Matrix Analysis of Structures [3 ed.] 9780357448304

Table of contents :
Cover
Dedication
Contents
Preface
Chapter 1: Introduction
1.1: Historical Background
1.2: Classical, Matri X, and Finite-Element Methods of Structural Analysis
1.3: Flexibility and Stiffness Methods
1.4: Classification of Framed Structures
1.5: Analytical Models
1.6: Fundamental Relationships for Structural Analysis
1.7: Linear Versus Nonlinear Analysis
1.8: Software
Summary
Chapter 2: Matrix Algebra
2.1: Definition of a Matrix
2.2: Types of Matrices
2.3: Matrix Operations
2.4: Gauss-Jordan Elimination Method
Summary
Chapter 3: Plane Trusses
3.1: Global and Local Coordinate Systems
3.2: Degrees of Freedom
3.3: Member Stiffness Relations in the Local Coordinate System
3.4: Finite-Element Formulation Using Virtual Work*
3.5: Coordinate Transformations
3.6: Member Stiffness Relations in the Global Coordinate System
3.7: Structure Stiffness Relations
3.8: Procedure for Analysis
Summary
Chapter 4: Computer Program for Analysis of Plane Trusses
4.1: Data Input
4.2: Assignment of Structure Coordinate Numbers
4.3: Generation of the Structure Stiffness Matrix
4.4: Formation of the Joint Load Vector
4.5: Solution for Joint Displacements
4.6: Calculation of Member Forces and Support Reactions
Summary
Chapter 5: Beams
5.1: Analytical Model
5.2: Member Stiffness Relations
5.3: Finite-Element Formulation Using Virtual Work*
5.4: Member Fixed-End Forces Due to Loads
5.5: Structure Stiffness Relations
5.6: Structure Fixed-Joint Forces and Equivalent Joint Loads
5.7: Procedure for Analysis
5.8: Computer Program
Summary
Chapter 6: Plane Frames
6.1: Analytical Model
6.2: Member Stiffness Relations in the Local Coordinate System
6.3: Coordinate Transformations
6.4: Member Stiffness Relations in the Global Coordinate System
6.5: Structure Stiffness Relations
6.6: Procedure for Analysis
6.7: Computer Program
Summary
Chapter 7: Member Releases and Secondary Effects
7.1: Member Releases in Plane Frames and Beams
7.2: Computer Implementation of Analysis for Member Releases
7.3: Support Displacements
7.4: Computer Implementation of Support Displacement Effects
7.5: Temperaure Changes and Fabrication Errors
Summary
Chapter 8: Three-Dimensional Framed Structures
8.1: Space Trusses
8.2: Grids
8.3: Space Frames
Summary
Chapter 9: Special Topics and Modeling Techniques
9.1: the Structure Stiffness Matrix Including Restrained Coordinate-An Alternative Formulation of the Stiffness Method
9.2: Approximate Matrix Analysis of Rectangular Building Frames
9.3: Condensation of Degrees of Freedom, and Substructuring
9.4: Inclined Roller Supports
9.5: Offset Connections
9.6: Semirigid Connections
9.7: Shear Deformations
9.8: Nonprismatic Members
9.9: Solution of Large Systems of Stiffness Equations
Summary
Chapter 10: Introduction to Nonlinear Structural Analysis
10.1: Basic Concept of Geometrically Nonlinear Analysis
10.2: Geometrically Nonlinear Analysis of Plane Trusses
Summary
Computer Software
Flexibility Method
Bibliography
Answers to Selected Problems
Index

Citation preview

Matrix Analysis of Structures Third Edition, SI Version

Aslam Kassimali Southern Illinois University—Carbondale

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Matrix Analysis of Structures, Third Edition, SI Version Aslam Kassimali Product Director, Global Engineering: Timothy L. Anderson Vendor Content Manager: Charu Verma, MPS Limited

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Library of Congress Control Number: 2021904737 Student Edition: ISBN: 978-0-357-44830-4 Cengage 200 Pier 4 Boulevard Boston, MA 02210 USA Cengage is a leading provider of customized learning solutions with e ­ mployees residing in nearly 40 different countries and sales in more than 125 countries around the world. Find your local representative at www.cengage.com. To learn more about Cengage platforms and services, register or access your online learning solution, or purchase materials for your course, visit www.cengage.com.

Printed in the United States of America Print Number: 01   Print Year: 2021

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IN MEMORY OF MY FATHER, KASSIMALI B. ALLANA

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Contents

1



2 Matrix Algebra

Introduction1 1.1 Historical Background  2 1.2 Classical, Matrix, and Finite-Element Methods of Structural Analysis  3 1.3 Flexibility and Stiffness Methods  4 1.4 Classification of Framed Structures  5 1.5 Analytical Models  10 1.6 Fundamental Relationships for Structural Analysis  12 1.7 Linear versus Nonlinear Analysis  20 1.8 Software 21 Summary 21 2.1 2.2 2.3 2.4

23

Definition of a Matrix  24 Types of Matrices  25 Matrix Operations  27 Gauss-Jordan Elimination Method  39 Summary 45 Problems 46



3



4 Computer Program for Analysis of Plane Trusses

Plane Trusses 49 3.1 Global and Local Coordinate Systems  50 3.2 Degrees of Freedom  53 3.3 Member Stiffness Relations in the Local Coordinate System  59 3.4 Finite-Element Formulation Using Virtual Work  68 3.5 Coordinate Transformations  77 3.6 Member Stiffness Relations in the Global Coordinate System  86 3.7 Structure Stiffness Relations  90 3.8 Procedure for Analysis  106 Summary 123 Problems 124 4.1 4.2 4.3 4.4 4.5

129

Data Input  130 Assignment of Structure Coordinate Numbers  141 Generation of the Structure Stiffness Matrix  144 Formation of the Joint Load Vector  149 Solution for Joint Displacements  151

iv

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Contents v

4.6 Calculation of Member Forces and Support Reactions  153 Summary 160 Problems 161

5 Beams

163



6

252



7 Member Releases and Secondary Effects

345

8

422



5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8

Analytical Model  164 Member Stiffness Relations  172 Finite-Element Formulation Using Virtual Work  186 Member Fixed-End Forces Due to Loads  192 Structure Stiffness Relations  198 Structure Fixed-Joint Forces and Equivalent Joint Loads  207 Procedure for Analysis  215 Computer Program  228 Summary 248 Problems 248

Plane Frames 6.1 Analytical Model  253 6.2 Member Stiffness Relations in the Local Coordinate System  259 6.3 Coordinate Transformations  271 6.4 Member Stiffness Relations in the Global Coordinate System  279 6.5 Structure Stiffness Relations  287 6.6 Procedure for Analysis  302 6.7 Computer Program  322 Summary 339 Problems 339 7.1 Member Releases in Plane Frames and Beams  346 7.2 Computer Implementation of Analysis for Member Releases  366 7.3 Support Displacements  367 7.4 Computer Implementation of Support Displacement Effects  390 7.5 Temperature Changes and Fabrication Errors  395 Summary 415 Problems 416



Three-Dimensional Framed Structures 8.1 Space Trusses  423 8.2 Grids 438 8.3 Space Frames  461 Summary 499 Problems 499

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vi  Contents



9



10



Appendix



Appendix

Special Topics and Modeling Techniques 504 9.1 The Structure Stiffness Matrix Including Restrained Coordinates— An Alternative Formulation of the Stiffness Method  505 9.2 Approximate Matrix Analysis of Rectangular Building Frames  511 9.3 Condensation of Degrees of Freedom, and Substructuring  519 9.4 Inclined Roller Supports  535 9.5 Offset Connections  538 9.6 Semirigid Connections  542 9.7 Shear Deformations  546 9.8 Nonprismatic Members  553 9.9 Solution of Large Systems of Stiffness Equations  560 Summary 576 Problems 577 INTRODUCTION TO NONLINEAR STRUCTURAL ANALYSIS 10.1 Basic Concept of Geometrically Nonlinear Analysis  582 10.2 Geometrically Nonlinear Analysis of Plane Trusses  587 Summary 609 Problems 609

A Computer Software B Flexibility Method

580

611 613

Bibliography

620



Answers to Selected Problems

622



Index

637

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preface The objective of this book is to develop an understanding of the basic principles of the matrix methods of structural analysis, so that they can be efficiently implemented on modern computers. Focusing on the stiffness approach, Matrix Analysis of Structures, SI Version covers the linear analysis of two- and three-­dimensional framed structures in static equilibrium. It also presents an introduction to nonlinear structural analysis and contains the fundamentals of the flexibility approach. The book is divided into ten chapters. Chapter 1 presents a general ­introduction to the subject, and Chapter 2 reviews the basic concepts of matrix algebra relevant to matrix structural analysis. The next five chapters (Chapters 3 through 7) cover the analysis of plane trusses, beams, and plane rigid frames. The computer implementation of the stiffness method is initiated early in the text (beginning with Chapter 4), to allow students sufficient time to complete development of computer programs within the duration of a single course. Chapter 8 presents the analysis of space trusses, grids, and space rigid frames, Chapter 9 covers some special topics and modeling techniques, and Chapter 10 provides an introduction to nonlinear structural analysis. All the relationships necessary for matrix stiffness analysis are formulated using the basic principles of the mechanics of deformable bodies. Thus, a prior knowledge of the classical methods of structural analysis, while helpful, is not essential for understanding the material presented in the book. The format of the book is flexible enough to enable instructors to emphasize topics that are consistent with the goals of the course. Each chapter begins with a brief introduction that defines its objectives, and ends with a summary outlining its salient features. An important general feature of the book is the inclusion of step-by-step procedures for analysis, and detailed flowcharts, to enable students to make an easier transition from theory to problem solving and program development. Numerous solved examples are provided to clarify the fundamental concepts, and to illustrate the application of the procedures for analysis. A computer program for the analysis of two- and three-dimensional framed structures is available on the publisher’s website, https://login.cengage .com. This interactive software can be used by students to check their answers to text exercises, and to verify the correctness of their own computer programs. The MATLAB® code for various flowcharts given in the book is available to instructors for distribution to students (if they so desire). A solutions manual, containing complete solutions to text exercises, is also available for instructors. vii

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viii  Preface

A NOTE ON THE REVISED EDITION While maintaining the original theme of this book, that detailed explanation of concepts is the most effective way of teaching Matrix Analysis of Structures, SI Version the following improvements and changes have been made in this third edition: ●●

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●●

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Over 20% of the problems from the previous edition have been replaced with new ones. About 10% of the examples from the previous edition have been changed/upgraded. The shear, bending moment, and axial force diagrams for beams and frames have been added in Chapters 5 and 6 to illustrate the transition from the sign convention used in matrix analysis to the classical beam sign convention used for constructing such diagrams for design purposes. The coverage of shear deformations is expanded (Chapter 9). Some photographs have been replaced with new ones, and the page layout has been redesigned to enhance clarity. The computer software has been upgraded and recompiled to make it compatible with the latest versions of Microsoft Windows.

ACKNOWLEDGMENTS I wish to express my thanks to Timothy Anderson of Cengage Learning for his constant support and encouragement throughout this project, and to Charu Verma and Rose Kernan for all their help during the production phase. The comments and suggestions for improvement from colleagues and students who have used previous editions are gratefully acknowledged. All of their suggestions were carefully considered, and implemented whenever possible. Thanks are also due to the following reviewers for their careful reviews of the manuscripts of the various editions, and for their constructive suggestions: Riyad S. Aboutaha Georgia Institute of Technology Osama Abudayyeh Western Michigan University Amjad Aref University at Buffalo Sez Atamtuktur Clemson University George E. Blandford University of Kentucky Kenneth E. Buttry University of Wisconsin-Platteville Joel P. Conte University of California, San Diego

C. Armando Duarte University of Illinois, UrbanaChampaign Fouad Fanous Iowa State University Larry J. Feeser Rensselaer Polytechnic Institute Barry J. Goodno Georgia Institute of Technology George J. Kostyrko California State University Marc Levitan Louisiana State University

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Preface ix

Daniel G. Linzell The Pennsylvania State University Vernon C. Matzen North Carolina State University Everett E. McEwen University of Rhode Island Joel Moore California State University Ahmad Namini University of Miami

Nima Rahbar Worcester Polytechnic Institute Rudolf Seracino North Carolina State University Michael D. Symans Rensselaer Polytechnic Institute Richard L. Wood University of Nebraska-Lincoln

Finally, I would like to express my loving gratitude to my wife, Maureen, for her unfailing support and expertise in helping me prepare this manuscript, and to my sons, Jamil and Nadim, and my granddaughter, Flower, who are a never-ending source of love, pride, and inspiration for me. Aslam Kassimali

Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

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1

Introduction Learning Objectives At the end of this chapter, you will be able to: 1.1 Review the Historical Background of Matrix Structural Analysis 1.2 Differentiate Among the Classical, Matrix, and Finite-Element Methods of Structural Analysis 1.3 Define Flexibility and Stiffness Methods 1.4 Classify Common Types of Framed Structures 1.5 Define Analytical Models 1.6 Define Fundamental Relationships for Structural ­Analysis 1.7 Distinguish Between Linear and Nonlinear Analysis 1.8 Download the Accompanying Computer Software

Beijing National Olympic Stadium—Bird’s Nest (Eastimages/Shutterstock.com)

1

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2  Chapter 1   Introduction

Structural analysis, which is an integral part of any structural engineering project, is the process of predicting the performance of a given structure under a prescribed loading condition. The performance characteristics usually of interest in structural design are: (a) stresses or stress resultants (i.e., axial forces, shears, and bending moments); (b) deflections; and (c) support reactions. Thus, the analysis of a structure typically involves the determination of these quantities as caused by the given loads and/or other external effects (such as support displacements and temperature changes). This text is devoted to the analysis of framed structures— that is, structures composed of long straight members. Many commonly used structures such as beams, and plane and space trusses and rigid frames, are classified as framed structures (also referred to as skeletal structures). In most design offices today, the analysis of framed structures is routinely performed on computers, using software based on the matrix methods of structural analysis. It is therefore essential that structural engineers understand the basic principles of matrix analysis, so that they can develop their own computer programs and/or properly use commercially available software—and appreciate the physical significance of the analytical results. The objective of this text is to present the theory and computer implementation of matrix methods for the analysis of framed structures in static equilibrium. This chapter provides a general introduction to the subject of matrix computer analysis of structures. We start with a brief historical background in Section 1.1, followed by a discussion of how matrix methods differ from classical and finite-element methods of structural analysis (Section 1.2). Flexibility and stiffness methods of matrix analysis are described in Section 1.3, the six types of framed structures considered in this text (namely, plane trusses, beams, plane frames, space trusses, grids, and space frames) are discussed in Section 1.4, and the development of simplified models of structures for the purpose of analysis is considered in Section 1.5. The basic concepts of structural analysis necessary for formulating the matrix methods, as presented in this text, are reviewed in Section 1.6, and the roles and limitations of linear and nonlinear types of structural analysis are discussed in Section 1.7. Finally, we conclude the chapter with a brief note on the computer software that is provided on the publisher’s website for this book (Section 1.8) (https://cengage.com/engineering/matrixanalysis3e/kassimali/software).

1.1 

Historical background The theoretical foundation for matrix methods of structural analysis was laid by James C. Maxwell, who introduced the method of consistent deformations in 1864, and George A. Maney, who developed the slope-deflection method in 1915. These classical methods are considered to be the precursors of the matrix flexibility and stiffness methods, respectively. In the precomputer era, the main disadvantage of these earlier methods was that they required direct solution of simultaneous algebraic equations—a formidable task by hand calculations in cases of more than a few unknowns. The invention of computers in the late 1940s revolutionized structural analysis. As computers could solve large systems of simultaneous equations, the analysis methods yielding solutions in that form were no longer at a disadvantage,

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Section 1.2   Classical, Matrix, and Finite-Element Methods of Structural Analysis  3

but in fact were preferred, because simultaneous equations could be expressed in matrix form and conveniently programmed for solution on computers. S. Levy is generally considered to have been the first to introduce the flexibility method in 1947, by generalizing the classical method of consistent deformations. Among the subsequent researchers who extended the flexibility method and expressed it in matrix form in the early 1950s were H. ­Falkenheimer, B. Langefors, and P. H. Denke. The matrix stiffness method was developed by R. K. Livesley in 1954. In the same year, J. H. Argyris and S. Kelsey presented a formulation of matrix methods based on energy principles. In 1956, M. T. Turner, R. W. Clough, H. C. Martin, and L. J. Topp derived stiffness matrices for the members of trusses and frames using the finite-element approach, and introduced the now popular direct stiffness method for generating the structure stiffness matrix. In the same year, Livesley presented a nonlinear formulation of the stiffness method for stability analysis of frames. Since the mid-1950s, the development of matrix methods has continued at a tremendous pace, with research efforts in recent years directed mainly toward formulating procedures for the dynamic and nonlinear analysis of structures, and developing efficient computational techniques for analyzing large structures. Recent advances in these areas can be attributed to S.  S. Archer, C. Birnstiel, R. H. Gallagher, J. Padlog, J. S. Przemieniecki, C. K. Wang, and E. L. Wilson, among others.

1.2 Classical, Matrix, and Finite-Element Methods of Structural Analysis Classical versus Matrix Methods As we develop matrix methods in subsequent chapters of this book, readers who are familiar with classical methods of structural analysis will realize that both matrix and classical methods are based on the same fundamental principles—but that the fundamental relationships of equilibrium, compatibility, and member stiffness are now expressed in the form of matrix equations, so that the numerical computations can be efficiently performed on a computer. Most classical methods were developed to analyze particular types of structures, and since they were intended for hand calculations, they often involve certain assumptions (that are unnecessary in matrix methods) to reduce the amount of computational effort required for analysis. The application of these methods usually requires an understanding on the part of the analyst of the structural behavior. Consider, for example, the moment-distribution method. This classical method can be used to analyze only beams and plane frames undergoing bending deformations. Deformations due to axial forces in the frames are ignored to reduce the number of independent joint translations. While this assumption significantly reduces the computational effort, it complicates the analysis by requiring the analyst to draw a deflected shape of the frame corresponding to each degree of freedom of sidesway (independent joint translation), to estimate the relative magnitudes of member fixed-end moments; a difficult task even in the case of a few degrees of freedom of sidesway

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4  Chapter 1   Introduction

if the frame has inclined members. Because of their specialized and intricate nature, classical methods are generally not considered suitable for computer programming. In contrast to classical methods, matrix methods were specifically developed for computer implementation; they are systematic (so that they can be conveniently programmed), and general (in the sense that the same overall format of the analytical procedure can be applied to the various types of framed structures). It will become clear as we study matrix methods that, because of the latter characteristic, a computer program developed to analyze one type of structure (e.g., plane trusses) can be modified with relative ease to analyze another type of structure (e.g., space trusses or frames). As the analysis of large and highly redundant structures by classical methods can be quite time consuming, matrix methods are commonly used. However, classical methods are still preferred by many engineers for analyzing smaller structures because they provide a better insight into the behavior of structures. Classical methods may also be used for preliminary designs, for checking the results of computerized analyses, and for deriving the member force-displacement relations needed in the matrix analysis. Furthermore, a study of classical methods is considered to be essential for developing an understanding of structural behavior.

Matrix versus Finite-Element Methods Matrix methods can be used to analyze framed structures only. Finite-­element analysis, which originated as an extension of matrix analysis to surface structures (e.g., plates and shells), has now developed to the extent that it can be applied to structures and solids of practically any shape or form. From a theoretical viewpoint, the basic difference between the two is that, in matrix methods, the member force-displacement relationships are based on the exact solutions of the underlying (Euler-Bernoulli) differential equations, whereas in finite-element methods, such relations are generally derived by work-energy principles from assumed displacement or stress functions. Because of the approximate nature of its force-displacement relations, finite-element analysis generally yields approximate results. However, as will be shown in Chapters 3 and 5, in the case of linear analysis of framed structures composed of prismatic (uniform) members, both matrix and finite-element approaches yield identical results.

1.3 

Flexibility and Stiffness Methods Two different methods can be used for the matrix analysis of structures: the flexibility method, and the stiffness method. The flexibility method, which is also referred to as the force or compatibility method, is essentially a generalization in matrix form of the classical method of consistent deformations. In this approach, the primary unknowns are the redundant forces, which are calculated first by solving the structure’s compatibility equations. Once the redundant forces are known, the displacements can be evaluated by applying the equations of equilibrium and the appropriate member force-displacement relations.

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Section 1.4   Classification of Framed Structures  5

The stiffness method, which originated from the classical slope-deflection method, is also called the displacement or equilibrium method. In this approach, the primary unknowns are the joint displacements, which are determined first by solving the structure’s equations of equilibrium. With the joint displacements known, the unknown forces are obtained through compatibility considerations and the member force-displacement relations. Although either method can be used to analyze framed structures, the flexibility method is generally convenient for analyzing small structures with a few redundants. This method may also be used to establish member force-displacement relations needed to develop the stiffness method. The stiffness method is more systematic and can be implemented more easily on computers; therefore, it is preferred for the analysis of large and highly redundant structures. Most of the commercially available software for structural analysis is based on the stiffness method. In this text, we focus our attention mainly on the stiffness method, with emphasis on a particular version known as the direct stiffness method, which is currently used in professional practice. The fundamental concepts of the flexibility method are presented in Appendix B.

1.4 

Classification of Framed Structures Framed structures are composed of straight members whose lengths are significantly larger than their cross-sectional dimensions. Common framed structures can be classified into six basic categories based on the arrangement of their members, and the types of primary stresses that may develop in their members under major design loads.

Plane Trusses A truss is defined as an assemblage of straight members connected at their ends by flexible connections, and subjected to loads and reactions only at the joints (connections). The members of such an ideal truss develop only axial forces when the truss is loaded. In real trusses, such as those commonly used for supporting roofs and bridges, the members are connected by bolted or welded connections that are not perfectly flexible, and the dead weights of the members are distributed along their lengths. Because of these and other deviations from idealized conditions, truss members are subjected to some bending and shear. However, in most trusses, these secondary bending moments and shears are small in comparison to the primary axial forces, and are usually not considered in their designs. If large bending moments and shears are anticipated, then the truss should be treated as a rigid frame (discussed subsequently) for analysis and design. If all the members of a truss as well as the applied loads lie in a single plane, the truss is classified as a plane truss (Fig. 1.1). The members of plane trusses are assumed to be connected by frictionless hinges. The analysis of plane trusses is considerably simpler than the analysis of space (or three-dimensional) trusses. Fortunately, many commonly used trusses, such as bridge and roof trusses, can be treated as plane trusses for analysis (Fig. 1.2).

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6  Chapter 1   Introduction P2

P1

P1

P1

Fig. 1.1  Plane Truss

Fig. 1.2  Roof Truss (MP kullimratchai/Shutterstock.com)

Beams A beam is defined as a long straight structure that is loaded perpendicular to its longitudinal axis (Fig. 1.3). Loads are usually applied in a plane of symmetry of the beam’s cross-section, causing its members to be subjected only to bending moments and shear forces.

Plane Frames Frames, also referred to as rigid frames, are composed of straight members connected by rigid (moment resisting) and/or flexible connections (Fig. 1.4). Unlike trusses, which are subjected to external loads only at the joints, loads on frames may be applied on the joints as well as on the members.

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Section 1.4   Classification of Framed Structures  7 Plane of symmetry

P

w

M

Longitudinal axis

Fig. 1.3 Beam

Fig. 1.4  Skeleton of a Structural Steel Frame Building (Joe Gough/Shutterstock.com)

If all the members of a frame and the applied loads lie in a single plane, the frame is called a plane frame (Fig. 1.5). The members of a plane frame are, in general, subjected to bending moments, shears, and axial forces under the action of external loads. Many actual three-dimensional building frames can be subdivided into plane frames for analysis.

Space Trusses Some trusses (such as lattice domes, transmission towers, and certain aerospace structures (Fig. 1.6)) cannot be treated as plane trusses because of the arrangement of their members or applied loading. Such trusses, referred to as space trusses, are analyzed as three-dimensional structures subjected to three-dimensional force systems. The members of space trusses are assumed to be connected by frictionless ball-and-socket joints, and the trusses are subjected to loads and reactions only at the joints. Like plane trusses, the members of space trusses develop only axial forces.

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8  Chapter 1   Introduction w3

P3

w2

w2

w1

w1

P2

P1

Fig. 1.5  Plane Frame

Fig. 1.6  A Segment of the Integrated Truss Structure which Forms the Backbone of the International Space Station (Photo Courtesy of National Aeronautics and Space Administration 98-05165)

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Section 1.4   Classification of Framed Structures  9

Grids A grid, like a plane frame, is composed of straight members connected together by rigid and/or flexible connections to form a plane framework. The main difference between the two types of structures is that plane frames are loaded in the plane of the structure, whereas the loads on grids are applied in the direction perpendicular to the structure’s plane (Fig. 1.7). Members of grids may, therefore, be subjected to torsional moments, in addition to the bending moments and corresponding shears that cause the members to bend out of the plane of the structure. Grids are commonly used for supporting roofs covering large column-free areas in such structures as sports arenas, auditoriums, and aircraft hangars (Fig. 1.8). Y X w1

M2 M1

P2 P1

P3

w2

Z

Fig. 1.7 Grid

Fig. 1.8  National Air and Space Museum, Washington, D.C. (Under Construction) (Photo courtesy of Bethlehem Steel Corporation)

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10  Chapter 1   Introduction

Fig. 1.9  Space Frame (MNTravel/Alamy Stock Photo)

Space Frames Space frames constitute the most general category of framed structures. Members of space frames may be arranged in any arbitrary directions, and connected by rigid and/or flexible connections. Loads in any directions may be applied on members as well as on joints. The members of a space frame may, in general, be subjected to bending moments about both principal axes, shears in both principal directions, torsional moments, and axial forces (Fig. 1.9).

1.5 

Analytical Models The first (and perhaps most important) step in the analysis of a structure is to develop its analytical model. An analytical model is an idealized representation of a real structure for the purpose of analysis. Its objective is to simplify the analysis of a complicated structure by discarding much of the detail (about connections, members, etc.) that is likely to have little effect on the structure’s behavioral characteristics of interest, while representing, as accurately as practically possible, the desired characteristics. It is important to note that the structural response predicted from an analysis is valid only to the extent that the analytical model represents the actual structure. For framed structures, the establishment of analytical models generally involves consideration of issues such as whether the actual three-dimensional structure can be subdivided into plane structures for analysis and whether to idealize the actual bolted or welded connections as hinged, rigid, or semirigid joints. Thus, the development of accurate analytical models requires not only a thorough understanding of structural behavior and methods of analysis, but also experience and knowledge of design and construction practices.

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Section 1.5   Analytical Models  11

In matrix methods of analysis, a structure is modeled as an assemblage of straight members connected at their ends to joints. A member is defined as a part of the structure for which the member force-displacement relationships to be used in the analysis are valid. The member force-displacement relationships for the various types of framed structures will be derived in subsequent chapters. A joint is defined as a structural part of infinitesimal size to which the ends of the members are connected. In finite-element terminology, the members and joints of structures are generally referred to as elements and nodes, respectively. Supports for framed structures are commonly idealized as fixed supports, which do not allow any displacement; hinged supports, which allow rotation but prevent translation; or, roller or link supports, which prevent translation in only one direction. Other types of restraints, such as those which prevent rotation but permit translation in one or more directions, can also be considered in an analysis, as discussed in subsequent chapters.

Line Diagrams The analytical model of a structure is represented by a line diagram, on which each member is depicted by a line coinciding with its centroidal axis. The member dimensions and the size of connections are not shown. Rigid joints are usually represented by points, and hinged joints by small circles, at the intersections of members. Each joint and member of the structure is identified by a number. For example, the analytical model of the plane truss of Fig. 1.10(a) is shown in Fig. 1.10(b), in which the joint numbers are enclosed within circles to distinguish them from the member numbers enclosed within rectangles.

(a) Plane Truss 5

6

4

6

5

9 7

10

8

1

4 1

2

2

3

3

(b) Analytical Model

Fig. 1.10

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12  Chapter 1   Introduction

1.6 Fundamental Relationships for Structural Analysis

Structural analysis, in general, involves the use of three types of relationships: ●●

●●

●●

equilibrium equations, compatibility conditions, and constitutive relations.

