 45 The beam AB will fail if the maximum internal moment at D reaches 800 Nm or the normal force in member BC becomes 1500 N Determine the largest load w it can support 42 SHEAR AND MOMENT FUNCTIONS 147 4 46 Determine the internal normal force shear force and bending moment in the beam at points C and D Assume the support at A is a roller and B is a pin 49 Determine the internal normal force shear force and bending moment in the beam at point C The support at A is a roller and B is pinned 47 Determine the internal normal force shear force and bending moment at point C Assume the reactions at the supports A and B are vertical 410 Determine the internal normal force shear force and bending moment at point C Assume the reactions at the supports A and B are vertical A B 15 m 15 m 4 kNm C D 15 m 15 m Prob 46 D A B 05 kNm 15 kNm 3 m 6 m Prob 48 3 m C A B 05 kNm 15 kNm 6 m Prob 47 5 kN A C B 3 kNm 1 m 2 m 2 m Prob 49 6 ft A B 300 lbft 45 ft 400 lbft 45 ft E D 14 ft Prob 411 8 ft C A B 300 lbft 400 lbft 12 ft 9 ft Prob 410 48 Determine the internal normal force shear force and bending moment at point D Assume the reactions at the supports A and B are vertical 411 Determine the internal normal force shear force and bending moment at points D and EAssume the reactions at the supports A and B are vertical 412 Determine the shear and moment throughout the beam as a function of x 413 Determine the shear and moment in the floor girder as a function of x Assume the support at A is a pin and B is a roller 414 Determine the shear and moment throughout the beam as a function of x 415 Determine the shear and moment throughout the beam as a function of x 416 Determine the shear and moment throughout the beam as a function of x 417 Determine the shear and moment throughout the beam as a function of x 418 Determine the shear and moment throughout the beam as functions of x 419 Determine the shear and moment throughout the beam as functions of x 420 Determine the shear and moment in the beam as functions of x 421 Determine the shear and moment in the beam as a function of x 422 Determine the shear and moment throughout the tapered beam as a function of x 43 Shear and Moment Diagrams for a Beam If the variations of V and M as functions of x obtained in Sec 42 are plotted the graphs are termed the shear diagram and moment diagram respectively In cases where a beam is subjected to several concentrated forces couples and distributed loads plotting V and M versus x can become quite tedious since several functions must be plotted In this section a simpler method for constructing these diagrams is discusseda method based on differential relations that exist between the load shear and moment To derive these relations consider the beam AD in Fig 49a which is subjected to an arbitrary distributed loading w wx and a series of concentrated forces and couples In the following discussion the distributed load will be considered positive when the loading acts upward as shown We will consider the freebody diagram for a small segment of the beam having a length Δx Fig 49b Since this segment has been chosen at a point x along the beam that is not subjected to a concentrated force or couple any results obtained will not apply at points of concentrated loading The internal shear force and bending moment shown on the freebody diagram are assumed to act in the positive direction according to the established sign convention Fig 41 Note that both the shear force and moment acting on the right face must be increased by a small finite amount in order to keep the segment in equilibrium The distributed loading has been replaced by a concentrated force wx Δx that acts at a fractional distance εΔx from the right end where 0 ε 1 For example if wx is uniform or constant then wx Δx will act at 12 Δx so ε 12 Applying the equations of equilibrium we have ΣFy 0 V wx Δx V ΔV 0 ΔV wx Δx ΣMO 0 V Δx M wx Δx εΔx M ΔM 0 ΔM V Δx wx εΔx2 Fig 49 43 SHEAR AND MOMENT DIAGRAMS FOR A BEAM 153 4 Procedure for Analysis The following procedure provides a method for constructing the shear and moment diagrams for a beam using Eqs 41 through 46 Support Reactions Determine the support reactions and resolve the forces acting on the beam into components which are perpendicular and parallel to the beams axis Shear Diagram Establish the V and x axes and plot the values of the shear at the two ends of the beam Since the slope of the shear diagram at any point is equal to the intensity of the distributed loading at the point Note that w is positive when it acts upward If a numerical value of the shear is to be determined at the point one can find this value either by using the method of sections as discussed in Sec 41 or by using Eq 43 which states that the change in the shear force is equal to the area under the distributed loading diagram Since wx is integrated to obtain V if wx is a curve of degree n then Vx will be a curve of degree For example if wx is uniform Vx will be linear Moment Diagram Establish the M and x axes and plot the values of the moment at the ends of the beam Since the slope of the moment diagram at any point is equal to the intensity of the shear at the point At the point where the shear is zero and therefore this may be a point of maximum or minimum moment If the numerical value of the moment is to be determined at a point one can find this value either by using the method of sections as discussed in Sec 41 or by using Eq 44 which states that the change in the moment is equal to the area under the shear diagram Since Vx is integrated to obtain M if Vx is a curve of degree n then Mx will be a curve of degree For example if Vx is linear Mx will be parabolic n 1 dMdx 0 dMdx V n 1 dVdx w 160 CHAPTER 4 INTERNAL LOADINGS DEVELOPED IN STRUCTURAL MEMBERS 4 FUNDAMENTAL PROBLEMS F413 Draw the shear and moment diagrams for the beam Indicate values at the supports and at the points where a change in load occurs F417 Draw the shear and moment diagrams for the beam Indicate values at the supports and at the points where a change in load occurs F419 Draw the shear and moment diagrams for the beam Indicate values at the supports and at the points where a change in load occurs B 2 m 2 m 3 kN A 8 kN F413 B A 4 m 2 m 2 m 6 kN 6 kN m 8 kN F414 2 kft A 10 ft 30 kft F415 F414 Draw the shear and moment diagrams for the beam Indicate values at the supports and at the points where a change in load occurs F415 Draw the shear and moment diagrams for the beam Indicate values at the supports and at the points where a change in load occurs F416 Draw the shear and moment diagrams for the beam Indicate values at the supports and at the points where a change in load occurs 12 ft 12 ft 6 kft A B 18 k x F416 A B 2 kNm 2 kNm 45 m 45 m F417 A B 4 kNm 15 m 2 m 15 m F418 A B 6 kNm 2 m 2 m 2 m 6 kNm F419 F418 Draw the shear and moment diagrams for the beam Indicate values at the supports and at the points where a change in load occurs F420 Draw the shear and moment diagrams for the beam Indicate values at the supports and at the points where a change in load occurs A B 2 kft 6 ft 6 ft F420 44 SHEAR AND MOMENT DIAGRAMS FOR A FRAME 163 4 44 Shear and Moment Diagrams for a Frame Recall that a frame is composed of several connected members that are either fixed or pin connected at their ends The design of these structures often requires drawing the shear and moment diagrams for each of the members To analyze any problem we can use the procedure for analysis outlined in Sec 43 This requires first determining the reactions at the frame supports Then using the method of sections we find the axial force shear force and moment acting at the ends of each member Provided all loadings are resolved into components acting parallel and perpendicular to the members axis the shear and moment diagrams for each member can then be drawn as described previously When drawing the moment diagram one of two sign conventions is used in practice In particular if the frame is made of reinforced concrete designers often draw the moment diagram positive on the tension side of the frame In other words if the moment produces tension on the outer surface of the frame the moment diagram is drawn positive on this side Since concrete has a low tensile strength it will then be possible to tell at a glance on which side of the frame the reinforcement steel must be placed In this text however we will use the opposite sign convention and always draw the moment diagram positive on the compression side of the member This convention follows that used for beams discussed in Sec 41 The following examples illustrate this procedure numerically The simply supported girder of this concrete building frame was designed by first drawing its shear and moment diagrams 45 MOMENT DIAGRAMS CONSTRUCTED BY THE METHOD OF SUPERPOSITION 173 4 438 Draw the shear and moment diagrams for each of the three members of the frame Assume the frame is pin connected at A C and D and there is a fixed joint at B 440 Draw the shear and moment diagrams for each member of the frame Assume A is a rocker and D is pinned PROBLEMS 15 kNm 50 kN 40 kN A D B C 15 m 15 m 2 m 4 m 6 m Prob 438 A B C D 06 kft 08 kft 20 ft 16 ft Prob 439 2 kft 8 ft 4 ft 15 ft D A B C 4 k 3 k Prob 440 439 Draw the shear and moment diagrams for each member of the frameAssume the support at A is a pin and D is a roller 441 Draw the shear and moment diagrams for each member of the frameAssume the frame is pin connected at B C and D and A is fixed B C D A 3 k 6 k 6 k 3 k 15 ft 08 kft 8 ft 8 ft 8 ft Prob 441 178 CHAPTER 4 INTERNAL LOADINGS DEVELOPED IN STRUCTURAL MEMBERS 4 43P The idealized framing plan for a floor system located in the lobby of an office building is shown in the figure The floor is made using 4inthick reinforced stone concrete If the walls of the elevator shaft are made from 4inthick lightweight solid concrete masonry having a height of 10 ft determine the maximum moment in beam AB Neglect the weight of the members 8 ft 6 ft 6 ft 8 ft 8 ft 8 ft Elevator shaft A B I J C F H G E D Prob 43P Structural members subjected to planar loads support an internal normal force N shear force V and bend ing moment M To find these values at a specific point in a member the method of sections must be used This requires drawing a freebody diagram of a seg ment of the member and then applying the three equations of equilibriumAlways show the three inter nal loadings on the section in their positive directions The internal shear and moment can be expressed as a function of x along the member by establishing the origin at a fixed point normally at the left end of the member and then using the method of sections where the section is made a distance x from the origin For members subjected to several loads different x coor dinates must extend between the loads CHAPTER REVIEW M N V N V M positive sign convention P w x1 x2 x3 CHAPTER REVIEW 179 4 Shear and moment diagrams for structural members can be drawn by plotting the shear and moment functions They also can be plotted using the two graphical relationships Slope of Moment Diagramf 5Shear dM dx V Slope of Shear Diagram f eIntensity of Distributed Load dV dx wx Change in Momentf e Area under Shear Diagram M L Vx dx Change in Shearf L Area under Distributed Loading Diagram V L wx dx Note that a point of zero shear locates the point of maximum moment since V dMdx 0 A force acting downward on the beam will cause the shear diagram to jump downwards and a counterclockwise couple moment will cause the moment diagram to jump downwards Using the method of superposition the moment diagrams for a member can be represented by a series of simpler shapes The shapes represent the moment diagram for each of the separate loadings The resultant moment diagram is then the algebraic addition of the separate diagrams MR P VR ML ML VL VL V M VR x x MR M ML MR M P This parabolic arch bridge supports the deck above it 5 181 Cables and arches often form the main loadcarrying element in many types of structures and in this chapter we will discuss some of the important aspects related to their structural analysis The chapter begins with a general discussion of cables followed by an analysis of cables subjected to a concentrated load and to a uniform distributed load Since most arches are statically indeterminate only the special case of a threehinged arch will be considered The analysis of this structure will provide some insight regarding the fundamental behavior of all arched structures 51 Cables Cables are often used in engineering structures for support and to transmit loads from one member to another When used to support suspension roofs bridges and trolley wheels cables form the main loadcarrying element in the structure In the force analysis of such systems the weight of the cable itself may be neglected however when cables are used as guys for radio antennas electrical transmission lines and derricks the cable weight may become important and must be included in the structural analysis Two cases will be considered in the sections that follow a cable subjected to concentrated loads and a cable subjected to a distributed load Provided these loadings are coplanar with the cable the requirements for equilibrium are formulated in an identical manner Cables and Arches The deck of a cablestayed bridge is supported by a series of cables attached at various points along the deck and pylons 182 CHAPTER 5 CABLES AND ARCHES When deriving the necessary relations between the force in the cable and its slope we will make the assumption that the cable is perfectly flexible and inextensible Due to its flexibility the cable offers no resistance to shear or bending and therefore the force acting in the cable is always tangent to the cable at points along its length Being inextensible the cable has a constant length both before and after the load is appliedAs a result once the load is applied the geometry of the cable remains fixed and the cable or a segment of it can be treated as a rigid body 52 Cable Subjected to Concentrated Loads When a cable of negligible weight supports several concentrated loads the cable takes the form of several straightline segments each of which is subjected to a constant tensile force Consider for example the cable shown in Fig 51 Here specifies the angle of the cables cord AB and L is the cables span If the distances and and the loads and are known then the problem is to determine the nine unknowns consisting of the tension in each of the three segments the four components of reaction at A and B and the sags and at the two points C and D For the solution we can write two equations of force equilibrium at each of points A B C and D This results in a total of eight equations To complete the solution it will be necessary to know something about the geometry of the cable in order to obtain the necessary ninth equation For example if the cables total length is specified then the Pythagorean theorem can be used to relate to each of the three segmental lengths written in terms of and Unfortunately this type of problem cannot be solved easily by hand Another possibility however is to specify one of the sags either or instead of the cable length By doing this the equilibrium equations are then sufficient for obtaining the unknown forces and the remaining sag Once the sag at each point of loading is obtained can then be determined by trigonometry When performing an equilibrium analysis for a problem of this type the forces in the cable can also be obtained by writing the equations of equilibrium for the entire cable or any portion thereof The following example numerically illustrates these concepts l yD yC L3 u yC yD L1 L2 l l yD yC P2 P1 L3 L1 L2 u 5 yC yD C D A B L1 L2 L3 L P1 P2 u Fig 51 53 CABLE SUBJECTED TO A UNIFORM DISTRIBUTED LOAD 185 5 Performing a second integration with at yields 57 This is the equation of a parabola The constant may be obtained by using the boundary condition at Thus 58 Finally substituting into Eq 57 yields 59 From Eq 54 the maximum tension in the cable occurs when is maximum ie at Hence from Eqs 54 and 55 510 Or using Eq 58 we can express in terms of ie 511 Realize that we have neglected the weight of the cable which is uniform along the length of the cable and not along its horizontal projection Actually a cable subjected to its own weight and free of any other loads will take the form of a catenary curve However if the sagto span ratio is small which is the case for most structural applications this curve closely approximates a parabolic shape as determined here From the results of this analysis it follows that a cable will maintain a parabolic shape provided the dead load of the deck for a suspension bridge or a suspended girder will be uniformly distributed over the horizontal projected length of the cable Hence if the girder in Fig 54a is supported by a series of hangers which are close and uniformly spaced the load in each hanger must be the same so as to ensure that the cable has a parabolic shape Using this assumption we can perform the structural analysis of the girder or any other framework which is freely suspended from the cable In particular if the girder is simply supported as well as supported by the cable the analysis will be statically indeterminate to the first degree Fig 54b However if the girder has an internal pin at some intermediate point along its length Fig 54c then this would provide a condition of zero moment and so a determinate structural analysis of the girder can be performed Tmax w0 L21 1L2h22 w0 Tmax Tmax 2FH 2 1w0 L22 x L u y h L2 x2 FH w0 L2 2h x L y h FH y w0 2FH x2 x 0 y 0 Fig 54 a b c The VerrazanoNarrows Bridge at the entrance to New York Harbor has a main span of 4260 ft 130 km 53 CABLE SUBJECTED TO A UNIFORM DISTRIBUTED LOAD 187 5 Thus from Eqs 2 and 1 or Eq 56 we have 3 At point A Using Eq 54 Ans At point B Ans At point C Ans TC FH cos uC 36 4592 cos 440 507 k uC 440 tan uC dy dx x4142 002331141422 09657 x 4142 ft TB FH cos uB 36 4592 cos 0 365 k tan uB dy dx x0 0 uB 0 x 0 TA FH cos uA 36 4592 cos153792 617 k uA 5379 tan uA dy dx x 5858 002331158582 1366 x 1100 41422 5858 ft dy dx 850 36 4592 x 002331x FH 21251414222 36 4592 lb 53 CABLE SUBJECTED TO A UNIFORM DISTRIBUTED LOAD 189 5 If only half the suspended structure is considered Fig 56c then summing moments about the pin at C we have From these two equations To obtain the maximum tension in the cable we will use Eq 511 but first it is necessary to determine the value of an assumed uniform distributed loading from Eq 58 Thus using Eq 511 we have Ans 469 kN 3125112 m221 112 m218 m222 Tmax w0 L21 1L2h22 w0 2FH h L2 2128125 kN218 m2 112 m22 3125 kNm w0 FH 28125 kN 1875 0667FH Iy Ay 0667FH FH114 m2 FH16 m2 Iy112 m2 Ay112 m2 0 dMC 0 12 m 6 m 14 m C c Ay FH FH Cx Cy Iy Ax 190 CHAPTER 5 CABLES AND ARCHES 5 Prob 51 Prob 52 Prob 53 Probs 5455 51 Determine the tension in each segment of the cable and the cables total length 53 Determine the tension in each cable segment and the distance yD PROBLEMS 52 Cable ABCD supports the loading shown Determine the maximum tension in the cable and the sag of point B 54 The cable supports the loading shownDetermine the distance the force at point B acts from A Set 55 The cable supports the loading shown Determine the magnitude of the horizontal force P so that xB 6 ft P 40 lb xB 7 m B A D C 4 m 5 m 3 m 2 kN 4 kN yD 2 m 5 ft 2 ft 3 ft 30 lb D C B A xB 5 4 3 8 ft P 1 m A B C D yB 2 m 3 m 4 kN 6 kN 05 m 4 ft 5 ft A 3 ft B 7 ft 4 ft C D 50 lb 100 lb 53 CABLE SUBJECTED TO A UNIFORM DISTRIBUTED LOAD 191 5 Prob 56 57 The cable is subjected to the uniform loading If the slope of the cable at point O is zero determine the equation of the curve and the force in the cable at O and B 59 Determine the maximum and minimum tension in the cable 56 Determine the forces and needed to hold the cable in the position shown ie so segment CD remains horizontalAlso find the maximum loading in the cable P2 P1 58 The cable supports the uniform load of Determine the tension in the cable at each support A and B w0 600 lbft Prob 57 Prob 58 Prob 59 A P1 P2 5 kN 15 m 1 m B C D E 4 m 4 m 5 m 2 m 15 ft 8 ft y x A O B 15 ft 500 lbft 15 ft A B 10 ft 25 ft w0 10 m 16 kNm 2 m y x A B 10 m 54 Arches Like cables arches can be used to reduce the bending moments in longspan structures Essentially an arch acts as an inverted cable so it receives its load mainly in compression although because of its rigidity it must also resist some bending and shear depending upon how it is loaded and shaped In particular if the arch has a parabolic shape and it is subjected to a uniform horizontally distributed vertical load then from the analysis of cables it follows that only compressive forces will be resisted by the arch Under these conditions the arch shape is called a funicular arch because no bending or shear forces occur within the arch A typical arch is shown in Fig 57 which specifies some of the nomen clature used to define its geometry Depending upon the application several types of arches can be selected to support a loadingA fixed arch Fig 58a is often made from reinforced concrete Although it may require less material to construct than other types of arches it must have solid foundation abutments since it is indeterminate to the third degree and consequently additional stresses can be introduced into the arch due to relative settlement of its supportsA twohinged arch Fig 58b is commonly made from metal or timber It is indeterminate to the first degree and although it is not as rigid as a fixed arch it is somewhat insensitive to settlement We could make this structure statically determinate by replacing one of the hinges with a roller Doing so however would remove the capacity of the structure to resist bending along its span and as a result it would serve as a curved beam and not as an archA threehinged arch Fig 58c which is also made from metal or timber is statically determinate Unlike statically indeterminate arches it is not affected by settlement or temperature changes Finally if two and threehinged arches are to be constructed without the need for larger foundation abutments and if clearance is not a problem then the supports can be connected with a tie rod Fig 58d A tied arch allows the structure to behave as a rigid unit since the tie rod carries the horizontal component of thrust at the supports It is also unaffected by relative settlement of the supports 194 CHAPTER 5 CABLES AND ARCHES 5 Fig 57 extrados or back abutment intrados or soffit haunch centerline rise springline crown a fixed arch b twohinged arch d tied arch c threehinged arch Fig 58 55 THREEHINGED ARCH 195 5 55 ThreeHinged Arch To provide some insight as to how arches transmit loads we will now consider the analysis of a threehinged arch such as the one shown in Fig 59a In this case the third hinge is located at the crown and the supports are located at different elevations In order to determine the reactions at the supports the arch is disassembled and the freebody diagram of each member is shown in Fig 59b Here there are six unknowns for which six equations of equilibrium are available One method of solving this problem is to apply the moment equilibrium equations about points A and B Simultaneous solution will yield the reactions and The support reactions are then determined from the force equations of equilibrium Once obtained the internal normal force shear and moment loadings at any point along the arch can be found using the method of sections Here of course the section should be taken perpendicular to the axis of the arch at the point considered For example the freebody diagram for segment AD is shown in Fig 59c Threehinged arches can also take the form of two pinconnected trusses each of which would replace the arch ribs AC and CB in Fig 59a The analysis of this form follows the same procedure outlined aboveThe following examples numerically illustrate these concepts Cy Cx a A C B D P1 P2 b P1 P2 Ax Ay Cy Cx Cx Cy By Bx Ax Ay c VD ND MD Fig 59 b The threehinge truss arch is used to support a portion of the roof loading of this building a The closeup photo shows the arch is pinned at its top b a 198 CHAPTER 5 CABLES AND ARCHES 5 Fig 511 The threehinged tied arch is subjected to the loading shown in Fig 511a Determine the force in members CH and CB The dashed member GF of the truss is intended to carry no force EXAMPLE 55 SOLUTION The support reactions can be obtained from a freebody diagram of the entire arch Fig 511b A 15 kN 20 kN 15 kN E b Ay Ax Ey 3 m 3 m 3 m 3 m 4 m 1 m 1 m A 3 m 3 m 3 m 3 m B H D F G 15 kN 20 kN 15 kN C E a Ey112 m2 15 kN13 m2 20 kN16 m2 15 kN19 m2 0 dMA 0 5 m 3 m 3 m 15 kN 20 kN c Cx Cy 25 kN FAE 0 C The force components acting at joint C can be determined by consid ering the freebody diagram of the left part of the arch Fig 511c First we determine the force FAE 210 kN FAE15 m2 25 kN16 m2 15 kN13 m2 0 dMC 0 Ay 25 kN Ay 15 kN 20 kN 15 kN 25 kN 0 c Fy 0 Ax 0 Fx 0 Ey 25 kN 55 THREEHINGED ARCH 199 5 Then To obtain the forces in CH and CB we can use the method of joints as follows Joint G Fig 511d Joint C Fig 511e Thus Ans Ans FCH 474 kN 1T2 FCB 269 kN 1C2 FCBA 1 110B FCHA 1 110B 20 kN 10 kN 0 cFy 0 FCBA 3 110B 210 kN FCHA 3 110B 0 Fx 0 FGC 20 kN 1C2 FGC 20 kN 0 cFy 0 25 kN 15 kN 20 kN Cy 0 Cy 10 kN cFy 0 Cx 210 kN 0 Cx 210 kN Fx 0 FGC FHG 0 20 kN d G FCH 20 kN 3 1 FCB 10 kN 3 1 210 kN e C Note Tied arches are sometimes used for bridges Here the deck is supported by suspender bars that transmit their load to the arch The deck is in tension so that it supports the actual thrust or horizontal force at the ends of the arch 55 THREEHINGED ARCH 201 5 521 The tied threehinged arch is subjected to the loading shown Determine the components of reaction at A and C and the tension in the cable 523 The threehinged spandrel arch is subjected to the loading shown Determine the internal moment in the arch at point D PROBLEMS 522 Determine the resultant forces at the pins A B and C of the threehinged arched roof truss 524 The tied threehinged arch is subjected to the loading shown Determine the components of reaction A and C and the tension in the rod 3 m 5 m 3 m 3 m 1 m 1 m 2 m 2 m B C A 2 kN 3 kN 4 kN 4 kN 5 kN 10 kN 15 kN 2 m 2 m 05 m 2 m 1 m A B C Prob 521 Prob 522 Prob 523 Prob 524 A B C 3 m 4 kN 8 kN 8 kN 4 kN 3 kN 6 kN 6 kN 3 kN 5 m 2 m 2 m 2 m 2 m 2 m 2 m 3 m 5 m 8 m D A C B 4 k 3 k 5 k 6 ft 6 ft 8 ft 10 ft 10 ft 15 ft 5 CHAPTER REVIEW Cables support their loads in tension if we consider them perfectly flexible If the cable is subjected to concentrated loads then the force acting in each cable segment is determined by applying the equations of equilibrium to the freebody diagram of groups of segments of the cable or to the joints where the forces are applied If the cable supports a uniform load over a projected horizontal distance then the shape of the cable takes the form of a parabola Arches are designed primarily to carry a compressive force A parabolic shape is required to support a uniform loading distributed over its horizontal projection Threehinged arches are statically determinate and can be analyzed by separating the two members and applying the equations of equilibrium to each member L1 L2 L3 L P1 P2 w0 L h x y threehinged arch CHAPTER REVIEW 203 Moving loads caused by trains must be considered when designing the members of this bridge The influence lines for the members become an important part of the structural analysis 6 205 Influence lines have important application for the design of structures that resist large live loads In this chapter we will discuss how to draw the influence line for a statically determinate structure The theory is applied to structures subjected to a distributed load or a series of concentrated forces and specific applications to floor girders and bridge trusses are given The determination of the absolute maximum live shear and moment in a member is discussed at the end of the chapter 61 Influence Lines In the previous chapters we developed techniques for analyzing the forces in structural members due to dead or fixed loads It was shown that the shear and moment diagrams represent the most descriptive methods for displaying the variation of these loads in a member If a structure is subjected to a live or moving load however the variation of the shear and bending moment in the member is best described using the influence line An influence line represents the variation of either the reaction shear moment or deflection at a specific point in a member as a concentrated force moves over the member Once this line is constructed one can tell at a glance where the moving load should be placed on the structure so that it creates the greatest influence at the specified point Furthermore the magnitude of the associated reaction shear moment or deflection at the point can then be calculated from the ordinates of the influenceline diagram For these reasons influence lines play an important part in the design of bridges industrial crane rails conveyors and other structures where loads move across their span Influence Lines for Statically Determinate Structures 206 CHAPTER 6 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 6 Procedure for Analysis Either of the following two procedures can be used to construct the influence line at a specific point P in a member for any function reaction shear or moment For both of these procedures we will choose the moving force to have a dimensionless magnitude of unity Tabulate Values Place a unit load at various locations x along the member and at each location use statics to determine the value of the function reaction shear or moment at the specified point If the influence line for a vertical force reaction at a point on a beam is to be constructed consider the reaction to be positive at the point when it acts upward on the beam If a shear or moment influence line is to be drawn for a point take the shear or moment at the point as positive according to the same sign convention used for drawing shear and moment diagrams See Fig 41 All statically determinate beams will have influence lines that consist of straight line segments After some practice one should be able to minimize computations and locate the unit load only at points representing the end points of each line segment To avoid errors it is recommended that one first construct a table listing unit load at x versus the corresponding value of the function calculated at the specific point that is reaction R shear V or moment M Once the load has been placed at various points along the span of the member the tabulated values can be plotted and the influenceline segments constructed InfluenceLine Equations The influence line can also be constructed by placing the unit load at a variable position x on the member and then computing the value of R V or M at the point as a function of x In this manner the equations of the various line segments composing the influence line can be determined and plotted Although the procedure for constructing an influence line is rather basic one should clearly be aware of the difference between constructing an influence line and constructing a shear or moment diagram Influence lines represent the effect of a moving load only at a specified point on a member whereas shear and moment diagrams represent the effect of fixed loads at all points along the axis of the member The reason for this choice will be explained in Sec 62 63 Qualitative Influence Lines In 1886 Heinrich MüllerBreslau developed a technique for rapidly constructing the shape of an influence line Referred to as the Müller Breslau principle it states that the influence line for a function reaction shear or moment is to the same scale as the deflected shape of the beam when the beam is acted upon by the function In order to draw the deflected shape properly the capacity of the beam to resist the applied function must be removed so the beam can deflect when the function is applied For example consider the beam in Fig 612a If the shape of the influence line for the vertical reaction at A is to be determined the pin is first replaced by a roller guide as shown in Fig 612b A roller guide is necessary since the beam must still resist a horizontal force at A but no vertical force When the positive upward force is then applied at A the beam deflects to the dashed position which represents the general shape of the influence line for Fig 612c Numerical values for this specific case have been calculated in Example 61 If the shape of the influence line for the shear at C is to be determined Fig 613a the connection at C may be symbolized by a roller guide as shown in Fig 613b This device will resist a moment and axial force but no shear Applying a positive shear force to the beam at C and allowing the beam to deflect to the dashed position we find the influenceline shape as shown in Fig 613c Finally if the shape of the influence line for the moment at C Fig 614a is to be determined an internal hinge or pin is placed at C since this connection resists axial and shear forces but cannot resist a moment Fig 614b Applying positive moments to the beam the beam then deflects to the dashed position which is the shape of the influence line Fig 614c The proof of the MüllerBreslau principle can be established using the principle of virtual work Recall that work is the product of either a linear MC VC Ay Ay 216 CHAPTER 6 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 6 A a Ay A deflected shape b Ay x influence line for Ay c Throughout the discussion all deflected positions are drawn to an exaggerated scale to illustrate the concept Here the rollers symbolize supports that carry loads both in tension or compression See Table 21 support 2 Fig 612 Design of this bridge girder is based on influence lines that must be constructed for this train loading 63 QUALITATIVE INFLUENCE LINES 217 6 displacement and force in the direction of the displacement or a rotational displacement and moment in the direction of the displacement If a rigid body beam is in equilibrium the sum of all the forces and moments on it must be equal to zero Consequently if the body is given an imaginary or virtual displacement the work done by all these forces and couple moments must also be equal to zero Consider for example the simply supported beam shown in Fig 615a which is subjected to a unit load placed at an arbitrary point along its length If the beam is given a virtual or imaginary displacement at the support A Fig 615b then only the support reaction and the unit load do virtual work Specifically does positive work and the unit load does negative work The support at B does not move and therefore the force at B does no work Since the beam is in equilibrium and therefore does not actually move the virtual work sums to zero ie If is set equal to 1 then In other words the value of represents the ordinate of the influence line at the position of the unit load Since this value is equivalent to the displacement at the position of the unit load it shows that the shape of the influence line for the reaction at A has been established This proves the MüllerBreslau principle for reactions dy Ay Ay dy dy Ay dy 1 dy 0 1dy Ay dy Ay Ay dy C a VC x influence line for VC c C a MC influence line for MC c x C VC VC deflected shape b deflected shape C MC MC b B C 1 A a C 1 A B b Ay dy dy Fig 613 Fig 614 Fig 615 218 CHAPTER 6 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 6 In the same manner if the beam is sectioned at C and the beam undergoes a virtual displacement at this point Fig 615c then only the internal shear at C and the unit load do work Thus the virtual work equation is Again if then and the shape of the influence line for the shear at C has been established VC dy dy 1 VC dy 1 dy 0 dy Lastly assume a hinge or pin is introduced into the beam at point C Fig 615d If a virtual rotation is introduced at the pin virtual work will be done only by the internal moment and the unit load So Setting it is seen that which indicates that the deflected beam has the same shape as the influence line for the internal moment at point C see Fig 614 Obviously the MüllerBreslau principle provides a quick method for establishing the shape of the influence line Once this is known the ordinates at the peaks can be determined by using the basic method discussed in Sec 61 Also by simply knowing the general shape of the influence line it is possible to locate the live load on the beam and then determine the maximum value of the function by using statics Example 612 illustrates this technique MC dy df 1 MC df 1 dy 0 df A VC c VC 1 B dy dy C MC d MC 1 B A df dy Fig 615 63 QUALITATIVE INFLUENCE LINES 219 6 EXAMPLE 69 For each beam in Fig 616a through 616c sketch the influence line for the vertical reaction at A SOLUTION The support is replaced by a roller guide at A since it will resist but not The force Ay is then applied Ay Ax Again a roller guide is placed at A and the force is applied Ay A doubleroller guide must be used at A in this case since this type of support will resist both a moment at the fixed support and axial load Ax but will not resist Ay MA a A Ay A deflected shape Ay influence line for Ay x A c Ay A deflected shape Ay x influence line for Ay A b Ay A deflected shape Ay x influence line for Ay Fig 616 220 CHAPTER 6 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 6 For each beam in Figs 617a through 617c sketch the influence line for the shear at B SOLUTION The roller guide is introduced at B and the positive shear is applied Notice that the right segment of the beam will not deflect since the roller at A actually constrains the beam from moving vertically either up or down See support 2 in Table 21 VB EXAMPLE 610 Placing the roller guide at B and applying the positive shear at B yields the deflected shape and corresponding influence line Again the roller guide is placed at B the positive shear is applied and the deflected shape and corresponding influence line are shown Note that the left segment of the beam does not deflect due to the fixed support a A B VB VB A B deflected shape VB influence line for VB x B B VB VB deflected shape influence line for VB VB x c B b B VB VB deflected shape influence line for VB VB x Fig 617 63 QUALITATIVE INFLUENCE LINES 221 6 EXAMPLE 611 For each beam in Figs 618a through 618c sketch the influence line for the moment at B SOLUTION A hinge is introduced at B and positive moments are applied to the beam The deflected shape and corresponding influence line are shown MB Placing a hinge at B and applying positive moments to the beam yields the deflected shape and influence line MB With the hinge and positive moment at B the deflected shape and influence line are shown The left segment of the beam is constrained from moving due to the fixed wall at A B b B MB deflected shape MB MB x influence line for MB A B c B deflected shape A MB MB MB x influence line for MB a B B MB MB deflected shape x MB influence line for MB Fig 618 224 CHAPTER 6 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 6 FUNDAMENTAL PROBLEMS F61 Use the MüllerBreslau principle to sketch the influence lines for the vertical reaction at A the shear at C and the moment at C F65 Use the MüllerBreslau principle to sketch the influence lines for the vertical reaction at A the shear at C and the moment at C F62 Use the MüllerBreslau principle to sketch the influence lines for the vertical reaction at A the shear at D and the moment at B F66 Use the MüllerBreslau principle to sketch the influence lines for the vertical reaction at A the shear just to the left of the roller support at E and the moment at A F63 Use the MüllerBreslau principle to sketch the influence lines for the vertical reaction at A the shear at D and the moment at D F67 The beam supports a distributed live load of 15 kNm and single concentrated load of 8 kN The dead load is 2 kNm Determine a the maximum positive moment at C b the maximum positive shear at C F64 Use the MüllerBreslau principle to sketch the influence lines for the vertical reaction at A the shear at B and the moment at B F68 The beam supports a distributed live load of 2 kNm and single concentrated load of 6 kN The dead load is 4 kNm Determine a the maximum vertical positive reaction at C b the maximum negative moment at A C A B F61 D C A B F62 A C B D F63 C A B F64 D E C B A F 65 A C B D E F66 A C B 2 m 2 m 2 m F67 A C D B 3 m 3 m 3 m F68 63 QUALITATIVE INFLUENCE LINES 225 6 63 Draw the influence lines for a the vertical reaction at A b the moment at A and c the shear at B Assume the support at A is fixed Solve this problem using the basic method of Sec 61 64 Solve Prob 63 using the MüllerBreslau principle 69 Draw the influence line for a the vertical reaction at A b the shear at B and c the moment at BAssume A is fixed Solve this problem using the basic method of Sec 61 610 Solve Prob 69 using the MüllerBreslau principle 65 Draw the influence lines for a the vertical reaction at B b the shear just to the right of the rocker at A and c the moment at C Solve this problem using the basic method of Sec 61 66 Solve Prob 65 using MüllerBreslaus principle 611 Draw the influence lines for a the vertical reaction at A b the shear at C and c the moment at C Solve this problem using the basic method of Sec 61 612 Solve Prob 611 using MüllerBreslaus principle 61 Draw the influence lines for a the moment at C b the reaction at B and c the shear at C Assume A is pinned and B is a roller Solve this problem using the basic method of Sec 61 62 Solve Prob 61 using the MüllerBreslau principle 67 Draw the influence line for a the moment at B b the shear at C and c the vertical reaction at B Solve this problem using the basic method of Sec 61 Hint The support at A resists only a horizontal force and a bending moment 68 Solve Prob 67 using the MüllerBreslau principle PROBLEMS C A B 10 ft 10 ft 10 ft Probs 6162 B 5 ft 5 ft A Probs 6364 6 ft 6 ft A C 6 ft B Probs 6566 B C 4 m 4 m 4 m A Probs 6768 B 1 m 2 m A Probs 69610 6 ft 6 ft A C B 3 ft 3 ft Probs 611612 226 CHAPTER 6 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 6 613 Draw the influence lines for a the vertical reaction at A b the vertical reaction at B c the shear just to the right of the support at A and d the moment at CAssume the support at A is a pin and B is a roller Solve this problem using the basic method of Sec 61 614 Solve Prob 613 using the MüllerBreslau principle 617 A uniform live load of 300 lbft and a single live concentrated force of 1500 lb are to be placed on the beam The beam has a weight of 150 lbft Determine a the max imum vertical reaction at support B and b the maximum negative moment at point B Assume the support at A is a pin and B is a roller 615 The beam is subjected to a uniform dead load of 12 kNm and a single live load of 40 kN Determine a the maximum moment created by these loads at C and b the maximum positive shear at C Assume A is a pin and B is a roller 618 The beam supports a uniform dead load of 04 kft a live load of 15 kft and a single live concentrated force of 8 k Determine a the maximum positive moment at C and b the maximum positive vertical reaction at B Assume A is a roller and B is a pin 616 The beam supports a uniform dead load of 500 Nm and a single live concentrated force of 3000 N Determine a the maximum positive moment at Cand b the maximum positive shear at CAssume the support at A is a roller and B is a pin 619 The beam is used to support a dead load of 06 kft a live load of 2 kft and a concentrated live load of 8 k Determine a the maximum positive upward reaction at A b the maximum positive moment at C and c the maximum positive shear just to the right of the support at A Assume the support at A is a pin and B is a roller 2 m 2 m A B C 2 m 2 m Probs 613614 A 6 m 6 m B C 40 kN Prob 615 1 m 3 m C A B Prob 616 B A 20 ft 10 ft Prob 617 B C A 10 ft 10 ft 15 ft Prob 618 B C A 5 ft 10 ft 10 ft 10 ft Prob 619 228 CHAPTER 6 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 6 64 Influence Lines for Floor Girders Occasionally floor systems are constructed as shown in Fig 620a where it can be seen that floor loads are transmitted from slabs to floor beams then to side girders and finally supporting columns An idealized model of this system is shown in plane view Fig 620b Here the slab is assumed to be a oneway slab and is segmented into simply supported spans resting on the floor beams Furthermore the girder is simply supported on the columns Since the girders are main loadcarrying members in this system it is sometimes necessary to construct their shear and moment influence lines This is especially true for industrial buildings subjected to heavy concentrated loads In this regard notice that a unit load on the floor slab is transferred to the girder only at points where it is in contact with the floor beams ie points A B C and D These points are called panel points and the region between these points is called a panel such as BC in Fig 620b Fig 620 slab floor beam girder column A B C D a P A B C D x s s s panel b 1 P 1 d c FB FC B C F1 F2 d F1 P s FB MP VP d 64 INFLUENCE LINES FOR FLOOR GIRDERS 229 6 The influence line for a specified point on the girder can be determined using the same statics procedure as in Sec 61 ie place the unit load at various points x on the floor slab and always compute the function shear or moment at the specified point P in the girder Fig 620b Plotting these values versus x yields the influence line for the function at P In particular the value for the internal moment in a girder panel will depend upon where point P is chosen for the influence line since the magnitude of depends upon the points location from the end of the girder For example if the unit load acts on the floor slab as shown in Fig 620c one first finds the reactions and on the slab then calculates the support reactions and on the girder The internal moment at P is then determined by the method of sections Fig 620d This gives Using a similar analysis the internal shear can be determined In this case however will be constant throughout the panel and so it does not depend upon the exact location d of P within the panel For this reason influence lines for shear in floor girders are specified for panels in the girder and not specific points along the girderThe shear is then referred to as panel shear It should also be noted that since the girder is affected only by the loadings transmitted by the floor beams the unit load is generally placed at each floorbeam location to establish the necessary data used to draw the influence line The following numerical examples should clarify the force analysis BC1VP F1 FB2 VP VP MP F1 d FB1d s2 F2 F1 FC FB MP The design of the floor system of this warehouse building must account for critical locations of storage materials on the floor Influence lines must be used for this purpose Photo courtesy of Portland Cement Association 232 CHAPTER 6 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 6 65 Influence Lines for Trusses Trusses are often used as primary loadcarrying elements for bridges Hence for design it is important to be able to construct the influence lines for each of its members As shown in Fig 623 the loading on the bridge deck is transmitted to stringers which in turn transmit the loading to floor beams and then to the joints along the bottom cord of the truss Since the truss members are affected only by the joint loading we can therefore obtain the ordinate values of the influence line for a member by loading each joint along the deck with a unit load and then use the method of joints or the method of sections to calculate the force in the member The data can be arranged in tabular form listing unit load at joint versus force in member As a convention if the member force is tensile it is considered a positive value if it is compressive it is negative The influence line for the member is constructed by plotting the data and drawing straight lines between