Equilibrium Equations A structure is considered to be in equilibrium if, initially at rest, it remains at rest when subjected to a system of forces and couples. If a structure is in equilibrium, then all of its members and joints must also be in equilibrium. Recall from statics that for a plane (two-dimensional) structure lying in the XY plane and subjected to a coplanar system of forces and couples (Fig. 1.11), the necessary and sufficient conditions for equilibrium can be expressed in Cartesian (XY) coordinates as

oF



X

oF

50

50

Y

oM 5 0

(1.1)



These equations are referred to as the equations of equilibrium for plane structures. For a space (three-dimensional) structure subjected to a general threedimensional system of forces and couples (Fig. 1.12), the equations of equilibrium are expressed as

oF 5 0 oM 5 0

oF 5 0 oM 5 0

X



Y

X

Y

oF 5 0 oM 5 0 Z



(1.2)

Z

For a structure subjected to static loading, the equilibrium equations must be satisfied for the entire structure as well as for each of its members and joints. In structural analysis, equations of equilibrium are used to relate the forces (including couples) acting on the structure or one of its members or joints. F2 F3

M2 M3 F1

Y M4 M1

0

X

F4

Fig. 1.11 Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 1.6   Fundamental Relationships for Structural Analysis  13 M3 F2 F3 M1

F4

M2 F1 Y

M4

F5

0

X

Z

Fig. 1.12

Compatibility Conditions The compatibility conditions relate the deformations of a structure so that its various parts (members, joints, and supports) fit together without any gaps or overlaps. These conditions (also referred to as the continuity conditions) ensure that the deformed shape of the structure is continuous (except at the locations of any internal hinges or rollers), and is consistent with the support conditions. Consider, for example, the two-member plane frame shown in Fig. 1.13. The deformed shape of the frame due to an arbitrary loading is also depicted, using an exaggerated scale. When analyzing a structure, the compatibility conditions are used to relate member end displacements to joint displacements that, in turn, are related to the support conditions. For example, because joint 1 of the frame in Fig. 1.13 is attached to a roller support that cannot translate in the vertical direction, the vertical displacement of this joint must be zero. Similarly, because joint 3 is attached to a fixed support that can neither rotate nor translate in any direction, the rotation and the horizontal and vertical displacements of joint 3 must be zero. The displacements of the ends of members are related to the joint displacements by the compatibility requirement that the displacements of a member’s end must be the same as the displacements of the joint to which the member end is connected. Thus, as shown in Fig. 1.13, because joint 1 of the example frame displaces to the right by a distance d1 and rotates clockwise by an angle u1, the left end of the horizontal member (member 1) that is attached to joint 1 must also translate to the right by distance d1 and rotate clockwise by angle u1. Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

14  Chapter 1   Introduction d2

19

1

2

1

29

θ1

d3

θ2

d1

2 θ2

Undeformed shape

Deformed shape

3

Fig. 1.13

Similarly, because the displacements of joint 2 consist of the translations d2 to the right and d3 downward and the counterclockwise rotation u2, the right end of the horizontal member and the top end of the vertical member that are connected to joint 2 must also undergo the same displacements (i.e., d2, d3, and u2). The bottom end of the vertical member, however, is not subjected to any displacements because joint 3, to which this particular member end is attached, can neither rotate nor translate in any direction. Finally, compatibility requires that the deflected shapes of the members of a structure be continuous (except at any internal hinges or rollers) and be consistent with the displacements at the corresponding ends of the members.

Constitutive Relations The constitutive relations (also referred to as the stress-strain relations) describe the relationships between the stresses and strains of a structure in accordance with the stress-strain properties of the structural material. As discussed previously, the equilibrium equations provide relationships between the forces, whereas the compatibility conditions involve only deformations. The constitutive relations provide the link between the equilibrium equations and compatibility conditions that is necessary to establish the load-deformation relationships for a structure or a member. In the analysis of framed structures, the basic stress-strain relations are first used, along with the member equilibrium and compatibility equations, to establish relationships between the forces and displacements at the ends of a member. The member force-displacement relations thus obtained are then

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Section 1.6   Fundamental Relationships for Structural Analysis  15

treated as the constitutive relations for the entire structure, and are used to link the structure’s equilibrium and compatibility equations, thereby yielding the load-deformation relationships for the entire structure. These load-deformation relations can then be solved to determine the deformations of the structure due to a given loading. In the case of statically determinate structures, the equilibrium equations can be solved independently of the compatibility and constitutive relations to obtain the reactions and member forces. The deformations of the structure, if desired, can then be determined by employing the compatibility and constitutive relations. In the analysis of statically indeterminate structures, however, the equilibrium equations alone are not sufficient for determining the reactions and member forces. Therefore, it becomes necessary to satisfy simultaneously the three types of fundamental relationships (i.e., equilibrium, compatibility, and constitutive relations) to determine the structural response. Matrix methods of structural analysis are usually formulated by direct application of the three fundamental relationships as described in general terms in the preceding paragraphs. (The details of the formulations are presented in subsequent chapters.) However, matrix methods can also be formulated by using work-energy principles that satisfy the three fundamental relationships indirectly. Work-energy principles are generally preferred in the formulation of finite-element methods, because they can be more conveniently applied to derive the approximate force-displacement relations for the elements of surface structures and solids. The matrix methods presented in this text are formulated by the direct application of the equilibrium, compatibility, and constitutive relationships. However, to introduce readers to the finite-element method, and to familiarize them with the application of the work-energy principles, we also derive the member force-displacement relations for plane structures by a finite-element approach that involves a work-energy principle known as the principle of virtual work. In the following paragraphs, we review two statements of this principle pertaining to rigid bodies and deformable bodies, for future reference.

Principle of Virtual Work for Rigid Bodies The principle of virtual work for rigid bodies (also known as the principle of virtual displacements for rigid bodies) can be stated as follows: If a rigid body, which is in equilibrium under a system of forces (and couples), is subjected to any small virtual rigid-body displacement, the virtual work done by the external forces (and couples) is zero. In the foregoing statement, the term virtual simply means imaginary, not real. Consider, for example, the cantilever beam shown in Fig. 1.14(a). The free-body diagram of the beam is shown in Fig. 1.14(b), in which PX and PY are the components of the external load P in the X and Y directions, respectively, and R1, R2, and R3 represent the reactions at the fixed support 1. Note that the beam is in equilibrium under the action of the forces PX, PY, R1, and R2, and the couple R3. Now, imagine that the beam is given an arbitrary, small virtual rigid-body displacement from its initial equilibrium position 1–2 to another

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16  Chapter 1   Introduction P

1

2 L

(a) PY

Y

PX

R1

1

R3

2

R2 X

δdX

(b) 29

Virtual displaced position

19

(δθ)L

δθ PY

R1

R2

PX

2

1

R3

δdY

Initial equilibrium position

δdX

L

(c)

Fig. 1.14

position 19–29, as shown in Fig. 1.14(c). As this figure indicates, the total virtual displacement of the beam can be decomposed into rigid-body translations ddX and ddY in the X and Y directions, respectively, and a rigid-body rotation du about point 1. Note that the symbol d is used here to identify the virtual quantities. As the beam undergoes the virtual displacement from position 1–2 to position 19–29, the forces and the couple acting on it perform work, which is referred to as the virtual work. The total virtual work, dWe, can be expressed as the algebraic sum of the virtual work dWX and dWY, performed during translations in the X and Y directions, respectively, and the virtual work dWR, done during the rotation; that is,

(1.3)

dWe 5 dWX 1 dWY 1 dWR

During the virtual translation ddX of the beam, the virtual work performed by the forces can be expressed as follows (Fig 1.14c):

dWX 5 R1ddX 2 PX ddX 5 (R1 2 PX) ddX 5 _

o F + dd  X

X

(1.4)

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Section 1.6   Fundamental Relationships for Structural Analysis  17

Similarly, the virtual work done during the virtual translation ddY is given by

dWY 5 R2ddY 2 PY ddY 5 (R2 2 PY) ddY 5 _

o F + dd  Y

(1.5)

Y

and the virtual work done by the forces and the couple during the small virtual rotation du can be expressed as follows (Fig. 1.14c):

dWR 5 R3du 2 PY (L du) 5 (R3 2 PY L) du 5 _

o M➀ + du

(1.6)

 

The expression for the total virtual work can now be obtained by substituting Eqs. (1.4–1.6) into Eq. (1.3). Thus,

o M➀ + du However, because the beam is in equilibrium, o F o M➀ 5 0; therefore, Eq. (1.7) becomes dWe 5 _

o F + dd X

X

1_

o F + dd Y

Y

1_

X



(1.7)

 

5 0,

oF

Y

5 0, and

(1.8)

dWe 5 0 

which is the mathematical statement of the principle of virtual work for rigid bodies.

Principle of Virtual Work for Deformable Bodies The principle of virtual work for deformable bodies (also called the principle of virtual displacements for deformable bodies) can be stated as follows: If a deformable structure, which is in equilibrium under a system of forces (and couples), is subjected to any small virtual displacement consistent with the support and continuity conditions of the structure, then the virtual external work done by the real external forces (and couples) acting through the virtual external displacements (and rotations) is equal to the virtual strain energy stored in the structure. To demonstrate the validity of this principle, consider the two-member truss of Fig. 1.15(a), which is in equilibrium under the action of an external load P. The free-body diagram of joint 3 of the truss is shown in Fig. 1.15(b). Since joint 3 is in equilibrium, the external and internal forces acting on it must satisfy the following two equations of equilibrium:

oF 5 0 1 c oF 5 0 1S

X

Y

2F1 sin u1 1 F2 sin u2 5 0 F1 cos u1 1 F2 cos u2 2 P 5 0

(1.9)

in which F1 and F2 denote the internal (axial) forces in members 1 and 2, respectively; and u1 and u2 are, respectively, the angles of inclination of these members with respect to the vertical as shown in the figure. Now, imagine that joint 3 is given a small virtual compatible displacement, dd, in the downward direction, as shown in Fig. 1.15(a). It should be noted that this virtual displacement is consistent with the support conditions of the truss in the sense that joints 1 and 2, which are attached to supports, are

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18  Chapter 1   Introduction

2 1

1

Initial equilibrium position

2

θ2

θ1

Virtual displaced position δd

3 39 P (a)

θ1

θ2

F2

3 sθ co d) (θ

θ1

1

X

θ2

sθ

3

co

Y

d) (δ

2

F1

δd

P Real joint forces

Virtual joint displacements (b)

Fig. 1.15

not displaced. Because the reaction forces at joints 1 and 2 do not perform any work, the total virtual work for the truss, dW, is equal to the algebraic sum of the virtual work of the forces acting at joint 3. Thus, from Fig. 1.15(b),

dW 5 Pdd 2 F1(dd cos u1) 2 F2(dd cos u2) which can be rewritten as



dW 5 (P 2 F1 cos u1 2 F2 cos u2) dd

(1.10)

As indicated by Eq. (1.9), the term in parentheses on the right-hand side of Eq. (1.10) is zero. Therefore, the total virtual work, dW, is zero. By substituting dW 5 0 into Eq. (1.10) and rearranging terms, we write

P(dd) 5 F1(dd cos u1) 1 F2(dd cos u2)

(1.11)

in which the quantity on the left-hand side represents the virtual external work, dWe, performed by the real external force P acting through the virtual external displacement dd. Furthermore, because the terms (dd) cos u1 and (dd) cos u2 are equal to the virtual internal displacements (elongations) of members 1 and 2,

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Section 1.6   Fundamental Relationships for Structural Analysis  19

respectively, we can conclude that the right-hand side of Eq. (1.11) represents the virtual internal work, dWi, done by the real internal forces acting through the corresponding virtual internal displacements; that is,

(1.12)

dWe 5 dWi

Realizing that the internal work is also referred to as the strain energy, U, we can express Eq. (1.12) as

(1.13)

dWe 5 dU 

in which dU denotes the virtual strain energy. Note that Eq. (1.13) is the mathematical statement of the principle of virtual work for deformable bodies. For computational purposes, it is usually convenient to express Eq. (1.13) in terms of the stresses and strains in the members of the structure. For that purpose, let us consider a differential element of a member of an arbitrary structure subjected to a general loading (Fig. 1.16). The element is in equilibrium under a general three-dimensional stress condition, due to the real forces acting on the structure. Now, as the structure is subjected to a virtual displacement, virtual strains develop in the element and the internal forces due to the real stresses perform virtual internal work as they move through the internal displacements caused by the virtual strains. For example, the virtual internal work done by the real force due to the stress sx as it moves through the virtual displacement caused by the virtual strain d«x can be determined as follows:

real force 5 stress 3 area 5 sx (dy dz) virtual displacement 5 strain 3 length 5 (d«x) dx y

σy

τxy τyz

τxy

τyz

σx

τzx

τzx

x

dy σz

dz

dx z

Fig. 1.16

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20  Chapter 1   Introduction

Therefore,

virtual internal work 5 real force 3 virtual displacement 5 (sx dy dz) (d«x dx)



5 (d«x sx) dV in which dV  5 dx dy dz is the volume of the differential element. Thus, the vir­ tual internal work due to all six stress components is given by



virtual internal work in element dV 5 (d«xsx 1 d«ysy 1 d«zsz 1 dgxytxy 1 dgyztyz 1 dgzxtzx) dV

(1.14)

In Eq. (1.14), d«x, d«y, d«z, dgxy, dgyz, and dgzx denote, respectively, the virtual strains corresponding to the real stresses sx, sy, sz, txy, tyz, and tzx, as shown in Fig. 1.16. The total virtual internal work, or the virtual strain energy stored in the entire structure, can be obtained by integrating Eq. (1.14) over the volume V of the structure. Thus,

dU 5

# (d« s 1 d« s 1 d« s 1 dg t V

x

x

y

y

z

z

xy xy

1 dgyztyz 1 dgzxtzx ) dV(1.15)

Finally, by substituting Eq. (1.15) into Eq. (1.13), we obtain the statement of the principle of virtual work for deformable bodies in terms of the stresses and strains of the structure. dWe 5

1.7 

#

 (d«xsx 1 d«ysy 1 d«zsz 1 dgxytxy 1 dgyztyz 1 dgzxtzx) dV

(1.16)

V

Linear versus Nonlinear Analysis In this text, we focus our attention mainly on linear analysis of structures. Linear analysis of structures is based on the following two fundamental assumptions: 1. The structures are composed of linearly elastic material; that is, the stress-strain relationship for the structural material follows Hooke’s law. 2. The deformations of the structures are so small that the squares and higher powers of member slopes, (chord) rotations, and axial strains are negligible in comparison with unity, and the equations of equilibrium can be based on the undeformed geometry of the structure. The reason for making these assumptions is to obtain linear relationships between applied loads and the resulting structural deformations. An important advantage of linear force-deformation relations is that the principle of

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Summary  21

superposition can be used in the analysis. This principle states essentially that the combined effect of several loads acting simultaneously on a structure equals the algebraic sum of the effects of each load acting individually on the structure. Engineering structures are usually designed so that under service loads they undergo small deformations, with stresses within the initial linear portions of the stress-strain curves of their materials. Thus, linear analysis generally proves adequate for predicting the performance of most common types of structures under service loading conditions. However, at higher load levels, the accuracy of linear analysis generally deteriorates as the deformations of the structure increase and/or its material is strained beyond the yield point. Because of its inherent limitations, linear analysis cannot be used to predict the ultimate load capacities and instability characteristics (e.g., buckling loads) of structures. With the recent introduction of design specifications based on the ultimate strengths of structures, the use of nonlinear analysis in structural design is increasing. In a nonlinear analysis, the restrictions of linear analysis are removed by formulating the equations of equilibrium on the deformed geometry of the structure that is not known in advance, and/or taking into account the effects of inelasticity of the structural material. The load-deformation relationships thus obtained for the structure are nonlinear, and are usually solved using iterative techniques. An introduction to this still-evolving field of nonlinear structural analysis is presented in Chapter 10.

1.8 Software Software for the analysis of framed structures using the matrix stiffness method is provided on the publisher’s website for this book, https://cengage.com /engineering/matrixanalysis3e/kassimali/software. The software can be used by readers to verify the correctness of various subroutines and programs that they will develop during the course of study of this text, as well as to check the answers to the problems given at the end of each chapter. A description of the software, and information on how to install and use it, is presented in Appendix A.

Summary In this chapter, we discussed the topics summarized in the following list: 1. Structural analysis is the prediction of the performance of a given structure under prescribed loads and/or other external effects. 2. Both matrix and classical methods of structural analysis are based on the same fundamental principles. However, classical methods were developed to analyze particular types of structures, whereas matrix methods are more general and systematic so that they can be conveniently programmed on computers. 3.  Two different methods can be used for matrix analysis of structures; namely, the flexibility and stiffness methods. The stiffness method is more systematic and can be implemented more easily on computers, and is therefore currently preferred in professional practice.

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22  Chapter 1   Introduction

4. Framed structures are composed of straight members whose lengths are significantly larger than their cross-sectional dimensions. Framed structures can be classified into six basic categories: plane trusses, beams, plane frames, space trusses, grids, and space frames. 5. An analytical model is a simplified (idealized) representation of a real structure for the purpose of analysis. Framed structures are modeled as assemblages of straight members connected at their ends to joints, and these analytical models are represented by line diagrams. 6.  The analysis of structures involves three fundamental relationships: equilibrium equations, compatibility conditions, and constitutive relations. 7.  The principle of virtual work for deformable bodies states that if a deformable structure, which is in equilibrium, is subjected to a small compatible virtual displacement, then the virtual external work is equal to the virtual strain energy stored in the structure. 8.  Linear structural analysis is based on two fundamental assumptions: the stress-strain relationship for the structural material is linearly elastic, and the structure’s deformations are so small that the equilibrium equations can be based on the undeformed geometry of the structure.

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2

Matrix Algebra Learning Objectives At the end of this chapter, you will be able to: 2.1 2.2 2.3 2.4

Define a Matrix Identify the Different Types of Matrices Perform Common Matrix Operations Solve Simultaneous Equations Using the Gauss-Jordan Elimination Method

Somerset Corporate Center Office Building, New Jersey, and its Analytical Model (Photos courtesy of Ram International. Structural Engineer: The Cantor Seinuk Group, P.C.)

23

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24  Chapter 2   Matrix Algebra

In matrix methods of structural analysis, the fundamental relationships of equilibrium, compatibility, and member force-displacement relations are expressed in the form of matrix equations, and the analytical procedures are formulated by applying various matrix operations. Therefore, familiarity with  the basic concepts of matrix algebra is a prerequisite to understanding matrix structural analysis. The objective of this chapter is to concisely present the basic concepts of matrix algebra necessary for formulating the methods of structural analysis covered in the text. A general procedure for solving simultaneous linear equations, the Gauss-Jordan method, is also discussed. We begin with the basic definition of a matrix in Section 2.1, followed by brief descriptions of the various types of matrices in Section 2.2. The matrix operations of equality, addition and subtraction, multiplication, transposition, differentiation and integration, inversion, and partitioning are defined in Section 2.3; we conclude the chapter with a discussion of the Gauss-Jordan elimination method for solving simultaneous equations (Section 2.4).

2.1 

Definition of a Matrix A matrix is defined as a rectangular array of quantities arranged in rows and columns. A matrix with m rows and n columns can be expressed as follows.

3

A11 A12 A13 . . . A21 A22 A23 . . . A 5 [A] 5 A31 A32 A33 . . . . . . . . . . . . . . . Aij Am1 Am2 Am3 . . .

... ... ... ... ...

A1n A2n A3n Amn

4

 (2.1) ith row

jth column m 3 n As shown in Eq. (2.1), matrices are denoted either by boldface letters (A) or by italic letters enclosed within brackets ([A]). The quantities forming a matrix are referred to as its elements. The elements of a matrix are usually numbers, but they can be symbols, equations, or even other matrices (called submatrices). Each element of a matrix is represented by a double-subscripted letter, with the first subscript identifying the row and the second subscript identifying the column in which the element is located. Thus, in Eq. (2.1), A23 represents the element located in the second row and third column of matrix A. In general, Aij refers to an element located in the ith row and jth column of matrix A. The size of a matrix is measured by the number of its rows and columns and is referred to as the order of the matrix. Thus, matrix A in Eq. (2.1), which has m rows and n columns, is considered to be of order m 3 n (m by n). As an example, consider a matrix D given by



D5

3

3 8 12 7

5 26 23 29

4

37 0 2 21

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Section 2.2   Types of Matrices  25

The order of this matrix is 4 3 3, and its elements are symbolically denoted by  Dij with i  5 1 to 4 and j 5 1 to 3; for example, D13 5 37, D31 5 12, D42 5 29, etc.

2.2 

Types of Matrices We describe some of the common types of matrices in the following paragraphs.

Column Matrix (Vector) If all the elements of a matrix are arranged in a single column (i.e., n 5 1), it is called a column matrix. Column matrices are usually referred to as vectors, and are sometimes denoted by italic letters enclosed within braces. An example of a column matrix or vector is given by



B 5 {B} 5

34 35 9 12 3 26

Row Matrix A matrix with all of its elements arranged in a single row (i.e., m 5 1) is referred to as a row matrix. For example,

C 5 [9  35  212 7 22]

Square Matrix If a matrix has the same number of rows and columns (i.e., m 5 n), it is called a square matrix. An example of a 4 3 4 square matrix is given by



A5

3

6 15 224 40

4

12 0 20 29 237 3  13 8 1 0 11 25 main diagonal

(2.2)

As shown in Eq. (2.2), the main diagonal of a square matrix extends from the upper left corner to the lower right corner, and it contains elements with matching subscripts—that is, A11, A22, A33, . . . , Ann. The elements forming the main diagonal are referred to as the diagonal elements; the remaining elements of a square matrix are called the off-diagonal elements.

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26  Chapter 2   Matrix Algebra

Symmetric Matrix When the elements of a square matrix are symmetric about its main diagonal (i.e., Aij 5 Aji), it is termed a symmetric matrix. For example,

A 5

3

6 15 224 40

15 29 13 0

4

40 0 11 25

224 13 8 11

Lower Triangular Matrix If all the elements of a square matrix above its main diagonal are zero, (i.e., Aij 5 0 for j . i), it is referred to as a lower triangular matrix. An example of a 4 3 4 lower triangular matrix is given by

A 5

3

8 12 33 22

0 29 17 5

0 0 6 15

4

0 0 0 3

Upper Triangular Matrix When all the elements of a square matrix below its main diagonal are zero (i.e., Aij 5 0 for j , i), it is called an upper triangular matrix. An example of a 3 3 3 upper triangular matrix is given by

A 5

3

27 0 0

4

6 12 0

17 11 20

Diagonal Matrix A square matrix with all of its off-diagonal elements equal to zero (i.e., Aij 5 0 for i ? j), is called a diagonal matrix. For example,

A 5

3

6 0 0 0

0 23 0 0

0 0 11 0

0 0 0 27

4

Unit or Identity Matrix If all the diagonal elements of a diagonal matrix are equal to 1 (i.e., Iij 5 1 and Iij 5 0 for i ? j), it is referred to as a unit (or identity) matrix. Unit

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Section 2.3   Matrix Operations  27

­ atrices are commonly denoted by I or [I]. An example of a 3 3 3 unit m matrix is given by

3

1 I 5 0 0

4

0 1 0

0 0 1

Null Matrix If all the elements of a matrix are zero (i.e., Oij 5 0), it is termed a null matrix. Null matrices are usually denoted by O or [O]. An example of a 3 3 4 null matrix is given by

3

0 O5 0 0



2.3 

0 0 0

0 0 0

4

0 0 0

Matrix Operations Equality Matrices A and B are considered to be equal if they are of the same order and if their corresponding elements are identical (i.e., Aij  5 Bij). Consider, for example, matrices

3

6 A 5 27 3



4

3

4

2 6 8   and  B 5 27 29 3

2 8 29

Since both A and B are of order 3 3 2, and since each element of A is equal to the corresponding element of B, the matrices A and B are equal to each other; that is, A 5 B.

Addition and Subtraction Matrices can be added (or subtracted) only if they are of the same order. The addition (or subtraction) of two matrices A and B is carried out by adding (or  subtracting) the corresponding elements of the two matrices. Thus, if A 1 B 5 C, then Cij 5 Aij 1 Bij; and if A 2 B 5 D, then Dij 5 Aij 2 Bij . The matrices C and D have the same order as matrices A and B.

E x ample 2.1 Calculate the matrices C 5 A 1 B and D 5 A 2 B if

3 4

6 A 5 22 5

3

0 2 9   and  B 5 7 1 212

4

3 5 21

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28  Chapter 2   Matrix Algebra S olutio n

4 3 4 3

4

3

(0 1 3) 8 (9 1 5) 5 5 (1 2 1) 27

3 14  0

Ans

3

(0 2 3) 4 (9 2 5) 5 29 (1 1 1) 17

23 4  2

4

Ans

(6 1 2) C 5 A 1 B 5 (22 1 7) (5 2 12) (6 2 2) D 5 A 2 B 5 (22 2 7) (5 1 12)

Multiplication by a Scalar The product of a scalar c and a matrix A is obtained by multiplying each element of the matrix A by the scalar c. Thus, if cA 5 B, then Bij 5 cAij.

E x ample 2.2 Calculate the matrix B 5 cA if c 5 26 and A5

3

3 0 12

7 8 24

4

22 1 10

S olutio n

B 5 cA 5

3

26(3) 26(0) 26(12)

26(7) 26(8) 26(24)

4 3

26(22) 218 26(1) 5 0 272 26(10)

242 248 24

4

12 26  260

Ans

Multiplication of Matrices Two matrices can be multiplied only if the number of columns of the first matrix equals the number of rows of the second matrix. Such matrices are said to be conformable for multiplication. Consider, for example, the matrices



3

1 A5 4 25

4

8 6 22   and  B 5 21 3 (3 3 2)

3

4

27 2

(2.3)

(2 3 2)

The product AB of these matrices is defined because the first matrix, A, of the sequence AB has two columns and the second matrix, B, has two rows. However, if the sequence of the matrices is reversed, then the product BA does not exist, because now the first matrix, B, has two columns and the second matrix, A, has three rows. The product AB is referred to either as A postmultiplied by B, or as B premultiplied by A. Conversely, the product BA is referred to either as B postmultiplied by A, or as A premultiplied by B. When two conformable matrices are multiplied, the product matrix thus obtained has the number of rows of the first matrix and the number of columns of the

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Section 2.3   Matrix Operations  29

second matrix. Thus, if a matrix A of order l 3 m is postmultiplied by a matrix B of order m 3 n, then the product matrix C 5 AB has the order l 3 n; that is, 5 A B C (l 3 m) (m 3 n) (l 3 n) equal

(2.4)

3

Ai1

Ai2

. .. . ..

i th row

Aim

33 3 3 3 B1 j B2 j .. . .. . Bm j

Ci j

i th row

5

j th column

j th column





Any element Cij of the product matrix C can be determined by multiplying each element of the ith row of A by the corresponding element of the jth column o f B (see Eq. 2.4), and by algebraically summing the products; that is, Cij 5 Ai1B1j 1 Ai2B2j 1 . . . 1 AimBmj



(2.5)

Eq. (2.5) can be expressed as

m

Cij 5

oA B ik

k51



(2.6)

kj

in which m represents the number of columns of A, or the number of rows of B. Equation (2.6) can be used to determine all elements of the product matrix C 5 AB.



E x ample 2.3 Calculate the product C 5 AB of the matrices A and B given in Eq. (2.3).

S olutio n

C 5 AB 5

3

1 4 25

43

8 22 3

(3 3 2)

6 21

4

27 5 2

(2 3 2)

3

22 26 233

4

9 232  41

Ans

(3 3 2)

The element C11 of the product matrix C is determined by multiplying each element of the first row of A by the corresponding element of the first column of B and summing the resulting products; that is, C11 5 1(6) 1 8(21) 5 22

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30  Chapter 2   Matrix Algebra Similarly, the element C12 is obtained by multiplying the elements of the first row of A by the corresponding elements of the second column of B and adding the resulting products; that is, C12 5 1(27) 1 8(2) 5 9 The remaining elements of C are computed in a similar manner: C21 5 4(6) 1 (22)(21) 5 26 C22 5 4(27) 22(2) 5 232 C31 5 25(6) 1 3(21) 5 233 C32 5 25(27) 1 3(2) 5 41

A flowchart for programming the matrix multiplication procedure on a computer is given in Fig. 2.1. Any programming language (such as FORTRAN, BASIC, or C, among others) can be used for this purpose. The reader is encouraged to write this program in a general form (e.g., as a subroutine), so that it can be included in the structural analysis computer programs to be developed in later chapters. An important application of matrix multiplication is to express simultaneous equations in compact matrix form. Consider the following system of linear simultaneous equations:

A11x1 1 A12x2 1 A13x3 1 A14x4 5 P1 A21x1 1 A22x2 1 A23x3 1 A24x4 5 P2 A31x1 1 A32x2 1 A33x3 1 A34x4 5 P3 A41x1 1 A42x2 1 A43x3 1 A44x4 5 P4



(2.7)

in which xs are the unknowns and As and Ps represent the coefficients and constants, respectively. By using the definition of multiplication of matrices, this system of equations can be expressed in matrix form as







3

A11 A21 A31 A41

A12 A22 A32 A42

A13 A23 A33 A43

A14 A24 A34 A44

43 4 3 4 x1 x2 x3 x4

5

P1 P2  P3 P4

or, symbolically, as Ax 5 P Matrix multiplication is generally not commutative; that is, AB ? BA



(2.8)

(2.9)

(2.10)

Even when the orders of two matrices A and B are such that both products AB and BA are defined and are of the same order, the two products, in general, will

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Section 2.3   Matrix Operations  31 Start Input A(L, M), B(M, N) Dimension C(L, N) I51

I # L?

no

yes J51

J # N?

no

yes C(I, J) 5 0.0 K51

K # M?

no

yes C(I, J) 5 C(I, J) 1 A(I, K)*B(K, J) K5K11 J5J11 I5I11 Output C Stop

Fig. 2.1  Flowchart for Matrix Multiplication

not be equal. It is essential, therefore, to maintain the proper sequential order of matrices when evaluating matrix products.

E x ample 2.4 Calculate the product AB and BA if A5

3271

4

3

28 6   and  B 5 2 4

4

23 25

Are the products AB and BA equal?

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32  Chapter 2   Matrix Algebra S olutio n

AB 5

3271

BA 5

364

43

28 6   2 4

43

23 1   25 27

4 3

37  11

4 3

254  242

23 226 5 25 234 28 27 5 2 39

4

Ans

4

Ans

Comparing products AB and BA, we can see that AB ? BA.

Ans

Matrix multiplication is associative and distributive, provided that the sequential order in which the matrices are to be multiplied is maintained. Thus, ABC 5 (AB)C 5 A(BC)

(2.11)

and A(B 1 C) 5 AB 1 AC

(2.12)

The product of any matrix A and a conformable null matrix O equals a null matrix; that is, AO 5 O          and          OA 5 O

(2.13)

For example,

3262



4 3 004 5 300 004

24 0   8 0

The product of any matrix A and a conformable unit matrix I equals the original matrix A; thus, AI 5 A   and   IA 5 A

(2.14)

For example,

3262



4 3 014 5 3262

24 1   8 0

4

24 8

and

310 014  3262

4 3

24 2 5 8 26

4

24 8

We can see from Eqs. (2.13) and (2.14) that the null and unit matrices serve purposes in matrix algebra that are similar to those of the numbers 0 and 1, respectively, in scalar algebra.

Transpose of a Matrix The transpose of a matrix is obtained by interchanging its corresponding rows and columns. The transposed matrix is commonly identified by placing a superscript T on the symbol of the original matrix. Consider, for example, a 3 3 2 matrix

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Section 2.3   Matrix Operations  33

3

2 B 5 25 1



4

24 8 3

332 The transpose of B is given by

BT 5

3242

4

1 3

25 8



233 Note that the first row of B becomes the first column of BT. Similarly, the second and third rows of B become, respectively, the second and third columns of BT. The order of BT thus obtained is 2 3 3. As another example, consider the matrix

3

2 C 5 21 6



4

6 29 5

21 7 29

Because the elements of C are symmetric about its main diagonal (i.e., Cij 5 Cji for i ? j), interchanging the rows and columns of this matrix produces a matrix CT that is identical to C itself; that is, CT 5 C. Thus, the transpose of a symmetric matrix equals the original matrix. Another useful property of matrix transposition is that the transpose of a product of matrices equals the product of the transposed matrices in reverse order. Thus,



(AB)T 5 BTAT

(2.15)

Similarly, (ABC)T 5 CT BTAT

(2.16)

E x ample 2.5 Show that (AB)T 5 BTAT if A5

3

9 2 23

4

25 6 1   and  B 5 22 4

3

21 7

4

10 5

S olutio n

3

9 AB 5 2 23

43

25 6 1   22 4

21 7

4

10 5 5

3

64 10 226

244 5 31

4

65 25  210

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34  Chapter 2   Matrix Algebra

3 3

64 (AB) 5 244 65 T

4

10 5 25

226 31  210

43

6 BT AT 5 21 10

22 9 7   25 5

(1)

2 1

3

64 23 5 244 4 65

4

10 5 25

4

226 31  210

By comparing Eqs. (1) and (2), we can see that (AB)T 5 BTAT.

(2) Ans

Differentiation and Integration A matrix can be differentiated (or integrated) by differentiating (or integrating) each of its elements.

E x ample 2.6 Determine the derivative dA/dx if A5



S olutio n

3

3 sin x 2x cos2 x

x2 3 sin x 2x4

4

2x4 cos2 x 7x3

By differentiating the elements of A, we obtain d A11

A11 5 x2

dx d A21

A21 5 A12 5 3 sin x

dx d A31

A31 5 A13 5 2x4

dx d A22

A22 5 2x

dx d A32

A32 5 A23 5 cos2 x

dx

5 2x

5 5

d A12 dx d A13 dx

5 3 cos x 5 24x3

5 21

5

d A23 dx

5 22 cos x sin x

d A33

A33 5 7x3

5 21x2 dx Thus, the derivative dA/dx is given by

3

2x dA 5 3 cos x dx 24x3

3 cos x 21 22 cos x sin x

4

24x3 22 cos x sin x  21x2

Ans

E x ample 2.7 Determine the partial derivative −B/−y if

3

2y3 B 5 3xy2 2x2

2yz yz 22xz

4

22xz 2z2 3xy2

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Section 2.3   Matrix Operations  35 S olutio n We determine the partial derivative, −Bij  /−y, of each element of B to obtain

3

6y2 −B 5 6xy −y 0

4

0 0  6xy

2z z 0

Ans

E x ample 2.8 Calculate the integral eL0 AAT dx if

3 4 12

A5

S olutio n

x L

x L

First, we calculate the matrix product AAT as

B 5 AAT 5

3 4 31 1

12

x L

2

x L

2 4

x x 12 5 L L

3

1 2 L 11 2 L 2 x 1 2 L

x 2 L x x 12 L L 12

x

x

2

2

4

Next, we integrate the elements of B to obtain L

#B 0

L

11

dx 5

L

# 11 2 L2 dx 5 # 11 2 L 1 L 2dx x

2

0

2x

x2

2

0

1

2 53 x x x x # B dx 5 # B dx 5 # L11 2 L2dx 5 # 1L 2 L 2dx L L L x x 51 2 5 2 5 2L 3L 2 2 3 6 x2 x3 1 2 L 3L

5 x2

L

L

21

0

12

0

2

3

L

L 0

L

L

0

0

2

2

L

2

0

L

#B 0

L

22

dx 5

# 1L 2dx 5 13L 2 0

x

x3

2

2

L

5

2

0

L 3

Thus, L

# AA dx 5 T

0

3 4 L 3 L 6

L 6 L 3

5

3 124

L 2 6 1

Ans

Inverse of a Square Matrix The inverse of a square matrix A is defined as a matrix A21 with elements of such magnitudes that the product of the original matrix A and its inverse A21 equals a unit matrix I; that is,

AA21 5 A21A 5 I



(2.17)

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36  Chapter 2   Matrix Algebra

The operation of inversion is defined only for square matrices, with the inverse of such a matrix also being a square matrix of the same order as the original matrix. A procedure for determining inverses of matrices will be presented in the next section.

E x ample 2.9 Check whether or not matrix B is the inverse of matrix A, if A5

2 0.5   and  B 5 3 324 4 23 1 1.5

4

21 22

S olutio n

AB 5

323 14  31.5 24

2

0.5

4 3

21 (22 1 3) 5 22 (21.5 1 1.5)

4 3 14

(4 2 4) 1 5 (3 2 2) 0

0

Also, BA 5

31.5 0.5

43

21 24   22 23

4 3

2 (22 1 3) 5 1 (26 1 6)

4 3 14

(1 2 1) 1 5 (3 2 2) 0

0

Since AB 5 BA 5 I, B is the inverse of A; that is, B 5 A21

Ans

The operation of matrix inversion serves a purpose analogous to the o­ peration of division in scalar algebra. Consider a system of simultaneous linear equations expressed in matrix form as Ax 5 P in which A is the square matrix of known coefficients, x is the vector of the unknowns, and P is the vector of the constants. As the operation of division is not defined in matrix algebra, the equation cannot be solved for x by dividing P by A (i.e., x 5 P/A). However, we can determine x by premultiplying both sides of the equation by A21, to obtain A21Ax 5 A21P As A21A 5 I and Ix 5 x, we can write x 5 A21P which shows that a system of simultaneous linear equations can be solved by premultiplying the vector of constants by the inverse of the coefficient matrix. An important property of matrix inversion is that the inverse of a symmetric matrix is also a symmetric matrix.

Orthogonal Matrix If the inverse of a matrix is equal to its transpose, the matrix is referred to as an orthogonal matrix. In other words, a matrix A is orthogonal if

A21 5 AT

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Section 2.3   Matrix Operations  37

E x ample

2.10

Determine whether matrix A given below is an orthogonal matrix.

A5

3

0.8 20.6 0 0

0.6 0.8 0 0

4

0 0 0.8 20.6

0 0 0.6 0.8

S olutio n

AAT 5

5

5

3 3 3

0.8 20.6 0 0

0.6 0.8 0 0

(0.64 1 0.36) (20.48 1 0.48) 0 0

1 0 0 0

0 1 0 0

0 0 1 0

43

0 0 0.8 20.6

0 0   0.6 0.8

0.8 0.6 0 0

20.6 0.8 0 0

(20.48 1 0.48) (0.36 1 0.64) 0 0

0 0 0.8 0.6

0 0 20.6 0.8

0 0 (0.64 1 0.36) (20.48 1 0.48)

4 4

0 0 (20.48 1 0.48) (0.36 1 0.64)

4

0 0 0 1

which shows that AAT 5 I. Thus, A21 5 AT Therefore, matrix A is orthogonal.

Ans

Partitioning of Matrices In many applications, it becomes necessary to subdivide a matrix into a number of smaller matrices called submatrices. The process of subdividing a matrix into submatrices is referred to as partitioning. For example, a 4 3 3 matrix B is partitioned into four submatrices by drawing horizontal and vertical dashed partition lines:



B5

3

2 25 8 1

24 7 29 3

21 3 6 8

4

 5

3BB

11 21

4

B12  B22

(2.18)

in which the submatrices are

3

4

3 4



2 B11 5 25 8



B21 5 [1 3]      B22 5 [8]

24 21 7   B12 5 3 29 6

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38  Chapter 2   Matrix Algebra

Matrix operations (such as addition, subtraction, and multiplication) can be performed on partitioned matrices in the same manner as discussed previously by treating the submatrices as elements—provided that the ­matrices are partitioned in such a way that their submatrices are conformable for the particular operation. For example, suppose that the 4  3  3 matrix B of Eq. (2.18) is to be postmultiplied by a 3 3 2 matrix C, which is partitioned into two submatrices:

C5

4

3

9 4 23

26 C 2 5 11  C21 1

3 4

(2.19)

The product BC is expressed in terms of submatrices as BC 5

3BB

43 43

4

B12 C11 B11C11 1 B12C21     (2.20) B22 C21 B21C11 1 B22C21

11 21

It is important to realize that matrices B and C have been partitioned in such a way that their corresponding submatrices are conformable for multiplication; that is, the orders of the submatrices are such that the products B11C11, B12C21, B21C11, and B22C21 are defined. It can be seen from Eqs. (2.18) and (2.19) that this is achieved by partitioning the rows of the second matrix C of the product BC in the same way that the columns of the first matrix B are partitioned. The products of the submatrices are:

3

43



2 B11C11 5 25 8



B12C21 5



B21C11 5 [1 3] 5



B22C21 5 [8][23 1] 5 [224 8]

3 4

4

3

21 3 3 [23 1] 5 29 6 218

394

3

2 26 5 217 2 36

24 9 7   4 29

4

220 44 266

4

21 3 6

4

26 5 [21 0] 2

By substituting the numerical values of the products of submatrices into Eq. (2.20), we obtain

BC 5

33

2 217 36 [21 

4 3

44

220 3 21 44 1 29 3 5 266 218 6   0]  1  [224    8]

3

5 226 18 23

4

221 47 260 8

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Section 2.4   Gauss-Jordan Elimination Method  39

2.4  Gauss-Jordan Elimination Method The Gauss-Jordan elimination method is one of the most commonly used procedures for solving simultaneous linear equations, and for determining inverses of matrices.

Solution of Simultaneous Equations To illustrate the Gauss-Jordan method for solving simultaneous equations, consider the following system of three linear algebraic equations: 5x1 1 6x2 2 3x3 5 66 9x1 2 x2 1 2x3 5 8 8x1 2 7x2 1 4x3 5 239

(2.21a)

To determine the unknowns x1, x2, and x3, we begin by dividing the first equation by the coefficient of its x1 term to obtain   x1 1 1.2x2 2 0.6x3 5   13.2 9x1 2   x2 1  2x3 5  8 8x1 2  7x2 1  4x3 5 239

(2.21b)

Next, we eliminate the unknown x1 from the second and third equations by successively subtracting from each equation the product of the coefficient of its x1 term and the first equation. Thus, to eliminate x1 from the second equation, we multiply the first equation by 9 and subtract it from the second equation. Similarly, we eliminate x1 from the third equation by multiplying the first equation by 8 and subtracting it from the third equation. This yields the system of equations

x1 1 1.2x2 2 0.6x3 5   13.2  211.8x2 1 7.4x3 5 2110.8  216.6x2 1 8.8x3 5 2144.6

(2.21c)

With x1 eliminated from all but the first equation, we now divide the second equation by the coefficient of its x2 term to obtain

x1 1 1.2x2 2 0.6   x3 5       13.2     x2  2 0.6271x3   5    9.39  216.6x2 1 8.8     x3  5 2144.6

(2.21d)

Next, the unknown x2 is eliminated from the first and the third equations, successively, by multiplying the second equation by 1.2 and subtracting it from the first equation, and then by multiplying the second equation by 216.6 and subtracting it from the third equation. The system of equations thus obtained is x1  1  0.1525x3 5    1.932        x2 2 0.6271x3 5  9.39    2 1.61  x3 5 11.27

(2.21e)

Focusing our attention now on the unknown x3, we divide the third equation by the coefficient of its x3 term (which is 21.61) to obtain

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40  Chapter 2   Matrix Algebra



x1    

1  0.1525x3 5  1.932  x2 2 0.6271x3 5  9.39 x3 5 27

(2.21f)

Finally, we eliminate x3 from the first and the second equations, successively, by multiplying the third equation by 0.1525 and subtracting it from the first equation, and then by multiplying the third equation by 20.6271 and subtracting it from the second equation. This yields the solution of the given system of equations: x1    x2

 5 3  5 5 x3 5 27

(2.21g)

or, equivalently, (2.21h)

x1 5 3;   x2 5 5;   x3 5 27

To check that this solution is correct, we substitute the numerical values of x1, x2, and x3 back into the original equations (Eq. 2.21(a)): 5(3) 1 6(5) 2 3(27) 5    66

Checks

9(3) 2 5    1 2(27) 5       8

Checks

8(3) 2 7(5) 1 4(27) 5 239

Checks

As the foregoing example illustrates, the Gauss-Jordan method basically involves eliminating, in order, each unknown from all but one of the equations of the system by applying the following operations: dividing an equation by a scalar, and multiplying an equation by a scalar and subtracting the resulting equation from another equation. These operations (called the elementary operations) when applied to a system of equations yield another system of equations that has the same solution as the original system. In the Gauss-Jordan method, the elementary operations are performed repeatedly until a system with each equation containing only one unknown is obtained. The Gauss-Jordan elimination method can be performed more conveniently by using the matrix form of the simultaneous equations (Ax 5 P). In this approach, the coefficient matrix A and the vector of constants P are treated as submatrices of a partitioned augmented matrix,

G

5

[A

Á

P]

n 3 (n 1 1)   n 3 n  n 3 1

(2.22)

where n represents the number of equations. The elementary operations are then applied to the rows of the augmented matrix, until the coefficient matrix is reduced to a unit matrix. The elements of the vector, which initially contained the constant terms of the original equations, now represent the solution of the original system of equations; that is,

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Section 2.4   Gauss-Jordan Elimination Method  41

G5



5[A[I

P] x]

 

[elementary operations

(2.23)

This procedure is illustrated by the following example.



E x ample

2.11

S olutio n

Solve the system of simultaneous equations given in Eq. 2.21(a) by the Gauss-Jordan method. The given system of equations can be written in matrix form as Ax 5 P

3

5 9 8

6 21 27

x1 23 66 2   x2 5 8  x3 4 239

43 4 3 4

(1)

from which we form the augmented matrix

3

5 G 5 [A Á P] 5 9 8

6 21 27

4

66 8  239

23 2 4

(2)

We begin Gauss-Jordan elimination by dividing row 1 of the augmented matrix by G11 5 5 to obtain

3

1 G5 9 8

4

1.2 20.6 13.2 21 2 8  27 4 239

(3)

Next, we multiply row 1 by G21 5 9 and subtract it from row 2; then multiply row 1 by G31 5 8 and subtract it from row 3. This yields

3

1 G5 0 0

1.2 211.8 216.6

20.6 7.4 8.8

4

13.2 2110.8  2144.6

(4)

We now divide row 2 by G22 5 211.8 to obtain

3

1 G5 0 0

1.2 1 216.6

4

20.6 13.2 20.6271 9.39  8.8 2144.6

(5)

Next, we multiply row 2 by G12 5 1.2 and subtract it from row 1, and then multiply row 2 by G32 5 216.6 and subtract it from row 3. Thus,

3

1 G5 0 0

0 1 0

0.1525 20.6271 21.61

4

1.932 9.39  11.27

(6)

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42  Chapter 2   Matrix Algebra By dividing row 3 by G33 5 21.61, we obtain

3

1 G5 0 0

0 1 0

0.1525 20.6271 1

4

1.932 9.39  27

(7)

Finally, we multiply row 3 by G13 5 0.1525 and subtract it from row 1; then multiply row 3 by G23 5 20.6271 and subtract it from row 2 to obtain

3

1 G5 0 0

0 1 0

0 0 1

4

3 5 (8) 27

Thus, the solution of the given system of equations is

x5

3 4

3 5  27

Ans

To check our solution, we substitute the numerical value of x back into Eq. (1). This yields

3

5 9 8

6 21 27

43 4 3

3 15 1 30 1 21 5 66 23 5 27 2 5 2 14 5 8 5 2   24 2 35 2 28 5 239 27 4

4



Checks

The solution of large systems of simultaneous equations by the Gauss-­ Jordan method is usually carried out by computer, and a flowchart for programming this procedure is given in Fig. 2.2. The reader should write this program in a general form (e.g., as a subroutine), so that it can be conveniently included in the structural analysis computer programs to be developed in later chapters. It should be noted that the Gauss-Jordan method as described in the preceding paragraphs breaks down if a diagonal element of the coefficient matrix A becomes zero during the elimination process. This situation can be remedied by interchanging the row of the augmented matrix containing the zero diagonal element with another row, to place a nonzero element on the diagonal; the elimination process is then continued. However, when solving the systems of equations encountered in structural analysis, the condition of a zero diagonal element should not arise; the occurrence of such a condition would indicate that the structure being analyzed is unstable [2]*.

Matrix Inversion The procedure for determining inverses of matrices by the Gauss-Jordan method is similar to that described previously for solving simultaneous equations. The procedure involves forming an augmented matrix G composed of the matrix A that is to be inverted and a unit matrix I of the same order as A; that is,

*Numbers in brackets refer to items listed in the bibliography.

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Section 2.4   Gauss-Jordan Elimination Method  43 Start Input G(N, N 1 1) 5 [A P] I51

I # N?

no

yes C 5 G(I, I) J51

J # N 1 1?

no

yes G(I, J) 5 G(I, J) / C J5J11 K51

K # N?

no

yes K 5 I?

yes

no D 5 G(K, I ) M5I

M # N 1 1?

no

yes G(K, M) 5 G(K, M) 2 G(I, M)*D M5M11 K5K11 I5I11 Output G Stop

Fig. 2.2  Flowchart for Solution of Simultaneous Equations by the Gauss-Jordan Method Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

44  Chapter 2   Matrix Algebra

 G



  5

[A

I]

Á

(2.24)

n 3 2n   n 3 n   n 3 n

Elementary operations are then applied to the rows of the augmented matrix to reduce A to a unit matrix. Matrix I, which was initially the unit matrix, now represents the inverse matrix A21; thus, G5





E x ample

2.12

5[A[I

3

S olutio n

 

(2.25)

[elementary operations

Determine the inverse of the matrix shown using the Gauss-Jordan method. 13 A 5 26 6



I] A ] 21

4

6 21 9

26 12 21

The augmented matrix is given by

3

13 G 5 [A Á I] 5 26 6

26 12 21

6 21 9

1 0 0

0 1 0

4

0 0  1

(1)

We begin the Gauss-Jordan elimination process by dividing row 1 of the augmented matrix by G11 5 13:

3

1 G 5 26 6

20.4615 0.4615 0.07692 12 21 0 21 9 0

0 1 0

4

0 0  1

(2)

Next, we multiply row 1 by G21 5 26 and subtract it from row 2, and then multiply row 1 by G31 5 6 and subtract it from row 3. This yields

3

1 G5 0 0

20.4615 9.231 1.769

0.4615 1.769 6.231

0.07692 0 0.4615 1 20.4615 0

4

0 0  1

(3)

Dividing row 2 by G22 5 9.231, we obtain

3

1 G5 0 0

20.4615 1 1.769

0.4615 0.1916 6.231

0.07692 0.04999 20.4615

0 0.1083 0

4

0 0  1

(4)

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Summary  45

Next, we multiply row 2 by G12 5 20.4615 and subtract it from row 1; then multiply row 2 by G32 5 1.769 and subtract it from row 3. This yields

3

1 G5 0 0

0 1 0

0.5499 0.1916 5.892

4

0.09999 0.04999 20.5499

0.04998 0 0.1083 0  20.1916 1

(5)

Divide row 3 by G33 5 5.892:

3

1 G5 0 0

0 1 0

0.5499 0.1916 1

0.09999 0.04999 20.09333

0.04998 0.1083 20.03252

4

0 0  0.1697

(6)

Multiply row 3 by G13 5 0.5499 and subtract it from row 1; then multiply row 3 by G23 5 0.1916 and subtract it from row 2 to obtain

3

1 G5 0 0

0 1 0

0 0 1

0.1513 0.06787 20.09333

0.06787 0.1145 20.03252

4

20.09333 20.03252  0.1697

(7)

Thus, the inverse of the given matrix A is

A21 5

3

0.1513 0.06787 20.09333

0.06787 0.1145 20.03252

4

20.09333 20.03252  0.1697

Ans

Finally, we check our computations by using the relationship AA21 5 I:

AA



21

3

13 5 26 6

3

0.9997 5 0 0

26 12 21

43

6 0.1513 21   0.06787 9 20.09333

0.0002 0.9993 0

0 0 0.9998

4

< I

0.06787 0.1145 20.03252

4

20.09333 20.03252  0.1697

Checks

Summary In this chapter, we discussed the basic concepts of matrix algebra that are necessary for formulating the matrix methods of structural analysis: 1.  A matrix is defined as a rectangular array of quantities (elements) arranged in rows and columns. The size of a matrix is measured by its number of rows and columns, and is referred to as its order. 2.  Two matrices are considered to be equal if they are of the same order, and if their corresponding elements are identical. 3.  Two matrices of the same order can be added (or subtracted) by adding (or subtracting) their corresponding elements.

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46  Chapter 2   Matrix Algebra

4.  The matrix multiplication AB 5 C is defined only if the number of columns of the first matrix A equals the number of rows of the second matrix B. Any element Cij of the product matrix C can be evaluated by using the relationship m



Cij 5

oA B

k51

ik

(2.6)

kj

where m is the number of columns of A, or the number of rows of B. Matrix multiplication is generally not commutative; that is, AB ? BA. 5.  The transpose of a matrix is obtained by interchanging its corresponding rows and columns. If C is a symmetric matrix, then CT 5 C. Another useful property of matrix transposition is that (AB)T 5 BTAT

(2.15)

6.  A matrix can be differentiated (or integrated) by differentiating (or integrating) each of its elements. 7.  The inverse of a square matrix A is defined as a matrix A21, which satisfies the relationship: AA21 5 A21A 5 I

(2.17)

8.  If the inverse of a matrix equals its transpose, the matrix is called an orthogonal matrix. 9. The Gauss-Jordan method of solving simultaneous equations essentially involves successively eliminating each unknown from all but one of the equations of the system by performing the following operations: dividing an equation by a scalar; and multiplying an equation by a scalar and subtracting the resulting equation from another equation. These elementary operations are applied repeatedly until a system with each equation containing only one unknown is obtained.

P r o b lems

2.3  Determine the products C 5 AB and D 5 BA if

Section 2.3 2.1  Determine the matrices C 5 A 1 B and D 5 A 2 B if

3

9 3 A5 22



23 25 24

4 3

2 8 4   B 5 27 6 21

27 9 0

A 5

4

21 0 26

34

25 2   B 5 [3  27 1 26] 28 4

2.4  Determine the products C 5 AB and D 5 BA if

2.2  Determine the matrices C 5 2A 1 B and D 5 A 2 3B if

3

6 A 5 28 25

21 24 22

28 3 23

4 3

24 5 6   B5 4 7 21

22 5 2

4 26 8

4

1 7 25

3

1 A 5 3

210 28

22 212

3

2 26 6   B 5 8 22 5

4

4

210 21 10 9

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Problems  47

2.13  Determine the derivative dA/dx if

2.5  Determine the products C 5 AB and D 5 BA if

3

A5



5 7 22

4 3

7 9 0

22 2 0   B 5 28 8 1

28 3 5

4

1 5 6

3

sin2 x A 5 26x2 7

2.6  Determine the products C 5 AB if



A5

3

3 2 23

4 3

4 3 0

2 8 28   B 5 6 6 3

1 22 0

6 5 21

4

3

4

11 29

25 18

  B5

3

12 2 10

6 15 25

27 9 24

4

0 213 8

2.9  Show that (ABC)T 5 CTBTAT by using the following matrices

3

29 13 A 5 8 211

3

27 C 5 21 16

4

0 20 15   B5 23 6 25 10 2 12

3

6 28 2

21 16

4

24 9

3

3 0.6 B 5 3 20.8

300 2100

4

0 22 8

4 3

7 8 24

4

23 4 9

3

4x 2 A 5 25x2

4 3

4

25x3 25x2 6 2x   B 5 2x2 7

2 23x3 2x

2x 23x2 4x

2.16 Determine the partial derivatives −A/−x, −A/−y, and −A/−z, if

3

x2 A 5 2y2 2z2

3

6x2 A 5 2x3 9

2y2 3xy 2yz

2x3 2x4 4x

3

2x A 5 2sin x 2 cos2 x

22z2 2yz 4xz

4

4

9 4x 5x3

4

20.8 20.6

2.12  Develop a computer program to determine the matrix triple product C 5 BTAB, where A is a square matrix of any order. Check the program by solving Problems 2.10 and 2.11 and comparing the results to those determined by hand calculations.