the points The following examples illustrate the method of construction bottom cord panel floor beam portal end post stringers portal bracing sway bracing top cord deck lateral bracing Fig 623 The members of this truss bridge were designed using influence lines in accordance with the AASHTO specifications 65 INFLUENCE LINES FOR TRUSSES 237 6 C D E F B A 2 m 2 m 2 m 4 m Prob 632 B A C D E F 2 ft 2 ft 2 ft 2 ft 2 ft Prob 633 632 Draw the influence line for the moment at F in the girder Determine the maximum positive live moment in the girder at F if a single concentrated live force of 8 kN moves across the top floor beams Assume the supports for all members can only exert either upward or downward forces on the members 635 Draw the influence line for the shear in panel CD of the girder Determine the maximum negative live shear in panel CD due to a uniform live load of 500 lbft acting on the top beams 633 A uniform live load of 4 kft and a single concen trated live force of 20 k are placed on the floor beams If the beams also support a uniform dead load of 700 lbft determine a the maximum negative shear in panel DE of the girder and b the maximum negative moment in the girder at C 636 A uniform live load of 65 kNm and a single concentrated live force of 15 kN are placed on the floor beams If the beams also support a uniform dead load of 600 Nm determine a the maximum positive shear in panel CD of the girder and b the maximum positive moment in the girder at D 634 A uniform live load of 02 kft and a single concen trated live force of 4 k are placed on the floor beams Determine a the maximum positive shear in panel DE of the girder and b the maximum positive moment at H 637 A uniform live load of 175 kNm and a single concentrated live force of 8 kN are placed on the floor beams If the beams also support a uniform dead load of 250 Nm determine a the maximum negative shear in panel BC of the girder and b the maximum positive moment at B E H F G D C B A 6 ft 3 ft 3 ft 6 ft 6 ft 6 ft 6 ft Prob 634 8 ft D A B C 8 ft 8 ft 8 ft 8 ft E Prob 635 4 m 4 m 4 m 4 m E B C D A Prob 636 C 3 m 15 m 15 m A B D Prob 637 242 CHAPTER 6 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 6 When many concentrated loads act on the span as in the case of the E72 load of Fig 111 the trialanderror computations used above can be tedious Instead the critical position of the loads can be determined in a more direct manner by finding the change in shear which occurs when the loads are moved from Case 1 to Case 2 then from Case 2 to Case 3 and so on As long as each computed is positive the new position will yield a larger shear in the beam at C than the previous positionEach movement is investigated until a negative change in shear is computed When this occurs the previous position of the loads will give the critical value The change in shear for a load P that moves from position to over a beam can be determined by multiplying P by the change in the ordinate of the influence line that is If the slope of the influence line is s then and therefore 61 If the load moves past a point where there is a discontinuity or jump in the influence line as point C in Fig 627a then the change in shear is simply 62 Use of the above equations will be illustrated with reference to the beam loading and influence line for shown in Fig 628a Notice that the magnitude of the slope of the influence line is and the jump at C has a magnitude of Consider the loads of Case 1 moving 5 ft to Case 2 Fig 628bWhen this occurs the 1k load jumps down and all the loads move up the slope of the influence lineThis causes a change of shear Since is positive Case 2 will yield a larger value for than Case 1 Compare the answers for and previously computed where indeed Investigating which occurs when Case 2 moves to Case 3 Fig 628b we must account for the downward negative jump of the 4k load and the 5ft horizontal movement of all the loads up the slope of the influence lineWe have Since is negative Case 2 is the position of the critical loading as determined previously V23 V23 4112 11 4 42100252152 2875 k V23 1VC22 1VC21 0125 1VC22 1VC21 VC V12 V12 1112 1 4 4100252152 0125 k 112 075 025 1 02510 0025 s 075140 102 VC V P1y2 y12 Jump V Ps1x2 x12 Sloping Line 1y2 y12 s1x2 x12 1y2 y12 x2 x1 V V V 244 CHAPTER 6 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 6 Moment We can also use the foregoing methods to determine the critical position of a series of concentrated forces so that they create the largest internal moment at a specific point in a structure Of course it is first necessary to draw the influence line for the moment at the point and determine the slopes s of its line segments For a horizontal movement of a concentrated force P the change in moment is equivalent to the magnitude of the force times the change in the influenceline ordinate under the load that is 63 As an example consider the beam loading and influence line for the moment at point C in Fig 629a If each of the three concentrated forces is placed on the beam coincident with the peak of the influence line we will obtain the greatest influence from each force The three cases of loading are shown in Fig 629bWhen the loads of Case 1 are moved 4 ft to the left to Case 2 it is observed that the 2k load decreases since the slope is downward Fig 629a Likewise the 4k and 3k forces cause an increase of since the slope is upwardWe have Since is positive we must further investigate moving the loads 6 ft from Case 2 to Case 3 Here the change is negative so the greatest moment at C will occur when the beam is loaded as shown in Case 2 Fig 629cThe maximum moment at C is therefore The following examples further illustrate this method 1MC2max 21452 41752 31602 570 k ft M23 12 42a75 10 b162 3a 75 40 10 b162 225 k ft M12 M12 2a 75 10 b142 14 32a 75 40 10 b142 10 k ft 75140 102 M12 175102 M12 M Ps1x2 x12 Sloping Line M 1x2 x12 The girders of this bridge must resist the maximum moment caused by the weight of this jet plane as it passes over it 66 MAXIMUM INFLUENCE AT A POINT DUE TO A SERIES OF CONCENTRATED LOADS 245 6 MC A C B 10 ft 30 ft 2 k 4 k 3 k 6 ft 4 ft 75 10 40 a x influence line for MC A C 2 k 4 k 3 k 4 ft 10 ft 6 ft B Case 1 A C 2 k 4 k 3 k 4 ft 6 ft B Case 3 b A C 2 k 4 k 3 k 4 ft 6 ft 6 ft B Case 2 45 75 60 6 10 16 40 MC x c Fig 629 67 ABSOLUTE MAXIMUM SHEAR AND MOMENT 251 6 measured from Once this is done moments are summed about B which yields the beams left reaction that is If the beam is sectioned just to the left of the resulting freebody diagram is shown in Fig 636bThe moment under is therefore For maximum we require or Hence we may conclude that the absolute maximum moment in a simply supported beam occurs under one of the concentrated forces such that this force is positioned on the beam so that it and the resultant force of the system are equidistant from the beams centerline Since there are a series of loads on the span for example in Fig 636a this principle will have to be applied to each load in the series and the corresponding maximum moment computed By comparison the largest moment is the absolute maximumAs a general rule though the absolute maximum moment often occurs under the largest force lying nearest the resultant force of the system Envelope of Maximum InfluenceLine Values Rules or formulations for determining the absolute maximum shear or moment are difficult to establish for beams supported in ways other than the cantilever or simple support discussed here An elementary way to proceed to solve this problem however requires constructing influence lines for the shear or moment at selected points along the entire length of the beam and then computing the maximum shear or moment in the beam for each point using the methods of Sec 66 These values when plotted yield an envelope of maximums from which both the absolute maximum value of shear or moment and its location can be found Obviously a computer solution for this problem is desirable for complicated situations since the work can be rather tedious if carried out by hand calculations F2 F3 F1 x x 2 dM2 dx 2FR x L FRx L 0 M2 FR L 4 FRx 2 FR x2 L FR xx L F1 d1 1 L 1FR2cL 2 1x x2d a L 2 xb F1 d1 M2 AyaL 2 xb F1 d1 M 0 F2 M2 F2 Ay 1 L 1FR2cL 2 1x x2d MB 0 Ay F2 x The absolute maximum moment in this girder bridge is the result of the moving concentrated loads caused by the wheels of these train cars The cars must be in the critical position and the location of the point in the girder where the absolute maximum moment occurs must be identified 254 CHAPTER 6 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 6 The truck has a mass of 2 Mg and a center of gravity at G as shown in Fig 638a Determine the absolute maximum moment developed in the simply supported bridge deck due to the trucks weightThe bridge has a length of 10 m SOLUTION As noted in Fig 638a the weight of the truck and the wheel reactions have been calculated by statics Since the largest reaction occurs at the front wheel we will select this wheel along with the resultant force and position them equidistant from the centerline of the bridge Fig 638b Using the resultant force rather than the wheel loads the vertical reaction at B is then The maximum moment occurs under the front wheel loading Using the right section of the bridge deck Fig 638c we have Ans Ms 397 kN m 88291452 Ms 0 d Ms 0 By 8829 kN By1102 19621452 0 d MA 0 1962kN 211032kg1981 ms22 EXAMPLE 622 a 654 kN 1308 kN 1 m 2 m 1962 kN G 5 m 5 m b A B Ay By 654 kN 1308 kN 1308 kN 883 kN 05 m 05 m 1962 kN 45 m Vs Ms Fig 638 c 67 ABSOLUTE MAXIMUM SHEAR AND MOMENT 255 6 659 Determine the maximum moment at point C on the single girder caused by the moving dolly that has a mass of 2 Mg and a mass center at GAssume A is a roller 662 Determine the maximum positive moment at the splice C on the side girder caused by the moving load which travels along the center of the bridge PROBLEMS 660 Determine the maximum moment in the suspended rail at point B if the rail supports the load of 25 k on the trolley 663 Determine the maximum moment at C caused by the moving load 664 Draw the influence line for the force in member IH of the bridge truss Determine the maximum force tension or compression that can be developed in this member due to a 72k truck having the wheel loads shown Assume the truck can travel in either direction along the center of the deck so that half its load is transferred to each of the two side trussesAlso assume the members are pin connected at the gusset plates G 5 m 5 m 5 m C B A 15 m 05 m Prob 659 8 ft 8 ft 6 ft 6 ft A B C 25 k 2 ft 1 ft Prob 661 8 ft 8 ft 6 ft 6 ft A B C 25 k 2 ft 1 ft Prob 660 B C A 8 m 8 m 8 m 4 kN 4 m 8 kN Prob 662 15 ft 15 ft A C B 2 ft 1 ft 2400 lb Prob 663 J I H G A B C D E K L M 10 ft 10 ft F 32 k 32 k 8 k 20 ft 20 ft 20 ft 20 ft 20 ft 25 ft 15 ft Prob 664 661 Determine the maximum positive shear at point B if the rail supports the load of 25 k on the trolley 256 CHAPTER 6 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 6 665 Determine the maximum positive moment at point C on the single girder caused by the moving load 668 Draw the influence line for the force in member IC of the bridge truss Determine the maximum force tension or compression that can be developed in the member due to a 5k truck having the wheel loads shown Assume the truck can travel in either direction along the center of the deck so that half the load shown is transferred to each of the two side trusses Also assume the members are pin connected at the gusset plates 666 The cart has a weight of 2500 lb and a center of gravity at G Determine the maximum positive moment created in the side girder at C as it crosses the bridge Assume the car can travel in either direction along the center of the deck so that half its load is transferred to each of the two side girders 669 The truck has a mass of 4 Mg and mass center at and the trailer has a mass of 1 Mg and mass center at Determine the absolute maximum live moment developed in the bridge G2 G1 667 Draw the influence line for the force in member BC of the bridge truss Determine the maximum force tension or compression that can be developed in the member due to a 5k truck having the wheel loads shown Assume the truck can travel in either direction along the center of the deck so that half the load shown is transferred to each of the two side trusses Also assume the members are pin connected at the gusset plates 670 Determine the absolute maximum live moment in the bridge in Problem 669 if the trailer is removed 5 m A B 2 m 15 m 4 kN 6 kN 8 kN 5 m C Prob 665 8 ft 8 ft A B 15 ft 1 ft G C Prob 666 J I H G D C B E F 15 ft 3 k 2 k 8 ft A 20 ft 20 ft 20 ft 20 ft Probs 667668 8 m A B G1 G2 15 m 075 m 15 m Prob 669 8 m A B G1 G2 15 m 075 m 15 m Prob 670 67 ABSOLUTE MAXIMUM SHEAR AND MOMENT 257 6 671 Determine the absolute maximum live shear and absolute maximum moment in the jib beam AB due to the 10kN loading The end constraints require 01 m x 39 m 673 Determine the absolute maximum moment in the girder bridge due to the truck loading shown The load is applied directly to the girder 672 Determine the maximum moment at C caused by the moving loads 674 Determine the absolute maximum shear in the beam due to the loading shown 4 m x A B 10 kN Prob 671 20 ft 30 ft C A B 2 k 2 k 4 k6 k 3 ft 4 ft 3 ft Prob 672 B 80 ft 20 ft 8 ft 10 k 15 k 7 k 3 k 4 ft A Prob 673 12 m 20 kN 25 kN 40 kN 4 m A B 15 m Prob 675 12 m 20 kN 25 kN 40 kN 4 m A B 15 m Prob 674 675 Determine the absolute maximum moment in the beam due to the loading shown 258 CHAPTER 6 INFLUENCE LINES FOR STATICALLY DETERMINATE STRUCTURES 6 676 Determine the absolute maximum shear in the bridge girder due to the loading shown 679 Determine the absolute maximum shear in the beam due to the loading shown 678 Determine the absolute maximum moment in the girder due to the loading shown 681 The trolley rolls at C and D along the bottom and top flange of beam AB Determine the absolute maximum moment developed in the beam if the load supported by the trolley is 2 k Assume the support at A is a pin and at B a roller 30 ft 8 ft B A 10 k 6 k Prob 677 30 ft 8 ft B A 10 k 6 k Prob 676 25 ft 10 k 8 k 3 k 2 ft2 ft 3 ft 4 k Prob 678 30 ft 3 k 6 k 2 k 3 ft3 ft 5 ft 4 k Prob 680 30 ft 3 k 6 k 2 k 3 ft3 ft 5 ft 4 k Prob 679 A D B C 1 ft 05 ft 20 ft Prob 681 677 Determine the absolute maximum moment in the bridge girder due to the loading shown 680 Determine the absolute maximum moment in the bridge due to the loading shown PROJECT PROBLEMS 259 6 61P The chain hoist on the wall crane can be placed anywhere along the boom and has a rated capacity of 28 kN Use an impact factor of 03 and determine the absolute maximum bending moment in the boom and the maximum force developed in the tie rod BC The boom is pinned to the wall column at its left end A Neglect the size of the trolley at D 01 m 6 x 6 34 m 62P A simply supported pedestrian bridge is to be constructed in a city park and two designs have been proposed as shown in case a and case b The truss members are to be made from timberThe deck consists of 15mlong planks that have a mass of A local code states the live load on the deck is required to be 5 kPa with an impact factor of 02 Consider the deck to be simply supported on stringers Floor beams then transmit the load to the bottom joints of the truss See Fig 623 In each case find the member subjected to the largest tension and largest compression load and suggest why you would choose one design over the other Neglect the weights of the truss members 20 kgm2 PROJECT PROBLEMS 075 m 3 m x 05 m 01 m 28 kN A D B C Prob 61P E 125 m 125 m 125 m 125 m 125 m case b A B C D F G H 125 m 125 m 125 m 125 m 125 m case a A B C D E E F G H Prob 62P CHAPTER REVIEW 261 6 Influence lines for floor girders and trusses can be established by placing the unit load at each panel point or joint and calculating the value of the required reaction shear or moment When a series of concentrated loads pass over the member then the various positions of the load on the member have to be considered to determine the largest shear or moment in the member In general place the loadings so that each contributes its maximum influence as determined by multiplying each load by the ordinate of the influence line This process of finding the actual position can be done using a trialanderror procedure or by finding the change in either the shear or moment when the loads are moved from one position to another Each moment is investigated until a negative value of shear or moment occurs Once this happens the previous position will define the critical loading Vabs max Absolute maximum moment in a cantilevered beam occurs when the series of concentrated loads are placed at the farthest point away from the fixed support F1 F2 F3 FR By Ay L 2 2 L 2 x x 2 x Absolute maximum shear in a cantilever or simply supported beam will occur at a support when one of the loads is placed next to the support To determine the absolute maximum moment in a simply supported beam the resultant of the force system is first determined Then it along with one of the concentrated forces in the system is positioned so that these two forces are equidistant from the centerline of the beam The maximum moment then occurs under the selected force Each force in the system is selected in this manner and by comparison the largest for all these cases is the absolute maximum moment Mabs max Vabs max The portal to this bridge must resist loteral loads due to wind and traffic An approximate analysis can be made of the forces produced for a preliminary design of the members before a more exact structural analysis is done 7 263 In this chapter we will present some of the approximate methods used to analyze statically indeterminate trusses and frames These methods were developed on the basis of structural behavior and their accuracy in most cases compares favorably with more exact methods of analysis Although not all types of structural forms will be discussed here it is hoped that enough insight is gained from the study of these methods so that one can judge what would be the best approximations to make when performing an approximate force analysis of a statically indeterminate structure 71 Use of Approximate Methods When a model is used to represent any structure the analysis of it must satisfy both the conditions of equilibrium and compatibility of displacement at the jointsAs will be shown in later chapters of this text the compatibility conditions for a statically indeterminate structure can be related to the loads provided we know the materials modulus of elasticity and the size and shape of the members For an initial design however we will not know a members size and so a statically indeterminate analysis cannot be considered For analysis a simpler model of the structure must be developed one that is statically determinate Once this model is specified the analysis of it is called an approximate analysis By performing an approximate analysis a preliminary design of the members of a structure can be made and when this is complete the more exact indeterminate analysis can then be performed and the design refined An approximate analysis also provides insight as to a structures behavior under load and is beneficial when checking a more exact analysis or when time money or capability are not available for performing the more exact analysis Approximate Analysis of Statically Indeterminate Structures 268 CHAPTER 7 APPROXIMATE ANALYSIS OF STATICALLY INDETERMINATE STRUCTURES 7 71 Determine approximately the force in each member of the truss Assume the diagonals can support either a tensile or a compressive force 72 Solve Prob 71 assuming that the diagonals cannot support a compressive force 75 Determine approximately the force in each member of the truss Assume the diagonals can support either a tensile or a compressive force 76 Solve Prob 75 assuming that the diagonals cannot support a compressive force PROBLEMS 73 Determine approximately the force in each member of the truss Assume the diagonals can support either a tensile or a compressive force 74 Solve Prob 73 assuming that the diagonals cannot support a compressive force 77 Determine approximately the force in each member of the truss Assume the diagonals can support either a tensile or compressive force 78 Solve Prob 77 assuming that the diagonals cannot support a compressive force Probs 7576 Probs 7374 Probs 7172 3 m 3 m 3 m 50 kN A D B C 40 kN 20 kN F E 20 ft 20 ft 20 ft 20 ft 10 k H A D B C 10 k G 10 k F 10 k 5 k E 8 ft 6 ft 8 ft 8 ft 7 k H A D B C 14 k G 14 k F 7 k 2 k E 8 kN 4 kN F E D 15 m A B C 2 m 2 m Probs 7778 72 TRUSSES 269 7 79 Determine approximately the force in each member of the truss Assume the diagonals can support both tensile and compressive forces 711 Determine approximately the force in each member of the truss Assume the diagonals can support either a tensile or compressive force 15 ft 15 ft 2 k 2 k 15 k 15 ft 15 ft E F A B C G D 8 kN 15 m E F A B C D 10 kN 2 m 2 m Prob 712 Prob 710 15 ft 15 ft 2 k 2 k 15 k 15 ft 15 ft E F A B C G D 8 kN 15 m E F A B C D 10 kN 2 m 2 m Prob 711 Prob 79 710 Determine approximately the force in each member of the truss Assume the diagonals DG and AC cannot support a compressive force 712 Determine approximately the force in each member of the truss Assume the diagonals cannot support a compressive force 270 CHAPTER 7 APPROXIMATE ANALYSIS OF STATICALLY INDETERMINATE STRUCTURES 73 Vertical Loads on Building Frames Building frames often consist of girders that are rigidly connected to columns so that the entire structure is better able to resist the effects of lateral forces due to wind and earthquake An example of such a rigid framework often called a building bent is shown in Fig 74 In practice a structural engineer can use several techniques for performing an approximate analysis of a building bent Each is based upon knowing how the structure will deform under load One technique would be to consider only the members within a localized region of the structureThis is possible provided the deflections of the members within the region cause little disturbance to the members outside the region Most often however the deflection curve of the entire structure is considered From this the approximate location of points of inflection that isthe points where the member changes its curvaturecan be specified These points can be considered as pins since there is zero moment within the member at the points of inflection We will use this idea in this section to analyze the forces on building frames due to vertical loads and in Secs 75 and 76 an approximate analysis for frames subjected to lateral loads will be presented Since the frame can be subjected to both of these loadings simultaneously then provided the material remains elastic the resultant loading is determined by superposition Assumptions for Approximate Analysis Consider a typical girder located within a building bent and subjected to a uniform vertical load as shown in Fig 75a The column supports at A and B will each exert three reactions on the girder and therefore the girder will be statically indeterminate to the third degree 6 reactions 3 equations of equilibrium To make the girder statically determinate an approximate analysis will therefore require three assumptions If the columns are extremely stiff no rotation at A and B will occur and the deflection 7 typical building frame Fig 74 73 VERTICAL LOADS ON BUILDING FRAMES 271 curve for the girder will look like that shown in Fig 75b Using one of the methods presented in Chapters 9 through 11 an exact analysis reveals that for this case inflection points or points of zero moment occur at 021L from each support If however the column connections at A and B are very flexible then like a simply supported beam zero moment will occur at the supports Fig 75c In reality however the columns will provide some flexibility at the supports and therefore we will assume that zero moment occurs at the average point between the two extremes ie at from each support Fig 75d Furthermore an exact analysis of frames supporting vertical loads indicates that the axial forces in the girder are negligible In summary then each girder of length L may be modeled by a simply supported span of length 08L resting on two cantilevered ends each having a length of 01L Fig 75e The following three assumptions are incorporated in this model 1 There is zero moment in the girder 01L from the left support 2 There is zero moment in the girder 01L from the right support 3 The girder does not support an axial force By using statics the internal loadings in the girders can now be obtained and a preliminary design of their cross sections can be made The following example illustrates this numerically 1021L 022 L 01L 7 w column column girder A B L a w A B L points of zero moment 021L 021L fixed supported b w A B L simply supported c point of zero moment point of zero moment w L assumed points of zero moment 01L approximate case d 01L Fig 75 w 01L model e 01L 08L 278 CHAPTER 7 APPROXIMATE ANALYSIS OF STATICALLY INDETERMINATE STRUCTURES 7 Prob 713 6 m A B C D E F G H 8 m 6 m 6 m 3 kNm Prob 714 400 lbft F E A B C D 15 ft 20 ft Prob 715 8 m A C E B D F 5 kNm 9 kNm Prob 716 A B D E F K L G H I J C 5 kNm 5 kNm 3 kNm 8 m 8 m 8 m 713 Determine approximately the internal moments at joints A and B of the frame 715 Determine approximately the internal moment at A caused by the vertical loading PROBLEMS 714 Determine approximately the internal moments at joints F and D of the frame 716 Determine approximately the internal moments at A and B caused by the vertical loading 74 PORTAL FRAMES AND TRUSSES 279 7 Prob 717 20 ft 40 ft 30 ft A H I B G F K L E D J C 05 kft 15 kft 15 kft Prob 718 15 ft 20 ft A D F C H G E B 400 lbft 1200 lbft 718 Determine approximately the support actions at A B and C of the frame 720 Determine approximately the internal moment and shear at the ends of each member of the portal frame Assume the supports at A and D are partially fixed such that an inflection point is located at h3 from the bottom of each column 717 Determine approximately the internal moments at joints I and L Also what is the internal moment at joint H caused by member HG 719 Determine approximately the support reactions at A and B of the portal frame Assume the supports are a pinned and b fixed Prob 719 6 m 4 m 12 kN A D C B Prob 720 P B A C D b h 280 CHAPTER 7 APPROXIMATE ANALYSIS OF STATICALLY INDETERMINATE STRUCTURES 7 Probs 721722 7 ft 500 lb B A E G H F D C 6 ft 15 ft 15 ft 15 ft Probs 723724 8 ft 2 k 8 ft 6 ft 12 ft A B C D F E G 1 k Prob 725 2 m 2 m 15 m 5 m A B D G F C E 4 kN 8 kN Prob 726 2 m 2 m 15 m 5 m A B D G F C E 4 kN 8 kN 721 Draw approximately the moment diagram for member ACE of the portal constructed with a rigid member EG and knee braces CF and DH Assume that all points of connection are pins Also determine the force in the knee brace CF 722 Solve Prob 721 if the supports at A and B are fixed instead of pinned 725 Draw approximately the moment diagram for col umn AGF of the portal Assume all truss members and the columns to be pin connected at their ends Also determine the force in all the truss members 723 Determine approximately the force in each truss member of the portal frame Also find the reactions at the fixed column supports A and BAssume all members of the truss to be pin connected at their ends 724 Solve Prob 723 if the supports at A and B are pinned instead of fixed 726 Draw approximately the moment diagram for col umn AGF of the portal Assume all the members of the truss to be pin connected at their ends The columns are fixed at A and B Also determine the force in all the truss members 75 LATERAL LOADS ON BUILDING FRAMES PORTAL METHOD 283 7 In summary the portal method for analyzing fixedsupported building frames requires the following assumptions 1 A hinge is placed at the center of each girder since this is assumed to be a point of zero moment 2 A hinge is placed at the center of each column since this is assumed to be a point of zero moment 3 At a given floor level the shear at the interior column hinges is twice that at the exterior column hinges since the frame is considered to be a superposition of portals These assumptions provide an adequate reduction of the frame to one that is statically determinate yet stable under loading By comparison with the more exact statically indeterminate analysis the portal method is most suitable for buildings having low elevation and uniform framingThe reason for this has to do with the structures action under load In this regard consider the frame as acting like a cantilevered beam that is fixed to the ground Recall from mechanics of materials that shear resistance becomes more important in the design of short beams whereas bending is more important if the beam is long See Sec 76 The portal method is based on the assumption related to shear as stated in item 3 above The following examples illustrate how to apply the portal method to analyze a building bent The portal method of analysis can be used to approximately perform a lateralload analysis of this singlestory frame 284 CHAPTER 7 APPROXIMATE ANALYSIS OF STATICALLY INDETERMINATE STRUCTURES 7 Determine approximately the reactions at the base of the columns of the frame shown in Fig 713a Use the portal method of analysis EXAMPLE 75 SOLUTION Applying the first two assumptions of the portal method we place hinges at the centers of the girders and columns of the frame Fig 713a A section through the column hinges at I J K L yields the freebody diagram shown in Fig 713b Here the third assumption regarding the column shears appliesWe require Using this result we can now proceed to dismember the frame at the hinges and determine their reactions As a general rule always start this analysis at the corner or joint where the horizontal load is applied Hence the freebody diagram of segment IBM is shown in Fig 713c The three reaction components at the hinges and are determined by applying respectively The adjacent segment MJN is analyzed next Fig 713d followed by segment NKO Fig 713e and finally segment OGL Fig 713f Using these results the freebody diagrams of the columns with their support reactions are shown in Fig 713g MM 0 Fx 0 Fy 0 My Mx Iy Fx 0 1200 6V 0 V 200 lb B M D N F O G I J K L A C E H 1200 lb 12 ft 16 ft 16 ft 16 ft a Fig 713 1200 lb V Iy I 2V Jy J 2V Ky K V Ly L b 286 CHAPTER 7 APPROXIMATE ANALYSIS OF STATICALLY INDETERMINATE STRUCTURES 7 Determine approximately the reactions at the base of the columns of the frame shown in Fig 714a Use the portal method of analysis EXAMPLE 76 SOLUTION First hinges are placed at the centers of the girders and columns of the frame Fig 714aA section through the hinges at O P Q and J K L yields the freebody diagrams shown in Fig 714bThe column shears are calculated as follows 20 30 4V 0 V 125 kN Fx 0 20 4V 0 V 5 kN Fx 0 20 kN V Oy 25 m 2V Py V Qy 20 kN 30 kN G R H S I D M E N F O P Q J K L A C 5 m 6 m 8 m 8 m a B 20 kN 5 m 30 kN 3 m V Jy 2V Ky V Ly b Fig 714 288 CHAPTER 7 APPROXIMATE ANALYSIS OF STATICALLY INDETERMINATE STRUCTURES 7 Fig 715 76 Lateral Loads on Building Frames Cantilever Method The cantilever method is based on the same action as a long cantilevered beam subjected to a transverse load It may be recalled from mechanics of materials that such a loading causes a bending stress in the beam that varies linearly from the beams neutral axis Fig 715a In a similar manner the lateral loads on a frame tend to tip the frame over or cause a rotation of the frame about a neutral axis lying in a horizontal plane that passes through the columns between each floor To counteract this tipping the axial forces or stress in the columns will be tensile on one side of the neutral axis and compressive on the other side Fig 715b Like the cantilevered beam it therefore seems reasonable to assume this axial stress has a linear variation from the centroid of the column areas or neutral axis The cantilever method is therefore appropriate if the frame is tall and slender or has columns with different crosssectional areas P beam a building frame b 76 LATERAL LOADS ON BUILDING FRAMES CANTILEVER METHOD 289 7 The building framework has rigid connectionsA lateralload analysis can be performed approximately by using the cantilever method of analysis In summary using the cantilever method the following assumptions apply to a fixedsupported frame 1 A hinge is placed at the center of each girder since this is assumed to be a point of zero moment 2 A hinge is placed at the center of each column since this is assumed to be a point of zero moment 3 The axial stress in a column is proportional to its distance from the centroid of the crosssectional areas of the columns at a given floor level Since stress equals force per area then in the special case of the columns having equal crosssectional areas the force in a column is also proportional to its distance from the centroid of the column areas These three assumptions reduce the frame to one that is both stable and statically determinate The following examples illustrate how to apply the cantilever method to analyze a building bent 294 CHAPTER 7 APPROXIMATE ANALYSIS OF STATICALLY INDETERMINATE STRUCTURES 7 6 m A B C E D F 8 m 8 m 15 kN Prob 735 15 ft A B C E D J I H G F 18 ft 18 ft 18 ft 18 ft 4 k Prob 736 9 kN 5 m 5 m 5 m 4 m 4 m J K L G F E H D C B A I 12 kN Probs 737738 12 ft 15 ft 4 k 5 k A D E F C B 12 ft Probs 739740 735 Use the portal method of analysis and draw the moment diagram for girder FED 739 Use the portal method of analysis and draw the moment diagram for column AFE 740 Solve Prob 739 using the cantilever method of analysisAll the columns have the same crosssectional area PROBLEMS 737 Use the portal method and determine approximately the reactions at supports A B C and D 738 Use the cantilever method and determine approximately the reactions at supports A B C and D All columns have the same crosssectional area 736 Use the portal method of analysis and draw the moment diagram for girder JIHGF 741 Use the portal method and determine approximately the reactions at A 742 Use the cantilever method and determine approximately the reactions at AAll of the columns have the same crosssectional area 3 k 18 ft 20 ft 15 ft 15 ft G F E H D C B A I 4 k Probs 741742 The deflection of this arch bridge must be carefully monitored while it is under construction 8 299 In this chapter we will show how to determine the elastic deflections of a beam using the method of double integration and two important geometrical methods namely the momentarea theorems and the conjugatebeam method Double integration is used to obtain equations which define the slope and the elastic curve The geometric methods provide a way to obtain the slope and deflection at specific points on the beam Each of these methods has particular advantages or disadvantages which will be discussed when each method is presented 81 Deflection Diagrams and the Elastic Curve Deflections of structures can occur from various sources such as loads temperature fabrication errors or settlement In design deflections must be limited in order to provide integrity and stability of roofs and prevent cracking of attached brittle materials such as concrete plaster or glass Furthermore a structure must not vibrate or deflect severely in order to appear safe for its occupants More important though deflections at specified points in a structure must be determined if one is to analyze statically indeterminate structures The deflections to be considered throughout this text apply only to structures having linear elastic material response Under this condition a structure subjected to a load will return to its original undeformed position after the load is removedThe deflection of a structure is caused Deflections 302 CHAPTER 8 DEFLECTIONS 8 Draw the deflected shape of each of the beams shown in Fig 85 SOLUTION In Fig 85a the roller at A allows free rotation with no deflection while the fixed wall at B prevents both rotation and deflection The deflected shape is shown by the bold line In Fig 85b no rotation or deflection can occur at A and B In Fig 85c the couple moment will rotate end A This will cause deflections at both ends of the beam since no deflection is possible at B and C Notice that segment CD remains undeformed a straight line since no internal load acts within it In Fig 85d the pin internal hinge at B allows free rotation and so the slope of the deflection curve will suddenly change at this point while the beam is constrained by its supports In Fig 85e the compound beam deflects as shown The slope abruptly changes on each side of the pin at B Finally in Fig 85f span BC will deflect concave upwards due to the load Since the beam is continuous the end spans will deflect concave downwards EXAMPLE 81 Fig 85 D A C B d P B w A D C f C B e P A A w b B A a P B M D B A C c 81 DEFLECTION DIAGRAMS AND THE ELASTIC CURVE 303 8 EXAMPLE 82 Draw the deflected shapes of each of the frames shown in Fig 86 Fig 86 d A I H G F D E B C P P a A D B C P b A F E B C D P c A H G F B C D E w SOLUTION In Fig 86a when the load P pushes joints B and C to the right it will cause clockwise rotation of each column as shown As a result joints B and C must rotate clockwise Since the 90 angle between the connected members must be maintained at these joints the beam BC will deform so that its curvature is reversed from concave up on the left to concave down on the right Note that this produces a point of inflection within the beam In Fig 86b P displaces joints B C and D to the right causing each column to bend as shown The fixed joints must maintain their 90 angles and so BC and CD must have a reversed curvature with an inflection point near their midpoint In Fig 86c the vertical loading on this symmetric frame will bend beam CD concave upwards causing clockwise rotation of joint C and counterclockwise rotation of joint D Since the 90 angle at the joints must be maintained the columns bend as shown This causes spans BC and DE to be concave downwards resulting in counterclockwise rotation at B and clockwise rotation at EThe columns therefore bend as shown Finally in Fig 86d the loads push joints B and C to the right which bends the columns as shown The fixed joint B maintains its 90 angle however no restriction on the relative rotation between the members at C is possible since the joint is a pin Consequently only beam CD does not have a reverse curvature 304 CHAPTER 8 DEFLECTIONS 8 F81 Draw the deflected shape of each beam Indicate the inflection points FUNDAMENTAL PROBLEMS F83 Draw the deflected shape of each frame Indicate the inflection points F82 Draw the deflected shape of each frame Indicate the inflection points a F81 c F82 F83 a b c b a b 82 ELASTICBEAM THEORY 305 8 82 ElasticBeam Theory In this section we will develop two important differential equations that relate the internal moment in a beam to the displacement and slope of its elastic curve These equations form the basis for the deflection methods presented in this chapter and for this reason the assumptions and limitations used in their development should be fully understood To derive these relationships we will limit the analysis to the most common case of an initially straight beam that is elastically deformed by loads applied perpendicular to the beams x axis and lying in the plane of symmetry for the beams crosssectional area Fig 87a Due to the loading the deformation of the beam is caused by both the internal shear force and bending moment If the beam has a length that is much greater than its depth the greatest deformation will be caused by bending and therefore we will direct our attention to its effects Deflections caused by shear will be discussed later in the chapter When the internal moment M deforms the element of the beam each cross section remains plane and the angle between them becomes Fig 87b The arc dx that represents a portion of the elastic curve intersects the neutral axis for each cross section The radius of curvature for this arc is defined as the distance which is measured from the center of curvature to dx Any arc on the element other than dx is subjected to a normal strain For example the strain in arc ds located at a position y from the neutral axis is However and and so If the material is homogeneous and behaves in a linear elastic manner then Hookes law applies Also since the flexure formula applies Combining these equations and substituting into the above equation we have 81 Here the radius of curvature at a specific point on the elastic curve is referred to as the curvature the internal moment in the beam at the point where is to be determined the materials modulus of elasticity I the beams moment of inertia computed about the neutral axis E r M 1r r 1 r M EI s MyI P sE P 1r y2 du r du r du or 1 r P y ds 1r y2 du ds dx r du P 1ds ds2ds O r du xv a A B x dx v P w u b y y dx ds dx ds before deformation after deformation du M M O r r Fig 87 306 CHAPTER 8 DEFLECTIONS 8 The product EI in this equation is referred to as the flexural rigidity and it is always a positive quantity Since then from Eq 81 82 If we choose the axis positive upward Fig 87a and if we can express the curvature in terms of x and we can then determine the elastic curve for the beam In most calculus books it is shown that this curvature relationship is Therefore 83 This equation represents a nonlinear secondorder differential equation Its solution gives the exact shape of the elastic curve assuming of course that beam deflections occur only due to bending In order to facilitate the solution of a greater number of problems Eq 83 will be modified by making an important simplification Since the slope of the elastic curve for most structures is very smallwe will use small deflection theory and assume Consequently its square will be negligible compared to unity and therefore Eq 83 reduces to 84 It should also be pointed out that by assuming the original length of the beams axis x and the arc of its elastic curve will be approx imately the same In other words ds in Fig 87b is approximately equal to dx since This result implies that points on the elastic curve will only be displaced vertically and not horizontally Tabulated Results In the next section we will show how to apply Eq 84 to find the slope of a beam and the equation of its elastic curve The results from such an analysis for some common beam loadings often encountered in structural analysis are given in the table on the inside front cover of this book Also listed are the slope and displacement at critical points on the beam Obviously no single table can account for the many different cases of loading and geometry that are encountered in practice When a table is not available or is incomplete the displacement or slope of a specific point on a beam or frame can be determined by using the double integration method or one of the other methods discussed in this and the next chapter ds 2dx2 dv2 21 1dvdx22 dx L dx dvdx L 0 d2v dx2 M EI dvdx L 0 v f1x2 M EI d2vdx2 1 1dvdx2232 1 r d2vdx2 1 1dvdx2232 v 11r2 v du M EI dx dx r du 308 CHAPTER 8 DEFLECTIONS 8 Procedure for Analysis The following procedure provides a method for determining the slope and deflection of a beam or shaft using the method of double integration It should be realized that this method is suitable only for elastic deflections for which the beams slope is very small Furthermore the method considers only deflections due to bending Additional deflection due to shear generally represents only a few percent of the bending deflection and so it is usually neglected in engineering practice Elastic Curve Draw an exaggerated view of the beams elastic curve Recall that points of zero slope and zero displacement occur at a fixed support and zero displacement occurs at pin and roller supports Establish the x and v coordinate axes The x axis must be parallel to the undeflected beam and its origin at the left side of the beam with a positive direction to the right If several discontinuous loads are present establish x coordi nates that are valid for each region of the beam between the discontinuities In all cases the associated positive v axis should be directed upward Load or Moment Function For each region in which there is an x coordinate express the internal moment M as a function of x Always assume that M acts in the positive direction when apply ing the equation of moment equilibrium to determine Slope and Elastic Curve Provided EI is constant apply the moment equation which requires two integrations For each integration it is important to include a constant of integration The constants are determined using the boundary conditions for the supports and the continuity conditions that apply to slope and displacement at points where two functions meet Once the integration constants are determined and substituted back into the slope and deflection equations the slope and displacement at specific points on the elastic curve can be deter minedThe numerical values obtained can be checked graphically by comparing them with the sketch of the elastic curve Positive values for slope are counterclockwise and positive displacement is upward Mx EI d2vdx2 M fx EXAMPLE 83 Each simply supported floor joist shown in the photo is subjected to a uniform design loading of 4 kNm Fig 811a Determine the maximum deflection of the joist EI is constant Elastic Curve Due to symmetry the joists maximum deflection will occur at its center Only a single x coordinate is needed to determine the internal moment Moment Function From the freebody diagram Fig 811b we have Slope and Elastic Curve Applying Eq 84 and integrating twice gives Here at so that and at so that The equation of the elastic curve is therefore At note that The maximum deflection is therefore Ans v max 521 EI dvdx 0 x 5 m EIv 3333x3 01667x4 1667x C1 1667 x 10 v 0 C2 0 x 0 v 0 EIv 3333x3 01667x4 C1x C2 EI dv dx 10x2 06667x3 C1 EI d2v dx2 20x 2x2 M 20x 4xax 2 b 20x 2x2 4 kNm a 10 m x 20 kN 20 kN x 2 4 x N b x 20 kN M V Fig 811 83 THE DOUBLE INTEGRATION METHOD 309 8 310 CHAPTER 8 DEFLECTIONS 8 The cantilevered beam shown in Fig 812a is subjected to a couple moment at its end Determine the equation of the elastic curve EI is constant M0 EXAMPLE 84 SOLUTION Elastic Curve The load tends to deflect the beam as shown in Fig 89a By inspection the internal moment can be represented throughout the beam using a single x coordinate Moment Function From the freebody diagram with M acting in the positive direction Fig 812b we have Slope and Elastic Curve Applying Eq 84 and integrating twice yields 1 2 3 Using the boundary conditions at and at then Substituting these results into Eqs 2 and 3 with we get Ans v M0x2 2EI u M0x EI u dvdx C1 C2 0 x 0 v 0 x 0 dvdx 0 EIv M0x2 2 C1x C2 EIdv dx M0x C1 EId2v dx2 M0 M M0 L x A M0 a x M M0 b Fig 812 83 THE DOUBLE INTEGRATION METHOD 311 8 Maximum slope and displacement occur at for which 4 5 The positive result for indicates counterclockwise rotation and the positive result for indicates that is upward This agrees with the results sketched in Fig 812a In order to obtain some idea as to the actual magnitude of the slope and displacement at the end A consider the beam in Fig 812a to have a length of 12 ft support a couple moment of and be made of steel having If this beam were designed without a factor of safety by assuming the allowable normal stress is equal to the yield stress then a would be found to be adequate From Eqs 4 and 5 we get Since this justifies the use of Eq 84 rather than applying the more exact Eq 83 for computing the deflection of beams Also since this numerical application is for a cantilevered beam we have obtained larger values for maximum and than would have been obtained if the beam were supported using pins rollers or other supports v u u2 A 000297 rad2 V 1 vA 15 k ft112 inft2112 ft22112 in1 ft22 212911032 kin221164 in42 392 in uA 15 k ft112 inft2112 