2sin x 5 24x3

2 cos2 x 24x3 (1 2 x2)

4

2.19  Calculate the integral e0L AB dx if

3

2x3 A 5 2x

4

20.6 0.8

2x 8 23x2 6x

2.18  Calculate the integral e0L A dx if

2100 200 0.8 0.6

4

2x2 212x 2x3 21

2.15  Determine the derivative d(AB)/dx if

2.11  Determine the matrix triple product C 5 BTAB if A 5

4 3

23x 5 2 4x 2x3   B 5 A 5 27 5x 2x3 2x2



225 5 12   B 5 27 30 23

210 15 12

4

2.17  Calculate the integral e0L A dx if

2.10  Determine the matrix triple product C 5 BTAB if 40 A 5 210 225

3

22 8 0

2.8  Show that (AB)T 5 BTAT by using the following matrices 14 23

26x 5 tan x

2.14  Determine the derivative d(A 1 B)/dx if

2.7  Develop a computer program to determine the matrix product C 5 AB of two conformable matrices A and B of any order. Check the program by solving Problems 2.4–2.6 and comparing the computer-generated results to those determined by hand calculations.

   A 5

23x3 cos2 x 3 sin x

2x2 2x2

4

3 2x3

3

4

x2 22x 23

22x 5 B5 3x3

2.20  Determine whether the matrices A and B given below are orthogonal matrices.



A5

3

20.28 0.96 0 0

20.96 20.28 0 0

4

0 0 0 0 20.28 20.96 0.96 20.28

Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

48  Chapter 2   Matrix Algebra

B5



3

20.28 0.96 0 0

0.96 2 0.28 0 0

0 0 2 0.28 0.96

4

0 0 0.96 2 0.28

Section 2.4 2.21 through 2.25  Solve the following systems of simultaneous equations by the Gauss-Jordan method. 2.21  

2x1 2 3x2 1 x3 5 218 29x1 1 5x2 1 3x3 5 18 4x1 1 7x2 2 8x3 5 53

2.25

2x1 2 5x2 1 8x3 1 11x4 5 39 10x1 1 7x2 1 4x3 2 x4 5 127   2 3x1 1 9x2 1 5x3 2 6x4 5 58 x1 2 4x2 2 2x3 1 9x4 5 214

2.26  Develop a computer program to solve a system of simultaneous equations of any size by the Gauss-Jordan method. Check the program by solving Problems 2.21 through 2.25 and comparing the computer-generated results to those determined by hand calculations. 2.27 through 2.30  Determine the inverse of the matrices shown by the Gauss-Jordan method.

3

3 8 22

24 22 7

3

24 9 2

1 3 5

7 26 2.29  A 5 3 22

3 3

26 4 21 5

3 21 8 9

5 10 2.30  A5 21 29

27 26 12 7

23 213 8 25

2.27  A 5 2.22

20x1 2 9x2 1 15x3 5 354   29x1 1 16x2 2 5x3 5 2275 15x1 2 5x2 1 18x3 5 307

2.23 4x1 2 2x2 1 3x3 5 37.2   3x1 1 5x2 2 x3 5 27.2 x1 2 4x2 1 2x3 5 30.3 2.24  

6x1 1 15x2 2 24x3 1 40x4 5 190.9 15x1 1 9x2 2 13x3 5 69.8 224x1 2 13x2 1 8x3 2 11x4 5 296.3 2 11x3 1 5x4 5 119.35 40x1

5 3 24

6 2.28  A 5 21 4

4

4 22 5 9 2

4 4

11 2 24 6

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3

Plane Trusses Learning Objectives At the end of this chapter, you will be able to: 3.1 Establish global and local coordinate systems 3.2 Identify degrees of freedom 3.3 Evaluate member stiffness relations in local coordinate system 3.4 Apply the finite-element formulation using virtual work 3.5 Perform coordinate transformations 3.6 Calculate member stiffness relations in global coordinate system 3.7 Develop structure stiffness relations 3.8 Analyze plane trusses

Goethals Bridge, a Cantilever Truss Bridge between Staten Island, NY, and Elizabeth, NJ. (Photo courtesy of Port Authority of New York and New Jersey)

49

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50  Chapter 3   Plane Trusses

A plane truss is defined as a two-dimensional framework of straight prismatic members connected at their ends by frictionless hinged joints, and subjected to loads and reactions that act only at the joints and lie in the plane of the structure. The members of a plane truss are subjected to axial compressive or tensile forces only. The objective of this chapter is to develop the analysis of plane trusses based on the matrix stiffness method. This method of analysis is general, in the sense that it can be applied to statically determinate, as well as indeterminate, plane trusses of any size and shape. We begin the chapter with the definitions of the global and local coordinate systems to be used in the analysis. The concept of “degrees of freedom” is introduced in Section 3.2; and the member force-displacement relations are established in the local coordinate system, using the equilibrium equations and the principles of mechanics of materials, in Section 3.3. The finite-element formulation of member stiffness relations using the principle of virtual work is presented in Section 3.4; and transformation of member forces and displacements from a local to a global coordinate system, and vice versa, is considered in Section 3.5. Member stiffness relations in the global coordinate system are derived in Section 3.6; the formulation of the stiffness relations for the entire truss, by combining the member stiffness relations, is discussed in Section 3.7; and a step-by-step procedure for the analysis of plane trusses subjected to joint loads is developed in Section 3.8.

3.1 

Global and Local Coordinate Systems In the matrix stiffness method, two types of coordinate systems are employed to specify the structural and loading data and to establish the necessary forcedisplacement relations. These are referred to as the global (or structural) and the local (or member) coordinate systems.

Global Coordinate System The overall geometry and the load-deformation relationships for an entire structure are described with reference to a Cartesian or rectangular global coordinate system. The global coordinate system used in this text is a right-handed XYZ coordinate system with the plane structure lying in the XY plane.

When analyzing a plane (two-dimensional) structure, the origin of the global XY coordinate system can be located at any point in the plane of the structure, with the X and Y axes oriented in any mutually perpendicular directions in the structure’s plane. However, it is usually convenient to locate the origin at a lower left joint of the structure, with the X and Y axes oriented

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Section 3.1   Global and Local Coordinate Systems  51

in the horizontal (positive to the right) and vertical (positive upward) directions, respectively, so that the X and Y coordinates of most of the joints are positive. Consider, for example, the truss shown in Fig. 3.1(a), which is composed of six members and four joints. Figure 3.1(b) shows the analytical model of the truss as represented by a line diagram, on which all the joints and members are identified by numbers that have been assigned arbitrarily. The global coordinate system chosen for analysis is usually drawn on the line diagram of the structure, as shown in Fig. 3.1(b). Note that the origin of the global XY coordinate system is located at joint 1.

Local Coordinate System Since it is convenient to derive the basic member force-displacement relationships in terms of the forces and displacements in the directions along and perpendicular to members, a local coordinate system is defined for each member of the structure.

Y 225 kN

450 kN

225 kN

(4,000 mm2)

30˚

y2

225 kN

3

x2

389.7 kN

4

2

(6,

m

00

0m

0m

00

m2 )

(6,

6 (6,000 mm2)

(6,000 mm2)

)

8m

2

5

3

4

x5 x6

y1 y5 (4,000 mm2) 1

E = 200 GPa (a) Actual Truss

y4

x1

y3 6m

x4

x3

2

X

1 y6

(b) Analytical Model Showing Global and Local Coordinate Systems

Fig. 3.1

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52  Chapter 3   Plane Trusses Y

Y

d4 49

d2

3

39 d5

d3

5 2

3

4

4

2

4

3 Undeformed configuration

Deformed configuration

6

5

3

2

1

29

4

1

X

2

6

1

X

1 d1 (c) Degrees of Freedom

7 8 (d) Degrees of Freedom and Restrained Coordinate Numbers

Y

P5 , d5

P3 , d3

2

3

P 2 , d2

6

4

P4 , d4

5

3

4

1

2 P1, d1

R6

X

1

R7

R8

(e) Degrees of Freedom, Joint Loads, and Reactions

Fig. 3.1  (continued)

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Section 3.2   Degrees of Freedom  53

The origin of the local xyz coordinate system for a member may be arbitrarily located at one of the ends of the member in its undeformed state, with the x axis directed along the member’s centroidal axis in the undeformed state. The positive direction of the y axis is defined so that the coordinate system is right-handed, with the local z axis pointing in the positive direction of the global Z axis. On the line diagram of the structure, the positive direction of the x axis for each member is indicated by drawing an arrow along each member, as shown in Fig. 3.1(b). For example, this figure shows the origin of the local coordinate system for member 1 located at its end connected to joint 1, with the x1 axis directed from joint 1 to joint 2; the origin of the local coordinate system for member 4 located at its end connected to joint 2, with the x4 axis directed from joint 2 to joint 4, etc. The joint to which the member end with the origin of the local coordinate system is connected is termed the beginning joint for the member, and the joint adjacent to the opposite end of the member is referred to as the end joint. For example, in Fig. 3.1(b), member 1 begins at joint 1 and ends at joint 2, member 4 begins at joint 2 and ends at joint 4, and so on. Once the local x axis is specified for a member, its y axis can be established by applying the right-hand rule. The local y axes thus obtained for all six members of the truss are depicted in Fig. 3.1(b). It can be seen that, for each member, if we curl the fingers of our right hand from the direction of the x axis toward the direction of the corresponding y axis, then the extended thumb points out of the plane of the page, which is the positive direction of the global Z axis.

3.2 

Degrees of Freedom The degrees of freedom of a structure, in general, are defined as the independent joint displacements (translations and rotations) that are necessary to specify the deformed shape of the structure when subjected to an arbitrary loading. Since the joints of trusses are assumed to be frictionless hinges, they are not subjected to moments and, therefore, their rotations are zero. Thus, only joint translations must be considered in establishing the degrees of freedom of trusses. Consider again the plane truss of Fig. 3.1(a). The deformed shape of the truss, for an arbitrary loading, is depicted in Fig. 3.1(c) using an exaggerated scale. From this figure, we can see that joint 1, which is attached to the hinged support, cannot translate in any direction; therefore, it has no degrees of freedom. Because joint 2 is attached to the roller support, it can translate in the X direction, but not in the Y direction. Thus, joint 2 has only one degree of freedom, which is designated d1 in the figure. As joint 3 is not attached to a support, two displacements (namely, the translations d2 and d3 in the X and Y directions, respectively) are needed to completely specify its deformed position 39. Thus, joint 3 has two degrees of freedom. Similarly, joint 4, which is also a free joint, has two degrees of freedom, designated d4 and d5. Thus, the entire truss has a total of five degrees of freedom. As shown in Fig. 3.1(c), the joint displacements

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54  Chapter 3   Plane Trusses

are defined relative to the global coordinate system, and are considered to be positive when in the positive directions of the X and Y axes. Note that all the joint displacements are shown in the positive sense in Fig. 3.1(c). The five joint displacement of the truss can be collectively written in matrix form as



d5

34 d1 d2 d3 d4 d5

in which d is called the joint displacement vector, with the number of rows equal to the number of degrees of freedom of the structure. It is important to note that the five joint displacements d1 through d5 are necessary and sufficient to uniquely define the deformed shape of the truss under any arbitrary loading condition. Furthermore, the five joint displacements are independent, in the sense that each displacement can be varied arbitrarily and independently of the others. As the foregoing example illustrates, the degrees of freedom of all types of framed structures, in general, are the same as the actual joint displacements. Thus, the number of degrees of freedom of a framed structure can be determined by subtracting the number of joint displacements restrained by supports from the total number of joint displacements of the unsupported structure; that is,



1

number of degrees of freedom

2

5

1

21

number of joint displacements of the unsupported structure

2

2

number of joint displacements (3.1) restrained by supports

As the number of displacements of an unsupported structure equals the product of the number of degrees of freedom of a free joint of the structure and the total number of joints of the structure, we can express Eq. (3.1) as NDOF 5 NCJT (NJ) 2 NR

(3.2)

in which NDOF represents the number of degrees of freedom of the structure (sometimes referred to as the degree of kinematic indeterminacy of the structure); NCJT represents the number of degrees of freedom of a free joint (also called the number of structure coordinates per joint); NJ is the number of joints; and NR denotes the number of joint displacements restrained by supports. Since a free joint of a plane truss has two degrees of freedom, which are translations in the X and Y directions, we can specialize Eq. (3.2) for the case of plane trusses:

NCJT 5 2 for plane trusses NDOF 5 2(NJ) 2 NR f 

(3.3)

Let us apply Eq. (3.3) to the truss of Fig. 3.1(a). The truss has four joints (i.e., NJ 5 4), and the hinged support at joint 1 restrains two joint displacements,

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Section 3.2   Degrees of Freedom  55

namely, the translations of joint 1 in the X and Y directions; whereas the roller support at joint 2 restrains one joint displacement, which is the translation of joint 2 in the Y direction. Thus, the total number of joint displacements that are restrained by all supports of the truss is 3 (i.e., NR 5 3). Substituting the numerical values of NJ and NR into Eq. (3.3), we obtain NDOF 5 2(4) 2 3 5 5 which is the same as the number of degrees of freedom of the truss obtained previously. The degrees of freedom (or joint displacements) of a structure are also termed the structure’s free coordinates; the joint displacements restrained by supports are commonly called the restrained coordinates of the structure. The free and restrained coordinates are referred to collectively as simply the structure coordinates. It should be noted that each structure coordinate represents an unknown quantity to be determined by the analysis, with a free coordinate representing an unknown joint displacement, and a restrained coordinate representing an unknown support reaction. Realizing that NCJT (i.e., the number of structure coordinates per joint) equals the number of unknown joint displacements and/or support reactions per joint of the structure, the total number of unknown joint displacements and reactions for a structure can be expressed as

1

2

number of unknown joint displacements 5 NDOF 1 NR 5 NCJT(NJ) and support reactions

Numbering of Degrees of Freedom and Restrained Coordinates When analyzing a structure, it is not necessary to draw the structure’s deformed shape, as shown in Fig. 3.1(c), to identify its degrees of freedom. Instead, the degrees of freedom can be directly specified on the line diagram of the structure by assigning numbers to the arrows drawn at the joints in the directions of the joint displacements, as shown in Fig. 3.1(d). The restrained coordinates are identified in a similar manner. However, the arrows representing the restrained coordinates are usually drawn with a slash ( ) to distinguish them from the arrows identifying the degrees of freedom. The degrees of freedom of a plane truss are numbered starting at the lowest-numbered joint that has a degree of freedom, and proceeding sequentially to the highest-numbered joint. In the case of more than one degree of freedom at a joint, the translation in the X direction is numbered first, followed by the translation in the Y direction. The first degree of freedom is assigned the number one, and the last degree of freedom is assigned a number equal to NDOF. Once all the degrees of freedom of the structure have been numbered, we number the restrained coordinates in a similar manner, but begin with a number equal to NDOF 1 1. We start at the lowest-numbered joint that is attached to a support, and proceed sequentially to the highest-numbered joint. In the case of more than one restrained coordinate at a joint, the coordinate in the X direction

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56  Chapter 3   Plane Trusses

is numbered first, followed by the coordinate in the Y direction. Note that this procedure will always result in the last restrained coordinate of the structure being assigned a number equal to 2(NJ). The degrees of freedom and the restrained coordinates of the truss in Fig. 3.1(d) have been numbered using the foregoing procedure. We start numbering the degrees of freedom by examining joint 1. Since the displacements of joint 1 in both the X and Y directions are restrained, this joint does not have any degrees of freedom; therefore, at this point, we do not assign any numbers to the two arrows representing its restrained coordinates, and move on to the next joint. Focusing our attention on joint 2, we realize that this joint is free to displace in the X direction, but not in the Y direction. Therefore, we assign the number 1 to the horizontal arrow indicating that the X displacement of joint 2 will be denoted by d1. Note that, at this point, we do not assign any number to the vertical arrow at joint 2, and change our focus to the next joint. Joint 3 is free to displace in both the X and Y directions; we number the X displacement first by assigning the number 2 to the horizontal arrow, and then number the Y displacement by assigning the number 3 to the vertical arrow. This indicates that the X and Y displacements of joint 3 will be denoted by d2 and d3, respectively. Next, we focus our attention on joint 4, which is also free to displace in both the X and Y directions; we assign numbers 4 and 5, respectively, to its displacements in the X and Y directions, as shown in Fig. 3.1(d). Again, the arrow that is numbered 4 indicates the location and direction of the joint displacement d4; the arrow numbered 5 shows the location and direction of d5. Having numbered all the degrees of freedom of the truss, we now return to joint 1, and start numbering the restrained coordinates of the structure. As previously discussed, joint 1 has two restrained coordinates; we first assign the number 6 (i.e., NDOF 1 1 5 5 1 1 5 6) to the X coordinate (horizontal arrow), and then assign the number 7 to the Y coordinate (vertical arrow). Finally, we consider joint 2, and assign the number 8 to the vertical arrow representing the restrained coordinate in the Y direction at that joint. We realize that the displacements corresponding to the restrained coordinates 6 through 8 are zero (i.e., d6 5 d7 5 d8 5 0). However, we use these restrained coordinate numbers to specify the reactions at supports of the structure, as discussed subsequently in this section.

Joint Load Vector External loads applied to the joints of trusses are specified as force components in the global X and Y directions. These load components are considered positive when acting in the positive directions of the X and Y axes, and vice versa. Any loads initially given in inclined directions are resolved into their X and Y components, before proceeding with an analysis. For example, the 450 kN inclined load acting on a joint of the truss in Fig. 3.1(a) is resolved into its rectangular components as load component in X direction 5 450 cos 308 5 389.7 kN →   load component in Y direction 5 450 sin 308 5 225 kN ↓

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Section 3.2   Degrees of Freedom  57

These components are applied at joint 3 of the line diagram of the truss, as shown in Fig. 3.1(b). In general, a load can be applied to a structure at the location and in the direction of each of its degrees of freedom. For example, a five-degree-offreedom truss can be subjected to a maximum of five loads, P1 through P5, as shown in Fig. 3.1(e). As indicated there, the numbers assigned to the degrees of freedom are also used to identify the joint loads. In other words, a load corresponding to a degree of freedom di is denoted by the symbol Pi. The five joint loads of the truss can be collectively written in matrix form as



P5

34 3 4 P1 P2 P3 P4 P5

0 389.7 5 2225 0 2225

kN

in which P is called the joint load vector of the truss. The numerical form of P is obtained by comparing Figs. 3.1(b) and 3.1(e). This comparison shows that P1 5 0; P2 5 389.7 kN; P3 5 2225 kN; P4 5 0; and P5 5 2225 kN. The negative signs assigned to the magnitudes of P3 and P5 indicate that these loads act in the negative Y (i.e., downward) direction. The numerical values of P1 through P5 are then stored in the appropriate rows of the joint load vector P, as shown in the foregoing equation. It should be noted that the number of rows of P equals the number of degrees of freedom (NDOF) of the structure.

Reaction Vector A support that prevents the translation of a joint of a structure in a particular direction exerts a reaction force on the joint in that direction. Thus, when a truss is subjected to external loads, a reaction force component can develop at the location and in the direction of each of its restrained coordinates. For example, a truss with three restrained coordinates can develop up to three reactions, as shown in Fig. 3.1(e). As indicated there, the numbers assigned to the restrained coordinates are used to identify the support reactions. In other words, a reaction corresponding to an ith restrained coordinate is denoted by the symbol Ri. The three support reactions of the truss can be collectively expressed in matrix form as

34

R6 R 5 R7 R8

in which R is referred to as the reaction vector of the structure. Note that the number of rows of R equals the number of restrained coordinates (NR) of the structure. The procedure presented in this section for numerically identifying the degrees of freedom, joint loads, and reactions of a structure considerably simplifies the task of programming an analysis on a computer, as will become apparent in Chapter 4.

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58  Chapter 3   Plane Trusses

E x a m ple 3.1 Identify numerically the degrees of freedom and restrained coordinates of the tower



truss shown in Fig. 3.2(a). Also, form the joint load vector P for the truss.



S ol u tio n

The truss has nine degrees of freedom, which are identified by the numbers 1 through 9 in Fig. 3.2(c). The five restrained coordinates of the truss are identified by the numbers 10 through 14 in the same figure. Ans By comparing Figs. 3.2(b) and (c), we express the joint load vector as

FG

60 0 0 60 0 P5 0 2105 30 260

kN

Ans

Y Y 60 kN

9

60 kN 7

30 kN

30 kN

12

5

5m 105 kN 60 kN

7

8

5

105 kN 60 kN

7

7

5

6

4

6 5

3

5m

3 60 kN

11

10

2

4

6

60 kN

3

1 1

5m

6

4

8

9

14 3

4

2 1

1 5m

(a) Tower Truss

2

2

X 10

X

12

(b) Analytical Model 11

13

(c) Degrees of Freedom and Restrained Coordinates (NDOF = 9, NR = 5)

Fig. 3.2

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Section 3.3   Member Stiffness Relations in the Local Coordinate System   59

3.3 Member Stiffness Relations in the Local Coordinate System

In the stiffness method of analysis, the joint displacements, d, of a structure due to an external loading, P, are determined by solving a system of simultaneous equations, expressed in the form P 5 Sd

(3.4)

in which S is called the structure stiffness matrix. It will be shown subsequently that the stiffness matrix for the entire structure, S, is formed by assembling the stiffness matrices for its individual members. The stiffness matrix for a member expresses the forces at the ends of the member as functions of the displacements of those ends. In this section, we derive the stiffness matrix for the members of plane trusses in the local coordinate system. To establish the member stiffness relations, let us focus our attention on an arbitrary member m of the plane truss shown in Fig. 3.3(a). When the truss is subjected to external loads, m deforms and internal forces are induced at its ends. The initial and displaced positions of m are shown in Fig. 3.3(b), where L,  E, and A denote, respectively, the length, Young’s modulus of elasticity, and the cross-sectional area of m. The member is prismatic in the sense that its axial rigidity, EA, is constant. As Fig. 3.3(b) indicates, two displacements— translations in the x and y directions—are needed to completely specify the displaced position of each end of m. Thus, m has a total of four end displacements or degrees of freedom. As shown in Fig. 3.3(b), the member end displacements are denoted by u1 through u4, and the corresponding member end forces are denoted by Q1 through Q4. Note that these end displacements and forces are defined relative to the local coordinate system of the member, and are considered positive when in the positive directions of the local x and y axes. Y

y y

b9

e x

b

u3

Q1

Q3

Q2

L

x

e

m

b

X

u4

Initial position

u2

m

(a) Plane Truss

e9

Displaced position

u1

Q4

EA = constant (b) Member Forces and Displacements in Local Coordinate System

=

Fig. 3.3

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60  Chapter 3   Plane Trusses u1  1 k11 

EA L

b9

b

EA k31  2 L

e

k21  0

3

u1

k41  0 L (c)

1 b9

L9 ≈

u2  1

L

ρ

k12  0

k32  0

e

b k22  0

3 u2

k42  0 L (d)

1 u3  1 EA k13  2 L

b k23  0

k33 

e

L

e9

EA L

3 u3

k43  0

(e)

1 e9 u4  1

k14  0

e

b k24  0

L

k34  0

3 u4

k44  0

(f)

Fig. 3.3  (continued)

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Section 3.3   Member Stiffness Relations in the Local Coordinate System   61

As indicated in Fig. 3.3(b), the numbering scheme used to identify the member end displacements and forces is as follows: Member end displacements and forces are numbered by beginning at the end of the member designated “b,” where the origin of the local coordinate system is located, with the translation and force in the x direction numbered first, followed by the translation and force in the y direction. The displacements and forces at the opposite end of the member, designated “e,” are then numbered in the same sequential order. It should be remembered that our objective here is to determine the relationships between member end forces and end displacements. Such relationships can be conveniently established by subjecting the member, separately, to each of the four end displacements as shown in Figs. 3.3(c) through (f); and by expressing the total member end forces as the algebraic sums of the end forces required to cause the individual end displacements. Thus, from Figs. 3.3(b) through (f), we can see that Q1 5 k11u1 1 k12u2 1 k13u3 1 k14u4

(3.5a)

Q2 5 k21u1 1 k22u2 1 k23u3 1 k24u4

(3.5b)

Q3 5 k31u1 1 k32u2 1 k33u3 1 k34u4

(3.5c)

Q4 5 k41u1 1 k42u2 1 k43u3 1 k44u4

(3.5d)

in which kij represents the force at the location and in the direction of Qi required, along with other end forces, to cause a unit value of displacement uj, while all other end displacements are zero. These forces per unit displacement are called stiffness coefficients. It should be noted that a double-subscript notation is used for stiffness coefficients, with the first subscript identifying the force and the second subscript identifying the displacement. By using the definition of matrix multiplication, Eqs. (3.5) can be expressed in matrix form as



34 3 Q1 Q2 Q3 Q4

5

k11 k21 k31 k41

k12 k22 k32 k42

k13 k23 k33 k43

k14 k24 k34 k44

434

u1 u2  u3 u4

(3.6)

or, symbolically, as

Q 5 ku 

(3.7)

in which Q and u are the member end force and member end displacement vectors, respectively, in the local coordinate system; and k is called the member stiffness matrix in the local coordinate system.

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62  Chapter 3   Plane Trusses

The stiffness coefficients kij can be evaluated by subjecting the member, separately, to unit values of each of the four end displacements. The member end forces required to cause the individual unit displacements are then determined by applying the equations of equilibrium, and by using the principles of mechanics of materials. The member end forces thus obtained represent the stiffness coefficients for the member. Let us determine the stiffness coefficients corresponding to a unit value of the displacement u1 at end b of m, as shown in Fig. 3.3(c). Note that all other displacements of m are zero (i.e., u2 5 u3 5 u4 5 0). Since m is in equilibrium, the end forces k11, k21, k31, and k41 acting on it must satisfy the three equilibrium equations: Fx 5 0, Fy 5 0, and M 5 0. Applying the equations of equilibrium, we write

o



1S 1c

oF 5 0

1

x

oF 5 0 oM 5 0 y

[



e

o

k11 1 k31 5 0  k31 5 2k11 k21 1 k41 5 0  2k21(L) 5 0

o

(3.8) (3.9)

Since L is not zero, k21 must be zero; that is, k21 5 0 (3.10) By substituting Eq. (3.10) into Eq. (3.9), we obtain k41 5 0 (3.11) Equations (3.8), (3.10), and (3.11) indicate that m is in equilibrium under the action of two axial forces, of equal magnitude but with opposite senses, applied at its ends. Furthermore, since the displacement u1 5 1 results in the shortening of the member’s length, the two axial forces causing this displacement must be compressive; that is, k11 must act in the positive direction of the local x axis, and k31 (with a magnitude equal to k11) must act in the negative direction of the x axis. To relate the axial force k11 to the unit axial deformation (u1 5 1) of m, we use the principles of the mechanics of materials. Recall that in a prismatic member subjected to axial tension or compression, the normal stress s is given by







k11 axial force 5  cross { sectional area A and the normal strain, «, is expressed as s5

«5

change in length 1 5  original length L

(3.12)

(3.13)

For linearly elastic materials, the stress-strain relationship is given by Hooke’s law as s 5 E« (3.14) Substitution of Eqs. (3.12) and (3.13) into Eq. (3.14) yields k11 A

5E

1L1 2

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Section 3.3   Member Stiffness Relations in the Local Coordinate System   63

from which we obtain the expression for the stiffness coefficient k11,

k11 5

EA  L

(3.15)

The expression for k31 can now be obtained from Eq. (3.8) as

k31 5 2k11 5 2

EA  L

(3.16)

in which the negative sign indicates that this force acts in the negative x direction. Figure 3.3(c) shows the expressions for the four stiffness coefficients required to cause the end displacement u1 5 1 of m. By using a similar approach, it can be shown that the stiffness coefficients required to cause the axial displacement u3 5 1 at end e of m are as follows (Fig. 3.3(e)).

k13 5 2

EA L

k23 5 0

k33 5

EA L

k43 5 0

(3.17)

The deformed shape of m due to a unit value of displacement u2, while all other displacements are zero, is shown in Fig. 3.3(d). Applying the equilibrium equations, we write



1S 1c

oF 5 0 x

oF 5 0 oM 5 0 y

[



1

e

k12 1 k32 5 0  k32 5 2k12 k22 1 k42 5 0  2k22(L) 5 0

(3.18) (3.19)

from which we obtain k22 5 0

(3.20)

Substitution of Eq. (3.20) into Eq. (3.19) yields k42 5 0

(3.21)

Thus, the forces k22 and k42, which act perpendicular to the longitudinal axis of m, are both zero. As for the axial forces k12 and k32, Eq. (3.18) indicates that they must be of equal magnitude but with opposite senses. From Fig. 3.3(d), we can see that the deformed length of the member, L9, can be expressed in terms of its undeformed length L as

L9 5

L  cos r

(3.22)

in which the angle r denotes the rotation of the member due to the end displacement u2 5 1. Since the displacements are assumed to be small, cos r  0

Therefore, the member is in equilibrium.