ft2112 inft2 2911032 kin21164 in42 00545 rad 1I 164 in42 W6 9 sallow 36 ksi Est 2911032 ksi 15 k ft vA vA uA vA M0L2 2EI uA M0L EI A 1x L2 312 CHAPTER 8 DEFLECTIONS 8 The beam in Fig 813a is subjected to a load P at its end Determine the displacement at C EI is constant EXAMPLE 85 SOLUTION Elastic Curve The beam deflects into the shape shown in Fig 813a Due to the loading two x coordinates must be considered Moment Functions Using the freebody diagrams shown in Fig 813b we have Slope and Elastic Curve Applying Eq 84 for x1 1 2 EIv1 P 12x3 1 C1x1 C2 EIdv1 dx1 P 4 x2 1 C1 EId2v1 dx2 1 P 2 x1 Px2 3Pa 2a x2 3a M2 P 2 x2 3P 2 1x2 2a2 M1 P 2 x1 0 x1 2a x2 A P C B 2a a x1 a vC V2 M2 x2 P 2 3P 2 2a b Fig 813 x1 P 2 V1 M1 83 THE DOUBLE INTEGRATION METHOD 313 8 For x2 3 4 The four constants of integration are determined using three bound ary conditions namely at at and at and one continuity equation Here the continuity of slope at the roller requires at Note that continuity of displacement at B has been indirectly considered in the boundary conditions since at Applying these four conditions yields Solving we obtain Substituting and into Eq 4 gives The displacement at C is determined by setting We get Ans vC Pa3 EI x2 3a v2 P 6EIx3 2 3 2 Pa EIx2 2 10Pa2 3EI x2 2Pa3 EI C4 C3 C1 Pa2 3 C2 0 C3 10 3 Pa2 C4 2Pa3 P 4 12a22 C1 P 2 12a22 3Pa12a2 C3 dv112a2 dx1 dv212a2 dx2 0 P 6 12a23 3 2 Pa12a22 C312a2 C4 v2 0 at x2 2a 0 P 12 12a23 C112a2 C2 v1 0 at x1 2a 0 0 0 C2 v1 0 at x1 0 x1 x2 2a v1 v2 0 x1 x2 2a dv1dx1 dv2dx2 x2 2a v2 0 x1 2a x1 0 v1 0 v1 0 EIv2 P 6 x3 2 3 2Pax2 2 C3x2 C4 EIdv2 dx2 P 2 x2 2 3Pax2 C3 EId2v2 dx2 2 Px2 3Pa 314 CHAPTER 8 DEFLECTIONS 8 F84 Determine the equation of the elastic curve for the beam using the x coordinate that is valid for EI is constant 0 6 x 6 L FUNDAMENTAL PROBLEMS F87 Determine the equation of the elastic curve for the beam using the x coordinate that is valid for EI is constant 0 6 x 6 L F85 Determine the equation of the elastic curve for the beam using the x coordinate that is valid for EI is constant 0 6 x 6 L F88 Determine the equation of the elastic curve for the beam using the x coordinate that is valid for EI is constant 0 6 x 6 L F86 Determine the equation of the elastic curve for the beam using the x coordinate that is valid for EI is constant 0 6 x 6 L F89 Determine the equation of the elastic curve for the beam using the x coordinate that is valid for EI is constant 0 6 x 6 L L 2 L 2 x A B P L 2 L 2 x A B M0 A L B M0 x L x P L x w0 F84 F85 F86 F87 F88 F89 w L x 83 THE DOUBLE INTEGRATION METHOD 315 8 82 The bar is supported by a roller constraint at B which allows vertical displacement but resists axial load and moment If the bar is subjected to the loading shown determine the slope at A and the deflection at C EI is constant 83 Determine the deflection at B of the bar in Prob 82 87 Determine the elastic curve for the simply supported beam using the x coordinate Also determine the slope at A and the maximum deflection of the beam EI is constant 0 x L2 84 Determine the equations of the elastic curve using the coordinates and specify the slope and deflection at B EI is constant 85 Determine the equations of the elastic curve using the coordinates and and specify the slope and deflection at point B EI is constant x3 x1 x2 x1 88 Determine the equations of the elastic curve using the coordinates and and specify the slope at C and displacement at B EI is constant 89 Determine the equations of the elastic curve using the coordinates and and specify the slope at B and deflection at C EI is constant x3 x1 x2 x1 A B P P L x2 x1 a a w A B L L x1 x2 C L x A B w0 L A B a w x1 x2 x3 C B A a x1 x3 x2 a w C P A C B L 2 L 2 81 Determine the equations of the elastic curve for the beam using the and coordinates Specify the slope at A and the maximum deflection EI is constant x2 x1 86 Determine the maximum deflection between the supports A and B EI is constant Use the method of integration PROBLEMS Prob 81 Probs 8283 Probs 8485 Prob 87 Prob 86 Probs 8889 316 CHAPTER 8 DEFLECTIONS 8 84 MomentArea Theorems The initial ideas for the two momentarea theorems were developed by Otto Mohr and later stated formally by Charles E Greene in 1873These theorems provide a semigraphical technique for determining the slope of the elastic curve and its deflection due to bending They are particularly advantageous when used to solve problems involving beams especially those subjected to a series of concentrated loadings or having segments with different moments of inertia To develop the theorems reference is made to the beam in Fig 814a If we draw the moment diagram for the beam and then divide it by the flexural rigidity EI the MEI diagram shown in Fig 814b results By Eq 82 Thus it can be seen that the change in the slope of the tangents on either side of the element dx is equal to the lightershaded area under the MEI diagram Integrating from point A on the elastic curve to point B Fig 814c we have 85 This equation forms the basis for the first momentarea theorem Theorem 1The change in slope between any two points on the elastic curve equals the area of the MEI diagram between these two points The notation is referred to as the angle of the tangent at B measured with respect to the tangent at A From the proof it should be evident that this angle is measured counterclockwise from tangent A to tangent B if the area of the MEI diagram is positive Fig 814c Conversely if this area is negative or below the x axis the angle is measured clockwise from tangent A to tangent B Furthermore from the dimensions of Eq 85 is measured in radians uBA uBA uBA uBA L B A M EI dx du du a M EIb dx a A B w x dx M EI M EI A B x dx b x A B elastic curve tan B tan A uBA c Fig 814 84 MOMENTAREA THEOREMS 317 8 The second momentarea theorem is based on the relative deviation of tangents to the elastic curve Shown in Fig 815c is a greatly exaggerated view of the vertical deviation dt of the tangents on each side of the differential element dx This deviation is measured along a vertical line passing through point A Since the slope of the elastic curve and its deflection are assumed to be very small it is satisfactory to approximate the length of each tangent line by x and the arc by dt Using the circulararc formula where r is of length xwe can write Using Eq 82 the vertical deviation of the tangent at A with respect to the tangent at B can be found by integration in which case 86 Recall from statics that the centroid of an area is determined from Since represents an area of the MEI dia gram we can also write 87 Here is the distance from the vertical axis through A to the centroid of the area between A and B Fig 815b The second momentarea theorem can now be stated as follows Theorem 2 The vertical deviation of the tangent at a point A on the elastic curve with respect to the tangent extended from another point B equals the moment of the area under the MEI diagram between the two points A and B This moment is computed about point A the point on the elastic curve where the deviation is to be determined Provided the moment of a positive MEI area from A to B is computed as in Fig 815b it indicates that the tangent at point A is above the tangent to the curve extended from point B Fig 815c Similarly negative MEI areas indicate that the tangent at A is below the tangent extended from B Note that in general is not equal to which is shown in Fig 815d Specifically the moment of the area under the MEI diagram between A and B is computed about point A to determine Fig 815b and it is computed about point B to determine It is important to realize that the momentarea theorems can only be used to determine the angles or deviations between two tangents on the beams elastic curve In general they do not give a direct solution for the slope or displacement at a point on the beamThese unknowns must first be related to the angles or vertical deviations of tangents at points on the elastic curve Usually the tangents at the supports are drawn in this regard since these points do not undergo displacement andor have zero slope Specific cases for establishing these geometric relationships are given in the example problems tBA tAB tBA tAB tAB x tAB x L B A M EI dx 1MEI dx x1dA 1x dA tAB L B A x M EI dx du 1MEI2 dx dt x du s ur ds a A B w x dx A B x b x M EI elastic curve x dx A B tan A ds dt tan B c tAB elastic curve A B tan A tAB tBA tan B d Fig 815 318 CHAPTER 8 DEFLECTIONS 8 Procedure for Analysis The following procedure provides a method that may be used to determine the displacement and slope at a point on the elastic curve of a beam using the momentarea theorems MEI Diagram Determine the support reactions and draw the beams MEI diagram If the beam is loaded with concentrated forces the MEI diagram will consist of a series of straight line segments and the areas and their moments required for the momentarea theorems will be relatively easy to compute If the loading consists of a series of concentrated forces and distributed loads it may be simpler to compute the required MEI areas and their moments by drawing the MEI diagram in parts using the method of superposition as discussed in Sec 45 In any case the MEI diagram will consist of parabolic or perhaps higherorder curves and it is suggested that the table on the inside back cover be used to locate the area and centroid under each curve Elastic Curve Draw an exaggerated view of the beams elastic curve Recall that points of zero slope occur at fixed supports and zero displacement occurs at all fixed pin and roller supports If it becomes difficult to draw the general shape of the elastic curve use the moment or MEI diagram Realize that when the beam is subjected to a positive moment the beam bends concave up whereas negative moment bends the beam concave down Furthermore an inflection point or change in curvature occurs where the moment in the beam or MEI is zero The displacement and slope to be determined should be indicated on the curve Since the momentarea theorems apply only between two tangents attention should be given as to which tangents should be constructed so that the angles or deviations between them will lead to the solution of the problem In this regard the tangents at the points of unknown slope and displacement and at the supports should be considered since the beam usually has zero displacement andor zero slope at the supports MomentArea Theorems Apply Theorem 1 to determine the angle between two tangents and Theorem 2 to determine vertical deviations between these tangents Realize that Theorem 2 in general will not yield the displacement of a point on the elastic curve When applied properly it will only give the vertical distance or deviation of a tangent at point A on the elastic curve from the tangent at B After applying either Theorem 1 or Theorem 2 the algebraic sign of the answer can be verified from the angle or deviation as indicated on the elastic curve 84 MOMENTAREA THEOREMS 321 8 EXAMPLE 88 Determine the slope at point C of the beam in Fig 818a SOLUTION MEI Diagram Fig 818b Elastic Curve Since the loading is applied symmetrically to the beam the elastic curve is symmetric as shown in Fig 818c We are required to find This can easily be done realizing that the tangent at D is horizontal and therefore by the construction the angle between tan C and tan D is equal to that is MomentArea Theorem Using Theorem 1 is equal to the shaded area under the MEI diagram between points C and D We have Thus Ans uC 135 kN m2 20011062 kNm2611062110122 m4 0112 rad 135 kN m2 EI uC uDC 3 ma30 kN m EI b 1 2 13 m2a60 kN m EI 30 kN m EI b uDC uC uDC uC uDC uC I 6106 mm4 E 200 GPa uDC horizontal tan D tan C D C uC c 3 m b C D x M EI 60 EI 30 EI A B 3 m 6 m Fig 818 20 kN 3 m 3 m 6 m C D a A B 8 k 6 ft 12 ft 6 ft C a A B 322 CHAPTER 8 DEFLECTIONS 8 Determine the slope at point C of the beam in Fig 819a SOLUTION MEI Diagram Fig 819b Elastic Curve The elastic curve is shown in Fig 819c We are required to find To do this establish tangents at A B the supports and C and note that is the angle between the tangents at A and C Also the angle in Fig 819c can be found using This equation is valid since is actually very small so that can be approximated by the length of a circular arc defined by a radius of and sweep of Recall that From the geometry of Fig 819c we have 1 MomentArea Theorems Using Theorem 1 is equivalent to the area under the MEI diagram between points A and C that is Applying Theorem 2 is equivalent to the moment of the area under the MEI diagram between B and A about point B since this is the point where the tangential deviation is to be determinedWe have Substituting these results into Eq 1 we have so that Ans 000119 rad uC 144 k ft2 2911032 kin21144 in2ft22 600 in411 ft411224 in42 uC 4320 k ft3 124 ft2 EI 36 k ft2 EI 144 k ft2 EI 4320 k ft3 EI 2 3 16 ft2c1 2 16 ft2a36 k ft EI b d tBA c6 ft 1 3 118 ft2d c1 2 118 ft2a36 k ft EI b d tBA uCA 1 2 16 ft2a12 k ft EI b 36 k ft2 EI uCA uC f uCA tBA 24 uCA s ur f LAB 24 ft tBA tBA f tBALAB f uCA uC I 600 in4 E 29103 ksi EXAMPLE 89 b x M EI 36 EI 12 EI 6 ft 6 ft 12 ft tan B tan C C uC c A uCA tBA B tan A f Fig 819 326 CHAPTER 8 DEFLECTIONS 8 85 ConjugateBeam Method The conjugatebeam method was developed by H MüllerBreslau in 1865 Essentially it requires the same amount of computation as the momentarea theorems to determine a beams slope or deflection however this method relies only on the principles of statics and hence its application will be more familiar The basis for the method comes from the similarity of Eq 41 and Eq 42 to Eq 82 and Eq 84To show this similarity we can write these equations as follows Or integrating Here the shear V compares with the slope the moment M compares with the displacement and the external load w compares with the MEI diagram To make use of this comparison we will now consider a beam having the same length as the real beam but referred to here as the conjugate beam Fig 823 The conjugate beam is loaded with the MEI diagram derived from the load w on the real beam From the above comparisons we can state two theorems related to the conjugate beam namely Theorem 1 The slope at a point in the real beam is numerically equal to the shear at the corresponding point in the conjugate beam Theorem 2 The displacement of a point in the real beam is numerically equal to the moment at the corresponding point in the conjugate beam ConjugateBeam Supports When drawing the conjugate beam it is important that the shear and moment developed at the supports of the conjugate beam account for the corresponding slope and displacement of the real beam at its supports a consequence of Theorems 1 and 2 For v u V L w dx D D u L a M EIb dx 5 M L c L w dxd dx D D v L c L a M EIb dxd dx dV dx w du dx M EI 4 d2M dx2 w d2v dx2 M EI L L A B w real beam A B conjugate beam M EI Fig 823 328 CHAPTER 8 DEFLECTIONS 8 Fig 824 real beam conjugate beam Procedure for Analysis The following procedure provides a method that may be used to determine the displacement and slope at a point on the elastic curve of a beam using the conjugatebeam method Conjugate Beam Draw the conjugate beam for the real beam This beam has the same length as the real beam and has corresponding supports as listed in Table 82 In general if the real support allows a slope the conjugate support must develop a shear and if the real support allows a displacement the conjugate support must develop a moment The conjugate beam is loaded with the real beams MEI diagram This loading is assumed to be distributed over the conjugate beam and is directed upward when MEI is positive and downward when MEI is negative In other words the loading always acts away from the beam Equilibrium Using the equations of equilibrium determine the reactions at the conjugate beams supports Section the conjugate beam at the point where the slope and displacement of the real beam are to be determined At the section show the unknown shear and moment acting in their positive sense Determine the shear and moment using the equations of equilibrium and equal and respectively for the real beam In particular if these values are positive the slope is counterclockwise and the displacement is upward u M V M V u 85 CONJUGATEBEAM METHOD 333 8 e MB VBR 15 ft 75 ft 5 ft 36 EI 225 EI 450 EI B f 15 ft 75 ft 5 ft 225 EI 450 EI MB VBL B 2286 EI 36 EI Equilibrium The external reactions at and are calculated first and the results are indicated in Fig 828d In order to determine the conjugate beam is sectioned just to the right of and the shear force is computed Fig 828eThus Ans The internal moment at yields the displacement of the pinThus Ans The slope can be found from a section of beam just to the left of Fig 828fThus Ans Obviously for this segment is the same as previously calculatedsince the moment arms are only slightly different in Figs 828e and 828f B MB 1uB2L 1VB2L 0 1VB2L 2286 EI 225 EI 450 EI 36 EI 0 cFy 0 B uBL 0381 ft 458 in 2304 k ft3 291103211442 kft23011224 ft4 B MB 2304 k ft3 EI MB 225 EI 152 450 EI 1752 36 EI 1152 0 dMB 0 B 00378 rad 2286 k ft2 291103211442 kft23011224 ft4 1uB2R 1VB2R 2286 k ft2 EI 1VB2R 225 EI 450 EI 36 EI 0 cFy 0 VBR B uBR C B 85 CONJUGATEBEAM METHOD 335 8 812 Determine the slope and displacement at C EI is constant Use the momentarea theorems 813 Solve Prob 812 using the conjugatebeam method 810 Determine the slope at B and the maximum displacement of the beam Use the momentarea theorems Take 811 Solve Prob 810 using the conjugatebeam method E 29103 ksi I 500 in4 814 Determine the value of a so that the slope at A is equal to zero EI is constant Use the momentarea theorems 815 Solve Prob 814 using the conjugatebeam method 816 Determine the value of a so that the displacement at C is equal to zero EI is constant Use the momentarea theorems 817 Solve Prob 816 using the conjugatebeam method PROBLEMS 6 ft 6 ft A B C 15 k Probs 810811 A B C 15 ft 15 k 30 ft Probs 812813 A D P B C P a L 2 L 2 Probs 814815816817 a a a B A C P Probs 818819 818 Determine the slope and the displacement at C EI is constant Use the momentarea theorems 819 Solve Prob 818 using the conjugatebeam method 820 Determine the slope and the displacement at the end C of the beam Use the momentarea theorems 821 Solve Prob 820 using the conjugatebeam method E 200 GPa I 70106 mm4 B D A C 3 m 3 m 8 kN 4 kN 3 m Probs 820821 822 At what distance a should the bearing supports at A and B be placed so that the displacement at the center of the shaft is equal to the deflection at its ends The bearings exert only vertical reactions on the shaft EI is constant Use the momentarea theorems 823 Solve Prob 822 using the conjugatebeam method A B a L P P a Probs 822823 336 CHAPTER 8 DEFLECTIONS 8 829 Determine the force F at the end of the beam C so that the displacement at C is zero EI is constant Use the conjugatebeam method 826 Determine the displacement at C and the slope at B EI is constant Use the momentarea theorems 828 Determine the force F at the end of the beam C so that the displacement at C is zero EI is constant Use the momentarea theorems 824 Determine the displacement at C and the slope at B EI is constant Use the momentarea theorems 825 Solve Prob 824 using the conjugatebeam method A B C 3 m 15 m 15 m 4 kN 4 kN 3 m Probs 824825 A C B P P a a a a 2 P 2 Prob 826 A C B P P a a a a 2 P 2 Prob 827 a a a B D A C P F Prob 828 a a a B D A C P F Prob 829 a a a A C P P B Prob 830 830 Determine the slope at B and the displacement at C EI is constant Use the momentarea theorems a a a A C P P B Prob 831 831 Determine the slope at B and the displacement at C EI is constant Use the conjugatebeam method 827 Determine the displacement at C and the slope at B EI is constant Use the conjugatebeam method 338 CHAPTER 8 DEFLECTIONS 8 The deflection of a member or structure can always be established provided the moment diagram is known because positive moment will tend to bend the member concave upwards and negative moment will tend to bend the member concave downwards Likewise the general shape of the moment diagram can be determined if the deflection curve is known CHAPTER REVIEW Deflection of a beam due to bending can be determined by using double integration of the equation Here the internal moment M must be expressed as a function of the x coordinates that extend across the beam The constants of integration are obtained from the boundary conditions such as zero deflection at a pin or roller support and zero deflection and slope at a fixed support If several x coordinates are necessary then the continuity of slope and deflection must be considered where at and v1a v2a u1a u2a x1 x2 a d2v dx2 M EI beam P1 P2 P x1 x2 v1v2 a b v u M x moment diagram inflection point deflection curve The displacement at the ends of this bridge deck as it is being constructed can be determined using energy methods 9 341 In this chapter we will show how to apply energy methods to solve problems involving slope and deflection The chapter begins with a discussion of work and strain energy followed by a development of the principle of work and energy The method of virtual work and Castiglianos theorem are then developed and these methods are used to determine the displacements at points on trusses beams and frames 91 External Work and Strain Energy The semigraphical methods presented in the previous chapters are very effective for finding the displacements and slopes at points in beams subjected to rather simple loadings For more complicated loadings or for structures such as trusses and frames it is suggested that energy methods be used for the computations Most energy methods are based on the conservation of energy principle which states that the work done by all the external forces acting on a structure is transformed into internal work or strain energy which is developed when the structure deforms If the materials elastic limit is not exceeded the elastic strain energy will return the structure to its undeformed state when the loads are removed The conservation of energy principle can be stated mathematically as 91 Before developing any of the energy methods based on this principle howeverwe will first determine the external work and strain energy caused by a force and a moment The formulations to be presented will provide a basis for understanding the work and energy methods that follow Ue Ui Ui Ue Deflections Using Energy Methods 92 PRINCIPLE OF WORK AND ENERGY 345 9 92 Principle of Work and Energy Now that the work and strain energy for a force and a moment have been formulated we will illustrate how the conservation of energy or the principle of work and energy can be applied to determine the displacement at a point on a structure To do this consider finding the displacement at the point where the force P is applied to the cantilever beam in Fig 95 From Eq 93 the external work is To obtain the resulting strain energy we must first determine the internal moment as a function of position x in the beam and then apply Eq 911 In this case so that Equating the external work to internal strain energy and solving for the unknown displacement we have Although the solution here is quite direct application of this method is limited to only a few select problems It will be noted that only one load may be applied to the structure since if more than one load were applied there would be an unknown displacement under each load and yet it is possible to write only one work equation for the beam Furthermore only the displacement under the force can be obtained since the external work depends upon both the force and its corresponding displacement One way to circumvent these limitations is to use the method of virtual work or Castiglianos theorem both of which are explained in the following sections PL3 3EI 1 2 P 1 6 P2L3 EI Ue Ui Ui L L 0 M2 dx 2EI L L 0 1Px22 dx 2EI 1 6 P2L3 EI M Px Ue 1 2 P P L P V x M Fig 95 93 PRINCIPLE OF VIRTUAL WORK 347 9 Once the virtual loadings are applied then the body is subjected to the real loads and Fig 96b Point A will be displaced an amount causing the element to deform an amount dLAs a result the external virtual force and internal virtual load u ride along by and dL respectively and therefore perform external virtual work of on the body and internal virtual work of on the element Realizing that the external virtual work is equal to the internal virtual work done on all the elements of the body we can write the virtualwork equation as virtual loadings real displacements 913 where virtual unit load acting in the direction of virtual load acting on the element in the direction of dL displacement caused by the real loads deformation of the element caused by the real loads By choosing it can be seen that the solution for follows directly since In a similar mannerif the rotational displacement or slope of the tangent at a point on a structure is to be determined a virtual couple moment having a unit magnitude is applied at the point As a consequence this couple moment causes a virtual load in one of the elements of the body Assuming that the real loads deform the element an amount dL the rotation can be found from the virtualwork equation virtual loadings real displacements 914 where virtual unit couple moment acting in the direction of virtual load acting on an element in the direction of dL rotational displacement or slope in radians caused by the real loads deformation of the element caused by the real loads This method for applying the principle of virtual work is often referred to as the method of virtual forces since a virtual force is applied resulting in the calculation of a real displacement The equation of virtual work in this case represents a compatibility requirement for the structure Although not important here realize that we can also apply the principle dL internal u external uu internal u M 1 external 1 u uu dL u uU M u dL P 1 dL internal external u internal P 1 external 1 u dL u dL 1 P P3 P2 P1 g g g g g g g g 94 METHOD OF VIRTUAL WORK TRUSSES 349 9 Temperature In some cases truss members may change their length due to temperature If is the coefficient of thermal expansion for a member and is the change in its temperature the change in length of a member is Hence we can determine the displacement of a selected truss joint due to this temperature change from Eq 913 written as 916 where external virtual unit load acting on the truss joint in the stated direction of internal virtual normal force in a truss member caused by the external virtual unit load external joint displacement caused by the temperature change coefficient of thermal expansion of member change in temperature of member length of member Fabrication Errors and Camber Occasionallyerrors in fabricating the lengths of the members of a truss may occurAlso in some cases truss members must be made slightly longer or shorter in order to give the truss a camber Camber is often built into a bridge truss so that the bottom cord will curve upward by an amount equivalent to the downward deflection of the cord when subjected to the bridges full dead weight If a truss member is shorter or longer than intended the displacement of a truss joint from its expected position can be determined from direct application of Eq 913 written as 917 where external virtual unit load acting on the truss joint in the stated direction of internal virtual normal force in a truss member caused by the external virtual unit load external joint displacement caused by the fabrication errors difference in length of the member from its intended size as caused by a fabrication error A combination of the right sides of Eqs 915 through 917 will be necessary if both external loads act on the truss and some of the members undergo a thermal change or have been fabricated with the wrong dimensions L n 1 1 n L L T a n 1 1 na T L L a T L T a 350 CHAPTER 9 DEFLECTIONS USING ENERGY METHODS 9 Procedure for Analysis The following procedure may be used to determine a specific displacement of any joint on a truss using the method of virtual work Virtual Forces n Place the unit load on the truss at the joint where the desired displacement is to be determined The load should be in the same direction as the specified displacement eg horizontal or vertical With the unit load so placed and all the real loads removed from the truss use the method of joints or the method of sections and calculate the internal n force in each truss member Assume that tensile forces are positive and compressive forces are negative Real Forces N Use the method of sections or the method of joints to determine the N force in each member These forces are caused only by the real loads acting on the trussAgain assume tensile forces are positive and compressive forces are negative VirtualWork Equation Apply the equation of virtual work to determine the desired displacement It is important to retain the algebraic sign for each of the corresponding n and N forces when substituting these terms into the equation If the resultant sum is positive the displacement is in the same direction as the unit load If a negative value results is opposite to the unit load When applying realize that if any of the members undergoes an increase in temperature will be positive whereas a decrease in temperature results in a negative value for For when a fabrication error increases the length of a member is positive whereas a decrease in length is negative When applying any formula attention should be paid to the units of each numerical quantity In particular the virtual unit load can be assigned any arbitrary unit lb kip N etc since the n forces will have these same units and as a result the units for both the virtual unit load and the n forces will cancel from both sides of the equation L 1 n L T T 1 na TL nNLAE 95 CASTIGLIANOS THEOREM 355 9 95 Castiglianos Theorem In 1879 Alberto Castigliano an Italian railroad engineer published a book in which he outlined a method for determining the deflection or slope at a point in a structure be it a truss beam or frame This method which is referred to as Castiglianos second theorem or the method of least work applies only to structures that have constant temperature unyielding supports and linear elastic material response If the displacement of a point is to be determined the theorem states that it is equal to the first partial derivative of the strain energy in the structure with respect to a force acting at the point and in the direction of displacement In a similar manner the slope at a point in a structure is equal to the first partial derivative of the strain energy in the structure with respect to a couple moment acting at the point and in the direction of rotation To derive Castiglianos second theorem consider a body structure of any arbitrary shape which is subjected to a series of n forces Since the external work done by these loads is equal to the internal strain energy stored in the body we can write The external work is a function of the external loads Thus Now if any one of the forces say is increased by a differential amount the internal work is also increased such that the new strain energy becomes 918 This value however should not depend on the sequence in which the n forces are applied to the body For example if we apply to the body first then this will cause the body to be displaced a differential amount in the direction of By Eq 93 the increment of strain energy would be This quantity however is a secondorder differential and may be neglected Further application of the loads which displace the body yields the strain energy 919 Here as before is the internal strain energy in the body caused by the loads and is the additional strain energy caused by Eq 94 In summary then Eq 918 represents the strain energy in the body determined by first applying the loads then and Eq 919 represents the strain energy determined by first applying dPi and dPi P2 Á Pn P1 Ue P dPi dUi dPii P1 P2 Á Pn Ui Ui dUi Ui dPii 1 2 Á n Á Pn P2 P1 1 2 dPi di AUe 1 2 PB dPi di dPi Ui dUi Ui 0Ui 0Pi dPi dPi Pi Ui Ue f1P1 P2 Á Pn2 1Ue 1P dx2 Ui Ue P1 P2 Á Pn 356 CHAPTER 9 DEFLECTIONS USING ENERGY METHODS 9 then the loads Since these two equations must be equal we require 920 which proves the theorem ie the displacement in the direction of is equal to the first partial derivative of the strain energy with respect to It should be noted that Eq 920 is a statement regarding the structures compatibility Also the above derivation requires that only conservative forces be considered for the analysis These forces do work that is independent of the path and therefore create no energy loss Since forces causing a linear elastic response are conservative the theorem is restricted to linear elastic behavior of the material This is unlike the method of virtual force discussed in the previous section which applied to both elastic and inelastic behavior 96 Castiglianos Theorem for Trusses The strain energy for a member of a truss is given by Eq 99 Substituting this equation into Eq 920 and omitting the subscript i we have It is generally easier to perform the differentiation prior to summation In the general case L A and E are constant for a given member and therefore we may write 921 where external joint displacement of the truss external force applied to the truss joint in the direction of internal force in a member caused by both the force P and the loads on the truss length of a member crosssectional area of a member E modulus of elasticity of a member A L N P aNa0N 0P b L AE 0 0P a N2L 2AE Ui N2L2AE Pi Pi i i 0Ui 0Pi P2 Á Pn P1 Castiglianos first theorem is similar to his second theorem however it relates the load to the partial derivative of the strain energy with respect to the corresponding displacement that is The proof is similar to that given above and like the method of virtual displacementCastiglianos first theorem applies to both elastic and inelastic material behavior This theorem is another way of expressing the equilibrium requirements for a structure and since it has very limited use in structural analysis it will not be discussed in this book Pi 0Ui0i Pi 96 CASTIGLIANOS THEOREM FOR TRUSSES 357 9 This equation is similar to that used for the method of virtual work Eq 915 except n is replaced by Notice that in order to determine this partial derivative it will be necessary to treat P as a variable not a specific numerical quantity and furthermore each member force N must be expressed as a function of P As a result computing generally requires slightly more calculation than that required to compute each n force directly These terms will of course be the same since n or is simply the change of the internal member force with respect to the load P or the change in member force per unit load 0N0P 0N0P 0N0P 11 nNLAE2 Procedure for Analysis The following procedure provides a method that may be used to determine the displacement of any joint of a truss using Castiglianos theorem External Force P Place a force P on the truss at the joint where the desired displacement is to be determinedThis force is assumed to have a variable magnitude in order to obtain the change Be sure P is directed along the line of action of the displacement Internal Forces N Determine the force N in each member caused by both the real numerical loads and the variable force PAssume tensile forces are positive and compressive forces are negative Compute the respective partial derivative for each member After N and have been determined assign P its numerical value if it has replaced a real force on the truss Otherwise set P equal to zero Castiglianos Theorem Apply Castiglianos theorem to determine the desired displacement It is important to retain the algebraic signs for corresponding values of N and when substituting these terms into the equation If the resultant sum is positive is in the same direction as P If a negative value results is opposite to P N10N0P2LAE 0N0P 0N0P 0N 0P 0N0P 97 METHOD OF VIRTUAL WORK BEAMS AND FRAMES 365 9 Fig 916 When applying Eqs 922 and 923 it is important to realize that the definite integrals on the right side actually represent the amount of virtual strain energy that is stored in the beam If concentrated forces or couple moments act on the beam or the distributed load is discontinuous a single integration cannot be performed across the beams entire length Instead separate x coordinates will have to be chosen within regions that have no discontinuity of loadingAlso it is not necessary that each x have the same origin however the x selected for determining the real moment M in a particular region must be the same x as that selected for determining the virtual moment m or within the same region For example consider the beam shown in Fig 916 In order to determine the displacement of D four regions of the beam must be considered and therefore four integrals having the form must be evalu ated We can use to determine the strain energy in region AB for region BC for region DE and for region DC In any case each x coordinate should be selected so that both M and m or can be easily formulated Integration Using Tables When the structure is subjected to a relatively simple loading and yet the solution for a displacement requires several integrations a tabular method may be used to perform these integrations To do so the moment diagrams for each member are drawn first for both the real and virtual loadings By matching these diagrams for m and M with those given in the table on the inside front cover the integral can be determined from the appropriate formula Examples 98 and 910 illustrate the application of this method 1mM dx mu x4 x3 x2 x1 11mMEI2 dx mu x1 Apply virtual unit load a A x2 x3 x4 B C D E 1 x1 Apply real loads b A w x2 x3 x4 B C D E P 366 CHAPTER 9 DEFLECTIONS USING ENERGY METHODS 9 Procedure for Analysis The following procedure may be used to determine the displacement andor the slope at a point on the elastic curve of a beam or frame using the method of virtual work Virtual Moments m or mu Place a unit load on the beam or frame at the point and in the direction of the desired displacement If the slope is to be determined place a unit couple moment at the point Establish appropriate x coordinates that are valid within regions of the beam or frame where there is no discontinuity of real or virtual load With the virtual load in place and all the real loads removed from the beam or frame calculate the internal moment m or as a function of each x coordinate Assume m or acts in the conventional positive direction as indicated in Fig 41 Real Moments Using the same x coordinates as those established for m or determine the internal moments M caused only by the real loads Since m or was assumed to act in the conventional positive direction it is important that positive M acts in this same direction This is necessary since positive or negative internal work depends upon the directional sense of load defined by or and displacement defined by VirtualWork Equation Apply the equation of virtual work to determine the desired displacement or rotation It is important to retain the algebraic sign of each integral calculated within its specified region If the algebraic sum of all the integrals for the entire beam or frame is positive or is in the same direction as the virtual unit load or unit couple moment respectively If a negative value results the direction of or is opposite to that of the unit load or unit couple moment u u u M dxEI mu m mu mu mu mu 376 CHAPTER 9 DEFLECTIONS USING ENERGY METHODS 9 Torsion Often threedimensional frameworks are subjected to torsional loadings If the member has a circular crosssectional area no warping of its cross section will occur when it is loaded As a result the virtual strain energy in the member can easily be derivedTo do so consider an element dx of the member that is subjected to an applied torque T Fig 923 This torque causes a shear strain of Provided linear elastic material response occurs then where Thus the angle of twist If a virtual unit load is applied to the structure that causes an internal virtual torque t in the member then after applying the real loads the virtual strain energy in the member of length dx will be Integrating over the length L of the member yields 926 where internal virtual torque caused by the external virtual unit load internal torque in the member caused by the real loads shear modulus of elasticity for the material polar moment of inertia for the cross section where c is the radius of the crosssectional area members length The virtual strain energy due to torsion for members having noncircular crosssectional areas is determined using a more rigorous analysis than that presented here Temperature In Sec 94 we considered the effect of a uniform temperature change on a truss member and indicated that the member will elongate or shorten by an amount In some cases however a structural member can be subjected to a temperature difference across its depth as in the case of the beam shown in Fig 924a If this occurs it is possible to determine the displacement of points along the elastic curve of the beam by using the principle of virtual workTo do so we must first compute the amount of rotation of a differential element dx of the beam as caused by the thermal gradient that acts over the beams cross section For the sake of discussion we will choose the most common case of a beam having a neutral axis located at the middepth c of the beam If we plot the temperature profile Fig 924b it will be noted that the mean temperature is If the temperature difference at the top of the element causes strain elongation while that at the bottom causes strain contraction In both cases the difference in temperature is Tm T1 Tm Tm T2 T1 7 T2 Tm 1T1 T222 L a TL T L J pc42 J G T t Ut tTL GJ dUt t du tT dxGJ du 1g dx2c 1tGc2 dx 1TGJ2 dx t TcJ g tG g 1cdu2dx dx du g T T c Fig 923 98 VIRTUAL STRAIN ENERGY CAUSED BY AXIAL LOAD SHEAR TORSION AND TEMPERATURE 379 9 Bending The virtual strain energy due to bending has been deter mined in Example 910There it was shown that Axial load From the data in Fig 925b and 925c we have Shear Applying Eq 925 with for rectangular cross sections and using the shear functions shown in Fig 925b and 925c we have Applying the equation of virtual work we have Ans Including the effects of shear and axial load contributed only a 06 increase in the answer to that determined only from bending Ch 137 in 1 k Ch 1357 in k 0001616 in k 000675 in k 540 k2 ft112 inft2 1211032 kin2180 in22 000675 in k L 10 0 12112140 4x12 dx1 GA L 8 0 12112521252 dx2 GA Us L L 0 Ka vV GAb dx K 12 0001616 in k 125 k125 k21120 in2 80 in22911032 kin2 1 k102196 in2 80 in22911032 kin2 Ua a nNL AE Ub L L 0 mM dx EI 13 6667 k2 ft3 EI 13 6667 k2 ft3 1123 in31 ft32 2911032 kin21600 in42 1357 in k 99 CASTIGLIANOS THEOREM FOR BEAMS AND FRAMES 381 9 99 Castiglianos Theorem for Beams and Frames The internal bending strain energy for a beam or frame is given by Eq 911 Substituting this equation into Eq 920 and omitting the subscript i we have Rather than squaring the expression for internal moment M integrating and then taking the partial derivative it is generally easier to differentiate prior to integration Provided E and I are constant we have 928 where external displacement of the point caused by the real loads acting on the beam or frame external force applied to the beam or frame in the direction of internal moment in the beam or frame expressed as a function of x and caused by both the force P and the real loads on the beam modulus of elasticity of beam material moment of inertia of crosssectional area computed about the neutral axis If the slope at a point is to be determined we must find the partial derivative of the internal moment M with respect to an external couple moment acting at the point ie 929 The above equations are similar to those used for the method of virtual work Eqs 922 and 923 except and replace m and respectively As in the case for trusses slightly more calculation is generally required to determine the partial derivatives and apply Castiglianos theorem rather than use the method of virtual work Also recall that this theorem applies only to material having a linear elastic response If a more complete accountability of strain energy in the structure is desired the strain energy due to shear axial force and torsion must be included The derivations for shear and torsion follow the same development as Eqs 925 and 926 The strain energies and their derivatives are respectively mu 0M0M 0M0P u L L 0 Ma 0M 0M b dx EI M u I E M P L L 0 Ma0M 0P b dx EI 0 0P L L 0 M2 dx 2EI 1i 0Ui0Pi2 1Ui 1M2 dx2EI2 382 CHAPTER 9 DEFLECTIONS USING ENERGY METHODS 9 Procedure for Analysis The following procedure provides a method that may be used to determine the deflection andor slope at a point in a beam or frame using Castiglianos theorem External Force P or Couple Moment M Place a force P on the beam or frame at the point and in the direction of the desired displacement If the slope is to be determined place a couple moment at the point It is assumed that both P and have a variable magnitude in order to obtain the changes or Internal Moments M Establish appropriate x coordinates that are valid within regions of the beam or frame where there is no discontinuity of force distributed load or couple moment Calculate the internal moment M as a function of P or and each x coordinate Also compute the partial derivative or for each coordinate x After M and or have been determined assign P or its numerical value if it has replaced a real force or couple moment Otherwise set P or equal to zero Castiglianos Theorem Apply Eq 928 or 929 to determine the desired displacement or slope It is important to retain the algebraic signs for corresponding values of M and or If the resultant sum of all the definite integrals is positive or is in the same direction as P or M u 0M0M 0M0P u M M 0M0M 0M0P 0M0M 0M0P M 0M0M 0M0P M M These effects however will not be included in the analysis of the problems in this text since beam and frame deflections are caused mainly by bending strain energy Larger frames or those with unusual geometry can be analyzed by computer where these effects can readily be incorporated into the analysis Ut L L 0 T2 dx 2JG 0Ut 0P L L 0 T JG a 0T 0Pb dx Us K L L 0 V2 dx 2AG 0Us 0P L L 0 V AG a0V 0P b dx EXAMPLE 914 Determine the displacement of point B of the beam shown in Fig 927aTake I 50011062 mm4 E 200 GPa 99 CASTIGLIANOS THEOREM FOR BEAMS AND FRAMES 383 9 Fig 927 12 kNm 10 m b P x 12 kNm 10 m B A a P x x 12 x 2 M V c SOLUTION External Force P A vertical force P is placed on the beam at B as shown in Fig 927b Internal Moments M A single x coordinate is needed for the solution since there are no discontinuities of loading between A and B Using the method of sections Fig 927c we have Setting its actual value yields Castiglianos Theorem Applying Eq 928 we have or Ans The similarity between this solution and that of the virtualwork method Example 97 should be noted 0150 m 150 mm B 1511032 kN m3 20011062 kNm250011062 mm411012 m4mm42 1511032 kN m3 EI B L L 0 Ma0M 0P b dx EI L 10 0 16x221x2 dx EI M 6x2 0M 0P x P 0 M 6x2 Px 0M 0P x M 112x2ax 2 b Px 0 d M 0 384 CHAPTER 9 DEFLECTIONS USING ENERGY METHODS 9 3 kN x1 B A b C x2 M Determine the slope at point B of the beam shown in Fig 928aTake SOLUTION External Couple Moment M Since the slope at point B is to be determined an external couple is placed on the beam at this point Fig 928b Internal Moments M Two coordinates and must be used to determine the internal moments within the beam since there is a discontinuity at B As shown in Fig 928b ranges from A to B and ranges from B to C Using the method of sections Fig 928c the internal moments and the partial derivatives are computed as follows For For Castiglianos Theorem Setting its actual value and apply ing Eq 929 we have or Ans The negative sign indicates that is opposite to the direction of the couple moment Note the similarity between this solution and that of Example 98 