6.3 

Checks Ans

Coordinate Transformations Unlike beams, whose members all are oriented in the same direction, plane frames usually contain members oriented in various directions in the plane of the structure. Therefore, it becomes necessary to transform the stiffness relations of the members of a plane frame from their local coordinate systems to the global coordinate system before they can be combined to establish the stiffness relations for the entire frame. In this section, we extend the transformation relationships developed in Section 3.5 for plane truss members to include end moments and rotations, so that they can be used for the members of plane frames. The revised transformation relations thus obtained are then used in Section 6.4 to develop the member stiffness relations in the global coordinate system for plane frames. Consider an arbitrary member m of a plane frame, as shown in Fig. 6.7(a). The orientation of the member with respect to the global XY coordinate system is defined by an angle u, measured counterclockwise from the positive direction of the global X axis to the positive direction of the local x axis, as shown in Fig. 6.7(a). When the frame is subjected to external loads, member m deforms, and internal forces and moments develop at its ends. The displaced position of member m, due to an arbitrary loading applied to the frame, is shown in Figs. 6.7(b) and (c). In Fig. 6.7(b), the member end displacements, u, and end forces, Q, are measured relative to the local xy coordinate system of the

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272  Chapter 6   Plane Frames

­ ember; whereas, in Fig. 6.7(c), the member end displacements, v, and end m forces, F, are defined with respect to the global XY coordinate system of the frame. The local and global systems of member end displacements and forces are equivalent, in the sense that both systems cause the same translations and rotations of the member ends b and e, and produce the same state of strain and stress in the member. As shown in Fig. 6.7(c), the global member end forces, F, and end displacements, v, are numbered by beginning at member end b, with the force and translation in the X direction numbered first, followed by the Y

e

y x

m

θ

b

X (a) Frame

Displaced position u 6

Y

e9

u5

y u3

Q4 e

u2

θ Q1

b Q3 Q 2

x

Q6

b9 m

Q5

u4

Initial position

u1 X (b) Member End Forces and End Displacements in the Local Coordinate System

Fig. 6.7 Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 6.3   Coordinate Transformations  273

5 Displaced position y6

Y

e9 y5

y4

y y3

y1

F6

b9

y2

b

F1 F3

x F4

e

θ

m

F5

Initial position

F2

X (c) Member End Forces and End Displacements in the Global Coordinate System

Fig. 6.7  (continued)

force and translation in the Y direction, and then the moment and rotation. The forces and displacements at the member’s opposite end e are then numbered in the same sequential order. Now, suppose that the member’s global end forces and end displacements are specified, and we wish to determine the corresponding end forces and end displacements in the local coordinate system of the member. As discussed in Section 3.5, the local forces Q1 and Q2 must be equal to the algebraic sums of the components of the global forces F1 and F2 in the directions of the local x and y axes, respectively; that is, Q1 5 F1 cos u 1 F2 sin u

(6.16a)

Q2 5 2F1 sin u 1 F2 cos u

(6.16b)

Note that Eqs. (6.16a and b) are identical to Eqs. (3.58a and b), respectively, derived previously for the case of plane truss members. As for the relationship between the local end moment Q3 and the global end moment F3—because the local z axis and the global Z axis are oriented in the same direction (i.e., directed out of the plane of the page), the local moment Q3 must be equal to the global moment F3. Thus, Q3 5 F3

(6.16c)

Using a similar reasoning at end e of the member, we express the local forces in terms of the global forces as Q4 5 F4 cos u 1 F5 sin u

(6.16d)

Q5 5 2F4 sin u 1 F5 cos u

(6.16e)

Q6 5 F6

(6.16f)

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274  Chapter 6   Plane Frames

FG F

GF G

We can write Eqs. (6.16a through f) in matrix form as Q1 Q2 Q3 Q4 Q5 Q6

5

cos u 2sin u 0 0 0 0

sin u cos u 0 0 0 0

0 0 1 0 0 0

0 0 0 cos u 2sin u 0

0 0 0 sin u cos u 0

0 0 0 0 0 1

F1 F2 F3  F4 F5 F6

(6.17)

or, symbolically, as Q 5 TF

F

in which the transformation matrix T is given by

T5

cos u 2sin u 0 0 0 0

sin u cos u 0 0 0 0

0 0 1 0 0 0

0 0 0 cos u 2sin u 0

0 0 0 sin u cos u 0

G

0 0 0 0 0 1

(6.18)



(6.19)

The direction cosines (cos  u and sin  u) of the plane frame members can be evaluated using Eqs. (3.62a and b), given in Section 3.5. Because member end displacements, like end forces, are vectors, which are defined in the same directions as the corresponding forces, the transformation matrix T (Eq. (6.19)) can also be used to transform member end displacements from the global to the local coordinate system; that is, u 5 Tv



(6.20)

Next, we consider the transformation of member end forces and end displacements from the local to the global coordinate system. Returning our attention to Figs. 6.7(b) and (c), we realize that at end b of the member, the global forces F1 and F2 must be equal to the algebraic sums of the components of the local forces Q1 and Q2 in the directions of the global X and Y axes, respectively; that is, F1 5 Q1 cos u 2 Q2 sin u

(6.21a)

F2 5 Q1 sin u 2 Q2 cos u

(6.21b)

and, as discussed previously, the global moment F3 equals the local moment Q3, or F3 5 Q3

(6.21c)

In a similar manner, the global forces at end e of the member can be expressed in terms of the local forces as F4 5 Q4 cos u 2 Q5 sin u

(6.21d)

F5 5 Q4 sin u 1 Q5 cos u

(6.21e)

F6 5 Q6

(6.21f)

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FG F

Section 6.3   Coordinate Transformations  275

We can write Eqs. (6.21a through f) in matrix form as F1 F2 F3 F4 F5 F6

5

cos u 2sin u sin u cos u 0 0 0 0 0 0 0 0

0 0 1 0 0 0

0 0 0 cos u sin u 0

0 0 0 2sin u cos u 0

GF G

0 0 0 0 0 1

Q1 Q2 Q3  (6.22) Q4 Q5 Q6

By comparing Eq. (6.22) to Eq. (6.17), we realize that the transformation matrix in Eq. (6.22), which transforms the forces from the local to the global coordinate system, is the transpose of the transformation matrix T in Eq. (6.17), which transforms the forces from the global to the local coordinate system. Therefore, Eq. (6.22) can be written as F 5 T TQ

(6.23)

Also, a comparison of Eqs. (6.18) and (6.23) indicates that the inverse of T equals its transpose; that is, T21 5 TT

(6.24)

which indicates that the transformation matrix T is orthogonal. As discussed previously, because the member end displacements are also vectors defined in the directions of their corresponding forces, the matrix TT also defines the transformation of member end displacements from the local to the global coordinate system; that is, v 5 TT u (6.25) By comparing the transformation matrix T derived herein for plane frame members (Eq. (6.19)) with the one developed in Section 3.5 for plane truss members (Eq. (3.61)), we observe that the T matrix for plane trusses can be obtained by deleting the third and sixth columns and the third and sixth rows from the T matrix for plane frame members. This is because there are no moments and rotations induced at the ends of plane truss members, which are subjected to axial forces only. E x ample 6.3

The displaced position of member 2, of the frame of Fig. 6.8(a), is given in Fig. 6.8(b). Calculate the end displacements and end forces for this member in the global coordinate system. Is the member in equilibrium under the global end forces?

F G

S ol u t i o n Member Local End Displacements and Forces:  In Example 6.2, we obtained the local

end displacement and force vectors for the member under consideration as

u2 5

0.030599 m 0.023897 m 20.0029702 rad  0.029582 m 0.021352 m 20.010447 rad

(1)

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276  Chapter 6   Plane Frames Y

2 400 kN

48 kN/m 2 3 6m

200 kN

1

3

3m

1

4 X 4m E, A, I 5 constant E 5 200 GPa A 5 12,500 mm2 I 5 275(106) mm4 (a)

y 23

.89

7m

m

29

0.0

02

2

30

.59

97

02

9m

Displaced position

rad

m 0.0

10

Initial position

44

7r

ad 39

3

29

.58

21

.35

2m

m

2m

m x

(b) Displaced Position of Member 2

Fig. 6.8

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Section 6.3   Coordinate Transformations  277

29 0.0029702 rad

2

Displaced position

0.7582 mm 38.818 mm 0.010447 rad

Y Initial position

39

3 0.6676 mm

36.477 mm

X (c) End Displacements in the Global Coordinate System for Member 2 246.92 kN • m 308.6 kN

2 48 kN/m 316.04 kN L5

Y

5m

308.6 kN

3

571.4 kN • m 556.04 kN X (d) End Forces in the Global Coordinate System for Member 2

Fig. 6.8  (continued) and

F G

436.5 kN 267.665 kN 2246.92 kN ? m Q2 5 k2u2 1 Qf 2 5 2580.5 kN 259.67 kN 2571.4 kN ? m



(2)

Transformation Matrix:  From Fig. 6.8(a), we can see that joint 2 is the beginning joint and joint 3 is the end joint for member 2. By applying Eqs. (3.62), we determine the member’s direction cosines as

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278  Chapter 6   Plane Frames

cos u 5 sin u 5

X3 2 X2 L Y3 2 Y2 L

F

5

420 5 0.8 5

5

326 5 20.6 5

G

The transformation matrix for member 2 can now be evaluated, using Eq. (6.19).

T2 5

0.8 20.6 0.6 0.8 0 0 0 0 0 0 0 0

0 0 1 0 0 0

0 0 0 0 0 0 0.8 20.6 0.6 0.8 0 0

0 0 0  0 0 1

F G

(3)

Member Global End Displacements:  By substituting the transpose of T2 from Eq. (3), and u2 from Eq. (1), into Eq. (6.25), we obtain

v2 5 TT2 u2 5

0.038818 m 0.0007582 m 20.0029702 rad  0.036477 m 20.0006676 m 20.010447 rad

Ans

These end displacements are depicted in Fig. 6.8(c). Member Global End Forces: Similarly, by substituting the transpose of T2 from Eq. (3), and Q2 from Eq. (2), into Eq. (6.23), we determine the global end forces for member 2 to be

F2 5 TT2 Q2 5

F G 308.6 kN 2316.07 kN 2246.92 kN ? m 2308.6 kN 556.04 kN 2571.4 kN ? m



Ans

The global member end forces are shown in Fig. 6.8(d). Equilibrium Check:  See Fig. 6.8(d).

o F 5 0 2308.6 1 308.6 5 0 1c o F 5 0 2316.04 2 48(5) 1 556.04 5 0 4 1 o M➁ 5 0  2246.92 2 48(5) 1 2 2 571.4 2 308.6(3) 2 1S

X

Checks

[

Y

Checks

1 556.04(4) 5 20.04 kN ? m > 0 Therefore, the member is in equilibrium.

Checks Ans

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Section 6.4   Member Stiffness Relations in the Global Coordinate System   279

6.4 Member Stiffness Relations in

the Global Coordinate System The process of establishing the stiffness relationships for plane frame members in the global coordinate system is similar to that for the members of plane trusses (Section 3.6). We first substitute the local stiffness relations Q 5 ku 1 Qf (Eq. (6.4)) into the force transformation relations F 5 TTQ (Eq. (6.23)) to obtain F 5 TT Q 5 TT ku 1 TT Qf

(6.26)

Then, we substitute the displacement transformation relations u 5 Tv (Eq. (6.20)) into Eq. (6.26) to determine the desired relationships between the member end forces F and end displacements v, in the global coordinate system: F 5 TT k Tv 1 TT Qf

(6.27)

Equation (6.27) can be conveniently expressed as F 5 Kv 1 Ff



(6.28)

with K 5 TT kT



Ff 5 TT Qf

(6.29) (6.30)

The matrix K represents the member stiffness matrix in the global coordinate system; Ff is called the member fixed-end force vector in the global coordinate system.

Member Global Stiffness Matrix K The expression of the member global stiffness matrix K given in Eq. (6.29), as a product of the three matrices TT, k, and T, is sometimes referred to as the matrix triple product form of K. The explicit form of K, in terms of L, E, A, I, and u of the member, can be determined by substituting the explicit forms of the member local stiffness matrix k from Eq. (6.6) and the member transformation matrix T from Eq. (6.19) into Eq. (6.29), and by multiplying the matrices TT, k, and T, in that order. The explicit form of the member global stiffness matrix K thus obtained is given in Eq. (6.31). From a computer programming viewpoint, it is usually more convenient to evaluate K using the numerical values of k and T in the matrix triple product given in Eq. (6.29), rather than the explicit form of K given in Eq. (6.31). In Section 6.7, we will develop a computer subroutine to generate K

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280  Chapter 6   Plane Frames

by ­multiplying the numerical forms of TT, k, and T, in sequence. The explicit form of K (Eq. (6.31)), however, provides insight into the physical interpretation of the member global stiffness matrix, and proves convenient for evaluating K by hand calculations.

K5

EI L3

3

AL2 cos2 u 1 12 sin2 u I AL2 2 12 cos u sin u I 26L sin u AL2 cos2 u 1 12 sin2 u 2 I AL2 2 12 cos u sin u 2 I 2 6L sin u

1

1 1

2

2

2

1ALI 2 122 cos u sin u 2

AL sin2 u 1 12 cos2 u I 6L cos u AL2 2 2 12 cos u sin u 3 I 2 AL sin2 u 1 12 cos2 u 2 I 6L cos u 2

1 1

2

1ALI cos u 1 12 sin u2 21ALI 2 122 cos u sin u AL AL 21 2 122 cos u sin u 21 sin u 1 12 cos u2 I I 2

26L sin u

2

2

2

2

2

6L cos u 4L

2

26L cos u 2L2

6L cos u

26L cos u

2L2

AL cos2 u 1 12 sin2 u I AL2 2 12 cos u sin u I 6L sin u

1ALI 2 122 cos u sin u

6L sin u

2

2

2

6L sin u

2

6L sin u

26L sin u

2

1

2

2

AL sin2 u 1 12 cos2 u I 26 L cos u 2

26L cos u 4L2



4

26L

(6.31)

The physical interpretation of the member global stiffness matrix K for plane frame members is similar to that of K for members of plane trusses. A stiffness coefficient Kij represents the force at the location and in the direction Fi required, along with other global end forces, to cause a unit value of displacement yj, while all other global end displacements are 0, and the member is not subjected to any external loads between its ends. In other words, as depicted in Figs. 6.9(a) through (f), the jth column of K ( j 5 1 through 6) represents the member end forces, in the global coordinate system, required to cause a unit value of the global end displacement yj , while all other end displacements are 0, and the member is not subjected to any external loads. K51 K61

K41 e

Initial position

Y

θ

b

K11 K31

y1 5 1

Displaced position

b9

X

K21

(a) First Column of K (y1 5 1, y2 5 y3 5 y4 5 y5 5 y6 5 0)

Fig. 6.9

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Section 6.4   Member Stiffness Relations in the Global Coordinate System   281 K52 K62

K42 e

Y

b9

y2 5 1 K12

θ X

b K32 K22

(b) Second Column of K (y2 5 1, y1 5 y3 5 y4 5 y5 5 y6 5 0) K53 K63

K43 e

y3 5 1 Y

θ K13

X

b K33 K23

(c) Third Column of K (y3 5 1, y1 5 y2 5 y4 5 y5 5 y6 5 0) K54

y4 5 1

K64 K44

e9 e

Y

θ

K14

b

K34

X

K24

(d) Fourth Column of K (y4 5 1, y1 5 y2 5 y3 5 y5 5 y6 5 0)

Fig. 6.9  (continued)

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282  Chapter 6   Plane Frames e9

y5 5 1

K65 K45

e K55

Y

θ K15

X

b K35 K25

(e) Fifth Column of K (y5 5 1, y1 5 y2 5 y3 5 y4 5 y6 5 0) K56 K66

K46 e

Y y6 5 1

θ K16

b

X

K36 K26

(f) Sixth Column of K (y6 5 1, y1 5 y2 5 y3 5 y4 5 y5 5 0)

Fig. 6.9  (continued)

We can use the foregoing interpretation of the member global stiffness matrix to check the explicit form of K given in Eq. (6.31). For example, to determine the first column of K, we subject the member to a unit end displacement y1 5 1, while all other end displacements are held at 0. As shown in Fig. 6.10(a), the components of this global end displacement in the directions along, and perpendicular to, the member’s longitudinal axis, respectively, are

ua 5 y1 cos u 5 1 cos u 5 cos u up 5 y1 sin u 5 1  sin u 5 sin u The axial compressive force in the member caused by the axial deformation ua is shown in Fig. 6.10(b), and the member end shears and moments due to the perpendicular displacement up are given in Fig. 6.10(c). Note that these member end shears and moments are obtained by multiplying the member end Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 6.4   Member Stiffness Relations in the Global Coordinate System   283 K51 K61

Initial position

K41 e

Y up 5 sin θ Displaced position

ua 5 cos θ

b K11

θ K31

X

b9

y1 5 1 K21

(a)

5 sθ

o EA c L

e

os EA c L

sθ

ua

θ

θ

b9

b 5

co

1

I in 6E 2 s L

(b)

e

n EI si 3 L

12

12

nθ EI si 3 L

I in 6E 2 s L

b up

5

sin

θ

θ

θ

θ

θ b9 (c)

Fig. 6.10 Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

284  Chapter 6   Plane Frames

forces developed previously (Fig. 6.3(d)) by the negative of up (or by setting u2 5 2up 5 2sin u in Fig. 6.3(d)). By comparing Figs. 6.10(a), (b), and (c), we realize that the global stiffness coefficients K11 and K21, at end b of the member, must be equal to the algebraic sums in the global X and Y directions, respectively, of the member end axial force and shear at end b; that is, K11 5 5

sin u2 sin u 1EAL cos u2 cos u 1 112EI L 3

EA 12EI cos2 u 1 3 sin2 u L L

(6.32a)

and sin u2 cos u 1EAL cos u2 sin u 2 112EI L EA 12EI 51 2 cos u sin u L L 2

K21 5

3

(6.32b)

3

Also, the global stiffness coefficient K31 in Fig. 6.10(a) must be equal to the member end moment in Fig. 6.10(c); that is, K31 5 2

6EI sin u L2

(6.32c)

Similarly, the global stiffness coefficients at end e of the member can be expressed as (see Figs. 6.10(a) through (c)) sin u2 sin u 1EAL cos u2 cos u 2 112EI L

K41 5 2 52

3

EA 12EI cos2 u 2 3 sin2 u L L

(6.32d)

sin u2 cos u 1EAL cos u2 sin u 1 112EI L EA 12EI 5 21 2 cos u sin u L L 2

K51 5 2

3

3

(6.32e)

and

K61 5 2

6EI sin u L2

(6.32f)

Note that the expressions for the member global stiffness coefficients, in Eqs. 6.32(a) through (f), are identical to those in the first column of the explicit form of K given in Eq. (6.31). The remaining columns of K can be verified in a similar manner.

Member Global Fixed-End Force Vector Ff

The explicit form of the member global fixed-end force vector Ff can be obtained by substituting Eqs. (6.19) and (6.15) into the relationship Ff 5 TTQf (Eq. (6.30)). This yields

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F G

Section 6.4   Member Stiffness Relations in the Global Coordinate System   285

Ff 5



FAb cos u 2 FSb sin u FAb sin u 1 FSb cos u FMb  FAe cos u 2 FSe sin u FAe sin u 1 FSe cos u FMe

(6.33)

The member global fixed-end forces Ff , like the local fixed-end forces Qf , represent the forces that would develop at the member ends due to external loads, if both member ends were restrained against translations and ­rotations. However, the global fixed-end forces Ff are oriented in the global X and Y directions of the structure (Fig. 6.11(a)), whereas the local fixed-end forces Qf are oriented in the local x and y directions of the member (Fig. 6.11(b)). Ff 6 e

Ff 4

W Y

w Ff 5

θ Ff1

X

b Ff 3

Ff2 (a) Member Global Fixed-End Force Vector Ff

5 x Q f6 W

Q f4

e

y w

Q f5

θ b

Q f1

Q f3 Q f2

Fig. 6.11

(b) Member Local Fixed-End Force Vector Qf

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286  Chapter 6   Plane Frames

E x ample 6.4 In Example 6.3, the global end displacement vector for member 2 of the frame of

F G

Fig. 6.8 was found to be

v2 5

0.038818 m 0.0007582 m 20.0029702 rad 0.036477 m 20.0006676 m 20.010447 rad

Calculate the end forces for this member in the global coordinate system using the member global stiffness relationship F 5 Kv 1 Ff . S ol u t i o n Member Global Stiffness Matrix:  It was shown in Example 6.3 that, for the member

under consideration, cos u 5 0.8  and  sin u 5 20.6

By substituting these direction cosines, and the numerical values of E 5 200 GPa, A 5 12,500(1026) m2, I 5 275(1026) m4, and L 5 5 m, into Eq. (6.31), we evaluate the global stiffness matrix for member 2 as

K2 5

F

321,900.8 2237,465.6 7,920 2321,900.8 237,465.6 7,920

2237,465.6 183,379.2 10,560 237,465.6 2183,379.2 10,560

7,920 10,560 44,000 27,920 210,560 22,000

2321,900.8 237,465.6 27,920 321,900.8 2237,465.6 27,920

237,465.6 2183,379.2 210,560 2237,465.6 183,379.2 210,560

G

7,920 10,560 22,000 27,920 210,560 44,000

The matrix K2 can be obtained alternatively by substituting the numerical forms of k2 (Eq. (1) of Example 6.2) and T2 (Eq. (3) of Example 6.3) into the relationship K 5 TT kT (Eq. (6.29)), and by evaluating the matrix triple product. The reader is encouraged to use this alternative approach to verify the foregoing K2 matrix. Member Global Fixed-End Force Vector:  From Example 6.2: FAb 5 FAe 5 272 kN; FSb 5 FSe 5 96 kN; and FMb 5 2FMe 5 80 kN ? m. By substituting these numerical values, and cos u 5 0.8 and sin u 5 20.6, into Eq. (6.33), we obtain

Ff 2 5

FG 0 120 80 0 120 280

Again, the reader is urged to verify this Ff 2 vector by substituting the numerical values of Qf 2 (Eq. (3) of Example 6.2) and T2 (Eq. (3) of Example 6.3) into the relationship Ff 5 TT Qf (Eq. (6.30)), and by performing the matrix multiplication.

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Section 6.5   Structure Stiffness Relations  287

Member Global End Forces: The global end forces for member 2 can now be determined by applying Eq. (6.28):

F2 5 K2v2 1 Ff 2 5

F G

308.7 kN 2316.13 kN 2246.92 kN ? m  2308.7 kN 556.13 kN 2571.4 kN ? m

Ans

Note that this F2 vector is the same as the one obtained in Example 6.3 by transforming the member end forces from the local to the global coordinate system. Equilibrium check:  See Example 6.3.

6.5 

Structure Stiffness Relations The process of establishing the structure stiffness relations for plane frames is essentially the same as that for beams (Section 5.5), except that the member global (instead of local) stiffness relations must now be used to assemble the structure stiffness matrices and the fixed-joint force vectors. Consider, for example, an arbitrary plane frame as shown in Fig. 6.12(a). As the analytical model of the frame in Fig. 6.12(b) indicates, the frame has three degrees of freedom, d1, d2, and d3, with the corresponding joint loads designated P1, P2, Y

P2, d2

W1

P3, d3

w

P1, d1

2 W2

1

R4

1 R6

2 R5

3

R7

X

R9

(a) Plane Frame R8 (b) Analytical Model

Fig. 6.12

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288  Chapter 6   Plane Frames P2 F6(1) (1),

F5

(1)

P3 P1

(1)

F4

y5

2

F6(1), y6(1) 2 1 (1),

F1

F4(1), y4(1)

F5(1)

F3 (2)

F2

F2(2), y2(2)

1

(1)

y1

F1(2) (2)

F1(2), y1(2)

F3(1), y3(1)

2

F3(2), y3(2)

F2(1), y2(1) Y

2

F6(2), y6(2) X

3

F4(2), y4(2)

F5(2), y5(2) (c)

Fig. 6.12  (continued)

and P3, respectively. Remember that our objective is to relate the known external joint (and member) loads to the as yet unknown joint displacements d. To achieve our objective, we first relate the joint loads P to the member global end forces F by writing the joint equilibrium equations. By applying the three equations of equilibrium, FX 5 0, FY 5 0, and M 5 0, to the free body of joint 2 drawn in Fig. 6.12(c), we obtain

o

o

o

P1 5 F 4(1) 1 F 1(2)

(6.34a)

P2 5 F 5(1) 1 F (2) 2

(6.34b)

P3 5 F6(1) 1 F3(2)

(6.34c)

in which the superscript (i) denotes the member number. Next, we relate the joint displacements d to the member global end displacements v by applying the compatibility conditions that the member end displacements must be the same as the corresponding joint displacements. Thus, by comparing Figs. 6.12(b) and (c), we write the compatibility equations for members 1 and 2, respectively, as y (1) 5 y (1) 5 y (1) 5 0   y (1) 5 d1  y (1) 5 d2  y (1) 5 d3 1 2 3 4 5 6

(6.35)

y (2) 5 d1  1

(6.36)

  y (2) 5 d2  y (2) 5 d3  y (2) 5 y (2) 5 y (2) 5 0 2 3 4 5 6

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Section 6.5   Structure Stiffness Relations  289

With the relationships between P and F, and v and d, now established, we express the member end forces F that appear in the equilibrium equations (Eqs. (6.34)) in terms of the member end displacements v, using the member global stiffness relations F 5 Kv 1 Ff (Eq. (6.28)). By writing this equation in expanded form for an arbitrary member i (i 5 1 or 2), we obtain

FGF F1(i) F2(i) F3(i) F4(i) F5(i) F6(i)

5

K11(i) K21(i) K31(i) K41(i) K51(i) K61(i)

K12(i) K22(i) K32(i) K42(i) K52(i) K62(i)

K13(i) K23(i) K33(i) K43(i) K53(i) K63(i)

K14(i) K24(i) K34(i) K44(i) K54(i) K64(i)

K15(i) K25(i) K35(i) K45(i) K55(i) K65(i)

GF G F G

K16(i) K26(i) K36(i) K46(i) K56(i) K66(i)

y1(i) y2(i) y3(i) y4(i) y5(i) y6(i)

1

Ff1(i) Ff (i)2 Ff3(i) Ff (i)4 Ff (i)5 Ff (i)6

(6.37)

From this, we determine the expressions for forces at end 2 of member 1 (i.e., i 5 1) to be F4(1) 5 K41(1) y1(1) 1 K42(1) y2(1) 1 K43(1) y3 (1) 1 K44(1) y4(1) 1 K45(1) y5(1) 1 K46(1) y6(1) 1 Ff 4(1)

(6.38a)

F5(1) 5 K51(1) y1(1) 1 K52(1) y2(1) 1 K53(1) y3(1) 1 K54(1) y4(1) 1 K55(1) y5(1) 1 K56(1) y6(1) 1 Ff (1) 5

(6.38b)

F6(1) 5 K61(1) y1(1) 1 K62(1) y2(1) 1 K63(1) y3(1) 1 K64(1) y4(1) 1 K65(1) y5(1) 1 K66(1) y6(1) 1 F f6(1)

(6.38c)

Similarly, from Eq. (6.37), we determine the expressions for forces at end 2 of member 2 (i.e., i 5 2) to be F1(2) 5 K11(2) y1(2) 1 K12(2) y2(2) 1 K13(2) y3(2) 1 K14(2) y4(2) 1 K15(2) y5(2) 1 K16(2) y6(2) 1 F (2) f1

(6.39a)

F2(2) 5 K21(2) y1(2) 1 K22(2) y2(2) 1 K23(2) y3(2) 1 K24(2) y4(2) 1 K25(2) y5(2) 1 K26(2) y6(2) 1 F (2)f 2

(6.39b)

F3(2) 5 K31(2) y1(2) 1 K32(2) y2(2) 1 K33(2) y3(2) 1 K34(2) y4(2) 1 K35(2) y5(2) 1 K36(2) y6(2) 1 Ff3(2)