M uB 000938 rad uB 1125 kN m2 20011062 kNm26011062 mm411012 m4mm42 L 5 0 13x12102 dx1 EI L 5 0 315 x22112 dx2 EI 1125 kN m2 EI uB L L 0 Ma 0M 0M b dx EI M 0 0M2 0M 1 M2 M 315 x22 M2 M 315 x22 0 d M 0 x2 0M1 0M 0 M1 3x1 M1 3x1 0 d M 0 x1 x2 x1 M x2 x1 M œ I 6011062 mm4 E 200 GPa EXAMPLE 915 3 kN x1 M1 V1 x2 5 m 3 kN M M2 V2 c Fig 928 3 kN 5 m B A a 5 m C 99 CASTIGLIANOS THEOREM FOR BEAMS AND FRAMES 389 9 935 Determine the slope and displacement at point B Assume the support at A is a pin and C is a roller Take Use the method of virtual work 936 Solve Prob 935 using Castiglianos theorem I 300 in4 E 29103 ksi 940 Determine the slope and displacement at point A Assume C is pinned Use the principle of virtual work EI is constant 941 Solve Prob 940 using Castiglianos theorem 938 Determine the displacement of point C Use the method of virtual work EI is constant 939 Solve Prob 938 using Castiglianos theorem 942 Determine the displacement at point D Use the principle of virtual work EI is constant B A C 4 kft 10 ft 5 ft B C 6 kNm 3 m 3 m A 8 k 4 ft 4 ft 3 kft B A C 4 ft 4 ft D B A C w0 L 2 L 2 Prob 937 B A C 4 kft 10 ft 5 ft Probs 935936 Probs 938939 Probs 940941 Prob 943 8 k 4 ft 4 ft 3 kft B A C 4 ft 4 ft D Prob 942 937 Determine the slope and displacement at point B Assume the support at A is a pin and C is a roller Account for the additional strain energy due to shear Take and assume AB has a crosssectional area of Use the method of virtual work A 750 in2 I 300 in4 G 12103 ksi E 29103 ksi 943 Determine the displacement at point D Use Castiglianos theorem EI is constant The fixedconnected joints of this concrete framework make this a statically indeterminate structure 10 395 In this chapter we will apply the force or flexibility method to analyze statically indeterminate trusses beams and frames At the end of the chapter we will present a method for drawing the influence line for a statically indeterminate beam or frame 101 Statically Indeterminate Structures Recall from Sec 24 that a structure of any type is classified as statically indeterminate when the number of unknown reactions or internal forces exceeds the number of equilibrium equations available for its analysis In this section we will discuss the merits of using indeterminate structures and two fundamental ways in which they may be analyzed Realize that most of the structures designed today are statically indeterminate This indeterminacy may arise as a result of added supports or members or by the general form of the structure For example reinforced concrete buildings are almost always statically indeterminate since the columns and beams are poured as continuous members through the joints and over supports Analysis of Statically Indeterminate Structures by the Force Method Advantages and Disadvantages Although the analysis of a statically indeterminate structure is more involved than that of a statically determinate one there are usually several very important reasons for choosing this type of structure for design Most important for a given loading the maximum stress and deflection of an indeterminate structure are generally smaller than those of its statically determinate counterpart For example the statically indeterminate fixedsupported beam in Fig 101a will be subjected to a maximum moment of whereas the same beam when simply supported Fig 101b will be subjected to twice the moment that is As a result the fixedsupported beam has one fourth the deflection and one half the stress at its center of the one that is simply supported Another important reason for selecting a statically indeterminate structure is because it has a tendency to redistribute its load to its redundant supports in cases where faulty design or overloading occurs In these cases the structure maintains its stability and collapse is prevented This is particularly important when sudden lateral loadssuch as wind or earthquake are imposed on the structure To illustrate consider again the fixedend beam in Fig 101aAs P is increased the beams material at the walls and at the center of the beam begins to yield and forms localized plastic hinges which causes the beam to deflect as if it were hinged or pin connected at these pointsAlthough the deflection becomes large the walls will develop horizontal force and moment reactions that will hold the beam and thus prevent it from totally collapsing In the case of the simply supported beam Fig 101b an excessive load P will cause the plastic hinge to form only at the center of the beam and due to the large vertical deflection the supports will not develop the horizontal force and moment reactions that may be necessary to prevent total collapse Although statically indeterminate structures can support a loading with thinner members and with increased stability compared to their statically determinate counterparts there are cases when these advantages may instead become disadvantages The cost savings in material must be compared with the added cost necessary to fabricate the structure since oftentimes it becomes more costly to construct the supports and joints of an indeterminate structure compared to one that is determinate More important though because statically indeterminate structures have redundant support reactions one has to be very careful to prevent differential displacement of the supports since this effect will introduce internal stress in the structure For example if the wall at one end of the fixedend beam in Fig 101a were to settle stress would be developed in the beam because of this forced deformation On the other hand if the beam were simply supported or statically determinate Fig 101b then any settlement of its end would not cause the beam to deform and therefore no stress would be developed in the beam In general then any deformation such as that caused by relative support displacement or changes in member lengths caused by temperature or fabrication errors will introduce additional stresses in the structurewhich must be considered when designing indeterminate structures Mmax PL4 Mmax PL8 396 CHAPTER 10 ANALYSIS OF STATICALLY INDETERMINATE STRUCTURES BY THE FORCE METHOD 10 101 STATICALLY INDETERMINATE STRUCTURES 397 10 Methods of Analysis When analyzing any indeterminate structure it is necessary to satisfy equilibrium compatibility and forcedisplacement requirements for the structure Equilibrium is satisfied when the reactive forces hold the structure at rest and compatibility is satisfied when the various segments of the structure fit together without intentional breaks or overlaps The forcedisplacement requirements depend upon the way the material responds in this text we have assumed linear elastic response In general there are two different ways to satisfy these requirements when analyzing a statically indeterminate structurethe force or flexibility method and the displacement or stiffness method Force Method The force method was originally developed by James Clerk Maxwell in 1864 and later refined by Otto Mohr and Heinrich MüllerBreslau This method was one of the first available for the analysis of statically indeterminate structures Since compatibility forms the basis for this method it has sometimes been referred to as the compatibility method or the method of consistent displacements This method consists of writing equations that satisfy the compatibility and forcedisplacement requirements for the structure in order to determine the redundant forces Once these forces have been determined the remaining reactive forces on the structure are determined by satisfying the equilibrium requirements The fundamental principles involved in applying this method are easy to understand and develop and they will be discussed in this chapter Displacement Method The displacement method of analysis is based on first writing forcedisplacement relations for the members and then satisfying the equilibrium requirements for the structure In this case the unknowns in the equations are displacementsOnce the displacements are obtained the forces are determined from the compatibility and force displacement equations We will study some of the classical techniques used to apply the displacement method in Chapters 11 and 12 Since almost all present day computer software for structural analysis is developed using this method we will present a matrix formulation of the displacement method in Chapters 14 15 and 16 Each of these two methods of analysis which are outlined in Fig 102 has particular advantages and disadvantages depending upon the geometry of the structure and its degree of indeterminacy A discussion of the usefulness of each method will be given after each has been presented Fig 101 L 2 L 2 P b L 2 L 2 P a 102 FORCE METHOD OF ANALYSIS GENERAL PROCEDURE 401 10 Procedure for Analysis The following procedure provides a general method for determining the reactions or internal loadings of statically indeterminate structures using the force or flexibility method of analysis Principle of Superposition Determine the number of degrees n to which the structure is indeterminate Then specify the n unknown redundant forces or moments that must be removed from the structure in order to make it statically determinate and stable Using the principle of superposition draw the statically indeterminate structure and show it to be equal to a series of corresponding statically determinate structures The primary structure supports the same external loads as the statically indeterminate structureand each of the other structures added to the primary structure shows the structure loaded with a separate redundant force or moment Also sketch the elastic curve on each structure and indicate symbolically the displacement or rotation at the point of each redundant force or moment Compatibility Equations Write a compatibility equation for the displacement or rotation at each point where there is a redundant force or moment These equations should be expressed in terms of the unknown redundants and their corresponding flexibility coefficients obtained from unit loads or unit couple moments that are collinear with the redundant forces or moments Determine all the deflections and flexibility coefficients using the table on the inside front cover or the methods of Chapter 8 or 9 Substitute these loaddisplacement relations into the compatibility equations and solve for the unknown redundants In particular if a numerical value for a redundant is negative it indicates the redundant acts opposite to its corresponding unit force or unit couple moment Equilibrium Equations Draw a freebody diagram of the structureSince the redundant forces andor moments have been calculatedthe remaining unknown reactions can be determined from the equations of equilibrium It should be realized that once all the support reactions have been obtained the shear and moment diagrams can then be drawn and the deflection at any point on the structure can be determined using the same methods outlined previously for statically determinate structures It is suggested that if the MEI diagram for a beam consists of simple segments the momentarea theorems or the conjugatebeam method be used Beams with complicated MEI diagrams that is those with many curved segments parabolic cubic etc can be readily analyzed using the method of virtual work or by Castiglianos second theorem 103 Maxwells Theorem of Reciprocal Displacements Bettis Law When Maxwell developed the force method of analysis he also published a theorem that relates the flexibility coefficients of any two points on an elastic structurebe it a truss a beam or a frame This theorem is referred to as the theorem of reciprocal displacements and may be stated as follows The displacement of a point B on a structure due to a unit load acting at point A is equal to the displacement of point A when the unit load is acting at point B that is Proof of this theorem is easily demonstrated using the principle of virtual work For example consider the beam in Fig 106 When a real unit load acts at A assume that the internal moments in the beam are represented by To determine the flexibility coefficient at B that is a virtual unit load is placed at B Fig 107 and the internal moments are computedThen applying Eq 918 yields Likewise if the flexibility coefficient is to be determined when a real unit load acts at B Fig 107 then represents the internal moments in the beam due to a real unit load Furthermore represents the internal moments due to a virtual unit load at A Fig 106 Hence fAB L mA mB EI dx mA mB fAB fBA L mB mA EI dx mB fBA mA fBA fAB 402 CHAPTER 10 ANALYSIS OF STATICALLY INDETERMINATE STRUCTURES BY THE FORCE METHOD 10 B A 1 fBA A fAB 1 B Fig 106 Fig 107 104 FORCE METHOD OF ANALYSIS BEAMS 403 10 Both integrals obviously give the same result which proves the theorem The theorem also applies for reciprocal rotations and may be stated as follows The rotation at point B on a structure due to a unit couple moment acting at point A is equal to the rotation at point A when the unit couple moment is acting at point B Furthermoreusing a unit force and unit couple moment applied at separate points on the structure we may also state The rotation in radians at point B on a structure due to a unit load acting at point A is equal to the displacement at point A when a unit couple moment is acting at point B As a consequence of this theorem some work can be saved when applying the force method to problems that are statically indeterminate to the second degree or higher For example only one of the two flexibility coefficients or has to be calculated in Eqs 101 since Furthermore the theorem of reciprocal displacements has applications in structural model analysis and for constructing influence lines using the MüllerBreslau principle see Sec 1010 When the theorem of reciprocal displacements is formalized in a more general sense it is referred to as Bettis law Briefly stated The virtual work done by a system of forces that undergo a displacement caused by a system of forces is equal to the virtual work caused by the forces when the structure deforms due to the system of forces In other words The proof of this statement is similar to that given above for the reciprocaldisplacement theorem 104 Force Method of Analysis Beams The force method applied to beams was outlined in Sec 102 Using the procedure for analysis also given in Sec 102 we will now present several examples that illustrate the application of this technique dUAB dUBA PB PA dUBA PA PB dUAB fBC fCB fCB fBC These bridge girders are statically indeterminate since they are continuous over their piers 105 FORCE METHOD OF ANALYSIS FRAMES 411 10 105 Force Method of Analysis Frames The force method is very useful for solving problems involving statically indeterminate frames that have a single story and unusual geometry such as gabled frames Problems involving multistory frames or those with a high degree of indeterminacy are best solved using the slopedeflection momentdistribution or the stiffness method discussed in later chapters The following examples illustrate the application of the force method using the procedure for analysis outlined in Sec 102 EXAMPLE 105 The frame or bent shown in the photo is used to support the bridge deck Assuming EI is constant a drawing of it along with the dimensions and loading is shown in Fig 1012a Determine the support reactions 10 m 5 m 5 m 5 m 40 kNm a A B Fig 1012 SOLUTION Principle of Superposition By inspection the frame is statically indeterminate to the first degree We will choose the horizontal reaction at A to be the redundant Consequently the pin at A is replaced by a rocker since a rocker will not constrain A in the horizontal direction The principle of superposition applied to the idealized model of the frame is shown in Fig 1012b Notice how the frame deflects in each case 105 FORCE METHOD OF ANALYSIS FRAMES 413 10 1571 kN 1571 kN 5 m 5 m e 40 kNm 200 kN 200 kN 10 m 5 m For we require application of a real unit load and a virtual unit load acting at A Fig 1012dThus Substituting the results into Eq 1 and solving yields Ans Equilibrium Equations Using this result the reactions on the idealized model of the frame are shown in Fig 1012e Ax 157 kN 0 91 6667 EI Axa 58333 EI b 58333 EI fAA L L 0 mm EI dx 2 L 5 0 11x122dx1 EI 2 L 5 0 1522dx2 2 L 5 0 1522dx3 fAA 416 CHAPTER 10 ANALYSIS OF STATICALLY INDETERMINATE STRUCTURES BY THE FORCE METHOD 10 F106 F102 F103 F104 F105 F101 FUNDAMENTAL PROBLEMS F101 Determine the reactions at the fixed support at A and the roller at B EI is constant F104 Determine the reactions at the pin at A and the rollers at B and C F102 Determine the reactions at the fixed supports at A and the roller at B EI is constant F103 Determine the reactions at the fixed support at A and the roller at B Support B settles 5 mm Take and I 30011062 mm4 E 200 GPa F105 Determine the reactions at the pin A and the rollers at B and C on the beam EI is constant F106 Determine the reactions at the pin at A and the rollers at B and C on the beam Support B settles 5 mm Take E 200 GPa I 30011062 mm4 A B 40 kN 2 m 2 m A B C L L M0 A B C 50 kN 4 m 2 m 2 m A B L w0 A B C 6 m 6 m 10 kNm 6 m 10 kNm A B 105 FORCE METHOD OF ANALYSIS FRAMES 417 10 101 Determine the reactions at the supports A and B EI is constant 102 Determine the reactions at the supports A B and C then draw the shear and moment diagrams EI is constant 104 Determine the reactions at the supports A B and C then draw the shear and moment diagram EI is constant 105 Determine the reactions at the supports then draw the shear and moment diagram EI is constant 103 Determine the reactions at the supports A and B EI is constant 106 Determine the reactions at the supports then draw the moment diagram Assume B and C are rollers and A is pinned The support at B settles downward 025 ft Take I 500 in4 E 29103 ksi L A w0 B L A B P L A C B 12 ft 3 kft 12 ft C A B P P L 2 L 2 L 2 L 2 A B w L 2 L 2 6 ft 12 ft 3 kipft A B C 6 ft 12 kip Prob 101 Prob 102 Prob 103 Prob 104 Prob 105 Prob 106 PROBLEMS 424 CHAPTER 10 ANALYSIS OF STATICALLY INDETERMINATE STRUCTURES BY THE FORCE METHOD 10 Determine the force in each member of the truss shown in Fig 1015a if the turnbuckle on member AC is used to shorten the member by 05 in Each bar has a crosssectional area of and E 2911062 psi 02 in2 EXAMPLE 108 SOLUTION Principle of Superposition This truss has the same geometry as that in Example 107 Since AC has been shortened we will choose it as the redundant Fig 1015b Compatibility Equation Since no external loads act on the primary structure truss there will be no relative displacement between the ends of the sectioned member caused by load that is The flexibility coefficient has been determined in Example 107 so Assuming the amount by which the bar is shortened is positive the compatibility equation for the bar is therefore Realizing that is a measure of displacement per unit force we have Thus Ans Since no external forces act on the truss the external reactions are zero Therefore using and analyzing the truss by the method of joints yields the results shown in Fig 1015c FAC FAC 6993 lb 699 k 1T2 05 in 0 3456 ft112 inft2 102 in222911062 lbin2 FAC fAC AC 05 in 0 3456 AE FAC fAC AC 3456 AE fAC AC AC 0 C B D A c 559 k C 559 k C 420 k C 420 k C 699 k T 699 k T C B 6 ft 8 ft actual truss a C B D FAC FAC fACAC FAC redundant FAC applied C B D A ΔAC 0 primary structure b A D A Fig 1015 107 COMPOSITE STRUCTURES 427 10 For we require application of the real loads Fig 1016c and a virtual unit load applied to the cut ends of member CE Fig 1016d Here we will use symmetry of both loading and geometryand only con sider the bending strain energy in the beam and of course the axial strain energy in the truss membersThus For we require application of a real unit load and a virtual unit load at the cut ends of member CE Fig 1016dThus Substituting the data into Eq 1 yields Ans FCE 785 kN 0 733311032 m FCE10934511032 mkN2 0934511032 mkN 333311032 20011092120211062 809011032 400110621200110922 13333 EI 2 EI 5590 AE 05 AE 2 AE a 1122122 AE b 2a 11118221252 AE b 2a 10522112 AE b fCE CE L L 0 m2dx EI a n2L AE 2 L 2 0 105x122dx1 EI 2 L 3 2 1122dx2 EI fCE CE 293311032 20011092120211062 733311032 m 12 EI 1733 EI 0 0 0 a 11022 AE b 2a 1052102112 AE b 2a 1111821021252 AE b 2 L 3 2 16x2 x2 22112dx2 EI CE L L 0 Mm EI dx a nNL AE 2 L 2 0 16x1 x2 12105x12dx1 EI CE 109 SYMMETRIC STRUCTURES 429 10 Although the details for applying the force method of analysis using computer methods will also be omitted here we can make some general observations and comments that apply when using this method to solve problems that are highly indeterminate and thus involve large sets of equations In this regard numerical accuracy for the solution is improved if the flexibility coefficients located near the main diagonal of the f matrix are larger than those located off the diagonal To achieve this some thought should be given to selection of the primary structureTo facilitate computations of it is also desirable to choose the primary structure so that it is somewhat symmetric This will tend to yield some flexibility coefficients that are similar or may be zero Lastly the deflected shape of the primary structure should be similar to that of the actual structure If this occurs then the redundants will induce only small corrections to the primary structure which results in a more accurate solution of Eq 102 109 Symmetric Structures A structural analysis of any highly indeterminate structure or for that matter even a statically determinate structure can be simplified provided the designer or analyst can recognize those structures that are symmetric and support either symmetric or antisymmetric loadings In a general sense a structure can be classified as being symmetric provided half of it develops the same internal loadings and deflections as its mirror image reflected about its central axis Normally symmetry requires the material composition geometry supports and loading to be the same on each side of the structure However this does not always have to be the case Notice that for horizontal stability a pin is required to support the beam and truss in Figs 1017a and 1017b Here the horizontal reaction at the pin is zero and so both of these structures will deflect and produce the same internal loading as their reflected counterpart As a result they can be classified as being symmetric Realize that this would not be the case for the frame Figs 1017c if the fixed support at A was replaced by a pin since then the deflected shape and internal loadings would not be the same on its left and right sides fij b w axis of symmetry P2 P1 a axis of symmetry Fig 1017 109 SYMMETRIC STRUCTURES 431 10 2 in2 2 in2 2 in2 2 in2 2 in2 3 in2 3 in2 3 in2 A 3 ft 4 ft 5 k B C D E 4 ft 4 k A B C 4 ft 8 k 6 k 3 ft 1 in2 1 in2 1 in2 1 in2 2 in2 2 in2 D 10 kN D C B E A 3 m 3 m 4 m A C D B 3 ft 800 lb 3 ft 4 ft Prob 1025 Prob 1026 Prob 1027 Prob 1028 1028 Determine the force in member AD of the truss The crosssectional area of each member is shown in the fig ure Assume the members are pin connected at their ends Take E 29103 ksi 1026 Determine the force in each member of the truss The crosssectional area of each member is indicated in the figure Assume the members are pin connected at their ends E 29103 ksi 1027 Determine the force in member AC of the truss AE is constant 1025 Determine the force in each member of the truss AE is constant PROBLEMS 432 CHAPTER 10 ANALYSIS OF STATICALLY INDETERMINATE STRUCTURES BY THE FORCE METHOD 10 C D E A B 10 kN 20 kN 15 kN 2 m 2 m 2 m B A C 4 m 3 m 9 kN 4 m 4 m D 10 ft 10 k 10ft 10 ft 10 ft 15 k 5 k H B C D E A G F 10 ft 2 k 2 k D A B C 3 ft 3 ft Prob 1029 Prob 1030 Prob 1031 Prob 1032 1029 Determine the force in each member of the truss Assume the members are pin connected at their ends AE is constant 1031 Determine the force in member CD of the truss AE is constant 1030 Determine the force in each member of the pin connected truss AE is constant 1032 Determine the force in member GB of the truss AE is constant 109 SYMMETRIC STRUCTURES 433 10 1033 The cantilevered beam AB is additionally supported using two tie rods Determine the force in each of these rods Neglect axial compression and shear in the beam For the beam and for each tie rod Take E 200 GPa A 100 mm2 Ib 200106 mm4 1035 The trussed beam supports the uniform distributed loading If all the truss members have a crosssectional area of 125 in2 determine the force in member BC Neglect both the depth and axial compression in the beam Take for all members Also for the beam IAD 750 in4Assume A is a pin and D is a rocker E 29103 ksi 1034 Determine the force in member AB BC and BD which is used in conjunction with the beam to carry the 30k loadThe beam has a moment of inertia of the members AB and BC each have a crosssectional area of 2 in2 and BD has a crosssectional area of 4 in2 Take ksi Neglect the thickness of the beam and its axial compression and assume all members are pin connected Also assume the support at A is a pin and E is a roller E 2911032 I 600 in4 1036 The trussed beam supports a concentrated force of 80 k at its center Determine the force in each of the three struts and draw the bendingmoment diagram for the beam The struts each have a crosssectional area of 2 in2 Assume they are pin connected at their end points Neglect both the depth of the beam and the effect of axial compression in the beam Take ksi for both the beam and struts Also for the beam I 400 in4 E 2911032 4 m 80 kN 3 m A B C D A B D C E 4 ft 4 ft 4 ft 4 ft 3 ft 5 kft A B C D 80 k 12 ft 5 ft 12 ft E D 3 ft C A B 3 ft 4 ft 30 k Prob 1033 Prob 1034 Prob 1035 Prob 1036 436 CHAPTER 10 ANALYSIS OF STATICALLY INDETERMINATE STRUCTURES BY THE FORCE METHOD 10 Shear at E If the influence line for the shear at point E of the beam in Fig 1021a is to be determined then by the MüllerBreslau principle the beam is imagined cut open at this point and a sliding device is inserted at E Fig 1021b This device will transmit a moment and normal force but no shear When the beam deflects due to positive unit shear loads acting at E the slope on each side of the guide remains the same and the deflection curve represents to some scale the influence line for the shear at E Fig 1021c Had the basic method for establishing the influence line for the shear at E been applied it would then be necessary to apply a unit load at each point D and compute the internal shear at E Fig 1021a This value would represent the ordinate of the influence line at D Using the force method and Maxwells theorem of reciprocal displacements as in the previous case it can be shown that This again establishes the validity of the MüllerBreslau principle namely a positive unit shear load applied to the beam at E Fig 1021c will cause the beam to deflect into the shape of the influ ence line for the shear at E Here the scale factor is 11fEE2 VE a 1 fEE bfDE VE a D E A C b E 1 1 c E 1 1 fEE D fDE Fig 1021 1010 INFLUENCE LINES FOR STATICALLY INDETERMINATE BEAMS 437 10 Moment at E The influence line for the moment at E in Fig 1022a can be determined by placing a pin or hinge at E since this connection transmits normal and shear forces but cannot resist a moment Fig 1022bApplying a positive unit couple moment the beam then deflects to the dashed position in Fig 1022c which yields to some scale the influence line again a consequence of the MüllerBreslau principle Using the force method and Maxwells reciprocal theorem we can show that The scale factor here is 11aEE2 ME a 1 aEE bfDE Fig 1022 a D E A C b E 1 1 c E 1 1 D fDE aEE 438 CHAPTER 10 ANALYSIS OF STATICALLY INDETERMINATE STRUCTURES BY THE FORCE METHOD 10 Influence lines for the continuous girder of this trestle were constructed in order to prop erly design the girder Procedure for Analysis The following procedure provides a method for establishing the influence line for the reaction shear or moment at a point on a beam using the MüllerBreslau technique Qualitative Influence Line At the point on the beam for which the influence line is to be determined place a connection that will remove the capacity of the beam to support the function of the influence line If the function is a vertical reaction use a vertical roller guide if the function is shear use a sliding device or if the function is moment use a pin or hinge Place a unit load at the connection acting on the beam in the positive direction of the function Draw the deflection curve for the beam This curve represents to some scale the shape of the influence line for the beam Quantitative Influence Line If numerical values of the influence line are to be determined compute the displacement of successive points along the beam when the beam is subjected to the unit load placed at the connection mentioned above Divide each value of displacement by the displacement determined at the point where the unit load acts By applying this scalar factor the resulting values are the ordinates of the influence line 1011 QUALITATIVE INFLUENCE LINES FOR FRAMES 439 10 Fig 1023 1011 Qualitative Influence Lines for Frames The MüllerBreslau principle provides a quick method and is of great value for establishing the general shape of the influence line for building frames Once the influenceline shape is known one can immediately specify the location of the live loads so as to create the greatest influence of the function reaction shear or moment in the frame For example the shape of the influence line for the positive moment at the center I of girder FG of the frame in Fig 1023a is shown by the dashed lines Thus uniform loads would be placed only on girders AB CD and FG in order to create the largest positive moment at I With the frame loaded in this manner Fig 1023b an indeterminate analysis of the frame could then be performed to determine the critical moment at I A E B F C G D H I a A E B F C G D H I b 1011 QUALITATIVE INFLUENCE LINES FOR FRAMES 441 10 x Ay A 1 C 0852 D 0481 B 0 For since no moment exists on the conjugate beam at Fig 1024d then For Fig 1024e For Fig 1024f For Fig 1024d Since a vertical 1k load acting at A on the beam in Fig 1024a will cause a vertical reaction at A of 1 k the displacement at A should correspond to a numerical value of 1 for the influenceline ordinate at A Thus dividing the other computed displacements by this factor we obtain A 1944EI A MA 1944 EI A C MC 162 EI 1122 1 2 a 12 EIb1122142 1656 EI MC 0 C D MD 162 EI 162 1 2 a 6 EIb162122 936 EI MD 0 D B MB 0 B B A plot of these values yields the influence line shown in Fig 1024g 6 12 18 x 0852 0481 1 Ay quantitative influence line for reaction at A g 6 ft MD VD 162 EI 6 EI e 12 ft 12 ft MC MC VC VC 12 EI 12 EI f f 162 EI 162 EI 446 CHAPTER 10 ANALYSIS OF STATICALLY INDETERMINATE STRUCTURES BY THE FORCE METHOD 10 Prob 1046 Prob 1047 Prob 1041 Probs 10421043 Prob 1044 Prob 1045 A C 6 m 6 m B A B C 15 ft 15 ft A B C 3 m 3 m 3 m A B C D E 3 m 6 m 6 m 2 m A B C D E F 2 m 4 m 2 m 2 m A B 3 m 3 m 1047 Sketch the influence line for a the vertical reaction at C b the moment at B and c the shear at E In each case indicate on a sketch of the beam where a uniform distributed live load should be placed so as to cause a maximum positive value of these functions Assume the beam is fixed at F 1045 Draw the influence line for the reaction at C Plot the numerical values every 5 ft EI is constant 1042 Draw the influence line for the moment at A Plot numerical values at the peaks Assume A is fixed and the support at B is a roller EI is constant 1043 Draw the influence line for the vertical reaction at B Plot numerical values at the peaksAssume A is fixed and the support at B is a roller EI is constant 1044 Draw the influence line for the shear at C Plot numerical values every 15 m Assume A is fixed and the support at B is a roller EI is constant 1046 Sketch the influence line for a the moment at E b the reaction at C and c the shear at E In each case indicate on a sketch of the beam where a uniform distributed live load should be placed so as to cause a maximum positive value of these functions Assume the beam is fixed at D 1041 Draw the influence line for the reaction at C Plot numerical values at the peaksAssume A is a pin and B and C are rollers EI is constant PROBLEMS 1011 QUALITATIVE INFLUENCE LINES FOR FRAMES 447 10 Prob 1050 Prob 1051 Prob 1048 Prob 1049 A B B A A B A C B 1050 Use the MüllerBreslau principle to sketch the general shape of the influence line for a the moment at A and b the shear at B 1049 Use the MüllerBreslau principle to sketch the general shape of the influence line for a the moment at A and b the shear at B 1048 Use the MüllerBreslau principle to sketch the general shape of the influence line for a the moment at A and b the shear at B 1051 Use the MüllerBreslau principle to sketch the general shape of the influence line for a the moment at A and b the shear at B The members of this building frame are all fixed connected so the framework is statically indeterminate 11 451 In this chapter we will briefly outline the basic ideas for analyzing structures using the displacement method of analysis Once these concepts have been presented we will develop the general equations of slope deflection and then use them to analyze statically indetermi nate beams and frames 111 Displacement Method of Analysis General Procedures All structures must satisfy equilibrium loaddisplacement and compatibility of displacements requirements in order to ensure their safety It was stated in Sec 101 that there are two different ways to satisfy these requirements when analyzing a statically indeterminate structure The force method of analysis discussed in the previous chapter is based on identifying the unknown redundant forces and then satisfying the structures compatibility equations This is done by expressing the displacements in terms of the loads by using the loaddisplacement relations The solution of the resultant equations yields the redundant reactions and then the equilibrium equations are used to determine the remaining reactions on the structure The displacement method works the opposite way It first requires satisfying equilibrium equations for the structure To do this the unknown displacements are written in terms of the loads by using the loaddisplacement relations then these equations are solved for the displacements Once the displacements are obtained the unknown loads are determined from the compatibility equations using the loaddisplacement relations Every displacement method follows this Displacement Method of Analysis Slope Deflection Equations 112 SLOPEDEFLECTION EQUATIONS 453 11 In summary specifying the kinematic indeterminacy or the number of unconstrained degrees of freedom for the structure is a necessary first step when applying a displacement method of analysis It identifies the number of unknowns in the problem based on the assumptions made regarding the deformation behavior of the structure Furthermore once these nodal displacements are known the deformation of the structural members can be completely specified and the loadings within the members obtained 112 SlopeDeflection Equations As indicated previously the method of consistent displacements studied in Chapter 10 is called a force method of analysis because it requires writing equations that relate the unknown forces or moments in a structure Unfortunately its use is limited to structures which are not highly indeterminate This is because much work is required to set up the compatibility equations and furthermore each equation written involves all the unknowns making it difficult to solve the resulting set of equations unless a computer is available By comparison the slopedeflection method is not as involved As we shall see it requires less work both to write the necessary equations for the solution of a problem and to solve these equations for the unknown displacements and associated internal loads Also the method can be easily programmed on a computer and used to analyze a wide range of indeterminate structures The slopedeflection method was originally developed by Heinrich Manderla and Otto Mohr for the purpose of studying secondary stresses in trusses Later in 1915 GA Maney developed a refined version of this technique and applied it to the analysis of indeterminate beams and framed structures General Case The slopedeflection method is so named since it relates the unknown slopes and deflections to the applied load on a structure In order to develop the general form of the slopedeflection equations we will consider the typical span AB of a continuous beam as shown in Fig 112 which is subjected to the arbitrary loading and has a constant EI We wish to relate the beams internal end moments and in terms of its three degrees of freedom namely its angular displacements and and linear displacement which could be caused by a relative settlement between the supports Since we will be developing a formula moments and angular displacements will be considered positive when they act clockwise on the span as shown in Fig 112 Furthermore the linear displacement is considered positive as shown since this displacement causes the cord of the span and the spans cord angle to rotate clockwise The slopedeflection equations can be obtained by using the principle of superposition by considering separately the moments developed at each support due to each of the displacements and and then the loads uB uA c uB uA MBA MAB uA c uB deflection curve MBA MAB EI is constant positive sign convention cord L w P A B Fig 112 FixedEnd Moments In the previous cases we have considered relationships between the displacements and the necessary moments and acting at nodes A and B respectively In general however the linear or angular displacements of the nodes are caused by loadings acting on the span of the member not by moments acting at its nodes In order to develop the slopedeflection equations we must transform these span loadings into equivalent moments acting at the nodes and then use the loaddisplacement relationships just derivedThis is done simply by finding the reaction moment that each load develops at the nodes For example consider the fixedsupported member shown in Fig 116a which is subjected to a concentrated load P at its center The conjugate beam for this case is shown in Fig 116b Since we require the slope at each end to be zero This moment is called a fixedend moment FEM Note that according to our sign convention it is negative at node A counterclockwise and positive at node B clockwise For convenience in solving problems fixedend moments have been calculated for other loadings and are tabulated on the inside back cover of the book Assuming these FEMs have been determined for a specific problem Fig 117 we have 116 MAB 1FEM2AB MBA 1FEM2BA M PL 8 c1 2 a PL 4EIbL d 2c1 2 a M EIbLd 0 cFy 0 MBA MAB 456 CHAPTER 11 DISPLACEMENT METHOD OF ANALYSIS SLOPEDEFLECTION EQUATIONS 11 conjugate beam b PL 4EI A B M EI M EI P A B V M V M real beam a L 2 L 2 Fig 116 A w P FEMAB B FEMBA Fig 117 112 SLOPEDEFLECTION EQUATIONS 457 11 SlopeDeflection Equation If the end moments due to each displacement Eqs 111 through 115 and the loading Eq 116 are added together the resultant moments at the ends can be written as 117 Since these two equations are similar the result can be expressed as a single equation Referring to one end of the span as the near end N and the other end as the far end F and letting the member stiffness be represented as and the spans cord rotation as we can write 118 where internal moment in the near end of the span this moment is positive clockwise when acting on the span E modulus of elasticity of material and span stiffness and farend slopes or angular displacements of the span at the supports the angles are measured in radians and are positive clockwise rotation of its cord due to a linear displacement that is this angle is measured in radians and is positive clockwise moment at the nearend support the moment is positive clockwise when acting on the span refer to the table on the inside back cover for various loading conditions From the derivation Eq 118 is both a compatibility and load displacement relationship found by considering only the effects of bending and neglecting axial and shear deformations It is referred to as the general slopedeflection equation When used for the solution of problems this equation is applied twice for each member span AB that is application is from A to B and from B to A for span AB in Fig 112 1FEM2N fixedend c L c span uN uF near k IL k MN MN 2Ek12uN uF 3c2 1FEM2N For Internal Span or End Span with Far End Fixed c 1psi2 L k IL MBA 2Ea I Lb c2uB uA 3a L b d 1FEM2BA MAB 2Ea I Lb c2uA uB 3a L b d 1FEM2AB This pedestrian bridge has a reinforced concrete deck Since it extends over all its supports it is indeterminate to the second degree The slope deflection equations provide a convenient method for finding the internal moments in each span 113 ANALYSIS OF BEAMS 459 11 113 Analysis of Beams Procedure for Analysis Degrees of Freedom Label all the supports and joints nodes in order to identify the spans of the beam or frame between the nodes By drawing the deflected shape of the structure it will be possible to identify the number of degrees of freedom Here each node can possibly have an angular displacement and a linear displacement Compatibility at the nodes is maintained provided the members that are fixed connected to a node undergo the same displacements as the node If these displacements are unknown and in general they will be then for convenience assume they act in the positive direction so as to cause clockwise rotation of a member or joint Fig 112 SlopeDeflection Equations The slopedeflection equations relate the unknown moments applied to the nodes to the displacements of the nodes for any span of the structure If a load exists on the span compute the FEMs using the table given on the inside back cover Also if a node has a linear displacement compute for the adjacent spans Apply Eq 118 to each end of the span thereby generating two slope deflection equations for each span However if a span at the end of a continuous beam or frame is pin supported apply Eq 1110 only to the restrained end thereby generating one slopedeflection equation for the span Equilibrium Equations Write an equilibrium equation for each unknown degree of freedom for the structure Each of these equations should be expressed in terms of unknown internal moments as specified by the slopedeflection equations For beams and frames write the moment equation of equilibrium at each support and for frames also write joint moment equations of equilibriumIf the frame sidesways or deflects horizontally column shears should be related to the moments at the ends of the columnThis is discussed in Sec 115 Substitute the slopedeflection equations into the equilibrium equations and solve for the unknown joint displacements These results are then substituted into the slopedeflection equations to determine the internal moments at the ends of each member If any of the results are negative they indicate counterclockwise rotation whereas positive moments and displacements are applied clockwise c L 460 CHAPTER 11 DISPLACEMENT METHOD OF ANALYSIS SLOPEDEFLECTION EQUATIONS 11 Draw the shear and moment diagrams for the beam shown in Fig 1110a EI is constant EXAMPLE 111 SOLUTION SlopeDeflection Equations Two spans must be considered in this problem Since there is no span having the far end pinned or roller supported Eq 118 applies to the solution Using the formulas for the FEMs tabulated for the triangular loading given on the inside back cover we have Note that is negative since it acts counterclockwise on the beam at BAlso since there is no load on span AB In order to identify the unknowns the elastic curve for the beam is shown in Fig 1110b As indicated there are four unknown internal moments Only the slope at B is unknown Since A and C are fixed supports Also since the supports do not settle nor are they displaced up or down For span AB considering A to be the near end and B to be the far end we have 1 Now considering B to be the near end and A to be the far end we have 2 In a similar manner for span BC we have 3 4 MCB 2Ea I 6 b2102 uB 3102 108 EI 3 uB 108 MBC 2Ea I 6 b2uB 0 3102 72 2EI 3 uB 72 MBA 2Ea I 8 b2uB 0 3102 0 EI 2 uB MAB 2Ea I 8 b2102 uB 3102 0 EI 4 uB MN 2Ea I Lb12uN uF 3c2 1FEM2N cAB cBC 0 uA uC 0 uB 1FEM2AB 1FEM2BA 0 1FEM2BC 1FEM2BC wL2 30 61622 30 72 kN m 1FEM2CB wL2 20 61622 20 108 kN m A B a 8 m 6 m C 6 kNm A B C MBA MBC uB uB MCB MAB b Fig 1110 462 CHAPTER 11 DISPLACEMENT METHOD OF ANALYSIS SLOPEDEFLECTION EQUATIONS 11 SOLUTION SlopeDeflection Equations Two spans must be considered in this problem Equation 118 applies to span ABWe can use Eq 1110 for span BC since the end C is on a roller Using the formulas for the FEMs tabulated on the inside back cover we have Note that and are negative since they act counterclockwise on the beam at A and B respectivelyAlso since the supports do not settle Applying Eq 118 for span AB and realizing that we have 1 2 Applying Eq1110 with B as the near end and C as the far endwe have 3 Remember that Eq 1110 is not applied from C near end to B far end MBC 0375EIuB 18 MBC 3Ea I 8 b1uB 02 18 MN 3Ea I L b1uN c2 1FEM2N MBA 01667EIuB 96 MBA 2Ea I 24 b2uB 0 3102 96 MAB 008333EIuB 96 MAB 2Ea I 24 b2102 uB 3102 96 MN 2Ea I L b12uN uF 3c2 1FEM2N uA 0 cAB cBC 0 1FEM2BC 1FEM2AB 1FEM2BC 3PL 16 31122182 16 18 k ft 1FEM2BA wL2 12 1 12 12212422 96 k ft 1FEM2AB wL2 12 1 12 12212422 96 k ft EXAMPLE 112 Draw the shear and moment diagrams for the beam shown in Fig1111a EI is constant 8 ft 24 ft 4 ft 2 kft B A C 12 k a Fig 1111 468 CHAPTER 11 DISPLACEMENT METHOD OF ANALYSIS SLOPEDEFLECTION EQUATIONS 11 117 Determine the moment at B then draw the moment diagram for the beam Assume the supports at A and C are pins and B is a roller EI is constant 1110 Determine the moments at A and B then draw the moment diagram for the beam EI is constant 118 Determine the moments at A B and C then draw the moment diagram EI is constant Assume the support at B is a roller and A and C are fixed 1112 Determine the moments acting at A and B Assume A is fixed supported B is a roller and C is a pin EI is constant 1111 Determine the moments at A B and C then draw the moment diagram for the beam Assume the support at A is fixed B and C are rollers and D is a pin EI is constant 119 Determine the moments at each support then draw the moment diagramAssume A is fixed EI is constant A B C 4 m 2 m 4 m 6 m 20 kN 40 kN B A C 3 m 9 m 3 m 80 kN 20 kNm A B C D 4 ft 4 ft 4 ft 12 ft 12 ft 6 k 6 k 3 kft 200 lbft 2400 lb 30 ft 10 ft A B C A 4 kft 20 ft 15 ft 8 ft 8 ft B C D 12 k 8 ft 8 ft 18 ft A B C 6 k 05 kft Prob 117 Prob 118 Prob 119 Prob 1110 Prob 1111 Prob 1112 114 ANALYSIS OF FRAMES NO SIDESWAY 469 11 114 Analysis of Frames No Sidesway A frame will not sidesway or be displaced to the left or right provided it is properly restrained Examples are shown in Fig 1114 Also no sidesway will occur in an unrestrained frame provided it is symmetric with respect to both loading and geometry as shown in Fig 1115 For both cases the term in the slopedeflection equations is equal to zero since bending does not cause the joints to have a linear displacement The following examples illustrate application of the slopedeflection equations using the procedure for analysis outlined in Sec 113 for these types of frames c w P w w w Fig 1114 Fig 1115 470 CHAPTER 11 DISPLACEMENT METHOD OF ANALYSIS SLOPEDEFLECTION EQUATIONS 11 Determine the moments at each joint of the frame shown in Fig 1116a EI is constant SOLUTION SlopeDeflection Equations Three spans must be considered in this problem AB BC and CD Since the spans are fixed supported at A and D Eq 118 applies for the solution From the table on the inside back cover the FEMs for BC are Note that and since no sidesway will occur Applying Eq 118 we have 1 2 3 4 5 6 MDC 01667EIuC MDC 2Ea I 12 b2102 uC 3102 0 MCD 0333EIuC MCD 2Ea I 12 b2uC 0 3102 0 MCB 05EIuC 025EIuB 80 MCB 2Ea I 8 b2uC uB 3102 80 MBC 05EIuB 025EIuC 80 MBC 2Ea I 8 b2uB uC 3102 80 MBA 0333EIuB MBA 2Ea I 12 b2uB 0 3102 0 MAB 01667EIuB MAB 2Ea I 12 b2102 uB 3102 0 MN 2Ek12uN uF 3c2 1FEM2N cAB cBC cCD 0 uA uD 0 1FEM2CB 5wL2 96 512421822 96 80 kN m 1FEM2BC 5wL2 96 512421822 96 80 kN m EXAMPLE 115 B 24 kNm C A D 8 m 12 m a Fig 1116 472 CHAPTER 11 DISPLACEMENT METHOD OF ANALYSIS SLOPEDEFLECTION EQUATIONS 11 Determine the internal moments at each joint of the frame shown in Fig 1117a The moment of inertia for each member is given in the figureTake E 2911032 ksi EXAMPLE 116 SOLUTION SlopeDeflection Equations Four spans must be considered in this problem Equation 118 applies to spans AB and BC and Eq 1110 will be applied to CD and CE because the ends at D and E are pinned Computing the member stiffnesses we have The FEMs due to the loadings are Applying Eqs 118 and 1110 to the frame and noting that since no sidesway occurs we have 1 MAB 107407uB MAB 22911032112221000128622102 uB 3102 0 MN 2Ek12uN uF 3c2 1FEM2N cAB cBC cCD cCE 0 uA 0 1FEM2CE wL2 8 311222 8 54 k ft 1FEM2CB PL 8 61162 8 12 k ft 1FEM2BC PL 8 61162 8 12 k ft kBC 800 1611224 0002411 ft3 kCE 650 1211224 0002612 ft3 kAB 400 1511224 0001286 ft3 kCD 200 1511224 0000643 ft3 3 kft 6 k 8 ft 8 ft 15 ft B A C D E 400 in4 200 in4 800 in4 650 in4 a 12 ft Fig 1117 114 ANALYSIS OF FRAMES NO SIDESWAY 473 11 2 3 4 5 6 Equations of Equilibrium These six equations contain eight unknowns Two moment equilibrium equations can be written for joints B and C Fig 1117bWe have 7 8 In order to solvesubstitute Eqs2 and 3 into Eq7and Eqs46 into Eq 8This gives Solving these equations simultaneously yields These values being clockwise tend to distort the frame as shown in Fig 1117a Substituting these values into Eqs 16 and solving we get Ans Ans Ans Ans Ans Ans MCE 373 k ft MCD 412 k ft MCB 331 k ft MBC 0592 k ft MBA 0592 k ft MAB 0296 k ft uB 275811052 rad uC 511311042 rad 20 1389uB 81 0590uC 42 61 7593uB 20 1389uC 12 MCB MCD MCE 0 MBA MBC 0 MCE 32 7257uC 54 MCE 3291103211222100026122uC 0 54 MCD 80556uC MCD 3291103211222100006432uC 0 0 MN 3Ek1uN c2 1FEM2N MCB 20 1389uB 40 2778uC 12 MCB 22911032112221000241122uC uB 3102 12 MBC 40 2778uB 20 1389uC 12 MBC 22911032112221000241122uB uC 3102 12 MBA 21 4815uB MBA 22911032112221000128622uB 0 3102 0 MBC MBC MBA MBA B C MCE MCE MCD MCD MCB MCB b b 115 ANALYSIS OF FRAMES SIDESWAY 475 11 B MBC MBC MBA MBA C MCB MCB MCD MCD b b 40 k 40 k B MBA MAB VA 12 ft C MCD 18 ft MDC VD c 4 5 6 Equations of Equilibrium The six equations contain nine unknowns Two moment equilibrium equations for joints B and C Fig 1119b can be written namely 7 8 Since a horizontal displacement occurs we will consider summing forces on the entire frame in the x directionThis yields The horizontal reactions or column shears and can be related to the internal moments by considering the freebody diagram of each column separately Fig 1119cWe have Thus 9 In order to solve substitute Eqs 2 and 3 into Eq 7 Eqs 4 and 5 into Eq 8 and Eqs 1 2 5 6 into Eq 9 This yields Solving simultaneously we have Finally using these results and solving Eqs 16 yields Ans Ans Ans Ans Ans Ans MDC 110 k ft MCD 948 k ft MCB 948 k ft MBC 135 k ft MBA 135 k ft MAB 208 k ft EIuB 43881 EIuC 13618 EIcDC 37526 05uB 0222uC 1944cDC 480 EI 0133uB 0489uC 0333cDC 0 06uB 0133uC 075cDC 0 40 MAB MBA 12 MDC MCD 18 0 VD MDC MCD 18 MC 0 VA MAB MBA 12 MB 0 VD VA 40 VA VD 0 Fx 0 MCB MCD 0 MBA MBC 0 MDC 2Ea I 18 b2102 uC 3cDC 0 EI10111uC 0333cDC2 MCD 2Ea I 18 b2uC 0 3cDC 0 EI10222uC 0333cDC2 MCB 2Ea I 15 b2uC uB 3102 0 EI10267uC 0133uB2 115 ANALYSIS OF FRAMES SIDESWAY 479 11 Equilibrium Equations Moment equilibrium of joints B C D and E Fig 1121b requires 13 14 15 16 As in the preceding examples the shear at the base of all the columns for any story must balance the applied horizontal loads Fig 1121c This yields 17 18 Solution requires substituting Eqs 112 into Eqs 1318 which yields six equations having six unknowns and These equations can then be solved simultaneously The results are resubstituted into Eqs 112 which yields the moments at the joints uE c2 uB uC uD c1 120 MAB MBA 5 MEF MFE 5 0 40 80 VAB VFE 0 Fx 0 40 MBC MCB 5 MED MDE 5 0 40 VBC VED 0 Fx 0 MEF MEB MED 0 MDC MDE 0 MCB MCD 0 MBA MBE MBC 0 MBC MBC MBE MBE MBA MBA B C MCB MCB MCD MCD MDC MDC MDE MDE D MED MED MEB MEB MEF MEF E b b 40 kN 40 kN VBC VBC VED VED 40 kN 40 kN 80 kN 80 kN VAB VAB VFE VFE c c 115 ANALYSIS OF FRAMES SIDESWAY 483 11 A B C 6 ft 15 ft 2 kft 6 ft 10 k A D C B 6 m 8 m 6 m 12 kNm A B D E C 3 m 4 m 4 m 10 kN 15 kN 12 kNm 16 kNm 3 kft 12 ft B A D C 5 ft 5 ft 10 ft 1117 Determine the moment that each member exerts on the joint at B then draw the moment diagram for each member of the frameAssume the support at A is fixed and C is a pin EI is constant 1119 Determine the moment at joints D and C then draw the moment diagram for each member of the frame Assume the supports at A and B are pins EI is constant 1118 Determine the moment that each member exerts on the joint at B then draw the moment diagram for each member of the frame Assume the supports at A C and D are pins EI is constant 1120 Determine the moment that each member exerts on the joints at B and D then draw the moment diagram for each member of the frame Assume the supports at A C and E are pins EI is constant Prob 1117 Prob 1118 Prob 1119 Prob 1120 484 CHAPTER 11 DISPLACEMENT METHOD OF ANALYSIS SLOPEDEFLECTION EQUATIONS 11 Prob 1121 Prob 1122 Prob 1123 Prob 1124 1121 Determine the moment at joints C and D then draw the moment diagram for each member of the frame Assume the supports at A and B are pins EI is constant 1123 Determine the moments acting at the supports A and D of the batteredcolumn frame Take I 600 in4 E 29103 ksi 1122 Determine the moment at joints A B C and D then draw the moment diagram for each member of the frame Assume the supports at A and B are fixed EI is constant 1124 Wind loads are transmitted to the frame at joint E If A B E D and F are all pin connected and C is fixed connected determine the moments at joint C and draw the bending moment diagrams for the girder BCE EI is constant B A D C 5 m 6 m 8 kNm A D B C 3 m 3 m 30 kNm A B C E F 6 m 4 m 8 m 12 kN D 4 kft 20 ft 6 k B A D C 15 ft 15 ft 20 ft CHAPTER REVIEW 485 11 3 ft A B C 3 ft 3 ft 3 ft 3 ft 3 ft 3 ft 3 ft Project Prob 111P CHAPTER REVIEW The unknown displacements of a structure are referred to as the degrees of freedom for the structureThey consist of either joint displacements or rotations The slopedeflection equations relate the unknown moments at each joint of a structural member to the unknown rotations that occur there The following equation is applied twice to each member or span considering each side as the near end and its counterpart as the far end For Internal Span or End Span with Far End Fixed This equation is only applied once where the far end is at the pin or roller support Only for End Span with Far End Pinned or Roller Supported Once the slopedeflection equations are written they are substituted into the equations of moment equilibrium at each joint and then solved for the unknown displacements If the structure frame has sidesway then an unknown horizontal displacement at each floor level will occur and the unknown column shears must be related to the moments at the joints using both the force and moment equilibrium equations Once the unknown displacements are obtained the unknown reactions are found from the loaddisplacement relations MN 3EkuN c FEMN MN 2Ek2uN uF 3c FEMN 111P The roof is supported by joists that rest on two girders Each joist can be considered simply supported and the front girder can be considered attached to the three columns by a pin at A and rollers at B and C Assume the roof will be made from 3 inthick cinder concrete and each joist has a weight of 550 lb According to code the roof will be subjected to a snow loading of 25 psf The joists have a length of 25 ft Draw the shear and moment diagrams for the girderAssume the supporting columns are rigid PROJECT PROBLEM The girders of this concrete building are all fixed connected so the statically indeterminate analysis of the framework can be done using the moment distribution method 12 487 The momentdistribution method is a displacement method of analysis that is easy to apply once certain elastic constants have been determined In this chapter we will first state the important definitions and concepts for moment distribution and then apply the method to solve problems involving statically indeterminate beams and frames Application to multistory frames is discussed in the last part of the chapter 121 General Principles and Definitions The method of analyzing beams and frames using moment distribution was developed by Hardy Cross in 1930 At the time this method was first published it attracted immediate attention and it has been recognized as one of the most notable advances in structural analysis during the twentieth century As will be explained in detail later moment distribution is a method of successive approximations that may be carried out to any desired degree of accuracy Essentially the method begins by assuming each joint of a structure is fixedThen by unlocking and locking each joint in succession the internal moments at the joints are distributed and balanced until the joints have rotated to their final or nearly final positions It will be found that this process of calculation is both repetitive and easy to apply Before explaining the techniques of moment distribution however certain definitions and concepts must be presented Displacement Method of Analysis Moment Distribution Sign Convention We will establish the same sign convention as that established for the slopedeflection equations Clockwise moments that act on the member are considered positive whereas counterclockwise moments are negative Fig 121 FixedEnd Moments FEMs The moments at the walls or fixed joints of a loaded member are called fixedend moments These moments can be determined from the table given on the inside back cover depending upon the type of loading on the member For example the beam loaded as shown in Fig 122 has fixedend moments of Noting the action of these moments on the beam and applying our sign convention it is seen that and Member Stiffness Factor Consider the beam in Fig 123 which is pinned at one end and fixed at the otherApplication of the moment M causes the end A to rotate through an angle In Chapter 11 we related M to using the conjugatebeam methodThis resulted in Eq 111 that is The term in parentheses 121 is referred to as the stiffness factor at A and can be defined as the amount of moment M required to rotate the end A of the beam uA 1 rad K 4EI L Far End Fixed M 14EIL2 uA uA uA MBA 1000 N m MAB 1000 N m FEM PL8 80011028 1000 N m 488 CHAPTER 12 DISPLACEMENT METHOD OF ANALYSIS MOMENT DISTRIBUTION 12 Fig 121 Fig 122 MAB B A MBA P w 800 N MBA MAB 5 m 5 m B A M M A B uA Fig 123 Member RelativeStiffness Factor Quite often a continuous beam or a frame will be made from the same material so its modulus of elasticity E will be the same for all the members If this is the case the common factor 4E in Eq 121 will cancel from the numerator and denominator of Eq 122 when the distribution factor for a joint is determined Hence it is easier just to determine the members relativestiffness factor 123 and use this for the computations of the DF CarryOver Factor Consider again the beam in Fig 123 It was shown in Chapter 11 that Eq 111 and Eq 112 Solving for and equating these equations we get In other words the moment M at the pin induces a moment of at the wall The carryover factor represents the fraction of M that is carried over from the pin to the wall Hence in the case of a beam with the far end fixed the carryover factor is The plus sign indicates both moments act in the same direction 1 2 M 1 2 M MBA MAB2 uA MBA 12EIL2 uA MAB 14EIL2 uA KR I L Far End Fixed 490 CHAPTER 12 DISPLACEMENT METHOD OF ANALYSIS MOMENT DISTRIBUTION 12 The statically indeterminate loading in bridge girders that are continuous over their piers can be determined using the method of moment distribution M M A B uA Fig 123 122 MOMENT DISTRIBUTION FOR BEAMS 495 12 fundamental process of moment distribution follows the same procedure as any displacement method There the process is to establish load displacement relations at each joint and then satisfy joint equilibrium requirements by determining the correct angular displacement for the joint compatibility Here however the equilibrium and compatibility of rotation at the joint is satisfied directly using a moment balance process that incorporates the loaddeflection relations stiffness factors Further simplification for using moment distribution is possible and this will be discussed in the next section Procedure for Analysis The following procedure provides a general method for determining the end moments on beam spans using moment distribution Distribution Factors and FixedEnd Moments The joints on the beam should be identified and the stiffness factors for each span at the joints should be calculated Using these values the distribution factors can be determined from Remember that for a fixed end and for an end pin or roller support The fixedend moments for each loaded span are determined using the table given on the inside back cover Positive FEMs act clockwise on the span and negative FEMs act counterclockwise For convenience these values can be recorded in tabular form similar to that shown in Fig 126c Moment Distribution Process Assume that all joints at which the moments in the connecting spans must be determined are initially lockedThen 1 Determine the moment that is needed to put each joint in equilibrium 2 Release or unlock the joints and distribute the counterbalancing moments into the connecting span at each joint 3 Carry these moments in each span over to its other end by multiplying each moment by the carryover factor By repeating this cycle of locking and unlocking the joints it will be found that the moment corrections will diminish since the beam tends to achieve its final deflected shapeWhen a small enough value for the corrections is obtained the process of cycling should be stopped with no carryover of the last moments Each column of FEMs distributed moments and carryover moments should then be added If this is done correctly moment equilibrium at the joints will be achieved 1 2 DF 1 DF 0 DF KK 496 CHAPTER 12 DISPLACEMENT METHOD OF ANALYSIS MOMENT DISTRIBUTION 12 Determine the internal moments at each support of the beam shown in Fig 127a EI is constant EXAMPLE 121 1FEM2CD PL 8 250182 8 250 kN m 1FEM2DC PL 8 250182 8 250 kN m 1FEM2BC wL2 12 2011222 12 240 kN m 1FEM2CB wL2 12 2011222 12 240 kN m SOLUTION The distribution factors at each joint must be computed first The stiffness factors for the members are Therefore The fixedend moments are DFCB 4EI12 4EI12 4EI8 04 DFCD 4EI8 4EI12 4EI8 06 DFAB DFDC 0 DFBA DFBC 4EI12 4EI12 4EI12 05 KAB 4EI 12 KBC 4EI 12 KCD 4EI 8 12 m 12 m 4 m 4 m A B C D 20 kNm 250 kN a Fig 127 Starting with the FEMs line 4 Fig 127b the moments at joints B and C are distributed simultaneously line 5 These moments are then carried over simultaneously to the respective ends of each span line 6 The resulting moments are again simultaneously distributed and carried over lines 7 and 8The process is continued until the resulting moments are diminished an appropriate amount line 13The resulting moments are found by summation line 14 Placing the moments on each beam span and applying the equations of equilibrium yields the end shears shown in Fig 127c and the bendingmoment diagram for the entire beam Fig 127d Here we have used the stiffness factor 4EIL however the relative stiffness factor IL could also have been used 123 STIFFNESSFACTOR MODIFICATIONS 501 12 Fig 1211 Symmetric Beam and Loading If a beam is symmetric with respect to both its loading and geometry the bendingmoment diagram for the beam will also be symmetric As a result a modification of the stiffness factor for the center span can be made so that moments in the beam only have to be distributed through joints lying on either half of the beam To develop the appropriate stiffnessfactor modification consider the beam shown in Fig 1211a Due to the symmetry the internal moments at B and C are equalAssuming this value to be M the conjugate beam for span BC is shown in Fig 1211bThe slope at each end is therefore or The stiffness factor for the center span is therefore 125 Thus moments for only half the beam can be distributed provided the stiffness factor for the center span is computed using Eq 125 By comparison the center spans stiffness factor will be one half that usually determined using K 4EIL K 2EI L Symmetric Beam and Loading M 2EI L u VB u ML 2EI VB1L2 M EI 1L2a L 2 b 0 dMCœ 0 u A D L L L P C B P real beam a u u B C VB VC L 2 L 2 M L EI M EI conjugate beam b 502 CHAPTER 12 DISPLACEMENT METHOD OF ANALYSIS MOMENT DISTRIBUTION 12 Symmetric Beam with Antisymmetric Loading If a symmetric beam is subjected to antisymmetric loading the resulting moment diagram will be antisymmetric As in the previous case we can modify the stiffness factor of the center span so that only one half of the beam has to be considered for the momentdistribution analysis Consider the beam in Fig 1212aThe conjugate beam for its center span BC is shown in Fig 1212b Due to the antisymmetric loading the internal moment at B is equal but opposite to that at C Assuming this value to be M the slope at each end is determined as follows u Fig 1212 A D L L L C B P real beam a P u u B C VB VC 5 L 6 1 L 6 conjugate beam b 1 M L 2 EI 2 1 M L 2 EI 2 M EI M EI VB1L2 1 2 a M EIb aL 2 b a5L 6 b 1 2 a M EIb aL 2 b aL 6 b 0 dMC 0 or The stiffness factor for the center span is therefore 126 Thus when the stiffness factor for the beams center span is computed using Eq 126 the moments in only half the beam have to be distributed Here the stiffness factor is one and a half times as large as that determined using K 4EIL Symmetric Beam with Antisymmetric Loading K 6EI L M 6EI L u VB u ML 6EI 506 CHAPTER 12 DISPLACEMENT METHOD OF ANALYSIS MOMENT DISTRIBUTION 12 Prob 121 Prob 122 Prob 123 121 Determine the moments at B and C EI is constant Assume B and C are rollers and A and D are pinned 124 Determine the reactions at the supports and then draw the moment diagramAssume A is fixed EI is constant PROBLEMS 122 Determine the moments at A B and CAssume the support at B is a roller and A and C are fixed EI is constant 125 Determine the moments at B and C then draw the moment diagram for the beamAssume C is a fixed support EI is constant 123 Determine the moments at A B and C then draw the moment diagram Assume the support at B is a roller and A and C are fixed EI is constant 126 Determine the moments at B and C then draw the moment diagram for the beam All connections are pins Assume the horizontal reactions are zero EI is constant A B C D 8 ft 8 ft 20 ft 3 kft 15 ft 20 ft 20 ft A D C B 500 lb 800 lbft 6 m 4 m 4 m B A C 8 kNm 12 kN A B C D 4 m 12 kNm 12 kNm 4 m 4 m A B C 36 ft 24 ft 2 kft 3 kft B 6 ft 6 ft 6 ft 10 ft 10 ft A C 900 lb 900 lb 400 lb Prob 124 Prob 125 Prob 126 CHAPTER REVIEW 521 12 1225 Determine the moments at joints B and C then draw the moment diagram for each member of the frame The supports at A and D are pinned EI is constant 1226 Determine the moments at C and D then draw the moment diagram for each member of the frame Assume the supports at A and B are pins EI is constant B C A D 12 ft 5 ft 10 ft 5 ft 8 k B C A D 12 ft 6 ft 8 ft 3 k Prob 1225 Prob 1226 CHAPTER REVIEW The process of moment distribution is conveniently done in tabular form Before starting the fixedend moment for each span must be calculated using the table on the inside back cover of the bookThe distribution factors are found by dividing a members stiffness by the total stiffness of the joint For members having a far end fixed use for a farend pinned or roller supported member for a symmetric span and loading and for an antisymmetric loading Remember that the distribution factor for a fixed end is and for a pin or rollersupported end DF 1 DF 0 K 6EIL K 2EIL K 3EIL K 4EIL Moment distribution is a method of successive approximations that can be carried out to any desired degree of accuracy It initially requires locking all the joints of the structure The equilibrium moment at each joint is then determined the joints are unlocked and this moment is distributed onto each connecting member and half its value is carried over to the other side of the span This cycle of locking and unlocking the joints is repeated until the carryover moments become acceptably smallThe process then stops and the moment at each joint is the sum of the moments from each cycle of locking and unlocking The use of variablemomentofinertia girders has reduced considerably the deadweight loading of each of these spans 13 523 In this chapter we will apply the slopedeflection and momentdistribution methods to analyze beams and frames composed of nonprismatic members We will first discuss how the necessary carryover factors stiffness factors and fixedend moments are obtained This is followed by a discussion related to using tabular values often published in design literature Finally the analysis of statically indeterminate structures using the slopedeflection and momentdistribution methods will be discussed 131 Loading Properties of Nonprismatic Members Often to save material girders used for long spans on bridges and buildings are designed to be nonprismatic that is to have a variable moment of inertia The most common forms of structural members that are nonprismatic have haunches that are either stepped tapered or parabolic Fig 131 Provided we can express the members moment of inertia as a function of the length coordinate x then we can use the principle of virtual work or Castiglianos theorem as discussed in Chapter 9 to find its deflectionThe equations are If the members geometry and loading require evaluation of an integral that cannot be determined in closed form then Simpsons rule or some other numerical technique will have to be used to carry out the integration L l 0 Mm EI dx or L l 0 0M 0P M EI dx Beams and Frames Having Nonprismatic Members Fig 131 stepped haunches tapered haunches parabolic haunches 131 LOADING PROPERTIES OF NONPRISMATIC MEMBERS 525 13 These properties can be obtained using for example the conjugate beam method or an energy method However considerable labor is often involved in the process As a result graphs and tables have been made available to determine this data for common shapes used in structural design One such source is the Handbook of Frame Constants published by the Portland Cement Association A portion of these tables taken from this publication is listed here as Tables 131 and 132 A more complete tabular form of the data is given in the PCA handbook along with the relevant derivations of formulas used The nomenclature is defined as follows ratio of the length of haunch at ends A and B to the length of span ratio of the distance from the concentrated load to end A to the length of span carryover factors of member AB at ends A and B respec tively depth of member at ends A and B respectively depth of member at minimum section moment of inertia of section at minimum depth stiffness factor at ends A and B respectively length of member fixedend moment at ends A and B respectively specified in tables for uniform load w or concentrated force P ratios for rectangular crosssectional areas where As noted the fixedend moments and carryover factors are found from the tables The absolute stiffness factor can be determined using the tabulated stiffness factors and found from 132 Application of the use of the tables will be illustrated in Example 131 KA kABEIC L KB kBAEIC L rA 1hA hC2hC rB 1hB hC2hC rA rB MAB MBA L kAB kBA IC hC hB hA CBA CAB b aB aA Timber frames having a variable moment of inertia are often used in the construction of churches uA 1 rad KA CABKA a A B uB 1 rad KB B A b CBAKB Fig 133 Handbook of Frame Constants Portland Cement Association Chicago Illinois 13 526 TABLE 131 Straight HaunchesConstant Width Note All carryover factors are negative and all stiffness factors are positive Concentrated Load Haunch Load at b Left Right FEMCoef PL rAhC bL P aAL aBL L rBhC hC A B Right Haunch Carryover Factors Stiffness Factors Unif Load FEM Coef wL2 01 03 05 07 09 FEM Coef wAL2 FEM Coef wBL2 04 0543 0766 919 652 01194 00791 00935 00034 02185 00384 01955 01147 00889 01601 00096 00870 00133 00008 00006 00058 06 0576 0758 953 724 01152 00851 00934 00038 02158 00422 01883 01250 00798 01729 00075 00898 00133 00009 00005 00060 02 10 0622 0748 1006 837 01089 00942 00931 00042 02118 00480 01771 01411 00668 01919 00047 00935 00132 00011 00004 00062 15 0660 0740 1052 938 01037 01018 00927 00047 02085 00530 01678 01550 00559 02078 00028 00961 00130 00012 00002 00064 20 0684 0734 1083 1009 01002 01069 00924 00050 02062 00565 01614 01645 00487 02185 00019 00974 00129 00013 00001 00065 04 0579 0741 947 740 01175 00822 00934 00037 02164 00419 01909 01225 00856 01649 00100 00861 00133 00009 00022 00118 06 0629 0726 998 864 01120 00902 00931 00042 02126 00477 01808 01379 00747 01807 00080 00888 00132 00010 00018 00124 03 10 0705 0705 1085 1085 01034 01034 00924 00052 02063 00577 01640 01640 00577 02063 00052 00924 00131 00013 00013 00131 15 0771 0689 1170 1310 00956 01157 00917 00062 02002 00675 01483 01892 00428 02294 00033 00953 00129 00015 00008 00137 20 0817 0678 1233 1485 00901 01246 00913 00069 01957 00750 01368 02080 00326 02455 00022 00968 00128 00017 00006 00141 04 0569 0714 797 635 01166 00799 00966 00019 02186 00377 01847 01183 00821 01626 00088 00873 00064 00001 00006 00058 06 0603 0707 826 704 01127 00858 00965 00021 02163 00413 01778 01288 00736 01752 00068 00901 00064 00001 00005 00060 02 10 0652 0698 870 812 01069 00947 00963 00023 02127 00468 01675 01449 00616 01940 00043 00937 00064 00002 00004 00062 15 0691 0691 908 908 01021 01021 00962 00025 02097 00515 01587 01587 00515 02097 00025 00962 00064 00002 00002 00064 20 0716 0686 934 975 00990 01071 00960 00028 02077 00547 01528 01681 00449 02202 00017 00975 00064 00002 00001 00065 04 0607 0692 821 721 01148 00829 00965 00021 02168 00409 01801 01263 00789 01674 00091 00866 00064 00002 00020 00118 06 0659 0678 865 840 01098 00907 00964 00024 02135 00464 01706 01418 00688 01831 00072 00892 00064 00002 00017 00123 03 10 0740 0660 938 1052 01018 01037 00961 00028 02078 00559 01550 01678 00530 02085 00047 00927 00064 00002 00012 00130 15 0809 0645 1009 1266 00947 01156 00958 00033 02024 00651 01403 01928 00393 02311 00029 00950 00063 00003 00008 00137 20 0857 0636 1062 1432 00897 01242 00955 00038 01985 00720 01296 02119 00299 02469 00020 00968 00063 00003 00005 00141 rB variable rA 15 aB variable aA 02 rB variable rA 10 aB variable aA 03 MBA MAB MBA MAB MBA MAB MBA MAB MBA MAB MBA MAB MBA MAB MBA MAB kBA kAB CBA CAB rB aB 13 TABLE 132 Parabolic HaunchesConstant Width Note All carryover factors are negative and all stiffness factors are positive Concentrated Load Haunch Load at b Left Right FEMCoef PL rAhC bL P aAL aBL L rBhC A B hC Right Haunch Carryover Factors Stiffness Factors Unif Load FEM Coef wL2 01 03 05 07 09 FEM Coef wAL2 FEM Coef wBL2 04 0558 0627 608 540 01022 00841 00938 00033 01891 00502 01572 01261 00715 01618 00073 00877 00032 00001 00002 00030 06 0582 0624 621 580 00995 00887 00936 00036 01872 00535 01527 01339 00663 01708 00058 00902 00032 00001 00002 00031 02 10 0619 0619 641 641 00956 00956 00935 00038 01844 00584 01459 01459 00584 01844 00038 00935 00032 00001 00001 00032 15 0649 0614 659 697 00921 01015 00933 00041 01819 00628 01399 01563 00518 01962 00025 00958 00032 00001 00001 00032 20 0671 0611 671 738 00899 01056 00932 00044 01801 00660 01358 01638 00472 02042 00017 00971 00032 00001 00000 00033 04 0588 0616 622 593 01002 00877 00937 00035 01873 00537 01532 01339 00678 01686 00073 00877 00032 00001 00007 00063 06 0625 0609 641 658 00966 00942 00935 00039 01845 00587 01467 01455 00609 01808 00057 00902 00032 00001 00005 00065 03 10 0683 0598 673 768 00911 01042 00932 00044 01801 00669 01365 01643 00502 02000 00037 00936 00031 00001 00004 00068 15 0735 0589 702 876 00862 01133 00929 00050 01760 00746 01272 01819 00410 02170 00023 00959 00031 00001 00003 00070 20 0772 0582 725 961 00827 01198 00927 00054 01730 00805 01203 01951 00345 02293 00016 00972 00031 00001 00002 00072 04 0488 0807 985 597 01214 00753 00929 00034 02131 00371 02021 01061 00979 01506 00105 00863 00171 00017 00003 00030 06 0515 0803 1010 645 01183 00795 00928 00036 02110 00404 01969 01136 00917 01600 00083 00892 00170 00018 00002 00030 02 10 0547 0796 1051 722 01138 00865 00926 00040 02079 00448 01890 01245 00809 01740 00056 00928 00168 00020 00001 00031 15 0571 0786 1090 790 01093 00922 00923 00043 02055 00485 01818 01344 00719 01862 00035 00951 00167 00021 00001 00032 20 0590 0784 1117 840 01063 00961 00922 00046 02041 00506 01764 01417 00661 01948 00025 00968 00166 00022 00001 00032 04 0554 0753 1042 766 01170 00811 00926 00040 02087 00442 01924 01205 00898 01595 00107 00853 00169 00020 00042 00145 06 0606 0730 1096 912 01115 00889 00922 00046 02045 00506 01820 01360 00791 01738 00086 00878 00167 00022 00036 00152 05 10 0694 0694 1203 1203 01025 01025 00915 00057 01970 00626 01639 01639 00626 01970 00057 00915 00164 00028 00028 00164 15 0781 0664 1312 1547 00937 01163 00908 00070 01891 00759 01456 01939 00479 02187 00039 00940 00160 00034 00021 00174 20 0850 0642 1409 1864 00870 01275 00901 00082 01825 00877 01307 02193 00376 02348 00027 00957 00157 00039 00016 00181 rB variable rA 10 aB variable aA 05 rB variable rA 10 aB variable aA 02 MBA MAB MBA MAB MBA MAB MBA MAB MBA MAB MBA MAB MBA MAB MBA MAB kBA kAB CBA CAB rB aB 527 532 CHAPTER 13 BEAMS AND FRAMES HAVING NONPRISMATIC MEMBERS 13 SOLUTION Since the haunches are parabolic we will use Table 132 to obtain the momentdistribution properties of the beam Span AB Entering Table 132 with these ratios we find Using Eqs 132 Since the far end of span BA is pinned we will modify the stiffness factor of BA using Eq 133We have Uniform loadTable 132 1FEM2BA 11950 k ft 1FEM2AB 100956212212522 11950 k ft KBA œ KBA11 CABCBA2 0171E1 0619106192 0105E KAB KBA kEIC L 641EA 1 12B1121223 25 0171E kAB kBA 641 CAB CBA 0619 aA aB 5 25 02 rA rB 4 2 2 10 Determine the internal moments at the supports of the beam shown in Fig 139a The beam has a thickness of 1 ft and E is constant EXAMPLE 131 3 ft 30 k 2 kft 5 ft C 25 ft 5 ft 15 ft 5 ft 5 ft 10 ft a 5 ft 4 ft A B 2 ft 4 ft 2 ft Fig 139 133 SlopeDeflection Equations for Nonprismatic Members The slopedeflection equations for prismatic members were developed in Chapter 11 In this section we will generalize the form of these equations so that they apply as well to nonprismatic membersTo do this we will use the results of the previous section and proceed to formulate the equations in the same manner discussed in Chapter 11 that is considering the effects caused by the loads relative joint displacement and each joint rotation separately and then superimposing the results Loads Loads are specified by the fixedend moments and acting at the ends A and B of the span Positive moments act clockwise Relative Joint Translation When a relative displacement between the joints occurs the induced moments are determined from Eq 136At end A this moment is and at end B it is Rotation at A If end A rotates the required moment in the span at A is Also this induces a moment of at end B Rotation at B If end B rotates a moment of must act at end B and the moment induced at end A is The total end moments caused by these effects yield the generalized slopedeflection equations which can therefore be written as Since these two equations are similar we can express them as a single equation Referring to one end of the span as the near end N and the other end as the far end F and representing the member rotation as we have 138 Here internal moment at the near end of the span this moment is positive clockwise when acting on the span absolute stiffness of the near end determined from tables or by calculation KN MN MN KN1uN CNuF c11 CN22 1FEM2N c L MBA KBBuB CBAuA L 11 CBA2R 1FEM2BA MAB KABuA CABuB L 11 CAB2R 1FEM2AB CBA KBuB CAB KAuB KBuB uB CAB KAuA CBA KBuA KAuA uA KBL11 CBA2 KAL11 CAB2 1FEM2BA 1FEM2AB 534 CHAPTER 13 BEAMS AND FRAMES HAVING NONPRISMATIC MEMBERS 13 near and farend slopes of the span at the supports the angles are measured in radians and are positive clockwise span cord rotation due to a linear displacement this angle is measured in radians and is positive clockwise fixedend moment at the nearend support the moment is positive clockwise when acting on the span and is obtained from tables or by calculations Application of the equation follows the same procedure outlined in Chapter 11 and therefore will not be discussed here In particular note that Eq 138 reduces to Eq 118 when applied to members that are prismatic 1FEM2N c L c uN uF 133 SLOPEDEFLECTION EQUATIONS FOR NONPRISMATIC MEMBERS 535 13 A continuous reinforcedconcrete highway bridge Lightweight metal buildings are often designed using frame members having variable moments of inertia 536 CHAPTER 13 BEAMS AND FRAMES HAVING NONPRISMATIC MEMBERS 13 131 Determine the moments at A B and C by the momentdistribution method Assume the supports at A and C are fixed and a roller support at B is on a rigid base The girder has a thickness of 4 ft Use Table 131 E is constantThe haunches are straight 132 Solve Prob 131 using the slopedeflection equations 135 Use the momentdistribution method to determine the moment at each joint of the symmetric bridge frame Supports at F and E are fixed and B and C are fixed connected Use Table 132 Assume E is constant and the members are each 1 ft thick 136 Solve Prob 135 using the slopedeflection equations PROBLEMS 133 Apply the momentdistribution method to determine the moment at each joint of the parabolic haunched frame Supports A and B are fixed Use Table 132 The members are each 1 ft thick E is constant 134 Solve Prob133 using the slopedeflection equations 137 Apply the momentdistribution method to determine the moment at each joint of the symmetric parabolic haunched frameSupports A and D are fixedUse Table 132 The members are each 1 ft thick E is constant 138 Solve Prob137 using the slopedeflection equations 6 ft 4 ft 4 ft 2 ft A C 4 ft 4 ft 4 ft 6 ft B 20 ft 20 ft 8 kft 9 ft 6 ft 8 ft 2 ft 4 ft 4 ft 25 ft 2 ft A B C D F E 30 ft 40 ft 30 ft 4 kft 5 ft 40 ft 40 ft 2 ft 2 ft 8 ft 12 ft C B A 15 kft 4 ft 32 ft 32 ft 15 ft 5 ft 25 ft 2 k 8 ft 8 ft A D B C 3 ft 3 ft Probs 131132 Probs 133134 Probs 135136 Probs 137138 CHAPTER REVIEW 537 13 20 ft 25 ft 6 ft 1 ft 6 ft 18 ft 25 ft 1 ft 1 ft A B C D 500 lbft 4 ft 3 ft 2 ft 2 ft 30 ft 30 ft 30 ft 30 ft 3 ft D C B E F A 2 kft 40 ft 40 ft 40 ft 12 ft 139 Use the momentdistribution method to determine the moment at each joint of the frame The supports at A and C are pinned and the joints at B and D are fixed connected Assume that E is constant and the members have a thickness of 1 ft The haunches are straight so use Table 131 1310 Solve Prob139 using the slopedeflection equations 1311 Use the momentdistribution method to determine the moment at each joint of the symmetric bridge frame Supports F and E are fixed and B and C are fixed connected The haunches are straight so use Table 132 Assume E is constant and the members are each 1 ft thick 1312 Solve Prob 1311 using the slopedeflection equations Nonprismatic members having a variable moment of inertia are often used on longspan bridges and building frames to save material A structural analysis using nonprismatic members can be performed using either the slopedeflection equations or moment distribution If this is done it then becomes necessary to obtain the fixedend moments stiffness factors and carryover factors for the member One way to obtain these values is to use the conjugate beam method although the work is somewhat tedious It is also possible to obtain these values from tabulated data such as published by the Portland Cement Association If the moment distribution method is used then the process can be simplified if the stiffness of some of the members is modified CHAPTER REVIEW Probs 1391310 Probs 13111312 The spacetruss analysis of electrical transmission towers can be performed using the stiffness method 14 539 In this chapter we will explain the basic fundamentals of using the stiffness method for analyzing structures It will be shown that this method although tedious to do by hand is quite suited for use on a computer Examples of specific applications to planar trusses will be given The method will then be expanded to include spacetruss analysis Beams and framed structures will be discussed in the next chapters 141 Fundamentals of the Stiffness Method There are essentially two ways in which structures can be analyzed using matrix methodsThe stiffness method to be used in this and the following chapters is a displacement method of analysisA force method called the flexibility method as outlined in Sec 91 can also be used to analyze structures however this method will not be presented in this text There are several reasons for this Most important the stiffness method can be used to analyze both statically determinate and indeterminate structures whereas the flexibility method requires a different procedure for each of these two cases Also the stiffness method yields the dis placements and forces directly whereas with the flexibility method the displacements are not obtained directly Furthermore it is generally much easier to formulate the necessary matrices for the computer operations using the stiffness method and once this is done the computer calculations can be performed efficiently Truss Analysis Using the Stiffness Method Application of the stiffness method requires subdividing the structure into a series of discrete finite elements and identifying their end points as nodes For truss analysis the finite elements are represented by each of the members that compose the truss and the nodes represent the joints The forcedisplacement properties of each element are determined and then related to one another using the force equilibrium equations written at the nodes These relationships for the entire structure are then grouped together into what is called the structure stiffness matrix K Once it is established the unknown displacements of the nodes can then be determined for any given loading on the structure When these displacements are known the external and internal forces in the structure can be calculated using the force displacement relations for each member Before developing a formal procedure for applying the stiffness method it is first necessary to establish some preliminary definitions and concepts Member and Node Identification One of the first steps when applying the stiffness method is to identify the elements or members of the structure and their nodesWe will specify each member by a number enclosed within a square and use a number enclosed within a circle to identify the nodes Also the near and far ends of the member must be identified This will be done using an arrow written along the member with the head of the arrow directed toward the far end Examples of member node and direction identification for a truss are shown in Fig 141a These assignments have all been done arbitrarily Global and Member Coordinates Since loads and displacements are vector quantities it is necessary to establish a coordinate system in order to specify their correct sense of direction Here we will use two different types of coordinate systems A single global or structure coordinate system x y will be used to specify the sense of each of the external force and displacement components at the nodes Fig 141a A local or member coordinate system will be used for each member to specify the sense of direction of its displacements and internal loadings This system will be identified using axes with the origin at the near node and the axis extending toward the far node An example for truss member 4 is shown in Fig 141b x y x 540 CHAPTER 14 TRUSS ANALYSIS USING THE STIFFNESS METHOD 14 For large trusses matrix manipulations using K are actually more efficient using selective numbering of the members in a wave pattern that is starting from top to bottom then bottom to top etc 141 FUNDAMENTALS OF THE STIFFNESS METHOD 541 14 Kinematic Indeterminacy As discussed in Sec 111 the unconstrained degrees of freedom for the truss represent the primary unknowns of any displacement method and therefore these must be identified As a general rule there are two degrees of freedom or two possible displacements for each joint node For application each degree of freedom will be specified on the truss using a code numbershown at the joint or node and referenced to its positive global coordinate direction using an associated arrow For example the truss in Fig 141a has eight degrees of freedom which have been identified by the code numbers 1 through 8 as shown The truss is kinematically indeterminate to the fifth degree because of these eight possible displacements 1 through 5 represent unknown or unconstrained degrees of freedom and 6 through 8 represent constrained degrees of freedom Due to the constraints the displacements here are zero For later application the lowest code numbers will always be used to identify the unknown displacements unconstrained degrees of freedom and the highest code numbers will be used to identify the known displacements constrained degrees of freedomThe reason for choosing this method of identification has to do with the convenience of later partitioning the structure stiffness matrix so that the unknown displacements can be found in the most direct manner Once the truss is labeled and the code numbers are specified the structure stiffness matrix K can then be determined To do this we must first establish a member stiffness matrix for each member of the truss This matrix is used to express the members loaddisplacement relations in terms of the local coordinates Since all the members of the truss are not in the same direction we must develop a means for transforming these quantities from each members local coordinate system to the structures global x y coordinate systemThis can be done using force and displacement transformation matrices Once established the elements of the member stiffness matrix are transformed from local to global coordinates and then assembled to create the structure stiffness matrix Using K as stated previously we can determine the node displacements first followed by the support reactions and the member forcesWe will now elaborate on the development of this method y x k a 3 4 2 1 1 3 5 2 4 2 1 5 6 4 3 8 7 y x 4 3 2 y x b Fig 141 143 DISPLACEMENT AND FORCE TRANSFORMATION MATRICES 543 14 the force at joint i when a unit displacement is imposed at joint j For exampleif then is the force at the near joint when the far joint is held fixedand the near joint undergoes a displacement of ie Likewise the force at the far joint is determined from so that These two terms represent the first column of the member stiffness matrix In the same manner the second column of this matrix represents the forces in the member only when the far end of the member under goes a unit displacement 143 Displacement and Force Transformation Matrices Since a truss is composed of many members elements we will now develop a method for transforming the member forces q and displacements d defined in local coordinates to global coordinates For the sake of convention we will consider the global coordinates positive x to the right and positive y upward The smallest angles between the positive x y global axes and the positive local axis will be defined as and as shown in Fig 143 The cosines of these angles will be used in the matrix analysis that follows These will be identified as Numerical values for and can easily be generated by a computer once the x y coordinates of the near end N and far end F of the member have been specified For example consider member NF of the truss shown in Fig 144 Here the coordinates of N and F are and respectively Thus 145 146 The algebraic signs in these generalized equations will automatically account for members that are oriented in other quadrants of the xy plane ly cos uy yF yN L yF yN 21xF xN22 1yF yN22 lx cos ux xF xN L xF xN 21xF xN22 1yF yN22 1xF yF2 1xN yN2 ly lx ly cos uy lx cos ux uy ux x qF kœ 21 AE L j 1 i 2 qN kœ 11 AE L dN 1 kœ 11 i j 1 Fig 143 Fig 144 The origin can be located at any convenient point Usually however it is located where the x y coordinates of all the nodes will be positive as shown in Fig 144 y x x F y N ux uy x F far N near x y yF yN xN xF uy ux Displacement Transformation Matrix In global coordinates each end of the member can have two degrees of freedom or independent displacements namely joint N has and Figs 145a and 145b and joint F has and Figs 145c and 145d We will now consider each of these displacements separately in order to determine its component displacement along the member When the far end is held pinned and the near end is given a global displacement Fig 145a the corresponding displacement deformation along the member is Likewise a displacement will cause the member to be displaced along the axis Fig 145b The effect of both global displacements causes the member to be displaced In a similar manner positive displacements and successively applied at the far end F while the near end is held pinned Figs 145c and 145d will cause the member to be displaced Letting and represent the direction cosines for the member we have which can be written in matrix form as 147 or 148 where 149 From the above derivation T transforms the four global x y displace ments D into the two local displacements d Hence T is referred to as the displacement transformation matrix x T clx ly 0 0 0 0 lx ly d d TD cdN dF d clx ly 0 0 0 0 lx ly d D DNx DNy DFx DFy T dF DFxlx DFyly dN DNxlx DNyly ly cos uy lx cos ux dF DFx cos ux DFy cos uy DFy DFx dN DNx cos ux DNy cos uy x DNy cos uy DNy DNx cos ux DNx DFy DFx DNy DNx 544 CHAPTER 14 TRUSS ANALYSIS USING THE STIFFNESS METHOD 14 The change in or uy will be neglected since it is very small ux Fig 145 a F x y x DNx DNx cos ux N ux b DNy DNy cos uy F x y x N uy c y x F N DFx cos ux DFx x ux d N y x F DFy cos uy DFy uy x 14 Force Transformation Matrix Consider now application of the force to the near end of the member the far end held pinned Fig 146a Here the global force components of at N are Likewiseif is applied to the barFig146bthe global force components at F are Using the direction cosines these equations become which can be written in matrix form as 1410 or 1411 where 1412 In this case transforms the two local forces q acting at the ends of the member into the four global x y force components Q By comparison this force transformation matrix is the transpose of the displacement transformation matrix Eq 149 1x2 TT TT D lx 0 ly 0 0 lx 0 ly T Q TTq D QNx QNy QFx QFy T D lx 0 ly 0 0 lx 0 ly T cqN qF d QFx qFlx QFy qFly QNx qNlx QNy qNly ly cos uy lx cos ux QFx qF cos ux QFy qF cos uy qF QNx qN cos ux QNy qN cos uy qN qN x QNy qN QNx y x F N a uy ux QFy qF QFx x y F N x uy b ux Fig 146 143 DISPLACEMENT AND FORCE TRANSFORMATION MATRICES 545 144 Member Global Stiffness Matrix We will now combine the results of the preceding sections and determine the stiffness matrix for a member which relates the members global force components Q to its global displacements D If we substitute Eq 148 into Eq 143 we can determine the members forces q in terms of the global displacements D at its end points namely 1413 Substituting this equation into Eq 1411 yields the final result or 1414 where 1415 The matrix k is the member stiffness matrix in global coordinates Since T and are known we have