(6.39c)

By substituting the compatibility equations for members 1 and 2 (Eqs. (6.35) and (6.36)) into Eqs. (6.38) and (6.39), respectively, we obtain F4(1) 5 K44(1) d1 1 K45(1) d2 1 K46(1) d3 1 Ff (1) 4

(6.40a)

F5(1) 5 K54(1) d1 1 K55(1) d2 1 K56(1) d3 1 Ff (1) 5

(6.40b)

F6(1) 5 K64(1) d1 1 K65(1) d2 1 K66(1) d3 1 Ff (1) 6

(6.40c)

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290  Chapter 6   Plane Frames

F1(2) 5 K11(2) d1 1 K12(2) d2 1 K13(2) d3 1 Ff1(2)

(6.40d)

F2(2) 5 K21(2) d1 1 K22(2) d2 1 K23(2) d3 1 Ff (2) 2

(6.40e)

F3(2) 5 K31(2) d1 1 K32(2) d2 1 K33(2) d3 1 Ff3(2)

(6.40f)

Finally, by substituting Eqs. (6.40) into the joint equilibrium equations (Eqs. (6.34)), we obtain the desired structure stiffness relations for the plane frame: P1 5 _K44(1) 1 K11(2) + d1 1 _K45(1) 1 K12(2)+ d2 



(6.41a)



(6.41b)



(6.41c)

1 _K46(1) 1 K13(2)+ d3 1 _Ff (1) 1 Ff1(2)+ 4 P2 5 _K54(1) 1 K21(2)+ d1 1 _K55(1) 1 K22(2)+ d2 1 _K56(1) 1 K23(2)+ d3 1 _Ff (1) 1 Ff (2) 5 2+ P3 5 _K64(1) 1 K31(2)+ d1 1 _K65(1) 1 K32(2)+ d2 1 _K66(1) 1 K33(2)+ d3 1 _Ff (1) 1 Ff (2) 6 3+

The foregoing equations can be symbolically expressed as P 5 Sd 1 Pf or P 2 Pf 5 Sd

(6.42)

in which S represents the NDOF 3 NDOF structure stiffness matrix, and Pf is the NDOF 3 1 structure fixed-joint force vector, for the plane frame with

3

K44(1) 1 K11(2) S 5 K54(1) 1 K21(2) K64(1) 1 K31(2)

K45(1) 1 K12(2) K55(1) 1 K22(2) K65(1) 1 K32(2)

4

K46(1) 1 K13(2) K56(1) 1 K23(2) (6.43) K66(1) 1 K33(2)

and

3

4

Ff (1) 1 Ff (2) 4 1 (1) Pf 5 Ff 5 1 Ff (2) (6.44) 2 (2) Ff (1) 1 F 6 f3

Structure Stiffness Matrix S As discussed in Chapters 3 and 5, an element Sij of the structure stiffness matrix S represents the force at the location and in the direction of Pi required, along with other joint forces, to cause a unit value of the displacement dj , while all other joint displacements are 0, and the frame is subjected to no external loads. In other words, the jth column of S consists of joint forces required, at

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Section 6.5   Structure Stiffness Relations  291

the ­locations and in the directions of all the degrees of freedom of the frame, to cause a unit value of the displacement dj while all other joint displacements are 0. We can use the foregoing interpretation to verify the S matrix given in Eq. (6.43) for the frame of Fig. 6.12. To obtain the first column of S, we subject the frame to a unit value of the joint displacement d1 5 1 (d2 5 d3 5 0), as shown in Fig. 6.13(a). As depicted there, this unit joint displacement induces unit global end displacements y (1)4 at the end of member 1, and y (2)1

Y

S21 S31 S11

d1 5 1 2

29

1

3 X

5 y4(1) 5 1

y1(2) 5 1 K21(2)

K54(1) (1)

K64 (1)

K44 1

K31(2)

2

29

(2)

K11

29 2

K14(1) 1 K34(1)

2

K24(1)

3 K41(2) (2)

K61 (2)

K51 (a) First Column of S (d1 5 1, d2 5 d3 5 0)

Fig. 6.13

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292  Chapter 6   Plane Frames Y 29 d2 5 1 S12

S32 2 S22

1

3 X

5

29 K45(1) 2

y5(1) 5 1 K65(1)

1 K15(1)

29

K35(1)

1

(2)

K12

2

y2(2) 5 1

K32(2)

K55(1)

K22(2) 2

K25(1) 3

K42(2) K62(2)

(2)

K52 (b) Second Column of S (d2 5 1, d1 5 d3 5 0)

Fig. 6.13  (continued)

at the beginning of member 2. The member global stiffness coefficients, necessary to cause the foregoing end displacements, are also given in Fig. 6.13(a). From this figure, we can see that the structure stiffness coefficients (or joint forces) S11 and S21 at joint 2 must be equal to the algebraic sums of the forces in the X and Y directions, respectively, at the two member ends connected to the joint; that is, S11 5 K44(1) 1 K11(2)

(6.45a)

S21 5 K54(1) 1 K21(2)

(6.45b)

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Section 6.5   Structure Stiffness Relations  293 Y S33

S23

2

S13

1 d3 5 1

3 X

5

K56(1)

K23(2) (2)

K33

K66(1) 1

2

K46(1)

K13(2) 2

1 K16(1)

K36(1) K26(1)

y6(1) 5 1

y3(2) 5 1 2

3 K43(2) K63(2) K53(2) (c) Third Column of S (d3 5 1, d1 5 d2 5 0)

Fig. 6.13  (continued)

Similarly, the structure stiffness coefficient (or joint moment) S31 at joint 2 must be equal to the algebraic sum of the moments at the two member ends connected to the joint; thus, S31 5 K(1) 1 K(2) 64 31

(6.45c)

Note that the expressions for Si1 (i 5 1 to 3) given in Eqs. (6.45) are identical to those listed in the first column of S in Eq. (6.43). The second and third columns of S can be verified in a similar manner using Figs. 6.13(b) and (c), respectively. It should be noted that the structure stiffness matrix S in Eq. (6.43) is symmetric, because of the symmetry of the

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294  Chapter 6   Plane Frames

member global stiffness matrices (i.e., Ki j 5 Kj i). The structure stiffness matrices of linear elastic structures must always be symmetric.

Structure Fixed-Joint Force Vector Pf and Equivalent Joint Loads As discussed in Chapter 5, the structure fixed-joint forces represent the reactions that would develop at the locations and in the directions of the frame’s degrees of freedom, due to member loads, if all the joints of the frame were fixed against translations and rotations. This definition enables us to directly express the structure fixed-joint forces in terms of the member global fixed-end forces (instead of deriving such expressions by combining the frame’s equilibrium, compatibility, and member force-­ displacement relations, as was done in the earlier part of this section—see Eqs. (6.34) through (6.44)). Let us verify the Pf vector, given in Eq. (6.44) for the frame of Fig. 6.12, using this direct approach. The frame is redrawn in Fig. 6.14(a) with its joint 2, which is actually free to translate in the X and Y directions and rotate, now restrained against these displacements by an imaginary restraint. When this hypothetical completely fixed frame is subjected to member loads only (note that the joint load W1 shown in Fig. 6.12(a) is not drawn in Fig. 6.14(a)), the structure fixed-joint forces (or reactions) Pf1, Pf 2, and Pf 3 develop at the imaginary restraint at joint 2. As shown in Fig. 6.14(a), the structure fixed-joint force at the location and in the direction of an ith degree of freedom is denoted by Pf i . To relate the structure fixed-joint forces Pf to the member global fixedend forces Ff , we draw the free-body diagrams of the two members of the hypothetical fixed frame, as shown in Fig. 6.14(b). Note that, because all the joints of the frame are restrained, the member ends, which are rigidly connected to the joints, are also fixed against any displacements. Therefore, only the fixed-end forces due to member loads, Ff , can develop at the ends of the members. By comparing Figs. 6.14(a) and (b), we realize that the structure fixedjoint forces Pf 1 and Pf 2 at joint 2 must be equal to the algebraic sums of the fixed-end forces in the X and Y directions, respectively, at the two member ends connected to the joint; that is, Pf1 5 F(1) 1 F(2) (6.46a) f4 f1 Pf 2 5 F(1) 1 F(2) (6.46b) f5 f2 Similarly, the structure fixed-joint moment Pf 3 at joint 2 must be equal to the algebraic sum of the fixed-end moments at the two member ends connected to the joint. Thus, Pf 3 5 F(1) 1 F(2) f6 f3

(6.46c)

Note that the expressions for Pf i (i 5 1 to 3) given in Eqs. (6.46) are the same as those listed in the Pf vector in Eq. (6.44).

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Section 6.5   Structure Stiffness Relations  295 Pf 2

Y Pf3 w

Pf1 2

Rf4

W2

1

Rf6

Rf5

3

Rf 7

X

Rf 9 Rf 8 (a) Fixed Frame Subjected to Member Loads

5 Ff 2(2) Ff6(1) w

Ff3(2) Ff4(1)

Ff1(2)

2

2 Ff 1(1)

1 Ff3(1)

1

W2 2

(1)

Ff5

3

Ff 2(1)

Ff4(2) Ff6(2) Ff5(2) (b) Member Global Fixed-End Forces

Fig. 6.14

It may be recalled from Section 5.6 that another interpretation of the structure fixed-joint forces due to member loads is that when they are applied to the structure with their directions reversed, the fixed-joint forces cause the same joint displacements as the actual member loads. The negatives of the structure fixed-joint forces are, therefore, referred to as the equivalent joint loads. We can show the validity of this interpretation by setting the joint

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296  Chapter 6   Plane Frames

loads equal to 0 (i.e., P 5 0) in Eq. (6.42), thereby reducing the structure stiffness relationship to 2Pf 5 Sd

(6.47)

in which d now represents the joint displacements due to the negatives of the structure fixed-joint forces applied to the joints of the structure. However, since member loads are now the only external effects acting on the structure, the d vector in Eq. (6.47) must also represent the joint displacements due to member loads. Thus, we can conclude that the negatives of the structure fixed-joint forces must cause the same joint displacements as the actual member loads. The validity of this interpretation can also be demonstrated using the principle of superposition. Figure 6.15(a) shows the two-member frame considered previously (Fig. 6.12), subjected to arbitrary member loads w and W2. In Fig. 6.15(b), joint 2 of the frame is fixed by an imaginary restraint so that, when the fixed frame is subjected to member loads, the structure fixed-joint forces Pf 1, Pf 2, and Pf 3 develop at the imaginary restraint. Lastly, in Fig. 6.15(c), the actual frame is subjected to joint loads, which are equal in magnitude to the structure fixed-joint forces Pf 1, Pf 2, and Pf 3, but reversed in direction. By comparing Figs. 6.15(a) through (c), we realize that the actual loading applied to the actual frame in Fig. 6.15(a) equals the algebraic sum of the loadings in Figs. 6.15(b) and (c), because the reactions Pf 1, Pf 2, and Pf 3 in Fig. 6.15(b) cancel the corresponding applied loads in Fig. 6.15(c). Thus, in accordance with the superposition principle, any joint displacement of the actual frame due to the member loads (Fig. 6.15(a)) must be equal to the algebraic sum of the corresponding joint displacement of the fixed frame due to the member loads (Fig. 6.15(b)) and the corresponding joint displacement of the actual frame subjected to no member loads, but to the structure fixed-joint forces with their directions reversed. However, since the joint displacements of  the fixed frame (Fig. 6.15(b)) are 0, the joint displacements of the frame due to the member loads (Fig. 6.15(a)) must be equal to the corresponding joint displacements of the frame due to the negatives of the fixed-joint forces (Fig. 6.15(c)). Thus, the negatives of the structure fixed-joint forces can be considered to be equivalent to member loads in terms of joint displacements. It should be noted that this equivalency is valid only for joint displacements, and it cannot be generalized to member end forces and support reactions.

Assembly of S and Pf , Using Member Code Numbers In the preceding paragraphs of this section, we have demonstrated that the structure stiffness matrix S for plane frames can be formulated directly by algebraically adding the appropriate elements of the member global stiffness matrices K (see, for example, Eqs. (6.43) and (6.45), and Fig. 6.13). Furthermore, it has been shown that the structure fixed-joint force vector Pf for plane frames can also be established directly by algebraically adding the member global fixed-end forces Ff at the location, and in the direction, of each of the structure’s degrees of freedom (see, for example, Eqs. (6.44) and (6.46), and Fig. 6.14).

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Section 6.5   Structure Stiffness Relations  297 w 2 W2

1

3 (a) Actual Frame Subjected to Member Loads

5 Pf 2 Pf 3 w

Pf1 2 W2

1

3 (b) Fixed Frame Subjected to Member Loads

1 Pf2 Pf3 Pf1 2 1

3 (c) Actual Frame Subjected to Equivalent Joint Loads

Fig. 6.15

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298  Chapter 6   Plane Frames

The foregoing process of directly generating S and Pf can be conveniently implemented by employing the member code number technique described in detail in Chapters 3 and 5. The application of this technique for plane frames remains essentially the same as that for the case of beams, except that the member global (instead of local) stiffness matrices K and the member global fixedend force vectors Ff must now be used to form S and Pf , respectively. It should also be realized that each member of the plane frame has six code numbers, arranged in the sequential order of the member’s global end displacements v. The application of the member code number technique for plane frames is illustrated in the following example.

E x ample 6.5 Determine the structure stiffness matrix, the fixed-joint force vector, and the equivalent joint loads for the frame shown in Fig. 6.16(a).

S ol u t i o n Analytical Model:  See Fig. 6.16(b). The frame has four degrees of freedom and five re-

strained coordinates, as shown. Structure Stiffness Matrix:  The 4 3 4 structure stiffness matrix will be generated by evaluating each member’s global stiffness matrix K, and storing its pertinent elements in S using the member code numbers. 75 kN

10 m

24 kN/m

4m

4m

E, A, I 5 constant E 5 200 GPa A 5 4,740 mm2 I 5 22.2(106) mm4 (a) Frame

Fig. 6.16

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Section 6.5   Structure Stiffness Relations  299

Y

2 3 4 1 2

2

8

3

9 1

1 5

X 7

6 (b) Analytical Model

37.5 kN 75 kN m 120 kN 125 kN m

2 3

1 (c) Equivalent Joint Loads

Fig. 6.16  (continued)

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300  Chapter 6   Plane Frames Member 1  As shown in Fig. 6.16(b), joint 1 is the beginning joint and joint 2 is the end joint for this member. Thus, cos u 5 sin u 5

X2 2 X1 L Y2 2 Y1 L

5

5

020 50 10

10 2 0 51 10

By substituting E 5 200(10)6 kN/m2, A 5 0.00474 m2, I 5 0.0000222 m4, L 5 10 m, and the foregoing values of the direction cosines into the expression for K given in Eq. (6.31), we obtain

F



K1 5

5 6 7 1 2 3 53.28 0 0 94,800 2266.4 0 253.28 0 0 294,800 2266.4 0

2266.4 0 1,776 266.4 0 888

253.28 0 0 294,800 266.4 0 53.28 0 0 94,800 266.4 0

G

2266.4 0 888 266.4 0 1,776

5 6 7  1 2 3

(1)

From Fig. 6.16(b), we observe that the code numbers for member 1 are 5, 6, 7, 1, 2, 3. These numbers are written on the right side and at the top of K1 (Eq. (1)) to indicate the rows and columns of S in which the elements of K1 are to be stored. Thus,

S 5

1 2

3

53.28 0 0 94,800 266.4 0 0 0

3 4

266.4 0 1,776 0

0 0 0 0

4

1 2  3 4

(2)

Member 2  From Fig. 6.16(b), we can see that this member is horizontal, with its leftend joint 2 selected as the beginning joint, thereby orienting the positive directions of the member’s local x and y axes in the positive directions of the global X and Y axes, respectively. Thus, no coordinate transformations are needed for this member (i.e., cos u 5 1, sin u 5 0, and T 5 I); and its stiffness relations, and fixed-end forces, are the same in the local and global coordinate systems. By substituting the numerical values of E, A, and I, and L 5 8 m into Eq. (6.6), we obtain

F

1 2 3 8 9 4

K2 5 k2 5

118,500 0 0 2118,500 0 0

0 104.06 416.25 0 2104.06 416.25

0 416.25 2,220 0 2416.25 1,110

2118,500 0 0 118,500 0 0

0 2104.06 2416.25 0 104.06 2416.25

G

0 416.25 1,110 0 2 416.25 2,220

1 2 3 8 9 4

The code numbers for this member—1, 2, 3, 8, 9, 4 (see Fig. 6.16(b))—are now used to add the pertinent elements of K2 in their proper positions in the structure stiffness matrix S given in Eq. (2), which now becomes

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Section 6.5   Structure Stiffness Relations  301



S5

1

3

53.28 1 118,500 0 266.4 0

2

0 94,800 1 104.06 416.25 416.25

3 4

266.4 416.25 1,776 1 2,220 1,110

0 416.25 1,110 2,220

4

1 2 3 4

Because the stiffnesses of both members of the frame have now been stored in S, the structure stiffness matrix for the given frame is

S5

1 2 3 4

3

118,553 0 266.4 0

0 94,904 416.25 416.25

266.4 416.25 3,996 1,110

0 416.25 1,110 2,220

4

1 2  3 4

Ans

Note that the structure stiffness matrix is symmetric. Structure Fixed-Joint Force Vector:  We will generate the 4 3 1 structure fixed-joint force vector by evaluating, for each member, the global fixed-end force vector Ff , and storing its pertinent elements in Pf using the member code numbers. Member 1  The 24 kN/m uniformly distributed load acting on this member is positive, because it acts in the negative direction of the member’s local y axis. By substituting w 5 24 kN/m, L 5 10 m, and l1 5 l2 5 0 into the fixed-end force expressions for loading type 3 listed inside the front cover, we evaluate 24(10) 5 120 kN 2 24(10)2 5 200 kN ? m FMb 5 2FMe 5 12 FSb 5 FSe 5

As the member is not subjected to any axial loads, F Ab 5 F Ae 5 0 By substituting the foregoing values of the member fixed-end forces, along with cos u 5 0 and sin u 5 1, into the explicit form of Ff given in Eq. (6.33), we obtain

Ff1 5

FG

2120 5 0 6 200 7  2120 1 0 2 2200 3

(3)

The code numbers of the member, 5, 6, 7, 1, 2, 3, are written on the right side of Ff 1 in Eq. (3) to indicate the rows of Pf in which the elements of Ff 1 are to be stored. Thus,

Pf 5

3 4 2120 0 2200 0

1 2  3 4

(4)

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302  Chapter 6   Plane Frames Member 2  By substituting P 5 75 kN, L 5 8 m, and l1 5 l2 5 4 m into the fixed-end force expressions for loading type 1, we determine the member fixed-end shears and moments to be 75 5 37.5 kN 2 75(8) 5 75 kN ? m FMb 5 2FMe 5 8

FSb 5 FSe 5

As no axial loads are applied to this member, F Ab 5 F Ae 5 0

FG

Thus,

0 37.5 75 Ff 2 5 Qf 2 5 0 37.5 275

1 2 3 8 9 4

Using the member code numbers 1, 2, 3, 8, 9, 4, we add the pertinent elements of Ff 2 in their proper positions in Pf (as given in Eq. (4)), which now becomes Pf 5

3

4

2120 37.5 2200 1 75 275

1 2 3 4

Because the fixed-end forces for both members of the frame have now been stored in Pf , the structure fixed-joint force vector for the given frame is

3 4

2120 37.5 Pf 5 2125 275

1 2  3 4

Ans

Equivalent Joint Loads:

Pe 5 2Pf 5

3 4 120 237.5 125 75

1 2  3 4

Ans

The equivalent joint loads are depicted in Fig. 6.16(c). These equivalent joint loads cause the same joint displacements of the frame as the actual member loads of Fig. 6.16(a).

6.6 

Procedure for Analysis Using the concepts discussed in the previous sections, we can now develop the following step-by-step procedure for the analysis of plane frames by the matrix stiffness method.

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Section 6.6   Procedure for Analysis  303

1. Prepare an analytical model of the structure, identifying its degrees of freedom and restrained coordinates (as discussed in Section 6.1). Recall that for horizontal members, the coordinate transformations can be avoided by selecting the left-end joint of the member as the beginning joint. 2. Evaluate the structure stiffness matrix S(NDOF 3 NDOF) and fixedjoint force vector Pf (NDOF 3 1). For each member of the structure, perform the following operations: a. Calculate the length and direction cosines (i.e., cos u and sin u) of the member (Eqs. (3.62)). b. Compute the member stiffness matrix in the global coordinate system, K, using its explicit form given in Eq. (6.31). The member global stiffness matrix alternatively can be obtained by first forming the member local stiffness matrix k (Eq. (6.6)) and the transformation matrix T (Eq. (6.19)), and then evaluating the matrix triple product, K = TTkT (Eq. (6.29)). The matrix K must be symmetric. c. If the member is subjected to external loads, then evaluate the member fixed-end force vector in the global coordinate system, Ff , using the expressions for fixed-end forces given inside the front cover, and the explicit form of Ff given in Eq. (6.33). The member global fixedend force vector can also be obtained by first forming the member local fixed-end force vector Qf (Eq. (6.15)), and then using the relationship Ff = TTQf (Eq. (6.30)). d. Identify the member code numbers and store the pertinent elements of K and Ff in their proper positions in the structure stiffness matrix S and the fixed-joint force vector Pf , respectively. The complete structure stiffness matrix S, obtained by assembling the stiffness coefficients of all the members of the structure, must be symmetric. 3. If the structure is subjected to joint loads, then form the joint load vector P(NDOF 3 1). 4. Determine the joint displacements d. Substitute P, Pf , and S into the structure stiffness relationship, P 2 Pf 5 Sd (Eq. (6.42)), and solve the resulting system of simultaneous equations for the unknown joint displacements d. To check the solution for correctness, substitute the numerical values of the joint displacements d back into the stiffness relationship P 2 Pf 5 Sd. If the solution is correct, then the stiffness relationship should be satisfied. Note that joint translations are considered positive when in the positive directions of the global X and Y axes, and joint rotations are considered positive when counterclockwise. 5. Compute member end displacements and end forces, and support reactions. For each member of the structure, carryout the following steps: a. Obtain member end displacements in the global coordinate system, v, from the joint displacements, d, using the member code numbers. b. Form the member transformation matrix T (Eq. (6.19)), and determine the member end displacements in the local coordinate system, u, using the transformation relationship u 5 Tv (Eq. (6.20)). Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

304  Chapter 6   Plane Frames

c. Form the member local stiffness matrix k (Eq. (6.6)) and local fixed-end force vector Qf (Eq. (6.15)); then calculate the member end forces in the local coordinate system, Q, using the stiffness relationship Q 5 ku 1 Qf (Eq. (6.4)). d. Determine the member end forces in the global coordinate system, F, using the transformation relationship F = TTQ (Eq. (6.23)). e. If the member is attached to a support joint, then use the member code numbers to store the pertinent elements of F in their proper positions in the support reaction vector R. 6.  Check the calculation of member end forces and support reactions by apM 5 0+ plying the equilibrium equations _ FX 5 0, FY 5 0, and to the free body of the entire structure. If the calculations have been carried out correctly, then the equilibrium equations should be satisfied.

o

o

o

Instead of following steps 5(c) and (d) of this procedure, the member end forces alternatively can be obtained by first calculating the global forces F using the global stiffness relationship F 5 Kv 1 Ff (Eq. (6.28)), and then evaluating the local forces Q from the transformation relationship Q 5 TF (Eq. (6.18)). It should also be noted that it is usually not necessary to determine the global end forces for all the members of the structure because such forces are not used for design purposes. However, F vectors for the members that are attached to supports are always evaluated, so that they can be used to form the support reaction vector R.

Shear and Bending Moment Diagrams Once the member end forces in the local coordinate systems have been determined, the shear, bending moment, and axial force diagrams for the members can be drawn using the beam sign convention shown in Fig. 5.17. E x ample 6.6 Determine the joint displacements, member end forces, and support reactions for the two-member frame shown in Fig. 6.17(a) on the next page, using the matrix stiffness method. Draw the shear, bending moment, and axial force diagrams for the frame.

S ol u t i o n Analytical Model:  See Fig. 6.17(b). The frame has three degrees of freedom—the trans-

lations in the X and Y directions, and the rotation, of joint 2—which are numbered 1, 2, and 3, respectively. The six restrained coordinates of the frame are identified by numbers 4 through 9, as shown in Fig. 6.17(b). Structure Stiffness Matrix and Fixed-Joint Force Vector: Member 1  As shown in Fig. 6.17(b), we have selected joint 1 as the beginning joint, and joint 2 as the end joint for this member. By applying Eqs. (3.62), we determine L 5 Ï(X2 2 X1)2 1 (Y2 2 Y1)2 5 Ï(3 2 0)2 1 (6 2 0)2  5 6.7082 m cos u 5

sin u 5

X2 2 X1 L Y2 2 Y1 L

5

5

320 5 0.44721  6.7082

620 5 0.89443 6.7082

(1a) (1b)

(1c)

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Section 6.6   Procedure for Analysis  305

Using the units of kN and meters, we evaluate the member global stiffness matrix as (Eq. (6.31))

F



K1 5

4 5 6 1 2 3 46,137 90,225 23,077 246,137 290,225 23,077

90,225 181,477 1,538 290,225 2181,477 1,538

23,077 1,538 15,384 3,077 21,538 7,692

246,137 290,225 3,077 46,137 90,225 3,077

290,225 2181,477 21,538 90,225 181,477 21,538

23,077 1,538 7,692 3,077 21,538 15,384

G

4 5 6 1 2 3

Note that K1 is symmetric. 125 kN • m

x

16.667 kN/m

3m

2

3m

300

300 kN

sin

θ

300 kN

m

θ

2

3.3

y

541

m

E, A, I 5 constant E 5 200 GPa A 5 7,600 mm2 I 5 129(106) mm4 (a) Frame Y

082

300

6m

6.7

3m

cos

θ

θ

1 9

3 1 2

2

(c) Loading on Member 1

7

3

8

1 S5

1

2

1 5

1 4

X 6

Pf 5

5 (b) Analytical Model

3

46,137 1 253,333 90,225 3,077 90,225 181,477 1 1,433.33 21,538 1 4,300 3,077 21,538 1 4,300 15,384 1 17,200 299,470 90,225 3,077

2

0 150 150 2112.5 1 50

3 3,077 2,762 32,584

1 2 3

1 0 2 5 200 3 262.5

1 2 3

90,225 182,910 2,762

1 2 3

(d) Structure Stiffness Matrix and Fixed-Joint Force Vector

Fig. 6.17

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81. 24 kN 137 .77 72. kN 49 •m kN 2

y 16.667 kN/m 101.18 kN 2 12.77 kN • m 40.24 kN

3

2

101.18 kN x

71.33 kN • m 59.76 kN

134

.16

kN

268 .33 kN

x

306  Chapter 6   Plane Frames

N• .5 k

101

349

.56

kN

m 61.

67

1

kN

y

1

(e) Member End Forces in Local Coordinate Systems

R5

3 3 101.17 kN 340.24 kN 101.50 kN • m 2101.18 kN

59.76 kN 271.33 kN • m

4 5 6 7 8 9

(f) Support Reaction Vector 125 kN • m

Y

2

16.667 kN/m 101.18 kN

2

3

300 kN

71.33 kN • m

59.76 kN 1

1 101.17 kN

X 101.5 kN • m 340.24 kN

Fig. 6.17  (continued)

(g) Support Reactions

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Section 6.6   Procedure for Analysis  307

40.24 3.586 m

2

27 2.4 9

259.76

61. 67

Shear diagrams (kN)

1

212.77

2

.35

21

37.

77

35.8

105

271.33 Bending moment diagrams (kN • m) 1

21

01.