Performing the matrix operations yields 1416 k AE L Nx Ny Fx Fy D lx 2 lxly lx 2 lxly lxly ly 2 lxly ly 2 lx 2 lxly lx 2 lxly lxly ly 2 lxly ly 2 T Nx Ny Fx Fy k D lx 0 ly 0 0 lx 0 ly T AE L c 1 1 1 1 d clx ly 0 0 0 0 lx ly d k TT k TTkœT Q kD Q TTkTD Q TTq q kTD 1q kd2 1d TD2 546 CHAPTER 14 TRUSS ANALYSIS USING THE STIFFNESS METHOD 14 145 TRUSS STIFFNESS MATRIX 547 14 The location of each element in this symmetric matrix is referenced with each global degree of freedom associated with the near end N followed by the far end F This is indicated by the code number notation along the rows and columns that is Here k represents the forcedisplacement relations for the member when the components of force and displacement at the ends of the member are in the global or x y directions Each of the terms in the matrix is therefore a stiffness influence coefficient which denotes the x or y force component at i needed to cause an associated unit x or y displacement component at j As a result each identified column of the matrix represents the four force components developed at the ends of the member when the identified end undergoes a unit displacement related to its matrix column For example a unit displacement will create the four force components on the member shown in the first column of the matrix 145 Truss Stiffness Matrix Once all the member stiffness matrices are formed in global coordinates it becomes necessary to assemble them in the proper order so that the stiffness matrix K for the entire truss can be found This process of combining the member matrices depends on careful identification of the elements in each member matrixAs discussed in the previous section this is done by designating the rows and columns of the matrix by the four code numbers used to identify the two global degrees of freedom that can occur at each end of the member see Eq 1416 The structure stiffness matrix will then have an order that will be equal to the highest code number assigned to the truss since this represents the total number of degrees of freedom for the structureWhen the k matrices are assembled each element in k will then be placed in its same row and column designation in the structure stiffness matrix K In particular when two or more members are connected to the same joint or node then some of the elements of each members k matrix will be assigned to the same position in the K matrix When this occurs the elements assigned to the common location must be added together algebraically The reason for this becomes clear if one realizes that each element of the k matrix represents the resistance of the member to an applied force at its end In this way adding these resistances in the x or y direction when forming the K matrix determines the total resistance of each joint to a unit displacement in the x or y direction This method of assembling the member matrices to form the structure stiffness matrix will now be demonstrated by two numerical examples Although this process is somewhat tedious when done by hand it is rather easy to program on a computer Ny Fx Fy Nx DNx 1 kij Ny Fx Fy Nx 4 4 Determine the structure stiffness matrix for the twomember truss shown in Fig 147a AE is constant 548 CHAPTER 14 TRUSS ANALYSIS USING THE STIFFNESS METHOD 14 EXAMPLE 141 SOLUTION By inspection ② will have two unknown displacement components whereas joints ① and ③ are constrained from displacement Consequently the displacement components at joint ② are code numbered first followed by those at joints ③ and ① Fig 147b The origin of the global coordinate system can be located at any point For convenience we will choose joint ② as shown The members are identified arbitrarily and arrows are written along the two members to identify the near and far ends of each member The direction cosines and the stiffness matrix for each member can now be determined Member 1 Since ② is the near end and ③ is the far end then by Eqs 145 and 146 we have Using Eq 1416 dividing each element by we have The calculations can be checked in part by noting that is symmetric Note that the rows and columns in are identified by the x y degrees of freedom at the near end followed by the far end that is 1 2 3 4 respectively for member 1 Fig 147bThis is done in order to identify the elements for later assembly into the K matrix k1 k1 k1 AE 1 2 3 4 D 0333 0 0333 0 0 0 0 0 0333 0 0333 0 0 0 0 0 T 1 2 3 4 L 3 ft lx 3 0 3 1 ly 0 0 3 0 Fig 147 3 ft 4 ft 1 2 3 a 3 ft 4 ft 2 3 1 6 5 2 1 4 3 1 2 x y b 14 Member 2 Since ② is the near end and ① is the far end we have Thus Eq 1416 with becomes Here the rows and columns are identified as 1 2 5 6 since these numbers represent respectively the x y degrees of freedom at the near and far ends of member 2 Structure Stiffness Matrix This matrix has an order of since there are six designated degrees of freedom for the truss Fig 147b Corresponding elements of the above two matrices are added algebraically to form the structure stiffness matrixPerhaps the assembly process is easier to see if the missing numerical columns and rows in and are expanded with zeros to form two matrices Then K k1 k2 6 6 k2 k1 6 6 k2 AE 1 2 5 6 D 0072 0096 0072 0096 0096 0128 0096 0128 0072 0096 0072 0096 0096 0128 0096 0128 T 1 2 5 6 L 5 ft lx 3 0 5 06 ly 4 0 5 08 K AE F 0405 0096 0333 0 0072 0096 0096 0128 0 0 0096 0128 0333 0 0333 0 0 0 0 0 0 0 0 0 0072 0096 0 0 0072 0096 0096 0128 0 0 0096 0128 V 1 2 3 4 5 6 F 0072 0096 0 0 0072 0096 0096 0128 0 0 0096 0128 0 0 0 0 0 0 0 0 0 0 0 0 0072 0096 0 0 0072 0096 0096 0128 0 0 0096 0128 V 1 2 3 4 5 6 K AE 1 2 3 4 5 6 F 0333 0 0333 0 0 0 0 0 0 0 0 0 0333 0 0333 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 V 1 2 3 4 5 6 AE If a computer is used for this operation generally one starts with K having all zero elements then as the member global stiffness matrices are generated they are placed directly into their respective element positions in the K matrix rather than developing the member stiffness matrices storing them then assembling them 145 TRUSS STIFFNESS MATRIX 549 EXAMPLE 142 Determine the structure stiffness matrix for the truss shown in Fig 148a AE is constant SOLUTION Although the truss is statically indeterminate to the first degree this will present no difficulty for obtaining the structure stiffness matrix Each joint and member are arbitrarily identified numerically and the near and far ends are indicated by the arrows along the members As shown in Fig 148b the unconstrained displacements are code numbered first There are eight degrees of freedom for the truss and so K will be an matrix In order to keep all the joint coordinates positive the origin of the global coordinates is chosen at ① Equations 145 146 and 1416 will now be applied to each member Member 1 Here so that Member 2 Here so that Member 3 Here so that k3 AE 1 2 3 4 D 0 0 0 0 0 01 0 01 0 0 0 0 0 01 0 01 T 1 2 3 4 lx 0 0 10 0 ly 10 0 10 1 L 10 ft k2 AE 1 2 7 8 D 0035 0035 0035 0035 0035 0035 0035 0035 0035 0035 0035 0035 0035 0035 0035 0035 T 1 2 7 8 lx 10 0 1022 0707 ly 10 0 1022 0707 L 1022 ft k1 AE 1 2 6 5 D 01 0 01 0 0 0 0 0 01 0 01 0 0 0 0 0 T 1 2 6 5 lx 10 0 10 1 ly 0 0 10 0 L 10 ft 8 8 550 CHAPTER 14 TRUSS ANALYSIS USING THE STIFFNESS METHOD 14 Fig 148 10 ft 10 ft a 10 ft 10 ft 4 y 3 4 4 8 3 6 2 5 5 x 6 2 1 1 2 3 b 1 7 6 14 Member 4 Here so that Member 5 Here so that Member 6 Here so that Structure Stiffness Matrix The foregoing six matrices can now be assembled into the matrix by algebraically adding their corresponding elements For example since then and so on The final result is thus K11 AE101 00352 AE101352 1k1123 1k1124 1k1125 1k1126 0 1k1122 AE100352 1k1121 AE1012 8 8 K k6 AE 6 5 7 8 D 0 0 0 0 0 01 0 01 0 0 0 0 0 01 0 01 T 6 5 7 8 lx 10 10 10 0 ly 10 0 10 1 L 10 ft k5 AE 3 4 6 5 D 0035 0035 0035 0035 0035 0035 0035 0035 0035 0035 0035 0035 0035 0035 0035 0035 T 3 4 6 5 lx 10 0 1022 0707 ly 0 10 1022 0707 L 1022 ft k4 AE 3 4 7 8 D 01 0 01 0 0 0 0 0 01 0 01 0 0 0 0 0 T 3 4 7 8 lx 10 0 10 1 ly 10 10 10 0 L 10 ft K AE 1 2 3 4 5 6 7 8 H 0135 0035 0 0 0 01 0035 0035 0035 0135 0 01 0 0 0035 0035 0 0 0135 0035 0035 0035 01 0 0 01 0035 0135 0035 0035 0 0 0 0 0035 0035 0135 0035 0 01 01 0 0035 0035 0035 0135 0 0 0035 0035 01 0 0 0 0135 0035 0035 0035 0 0 01 0 0035 0135 X 1 2 3 4 5 6 7 8 Ans 145 TRUSS STIFFNESS MATRIX 551 146 Application of the Stiffness Method for Truss Analysis Once the structure stiffness matrix is formed the global force components Q acting on the truss can then be related to its global displacements D using 1417 This equation is referred to as the structure stiffness equation Since we have always assigned the lowest code numbers to identify the unconstrained degrees of freedom this will allow us now to partition this equation in the following form 1418 Here known external loads and displacements the loads here exist on the truss as part of the problem and the displacements are generally specified as zero due to support constraints such as pins or rollers unknown loads and displacements the loads here represent the unknown support reactions and the displacements are at joints where motion is unconstrained in a particular direction structure stiffness matrix which is partitioned to be compati ble with the partitioning of Q and D Expanding Eq 1418 yields 1419 1420 Most often since the supports are not displaced Provided this is the case Eq 1419 becomes Since the elements in the partitioned matrix represent the total resistance at a truss joint to a unit displacement in either the x or y direction then the above equation symbolizes the collection of all the force equilibrium equations applied to the joints where the external loads are zero or have a known value Solving for we have 1421 From this equation we can obtain a direct solution for all the unknown joint displacements then using Eq 1420 with yields 1422 from which the unknown support reactions can be determined The member forces can be determined using Eq 1413 namely q kTD Qu K21Du Dk 0 Du K111Qk Du 1Qk2 K11 Qk K11Du Dk 0 Qu K21Du K22Dk Qk K11Du K12Dk K Du Qu Dk Qk cQk Qu d cK11 K12 K21 K22 d cDu Dk d Q KD 552 CHAPTER 14 TRUSS ANALYSIS USING THE STIFFNESS METHOD 14 This partitioning scheme will become obvious in the numerical examples that follow 146 APPLICATION OF THE STIFFNESS METHOD FOR TRUSS ANALYSIS 553 14 Expanding this equation yields Since for equilibrium only one of the forces has to be found Here we will determine the one that exerts tension in the member Fig 142c 1423 In particular if the computed result using this equation is negative the member is then in compression qF AE L lx ly lx lyD DNx DNy DFx DFy T qF qN qF cqN qF d AE L c 1 1 1 1 d clx ly 0 0 0 0 lx ly d D DNx DNy DFx DFy T Procedure for Analysis The following method provides a means for determining the unknown displacements and support reactions for a truss using the stiffness method Notation Establish the x y global coordinate system The origin is usually located at the joint for which the coordinates for all the other joints are positive Identify each joint and member numerically and arbitrarily specify the near and far ends of each member symbolically by directing an arrow along the member with the head directed towards the far end Specify the two code numbers at each joint using the lowest numbers to identify unconstrained degrees of freedom followed by the highest numbers to identify the constrained degrees of freedom From the problem establish and Structure Stiffness Matrix For each member determine and and the member stiffness matrix using Eq 1416 Assemble these matrices to form the stiffness matrix for the entire truss as explained in Sec 145 As a partial check of the calculations the member and structure stiffness matrices should be symmetric Displacements and Loads Partition the structure stiffness matrix as indicated by Eq 1418 Determine the unknown joint displacements using Eq 1421 the support reactions using Eq 1422 and each member force using Eq 1423 qF Qu Du ly lx Qk Dk 554 CHAPTER 14 TRUSS ANALYSIS USING THE STIFFNESS METHOD 14 Determine the force in each member of the twomember truss shown in Fig 149a AE is constant SOLUTION Notation The origin of x y and the numbering of the joints and members are shown in Fig 149b Also the near and far ends of each member are identified by arrows and code numbers are used at each joint By inspection it is seen that the known external displacements are Also the known external loads are Hence Structure Stiffness Matrix Using the same notation as used here this matrix has been developed in Example 141 Displacements and Loads Writing Eq 1417 for the truss we have Q KD Dk D 0 0 0 0 T 3 4 5 6 Qk c 0 2 d 1 2 Q1 0 Q2 2 k D3 D4 D5 D6 0 EXAMPLE 143 1 F 0 2 Q3 Q4 Q5 Q6 V AE F 0405 0096 0333 0 0072 0096 0096 0128 0 0 0096 0128 0333 0 0333 0 0 0 0 0 0 0 0 0 0072 0096 0 0 0072 0096 0096 0128 0 0 0096 0128 V F D1 D2 0 0 0 0 V From this equation we can now identify and thereby determine It is seen that the matrix multiplication like Eq 1419 yields Here it is easy to solve by a direct expansion Physically these equations represent and applied to joint ② Solving we get D1 4505 AE D2 19003 AE Fy 0 Fx 0 2 AE10096D1 0128D22 0 AE10405D1 0096D22 c 0 2 d AE c0405 0096 0096 0128 d cD1 D2 d c0 0 d Du K11 Fig 149 3 ft 4 ft a 2 k 14 By inspection of Fig 149b one would indeed expect a rightward and downward displacement to occur at joint ② as indicated by the positive and negative signs of these answers Using these results the support reactions are now obtained from Eq 1 written in the form of Eq 1420 or Eq 1422 as Expanding and solving for the reactions The force in each member is found from Eq 1423 Using the data for and in Example 141 we have Member 1 Ans Member 2 Ans These answers can of course be verified by equilibrium applied at joint ② 1 5 06145052 081190032 25 k q2 AE 5 1 2 5 6 C 06 08 06 08D 1 AE D 4505 19003 0 0 T 1 2 5 6 ly 08 L 5 ft lx 06 1 3 4505 15 k q1 AE 3 1 2 3 4 C 1 0 1 0D 1 AE D 4505 19003 0 0 T 1 2 3 4 ly 0 L 3 ft lx 1 ly lx Q6 0096145052 01281190032 20 k Q5 0072145052 00961190032 15 k Q4 0 Q3 0333145052 15 k D Q3 Q4 Q5 Q6 T AE D 0333 0 0 0 0072 0096 0096 0128 T 1 AE c 4505 19003 d D 0 0 0 0 T 2 3 1 6 5 2 1 4 3 1 2 x y b 2 k 146 APPLICATION OF THE STIFFNESS METHOD FOR TRUSS ANALYSIS 555 EXAMPLE 144 Determine the support reactions and the force in member 2 of the truss shown in Fig 1410a AE is constant SOLUTION Notation The joints and members are numbered and the origin of the x y axes is established at ① Fig 1410b Also arrows are used to reference the near and far ends of each member Using the code numbers where the lowest numbers denote the unconstrained degrees of freedom Fig 1410b we have Structure Stiffness Matrix This matrix has been determined in Example 142 using the same notation as in Fig 1410b Displacements and Loads For this problem is Q KD Dk C 0 0 0 S 6 7 8 Qk E 0 0 2 4 0 U 1 2 3 4 5 556 CHAPTER 14 TRUSS ANALYSIS USING THE STIFFNESS METHOD 14 1 H 0 0 2 4 0 Q6 Q7 Q8 X AE H 0135 0035 0 0 0 01 0035 0035 0035 0135 0 01 0 0 0035 0035 0 0 0135 0035 0035 0035 01 0 0 01 0035 0135 0035 0035 0 0 0 0 0035 0035 0135 0035 0 01 01 0 0035 0035 0035 0135 0 0 0035 0035 01 0 0 0 0135 0035 0035 0035 0 0 01 0 0035 0135 X H D1 D2 D3 D4 D5 0 0 0 X 4 y 3 4 4 8 3 6 2 5 5 x 2 1 1 2 3 b 4 k 2 k 1 7 6 10 ft 10 ft a 2 k 4 k 2 Fig 1410 Multiplying so as to formulate the unknown displacement equation 1418 we get E 0 0 2 4 0 U AE E 0135 0035 0 0 0 0035 0135 0 01 0 0 0 0135 0035 0035 0 01 0035 0135 0035 0 0 0035 0035 0135 U E D1 D2 D3 D4 D5 U E 0 0 0 0 0 U 14 Expanding and solving the equations for the displacements yields Developing Eq 1420 from Eq 1 using the calculated results we have E D1 D2 D3 D4 D5 U 1 AE E 1794 6920 206 8714 2206 U Expanding and computing the support reactions yields C Q6 Q7 Q8 S AE C 01 0 0035 0035 0035 0035 0035 01 0 0 0035 0035 0 0 01 S 1 AE E 1794 6920 206 8714 2206 U C 0 0 0 S Ans Ans Ans The negative sign for indicates that the rocker support reaction acts in the negative x direction The force in member 2 is found from Eq 1423 where from Example 142 Thus Ans 256 k q2 AE 1022 0707 0707 0707 0707 1 AE D 1794 6920 0 0 T L 1012 ft ly 0707 lx 0707 Q6 Q8 40 k Q7 20 k Q6 40 k 146 APPLICATION OF THE STIFFNESS METHOD FOR TRUSS ANALYSIS 557 EXAMPLE 145 Determine the force in member 2 of the assembly in Fig 1411a if the support at joint ① settles downward 25 mm Take SOLUTION Notation For convenience the origin of the global coordinates in Fig 1411b is established at joint ③ and as usual the lowest code num bers are used to reference the unconstrained degrees of freedom Thus Structure Stiffness Matrix Using Eq 1416 we have Member 1 so that Member 2 so that Member 3 so that By assembling these matrices the structure stiffness matrix becomes K AE 1 2 3 4 5 6 7 8 H 0378 0096 0 0 0128 0096 025 0 0096 0405 0 0333 0096 0072 0 0 0 0 0 0 0 0 0 0 0 0333 0 0333 0 0 0 0 0128 0096 0 0 0128 0096 0 0 0096 0072 0 0 0096 0072 0 0 025 0 0 0 0 0 025 0 0 0 0 0 0 0 0 0 X 1 2 3 4 5 6 7 8 k3 AE 7 8 1 2 D 025 0 025 0 0 0 0 0 025 0 025 0 0 0 0 0 T 7 8 1 2 ly 0 L 4 m lx 1 k2 AE 1 2 5 6 D 0128 0096 0128 0096 0096 0072 0096 0072 0128 0096 0128 0096 0096 0072 0096 0072 T 1 2 5 6 ly 06 L 5 m lx 08 k1 AE 3 4 1 2 D 0 0 0 0 0 0333 0 0333 0 0 0 0 0 0333 0 0333 T 3 4 1 2 ly 1 L 3 m lx 0 Dk F 0 0025 0 0 0 0 V 3 4 5 6 7 8 Qk c0 0 d 1 2 AE 811032 kN 558 CHAPTER 14 TRUSS ANALYSIS USING THE STIFFNESS METHOD 14 7 1 3 2 1 2 1 2 4 3 4 8 6 5 3 b x y 1 3 2 1 4 m 3 m a 2 4 3 Fig 1411 14 Displacements and Loads Here yields Developing the solution for the displacements Eq 1419 we have c0 0d AE c0378 0096 0096 0405d cD1 D2 d AE c0 0 0128 0096 025 0 0 0333 0096 0072 0 0 d F 0 0025 0 0 0 0 V H 0 0 Q3 Q4 Q5 Q6 Q7 Q8 X AE H 0378 0096 0 0 0128 0096 025 0 0096 0405 0 0333 0096 0072 0 0 0 0 0 0 0 0 0 0 0 0333 0 0333 0 0 0 0 0128 0096 0 0 0128 0096 0 0 0096 0072 0 0 0096 0072 0 0 025 0 0 0 0 0 025 0 0 0 0 0 0 0 0 0 X H D1 D2 0 0025 0 0 0 0 X Q KD which yields Solving these equations simultaneously gives Although the support reactions do not have to be calculated if needed they can be found from the expansion defined by Eq 1420 Using Eq 1423 to determine the force in member 2 yields Member 2 so that Ans Using the same procedure show that the force in member 1 is and in member 3 The results are shown on the freebody diagram of joint ② Fig 1411c which can be checked to be in equilibrium q3 111 kN q1 834 kN 811032 5 1000444 001312 139 kN q2 811032 5 08 06 08 06D 000556 0021875 0 0 T ly 06 L 5 m AE 811032 kN lx 08 D2 0021875 m D1 000556 m 0 AE10096D1 0405D22 000833 0 AE10378D1 0096D22 0 2 111 kN 834 kN 139 kN 3 4 5 c 146 APPLICATION OF THE STIFFNESS METHOD FOR TRUSS ANALYSIS 559 147 Nodal Coordinates On occasion a truss can be supported by a roller placed on an incline and when this occurs the constraint of zero deflection at the support node cannot be directly defined using a single horizontal and vertical global coordinate system For example consider the truss in Fig 1412a The condition of zero displacement at node ① is defined only along the axis and because the roller can displace along the axis this node will have displacement components along both global coordinate axes x y For this reason we cannot include the zero displacement condition at this node when writing the global stiffness equation for the truss using x y axes without making some modifications to the matrix analysis procedure To solve this problem so that it can easily be incorporated into a computer analysis we will use a set of nodal coordinates located at the inclined support These axes are oriented such that the reactions and support displacements are along each of the coordinate axes Fig 1412a In order to determine the global stiffness equation for the truss it then becomes necessary to develop force and displacement transformation matrices for each of the connecting members at this support so that the results can be summed within the same global x y coordinate system To show how this is done consider truss member 1 in Fig 1412b having a global coordinate system x y at the near node and a nodal coordinate system at the far node When displacements D occur so that they have components along each of these axes as shown in Fig 1412c the displacements d in the direction along the ends of the member become dF DFxfl cos uxfl DFyfl cos uyfl dN DNx cos ux DNy cos uy x F x y N y x x y 560 CHAPTER 14 TRUSS ANALYSIS USING THE STIFFNESS METHOD 14 y x x y a 2 3 3 2 1 1 Fig 1412 147 NODAL COORDINATES 561 14 These equations can be written in matrix form as Likewise forces q at the near and far ends of the member Fig 1412d have components Q along the global axes of which can be expressed as The displacement and force transformation matrices in the above equations are used to develop the member stiffness matrix for this situationApplying Eq 1415 we have Performing the matrix operations yields 1424 This stiffness matrix is then used for each member that is connected to an inclined roller support and the process of assembling the matrices to form the structure stiffness matrix follows the standard procedure The following example problem illustrates its application k AE L D lx 2 lxly lxlxfl lxlyfl lxly ly 2 lylxfl lylyfl lxlxfl lylxfl lxfl 2 lxfllyfl lxlyfl lylyfl lxfllyfl lyfl 2 T k D lx 0 ly 0 0 lxfl 0 lyfl T AE L c 1 1 1 1 d clx ly 0 0 0 0 lxfl lyfl d k TTkT D QNx QNy QFxfl QFyfl T D lx 0 ly 0 0 lxfl 0 lyfl T cqN qF d QFxfl qF cos uxfl QFyfl qF cos uyfl QNx qN cos ux QNy qN cos uy cdN dF d clx ly 0 0 0 0 lxfl lyfl d D DNx DNy DFxfl DFyfl T y y x x y x global coordinates nodal coordinates local coordinates b F N y x F N x y x global coordinates c DNy DFy DFx DNy cos uy DNx cos ux DFx cos ux DFy cos uy DNx uy ux uy ux y x y x x d qN qF F N QNy QFx QFy QNx uy ux ux uy 14 Structure Stiffness Matrix Assembling these matrices to determine the structure stiffness matrix we have 1 F 30 0 0 Q4 Q5 Q6 V AE F 0128 0096 0 0 0128 0096 0096 04053 02357 02357 0096 0072 0 02357 02917 00417 017675 0 0 02357 00417 02917 017675 0 0128 0096 017675 017675 0378 0096 0096 0072 0 0 0096 0072 V F D1 D2 D3 0 0 0 V Carrying out the matrix multiplication of the upper partitioned matrices the three unknown displacements D are determined from solving the resulting simultaneous equations ie The unknown reactions Q are obtained from the multiplication of the lower partitioned matrices in Eq 1 Using the computed displacements we have Ans Ans Ans 225 kN Q6 0096135252 0072115752 0112732 75 kN Q5 0128135252 0096115752 017675 112732 318 kN Q4 0135252 02357115752 00417112732 D3 1273 AE D2 1575 AE D1 3525 AE 147 NODAL COORDINATES 563 148 Trusses Having Thermal Changes and Fabrication Errors If some of the members of the truss are subjected to an increase or decrease in length due to thermal changes or fabrication errors then it is necessary to use the method of superposition to obtain the solution This requires three steps First the fixedend forces necessary to prevent movement of the nodes as caused by temperature or fabrication are calculated Second the equal but opposite forces are placed on the truss at the nodes and the displacements of the nodes are calculated using the matrix analysis Finally the actual forces in the members and the reactions on the truss are determined by superposing these two results This procedure of course is only necessary if the truss is statically indeterminate If the truss is statically determinate the displacements at the nodes can be found by this method however the temperature changes and fabrication errors will not affect the reactions and the member forces since the truss is free to adjust to these changes of length Thermal Effects If a truss member of length L is subjected to a temperature increase the member will undergo an increase in length of where is the coefficient of thermal expansion A compressive force applied to the member will cause a decrease in the members length of If we equate these two displacements then This force will hold the nodes of the member fixed as shown in Fig 1414 and so we have Realize that if a temperature decrease occurs then becomes negative and these forces reverse direction to hold the member in equilibrium We can transform these two forces into global coordinates using Eq 1410 which yields 1425 Fabrication Errors If a truss member is made too long by an amount before it is fitted into a truss then the force needed to keep the member at its design length L is and so for the member in Fig 1414 we have 1qF20 AEL L 1qN20 AEL L q0 AELL q0 L D 1QNx20 1QNy20 1QFx20 1QFy20 T D lx 0 ly 0 0 lx 0 ly TAEaTc 1 1 d AEaTD lx ly lx ly T T 1qF20 AEaT 1qN20 AEaT q0 AEaT L q0LAE q0 a L aTL T 564 CHAPTER 14 TRUSS ANALYSIS USING THE STIFFNESS METHOD 14 Fig 1414 y x qN0 qF0 L F N 148 TRUSSES HAVING THERMAL CHANGES AND FABRICATION ERRORS 565 14 If the member is originally too short then becomes negative and these forces will reverse In global coordinates these forces are 1426 Matrix Analysis In the general case with the truss subjected to applied forces temperature changes and fabrication errors the initial forcedisplacement relationship for the truss then becomes 1427 Here is a column matrix for the entire truss of the initial fixedend forces caused by the temperature changes and fabrication errors of the members defined in Eqs 1425 and 1426We can partition this equation in the following form Carrying out the multiplication we obtain 1428 1429 According to the superposition procedure described above the unknown displacements are determined from the first equation by subtracting and from both sides and then solving for This yields Once these nodal displacements are obtained the member forces are then determined by superposition ie If this equation is expanded to determine the force at the far end of the member we obtain 1430 This result is similar to Eq 1423 except here we have the additional term which represents the initial fixedend member force due to temperature changes andor fabrication error as defined previously Re alize that if the computed result from this equation is negative the mem ber will be in compression The following two examples illustrate application of this procedure 1qF20 qF AE L lx ly lx ly D DNx DNy DFx DFy T 1qF20 q kTD q0 Du K11 11Qk K12Dk 1Qk202 Du 1Qk20 K12Dk Du Qu K21Du K22Dk 1Qu20 Qk K11Du K12Dk 1Qk20 cQk Qu d cK11 K12 K21 K22 d cDu Dk d c1Qk20 1Qu20 d Q0 Q KD Q0 D 1QNx20 1QNy20 1QFx20 1QFy20 T AEL L D lx ly lx ly T L EXAMPLE 147 Determine the force in members 1 and 2 of the pinconnected assembly of Fig 1415 if member 2 was made 001 m too short before it was fitted into placeTake AE 811032 kN 566 CHAPTER 14 TRUSS ANALYSIS USING THE STIFFNESS METHOD 14 SOLUTION Since the member is short then and therefore applying Eq 1426 to member 2 with we have The structure stiffness matrix for this assembly has been established in Example 145Applying Eq 1427 we have D 1Q120 1Q220 1Q520 1Q620 T AE10012 5 D 08 06 08 06 T AE D 00016 00012 00016 00012 T 1 2 5 6 lx 08 ly 06 L 001 m 1 H 0 0 Q3 Q4 Q5 Q6 Q7 Q8 X AE H 0378 0096 0 0 0128 0096 025 0 0096 0405 0 0333 0096 0072 0 0 0 0 0 0 0 0 0 0 0 0333 0 0333 0 0 0 0 0128 0096 0 0 0128 0096 0 0 0096 0072 0 0 0096 0072 0 0 025 0 0 0 0 0 025 0 0 0 0 0 0 0 0 0 X H D1 D2 0 0 0 0 0 0 X AE H 00016 00012 0 0 00016 00012 0 0 X Fig 1415 1 3 2 1 4 m 3 m 2 4 3 2 8 6 y 1 3 5 x 4 7 14 Partitioning the matrices as shown and carrying out the multiplication to obtain the equations for the unknown displacements yields c0 0 d AE c0378 0096 0096 0405 d cD1 D2 d AE c0 0 0128 0096 025 0 0 0333 0096 0072 0 0d F 0 0 0 0 0 0 V AE c00016 00012 d which gives Solving these equations simultaneously Although not needed the reactions Q can be found from the expan sion of Eq 1 following the format of Eq 1429 In order to determine the force in members 1 and 2 we must apply Eq 1430 in which case we have Member 1 so that Ans Member 2 ly 06 L 5 m AE 811032 kN so lx 08 q1 556 kN q1 811032 3 0 1 0 1D 0 0 0003704 0002084 T 0 ly 1 L 3 m AE 811032 kN lx 0 D2 0002084 m D1 0003704 m 0 AE0096D1 0405D2 AE0 AE00012 0 AE0378D1 0096D2 AE0 AE00016 Ans q2 926 kN q2 811032 5 08 06 08 06D 0003704 0002084 0 0 T 811032 10012 5 148 TRUSSES HAVING THERMAL CHANGES AND FABRICATION ERRORS 567 EXAMPLE 148 Member 2 of the truss shown in Fig 1416 is subjected to an increase in temperature of 150F Determine the force developed in member 2 Take Each member has a cross sectional area of A 075 in2 E 2911062 lbin2 a 6511062F 568 CHAPTER 14 TRUSS ANALYSIS USING THE STIFFNESS METHOD 14 SOLUTION Since there is a temperature increase Applying Eq 1425 to member 2 where we have The stiffness matrix for this truss has been developed in Example 142 D 1Q120 1Q220 1Q720 1Q820 T AE1652 11062 11502D 07071 07071 07071 07071 T AE D 0000689325 0000689325 0000689325 0000689325 T 1 2 7 8 lx 07071 ly 07071 T 150F 10 ft 4 3 1 4 y 8 3 10 ft 2 1 3 4 2 6 2 5 5 x 6 7 1 1 H 0 0 0 0 0 Q6 Q7 Q8 X AE H 0135 0035 0 0 0 01 0035 0035 0035 0135 0 01 0 0 0035 0035 0 0 0135 0035 0035 0035 01 0 0 01 0035 0135 0035 0035 0 0 0 0 0035 0035 0135 0035 0 01 01 0 0035 0035 0035 0135 0 0 0035 0035 01 0 0 0 0135 0035 0035 0035 0 0 01 0 0035 0135 X H D1 D2 D3 D4 D5 0 0 0 X AE H 0000689325 0000689325 0 0 0 0 0000689325 0000689325 X 1 2 3 4 5 6 7 8 Fig 1416 14 Expanding to determine the equations of the unknown displacements and solving these equations simultaneously yields Using Eq 1430 to determine the force in member 2 we have D5 0002027 ft D4 0009848 ft D3 0002027 ft D2 001187 ft D1 0002027 ft Ans 6093 lb 609 k q2 0752911062 1022 0707 0707 0707 0707D 0002027 001187 0 0 T 0752911062651106211502 Note that the temperature increase of member 2 will not cause any reactions on the truss since externally the truss is statically determinate To show this consider the matrix expansion of Eq 1 for determining the reactions Using the results for the displacements we have 0 01100020272 AE0000689325 0 Q8 AE0035100020272 003510011872 0 01100020272 0 0 AE0000689325 0 Q7 AE0035100020272 003510011872 0035100098482 0035100020272 AE0 0 Q6 AE01100020272 0 0035100020272 148 TRUSSES HAVING THERMAL CHANGES AND FABRICATION ERRORS 569 149 SpaceTruss Analysis The analysis of both statically determinate and indeterminate space trusses can be performed by using the same procedure discussed previously To account for the threedimensional aspects of the problem however additional elements must be included in the transformation matrix T In this regard consider the truss member shown in Fig 1417 The stiffness matrix for the member defined in terms of the local coordinate is given by Eq 144 Furthermore by inspection of Fig 1417 the direction cosines between the global and local coordinates can be found using equations analogous to Eqs 145 and 146 that is 1431 1432 1433 As a result of the third dimension the transformation matrix Eq 149 becomes Substituting this and Eq 144 into Eq 1415 yields k F lx 0 ly 0 lz 0 0 lx 0 ly 0 lz V AE L c 1 1 1 1 d clx ly lz 0 0 0 0 0 0 lx ly lz d k TTkT T clx ly lz 0 0 0 0 0 0 lx ly lz d zF zN 21xF xN22 1yF yN22 1zF zN22 lz cos uz zF zN L yF yN 21xF xN22 1yF yN22 1zF zN22 ly cos uy yF yN L xF xN 21xF xN22 1yF yN22 1zF zN22 lx cos ux xF xN L x 570 CHAPTER 14 TRUSS ANALYSIS USING THE STIFFNESS METHOD 14 F x L 6 5 4 3 2 1 N x z y yN xN yF xF zF zN uy ux uz Fig 1417 Carrying out the matrix multiplication yields the symmetric matrix k AE L Nx Ny Nz Fx Fy Fz F lx 2 lxly lxlz lylx ly 2 lylz lzlx lzly lz 2 lx 2 lxly lxlz lylx ly 2 lylz lzlx lzly lz 2 lx 2 lxly lxlz lylx ly 2 lylz lzlx lzly lz 2 lx 2 lxly lxlz lylx ly 2 lylz lzlx lzly lz 2 V Nx Ny Nz Fx Fy Fz 14 1434 This equation represents the member stiffness matrix expressed in global coordinatesThe code numbers along the rows and columns reference the x y z directions at the near end followed by those at the far end For computer programming it is generally more efficient to use Eq 1434 than to carry out the matrix multiplication for each member Computer storage space is saved if the structure stiffness matrix K is first initialized with all zero elements then as the elements of each member stiffness matrix are generated they are placed directly into their respective positions in K After the structure stiffness matrix has been developed the same procedure outlined in Sec 146 can be followed to determine the joint displacements support reactions and internal member forces TTkT Fx Fy Fz Ny Nz Nx CHAPTER REVIEW The stiffness method is the preferred method for analyzing structures using a computer It first requires identifying the number of structural elements and their nodes The global coordinates for the entire structure are then established and each members local coordinate system is located so that its origin is at a selected near end such that the positive axis extends towards the far end Formulation of the method first requires that each member stiffness matrix be constructed It relates the loads at the ends of the member q to their displacements d where Then using the transformation matrix T the local displacements d are related to the global displacements D where Also the local forces q are transformed into the global forces Q using the transformation matrix T ie When these matrices are combined one obtains the members stiffness matrix K in global coordinates Assembling all the member stiffness matrices yields the stiffness matrix K for the entire structure The displacements and loads on the structure are then obtained by partitioning such that the unknown displacements are obtained from provided the supports do not displace Finally the support reactions are obtained from and each member force is found from q kTD Qu K21 Du Du 3K1141Qk Q KD k TTkT Q TTq d TD q kd k x The structural framework of this aircraft hangar is constructed entirely of trusses in order to reduce significantly the weight of the structure Courtesy of Bethlehem Steel Corporation CHAPTER REVIEW 571 572 CHAPTER 14 TRUSS ANALYSIS USING THE STIFFNESS METHOD 14 144 Determine the stiffness matrix K for the trussTake ksi 145 Determine the horizontal displacement of joint ① and the force in member Take ksi 146 Determine the force in member if its temperature is increased by Take a 65106F A 075 in2 E 29103 ksi 100F ƒ 2 ƒ E 29103 A 075 in2 ƒ 2 ƒ A 075 in2 E 29103 149 Determine the stiffness matrix K for the truss Take and GPa for each member 1410 Determine the force in member Take and GPa for each member 1411 Determine the vertical displacement of node ② if member was 10 mm too long before it was fitted into the truss For the solution remove the 20k load Take and E 200 GPa for each member A 00015 m2 ƒ 6 ƒ E 200 A 00015 m2 ƒ 5 ƒ E 200 A 00015 m2 147 Determine the stiffness matrix K for the truss Take and GPa for each member 148 Determine the vertical displacement at joint ② and the force in member Take and E 200 GPa A 00015 m2 ƒ 5 ƒ E 200 A 00015 m2 141 Determine the stiffness matrix K for the assembly Take and ksi for each member 142 Determine the horizontal and vertical displacements at joint ③ of the assembly in Prob 141 143 Determine the force in each member of the assem bly in Prob 141 E 29103 A 05 in2 PROBLEMS Probs 144145146 Probs 147148 Probs 14914101411 Probs 141142143 10 9 3 3 6 5 4 4 3 m 4 m 4 m 20 kN 4 7 6 1 2 2 1 5 3 5 7 8 1 2 2 4 ft 3 ft 4 ft 1 2 500 lb 8 7 6 4 5 3 2 3 4 1 1 3 7 2 1 3 4 ft 6 ft 3 ft 3 ft 4 k 3 3 5 8 1 6 4 2 4 1 2 30 kN 10 9 3 3 6 5 4 4 2 m 2 m 2 m 4 6 1 1 2 5 3 5 7 1 2 2 8 The statically indeterminate loading in bridge girders that are continuous over their piers can be determined using the stiffness method 15 575 The concepts presented in the previous chapter will be extended here and applied to the analysis of beams It will be shown that once the member stiffness matrix and the transformation matrix have been developed the procedure for application is exactly the same as that for trusses Special consideration will be given to cases of differential settlement and temperature 151 Preliminary Remarks Before we show how the stiffness method applies to beams we will first discuss some preliminary concepts and definitions related to these members Member and Node Identification In order to apply the stiffness method to beams we must first determine how to subdivide the beam into its component finite elements In general each element must be free from load and have a prismatic cross section For this reason the nodes of each element are located at a support or at points where members are connected together where an external force is applied where the crosssectional area suddenly changes or where the vertical or rotational displacement at a point is to be determined For example consider the beam in Fig 151a Using the same scheme as that for trusses four nodes are specified numerically within a circle and the three elements are identified numerically within a squareAlso notice that the near and far ends of each element are identified by the arrows written alongside each element Beam Analysis Using the Stiffness Method a x y 2 3 1 1 3 5 6 2 4 7 1 8 3 P 2 4 Fig 151 Global and Member Coordinates The global coordinate system will be identified using x y z axes that generally have their origin at a node and are positioned so that the nodes at other points on the beam all have positive coordinates Fig 151a The local or member coordinates have their origin at the near end of each element and the positive axis is directed towards the far end Figure 151b shows these coordinates for element 2 In both cases we have used a righthanded coordinate system so that if the fingers of the right hand are curled from the axis towards the axis the thumb points in the positive direction of the axis which is directed out of the page Notice that for each beam element the x and axes will be collinear and the global and member coordinates will all be parallel Therefore unlike the case for trusses here we will not need to develop transformation matrices between these coordinate systems Kinematic Indeterminacy Once the elements and nodes have been identified and the global coordinate system has been established the degrees of freedom for the beam and its kinematic determinacy can be determined If we consider the effects of both bending and shear then each node on a beam can have two degrees of freedom namely a vertical displacement and a rotation As in the case of trusses these linear and rotational displacements will be identified by code numbers The lowest code numbers will be used to identify the unknown displacements unconstrained degrees of freedom and the highest numbers are used to identify the known displacements constrained degrees of freedomRecall that the reason for choosing this method of identification has to do with the convenience of later partitioning the structure stiffness matrix so that the unknown displacements can be found in the most direct manner To show an example of codenumber labeling consider again the con tinuous beam in Fig 151a Here the beam is kinematically indeterminate to the fourth degree There are eight degrees of freedom for which code numbers 1 through 4 represent the unknown displacements and numbers 5 through 8 represent the known displacements which in this case are all zero As another example the beam in Fig 152a can be subdivided into three elements and four nodes In particular notice that the internal hinge at node 3 deflects the same for both elements 2 and 3 however the rotation at the end of each element is different For this reason three code numbers are used to show these deflections Here there are nine degrees of freedom five of which are unknown as shown in Fig 152b and four known again they are all zero Finally consider the slider mechanism used on the beam in Fig 153a Here the deflection of the beam is shown in Fig 153b and so there are five unknown deflection components labeled with the lowest code numbers The beam is kinematically indeterminate to the fifth degree Development of the stiffness method for beams follows a similar procedure as that used for trusses First we must establish the stiffness matrix for each element and then these matrices are combined to form the beam or structure stiffness matrix Using the structure x z 1z2 y 1y2 x 1x2 x y z x 576 CHAPTER 15 BEAM ANALYSIS USING THE STIFFNESS METHOD 15 2 x y b 2 3 1 2 3 3 2 4 P 5 8 4 1 7 6 2 3 a 1 9 D1 D2 D4 D5 D3 b 1 2 3 1 3 2 4 P 4 6 5 7 9 3 8 12 a D4 D2 D3 D3 D1 D5 b Fig 152 Fig 153 Fig 151 154 APPLICATION OF THE STIFFNESS METHOD FOR BEAM ANALYSIS 579 15 153 BeamStructure Stiffness Matrix Once all the member stiffness matrices have been found we must assemble them into the structure stiffness matrix K This process depends on first knowing the location of each element in the member stiffness matrix Here the rows and columns of each k matrix Eq 151 are identified by the two code numbers at the near end of the member followed by those at the far end Therefore when assembling the matrices each element must be placed in the same location of the K matrix In this way K will have an order that will be equal to the highest code number assigned to the beam since this represents the total number of degrees of freedom Also where several members are connected to a node their member stiffness influence coefficients will have the same position in the K matrix and therefore must be algebraically added together to determine the nodal stiffness influence coefficient for the structure This is necessary since each coefficient represents the nodal resistance of the structure in a particular direction or when a unit displacement or occurs either at the same or at another node For example represents the load in the direction and at the location of code number 2 when a unit displacement occurs in the direction and at the location of code number 3 154 Application of the Stiffness Method for Beam Analysis After the structure stiffness matrix is determined the loads at the nodes of the beam can be related to the displacements using the structure stiffness equation Here Q and D are column matrices that represent both the known and unknown loads and displacements Partitioning the stiffness matrix into the known and unknown elements of load and displacement we have which when expanded yields the two equations 153 154 The unknown displacements are determined from the first of these equations Using these values the support reactions are computed for the second equation Qu Du Qu K21Du K22Dk Qk K11Du K12Dk cQk Qu d cK11 K12 K21 K22 d cDu Dk d Q KD K23 z y z y 1Fy Fz2 1Ny Nz2 154 APPLICATION OF THE STIFFNESS METHOD FOR BEAM ANALYSIS 581 15 Procedure for Analysis The following method provides a means of determining the displacements support reactions and internal loadings for the members or finite elements of a statically determinate or statically indeterminate beam Notation Divide the beam into finite elements and arbitrarily identify each element and its nodes Use a number written in a circle for a node and a number written in a square for a member Usually an element extends between points of support points of concentrated loads and joints or to points where internal loadings or displacements are to be determinedAlso E and I for the elements must be constants Specify the near and far ends of each element symbolically by directing an arrow along the element with the head directed toward the far end At each nodal point specify numerically the y and z code numbers In all cases use the lowest code numbers to identify all the unconstrained degrees of freedom followed by the remaining or highest numbers to identify the degrees of freedom that are constrained From the problem establish the known displacements and known external loads Include any reversed fixedend loadings if an element supports an intermediate load Structure Stiffness Matrix Apply Eq 151 to determine the stiffness matrix for each element expressed in global coordinates After each member stiffness matrix is determined and the rows and columns are identified with the appropriate code numbers assemble the matrices to determine the structure stiffness matrix K As a partial check the member and structure stiffness matrices should all be symmetric Displacements and Loads Partition the structure stiffness equation and carry out the matrix multiplication in order to determine the unknown displacements and support reactions The internal shear and moment q at the ends of each beam element can be determined from Eq 155 accounting for the additional fixedend loadings Qu Du Qk Dk EXAMPLE 151 582 CHAPTER 15 BEAM ANALYSIS USING THE STIFFNESS METHOD 15 k1 EI 6 4 5 3 D 15 15 15 15 15 2 15 1 15 15 15 15 15 1 15 2 T 6 4 35 k2 EI 5 3 2 1 D 15 15 15 15 15 2 15 1 15 15 15 15 15 1 15 2 T 5 3 2 1 Determine the reactions at the supports of the beam shown in Fig 158a EI is constant a 5 kN 2 m 2 m SOLUTION Notation The beam has two elements and three nodes which are identified in Fig 158b The code numbers 1 through 6 are indicated such that the lowest numbers 14 identify the unconstrained degrees of freedom The known load and displacement matrices are Qk D 0 5 0 0 T 1 2 3 4 Dk c0 0 d 5 6 b 1 2 6 5 2 1 3 4 3 1 5 kN 2 Member Stiffness Matrices Each of the two member stiffness matrices is determined from Eq 151 Note carefully how the code numbers for each column and row are established Fig 158 15 154 APPLICATION OF THE STIFFNESS METHOD FOR BEAM ANALYSIS 583 Displacements and Loads We can now assemble these elements into the structure stiffness matrix For example element etcThus The matrices are partitioned as shown Carrying out the multiplication for the first four rows we have Solving Using these results and multiplying the last two rows gives Ans Ans 5 kN Q6 0 0 15EIa 667 EI b 15EIa333 EI b 10 kN Q5 15EIa 1667 EI b 15EIa 2667 EI b 0 15EIa333 EI b D4 333 EI D3 667 EI D2 2667 EI D1 1667 EI 0 0 0 D3 2D4 0 D1 15D2 4D3 D4 5 EI 15D1 15D2 15D3 0 0 2D1 15D2 D3 0 F 0 5 0 0 Q5 Q6 V EI 1 2 3 4 5 6 F 2 15 1 0 15 0 15 15 15 0 15 0 1 15 4 1 0 15 0 0 1 2 15 15 15 15 0 15 3 15 0 0 15 15 15 15 V F D1 D2 D3 D4 0 0 V Q KD K55 15 15 3 K11 0 2 2 EXAMPLE 152 Determine the internal shear and moment in member 1 of the compound beam shown in Fig 159a EI is constant SOLUTION Notation When the beam deflects the internal pin will allow a single deflection however the slope of each connected member will be differentAlso a slope at the roller will occurThese four unknown degrees of freedom are labeled with the code numbers 1 2 3 and 4 Fig 159b Member Stiffness Matrices Applying Eq 151 to each member in accordance with the code numbers shown in Fig 159b we have Qk 0 0 0 M0 1 2 3 4 Dk C 0 0 0 S 5 6 7 584 CHAPTER 15 BEAM ANALYSIS USING THE STIFFNESS METHOD 15 k2 EI 3 2 5 4 H 12 L3 6 L2 12 L3 6 L2 6 L2 4 L 6 L2 2 L 12 L3 6 L2 12 L3 6 L2 6 L2 2 L 6 L2 4 L X 3 2 5 4 k1 EI 6 7 3 1 H 12 L3 6 L2 12 L3 6 L2 6 L2 4 L 6 L2 2 L 12 L3 6 L2 12 L3 6 L2 6 L2 2 L 6 L2 4 L X 6 7 3 1 Displacements and Loads The structure stiffness matrix is formed by assembling the elements of the member stiffness matricesApplying the structure matrix equation we have Q KD Fig 159 1 2 L L M0 b 1 2 L L 6 3 5 4 2 1 7 2 3 1 M0 a 15 154 APPLICATION OF THE STIFFNESS METHOD FOR BEAM ANALYSIS 585 Multiplying the first four rows to determine the displacement yields So that Using these results the reaction is obtained from the multiplication of the fifth row Ans This result can be easily checked by statics applied to member 2 Q5 M0 L Q5 6EI L2 a M0L 6EI b 12EI L3 a M0L2 3EI b 6EI L2 a 2M0L 3EI b Q5 D4 2M0L 3EI D3 M0L2 3EI D2 M0L 6EI D1 M0L 2EI M0 2 LD2 6 L2D3 4 LD4 0 6 L2D1 6 L2D2 24 L3D3 6 L2D4 0 4 LD2 6 L2D3 2 LD4 0 4 LD1 6 L2D3 EI 1 2 3 4 5 6 7 4 L 0 6 L2 0 0 6 L2 2 L 0 4 L 6 L2 2 L 6 L2 0 0 6 L2 6 L2 24 L3 6 L2 12 L3 12 L3 6 L2 0 2 L 6 L2 4 L 6 L2 0 0 0 6 L2 12 L3 6 L2 12 L3 0 0 6 L2 0 12 L3 0 0 12 L3 6 L2 2 L 0 6 L2 0 0 6 L2 4 L D1 D2 D3 D4 0 0 0 1 2 3 4 5 6 7 0 0 0 M0 Q5 Q6 Q7 1 2 3 4 5 6 7 15 154 APPLICATION OF THE STIFFNESS METHOD FOR BEAM ANALYSIS 587 Displacements and Loads Assembling the structure stiffness matrix and writing the stiffness equation for the structure yields Solving for the unknown displacements Substituting and solving Using these results the support reactions are therefore D3 0001580 rad D2 0 D1 0001580 rad EI 20011062122211062 4 EI 0D1 1D2 2D3 0 151000152 0 0 1D1 4D2 1D3 15102 0 0 4 EI 2D1 D2 0D3 15102 151000152 0 F 4 0 4 Q4 Q5 Q6 V EI 1 2 3 4 5 6 F 2 1 0 15 15 0 1 4 1 15 0 15 0 1 2 0 15 15 15 15 0 15 15 0 15 0 15 15 3 15 0 15 15 0 15 15 V F D1 D2 D3 0 00015 0 V k2 EI 5 2 4 1 D 15 15 15 15 15 2 15 1 15 15 15 15 15 1 15 2 T 5 2 4 1 k1 EI 6 3 5 2 D 15 15 15 15 15 2 15 1 15 15 15 15 15 1 15 2 T 6 3 5 2 Ans Ans Ans Q6 2001106222110620 15102 15100015802 0 151000152 15102 0525 kN Q5 20011062221106215100015802 0 15100015802 15102 31000152 15102 105 kN Q4 20011062221106215100015802 15102 0 15102 151000152 0 0525 kN 15 154 APPLICATION OF THE STIFFNESS METHOD FOR BEAM ANALYSIS 589 Displacements and Loads We require Q KD k2 5 2 6 1 D 200602 962891 200602 962891 962891 616 250 962891 308 125 200602 962891 200602 962891 962891 308 125 962891 616 250 T 5 2 6 1 2EI L 212921103215102 81122 308 125 4EI L 412921103215102 81122 616 250 F 144 1008 Q3 Q4 Q5 Q6 V 1 2 3 4 5 6 F 616 250 308 125 0 0 962891 962891 308 125 821 667 102 708 10699 855901 962891 0 102 708 205 417 10699 10699 0 0 10699 10699 7430 7430 0 962891 855901 10699 7430 20803 200602 962891 962891 0 0 200602 200602 V F D1 D2 0 0 0 0 V Solving in the usual manner Thus The actual moment at A must include the fixedsupported reaction of shown in Fig 1511c along with the calculated result for Thus Ans This result compares with that determined in Example 112 Although not required here we can determine the internal moment and shear at B by considering for example member 1 node 2 Fig 1511bThe result requires expanding D q4 q3 q5 q2 T 4 3 5 2 D 7430 10699 7430 10699 10699 205 417 10699 102 708 7430 10699 7430 10699 10699 102 708 10699 205 417 T D 0 0 0 140203 T 11032 D 24 1152 24 1152 T q1 k1d 1q021 MAB 12 k ft 96 k ft 108 k ftg Q3 96 k ft Q3 0 102 7081140203211032 144 k in 12 k ft D2 14020311032 in D1 0467311032 in 1008 308 125D1 821 667D2 144 616 250D1 308 125D2 Determine the deflection at ① and the reactions on the beam shown in Fig 1512a EI is constant 590 CHAPTER 15 BEAM ANALYSIS USING THE STIFFNESS METHOD 15 EXAMPLE 155 SOLUTION Notation The beam is divided into two elements and the nodes and members are identified along with the directions from the near to far ends Fig 1512b The unknown deflections are shown in Fig 1512c In particular notice that a rotational displacement does not occur because of the roller constraint D4 Member Stiffness Matrices Since EI is constant and the members are of equal length the member stiffness matrices are identical Using the code numbers to identify each column and row in accordance with Eq 151 and Fig 1512b we have k2 EI 1 2 5 6 D 15 15 15 15 15 2 15 1 15 15 15 15 15 1 15 2 T 1 2 5 6 k1 EI 3 4 1 2 D 15 15 15 15 15 2 15 1 15 15 15 15 15 1 15 2 T 3 4 1 2 2 m 2 m a 1 P 4 2 6 3 5 1 b 1 3 2 P 1 2 Fig 1512 15 154 APPLICATION OF THE STIFFNESS METHOD FOR BEAM ANALYSIS 591 Displacements and Loads Assembling the member stiffness matrices into the structure stiffness matrix and applying the structure stiffness matrix equation we have Solving for the displacements yields Ans Note that the signs of the results match the directions of the deflections shown in Fig 1512c Using these results the reactions therefore are D3 2667P EI D2 P EI D1 1667P EI 0 15D1 15D2 15D3 0 0D1 4D2 15D3 P EI 3D1 0D2 15D3 F P 0 0 Q4 Q5 Q6 V EI 1 2 3 4 5 6 F 3 0 15 15 15 15 0 4 15 1 15 1 15 15 15 15 0 0 15 1 15 2 0 0 15 15 0 0 15 15 15 1 0 0 15 2 V F D1 D2 D3 0 0 0 V Q KD c D3 D1 D2 Ans Ans Ans 15P Q6 15EIa 1667P EI b 1EIa P EIb 0a 2667P EI b P Q5 15EIa 1667P EI b 15EIa P EIb 0a 2667P EI b 05P Q4 15EIa 1667P EI b 1EIa P EIb 15EIa 2667P EI b 15 154 APPLICATION OF THE STIFFNESS METHOD FOR BEAM ANALYSIS 593 Prob 158 Prob 159 Prob 1510 Prob 1511 Prob 1513 Prob 1512 158 Determine the reactions at the supports EI is constant 1511 Determine the reactions at the supports There is a smooth slider at ① EI is constant 159 Determine the moments at ② and ③ EI is constant Assume ① ② and ③ are rollers and ④ is pinned 1512 Use a computer program to determine the reactions on the beamAssume A is fixed EI is constant 1510 Determine the reactions at the supportsAssume ② is pinned and ① and ③ are rollers EI is constant 1513 Use a computer program to determine the reactions on the beam Assume A and D are pins and B and C are rollers EI is constant 1 2 6 7 4 3 2 1 5 4 m 15 kNm 3 m 1 2 3 12 m 12 m 12 m 4 kNm 2 3 1 2 1 1 2 3 4 3 4 A B C D 8 ft 8 ft 20 ft 3 kft 8 ft 8 ft 4 ft 4 ft 3 kft 1 1 4 5 6 2 2 3 3 1 2 8 ft 8 ft 15 ft 20 ft A D C B 12 k 4 kft 1 2 1 3 1 2 4 30 kNm 4 m The frame of this building is statically indeterminate The force analysis can be done using the stiffness method 16 595 The concepts presented in the previous chapters on trusses and beams will be extended in this chapter and applied to the analysis of frames It will be shown that the procedure for the solution is similar to that for beams but will require the use of transformation matrices since frame members are oriented in different directions 161 FrameMember Stiffness Matrix In this section we will develop the stiffness matrix for a prismatic frame member referenced from the local coordinate system Fig 161 Here the member is subjected to axial loads shear loads and bending moments at its near and far ends respectively These loadings all act in the positive coordinate directions along with their associated displacements As in the case of beams the moments and are positive counterclockwise since by the righthand rule the moment vectors are then directed along the positive axis which is out of the page We have considered each of the loaddisplacement relationships caused by these loadings in the previous chaptersThe axial load was discussed in reference to Fig 142 the shear load in reference to Fig 155 and the bending moment in reference to Fig 156 By superposition if these z qFz qNz qNz qFz qNy qFy qNx qFx y z x Plane Frame Analysis Using the Stiffness Method 600 CHAPTER 16 PLANE FRAME ANALYSIS USING THE STIFFNESS METHOD 16 164 Application of the Stiffness Method for Frame Analysis Once the member stiffness matrices are established they may be assembled into the structure stiffness matrix in the usual manner By writing the structure matrix equation the displacements at the unconstrained nodes can be determined followed by the reactions and internal loadings at the nodes Lateral loads acting on a member fabrication errors temperature changes inclined supports and internal supports are handled in the same manner as was outlined for trusses and beams Procedure for Analysis The following method provides a means of finding the displacements support reactions and internal loadings for members of statically determinate and indeterminate frames Notation Divide the structure into finite elements and arbitrarily identify each element and its nodes Elements usually extend between points of support points of concentrated loads corners or joints or to points where internal loadings or displacements are to be determined Establish the xyzglobal coordinate systemusually for convenience with the origin located at a nodal point on one of the elements and the axes located such that all the nodes have positive coordinates At each nodal point of the frame specify numerically the three x y z coding components In all cases use the lowest code numbers to identify all the unconstrained degrees of freedom followed by the remaining or highest code numbers to identify the constrained degrees of freedom From the problem establish the known displacements and known external loads When establishing be sure to include any reversed fixedend loadings if an element supports an intermediate load Structure Stiffness Matrix Apply Eq 1610 to determine the stiffness matrix for each element expressed in global coordinates In particular the direction cosines and are determined from the x y coordinates of the ends of the element Eqs 145 and 146 After each member stiffness matrix is written and the six rows and columns are identified with the near and far code numbers merge the matrices to form the structure stiffness matrix K As a partial check the element and structure stiffness matrices should all be symmetric ly lx Qk Qk Dk 164 APPLICATION OF THE STIFFNESS METHOD FOR FRAME ANALYSIS 601 16 Displacements and Loads Partition the stiffness matrix as indicated by Eq 1418 Expansion then leads to The unknown displacements are determined from the first of these equations Using these values the support reactions are computed from the second equation Finally the internal loadings q at the ends of the members can be computed from Eq 167 namely If the results of any of the unknowns are calculated as negative quantities it indicates they act in the negative coordinate directions q kTD Qu Du Qu K21Du K22Dk Qk K11Du K12Dk Fig 164 EXAMPLE 161 Determine the loadings at the joints of the twomember frame shown in Fig 164a Take and for both members SOLUTION Notation By inspection the frame has two elements and three nodes which are identified as shown in Fig 164b The origin of the global coordinate system is located at ①The code numbers at the nodes are specified with the unconstrained degrees of freedom numbered first From the constraints at ① and ➂ and the applied loading we have Structure Stiffness Matrix The following terms are common to both element stiffness matrices 12EI L3 12291103215002 2011223 126 kin AE L 102911032 201122 12083 kin Dk D 0 0 0 0 T 6 7 8 9 Qk E 5 0 0 0 0 U 1 2 3 4 5 E 2911032 ksi A 10 in2 I 500 in4 b y x 5 6 4 2 1 8 1 2 2 3 7 3 5 k 9 1 20 ft 5 k 20 ft a 602 CHAPTER 16 PLANE FRAME ANALYSIS USING THE STIFFNESS METHOD 16 EXAMPLE 161 Continued Member 1 Substituting the data into Eq 1610 we have The rows and columns of this matrix are identified by the three x y z code numbers first at the near end and followed by the far end that is 4 6 5 1 2 3 respectively Fig 164b This is done for later assembly of the elements Member 2 Substituting the data into Eq 1610 yields As usual column and row identification is referenced by the three code numbers in x y z sequence for the near and far ends respectively that is 1 2 3 then 7 8 9 Fig 164b k2 1 2 3 7 8 9 F 126 0 15104 126 0 15104 0 12083 0 0 12083 0 15104 0 241711032 15104 0 1208311032 126 0 15104 126 0 15104 0 12083 0 0 12083 0 15104 0 1208311032 15104 0 241711032 V 1 2 3 7 8 9 lx 20 20 20 0 ly 20 0 20 1 6 6 k1 4 6 5 1 2 3 F 12083 0 0 12083 0 0 0 126 15104 0 126 15104 0 15104 241711032 0 15104 1208311032 12083 0 0 12083 0 0 0 126 15104 0 126 15104 0 15104 1208311032 0 15104 241711032 V 4 6 5 1 2 3 lx 20 0 20 1 ly 0 0 20 0 2EI L 2291103215002 201122 1208311032 k in 4EI L 4291103215002 201122 241711032 k in 6EI L2 6291103215002 2011222 15104 k EXAMPLE 162 Continued 606 CHAPTER 16 PLANE FRAME ANALYSIS USING THE STIFFNESS METHOD 16 Applying Eq 1610 we have k1 4 5 6 1 2 3 F 74518 55309 696 74518 55309 696 55309 42255 928 55309 42255 928 696 928 23211032 696 928 11611032 74518 55309 696 74518 55309 696 55309 42255 928 55309 42255 928 696 928 11611032 696 928 23211032 V 4 5 6 1 2 3 Member 2 Thus Eq 1610 becomes lx 40 20 20 1 ly 15 15 20 0 2EI L 232911034600 3201124 14511052 k in 4EI L 42911032600 201122 29011052 k in 6EI L2 62911032600 2011222 181250 k 12EI L3 122911032600 2011223 1510 kin AE L 122911032 201122 1450 kin k2 1 2 3 7 8 9 F 1450 0 0 1450 0 0 0 1510 181250 0 1510 181250 0 181250 29011032 0 181250 14511032 1450 0 0 1450 0 0 0 1510 181250 0 1510 181250 0 181250 14511032 0 181250 29011032 V 1 2 3 7 8 9 610 CHAPTER 16 PLANE FRAME ANALYSIS USING THE STIFFNESS METHOD 16 169 Determine the stiffness matrix K for the frameTake for each member 1610 Determine the support reactions at ① and ➂Take I 300 in4 A 10 in2 for each member E 2911032 ksi I 300 in4 A 10 in2 E 2911032 ksi 10 ft 20 ft 2 1 3 2 7 1 4 6 9 8 5 2 1 3 2 kft Probs 1691610 167 Determine the structure stiffness matrix K for the frame Take for each member I 650 in4 A 20 in2 E 2911032 ksi Prob 167 2 1 6 5 4 2 1 3 2 1 6 k 4 k 3 9 8 7 12 ft 10 ft 168 Determine the components of displacement at ① Take for each member A 20 in2 I 650 in4 E 2911032 ksi Prob 168 2 1 6 5 4 2 1 3 2 1 6 k 4 k 3 9 8 7 12 ft 10 ft Prob 166 9 1 2 3 1 2 2 1 3 6 4 4 m 2 m 2 m 60 kN 5 8 7 166 Determine the support reactions at pins ① and ➂ Take for each member E 200 GPa I 35011062 mm4 A 1511032 mm2 16 164 APPLICATION OF THE STIFFNESS METHOD FOR FRAME ANALYSIS 611 1613 Use a computer program to determine the reactions on the frame AE and EI are constant 1611 Determine the structure stiffness matrix K for the frame Take for each member I 700 in4 A 20 in2 E 2911032 ksi 1614 Use a computer program to determine the reactions on the frame Assume A B D and F are pins AE and EI are constant 1 2 9 8 3 2 1 3 16 ft 20 k 12 ft 12 ft 1 2 4 7 6 5 A B C E F 6 m 4 m 8 m 8 kN D A D 15 k B C 20 ft 24 ft 15 kft Prob 1611 1612 Determine the support reactions at the pins ① and ➂ Take for each member I 700 in4 A 20 in2 E 2911032 ksi 1 2 9 8 3 2 1 3 16 ft 20 k 12 ft 12 ft 1 2 4 7 6 5 Prob 1612 Prob 1613 Prob 1614 612 Matrix Algebra for Structural Analysis APPENDIXA A1 Basic Definitions and Types of Matrices With the accessibility of desk top computers application of matrix algebra for the analysis of structures has become widespread Matrix algebra provides an appropriate tool for this analysis since it is relatively easy to formulate the solution in a concise form and then perform the actual matrix manipulations using a computer For this reason it is important that the structural engineer be familiar with the fundamental operations of this type of mathematics Matrix A matrix is a rectangular arrangement of numbers having m rows and n columnsThe numberswhich are called elementsare assembled within brackets For example the A matrix is written as Such a matrix is said to have an order of m by n Notice that the first subscript for an element denotes its row position and the second subscript denotes its column position In general then is the element located in the ith row and jth column Row Matrix If the matrix consists only of elements in a single row it is called a row matrix For example a row matrix is written as Here only a single subscript is used to denote an element since the row subscript is always understood to be equal to 1 that is and so on a2 a12 a1 a11 A a1 a2 Á an 1 n aij m n A D a11 a12 Á a1n a21 a22 Á a2n o am1 am2 Á amn T Column Matrix A matrix with elements stacked in a single column is called a column matrix The column matrix is Here the subscript notation symbolizes and so on Square Matrix When the number of rows in a matrix equals the number of columns the matrix is referred to as a square matrix An square matrix would be Diagonal Matrix When all the elements of a square matrix are zero except along the main diagonal running down from left to right the matrix is called a diagonal matrix For example Unit or Identity Matrix The unit or identity matrix is a diagonal matrix with all the diagonal elements equal to unity For example Symmetric Matrix A square matrix is symmetric provided For example A C 3 5 2 5 1 4 2 4 8 S aij aji I C 1 0 0 0 1 0 0 0 1 S A C a11 0 0 0 a22 0 0 0 a33 S A D a11 a12 Á a1n a21 a22 Á a2n o an1 an2 Á ann T n n a2 a21 a1 a11 A D a1 a2 o am T m 1 A1 BASIC DEFINITIONS AND TYPES OF MATRICES 613 A 614 APPENDIX A MATRIX ALGEBRA FOR STRUCTURAL ANALYSIS A A2 Matrix Operations Equality of Matrices Matrices A and B are said to be equal if they are of the same order and each of their corresponding elements are equal that is For example if then Addition and Subtraction of Matrices Two matrices can be added together or subtracted from one another if they are of the same order The result is obtained by adding or subtracting the corresponding elements For example if then Multiplication by a Scalar When a matrix is multiplied by a scalar each element of the matrix is multiplied by the scalar For example if then Matrix Multiplication Two matrices A and B can be multiplied together only if they are conformable This condition is satisfied if the number of columns in A equals the number of rows in B For example if A1 then AB can be determined since A has two columns and B has two rows Notice however that BA is not possibleWhy A ca11 a12 a21 a22 d B cb11 b12 b13 b21 b22 b23 d kA c 24 6 36 12 d A c4 1 6 2 d k 6 A B c1 15 3 3 d A B c11 1 1 5 d A c6 7 2 1 d B c 5 8 1 4 d A B A c2 6 4 3 d B c2 6 4 3 d aij bij A2 MATRIX OPERATIONS 615 If matrix A having an order of is multiplied by matrix B having an order of it will yield a matrix C having an order of that is The elements of matrix C are found using the elements of A and of B as follows A2 The methodology of this formula can be explained by a few simple examples Consider By inspection the product is possible since the matrices are conformable that is A has three columns and B has three rows By Eq A2 the multiplication will yield matrix C having two rows and one columnThe results are obtained as follows Multiply the elements in the first row of A by corresponding elements in the column of B and add the results that is Multiply the elements in the second row of A by corresponding elements in the column of B and add the results that is Thus C c49 41 d c21 c2 1122 6162 1172 41 c21 c11 c1 2122 4162 3172 49 c11 C AB A c 2 4 3 1 6 1 d B C 2 6 7 S cij a n k1 aikbkj bij aij 1m n21n q2 1m q2 A B C 1m q2 1n q2 1m n2 A 616 APPENDIX A MATRIX ALGEBRA FOR STRUCTURAL ANALYSIS A As a second example consider Here again the product can be found since A has two columns and B has two rowsThe resulting matrix C will have three rows and two columnsThe elements are obtained as follows The scheme for multiplication follows application of EqA2Thus Notice also that BA does not exist since written in this manner the matrices are nonconformable The following rules apply to matrix multiplication 1 In general the product of two matrices is not commutative A3 2 The distributive law is valid A4 3 The associative law is valid A5 Transposed Matrix A matrix may be transposed by interchanging its rows and columns For example if A C a11 a12 a13 a21 a22 a23 a31 a32 a33 S B b1 b2 b3 A1BC2 1AB2C A1B C2 AB AC AB Z BA C C 1 47 5 32 28 18 S c32 2172 8142 18 1third row of A times second column of B2 c31 2122 8132 28 1third row of A times first column of B2 c22 4172 1142 32 1second row of A times second column of B2 c21 4122 1132 5 1second row of A times first column of B2 c12 5172 3142 47 1first row of A times second column of B2 c11 5122 3132 1 1first row of A times first column of B2 C AB A C 5 3 4 1 2 8 S B c 2 7 3 4 d A2 MATRIX OPERATIONS 617 Then Notice that AB is nonconformable and so the product does not exist A has three columns and B has one row Alternatively multiplication is possible since here the matrices are conformable A has three columns and has three rowsThe following properties for transposed matrices hold A6 A7 A8 This last identity will be illustrated by example If Then by EqA8 Matrix Partitioning A matrix can be subdivided into submatrices by partitioning For example Here the submatrices are A21 ca21 a31 d A22 ca22 a23 a24 a32 a33 a34 d A11 a11 A12 a12 a13 a14 A C a11 a12 a13 a14 a21 a22 a23 a24 a31 a32 a33 a34 S cA11 A12 A21 A22 d c28 2 28 12 d c28 2 28 12 d a c 28 28 2 12d b T c28 2 28 12 d a c6 2 1 3 d c4 3 2 5d b T c4 2 3 5 d c6 1 2 3 d A c6 2 1 3 d B c4 3 2 5 d 1AB2T BTAT 1kA2T kAT 1A B2T AT BT BT ABT AT C a11 a21 a31 a12 a22 a32 a13 a23 a33 S BT C b1 b2 b3 S A 618 APPENDIX A MATRIX ALGEBRA FOR STRUCTURAL ANALYSIS A The rules of matrix algebra apply to partitioned matrices provided the partitioning is conformable For example corresponding submatrices of A and B can be added or subtracted provided they have an equal number of rows and columns Likewise matrix multiplication is possible provided the respective number of columns and rows of both A and B and their submatrices are equal For instance if then the product AB exists since the number of columns of A equals the number of rows of B three Likewise the partitioned matrices are conformable for multiplication since A is partitioned into two columns and B is partitioned into two rows that is Multiplication of the submatrices yields A3 Determinants In the next section we will discuss how to invert a matrix Since this operation requires an evaluation of the determinant of the matrix we will now discuss some of the basic properties of determinants A determinant is a square array of numbers enclosed within vertical bars For example an nthorder determinant having n rows and n columns is A9 ƒAƒ 4 a11 a12 Á a1n a21 a22 Á a2n o an1 an2 Á ann 4 AB D c 8 4 4 2 d c 7 35 4 20 d 12 18 56 32 T C 1 0 39 18 68 50 S A22B21 87 4 56 32 A21B11 6 3c2 1 0 8 d 12 18 A12B21 c 1 5 d7 4 c c 7 4 35 20d A11B11 c 4 1 2 0 d c2 1 0 8 d c 8 4 4 2 d AB cA11 A12 A21 A22 d cB11 B21 d cA11B11 A12B21 A21B11 A22B21 d A C 4 1 1 2 0 5 6 3 8 S B C 2 1 0 8 7 4 S A3 DETERMINANTS 619 Evaluation of this determinant leads to a single numerical value which can be determined using Laplaces expansion This method makes use of the determinants minors and cofactors Specifically each element of a determinant of nth order has a minor which is a determinant of order This determinant minor remains when the ith row and jth column in which the element is contained is canceled out If the minor is multiplied by it is called the cofactor of and is denoted as A10 For example consider the thirdorder determinant The cofactors for the elements in the first row are Laplaces expansion for a determinant of order n Eq A9 states that the numerical value represented by the determinant is equal to the sum of the products of the elements of any row or column and their respective cofactors ie or A11 For applicationit is seen that due to the cofactors the number D is defined in terms of n determinants cofactors of order each These determinants can each be reevaluated using the same formula whereby one must then evaluate determinants of order and so onThe process of evaluation continues until the remaining determinants to be evaluated reduce to the second order whereby the cofactors of the elements are single elements of D Consider for example the following secondorder determinant We can evaluate D along the top row of elements which yields Or for example using the second column of elements we have D 511212112 211222132 11 D 311211122 511212112 11 D 3 5 1 2 1n 22 1n 12 n 1 D a1jC1j a2jC2j Á anjCnj 1j 1 2 Á or n2 D ai1Ci1 ai2Ci2 Á ainCin 1i 1 2 Á or n2 C13 11213 a21 a22 a31 a32 a21 a22 a31 a32 C12 11212 a21 a23 a31 a33 a21 a23 a31 a33 C11 11211 a22 a23 a32 a33 a22 a23 a32 a33 3 a11 a12 a13 a21 a22 a23 a31 a32 a33 3 Cij 112ijMij aij 112ij aij n 1 Mij aij A 620 APPENDIX A MATRIX ALGEBRA FOR STRUCTURAL ANALYSIS A Rather than using Eqs A11 it is perhaps easier to realize that the evaluation of a secondorder determinant can be performed by multiplying the elements of the diagonal from top left down to right and subtract from this the product of the elements from top right down to left ie follow the arrow Consider next the thirdorder determinant Using EqA11 we can evaluate using the elements either along the top row or the first column that is As an exercise try to evaluate using the elements along the second row A4 Inverse of a Matrix Consider the following set of three linear equations which can be written in matrix form as A12 A13 Ax C C a11 a12 a13 a21 a22 a23 a31 a32 a33 S C x1 x2 x3 S C c1 c2 c3 S a31 x1 a32 x2 a33 x3 c3 a21 x1 a22 x2 a23 x3 c2 a11 x1 a12 x2 a13 x3 c1 ƒDƒ 114 02 416 02 1118 22 40 D 111211 2 6 0 2 411221 3 1 0 2 11211231 3 1 2 6 114 02 318 62 110 22 40 D 11211211 2 6 0 2 13211212 4 6 1 2 11211213 4 2 1 0 ƒDƒ ƒDƒ 3 1 3 1 4 2 6 1 0 2 3 D 3 5 1 2 3122 5112 11 N A4 INVERSE OF A MATRIX 621 A One would think that a solution for x could be determined by dividing C by A however division is not possible in matrix algebra Instead one multiplies by the inverse of the matrix The inverse of the matrix A is another matrix of the same order and symbolically written as It has the following property where I is an identity matrix Multiplying both sides of EqA13 by we obtain Since we have A14 Provided can be obtained a solution for x is possible For hand calculation the method used to formulate can be developed using Cramers rule The development will not be given here instead only the results are given In this regard the elements in the matrices of EqA14 can be written as A15 Here is an evaluation of the determinant of the coefficient matrix A which is determined using the Laplace expansion discussed in Sec A3 The square matrix containing the cofactors is called the adjoint matrix By comparison it can be seen that the inverse matrix is obtained from A by first replacing each element by its cofactor then transposing the resulting matrixyielding the adjoint matrixand finally multiplying the adjoint matrix by To illustrate how to obtain numerically we will consider the solution of the following set of linear equations A16 Here A C 1 1 1 1 1 1 1 2 2 S x1 2x2 2x3 5 x1 x2 x3 1 x1 x2 x3 1 A1 1 ƒAƒ Cij aij A1 Cij ƒAƒ C x1 x2 x3 S 1 ƒAƒ C C11 C21 C31 C12 C22 C32 C13 C23 C33 S C c1 c2 c3 S x A1C A1 A1 x A1C A1Ax Ix x A1Ax A1C A1 AA1 A1A I A1 See Kreyszig E Advanced Engineering Mathematics John Wiley Sons Inc New York 622 APPENDIX A MATRIX ALGEBRA FOR STRUCTURAL ANALYSIS A The cofactor matrix for A is Evaluating the determinants and taking the transpose the adjoint matrix is Since The inverse of A is therefore Solution of EqsA16 yields Obviously the numerical calculations are quite expanded for larger sets of equations For this reason computers are used in structural analysis to determine the inverse of matrices x3 1 6132112 132112 102152 1 x2 1 6112112 132112 122152 1 x1 1 6142112 0112 122152 1 C x1 x2 x3 S 1 6 C 4 0 2 1 3 2 3 3 0 S C 1 1 5 S A1 1 6 C 4 0 2 1 3 2 3 3 0 S A 1 1 1 1 1 1 1 2 2 6 CT C 4 0 2 1 3 2 3 3 0 S C F 1 2 1 2 1 1 1 2 1 1 1 2 1 2 1 2 1 1 1 2 1 1 1 2 1 1 1 1 1 1 1 1 1 1 1 1 V A5 THE GAUSS METHOD FOR SOLVING SIMULTANEOUS EQUATIONS 623 A A5 The Gauss Method for Solving Simultaneous Equations When many simultaneous linear equations have to be solved the Gauss elimination method may be used because of its numerical efficiency Application of this method requires solving one of a set of n equations for an unknown say in terms of all the other unknowns Substituting this socalled pivotal equation into the remaining equations leaves a set of equations with unknowns Repeating the process by solving one of these equations for in terms of the remaining unknowns forms the second pivotal equation This equation is then substituted into the other equations leaving a set of equations with unknowns The process is repeated until one is left with a pivotal equation having one unknown which is then solved The other unknowns are then determined by successive back substitution into the other pivotal equationsTo improve the accuracy of solution when developing each pivotal equation one should always select the equation of the set having the largest numerical coefficient for the unknown one is trying to eliminateThe process will now be illustrated by an example Solve the following set of equations using Gauss elimination A17 A18 A19 We will begin by eliminating The largest coefficient of is in Eq A19 hence we will take it to be the pivotal equation Solving for we have A20 Substituting into EqsA17 and A18 and simplifying yields A21 A22 Next we eliminate Choosing Eq A21 for the pivotal equation since the coefficient of is largest here we have A23 Substituting this equation into Eq A22 and simplifying yields the final pivotal equation which can be solved for This yields Substituting this value into the pivotal EqA23 gives Finally from pivotal EqA20 we get x1 075 x2 025 x3 075 x3 x2 0727 0636x3 x2 x2 15x2 05x3 0 275x2 175x3 2 x1 1 125x2 075x3 x1 x1 x1 4x1 5x2 3x3 4 2x1 x2 x3 2 2x1 8x2 2x3 2 n 3 n 3 x3 x4 Á xn n 2 x2 n 1 n 1 x2 x3 Á xn x1 624 APPENDIX A MATRIX ALGEBRA FOR STRUCTURAL ANALYSIS A A1 If and determine and A2 If and determine and A3 If and determine AB A4 If and determine AB A5 If and determine AB A6 If and show that A7 If determine A8 If determine A9 If determine A10 If and determine AB A11 If and determine AB B C 2 5 1 S A c2 5 1 3 2 5 d B C 2 0 1 S A c 5 6 0 1 2 3 d AAT A c 2 8 1 5d AAT A c2 5 8 1d A AT A C 2 3 6 5 9 2 1 0 2 S A BT AT BT B C 1 4 4 S A C 2 5 6 S B 4 6 5 A C 2 5 6 S B c6 2 5 1 d A c6 3 4 2d B c4 1 2 2 d A 2 5 A 2B 3A 2B B C 6 4 3 3 2 2 5 1 6 S A C 3 5 2 4 3 1 1 1 7 S A 3B 2A B B C 1 2 5 8 2 1 S A C 3 6 2 7 4 2 S A12 If and determine AB A13 Show that the distributive law is valid ie if A14 Show that the associative law is valid ie if A15 Evaluate the determinants and A16 If determine A17 If determine A18 Solve the equations using the matrix equation A19 Solve the equations in Prob A18 using the Gauss elimination method A20 Solve the equations using the matrix equation A21 Solve the equations in Prob A20 using the Gauss elimination method x A1C x1 x2 x3 1 x1 x2 x3 1 x1 2x2 2x3 5 x A1C x1 2x2 x3 2 5x1 4x2 3x3 4 4x1 x2 x3 1 A1 A C 3 5 7 4 1 2 0 3 1 S A1 A c2 5 4 1 d 3 5 7 2 1 8 2 1 4 0 3 2 4 3 1 6 2 C 2 1 3 B C 1 1 4 S A c 2 5 1 5 6 0 d ABC ABC C C 4 2 1 S B C 2 1 0 S A c4 2 1 3 5 6 d AB C AB AC B C 2 1 1 3 2 5 2 4 6 S A C 6 5 1 0 3 2 2 1 4 S PROBLEMS APPENDIXB 625 General Procedure for Using Structural Analysis Software Popular structural analysis software programs currently available such as STAAD RISA SAP etc are all based on the stiffness method of matrix analysisdescribed in Chapters 13 through 15 Although each program has a slightly different interface they all require the operator to input data related to the structure A general procedure for using any of these programs is outlined below Preliminary Steps Before using any program it is first necessary to numerically identify the members and joints called nodes of the structure and establish both global and local coordinate systems in order to specify the structures geometry and loadingTo do this you may want to make a sketch of the structure and specify each member with a number enclosed within a square and use a number enclosed within a circle to identify the nodes In some programs the near and far ends of the member must be identified This is done using an arrow written along the member with the head of the arrow directed toward the far end Member node and direction identification for a plane truss beam and plane frame are shown in Figs B1 B2 and B3 In Fig B1 node ➁ is at the near end of member 4 and node ➂ is at its far end These assignments can all be done arbitrarily Notice however that the nodes on the truss are always at the joints since this is where the loads are applied and the displacements and member forces are to be determined For beams and frames the nodes are at the supports at a corner or joint at an internal pin or at a point where the linear or rotational displacement is to be determined Fig B2 and B3 Since loads and displacements are vector quantities it is necessary to establish a coordinate system in order to specify their correct sense of direction Here we must use two types of coordinate systems Global Coordinates A single global or structure coordinate system using righthanded x y z axes is used to specify the location of each node relative to the origin and to identify the sense of each of the external load and displacement components at the nodes It is convenient to locate the origin at a node so that all the other nodes have positive coordinates See each figure A more complete coverage of this method including the effects of torsion in three dimensional frames is given in books related to matrix analysis 3 4 2 1 1 3 5 2 4 y x 200 N 2 m 2 m 4 m 3 2 4 x y Fig B1 APPENDIX B GENERAL PROCEDURE FOR USING STRUCTURAL ANALYSIS SOFTWARE 627 B will then give the correct reactions and internal forces but not the correct displacements If an internal hinge or pin connects two members of a beam or frame then the release of moment must be specified at that node For example member 3 of the frame in Fig B3 has a pin at the far node 4 In a like manner this pin can also be identified at the near node of member 4 Support Data Enter in turn each node located at a support and specify the called for global coordinate directions in which restraint occurs For example since node 5 of the frame in Fig B3 is a fixed support a zero is entered for the x y and z rotational directions however if this support settles downward 0003 m then the value entered for y would be Load Data Loads are specified either at nodes or on members Enter the algebraic values of nodal loadings relative to the global coordinates For example for the truss in Fig B1 the loading at node 2 is in the y direction and has a value of For beam and frame members the loadings and their location are referenced using the local coordinates For example the distributed loading on member 2 of the frame in Fig B3 is specified with an intensity of Nm located 075 m from the near node 2 and Nm located 3 m from this node Results Once all the data is entered then the problem can be solved One obtains the external reactions on the structure and the displacements and internal loadings at each node As a partial check of the results a statics check is often given at each of the nodes It is very important that you never fully trust the results obtained Instead it would be wise to perform an intuitive structural analysis to further check the outputAfter all the structural engineer must take full responsibility for both the modeling and computing of final results 400 400 200 0003 Fundamental Problems Partial Solutions and Answers Chapter 2 F21 Ans Ans Ans F22 Ans Ans Ans Ans F23 Ans Ans Ans Ans F24 Member AC Ans Member BC Ans Ans Ans MB 210 kN m 250122 8122112 MB 0 dMB 0 By 185 kN By 250 8122 0 cFy 0 Bx 0 Fx 0 Cy 250 kN Cy142 10112 0 dMA 0 NA 750 kN 10132 NA142 0 dMC 0 By Cy a 5 sin 60 b1sin 602 500 kN Bx Cx a 5 sin 60 b1cos 602 289 kN Ax 289 kN a 5 sin 60 b1cos 602 Ax 0 Fx 0 Ay 150 kN 10122132 Ay142 0 dMB 0 FBC 5 sin 60 kN FBC sin 60142 10122112 0 dMA 0 By Cy a 20 sin 45 b1sin 452 200 kN Bx Cx a 20 sin 45 b1cos 452 200 kN Ax 200 kN Ax a 20 sin 45 b1cos 452 0 Fx 0 Ay 200 kN 10142122 Ay142 0 dMB 0 FBC 20 sin 45 kN FBC sin 45142 10142122 0 dMA 0 Bx Cx 250A4 5B 200 kN By Cy 250A3 5B 150 kN Ax 200 kN Ax 250A4 5B 0 Fx 0 Ay 150 kN 60 Ay142 0 dMB 0 FBC 250 kN 60 FBCA3 5B142 0 dMA 0 A FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 629 F25 Ans Ans Ans Ans F26 Ans Ans Ans F27 Cy 200 kN Cy 400 6 0 cFy 0 Cx 200 kN Cx 2 0 Fx 0 NA 400 kN 6122 2122 NA142 0 dMC 0 By Cy 12513 52 750 lb Bx Cx 12514 52 100 lb Ay 225 lb Ay 125A3 5B 300 0 cFy 0 Ax 100 lb Ax 12514 52 0 Fx 0 FBC 125 lb FBC 13 52142 FBC14 52132 300122 0 dMA 0 25 m 3 5 3 m 4 m 8 kN 8 kN 2 m 2 m 2 m 4 m Ax Ay Dx Dy Bx Bx By By 3 5 4 Member AB Member BCD Ans Member AB Ans Ans Member BCD Ans Ans Dy 14833 kN 148 kN Dy 1167 8 8 0 cFy 0 Dx 1025 kN 1025 Dx 0 Fx 0 Ay 10167 kN 102 kN Ay 132152A3 5B 1167 0 cFy 0 Ax 175 kN Ax 315214 52 1025 0 Fx 0 By 1167 kN 117 kN Bx 1025 kN dMD 0 8122 8142 Bx142 By162 0 dMA 0 Bx142 By132 31521252 0 630 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS F28 Ax Bx Bx Cx Cy Cy Cx By By 2 m 3 m 6 kN 4 kN 6 kN 3 m 2 m 2 m 6 m Ay Dy MD Dx Member AB Ans Ans Member BC Ans Ans Ans Member AB Ans Member CD Ans Ans Ans MD 120 kN m MD 200162 0 dMD 0 Dy 600 kN Dy 600 0 cFy 0 Dx 200 kN 200 Dx 0 Fx 0 Ay 600 kN Ay 600 0 cFy 0 Cy 600 kN Cy162 6122 6142 0 dMB 0 By 600 kN 6122 6142 By162 0 dMC 0 Cx 200 kN 200 Cx 0 Fx 0 Ax 200 kN 4132 Ax162 0 dMB 0 Bx 200 kN Bx162 4132 0 dMA 0 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 631 Ax Bx Bx Cx Cy Cy Cx By By 4 ft 3 ft 28 k 056 k 3 ft 4 ft 4 ft Ay Dy MD Dx F29 Member AB Ans Ans Member BC Ans Ans Ans Member AB Ans Member CD Ans Ans Ans MD 600 k ft MD 150142 0 dMD 0 Dy 800 k Dy 800 0 cFy 0 Dx 150 k 150 Dx 0 Fx 0 Ay 800 k Ay 800 0 cFy 0 Cx 150 k 150 Cx 0 Fx 0 Cy 800 k Cy182 2182142 0 dMB 0 By 800 k 2182142 By182 0 dMC 0 Ax 150 k 05162132 Ax162 0 dMB 0 Bx 150 k Bx162 05162132 0 dMA 0 632 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS F210 2 m 2 m 2 m 6 kN 8 kN 8 kN 6 kN Ax Bx Bx Cx Cy Cy Cx By By 3 m 3 m 15 6 kN Ay Dy MD Dx Member BC Ans Ans Member AB Ans Ans Ans Member BC Ans Member CD Ans Ans Ans MD 270 kN m 15162132 MD 0 dMD 0 Dy 140 kN Dy 140 0 cFy 0 Dx 900 kN Dx 15162 0 Fx 0 Cx 0 Fx 0 Ay 140 kN Ay 140 0 cFy 0 Ax 0 Fx 0 Bx 0 dMA 0 By 140 kN 8122 8142 6162 By162 0 dMC 0 Cy 140 kN Cy162 8122 8142 6162 0 dMB 0 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 639 F41 Segment CB Ans Ans Ans MC 20 kN m MC 10112 10132 0 dMC 0 VC 0 VC 10 10 0 cFy 0 NC 0 Fx 0 By 100 kN By122 20 10142 0 dMA 0 F42 Segment CB Ans Ans Ans MC 675 kN m 1051152 8115210752 MC 0 dMC 0 VC 150 kN VC 105 81152 0 cFy 0 NC 0 Fx 0 By 105 kN By132 4115210752 8115212252 0 dMA 0 F43 Segment AC Ans Ans Ans MC 124 kN m MC 1 213211521052 9001152 0 dMC 0 VC 675 kN 900 1 21321152 VC 0 cFy 0 NC 0 Fx 0 Ax 0 Fx 0 Ay 900 kN 1 2162162132 Ay162 0 dMB 0 F44 Segment AC Ans Ans Ans MC 1125 lb MC 300115210752 3001152 0 dMC 0 VC 150 lb 300 3001152 VC 0 cFy 0 NC 0 Fx 0 Ax 0 Fx 0 Ay 300 lb 3001321152 1 2 3002132112 Ay132 0 dMB 0 F45 Reactions Segment AC Ans Ans Ans MC 5625 kN m MC 5115210752 0 dMC 0 VC 750 kN 51152 VC 0 cFy 0 NC 300 kN NC 300 0 Fx 0 Ay 0 4243 sin 45 5162 Ay 0 cFy 0 Ax 300 kN 4243 cos 45 Ax 0 Fx 0 FB 4243 kN FB sin 45132 5162132 0 dMA 0 640 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS F46 Reactions Segment CB Ans Ans Ans MC 5370 lb ft 537 k ft 1345162 150162132 MC 0 dMC 0 VC 445 lb VC 1345 150162 0 cFy 0 NC 0 Fx 0 By 1345 lb By1152 15019211052 600162 800132 0 dMA 0 F47 Left segment Ans Ans M 5x3 6x6 kN m M 1 2 a18 3 xb1x2ax 3 b 6x 0 dMO 0 V 53x2 66 kN 6 1 2 a 18 3 xb1x2 V 0 cFy 0 F48 Reaction Left segment Ans Ans M e120x 1 3x3 f kN m M 1 2 a12 6 xb1x2a x 3 b 120x 0 dMO 0 V 5120 x26 kN 120 1 2 a 12 6 xb1x2 V 0 cFy 0 Ay 120 kN 1 21122162122 Ay162 0 dMB 0 F49 Reactions left segment Ans Ans right segment Ans Ans M 54x2 40x 646 kN m 24018 x2 818 x2a 8 x 2 b M 0 dMO 0 V 540 8x6 kN V 240 818 x2 0 cFy 0 4 m 6 x 6 8 m M 58x6 kN m M 800x 0 dMO 0 V 586 kN 800 V 0 cFy 0 0 x 6 4 m Ay 800 kN 8142122 Ay182 0 dMB 0 By 240 kN By182 8142162 0 dMA 0 F410 Ans Ans Ans Ans M e 5 2x2 10x 45 f kN m M 51x 22a x 2 2 b 15 20 0 dMO 0 V 510 5x6 kN 51x 22 V 0 cFy 0 2 m 6 x 4 m M 20 kN m M 20 0 dMO 0 V 0 cFy 0 0 x 6 2 m 646 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS F82 F83 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 647 F84 For 1 2 For 3 4 at From Eq 2 at From Eq 1 at From Eq 3 at From Eq 4 Ans Ans F85 1 2 at From Eq 1 at From Eq 2 Ans v Px 6EI x2 3Lx C2 0 x 0 v 0 C1 0 x 0 dv dx 0 EI v P 6 x3 PL 2 x2 C1x C2 EI dv dx P 2 x2 PLx C1 EI d2v dx2 Px PL M Px PL v2 P 48EI14x3 2 12Lx2 2 9L2x2 L32 v1 Px1 48EI14x2 1 3L22 C4 PL3 48 x2 L v2 0 C3 3PL2 16 x2 L 2 dv2 dx2 0 C1 PL2 16 x1 L 2 dv1 dx1 0 C2 0 x1 0 v1 0 EI v2 PL 4 x2 2 P 12x3 2 C3x2 C4 EI dv2 dx2 PL 2 x2 P 4 x2 2 C3 EI d2v2 dx2 2 PL 2 P 2 x2 M2 P 2 1L x22 PL 2 P 2 x2 L 2 6 x2 L EI v1 P 12x3 1 C1x1 C2 EI dv1 dx1 P 4 x2 1 C1 EI d2v1 dx2 1 P 2 x1 M1 P 2 x1 0 x1 6 L 2 648 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS F86 1 at From Eq 1 at From Eq 1 Ans F87 For 1 2 For 3 4 at From Eq 2 at From Eq 4 5 at From Eqs 1 and 3 6 at From Eqs 2 and 4 7 C1L C3L 2C4 M0L2 4 x1 x2 L 2 v1 v2 C1 C3 M0L 2 x1 x2 L 2 dv1 dx1 dv2 dx2 0 C3L C4 M0L2 3 x2 L v2 0 C2 0 x1 0 v1 0 EIv2 M0 2 x2 2 M0 6Lx3 2 C3x2 C4 EI dv2 dx2 M0x2 M0 2Lx2 2 C3 EI d2v2 dy2 2 M0 M0 L x2 M M0 M0 L x2 L 2 6 x2 L EIv1 M0 6Lx3 1 C1x1 C2 EI dv1 dx1 M0 2Lx2 1 C1 EI d2v1 dx2 1 M0 L x1 M M0 L x1 0 x1 6 L 2 v M0 6EIL1x3 3Lx2 2L2x2 C1 M0L 3 x L v 0 C2 0 x 0 v 0 EI v M0 2 x2 M0 6Lx3 C1x C2 EI dv dx M0x M0 2Lx2 C1 EI d2v dx2 M0 M0 L x M M0 M0 L x FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 649 Solving Eqs 5 6 and 7 Ans Ans F88 From Eq 1 From Eq 2 Ans F89 From Eq 1 From Eq 2 Ans v w0 120EIL1x5 5L4x 4L52 C2 w0L4 30 x L at v 0 C1 w0L3 24 x L at dv dx 0 EI v w0 120Lx5 C1x C2 122 EI dv dx w0 24Lx4 C1 112 EI d2v dx2 w0 6Lx3 M w0 6Lx3 v w 24EI1x4 4Lx3 6L2x22 C2 0 x 0 at v 0 C1 0 x 0 at dv dx 0 EI v w 24x4 wL 6 x3 wL2 4 x2 C1x C2 122 EI dv dx w 6 x3 wL 2 x2 wL2 2 x C1 1 EI d2v dx2 w 2 x2 wLx wL2 2 M w 2 x2 wLx wL2 2 v2 M0 24EIL14x3 2 12Lx2 2 11L2x2 3L32 v1 M0 24EIL14x3 1 L2x12 C1 M0L 24 C3 11M0L 24 C4 M0L2 8 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 655 F91 Member n lb N lb L ft AB 1667 250 10 416667 AC 1 150 6 90000 BC 1333 200 8 213333 7200 nNL 1lb2 ft2 Member N L ft AB 1667P 1667 250 10 416667 AC P 1 150 6 90000 BC 1333P 1333 200 8 213333 7200 N a dN dP bL 1lb ft2 N 1P 150 lb2 dN dP Member n kN N kN L m AB 1 4041 2 80829 AC 0 80829 2 0 BC 0 80829 2 0 CD 0 80829 1 0 80829 nNL 1k N2 m2 Member N kN L m AB P 4041 1 4041 2 8083 AC 8083 0 8083 2 0 BC 8083 0 8083 2 0 CD 8083 0 8083 1 0 8083 N adN dP bL 1k N m2 N 1P 02 1k N2 dN dP Thus Ans F92 Bv 7200 lb ft AE T 1 lb Bv a nNL AE 7200 lb2 ft AE Ans F93 Bv aNadN dP b L AE 7200 lb ft AE T Thus Ans F94 Ah 808 kN m AE 808 kN m AE 1 kN Ah a nNL AE 80829 kN2 m AE Ans Ah aNadN dP b L AE 8083 kN m AE 808 kN m AE 656 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS Ans Dv 2375 kN m AE T 1 kN Dv a nNL AE 2375 kN2 m AE Member n kN N kN L AB 0 0 3 0 AC 1414 8485 5091 BC 1 6 3 1800 AD 0 6 3 0 CD 1 0 3 0 6891 322 nNL 1kN2 m2 F95 Ans F96 Dh 689 kN m AE 1 kN Dh a nNL AE 6891 kN2 m AE Ans F97 689 kN m AE Dh NadN dP b L AE Member N kN L m AB 0 0 0 3 0 AC 5091 BC 1 6 3 1800 AD 6 0 6 3 0 CD P 1 0 3 0 6891 1P 62 322 622 22 221P 62 N adN dP bL 1kN m2 N 1P 02 kN dN dP Member n kN N kN L m AB 0375 1875 3 2109 BC 0375 1875 3 2109 AD 0625 3125 5 9766 CD 0625 3125 5 9766 BD 0 50 4 0 2375 nNL 1kN2 m2 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 657 Ans Bv aNadN dp b L AE 4475 kN m AE T Member N kN L m AB 0375 1875 3 2109 BC 0375 1875 3 2109 AD 0625 3125 5 9766 CD 0625 3125 5 9766 BD 50 0 50 4 0 2375 15 8P 31252 15 8P 31252 3 8P 1875 3 8P 1875 N a dN dP bL 1kN m2 N 1P 02 kN dN dP F98 Ans F99 Dv aNadN dP b L AE 2375 kN m AE T Ans F910 Bv 4475 kN m AE T 1 kN Bv a nNL AE 4475 kN2 m AE Member n kN N kN L m AB 0 6 15 0 BC 0 6 15 0 BD 1 0 2 0 CD 0 10 25 0 AD 125 10 25 3125 DE 075 12 15 135 4475 nNL 1k N2 m2 Member N kN L m AB 6 0 6 15 0 BC 6 0 6 15 0 BD P 1 0 2 0 CD 10 0 10 25 0 AD 125 10 25 3125 DE 075 12 15 135 4475 075P 12 1125P 102 NadN dP bL 1k N m2 N 1P 02 1k N2 dN dP 658 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS Member n kN N kN L m AB 05 50 2 5000 DE 05 50 2 5000 BC 05 50 2 5000 CD 05 50 2 5000 AH 07071 7071 14142 EF 07071 7071 14142 BH 0 30 2 0 DF 0 30 2 0 CH 07071 2828 5657 CF 07071 2828 5657 CG 0 0 2 0 GH 1 70 2 14000 FG 1 70 2 14000 87898 222 222 222 222 nNL 1k N2 m2 Member N kN L m AB 05 50 2 5000 DE 05 50 2 5000 BC 05 50 2 5000 CD 05 50 2 5000 AH 07071 7071 14142 EF 07071 7071 14142 BH 30 0 30 2 0 DF 30 0 30 2 0 CH 07071P 07071 2828 5657 CF 07071P 07071 2828 5657 CG 0 0 0 2 0 GH 1 70 2 14000 FG 1 70 2 14000 87598 1P 302 1P 302 222 222 222 107071P 42432 222 107071P 42432 05P 30 05P 30 05P 30 05P 30 N adN dP bL kN m N 1P 40 k N dN dP F911 Ans F912 Cv 876 kN m AE T 1 kN Cv a nNL AE 87598 kN2 m AE Ans Cv 876 kN m AE T Cv aNadN dP b L AE 87598 kN m AE FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 659 F913 For the slope Ans For the displacement Ans F914 For the slope Then Set Then Ans For the displacement Then Set Then F915 For the slope and Ans For the displacement and Ans F916 For the slope Then Set Then Ans For the displacement Then Set Then Ans Av L L 0 Ma0M 0P b dx EI L 3 m 0 41x2dx EI 18 kN m3 EI c M 4 kN m P 0 0M 0P x M 1Px 42 kN m uA L L 0 Ma 0M 0M b dx EI L 3 m 0 4112dx EI 12 kN m2 EI M 4 kN m M 4 kN m 0M 0M 1 M M Av 18 kN m3 EI c 1 kN Av L L 0 mM EI dx L 3 m 0 x142dx EI 18 kN2 m3 EI M 4 kN m m x kN m uA 12 kN m2 EI 1 kN m uA L L 0 muM EI dx L 3 m 0 112142dx EI 12 kN2 m3 EI M 4 kN m mu 1 kN m Av L L 0 Ma0M 0P b dx EI L 3 m 0 30xxdx EI 270 kN m3 EI T M 130x2 kN m P 30 kN 0M 0P x M Px uA L L 0 Ma 0M 0M b dx EI L 3 0 m130x2112dx EI 135 kN m2 EI M 130x2 kN m M 0 0M 0M 1 M 30x M Av 270 kN m3 EI T 1 kN Av L L 0 mM EI dx L 3 m 0 1x2130x2 EI dx 270 kN2 m3 EI uA 135 kN m2 EI 1 kN m uA L 2 0 muM EI dx L 3 m 0 112130x2 EI dx 135 kN2 m3 EI 660 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS F917 For the slope and Ans For the displacement and Ans F918 For the slope Then Set Then Ans For the displacement Then Set then Ans F919 For the slope and Ans For the displacement and Ans F920 For the slope Then Set then Ans 17067 kN m2 EI 171 kN m2 EI uA L L 0 Ma 0M 0M b dx EI L 8 m 0 132x 4x2211 0125x2 EI dx M 132x 4x22 kN m M 0 0M 0M 1 0125x M M 0125Mx 32x 4x2 Cv 427 kN m3 EI T 1 kN Cv L mM EI dx 2 L 4 m 0 05x132x 4x22 EI dx 42667 kN2 m3 EI M 132x 4x22 kN m m 105x2 kN m uA 171 kN m2 EI 1 kN m uA L L 0 muM EI dx L 8 m 0 11 0125x2132x 4x22 EI dx 17067 kN2 m3 EI M 132x 4x22 kN m mu 11 0125x2 kN m Bv L L 0 MadM 0P b dx EI L 3 m 0 1x321x2dx EI 486 kN m3 EI T M 1x32 kN m P 0 0M 0P x M 1Px x32 kN m uB L L 0 Ma 0M 0M b dx EI L 3 m 0 1x32112dx EI 2025 kN m2 EI M 1x32 kN m M 0 0M 0M 1 M 1M x32 kN m Bv 486 kN m3 EI T 1 kN Bv L L 0 mM EI dx L 3 m 0 1x21x32 EI dx 486 kN2 m3 EI M 1x32 kN m m 1x2 kN m uB 2025 kN m2 EI 1 kN m uB L L 0 muM EI dx L 3 m 0 1121x32 EI dx 2025 kN2 m3 EI M 1x32 kN m mu 1 kN m FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 661 For the displacement Then Set then Ans F921 For the slope and Ans For the displacement and Ans F922 For the slope and Thus and Set Ans For the displacement and Thus and Set 80 kN m3 EI T C L L 0 Ma0M 0P b dx EI L 2 m 0 112x12102 EI dx L 2 m 0 3121x2 2241x22 EI dx P 0 M2 121x2 22 kN m 0M2 0P x2 0M1 0P 0 M2 121x2 22 Px2 M1 112x12 kN m 72 kN m EI uC L L 0 Ma 0M 0M b dx EI L 2 m 0 12x1102 EI dx L 2 0 3121x2 224112 EI dx M 0 M2 121x2 22 0M2 0M 1 0M1 0M 0 M2 121x2 22 M M1 112x12 kN m Cv 80 kN m3 EI T 1 kN Cv 80 kN2 m3 EI 1 kN C L L 0 mM EI dx L 2 m 0 0112x12 EI dx L 2 m 0 1x223121x2 224 EI dx M2 121x2 22 kN m m1 0 m2 x2 M1 112x12 kN m uC 72 kN m2 EI 1 kN m uC 72 kN2 m3 EI 1 kN m uC L L 0 muM EI dx L 2 m 0 0112x12 EI dx L 2 m 0 1123121x2 224 EI dx M2 121x2 22 kN m 1mu21 0 1mu22 1 kN m M1 112x12 kN m 42667 kN m3 EI 427 kN m3 EI T Cv L Ma0M 0P b dx EI 2 L 4 m 0 132x 4x22105x2dx EI M 132x 4x22 kN m P 0 0M 0P 05x M 05Px 32x 4x2 662 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS F923 and Ans F924 Then Set and Ans F101 Superposition Ans Equilibrium Ans Ans Ans MA 40 kN m 100122 40142 MA 0 d gMA 0 Ay 60 kN 100 40 Ay 0 c gFy 0 Ax 0 Fx 0 By 100 kN 1 c2 0 26667 kN m3 EI Bya 2667 m3 EI b B B ByfBB fBB 1L223 3EI L3 24EI 43 24EI 2667 m3 EI c B Px2 6EI13L x2 401222 6EI 33142 24 26667 kN m3 EI T 1620 kN m3 EI T Cv L L 0 Ma0M 0P b dx EI L 6 m 0 a24x1 1 6x3 1b105x12 EI dx1 L 6 m 0 a48x2 6x2 2 1 6x3 2b105x22 EI dx2 M2 a48x2 6x2 2 1 6x3 2b kN m P 0 M1 a24x1 1 6x3 1b kN m 0M1 0P 05x1 0M2 0P 05x2 M1 05Px1 24x1 1 6x3 1 M2 05Px2 48x2 6x2 2 1 6x3 2 Cv 1620 kN m3 EI T 1620 kN2 m3 EI 1 kN Cv L L 0 mM EI dx L 6 m 0 105x12a24x1 1 6x3 1b EI dx1 L 6 m 0 105x22a48x2 6x2 2 1 6x3 2b EI dx2 M2 a48x2 6x2 2 1 6x2 2b kN m M1 05x1 M2 05x2 M1 a24x1 1 6x1 3b kN m FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS 663 F102 Superposition Ans Equilibrium Ans Ans F103 Superposition Ans Equilibrium Ans Ans Ans F104 Superposition Ans 1 c2 0 M0L2 4EI Bya L3 6EIb By 3M0 2L B B ByfBB fBB L3 AC 48EI 12L23 48EI L3 6EI c B M0x 6EILAC 1L2 AC x22 M01L2 6EI12L2312L22 L24 M0L2 4EI T MA 700 kN m MA 1833162 60132 0 dMA 0 Ay 4167 kN 417 kN Ay 1833 60 0 cFy 0 Ax 0 Fx 0 By 183311032 N 1833 kN 183 kN T 511032 m 0027 m By31211062 mN4 B B ByfBB fBB L3 3EI 63 3EI 72 m3 EI 72 m3 320011092 Nm24330011062 m44 1211062 mNc B wL4 8EI 101642 8EI 1620 kN m3 EI 162011032 N m3 320011042 Nm24330011062 m44 0027 mT MA w0L2 15 MA w0L 10 1L2 a1 2 w0Lb aL 3 b 0 dMA 0 Ay 2w0L 5 Ay 1 2 w0L w0L 10 0 cFy 0 Ax 0 Fx 0 T 0 w0L4 30 EI Bya L3 3EIb By w0L 10 w0L 10 c B B ByfBB fBB L L 0 mm EI dx L L 0 1x21x2 EI dx L3 3EI T B L L 0 mM EI dx L L 0 1x2a w0 6Lx3b EI dx w0L4 30 EI T 664 FUNDAMENTAL PROBLEMS PARTIAL SOLUTIONS AND ANSWERS Equilibrium Ans Ans Ans F105 Superposition Ans Equilibrium Ans Ans