5

101.18

81. 2

4

2 Axial force diagrams (kN)

349

.56

1

(h) Shear, Bending Moment, and Axial Force Diagrams

Fig. 6.17  (continued)

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308  Chapter 6   Plane Frames As the 300 kN load applied to this member is inclined with respect to the member’s local coordinate system, we evaluate the rectangular components of the load in the directions of the local x and y axes as (see Fig. 6.17(c)) Wx 5 300 sin u 5 300(0.89443) 5 268.33 kN Wy 5 300 cos u 5 300(0.44721) 5 134.16 kN Note that both Wx and Wy are considered positive because they act in the negative directions of the local x and y axes, respectively. The member’s fixed-end axial forces can now be evaluated by substituting W 5 Wx 5 268.33 kN, L 5 6.7082 m, and l1 5 l2 5 3.3541 m into the expressions for loading type 5 given inside the front cover. This yields FAb 5 FAe 5

268.33 5 134.16 kN(2a) 2

Similarly, by substituting W 5 Wy 5 134.16 kN, and the numerical values of L, l1, and l2 into the equations for loading type 1, we obtain the fixed-end shears and moments as FSb 5 FSe 5

134.16 5 67.08 kN(2b) 2

FMb 5 2FMe 5

134.16(6.7082) 5 112.5 kN ? m 8

FG

(2c)

By substituting the numerical values of the member fixed-end forces and direction cosines into Eq. (6.33), we calculate the member global fixed-end force vector as

F f15

0 150 112.5 0 150 2112.5

4 5 6 1 2 3

From Fig. 6.17(b), we observe that the code numbers for member 1 are 4, 5, 6, 1, 2, 3. Using these code numbers, we store the pertinent elements of K1 and Ff 1 in their proper positions in the 3 3 3 structure stiffness matrix S and the 3 3 1 structure fixed-joint force vector Pf , respectively, as shown in Fig. 6.17(d). Member 2  As this member is horizontal, with its left-end joint 2 selected as the beginning joint, no coordinate transformations are needed; that is, T2 5 I, K2 5 k2, and Ff 2 5 Qf 2. Thus, by substituting L 5 6 m, E 5 200 GPa, A 5 7,600(1026) m2, and I 5 129(1026) m4 into Eq. (6.6), we obtain.

K2 5 k2 5

F

1 253,333 0 0 2253,333 0 0

2 0 1,433.33 4,300 0 21,433.33 4,300

3 7 8 9 0 2253,333 0 0 4,300 0 21,433.33 4,300 17,200 0 24,300 8,600 0 253,333 0 0 24,300 0 1,433.33 24,300 8,600 0 24,300 17,200

G

1 2 3 7 8 9 (3)

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Section 6.6   Procedure for Analysis  309

Using the equations given for loading type 3 inside the front cover, we obtain the fixed-end forces due to the uniformly distributed load of magnitude 16.667 kN/m: F Ab 5 F Ae 5 0 FSb 5 FSe 5

16.667(6) 5 50 kN 2

FMb 5 2FMe 5 Thus (Eq. (6.15)),

F f 2 5 Qf 2 5

16.667(6)2 5 50 kN ? m 12

FG 0 50 50 0 50 250

1 2 3 (4) 7 8 9

The relevant elements of K2 and Ff 2 are stored in S and Pf , respectively, using the member code numbers 1, 2, 3, 7, 8, 9. The completed structure stiffness matrix S and structure fixed-joint force vector Pf  are given in Fig. 6.17(d). Note that S is symmetric. Joint Load Vector:  By comparing Figs. 6.17(a) and (b), we write the joint load vector, in kN and meters, as

3 4

0 1 P 5 0 2 2125 3 Joint Displacements: By substituting the numerical values of P, Pf  , and S into Eq. (6.42), we write the stiffness relations for the entire frame as P 2 Pf 5 Sd

3 4 3 4 3

0 0 299,470 0 2 200 5 90,225 2125 262.5 3,077

90,225 182,910 2,762

43 4

3,077 2,762 32,584

d1 d2 d3

or

3 4 3

0 299,470 2200 5 90,225 262.5 3,077

90,225 182,910 2,762

43 4

3,077 2,762 32,584

d1 d2 d3

Solving these equations, we determine the joint displacements to be

3

0.0003994 m d 5 20.0012625 m  20.0018488 rad

4

Ans

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310  Chapter 6   Plane Frames To check the foregoing solution, we substitute the numerical values of d back into the structure stiffness relationship, as

33 3 4

299,470 P 2 Pf 5 Sd 5 90,225 3,077 5

90,225 182,910 2,762

33

3,077 2,762 32,584

4

0.0003994 20.0012625 20.0018488

0 2200  262.5

Checks

Member End Displacements and End Forces: Member 1  As in the case of plane trusses, the global end displacements v for a plane frame member can be obtained by applying the member’s compatibility equations, using its code numbers. Thus, for member 1 of the frame under consideration, the global end displacement vector can be established as

v1 5

FG FG F G y1 y2 y3 y4 y5 y6

4 5 6 5 1 2 3

0 0 0 d1 d2 d3

0 0 0 5 (5) 0.0003994 m 20.0012625 m 20.0018488 rad

As shown in Eq. (5), the code numbers for the member (4, 5, 6, 1, 2, 3) are first written on the right side of v. The fact that the code numbers corresponding to y1, y2, and y3 are the restrained coordinate numbers 4, 5, and 6, respectively, indicates that y1 5 y2 5 y3 5 0. Similarly, the code numbers 1, 2, and 3 corresponding to y4, y5, and y6, respectively, indicate that y4 5 d1, y5 5 d2, and y6 5 d3. Note that these compatibility equations can be verified easily by a visual inspection of the frame’s line diagram given in Fig. 6.17(b). To determine the member local end displacements, u, we first evaluate the transformation matrix T (Eq. (6.19)), using the direction cosines given in Eqs. (1):

T1 5

F

0.44721 20.89443 0 0 0 0

0.89443 0.44721 0 0 0 0

0 0 1 0 0 0

F G

0 0 0 0.44721 20.89443 0

0 0 0 0.89443 0.44721 0

G

0 0 0 0 0 1

The member local end forces can now be calculated using the relationship u 5 Tv (Eq. (6.20)), as

u1 5 T1v1 5

0 0 0 20.0009506 m 20.0009218 m 20.0018488 rad

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Section 6.6   Procedure for Analysis  311

F FG

G

Before we can calculate the member’s local end forces Q, we need to determine its local stiffness matrix k and fixed-end force vector Qf . Thus, using Eq. (6.6):

k1 5

226,588 0 0 2226,588 0 0

0 1,025.6 3,440 0 21,025.6 3,440

0 3,440 15,384.2 0 23,440 7,692.1

2226,588 0 0 226,588 0 0

0 21,025.6 23,440 0 1,025.6 23,440

0 3,440 7,692.1 0 23,440 15,384.2

and, by substituting Eqs. (2) into Eq. (6.15): 134.16 67.08 112.5 134.16 67.08 2112.5

Qf1 5

F G

Now, using Eq. (6.4), we compute the member local end forces as 349.56 kN 61.67 kN 101.50 kN ? m Q1 5 k1u1 1 Qf1 5 281.24 kN 72.49 kN 2137.77 kN ? m



Ans

The local end forces for member 1 are depicted in Fig. 6.17(e), and we can check our calculations for these forces by considering the equilibrium of the free body of the member, as follows:

oF 5 0 1 a oF 5 0 1Q

x

y

oM

[

1

➁

5 0 

349.56 2 268.33 2 81.24 5 20.01 > 0 

Checks

61.67 2 134.16 1 72.49 5 0

Checks

 101.5 2 61.67(6.7082) 1 134.16(3.3541) 2 137.77 5 20.02 > 0

F G

Checks

The member global end forces F can now be determined by applying Eq. (6.23) as

F1 5 TT1 Q1 5

101.17 340.24 101.50 2101.17 240.25 2137.77

4 5 6 (6) 1 2 3

Next, to generate the support reaction vector R, we write the member code numbers (4, 5, 6, 1, 2, 3) on the right side of F1 as shown in Eq. (6), and store the pertinent elements of F1 in their proper positions in R by matching the code numbers on the side of F1 to the restrained coordinate numbers on the right side of R in Fig. 6.17(f).

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312  Chapter 6   Plane Frames

FG FG F G F G

Member 2  The global and local end displacements for this horizontal member are

u2 5 v2 5

y1 y2 y3 y4 y5 y6

1 2 3 5 7 8 9

d1 d2 d3 0 0 0

0.0003994 m 20.0012625 m 20.0018488 rad 5 0 0 0

By using k2 from Eq. (3) and Qf 2 from Eq. (4), we compute the member local and global end forces to be

F2 5 Q2 5 k2u2 1 Qf 2 5

101.18 kN 40.24 kN 12.77 kN ? m 2101.18 kN 59.76 kN 271.33 kN ? m

1 2 3  7 8 9

Ans

These end forces for member 2 are depicted in Fig. 6.17(e). To check our calculations, we apply equilibrium equations to the free body of member 2, as follows:

o o

Fx 5 0 1 S 1c Fy 5 0

oM

[

1

➁

5 0 

101.18 2 101.18 5 0 

Checks

40.24 2 16.667(6) 1 59.76 5 0

Checks

 12.77 2 16.667(6)(3) 1 59.76(6) 271.33 5 2 0.006 > 0

Checks

Next, we store the pertinent elements of F2 in their proper positions in the reaction vector R, using the member code numbers (1, 2, 3, 7, 8, 9), as shown in Fig. 6.17(f). Support Reactions:  The completed reaction vector R is given in Fig. 6.17(f), and the support reactions are depicted on a line diagram of the structure in Fig. 6.17(g). Ans Equilibrium Check:  We check our calculations by considering the equilibrium of the free body of the entire structure (Fig. 6.17(g)), as follows:

o o

FX 5 0 1 S 1c FY 5 0

oM

[

1

➀

5 0 

101.17 2 101.18 5 20.01 > 0 

Checks

340.24 2 300 2 16.667(6) 1 59.76 5 20.002 > 0

Checks

 101.5 2 300(1.5) 2 125 2 16.667(6)(6) 1 101.18(6) 1 59.76(9) 2 71.33 5 0.078 > 0

Checks

Shear, Bending Moment, and Axial Force Diagrams: Finally, using the member local end forces (Fig. 6.17(e)), we construct the shear, bending moment, and axial force diagrams for the members of the frame. These diagrams are depicted in Fig. 6.17(h).

E x ample 6.7

Determine the joint displacements, member local end forces, and support reactions for the two-story frame, subjected to a wind loading, shown in Fig. 6.18(a).

S ol u t i o n Analytical Model:  See Fig. 6.18(b). The frame has nine degrees of freedom, numbered

1 through 9; and six restrained coordinates, identified by the numbers 10 through 15.

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Section 6.6   Procedure for Analysis  313

40 kN 12 kN/m 6m

80 kN E, A, I 5 constant E 5 30 GPa A 5 75,000 mm2 I 5 4.8(108) mm4

6m

9m (a) Frame

Y

8 9 7

5 3

5

5

2

6

3

1

4 4

3 1

4 2

1

2

10

13 12

X

15

11

14 (b) Analytical Model

Fig. 6.18

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314  Chapter 6   Plane Frames

S5

3

1 2 3 4 5 6 7 8 9 251,600 0 0 2250,000 0 0 2800 0 22,400 1 0 750,237 1,066.7 0 2237.04 1,066.7 0 2375,000 0 2 0 1,066.7 25,600 0 21,066.7 3,200 2,400 0 4,800 3 2250,000 0 0 394,851 295,943 1,990.4 2144,051 95,943 2409.62 4 0 2237.04 21,066.7 295,943 439,335 21,681.1 95,943 264,098 2614.44 5 0 1,066.7 3,200 1,990.4 21,681.1 21,325 409.62 614.44 2,662.6 6 2800 0 2,400 2144,051 95,943 409.62 144,851 295,943 2,809.6 7 0 2375,000 0 95,943 264,098 614.44 295,943 439,098 614.44 8 22,400 0 4,800 2409.62 2614.44 2,662.6 2,809.6 614.44 14,925 9

4



93.398

(c) Structure Stiffness Matrix

90.061

Pf 5

110.6 70.73

90.06

12 kN/m

1.6114

0 0 0 236 254 117 236 254 2117

1 2 3 4 5 6 7 8 9

(d) Structure Fixed-Joint Force Vector

5 3

27.004 1.6114 80.392

93.398

59.07 110.6 157.03

49.027 195.47

275.86 24.5

106.05

4

46.429

1

24.5 195.38 222.38 46.429

85.948

2

360.44

106.05

85.948 320.31 49.027

157.03 (e) Member Local End Forces

Fig. 6.18  (continued)

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Section 6.6   Procedure for Analysis  315

R5

3 3 2106.05 kN 2157.03 kN 360.44 kN.m 285.948 kN 49.027 kN 320.31 kN.m

10 11 12 13 14 15

(f ) Support Reaction Vector

Y

40 kN

5 12 kN/m

6m

80 kN

4

9m

3 6m

2

1

85.948

106.05 360.44 157.03

X

320.31 49.027

(g) Support Reactions

Fig. 6.18  (continued) Structure Stiffness Matrix and Fixed-Joint Force Vector: Members 1, 2, and 3  E 5 30(106) kN/m2, A 5 0.075 m2, I 5 480(1026)m4, L 5 6 m, cos u 5 0, and sin u 5 1. The member global stiffness matrix, in units of kN and meters, is given by the following (see Eq. (6.31)): Member 3 S 1 2 3 7 8 9 Member 2 S 1 3 14 15 4 5 6 Member 1 S 10 11 12 1 2 3

F

800 0 0 375,000 22,400 0 K1 5 K2 5 K3 5 2800 0 0 2375,000 22,400 0

G

22,400 2800 0 22,400 0 0 2375,000 0 9,600 2,400 0 4,800 2,400 800 0 2,400 0 0 375,000 0 4,800 2,400 0 9,600

10 13 1 11 14 2 12 15 3 1 4 7 2 5 8 3 6 9

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316  Chapter 6   Plane Frames As these members are not subjected to any loads, their fixed-end forces are 0. Thus, Ff 1 5 Ff 2 5 Ff  3 5 0 Using the code numbers for member 1 (10, 11, 12, 1, 2, 3), member 2 (13, 14, 15, 4, 5, 6), and member 3 (1, 2, 3, 7, 8, 9), the relevant elements of K1, K2, and K3 are stored in their proper positions in the 9 3 9 structure stiffness matrix S (Fig. 6.18(c)). Member 4 Substituting L 5 9 m and the foregoing values of E, A, and I into Eq. (6.6), we obtain

K4 5 k4 5

F

G

1 2 3 4 5 6 250,000 0 0 2250,000 0 0 1 0 237.04 1,066.7 0 2237.04 1,066.7 2 0 1,066.7 6,400 0 21,066.7 3,200 3 2250,000 0 0 250,000 0 0 4 0 2237.04 21,066.7 0 237.04 21,066.7 5 0 1,066.7 3,200 0 21,066.7 6,400 6

(1) As no loads are applied to this member, Ff 4 5 Qf 4 5 0 The pertinent elements of K4 are stored in S using the member code numbers 1, 2, 3, 4, 5, 6. Member 5 L 5 Ï(X5 2 X4)2 1 (Y5 2 Y4)2 5 Ï(0 2 9)2 1 (12 2 6)2 5 10.817 m cos u 5 sin u 5

F

X5 2 X4 L Y5 2 Y4 L

5

5

029 5 20.83205 10.817

12 2 6 5 0.5547 10.817

(2a) (2b) (2c)

G

4 5 6 7 8 9 144,051 295,943 2409.62 2144,051 95,943 2409.62 295,943 64,098 2614.44 95,943 264,098 2614.44 2409.62 2614.44 5,325.1 409.62 614.44 2,662.6 K5 5 2144,051 95,943 409.62 144,051 295,943 409.62 95,943 264,098 614.44 295,943 64,098 614.44 2409.62 2614.44 2,662.6 409.62 614.44 5,325.1

4 5 6 7 8 9

From Figs. 6.18(a) and (b), we observe that the 12 kN/m uniformly distributed load applied to member 5 acts in the negative direction of the member’s local y axis; therefore, it is considered positive for the purpose of calculating fixed-end forces. Thus, F Ab 5 F Ae 5 0 wL 12(10.817) 5 5 64.9 kN 2 2 wL2 12(10.817)2 5 5 117 kN ? m FMb 5 2FMe 5 12 12 FSb 5 FSe 5

(3a) (3b) (3c)

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Section 6.6   Procedure for Analysis  317

FG

Using Eq. (6.33), we determine the global fixed-end force vector for the member to be

Ff 5 5

236 254 117 236 254 2117

4 5 6 7 8 9

Using the code numbers 4, 5, 6, 7, 8, 9, we store the pertinent elements of K5 and Ff 5 in their proper positions in the S matrix and the Pf vector, respectively. The complete structure stiffness matrix S and structure fixed-joint force vector Pf are shown in Figs. 6.18(c) and (d), respectively. Joint Load Vector:

P5

44 80 0 0 0 0 0 40 0 0

1 2 3 4 5(4) 6 7 8 9

Joint Displacements: By substituting the numerical values of S (Fig. 6.18(c)), Pf (Fig. 6.18(d)), and P (Eq. (4)) into the structural stiffness relationship P – Pf 5 Sd (Eq. (6.42)), and solving the resulting system of simultaneous equations, we obtain the following joint displacements:

d5

3 4 0.185422 m 0.000418736 m 20.0176197 rad 0.18552 m 20.000130738 m 20.0260283 rad 0.186622 m 0.000713665 m 0.0178911 rad

1 2 3 4 5 6 7 8 9

Ans

To check this solution, we evaluate the matrix product Sd, using the foregoing values of the joint displacements d, and substitute the results into the structure stiffness relationship, as P 2 Pf 5 Sd

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318  Chapter 6   Plane Frames

34 3 4 3 4 80 0 0 0 0 0 40 0 0

2

0 0 0 236 254 117 236 254 2117

>

79.939 20.001466 0.0013239 36.051 54.006 2116.992 76.007 53.994 116.994



Checks

Member End Displacements and End Forces:

FG FG F G F G F G F

Member 1

v1 5

y1 y2 y3 y4 y5 y6

10 11 12 5 1 2 3

0 0 0 d1 d2 d3

5

0 0 0 0.185422 0.000418736 20.0176197

cos u 5 0, sin u 5 1

T1 5 T2 5 T3 5

0 21 0 0 0 0

1 0 0 0 0 0

0 0 1 0 0 0

0 0 0 0 21 0

0 0 0 1 0 0

0 0 0  0 0 1

(5)

0 0 0 u1 5 T1v1 5 0.000418736 20.185422 20.0176197

k1 5 k2 5 k3 5

375,000 0 0 2375,000 0 0

0 800 2,400 0 2800 2,400

0 2375,000 2,400 0 9,600 0 0 375,000 22,400 0 4,800 0

0 2800 22,400 0 800 22,400

0 2,400 4,800 0 22,400 9,600

G

(6) Qf1 5 0

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Section 6.6   Procedure for Analysis  319

F G FG F G F G F G FG

Q1 5 k1u1 5

2157.03 kN 106.05 kN 360.44 kN ? m 157.03 kN 2106.05 kN 275.86 kN ? m

F1 5 TT1 Q1 5

2106.05 2157.03 360.44 106.05 157.03 275.86



10 11 12  1 2 3

Ans

(7)

Member 2

v2 5

0 13 0 14 0 15 0.18552 4 20.000130738 5 20.0260283 6

Using T2 from Eq. (5), we obtain

u2 5 T2v2 5

0 0 0 20.000130738 20.18552 20.0260283

Using k2 from Eq. (6), and realizing that Qf 2 5 0, we determine that

Q2 5 k2u2 5

49.027 kN 85.948 kN 320.31 kN ? m 249.027 kN 285.948 kN 195.38 kN ? m

285.948 13 49.027 14 320.31 15 F2 5 TT2 Q2 5  85.948 4 249.027 5 195.38 6



Ans

(8)

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320  Chapter 6   Plane Frames

F G F G F G

Member 3

v3 5

0.185422 0.000418736 20.0176197 0.186622 0.000713665 0.0178911

1 2 3 7 8 9

Using T3 from Eq. (5), we compute 0.000418736 20.185422 20.0176197 0.000713665 20.186622 0.0178911

u3 5 T3v3 5

Using k3 from Eq. (6), and realizing that Qf 3 5 0, we obtain

Q3 5 k3u3 5

2110.6 kN 1.6114 kN 280.392 kN ? m 110.6 kN 21.6114 kN 90.06 kN ? m



Ans

Note that it is not necessary to compute the member global end force vector F3, because this member is not attached to any supports (and, therefore, none of the elements of F3 will appear in the support reaction vector R).

F G F G

Member 4  T4 5 I

u4 5 v4 5

0.185422 0.000418736 20.0176197 0.18552 20.000130738 20.0260283

1 2 3 4 5 6

Using k4 from Eq. (1), and Qf 4 5 0, we obtain

Q4 5 k4u4 5

224.5 kN 246.429 kN 2195.47 kN ? m  24.5 kN 46.429 kN 2222.38 kN ? m

Ans

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Section 6.6   Procedure for Analysis  321

F G F F G

Member 5

v5 5

0.18552 20.000130738 20.0260283 0.186622 0.000713665 0.0178911

4 5 6 7 8 9

cos u 5 20.83205, sin u 5 0.5547   (Eqs. (2))

T5 5

u5 5 T5v5 5

k5 5

F

0.5547 20.83205 0 0 0 0

20.83205 20.5547 0 0 0 0

208,013 0 0 2208,013 0 0

0 0 1 0 0 0

0 0 0 20.83205 20.5547 0

0 0 0 0.5547 20.83205 0

20.154434 20.102799 20.0260283 20.154883 20.104113 0.0178911 0 136.54 738.46 0 2136.54 738.46

FG

0 2208,013 738.46 0 5,325.1 0 0 208,013 2738.46 0 2,662.6 0

0 2136.54 2738.46 0 136.54 2738.46

0 0 0 0 0 1

G G

0 738.46 2,662.6 0 2738.46 5,325.1

From Eqs. (3), we obtain 0 64.9 117 Qf 5 5 0 64.9 2117

F G

93.398 kN 59.07 kN 27.004 kN ? m Q5 5 k5u5 1 Qf 5 5 293.398 kN 70.73 kN 290.061 kN ? m



Ans

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322  Chapter 6   Plane Frames The member local end forces are shown in Fig. 6.18(e). Support Reactions:  The reaction vector R, as assembled from the appropriate elements of the member global end force vectors F1 and F2 (Eqs. (7) and (8), respectively), is given in Fig. 6.18(f). Also, Fig. 6.18(g) depicts these support reactions on a line diagram of the frame. Ans Equilibrium Check: Considering the equilibrium of the entire frame, we write (Fig. 6.18(g)) 1S

oF 5 0 x

40 1 80 1 (12Ï117)



Ï117

2 106.05 2 85.948 5 0.002 > 0 Checks

oF

[

1c  1

Y

50

o M➀ 5 0

(12Ï117)

9 Ï117



2 157.03 1 49.027 5 20.003 > 0

360.44 2 80(6) 2 40(12) 2 (12Ï117) 1 (12Ï117)

6.7 

6

Checks

1Ï1172 9 6

1Ï1172 4.5 1 320.31 1 49.027(9)

5 20.007 > 0

9

Checks

Computer Program The overall organization and format of the computer program for the analysis of plane frames remains the same as the plane truss and beam analysis programs developed previously. All the parts, and many subroutines, of the new program can be replicated from the previous programs with no, or relatively minor, modifications. In this section, we discuss the development of this program for the analysis of plane frames, with emphasis on the programming aspects not considered in previous chapters.

Input Module Joint Data  The part of the computer program for reading and storing the joint data for plane frames (i.e., the number of joints, NJ, and X and Y coordinates of each joint) remains the same as Part I of the plane truss analysis program (see flowchart in Fig. 4.3(a)). As discussed in Section 4.1, the program stores the joint coordinates in a NJ 3 2 joint coordinate matrix COORD in computer memory. As an example, let us consider the gable frame of Fig. 6.19(a), with its analytical model shown in Fig. 6.19(b). Since the frame has five joints, its COORD matrix has five rows, with the X and Y coordinates of a joint i stored in the first and second columns, respectively, of the ith row, as shown in Fig. 6.19(c). An example of the input data file for the gable frame is given in Fig. 6.20 on page 325.

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Section 6.7   Computer Program  323

/m 8 kN

80 k

N

4

2.69 25 m 180 kN

3

Y

2m

2 300 kN

3

2

4

1

5m

4

5

1

X 5m

5m

(b) Analytical Model

Columns: Girders: E 5 200 GPa E 5 70 GPa A 5 19,000 mm2 A 5 20,000 mm2 I 5 1,000(106) mm4 I 5 1,250(106) mm4 (a) Gable Frame

X Coordinate Y Coordinate

COORD 5

3 3 0 0 5 10 10

0 5 7 5 0

Joint 1 Joint 2 Joint 3 Joint 4 Joint 5

NJ 3 2 (c) Joint Coordinate Matrix

Restraint in X Direction (0 5 free, 1 5 restrained)

Restraint in Y Direction (0 5 free, 1 5 restrained) Rotational Restraint (0 5 free, 1 5 restrained)

Joint Number

MSUP 5

1 5

1 1

1 1

1 0 NS 3 (NCJT 1 1)

(d) Support Data Matrix

Fig. 6.19

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324  Chapter 6   Plane Frames Area Moment of Inertia EM 5

Material No. 1 Material No. 2

200,000,000 70,000,000

CP 5

0.019 0.00100 0.020 0.00125

Cross-Section Type No.1 Cross-Section Type No.2

NCP 3 2

NMP 3 1

(f ) Cross-Sectional Property Matrix

(e) Elastic Modulus Vector

Beginning Joint End Joint Material No. Cross-Section Type No.

MPRP 5

1 2 4 5

2 3 3 4

1 2 2 1

1 2 2 1 NM 3 4

Member Member Member Member

1 2 3 4

Force in X Direction Force in Y Direction Moment

Joint Number JP 5 [2]

PJ 5 [300 0

NJL 3 1

(g) Member Data Matrix

0] NJL 3 NCJT

(h) Joint Load Data Matrices

Member Number Load Type Number MP 5

2 3 3

3 1 5 NML 3 2

48 PM 5 2180 80

0 0 0

NML 3 4

W, M, w or w1 w2 (if Load Type 5 4) 0 (otherwise)

0 0 2.6925 0 2.6925 0

l1

l 2 (if Load Type 5 3, 4 or 6) 0 (otherwise) (i) Member Load Data Matrices

Fig. 6.19  (continued)

Support Data  The computer code written for Part II of the plane truss analysis program (see flowchart in Fig. 4.3(b)) can be used to input the support data for plane frames, provided that the number of structure coordinates per joint is set equal to 3 (i.e., NCJT 5 3) in the program. A three-digit code is now used to specify the restraints at a support joint, with the first two digits representing the translational restraint conditions in the global X and Y directions, respectively, and the third digit representing the rotational restraint condition at the joint. As in the case of plane trusses and beams, each digit of the restraint code can be either 0 (indicating no restraint) or 1 (indicating restraint). The restraint codes for some common types of supports for plane frames are given in Fig. 6.21.