Ans F106 Ans Equilibrium Ans Ans Ans Ax 0 Fx 0 Ay 2667 kN 267 kN Ay 2667 6667 120 0 cFy 0 Cy 2667 kN 267 kN Cy1122 6667162 120162 0 dMA 0 1 T2 511032 m 0045m By30611062 mN4 By 666711032 N 667 kN B B ByfBB fBB L3 AC 48EI 123 48EI 36 m3 EI 36 m3 320011092 Nm24330011062 m44 0611062 mN c B 5wL4 AC 384EI 5110211242 384EI 2700 kN m3 EI 270011032 N m3 320011092 Nm24330011062 m44 0045 m T Ax 0 Fx 0 Ay 203125 kN 203 kN Ay 34375 50 46875 0 cFy 0 Cy 46875 kN 469 kN 34375142 50122 Cy182 0 dMA 0 By 34375 kN 344 kN 1 c2 0 36667 kN m3 EI Bya 10667 m3 EI b B B ByfBB fBB L3 AC 48EI 83 48EI 10667 m3 EI c B Pbx 6EILAC 1L2 AC b2 x22 50122142 6EI182 182 22 422 36667 kN m3 EI T Ay 5M0 4L 3M0 2L M0 4L Ay 0 cFy 0 Cy M0 4L Cy12L2 3M0 2L 1L2 M0 0 dMA 0 Ax 0 Fx 0 665 210 211 a b c d e 213 a b c 214 a b c Stable and statically indeterminate to the second degree 215 a b Stable and statically indeterminate to the first degree c 217 a Unstable b Stable and statically indeterminate to the sixth degree c Stable and statically determinate d Unstable 218 219 221 222 223 Cy 293 k NB 854 k Cx 920 k NA 959 k Ax 0 Ay 700 k NB 150 k ND 600 k NF 400 k MB 84 kN m By 30 kN Bx 0 NA 12 kN Ay 500 k Ax 953 k FB 110 k Ax 100 kN Ay 160 kN By 480 kN Unstable Unstable Stable and statically determinate Unstable Statically indeterminate to 1 Statically indeterminate to 1 Statically determinate Unstable Statically determinate Statically determinate Unstable Indeterminate to 2 on FED force of 132 k on BE 220 kft Answers to Selected Problems Chapter 1 11 12 13 15 16 17 19 110 111 113 114 115 117 Windward Leeward 118 119 121 122 Chapter 2 21 22 23 25 26 27 peak triangular peak triangular 29 and Ey 129 k on FED triangular loads peaks 206 kft on BE trapezoidal loading peak 4125 kft on ABCD with 2 forces of 736 lb 184 lbft on BG 368 lbft on ABCD 2 forces of 1725 lb on BG 230 lbft on ABCDE 3 forces of 675 k on BF 0675 kft on EF 09 kft on ABCDE 3 forces of 135 k with concentrated force of 267 kN at E on FED triangular loadings peaks 107 kNm on BE trapezodial load peak 214 kNm on FED Ey 356 kN on BE 142 kNm pf 36 lbft2 pf 0816 kNm2 F 813 kN p 186 psf p 154 psf p20 218 psf p015 209 psf p30 311 psf p25 301 psf p20 291 psf p015 278 psf L 302 kNm2 L 170 kNm2 Fs 945 k Total dead load 106 lbft2 w 240 lbft 620 kNm w 468 lbft F 173 kN w 521 lbft F 246 k F 483 k 666 ANSWERS TO SELECTED PROBLEMS 242 243 Chapter 3 31 a Unstable b Statically indeterminate to 1 c Statically determinate d Statically determinate 32 a Statically determinate b Statically determinate c Unstable 33 a Internally and externally stable to 2 b Internally and externally stable to 1 c Internally and externally stable to 1 35 36 37 FBE 924 kN C FCB 924 kN T FCE 924 kN C FDE 462 kN C FDC 924 kN T FDC 0 FED 35 k C FEC 224 k T FGC 0 FGE 224 k C FFE 15 k C FFG 0 FHG 224 k C FHC 224 k C FBH 0 FBC 400 k T FAB 400 k T FAH 447 k C FBA 722 lb T FBE 297 lb T FDE 780 lb C FDB 0 FCB 720 lb T FCD 780 lb C Cy 319 k Cx 816 k By 221 k Bx 248 k Dy 700 kN Ax 450 kN Ay 830 kN Cy 700 kN Dx 450 kN Cx 450 kN Cy 197 k Cx 678 lb 225 226 227 229 230 231 233 234 235 237 238 239 241 Ay 147 k Ax 522 lb FCD 350 lb FBE 153 k Dy 170 k Dx 170 k Ax 188 k Ay 700 lb T 350 lb Cy 300 N Cx 300 N Ax 300 N Ay 300 N Ay 170 k Ax 290 k By 166 k By 410 k Bx 970 k NA 112 k Cy 667 kN Cx 100 kN Ay 667 kN Ax 300 kN w2 167 lbft w1 833 lbft w2 4P L w1 2P L MB 320 kN m Bx 0 By 120 kN Ay 200 kN Bx 0 By 170 kN MB 630 kN m Ay 400 kN Ax 0 Ay 750 kN MA 450 kN m By 750 kN Cy 0 Bx 200 kN Ay 147 kN By 512 kN Ay 398 lb Ay 474 lb C 948 lb ANSWERS TO SELECTED PROBLEMS 667 39 310 311 313 314 FKJ 429 kN C FKB 400 kN C FAB 411 kN T FAK 429 kN C FFA 620 kN T FBF 620 kN C FBA 311 kN T FCB 220 kN T FCF 877 kN T FEC 620 kN C FEA 885 kN C FDC 840 kN T FDE 163 kN C FFE 125 kN T FFB 750 kN T FBE 417 kN C FAB 100 kN C FAF 180 kN C FGA 15 kN T FGF 20 kN T FCE 5 kN T FBC 667 kN C FCD 667 kN C FED 833 kN T FCH 586 k T FGH 767 k C FGC 300 k C FFG 767 k C FFC 404 k C FBC 1375 k T FBH 0 FAB 1375 k T FAH 215 k C FDC 8875 k T FDF 0 FED 8875 k T FEF 117 k C FDE 167 k C FCD 133 k T FCE 300 k C FFE 133 k C FFC 500 k T FBC 267 k C FBF 900 k C FAB 267 k C FAF 333 k T FEA 462 kN C FBA 924 kN T 315 317 318 319 Members KN NL MB BL CL IO OH GE EH HD are zero force members 321 322 323 325 FCD 101 kN C FID 424 kN T FIH 600 kN T FCF 0 FCD 223 kN C FGF 178 kN T FBG 180 kN T FHG 101 kN C FBC 800 kN T FGC 0 FGF 125 kN C FCD 667 kN T FJK 403 k C FJN 250 k T FCD 200 k T FFC 486 k T FFG 166 k C FDC 117 k T FBC 231 kN C FBF FCF 231 kN C FGB FEC 0 FGF FEF 400 kN T FAB FDC 346 kN C FAG FDE 400 kN T FGC 20 kN T FHC FCF 833 kN C FHG FGF 167 kN C FBH FDF 10 kN T FBC FCD 20 kN T FAB FDE 20 kN T FAH FFE 25 kN C FFE 411 kN T FGE 400 kN C FFG 429 kN C FID 911 kN T FED 343 kN T FHG 429 kN C FHE 794 kN T FHD 600 kN C FIH 357 kN C FCD 274 kN T FCI 911 kN T FJC 600 kN C FJI 357 kN C FBC 343 kN T FBJ 794 kN T 668 ANSWERS TO SELECTED PROBLEMS 335 337 338 339 Chapter 4 41 42 43 45 46 47 MC 850 kN m VC 175 kN NC 0 MD 1875 kN m VD 125 kN ND 0 MC 0375 kN m VC 075 kN NC 0 w 100 Nm MC 8125 k ft NC 120 kip VC 0 MB 6325 k ft VB 850 lb NB 0 MA 1125 k ft VA 450 lb NA 0 MD 917 k ft VD 667 k ND 0 MC 583 k ft VC 333 k NC 0 MD 933 kN m VD 533 kN ND 0 MC 0667 kN m VC 0667 kN NC 0 FED 346 kN T FCD 231 kN T FCF 0 FBE 416 kN T FDF 416 kN C FBC 115 kN C FAC FAE FDE FDC FCE 0 FBD 200 kN C FBE 566 kN T FAB 400 kN T FED 0 FFE 0 FBE 180 kN T FBG 180 kN T 326 327 329 330 331 333 334 FAG FAE 101 kN T FAB 24 kN C FBC FBD 134 kN C FBE 480 kN T FBC FBD 370 kN C FAC FAD 150 kN C FAB 646 kN T FBE 115 kN T FCD 473 kN C FCF 141 kN C FAF 158 kN T FAD 424 kN T FBC 115 kN T FBA 115 kN T FED 346 kN C FEF 115 kN T FED 160 kN C FFE FCD 113 kN C FFD FCE 894 kN T FAF FBC 400 kN C FAD FBE 0 FJE 150 k C FHE 0707 k T FCE 0707 k C FDE 0500 k C FCD 0 FJH 212 k T FKJ 150 k C FKH 0707 k T FFH 212 k T FFC 0707 k T FBC 100 k C FBF 212 k T FIF 0707 k T FIK 0707 k C FLK 0500 k C FLI 0707 k T FGI 0707 k C FGL 0500 k C FGB 0707 k T FAG 150 k C FAB 0 FCD 975 kN T FCJ 270 kN T FKJ 115 kN C FCD 101 kN C FIC 600 kN C FJI 900 kN T ANSWERS TO SELECTED PROBLEMS 669 49 410 411 413 414 415 417 418 419 M 5250x 35006 lb ft 10 ft 6 x 14 ft V 250 lb M 575x2 1050x 40006 lb ft 4 ft 6 x 6 10 ft V 51050 150x6 lb M 5250x6 lb ft 0 x 6 4 ft V 250 lb M 58x 1206 k ft 6 ft 6 x 10 ft V 800 k M 5x2 30x 2166 k ft 0 x 6 6 ft V 30 2x k M 520x 246 kN m 2 m 6 x 3 m V 20 kN M 512x 86 kN m 1 m 6 x 6 2 m V 12 kN M 54x6 kN m 0 x 6 1 m V 4 kN M 5325x 26 kN m V 325 kN 4 m 6 x 6 8 m M 325x 14 kN m 2 m 6 x 6 4 m V 325 kN M 375x kN m 0 x 6 2 m V 375 kN M MO L L x a 6 x L V MO L M MO L x 0 x 6 a V MO L M 5550x 226 kN m 3 m 6 x 4 m V 550 kN M 505x 46 kN m 1 m 6 x 6 3 m V 0500 kN M 5450x6 kN m 0 x 6 1 m V 450 kN ME 0675 k ft VE 0450 k NE 0 MD 110 k ft VD 0930 k ND 0 MC 112 k ft VC 0870 k NC 0 MC 350 kN m VC 125 kN NC 0 421 422 423 425 426 427 429 430 431 433 434 435 437 438 439 441 442 443 445 446 447 449 Mmax 90 k ft Vmax 12 k Mmax 155 k ft Vmax 222 k Mmax 525 kN m Vmax 145 kN Mmax 267 kN m Vmax 133 kN Mmax 162 k ft Vmax 36 k Mmax 144 k ft Vmax 200 k Mmax 90 k ft Vmax 12 k Mmax 876 k ft Vmax 118 k Mmax 180 kN m Vmax 83 kN Mmax 345 kN m Vmax 245 kN Mmax 552 k ft Vmax 380 k Mmax 6400 lb ft Vmax 1200 lb Mmax 224 kN m Vmax 186 kN Mmax 9wL2 128 Vmax 3wL 8 Mmax 2401 lb ft Vmax 510 lb Mmax 0521 kN Vmax 125 kN Mmax 116 k ft Vmax 304 k Mmax 60 k ft Vmax 101 k Mmax 20 kN m Vmax 489 kN Mmax 2400 lb ft Vmax 386 lb M 0148x3 4x2 36x 108 kN m V 0444x2 8x 36 kN M 5333x3 800x 12006 lb ft V 510x2 8006 lb 670 ANSWERS TO SELECTED PROBLEMS Chapter 5 51 52 53 55 56 57 59 510 511 513 514 515 517 518 519 521 522 523 525 526 h3 9375 ft h2 7500 ft h1 4375 ft Cx 467 k Cy 85 k Ax 467 k Ay 950 k Bx 467 k By 500 k MD 108 kN m FC 117 k FA 108 k FB 677 k T 432 kN Ay 155 kN Cy 955 kN Ax 0 Thanger 131 kN TE TD 875 kN TF 70 kN Tmax 520 kN y 385x2 577x103 m Tmax 693 MN Tmin 625 MN Mmax 625 k ft Vmax 5 k Thanger 10 k Tmax 117 k Tmin 100 k Tmax 109 k Tmin 130 k Tmax 144 k w 519 lbft Tmax 431 kN Tmin 400 kN TB 103 k TO 703 k y 00356x2 Fmax 125 kN P2 625 kN P1 250 kN P 714 lb yD 210 m TAB 299 kN TCD 372 kN TBC 160 kN yB 243 m Tmax 641 kN l 202 ft FBC 467 lb FBA 830 lb FCD 881 lb 527 529 Chapter 6 615 617 618 619 621 622 623 625 626 627 630 631 633 634 635 637 657 658 659 661 662 663 MCmax 168 k ft MCmax 200 kN m VBmax 146 k MCmax 441 kN m FCFmax 754 k T FCDmax 120 k T MBmax 123 kN m VBCmax 821 kN VCDmax 6 k MHmax 192 k ft VDEmax 507 k MCmax 118 k ft VDEmax 529 k MCmax 105 k ft VBCmax 7 k MDmax 6125 k ft VABmax 273 k MGmax 467 kN m VBCmax 178 kN MGmax 981 kN m VBCmax 715 kN VC max 330 k MEmax 5125 k ft Aymax 205 k VCmax 236 kN MCmax 720 kN m Bymax 876 kN VDmax 540 k MAmax 864 k ft MDmax 4 k ft VA max 401 k MCmax 151 k ft Aymax 701 k Bymax 2475 k MCmax 1125 k ft MBmax 375 k ft Bymax 124 k VCmax 20 kN MCmax 142 kN m TAD 408 k Ay 194 k Dy 806 k Ax 3 k Cx 0276 k Cy 0216 k Ax 272 k Ay 378 k By 0216 k Bx 272 k 75 76 77 79 FFG 325 k C FCF 248 k T FDG 248 k C FAF 600 kN T FBE 400 kN T FAB 133 kN C FFE 133 kN T FAE 100 kN C FFB 100 kN T FCD 200 kN T FBC 267 kN C FED 267 kN T 333 kN C FBD 333 kN T FEC FDE 205 k C FCF 140 k C FBG 145 k C FAH 215 k C FCD 0 FEF 200 kN C 173 k T FBC FFG 200 k C 0833 k T FCG FAB 200 k C 193 k C FGH 242 k T FBH FDE 205 k C FCE 225 k T FAG FBF FDF 0 FDE 1375 k C FBG 700 k C FAH 1425 k C FCD 900 k T FEF 110 k C FCE 1125 k T FDF 1125 k C 177 k T FBC 197 k C FFG FCG 0417 k T FBF 0417 k C 967 k C FGH 767 k T FAB FBH 121 k T FAG 121 k C FFC 50 k C FFE 0833 k C FED 158 k C FCD 583 k T 665 666 667 669 670 671 673 674 675 677 678 679 681 Chapter 7 71 72 73 825 k C FDF 825 k T FEC FBC 125 k T FGB 50 k C FGF 75 k C 118 k T FBF 118 k C FGC FHG 417 k C FAH 142 k C FAB 917 k T 589 k C FAG 589 k T FHB FCD 400 kN C FBE 400 kN C FAF 700 kN C FBC 0 FDE 200 kN C 283 kN T FBD FAB 0 FEF 200 kN C 283 kN T FBF FCE 0 FAE FCD 300 kN C FBE 200 kN C FAF 600 kN C FBC 100 kN T FDE 100 kN C FCE FBD 141 kN C FAB 100 kN T FEF 100 kN C FAE FBF 141 kN C Mmax 105 k ft Vabs max 125 k Mmax 130 k ft Mmax 972 k ft Mmax 164 kN m Vabs max 675 kN Mmax 555 k ft Mmax 39 kN m Vmax 10 kN Mmax 645 kN m Mmax 678 kN m FBCmax 237 k T MCmax 4375 k ft MCmax 340 kN m ANSWERS TO SELECTED PROBLEMS 671 710 711 713 714 715 717 718 719 Pinned Ax 600 kN MA 162 k ft MB 9 k ft MC 72 k ft Ay 12 k By 16 k Cy 4 k Ax 0 Bx 0 Cx 0 MH 270 k ft ML 2025 k ft MI 900 k ft 403 kN m MA MD 720 k ft MF 405 k ft MB 378 kN m MA 486 kN m FAB 900 kN T FCF 500 kN C FDE 400 kN C 227 kN T FAF 227 kN C FBC 533 kN T FEF FBF 150 kN C FAC 150 kN T 533 kN C FCD FCE 667 kN C FDF 667 kN T FAB 0 FCG 550 k C FDF 200 k C FDE 150 k T 212 k C FEF FAG 105 k C FBC 500 k T 778 k T FBG FFG 500 k C FCD 150 k T 495 k T FCF FDG FAC 0 FAB 275 k T FCG 100 k C FDF 0250 k C FDE 150 k T 212 k C FEF FAG 775 k C FBC 775 k T FBG 389 k T FAC 389 k C FCD 325 k T Fixed 721 722 723 725 726 727 FEI 175 kN C FHI 400 kN C FEH 175 kN T FGH 170 kN T FEF 165 kN C FFH 175 kN C MB 300 kN m By 140 kN Bx 100 kN MA 300 kN m Ay 140 kN 100 kN Ax FDE 100 kN T FCE 150 kN C FEF 140 kN C FCG 400 kN C FEG 150 kN T FDE 200 kN T FCE 275 kN C FCG 400 kN C FEF 240 kN C FEG 275 kN T FDE 300 k C FDF 3125 k T FFG 100 k C FCD 200 k T FDG 3125 k C MB 900 k ft By 1875 k Bx 150 k MA 900 k ft Ay 1875 k 150 k Ax FCE 106 k T FCF 177 k T MB 180 kN m By 900 kN Bx 600 kN MA 180 kN m Ay 900 kN Ax 600 kN By 180 kN Bx 600 kN Ay 180 kN 672 ANSWERS TO SELECTED PROBLEMS 86 87 89 810 811 813 814 815 817 818 819 821 822 823 825 90 kN m3 EI T C 18 kN m2 EI uB a 0152 L a 0152 L 386 mm T C 000171 rad uC Pa3 4EIc C Pa2 4EI uC Pa3 4EI c C Pa2 4EI uC a L 3 a 0153 L a 0153 L 50625 k ft3 EI T C 39375 k ft2 EI uC 0322 in T max 000268 rad uB 0322 in T max 000268 rad uB v3 w 24EIx4 3 8ax3 3 24a2x2 3 4a3x3 a4 7wa4 12EI vC v1 wax1 12EI2x2 1 9ax1 7wa3 6EI uB w0L4 120EI vmax v w0x 960EIL40L2x2 16x4 25L4 uA 5w0L3 192EI v2max wL4 1823EI vB wa3 24EIa 4L uB wa3 6EI 729 730 731 733 734 Chapter 8 81 82 83 85 v3 wa3 24EI4x3 a 4L v1 w 24EIx1 4 4ax3 1 6a2x1 2 vB 11PL3 48EI vC PL3 6EI uA 3PL2 8EI vmax Pa 24EI4a2 3L2 v2 Pa 6EI3x2x2 L a2 v1 Px1 6EIx1 2 3aa L uA Paa L 2EI 299 kN C FHL 186 kN T FKL 252 kN C FHG 543 kN C FHL 529 kN T FKL 402 kN C FHG FFH 3125 k C FEH 0500 k T FFG 0 FJK 0500 k T FGF 0 FGK 3125 k C By 1875 k Bx 200 k Ay 1875 k 200 k Ax FJK 0500 k C FGF 0 FGK 1875 k C MB 120 k ft By 1125 k Bx 200 k MA 120 k ft Ay 1125 k 200 k Ax FCI 250 kN C FDI 175 kN T FDE 450 kN T ANSWERS TO SELECTED PROBLEMS 673 826 827 829 830 831 833 834 835 837 838 839 Chapter 9 91 92 93 95 295 mmT Ev 338 mmT Bv 0536 mmT Av 0536 mm T Av 10368 k ft3 EI T D 1008 k ft2 EI uD 10368 k ft3 EI T D 1008 k ft2 EI uD 169 kN m3 EI T C 75 kN m2 EI uD 25 Pa3 6EI T C 3Pa2 EI uC 25 Pa3 6EI T C 3Pa2 EI uC max 000802M0L2 EI T uA M0L 24EI 3Pa3 4EI T C 5Pa2 12EI uB 3Pa3 4EI T C 5Pa2 12EI uB F P 4 9Pa3 4EI T C 7Pa2 4EI uB 9Pa3 4EI T C 7Pa2 4EI uB 96 97 99 910 911 913 914 915 917 918 919 921 922 923 925 926 927 929 930 931 933 934 935 0455 in B 000448 rad uB 66375 N m3 EI T B 3150 N m2 EI uB B 66375 N m3 EI T uB 3150 N m2 EI 0282 in T C 000670 uC 0145 inT C 000156 rad uC 0145 inT C 000156 rad uC Pa2 6EI uA 5Pa2 6EI uC 5Pa2 6EI uC 2Pa3 3EI T C PL2 16EI uB PL3 48EIT C uB PL2 16EI C PL3 48EIT 0507 inc Av 00341 inT Av 00341 inT Av 491 mmT Cv 170 k ft AE Dh Dh 170 k ft AE 00582 inT Av Fv 00392 inT 00392 inT Fv 199 kN m AE T Dv 295 mmT Ev 674 ANSWERS TO SELECTED PROBLEMS 103 105 106 107 109 1010 1011 1013 1014 1015 1017 MA 450 kN m Ax 240 kN Ay 330 kN Cy 390 kN MC 104 k ft Cy 565 k Cx 0 Ay 435 k MA 625 k ft Ay 3125 k Ax 300 k Cy 1875 k 296 k Ay Cy 424 k Ax 2175 k 375 k Cx Cy 0900 k Ax 0 Ay 0900 k By 720 k MA 200 lb ft Ay 75 lb Ax 0 By 75 lb 2640 k ft3 EI C 150 mm B 171 k Cy 171 k Ay Ax 0 377 k By By 5P 2 MA PL 2 Ax 0 Ay 3P 2 MA 9wL2 128 By 7wL 128 Ay 57wL 128 Ax 0 937 938 939 941 942 943 945 946 947 949 950 951 953 954 955 957 958 959 961 Chapter 10 101 102 Ay 2625 k By 3075 k Cy 14625 k Cx 0 MA w0L2 15 By w0L 10 Ay 2w0L 5 Ax 0 Cv 417 k ft3 EI T 791 k ft3 EI Ch 791 k ft3 EI Ch 00401 in Ch 0414103 rad uA 0414103 rad uA 281 mmT Cv 281 mmT Cv 1148 k ft3 EI Ch Ch 1148 k ft3 EI By wL4 4EI 5wL4 8EI Ch 440 k ft3 EI T Dv 1397 k ft3 EI T D D 1397 k ft3 EI T A 2295 kN m3 EI T uA 9 kN m2 EI w0L4 120EI T C C w0L4 120EIT 0469 inT B 000448 rad uB ANSWERS TO SELECTED PROBLEMS 675 1018 1019 1021 1022 1023 1025 1026 1027 1029 1030 1031 FCD 463 kN T FDB 0586 k C FAD FAB 0414 k T FDC FCB 0414 k T FAC 141 k T FDE 396 kN C FBD 140 kN C FBE 561 kN T FCD 100 kN C FCB 141 kN T FAB 604 kN T FAE 604 kN T FAD 854 kN C FAC 791 kN C FDA 494 k T FAB 101 k C FDB 510 k T FDC 658 k T FAC 823 k C FCB 306 k C FBC 0 FBD 0667 k T 0667 k C FAB Ay 150 kN By 750 kN Ax 153 kN 153 kN Bx Ay 0 By 0 Ax 265 kN 265 kN Bx 465 k Ay Dy 465 k Ax 259 k Dx 541 k Ax 227 k Ay 225 k Dy 225 k Dx 227 k MD 195 k ft Dy 150 k Dx 2 k Ay 150 k 1033 1034 1035 1037 1038 1039 1045 Chapter 11 111 112 113 115 116 117 119 MCD 261 k ft MCB 261 k ft MBC 660 k ft MBA 660 k ft MAB 167 k ft MBC 4125 kN m MBA 4125 kN m MDC 405 k ft MCD 9 k ft MCB 9 k ft MBC 135 k ft MBA 135 k ft MAB 495 k ft MDC 409 kN m MCD 818 kN m MCB 818 kN m MBC 818 kN m MBA 818 kN m MAB 409 kN m MBC 1925 kN m MBA 1925 kN m MCB 204 kN m MAB 185 kN m MCB 48 k ft MBC 84 k ft MBA 84 k ft MAB 102 k ft MC 106 k ft MB 876 k ft MA 462 k ft Cy 0241 k FAC 280 k FCD 748 kip Cy P 3 Cx 0 FBC 281 k T FBC 163 k T FAB 184 k T FBD 228 k C FCB 534 kN 192 kN FDB 676 ANSWERS TO SELECTED PROBLEMS 127 1214 1215 1219 1222 1223 1225 1226 Chapter 13 131 MCB 348 k ft MBC 301 k ft MBA 301 k ft MAB 348 k ft MCB 754 k ft MCD 754 k ft MDC 142 k ft MDA 142 k ft MCD 240 k ft MCB 240 k ft MBC 240 k ft MBA 240 k ft MAB MDC 0 MCD 196 k ft MCB 196 k ft MBC 104 k ft MBA 104 k ft MDC 557 k ft MCD 175 k ft MCB 175 k ft MBC 218 k ft MBA 218 k ft MAB 128 k ft MD 206 kN m MC 411 kN m MB 411 kN m MA 206 kN m MD 146 k ft Dy 960 k Dx 293 k MA 146 k ft Ay 960 k Ax 293 k MCB 0 MBC 194 k ft MBA 194 k ft MAB 230 k ft Cy 6 kN MA 30 kN m By 33 kN Ay 33 kN Ax 0 1110 1111 1113 1114 1115 1117 1118 1119 1121 1122 1123 Chapter 12 121 122 MC 122 k ft MB 187 k ft MA 230 k ft MB MC 840 k ft MDC 567 k ft MAB 254 k ft MCB 632 kN m MBC 943 kN m MCD 632 kN m MDC 332 kN m MDA 332 kN m MAD 259 kN m MCB 800 kN m MDA 640 kN m MCD 800 kN m MDC 640 kN m MCB 134 k ft MDA 134 k ft MCD 134 k ft MDC 134 k ft MBD 349 kN m MBC 349 kN m MBA 698 kN m MBC 408 k ft MBA 408 k ft MAB 211 k ft MBC 0540 kN m MBA 0540 kN m MAB 198 kN m MCB 167 k ft MAB 429 k ft MCB 36 k ft MBC 72 k ft MBA 72 k ft MAB 126 k ft MCD 272 k ft MCB 272 k ft MBC 0923 k ft MBA 0923 k ft MAB 245 k ft MBA 24 k ft MAB 105 k ft ANSWERS TO SELECTED PROBLEMS 677 132 133 135 136 137 139 1310 MDB 283 k ft MBD 283 k ft MBA 283 k ft MAB MCD 0 MDC 283 k ft MDB 283 k ft MBD 283 k ft MBA 283 k ft MDC 175 k ft MCD 351 k ft MCB 351 k ft MBC 351 k ft MBA 351 k ft MAB 175 k ft MDC 0 MEC 277 k ft MCE 553 k ft MCD 604 k ft MCB 610 k ft MFB 277 k ft MBF 553 k ft MBC 610 k ft MBA 604 k ft MAB 0 MDC 0 MEC 277 k ft MCE 553 k ft MCD 604 k ft MCB 610 k ft MFB 277 k ft MBF 553 k ft MBC 610 k ft MBA 604 k ft MAB 0 MBC 369 k ft MCB 751 k ft MCA 751 k ft MAC 376 k ft MCB 348 k ft MBC 301 k ft MBA 301 k ft MAB 348 k ft 1311 Chapter 14 141 142 143 145 146 147 149 X106 1134 288 75 0 384 288 0 0 0 0 288 216 0 0 288 216 0 0 0 0 75 0 150 0 0 0 0 0 75 0 0 0 0 100 0 100 0 0 0 0 384 288 0 0 1518 0 0 75 384 288 288 216 0 100 0 1432 0 0 288 216 0 0 0 0 0 0 100 0 0 100 0 0 0 0 75 0 0 75 0 0 0 0 75 0 384 288 0 0 1134 288 0 0 0 0 288 216 100 0 288 1216 H K X106 203033 53033 53033 53033 150 0 0 0 0 0 53033 53033 53033 53033 0 0 0 0 0 0 53033 53033 256066 0 0 0 53033 53033 150 0 53033 53033 0 256066 0 150 53033 53033 0 0 150 0 0 0 300 0 150 0 0 0 0 0 0 150 0 150 0 0 0 0 0 0 53033 53033 150 0 203033 53033 0 0 0 0 53033 53033 0 0 53033 53033 0 0 0 0 150 0 0 0 0 0 150 0 0 0 0 0 0 0 0 0 0 0 H K q2 657 k C q2 127 lb C D1 000172 in q3 333 k T q2 0 q1 333 k C D2 00230 in D1 0 K H 51072 0 20139 0 15467 116 15467 116 0 174 0 0 116 870 116 870 20139 0 20139 0 0 0 0 0 0 0 0 0 0 0 0 0 15467 116 0 0 15467 116 0 0 116 870 0 0 116 870 0 0 15467 116 0 0 0 0 15467 116 116 870 0 0 0 0 116 870 X MFC MEB 473 k ft MCB MBC 274 k ft MCF MBE 946 k ft MCD MBA 180 k ft MAB MCD 0 MDC 283 k ft 678 ANSWERS TO SELECTED PROBLEMS Chapter 16 161 162 163 165 166 167 R9 245 kN R8 554 kN R7 355 kN R6 554 kN K I 763125 0 2625 2625 0 13125 0 750 0 0 763125 2625 0 2625 0 750 0 13125 2625 2625 140 35 35 2625 0 0 2625 2625 0 35 70 0 2625 0 0 0 0 2625 35 0 70 0 0 0 2625 13125 0 2625 2625 0 13175 0 0 0 0 750 0 0 0 0 750 0 0 750 0 0 0 0 0 0 750 0 0 13125 2625 0 2625 0 0 0 13125 Y106 K I 851250 0 22500 22500 11250 0 440000 0 0 0 1055760 14400 0 0 1050000 0 5760 14400 22500 14400 108000 30000 22500 0 0 14400 24000 22500 0 30000 60000 22500 0 0 0 0 11250 0 22500 22500 11250 0 0 0 0 0 1050000 0 0 0 1050000 0 0 0 840000 0 0 0 0 0 140000 0 0 0 5760 14400 0 0 0 0 5760 14400 0 14400 24000 0 0 0 0 14400 48000 Y R9 196 kN m g R8 264 kN c R7 679 kN R6 272 kN m b R5 216 kN c R4 321 kN I 51129 0 225 1125 0 225 500 0 0 0 51125 225 0 500 0 0 1125 225 225 225 120 225 0 30 0 225 30 1125 0 225 1125 0 225 0 0 0 0 500 0 0 500 0 0 0 0 225 0 30 225 0 60 0 0 0 500 0 0 0 0 0 500 0 0 0 1125 225 0 0 0 0 1125 225 0 225 30 0 0 0 0 225 60 Y106 K 1410 1411 1413 1414 1415 Chapter 15 151 152 153 155 156 157 159 1510 1511 R4 160 kN m R3 120 kN R2 80 kN m Q6 255 k Q5 210 k Q4 255 k M2 M3 442 kN m R6 308 kN m R5 289 kN R4 230 kN m R3 7725 kN R2 414 kN R6 140 kN m R5 220 kN R4 8575 kN R3 3225 kN R6 124 kN R5 345 kN R4 193 kN R6 396 kN R5 866 kN m R4 402 kN R3 785 kN M3 116 kN m M1 275 kN m M3 225 kN m M1 90 kN m F 040533 0096 001697 011879 033333 0 0096 0128 002263 015839 0 0 001697 002263 0129 0153 0 017678 011879 015839 0153 0321 0 017678 033333 0 0 0 033333 0 0 0 017678 017678 0 025 V K AE q3 355 k T q5 164 k C D5 000546 m D6 00133 m q5 333 kN ANSWERS TO SELECTED PROBLEMS 679 K I 483333 0 0 483333 0 0 0 0 0 0 13090 785417 0 13090 785417 0 0 0 0 785417 62833333 0 785417 32416667 0 0 0 483333 0 0 490901 0 545428 7575 0 545428 0 13090 785417 0 415868 785417 0 402778 0 0 785417 31416667 545428 785417 115196464 545428 0 26180555 0 0 0 7575 0 545428 7575 0 545428 0 0 0 0 402778 0 0 402778 0 0 0 0 545428 0 26180555 545428 0 52361111 Y 680 ANSWERS TO SELECTED PROBLEMS 169 1610 1611 Appendix K I 204831 0 330404 330404 0 344170 0 201389 0 0 303103 146846 0 146846 0 302083 0 101976 330404 146846 704861 211458 140972 330404 0 0 146846 330404 0 211458 422917 0 330404 0 0 0 0 146846 140972 0 281944 0 0 0 0 344170 0 330404 330404 0 344170 0 0 0 0 302083 0 0 0 0 302083 0 0 201389 0 0 0 0 0 0 201389 0 0 101976 146846 0 146846 0 0 0 101976 Y R9 20 k R8 0 R7 20 k K I 126875 0 3625 0 3625 120833 0 604167 0 0 242422 90625 90625 0 0 75521 0 241667 3625 90625 435000 72500 145000 0 90625 3625 0 0 90625 72500 145000 0 0 906025 0 0 3625 0 145000 0 290000 0 0 3625 0 120833 0 0 0 0 120833 0 0 0 0 75521 90625 90625 0 0 75521 0 0 604167 0 3625 0 3625 0 0 604167 0 0 241667 0 0 0 0 0 0 241667 Y A1 A2 A3 A5 A6 A7 A9 AT c68 38 38 26 d A AT C 4 8 5 8 18 2 5 2 4 S 31 9 104 AT BT A BT AB C 8 12 10 20 30 25 24 36 30 S AB 318 124 A 2B C 9 3 4 2 1 5 9 3 5 S 3A 2B C 3 7 0 6 5 7 7 5 9 S A 3B C 0 12 17 31 2 1 S 2A B C 7 10 1 6 10 5 S A10 A11 A15 A17 A18 A19 A21 x3 1 x2 1 x1 1 x3 4 3 x2 5 9 x1 4 9 x3 4 3 x2 5 9 x1 4 9 A1 1 43 C 7 16 17 4 3 22 12 9 23 S ƒBƒ 30 ƒAƒ 27 AB c30 11 d AB c 10 5 d 681 Index antisymmetric loading 502 530 axial loads N 375 bending 303 305313 338 cantilevered 169 250 261 Castiglianos theorem for 381386 393 concentrated forces loads and 213214 240254 260261 concrete 5 conjugatebeam method for 326333 339 deflection and 205 216223 260 298339 364386 393 degrees of freedom 452453 459 displacement method of analysis 459466 491505 528533 distributed loads and 150151 213214 260 double integration method for 307313 338 fixedend moments FEM 491495 flanges 4 floor joists 3839 force method of analysis 403410 435438 girders 45 38 228231 261 idealized structures 3839 identification of members and nodes for 575 influence lines for 213231 240254 260261 435438 internal bending M 303 305308 internal loadings in 132159 168172 178179 kinematic indeterminacy and 576577 laminated 5 loaddisplacement relations 577578 member stiffness matrix k 577578 moment diagrams for 168172 moment distribution 491505 528533 momentarea theorems for 316325 339 MüllerBreslau principle for 216223 260 nonprismatic 528533 pinsupported ends 3437 50 500 528 procedures for analysis of 134 140 153 308 318 328 366 382 459 495 581 relative joint translation 531 rotational displacement of 364386 393 shear and moment diagrams for 150159 178179 shear and moment functions of 139143 179 shear loads V 375 A Absolute maximum shear V and moment M 250254 261 Allowablestress design ASD 26 American Association of State and Highway Transportation Officials AASHTO 9 1516 American Concrete Institute ACI 9 41 American Forest and Paper Association AFPA 9 American Institute of Steel Construction AISC 9 35 American Railroad Engineers Association AREA 9 15 American Society of Civil Engineers ASCE 9 Angular displacement 454455 Antisymmetric loads 430 502 530 Approximate methods of analysis 262297 assumptions for 264 270271 283 289 building frames 270272 282293 296 cantilever method for 288293 297 lateral loads 282293 297 portal frames 273274 282287 297 portal method for 282287 297 statically indeterminate structures 262297 trusses 264267 273277 296297 vertical loads 270272 296 Arches 7 31 194203 compressive forces and 194203 fixed 194 funicular 194 parabolic shape of 194 structural uses of 7 31 194 threehinged 194200 203 tied 194 twohinged 194 Axial forces N rotational displacement deflections and 303 344 375 B Ballandsocket connections 120121 Bays 80 Beam column 6 31 Beams 45 31 3439 50 132159 168172 178179 213231 240254 260261 298339 364386 393 403410 435438 452453 459466 491505 528533 574593 u portals 274275 297 trusses 275 297 Fixedend moments FEM 456458 485 491495 524525 531 534535 moment distribution of 488 491495 531 nonprismatic members 524525 534535 relative joint translation and 531 534535 slopedeflection equations and 456458 485 534535 Flanges 4 Flexibility matrix 428429 Flexibility of cables 182 203 Flexural rigidity EI 305306 Floors 3845 68 82 228231 261 beams 82 framing plans 3839 girders 38 228231 261 idealized structures 3845 influence lines for 228231 261 joists 3839 oneway slab system 4041 68 panel points 228229 tributary loadings 4043 68 truss bridges 82 twoway slab system 4243 68 Force F 3637 84 9495 104105 122123 130 194203 303 305313 338 342344 355362 375 381383 See also Loads Shear force arch structures 194203 axial N of 303 344 375 bending M 303 305313 338 344 by inspection 95 compressive C 84 9495 104105 130 194203 deflection rotational displacement and 303 305313 338 342344 355362 375 381383 external force P 355362 381383 internal force N 356362 magnitude 9495 support reactions 3637 tensile T 84 9495 104105 130 truss analysis and 84 9495 104105 122123 130 virtual work and 375 work and 342343 x y z components 122 zeroforce truss members 9899 122123 Force method of analysis 394449 antisymmetric loads 430 beams 403410 435438 flexural rigidity EI 305306 momentarea theorems for 316325 339 radius of curvature 305306 slope and 300301 307308 316 tangent deviations 317 Elements of a matrix 612 Energy methods of analysis 341393 Castiglianos theorem 355360 381386 393 conservation of energy principle 341 392 deflections 341393 external work Ue 341344 355 392 force F and 342343 principle of work and 346348 392 rotational displacements 341393 strain energy Ui 341 344345 355356 375380 392 virtual work 346354 364380 392393 work and 341393 Envelope of maximum influence line values 251 Equality of matrices 614 Equilibrium 4751 5967 69 182185 398401 459 cable analysis and 182185 determinacy and 4851 69 displacement and 397 459 equations of 4751 5967 69 182185 398401 459 force method of analysis and 397401 freebody diagrams for 4751 5960 requirements of 397 statically determinate applications 5967 unknowns 397 External stability trusses 87 120 131 External virtual work 348 364 External work Ue 341344 392 F Fabrication errors 349 392 564567 deflection and 349 392 force transformation matrix Q for 564565 stiffness method analysis for 564567 trusses 349 392 564567 Fan truss 8081 Fink truss 8081 Fixed arches 194 Fixed loads see Dead loads Fixed supports 3439 274275 282283 289 297 frames 274 282283 289 297 joint connections 3439 lateral loads 282283 289 297 r INDEX 685 Force method of analysis continued Bettis law 403 comparison of determinacy 396397 compatibility and 48 397407 composite structures 425427 displacements and 397398 428 equilibrium and 397401 flexibility matrix 428429 frames 411415 439445 influence lines for 435445 449 Maxwells theorem of reciprocal displacements 402403 448 principle of superposition for 400401 procedures for analysis of 401 438 statically indeterminate structures 394449 symmetric structures 429430 449 trusses 422425 Force transformation matrix Q 545 564569 598 Frames 8 31 163167 270274 282293 296297 364386 393 411415 439445 452453 459 469481 495 508517 594611 approximate analysis of 270274 282293 296297 axial loads N 375 building 270272 282293 296297 cantilever method for 288293 297 Castiglianos theorem for 381386 393 deflections and 270274 282283 297 364386 393 degrees of freedom 452453 459 displacement method of analysis 452453 459 469481 495 508517 displacement transformation matrix T 597 fixedsupported 274 282283 289 297 force transformation matrix Q 598 forced method of analysis 411415 439445 global member stiffness matrix k 599 hinges 282283 289 297 inflection points 274275 282 297 influence lines and 439445 loaddisplacement relation for 595596 member stiffness matrix k 595596 599 moment distribution 495 508517 multistory 510511 no sidesway of 469473 508509 partial fixity supports 274 297 pinsupported 273 297 portal method for 282287 297 portals 273274 297 procedure 600601 procedures for analysis of 366 382 459 495 rotational displacement of 364386 393 shear and moment diagrams for 163167 sidesway of 474481 510517 stiffness method of analysis 594611 slopedisplacement equations 459 469481 stiffness matrices 595596 599600 strain energy and 375380 structural use of 8 31 structure stiffness matrix K 600 symmetric member stiffness matrix 599 temperature T effects on 376377 transformation matrices for 597598 vertical loads on 270272 296 virtual work method of 364380 393 Framing plans 3839 Freebody diagrams 4751 5960 Funicular arches 194 G Gauss method for simultaneous solutions 623 Girders 45 38 228231 261 idealized structures 38 influence lines for 228231 261 plate 45 structural use of 45 Global member stiffness matrix k 546547 599 Global structure coordinates 540 576 625 H Highway bridges 15 Hinges 282283 289 297 437 Howe truss 8083 Hydrostatic pressure effects 25 I Idealized structures 3345 68 framing plans 3839 joints 3437 models 3845 oneway systems 4041 support connections for 3437 68 tributary loadings 4043 68 twoway system 4243 Identity matrix 613 Impact load factor I 16 686 INDEX moment diagrams for 168172 normal force N and 133135 178 procedures for analysis of 135 140 153 shear and moment diagrams for 150159 178179 shear and moment functions of 139143 178179 shear force V and 133138 178 sign convention for 134 specific points forces at 133138 178 structural members 132179 superposition method of for 168172 Internal stability trusses 8889 120 131 Internal virtual work 364365 Inverse of a matrix 620622 J Joints 3439 50 5967 68 84 9497 123 130131 489 531 534535 compressive force C applied to 84 130 equilibrium equations applied to 5967 fixedconnected 3439 fixedend moments FEM and 531 534535 force F reactions 3637 idealized structures 3339 68 member stresses and 84 method of 9497 123 131 nonprismatic members 531 534535 pinconnected 3437 50 5961 84 130 relative joint translation 531 534535 rollerconnected 3437 stiffness factor K 489 support connections for 3437 68 tensile force T applied to 84 130 truss analysis and 84 9497 13 130131 K Kinematic indeterminacy 541 576577 L Laminated beams 5 Lateral loads 282293 297 approximate analysis for 282293 297 building frames 282293 297 cantilever method for 288293 297 deflection by 282283 297 fixed supports for 282283 289 297 portal method for 282287 297 Line of action 94 Inflection points 274275 282 297 301 338 Influence area live loads 13 Influence lines 204261 435445 449 absolute maximum shear V and moment M 250254 261 beams 213231 240254 260261 435438 bridge design and 240254 261 building design and 228231 261 concentrated forces loads and 213214 240254 260261 construction of 205212 curve reactions for 435436 449 deflection and 205 216223 260 envelope of maximum values 251 equations 206212 floor girders 228231 261 frames 215 439445 live loads and 204261 maximum at a point 240249 Maxwells theorem of reciprocal displacements for 435437 moments M and 216219 221223 244245 250254 261 437 MüllerBreslau principle for 216223 260 pin or hinge for 437 procedures for analysis of 206 438 qualitative 216223 438445 quantitative 438 series of concentrated loads 240249 261 shear V and 216220 240243 250254 261 436 shear and moment diagrams compared to 205206 sliding devices for 436 statically determinate structures 204261 statically indeterminate structures 435445 449 trusses 232235 261 uniform distributed loads and 213214 260 unit load positions for 206212 260261 Integration for virtual work 364365 Internal loads 47 132179 303 305308 beams 132159 178179 bending moment force M 133138 178 303 305308 deflections and 303 305308 distributed loads and 150151 frames 163167 method of sections for 47 133138 178 method of superposition for 168172 INDEX 687 Linear displacement 453 455 Linear elastic material response 355356 375376 Live loads 1226 31 204261 bridge design and 1516 240254 261 building design and 1214 1626 228231 261 earthquake loads 2425 hydrostatic and soil pressure effects 25 impact factor 16 impact loads 16 influence area 13 influence lines for 204261 natural 26 reduced equation for 1314 snow loads 2324 uniform 1214 wind loads 1622 Load and resistance factor design LRFD 26 Load data structural software analysis 627 Loaddisplacement relations 542543 577578 595596 Loads 231 4043 47 68 132179 181183 203 204261 270272 282293 296297 430 501503 523527 529530 antisymmetric 430 502 530 building codes general 9 building design and 1214 1626 270272 296 cable structures 181193 203 concentrated force 182183 203 213214 240249 260261 dead 1012 31 205206 design codes 9 distributed 150151 184189 203 earthquake 2325 fixed 205206 highway bridges 15 hydrostatic pressure effects 25 idealized structures 4043 68 impact factor I 16 influence lines for 204261 internal 47 132179 lateral 282293 297 live 1226 31 204261 natural 26 nonprismatic members 523527 529530 Portland Cement Association publications for 525527 railroad bridges 15 series of 244245 261 snow 2224 soil pressure effects 25 structural members in 132179 structures and 231 symmetric 501 503 529 tributary 4043 68 uniform 1415 184189 203 213214 260 unit 206212 260261 vertical 270272 296 wind 1622 M Magnitude 9495 Matrices 428429 540551 570571 577579 597599 612624 addition and subtraction of 614 algebra using 612624 column 613 determinants for 618620 diagonal 613 displacement transformation T 544 597 elements 612 equality of 614 flexibility 428429 force transformation Q 545 564569 598 Gauss method for simultaneous solutions 623 identity 613 inverse of 620622 loaddisplacement relations and 542543 577578 595596 multiplication of 614616 order of 612 partitioning 617618 row 612 scalars and 614 square 613 stiffness 540543 546551 570571 577579 599 symmetric 578 599 613 transformation 543545 570 597598 transposed 616617 unit 613 Matrix analysis 539 565 See also Stiffness method of analysis Maxwells theorem of reciprocal displacements 402403 435437 448 Member local coordinates 540 576 627 Member data structural software analysis 626627 688 INDEX Pin supports continued moment distribution 500 528 nonprismatic members 528 portals 273 275 297 slopedeflection equations for 458 845 statically determinate structures 50 5961 statically indeterminate structures 273 275 297 437 458 485 500 528 stiffness factors for 458 500 trusses 84 130 275 297 Planar trusses 6 Portal method for analysis 282287 297 Portals 82 273277 282287 297 deflection of 270277 296297 fixedsupported 274 275 297 frames 273274 282287 297 lateral load analysis 282287 297 partial fixity 274 pinsupported 273 275 297 stability of 82 trusses 82 275277 297 Portland Cement Association 525527 Pratt truss 8083 Primary stress 84 Principle of virtual work 346348 392 Principle of work and energy 345 Purlins 80 Q Qualitative influence lines 216223 438445 Quantitative influence lines 438 R Radius of curvature 305306 Railroad bridges 15 Reduced live loads equation for 1314 Relative joint translation 531 534535 Relativestiffness factor KR 490 Rollerconnected joints 3437 120121 216217 Roofs 2324 4045 idealized structures 4045 snow loads 2324 tributary loads 4043 Rotation pinsupported end spans 457 Rotational displacement 341393 See also Deflection Row matrix 612 c r S Sag cables 182 Sawtooth truss 8081 Scalars matrix multiplication and 614 Scissors truss 8081 Secondary stress 84 Sections method of analysis 104109 131 Shear and moment diagrams 150159 163167 178179 205206 beams 150159 178179 dead loads and 205206 distributed loads and 150151 frames 163167 internal loads and 150159 163167 178179 Shear and moment functions 139143 178179 Shear force V 45 133138 178 216220 240243 250254 261 375 436 absolute maximum 250254 261 applied 45 cantilevered beams 250 261 concentrated loads and 240243 250254 261 envelope of maximum influence line values 251 influence lines and 216220 240243 250254 261 436 internal loads and 133138 178 live loads and 216220 260 MüllerBreslau principle for 216220 260 rotational displacement deflections and 375 series of concentrated loads 240243 261 simply supported beams 250251 261 virtual strain energy caused by 375 Shells surface structures 8 Sidesway 469481 485 508517 displacement method of analysis for 469481 485 508517 frames without 469473 508509 frames with 474481 510517 moment distribution for 508517 slopedeflection equations for 469480 Simple trusses 85 130 Slabs tributary loads and 4043 68 Sliding devices 436 Slopedeflection equations 450485 534535 angular displacement 454455 beams 459466 conjugatebeam method for 454457 displacement method of analysis using 450485 534535 u 690 INDEX determinacy of 4854 equilibrium equations applied to 5967 frames 51 idealized analysis of 3345 68 improper constraints for 5253 influence lines for 204261 partial constraints for 52 pinconnected 50 5961 procedures for analysis of 61 206 stability of 4854 statically indeterminate structures compared to 396397 trusses 79131 Statically indeterminate structures 4851 262297 394449 450485 486521 522537 approximate methods of analysis 262297 beams 403410 452453 435438 459466 491505 528533 Bettis law 403 building frames 270272 282293 296 cantilever method for 288293 297 composite structures 425427 deflection of 270277 282283 296297 degrees of freedom 452453 459 485 determinacy of 4851 395 452453 displacement method of analysis 450485 486521 522537 force method of analysis 394449 frames 270274 282293 296297 411415 439445 452453 459 469481 495 508517 inflection points 274275 282 297 influence lines for 435445 449 lateral loads 282293 297 Maxwells theorem of reciprocal displacements 402403 448 moment distribution for 486521 528533 nonprismatic members 522537 portal method for 282287 297 portals 273277 282287 297 procedures for analysis of 401 438 459 sidesway and 469481 slopedeflection equations for 450485 534535 statically determinate structures compared to 396397 supports and 273277 282283 289 296297 symmetric structures 429430 449 trusses 264267 275277 296297 422425 vertical loads 270272 296 fixedend moments FEM 456458 485 534535 frames 469481 linear displacement Δ 453 455 member stiffness k 457 nonprismatic members 534535 pinsupported end spans 458 485 principle of superposition for 453 procedure for analysis using 459 relative joint translation 534535 sidesway and 469481 485 sign convention for 453 span rotation 457 statically indeterminate structures 450485 stiffness factor k 457458 Slopes deflection and 300301 307308 316 Snow loads 2324 Software analysis procedure for 625627 Soil pressure effects on structures 25 Space trusses 6 120126 570571 design assumptions for 120 determinacy of 120 procedure for analysis 123 stability of 120 stiffness method of analysis 570571 supports for 120121 transformation matrices for 570 x y z force components of 122 zeroforce members in 122123 Span rotation 457 Span stiffness factor k 457458 Square matrix 613 Stability 4854 69 82 8791 120 131 by inspection 53 determinacy and 4854 69 equations of equilibrium and 4851 external 87 120 131 improper constraints and 5253 internal 8889 120 131 partial constraints and 52 space trusses 120 support reactions 52 trusses 82 8791 120 131 Static analysis earthquake loads 25 Statically determinate structures 3277 79131 212261 396397 analysis 79131 beams 49 c c INDEX 691 Stiffness factors 457458 488490 500505 524525 antisymmetric loading 502 beam member K 488 500505 joint 489 modification 500505 moment distribution and 488490 500505 nonprismatic members 524525 pinsupported ends 458 500 relative KR 490 slopedeflection equations 457458 span k 457458 symmetric beams 501503 symmetric loading 501 503 total KT 489 Stiffness matrices 540543 546551 570571 577579 599 beams 576579 frames 595596 599600 global member 546547 599 kinematic indeterminacy 541 576577 loaddisplacement relations and 542543 577578 595596 member k 541543 546551 577578 595596 599 structure K 540 547551 579 600 symmetric 578 581 trusses 540543 546551 Stiffness method of analysis 538573 574593 594611 applications of 552559 579591 600608 beams 574593 coordinate systems 540 543545 560563 576 displacement transformation matrix T 544 597 fabrication errors and 564567 force transformation matrix Q 545 564569 598 frames 594611 global member stiffness matrix k 546547 599 global structure coordinates 540 576 identification of members and nodes for 540 575 kinematic indeterminacy 541 576577 matrix analysis 539 565 member local coordinates 540 576 member stiffness matrix k 541543 546551 577578 595596 599 nodal coordinates 560563 nodes 540 575 procedures for analysis using 553 581 600601 space trusses 570571 stiffness matrices 540 542543 546559 570571 576579 595596 599600 structure stiffness equation 552 structure stiffness matrix K 540 547551 579 600 symmetric member stiffness matrix 578 599 thermal temperature effects and 564565 568569 transformation matrices for 543545 570 597598 trusses 538573 Strain energy Ui 341 344345 355356 375380 392 axial force N of 344 375 bending moment M from 344 Castiglianos theorem for 365366 393 circular members 376 deflection and 341 344 375380 392 principle of work and energy using 345 shear V and 375 temperature T changes and 376377 torsion T and 375 virtual work and 375380 Stresses joint members and 84 Stringers bridge loads and 82 Structural members see Beams Nonprismatic members Structure stiffness equation 552 Structure stiffness matrix K 540 547551 579 600 Structures 231 3277 79131 132179 180203 204261 262297 394449 450485 486521 522537 538573 574593 594611 625627 allowablestress design ASD 26 analysis of 34 79131 132179 180203 approximate methods of analysis 262297 arches 7 31 194203 beams 45 31 3839 132179 building codes general 9 cables 7 31 181193 203 classification of 48 columns 6 31 compatibility equations for 48 composite 425427 design of 9 26 determinacy of 4854 69 displacement method of analysis 397 450485 486521 522537 elements for 46 equilibrium equations of 4751 5967 69 force method of analysis 394449 frames 8 31 freebody diagrams for 4751 5960 girders 45 38 idealized 3345 68 692 INDEX rollerconnected joints 3437 120121 short links 36 121 space trusses 120121 statically indeterminate structures 273277 282283 289 296297 trusses 84 120121 130 275 297 Support data structural software analysis 627 Surface structures 8 Sway bracing truss stability 82 Symmetric matrices 578 599 Symmetric structures 429430 449 501503 529530 antisymmetric loads 430 502 530 beams 501503 529530 displacement method of analysis 501503 529530 force method of analysis 429430 449 loads 501 503 529 nonprismatic members 529530 T Temperature T 349 376377 564565 568569 effects on trusses 349 376377 564565 568569 force transformation matrix Q for 564565 rotational displacement deflections and 349 376377 stiffness method analysis for 564565 568569 Tensile force T 4 84 9495 104105 130 Thinplate structures 8 Threehinged arches 8081 194200 203 Tie rods 4 31 Tied arches 194 Torsional displacement circular members 376 Total stiffness factor KT 489 Transformation matrices 543545 570 597598 displacement T 544 597 force Q 545 564569 598 frames 597598 trusses 543545 570 Transposed matrix 616617 Tributary loads 4043 68 oneway slab system 4041 68 twoway slab system 4243 68 Trusses 67 31 79131 232235 261 264267 275277 296297 300 348360 376377 392393 422425 538573 approximate analysis of 264267 273277 296297 bridge 8283 camber of 349 improper constraints for 5253 influence lines for 204261 internal loadings in members 132179 load and resistance factor design LRFD 26 loads and 231 132179 204261 nonprismatic members 397 450485 486521 522537 partial constraints for 52 procedure for analysis of 61 software analysis 625627 stability of 4854 69 statically determinate 3277 79131 204261 statically indeterminate 4851 262297 394449 450485 486521 522537 stiffness method of analysis 538573 574593 594611 superposition principle of 46 69 support connections for 3437 68 surface 8 symmetric 429430 449 systems types of 66 thinplate shell 8 tie rods 4 31 tributary loadings 4043 68 trusses 67 31 79131 Subdivided trusses 82 Substitute members method of analysis 116119 Superposition 46 69 168172 400401 beams 168172 force method of analysis using 400401 moment diagrams constructed by method of 168172 principle of 46 69 400401 Support connections 3437 68 120121 181193 273277 282283 289 297 300303 326333 339 ballandsockets 120121 cables 37 181193 conjugatebeam method and 326333 339 deflection and 300303 326333 339 fixed 3437 274 275 282283 289 297 force F reactions 52 frames 273275 282283 289 297 hinges 282283 289 297 idealized structures 3437 joints 3437 68 partial fixity 274 pinned 3437 68 84 130 273 275 297 portals 273277 297 INDEX 693 Trusses continued Castiglianos theorem for 356360 390 classification of 8594 complex 86 116119 130 compound 8687 110112 130 coordinate systems 540 543545 560563 570 coplanar 8594 cross bracing for 264267 deflections of 275277 297 300 348360 376377 392393 design assumptions for 84 120 130 determinacy of 87 120 130 displacement transformation matrix T for 544 external loading and 348 fabrication errors 349 392 564567 fixed connections 275 297 force method of analysis 422425 force transformation matrix Q for 545 564569 global member stiffness matrix k 546547 gusset plate 79 identification of members and nodes for 540 influence lines for 232235 261 joint loadings 84 130 kinematic indeterminacy 541 loaddisplacement relations 542543 member stiffness matrix k 541543 546551 method of joints for 9497 123 131 method of sections for 104109 123 131 method of substitute members for 116119 nodal coordinates 560563 nodes 540 pin connections 84 130 275 297 planar 6 79 portals of 275277 297 procedures for analysis of 95 106 116117 123 350 357 553 roof 8081 rotational displacement of 300 348360 376377 392393 simple 85 130 space 6 120126 570571 stability of 82 8791 120 131 statically determinate 79131 statically indeterminate 264267 275277 296297 422425 stiffness matrices for 540 542543 546559 570571 stiffness method of analysis 538573 structural use of 67 31 79 structure stiffness matrix K 540 547551 supports for 275277 297 temperature thermal effects on 349 392 564565 568569 transformation matrices for 543545 570 types of 8083 vertical components 264 virtual work method of for 346354 392 zeroforce members 9899 122123 264 Twohinged arches 194 U Uniform loads 1214 184189 203 213214 260 beams 213214 260 cables and 184189 203 distributed 184189 203 213214 260 influence lines and 213214 260 live 1214 213214 260 Unit loads influence lines and 206212 260261 Unit matrix 613 V Vertical components trusses 264 Vertical loads building frame analysis and 270272 296 Virtual work 346354 364380 392393 axial force N and 375 beams 364380 393 deflection rotational displacement and 346354 364374 392 external 348 364 392 fabrication errors and 349 392 frames 364380 393 integration for 364365 internal 364365 principle of 346348 392 procedures for analysis using 350 366 shear V and 375 strain energy and 375380 temperature T and 349 392 temperature changes and 376377 torsion T and 375 truss displacements and 348354 392 W Warren truss 8083 Webs 4 694 INDEX strain energy Ui and 341 344 355356 375380 392 virtual 346354 364374 392 X x y z force components space trusses 122 Z Zero displacement and moments 327 Zeroforce truss members 9899 122123 Wind loads 1622 Work 341393 Castiglianos theorem for 355360 381386 393 conservation of energy principle 341 392 deflection rotational displacement and 341393 external Ue 341344 355 392 force F and 342343 moment M of 343 principle of energy and 345 principle of virtual 346348 INDEX 695 A L2 L2 B FEMBA PL 8 FEMAB PL 8 P A L2 L2 B FEMAB 3PL 16 P L A B FEMBA Pa2b L2 FEMAB Pb2a L2 P L A B FEMAB P L2 P a b a b b2a a2b 2 A B FEMAB 2PL 9 A B P L3 P L3 L3 FEMBA 2PL 9 P L3 P L3 L3 FEMAB PL 3 A B FEMAB 5PL 16 A B FEMBA 5PL 16 FEMAB 45PL 96 P L4 P L4 P L4 L4 P L4 P L4 P L4 L4 L A B FEMAB wL2 12 w FEMBA wL2 12 L A B FEMAB wL2 8 w A B FEMAB 11wL2 192 w FEMBA 5wL2 192 A B FEMAB 9wL2 128 L2 L2 w L2 L2 L A B FEMAB wL2 20 w FEMBA wL2 30 L A B FEMAB wL2 15 w A B w FEMAB 5wL2 96 FEMBA 5wL2 96 L2 L2 A B FEMAB 5wL2 64 w L2 L2 A B L FEMAB 6EIΔ L2 FEMBA 6EIΔ L2 A L FEMAB 3EIΔ L2 Δ Δ B Fixed End Moments