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Section 6.7   Computer Program  325

Fig. 6.20  An Example of an Input Data File

5 0, 0 0, 5 5, 7 10, 5 10, 0 2 1, 1, 1, 1 5, 1, 1, 0 2 200,000,000 70,000,000 2 0.019, 0.00100 0.020, 0.00125 4 1, 2, 1, 1 2, 3, 2, 2 4, 3, 2, 2 5, 4, 1, 1 1 2, 300, 0, 0 3 2, 3, 48, 0, 0 3, 1, 2180, 2.6925 3, 5, 80, 2.6925

Joint data

Support data Material property data Cross-sectional property data

Member data

Joint load data Member load data

Type of Support

Restraint Code

Free joint (no support)

0, 0, 0 RX

Roller with horizontal reaction

1, 0, 0 RX

Roller with vertical reaction

0, 1, 0 RY

Hinge

RY

RX RY

Support which prevents rotation, but not translation

Fixed

Fig. 6.21  Restraint Codes for Plane Frames

1, 1, 0

RX RY

0, 0, 1

MR

RX

1, 1, 1

RX MR RY

MR RY

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326  Chapter 6   Plane Frames

The program stores the support data in a NS 3 4 MSUP matrix, as shown in Fig. 6.19(d) for the gable frame, and an example of how this data may appear in an input file is given in Fig. 6.20. Material Property Data  This part of the program remains the same as Part III of the plane truss analysis program (see flowchart in Fig. 4.3(c)). The program stores the moduli of elasticity in a NMP 3 1 EM vector, as shown in Fig. 6.19(e) for the example gable frame; Fig. 6.20 illustrates how this data may appear in an input data file. Cross-Sectional Property Data  As two cross-sectional properties (namely, area and moment of inertia) are needed in the analysis of plane frames, the code written previously for Part IV of the plane truss program should be modified to increase the number of columns of the cross-sectional property matrix CP from one to two, as indicated by the flowchart in Fig. 6.22(a). As before, the number of rows of CP equals the number of cross-section types (NCP), with the area and moment of inertia of the cross-section i now stored in the first and second columns, respectively, of the ith row of the CP matrix of order NCP 3 2. For example, the CP matrix for the gable frame of Fig. 6.19(a) is shown in Fig. 6.19(f), and Fig. 6.20 shows how this data may appear in an input data file. Member Data  This part of the computer program remains the same as Part V of the plane truss analysis program (see flowchart in Fig. 4.3(e)). As discussed in Section 4.1, the program stores the member data in an integer matrix MPRP of order NM 3 4. The MPRP matrix for the example gable frame is shown in Fig. 6.19(g). Joint Load Data  The code written for Part VIa of the beam analysis program (see flowchart in Fig. 5.21(b)) can be used for inputting joint load data for plane frames, provided that NCJT is set equal to 3. The program stores the numbers of the loaded joints in an integer vector JP of order NJL 3 1, with the corresponding loads in the X and Y directions and the couple being stored in the first, second, and third columns, respectively, of a real matrix PJ of order NJL 3 3. The joint load matrices for the example gable frame are shown in Fig. 6.19(h); Fig. 6.20 illustrates how this data may appear in an input data file. Member Load Data  As members of plane frames may be subjected to both axial and perpendicular loads, the code written for Part VIb of the beam analysis program should be modified to include inputting of the member axial loads. The flowchart shown in Fig. 6.22(b) can be used for programming the input of the four perpendicular, and two axial, member load types (i.e., load types 1 through 6) given inside the front cover. The format for reading and storing the member load data for plane frames remains the same as that for beams, as discussed in Section 5.8. The member load matrices, MP and PM, for the example gable frame are shown in Fig. 6.19(i); Fig. 6.20 shows this member load data in an input file that can be read by the program. An example of a computer printout of the input data for the gable frame of Fig. 6.19 is given in Fig. 6.23.

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Section 6.7   Computer Program  327 Start Part VIb Read NML

NML . 0?

no

yes Dimension MP(NML, 2), PM(NML, 4) I  5 1

Start Part IV Read NCP

I # NML? Dimension CP(NCP, 2)

yes Initialize all elements of PM to zero

I51

I # NCP?

no

no

yes

Read MP(I, 1), MP(I, 2) If MP(I, 2) 5 1, 2, or 5 then read PM(I, 1), PM(I, 3) If MP(I, 2) 5 3 or 6 then read PM(I, 1), PM(I, 3), PM(I, 4) If MP(I, 2) 5 4 then read PM(I, 1), PM(I, 2), PM(I, 3), PM(I, 4)

Read CP(I, 1), CP(I, 2)

I 5 I 1 1

I5I11

Print All Input Data

Continue to Part V (a) Flowchart for Reading and Storing Cross-Sectional Property Data

Continue to Part VII (b) Flowchart for Reading and Storing Member Load Data

Fig. 6.22

Analysis Module Assignment of Structure Coordinate Numbers   The parts of the program for determining the number of degrees of freedom (NDOF) and forming the structure coordinate number vector (NSC), for plane frames, remain the same as Parts VII and VIII of the plane truss analysis program (see flowcharts in Figs. 4.8(a) and (b)), provided that NCJT is set equal to 3 in these programs. Generation of the Structure Stiffness Matrix and Equivalent Joint Load ­Vector  A flowchart for writing this part of the plane frame analysis program is given in Fig. 6.24. Comparing this flowchart with that for Part IX of the beam analysis program in Fig. 5.25, we can see that the two programs are similar; the present program, however, transforms the stiffness matrix and fixed-end force vector of each member from its local to the global coordinate system before storing their elements in the structure stiffness matrix and equivalent joint load

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328  Chapter 6   Plane Frames

******************************** * Computer Software * * * for * * Analysis of Structures Matrix * * Third Edition * * by * * * Aslam Kassimali * * ******************************* * General Structural Data Project Title: Figure 6–19 Structure Type: Plane Frame Number of Joints: 5 Number of Members: 4 Number of Material Property Sets (E): 2 Number of Cross-Sectional Property Sets: 2 Joint Coordinates Joint No. 1 2 3 4 5

X Coordinate 0.0000E+00 0.0000E+00 5.0000E+00 1.0000E+01 1.0000E+01

Y Coordinate  0.0000E+00   5.0000E+00   7.0000E+00   5.0000E+00   0.0000E+00

Supports Rotational Joint No. X Restraint Y Restraint   Restraint 1 Yes Yes Yes 5 Yes Yes No Material Properties Material Modulus of Coefficient of No. Elasticity (E) Thermal Expansion 1 2.0000E+08 0.0000E+00 2 7.0000E+07 0.0000E+00 Cross-Sectional Properties Property No. Area (A) 1 1.9000E-02 2 2.0000E-02

Moment of Inertia (I) 1.0000E-03 1.2500E-03

Fig. 6.23  A Sample Printout of Input Data

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Section 6.7   Computer Program  329

Member Data Member Beginning End Material Cross-Sectional No. Joint Joint No. Property No. 1 1 2 1 1 2 2 3 2 2 3 4 3 2 2 4 5 4 1 1 Joint Loads Moment Joint No. X Force Y Force 2 3.0000E+02 0.0000E+00 0.0000E+00 Member Loads Load Magnitude (W or M) or  Load Member Load Intensity Intensity Distance Distance No. Type (w or w1) w2 11 12 2 Uniform   4.800E+1 --- 0.0000E+0 0.00E+0 3 Axial–C   8.000E+1 --- 2.6925E+0 -- 3 Conc. -1.800E+2 --- 2.6925E+0 --************* End of Input Data ************* Fig. 6.23  (continued)

vector, respectively (see Fig. 6.24). Recall from Chapter 5 that such coordinate transformations are not necessary for beams, because the local and global coordinate systems of such structures are oriented in the same direction. From the flowchart in Fig. 6.24, we can see that for each member of the plane frame, the program first reads the member’s material and cross-sectional properties, and calculates its length and direction cosines. Next, the program calls the subroutines MSTIFFL and MTRANS to form the member local stiffness matrix BK and transformation matrix T, respectively. As the flowcharts given in Figs. 6.25 and 6.26 indicate, these subroutines calculate the matrices BK and T in accordance with Eqs. (6.6) and (6.19), respectively. The program then calls the subroutine MSTIFFG to obtain the member global stiffness matrix GK. A flowchart of this subroutine, which evaluates the member global stiffness matrix using the matrix triple product K 5 TTkT (Eq. (6.29)), is given in Fig. 6.27. As this flowchart indicates, the subroutine MSTIFFG uses two nested Do Loops to calculate the member global stiffness matrix GK (5 K). In the first loop, the member local stiffness matrix BK (5 K) is post-multiplied by its transformation matrix T to obtain an intermediate matrix TS; in the second, the matrix TS is premultiplied by the transpose of the transformation ­matrix

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330  Chapter 6   Plane Frames Start Part IX Dimension S(NDOF, NDOF), P(NDOF), GK(2*NCJT, 2*NCJT), BK(2*NCJT, 2*NCJT), FF(2*NCJT), QF(2*NCJT) Initialize all elements of S and P to zero IM 5 1

IM # NM?

no

yes JB 5 MPRP(IM, 1), JE 5 MPRP(IM, 2), I 5 MPRP(IM, 3), E 5 EM(I) I 5 MPRP(IM, 4), A 5 CP(I, 1), ZI 5 CP(I, 2) XB 5 COORD(JB, 1), YB 5 COORD(JB, 2), XE 5 COORD(JE, 1), YE 5 COORD(JE, 2) BL 5 SQR((XE 2 XB)^2 1 (YE 2 YB)^2), CX 5 (XE 2 XB)/BL, CY 5 (YE 2 YB)/BL Call Subroutine MSTIFFL Call Subroutine MTRANS Call Subroutine MSTIFFG Call Subroutine STORES

IM 5 IM 1 1

no

NML . 0? yes

Call Subroutine STOREPF

Initialize all elements of QF to zero IML 5 1

Call Subroutine MFEFG

no

IML # NML? yes IM 5 MP(IML, 1)?

no

IML 5 IML 1 1

yes Call Subroutine MFEFLL Continue to Part X

Fig. 6.24  Flowchart for Generating Structure Stiffness Matrix and Equivalent Joint Load Vector

Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 6.7   Computer Program  331 Start Subroutine MSTIFFL Arguments: E, A, ZI, BL, NCJT, BK I51

I # 2*NCJT?

Start Subroutine MTRANS no

Arguments: CX, CY, NCJT, T

yes

I51

J51

I5I11

no

I # 2*NCJT? J # 2*NCJT?

yes

yes BK(I, J ) 5 0

no

J51 J5J11

Z 5 E*A/BL BK(1, 1) 5 Z, BK(4, 1) 52Z BK(1, 4) 52Z, BK(4, 4) 5 Z Z 5 E*ZI/(BL^3) BK(2, 2) 5 12*Z, BK(3, 2) 5 6*BL*Z BK(5, 2) 5212*Z, BK(6, 2) 5 6*BL*Z BK(2, 3) 5 6*BL*Z, BK(3, 3) 5 4*(BL^2)*Z BK(5, 3) 526*BL*Z, BK(6, 3) 5 2*(BL^2)*Z BK(2, 5) 5212*Z, BK(3, 5) 526*BL*Z BK(5, 5) 5 12*Z, BK(6, 5) 526*BL*Z BK(2, 6) 5 6*BL*Z, BK(3, 6) 5 2*(BL^2)*Z BK(5, 6) 526*BL*Z, BK(6, 6) 5 4*(BL^2)*Z

I5I11

no

J # 2*NCJT? yes T(I, J) 5 0

J5J11

T(1, 1) 5 CX,  T(2, 1) 5 2CY  T(1, 2) 5 CY,  T(2, 2) 5 CX  T(4, 4) 5 CX,  T(5, 4) 5 2CY T(4, 5) 5 CY,  T(5, 5) 5 CX T(3, 3) 5 1,   T(6, 6) 5 1  End Subroutine MTRANS Return to calling program

End Subroutine MSTIFFL Return to calling program

Fig. 6.25  Flowchart of Subroutine MSTIFFL for Determining Member Local Stiffness Matrix for Plane Frames

Fig. 6.26  Flowchart of Subroutine MTRANS for Determining Member Transformation Matrix for Plane Frames

(i.e., TT) to obtain the desired member global stiffness matrix GK (5 K). Returning our attention to Fig. 6.24, we can see that the program then calls the subroutine STORES to store the pertinent elements of GK in the structure stiffness matrix S. This subroutine remains the same as the STORES subroutine of the plane truss analysis program (see flowchart in Fig. 4.11). After STORES has been executed, the program (Fig. 6.24) forms the member local fixed-end force vector QF (5 Qf) by calling the subroutine MFEFLL (Fig. 6.28), which calculates the values of the member fixed-end forces, for load types 1 through 6, using the equations given inside the front cover. Next, the program calls the subroutine MFEFG (Fig. 6.29), which evaluates the member global fixed-end force vector FF (5 Ff), using the relationship Ff 5 TTQf (Eq. (6.30)). Finally, the program calls the subroutine STOREPF (Fig. 6.30) to store the negative

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332  Chapter 6   Plane Frames Start Subroutine MSTIFFG Arguments: NCJT, BK, T, GK Dimension TS(2*NCJT, 2*NCJT) Initialize all elements of GK and TS to zero I51 no

I # 2*NCJT?

I51

yes

no

J # 2*NCJT?

yes J51

yes J # 2*NCJT?

K51

K # 2*NCJT?

no

J5J11 I5I11

no

yes K51

yes TS(I, J) 5 TS(I, J) 1 BK(I, K)*T(K, J) K5K11

no

I # 2*NCJT?

J51

K # 2*NCJT?

no

yes GK(I, J) 5 GK(I, J) 1 T(K, I )*TS(K, J) K5K11 J5J11 I5I11 End Subroutine MSTIFFG Return to calling program

Fig. 6.27  Flowchart of Subroutine MSTIFFG for Determining Member Global Stiffness Matrix

values of the pertinent elements of FF in their proper positions in the structure load vector P. When all the operations shown in Fig. 6.24 have been performed for each member of the frame, the structure stiffness matrix S is complete, and the structure load vector P equals the negative of the structure fixed-joint force vector (i.e., P 5 2Pf 5 Pe). Storage of Joint Loads into the Structure Load Vector  This is the same as Part X of the beam analysis program (see flowchart in Fig. 5.30). Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 6.7   Computer Program  333 Start Subroutine MFEFLL Arguments: IML, BL, MP, PM, QF Initialize FAB, FSB, FMB, FAE, FSE, FME to zero LDTYPE 5 MP(IML, 2)

If LDTYPE 5 1?

yes

BW 5 PM(IML, 1), BL1 5 PM(IML, 3) Use equations for load type 1 to calculate FSB, FMB, FSE, FME

yes

BM 5 PM(IML, 1), BL1 5 PM(IML, 3) Use equations for load type 2 to calculate FSB, FMB, FSE, FME

no If LDTYPE 5 2? no If LDTYPE 5 3?

yes

no If LDTYPE 5 4?

yes

no If LDTYPE 5 5?

yes

no no

If LDTYPE 5 6?

yes

W 5 PM(IML, 1), BL1 5 PM(IML, 3), BL2 5 PM(IML, 4) Use equations for load type 3 to calculate FSB, FMB, FSE, FME W1 5 PM(IML, 1), W2 5 PM(IML, 2), BL1 5 PM(IML, 3), BL2 5 PM(IML, 4) Use equations for load type 4 to calculate FSB, FMB, FSE, FME BW 5 PM(IML, 1), BL1 5 PM(IML, 3), Use equations for load type 5 to calculate FAB, FAE W 5 PM(IML, 1), BL1 5 PM(IML, 3), BL2 5 PM(IML, 4) Use equations for load type 6 to calculate FAB, FAE

QF(1) 5 QF(1) 1 FAB QF(2) 5 QF(2) 1 FSB QF(3) 5 QF(3) 1 FMB QF(4) 5 QF(4) 1 FAE QF(5) 5 QF(5) 1 FSE QF(6) 5 QF(6) 1 FME End Subroutine MFEFLL Return to calling program

Fig. 6.28  Flowchart of Subroutine MFEFLL for Determining Member Local Fixed-End Force Vector for Plane Frames

Solution for Joint Displacements  This part of the program remains the same as Part XI of the plane truss analysis program (see flowchart in Fig. 4.13). Recall that upon completion of this part, the vector P contains the values of the joint displacements d. Calculation of Member Forces and Support Reactions    A flowchart for writing this last part of the program is presented in Fig. 6.31. Note that this part of the plane frame analysis program is essentially a combination of the Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

334  Chapter 6   Plane Frames Start Subroutine MFEFG Arguments: NCJT, T, QF, FF I51

I # 2*NCJT?

no

yes I5I11

FF(I) 5 0 I51

I # 2*NCJT?

no

yes J51

I5I11

no

J # 2*NCJT? yes

FF(I ) 5 FF(I ) 1 T(J, I)*QF(J)

J5J11

End Subroutine MFEFG Return to calling program

Fig. 6.29  Flowchart of Subroutine MFEFG for Determining Member Global Fixed-End Force Vector

c­ orresponding parts (XII) of the plane truss and beam analysis programs developed previously. From the flowchart in Fig. 6.31, we can see that, for each member of the frame, the program first reads the member’s material and crosssectional properties, and calculates its length and direction cosines. Next, the program calls the subroutine MDISPG to form the member global end displacement vector V (5 v). This subroutine is the same as the MDISPG subroutine of the plane truss analysis program (see flowchart in Fig. 4.15). The program then calls the subroutine MTRANS (Fig. 6.26) to form the member transformation matrix T, and the subroutine MDISPL, which calculates the member local end displacement vector U (5 u), using the relationship u 5 Tv (Eq. (6.20)). The subroutine MDISPL remains the same as the corresponding subroutine of the plane truss program (see flowchart in Fig. 4.17). Next, the subroutine MSTIFFL (Fig. 6.25) is called by the program to form the member local stiffness matrix BK (5 k); if the member under consideration is

Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 6.7   Computer Program  335 Start Subroutine STOREPF Arguments: JB, JE, NCJT, NDOF, NSC, FF, P I51 no

I # 2*NCJT? yes yes

I # NCJT?

I1 5 (JB 2 1)*NCJT 1 I

no

I1 5 (JE 2 1)*NCJT 1 (I 2 NCJT)

N1 5 NSC(I1)

I5I11

no

N1 # NDOF? yes P(N1) 5 P(N1) 2 FF(I)

End Subroutine STOREPF Return to calling program

Fig. 6.30  Flowchart of Subroutine STOREPF for Storing Member Global FixedEnd Force Vector in Structure Load Vector

subjected to loads, then its local fixed-end force vector QF (5 Qf) is generated using the subroutine MFEFLL (Fig. 6.28). The program then calls the subroutines MFORCEL and MFORCEG, respectively, to calculate the member’s local and global end force vectors Q and F. The subroutine MFORCEL, which evaluates Q using the relationship Q 5 ku 1 Qf (Eq. (6.4)), is the same as the corresponding subroutine of the beam analysis program (see flowchart in Fig. 5.33); the subroutine MFORCEG, which computes F using the relationship F 5 TTQ (Eq. (6.23)), remains the same as the corresponding subroutine of the plane truss program (see flowchart in Fig. 4.20). Finally, the program stores the pertinent elements of F in the support reaction vector R by calling the subroutine STORER, which remains the same as the corresponding subroutine of the plane truss program (see flowchart in Fig. 4.21). The computational process depicted in Fig. 6.31 can be somewhat expedited by calling the subroutines MFORCEG and STORER for only those members of the frame that are attached to supports. To check whether or not a member is attached to a support, its beginning and end joint numbers (i.e., JB and JE) can be compared

Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

336  Chapter 6   Plane Frames Start Part XII Dimension BK(2*NCJT, 2*NCJT), T(2*NCJT, 2*NCJT), V(2*NCJT), U(2*NCJT), F(2*NCJT), Q(2*NCJT), QF(2*NCJT), R(NR) Initialize all elements of R to zero IM 5 1 no

IM # NM?

IM 5 IM 1 1

yes JB 5 MPRP(IM, 1), JE 5 MPRP(IM, 2), I 5 MPRP(IM, 3), E 5 EM(I) I 5 MPRP(IM, 4), A 5 CP(I, 1), ZI 5 CP(I, 2) XB 5 COORD(JB, 1), YB 5 COORD(JB, 2), XE 5 COORD(JE, 1), YE 5 COORD(JE, 2) BL 5 SQR((XE 2 XB)^2 1 (YE 2 YB)^2), CX 5 (XE 2 XB)/BL, CY 5 (YE 2 YB)/BL Call Subroutine MDISPG

Call Subroutine STORER

Call Subroutine MTRANS Call Subroutine MDISPL Call Subroutine MFORCEG Call Subroutine MSTIFFL Initialize all elements of QF to zero no

Call Subroutine MFORCEL

NML . 0? yes IML 5 1

no

IML # NML? yes IM 5 MP(IML, 1)?

no

IML 5 IML 1 1

yes Call Subroutine MFEFLL Print Support Reactions R End of Program

Fig. 6.31  Flowchart for Determination of Member Forces and Support Reactions for Plane Frames Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 6.7   Computer Program  337

*************************************************** * Results of Analysis * *************************************************** Joint Displacements Joint No. X Translation Y Translation Rotation (Rad) 1 2 3 4 5

0.0000E+00  0.0000E+00 4.3974E-02 21.7430E-04 5.0391E-02 21.7016E-02 5.6220E-02 24.0048E-04 0.0000E+00  0.0000E+00

 0.0000E+00 21.1970E-02  4.3117E-03 25.6746E-03 21.4029E-02

Member End Forces in Local Coordinates Member Joint Axial Force Shear Force Moment 1 1 1.3247E+02 2.6976E+02 1.1532E+03 2 -1.3247E+02 -2.6976E+02 1.9561E+02 2 3 4

2 3

7.7275E+01 1.1176E+02 -1.9561E+02 7.7275E+01 1.4672E+02 1.0148E+02 -

4 3

2.3715E+02 -2.3295E+02 -6.6834E+02 1.5715E+02 5.2955E+01 -1.0148E+02 -

5 4

3.0437E+02 1.3367E+02 1.0547E-14 -3.0437E+02 -1.3367E+02 6.6834E+02 Support Reactions

oint No. X Force Y Force J Moment 1 -2.6976E+02 1.3247E+02 1.1532E+03 5 -1.3367E+02 3.0437E+02 0.0000E+00 ************* End of Analysis ************* Fig. 6.32  A Sample Printout of Analysis Results

with the support joint numbers stored in the first column of the support data matrix MSUP. A sample printout, showing the results of analysis for the example gable frame of Fig. 6.19, is presented in Fig. 6.32, and the entire computer program for the analysis of plane frames is summarized in Table 6.1. As indicated in this table, the computer program consists of a main program (which is divided into twelve parts) and twelve subroutines. Of these, eight parts of the main program and six subroutines can be replicated from the previously developed plane truss and beam analysis programs without any modifications. Finally, it should be realized that the computer program, developed herein for the analysis of plane frames, can also be used to analyze beams, although it is not as efficient for beam analysis as the program developed specifically for that purpose in Chapter 5. Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

338  Chapter 6   Plane Frames Table 6.1  Computer Program for Analysis of Plane Frames Main program part Description

I Reads and stores joint data (Fig. 4.3(a)) II Reads and stores support data (Fig. 4.3(b)) III Reads and stores material properties (Fig. 4.3(c)) IV Reads and stores cross-sectional properties (Fig. 6.22(a)) V Reads and stores member data (Fig. 4.3(e)) Reads and stores joint loads (Fig. 5.21(b)) VIa VIb Reads and stores member loads (Fig. 6.22(b)) VII Determines the number of degrees of freedom NDOF of the structure (Fig. 4.8(a)) VIII Forms the structure coordinate number vector NSC (Fig. 4.8(b)) IX Generates the structure stiffness matrix S and the structure load vector P 5 Pe 5 2Pf due to member loads (Fig. 6.24) Subroutines called: MSTIFFL, MTRANS, MSTIFFG, STORES, MFEFLL, MFEFG, and STOREPF X Stores joint loads in the structure load vector P (Fig. 5.30) XI Calculates the structure’s joint displacements by solving the stiffness relationship, Sd 5 P, using Gauss-Jordan elimination. The vector P now contains joint displacements (Fig. 4.13). XII Determines the member end force vectors Q and F, and the support reaction vector R (Fig. 6.31) Subroutines called: MDISPG, MTRANS, MDISPL, MSTIFFL, MFEFLL, MFORCEL, MFORCEG, and STORER Subroutine Description

MDISPG Forms the member global displacement vector V from the joint displacement vector P (Fig. 4.15) MDISPL Evaluates the member local displacement vector U 5 TV (Fig. 4.17) MFEFG Determines the member global fixed-end force vector FF 5 TTQF (Fig. 6.29) MFEFLL Calculates the member local fixed-end force vector QF (Fig. 6.28) MFORCEG Evaluates the member global force vector F 5 TT Q (Fig. 4.20) MFORCEL Calculates the member local force vector Q = BK U 1 QF (Fig. 5.33) MSTIFFG Determines the member global stiffness matrix GK 5 TT BK T (Fig. 6.27) MSTIFFL Forms the member local stiffness matrix BK (Fig. 6.25) MTRANS Forms the member transformation matrix T (Fig. 6.26) STOREPF Stores the negative values of the pertinent elements of the member global fixed-end force vector FF in the structure load vector P (Fig. 6.30) STORER Stores the pertinent elements of the member global force vector F in the reaction vector R (Fig. 4.21) STORES Stores the pertinent elements of the member global stiffness matrix GK in the structure stiffness matrix S (Fig. 4.11)

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Problems  339

Summary In this chapter, we have developed the matrix stiffness method for the analysis of rigidly connected plane frames subjected to external loads. A block diagram summarizing the various steps of the analysis is shown in Fig. 6.33. Identify degrees of freedom d and restrained coordinates of the plane frame For each member: Evaluate K and Ff Store K in S and Ff in Pf Form joint load vector P Solve P 2 Pf 5 Sd for d For each member: Obtain v from d Calculate u 5 Tv, Q 5 ku 1 Qf and F 5 TTQ Store F in R

Fig. 6.33

P ro b lems Section 6.1 6.1 through 6.6  Identify by numbers the degrees of freedom and restrained coordinates of the frames shown in Figs. P6.1 through P6.6. Also, form the joint load vector P for these frames. 75 kN • m

2

1

125 kN 8m

37.5 kN/m

2

1 150 kN·m

3

24 kN/m 3

2m

2

200 kN

1

3m

3m

2 12 m

E, A, I 5 constant E 5 200 GPa A 5 13,000 mm2 I 5 762(106) mm4

2m 1

Fig. P6.2, P6.8, P6.17, P6.25, P6.33, P6.43 6m E, A, I 5 constant E 5 200 GPa A 5 4,000 mm2 I 5 80(106) mm4

Fig. P6.1, P6.7, P6.16, P6.24, P6.32, P6.42 Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

340  Chapter 6   Plane Frames 15 kN/m

27 kN/m 2

1

75 kN 2

45 kN

1

3

2

3m 2 3 6m

2m

1

3

8m

2m

E, A, I 5 constant E 5 70 GPa A 5 7,500 mm2 I 5 125(106) mm4

1

4

10 m

Fig. P6.3, P6.9, P6.18, P6.26, P6.34, P6.44

E, A, I 5 constant E 5 30 GPa A 5 90,000 mm2 I 5 675(106) mm4

Fig. P6.5, P6.11, P6.20, P6.28, P6.36, P6.48 3

25 kN/m

50 kN/m

20 kN/m 2

7m 2 2

A 5 2,600 mm2 I 5 31(106) mm4

3

A 5 2,900 mm2 I 5 24(106) mm4

25 kN/m

4m

2

1

1

16 m

1

1

24 m E, A, I 5 constant E 5 200 GPa A 5 16,000 mm2 I 5 1,186(106) mm4

Fig. P6.4, P6.10, P6.19, P6.27, P6.35, P6.45

2m

5m E 5 70 GPa

Fig. P6.6, P6.12, P6.21, P6.29, P6.37, P6.47 Section 6.2 6.7 through 6.12  Determine the local stiffness matrix k, and the fixed-end force vector Qf, for each member of the frames shown in Figs. P6.7 through P6.12. Use the fixed-end force equations given inside the front cover.

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Problems  341

6.13 Using the integration approach, derive the equations of fixed-end forces due to the concentrated axial member load shown in Fig. P6.13. Check the results using the fixed-end force expressions given inside the front cover. l1

l2

u1 5

W

b

e

L E, A, I 5 constant

Fig. P6.13

F G F G

6.14  Assume that the local end displacements for the members of the frame shown in Fig. P6.14 are

u1 5

0 0 0 ;    25.2507 mm 212.251 mm 20.12416 rad

u2 5

9.2888 mm 29.5586 mm 20.12416 rad 0 0 0

Calculate the member local end force vectors. Are the members in equilibrium under these forces?

F GF G F G

6.15  Assume that the local end displacements for the members of the frame shown in Fig. P6.15 are

300 kN

600 kN 3

0.625 m

2

2

2.5 m 64 kN/m

2

3.75 m

1 4

1

2

3m

1

;

Calculate the member local end force vectors. Are the members in equilibrium under these forces?

3

m kN/ 5m

160 kN

0.025799 m 0.0015313 m 0.0019382 rad ;   u2 5 0.02454 m 20.0091585 m 0.0048505 rad

0 0 0 u3 5 0.00092704 m 0.025827 m 0.0019382 rad

3

15

0 0 0 20.00098438 m 0.026175 m 0.0048505 rad

2m E, A, I 5 constant E 5 200 GPa A 5 6,500 mm2 I 5 120(106) mm4

10 m

Fig. P6.15, P6.23, P6.31 1

Section 6.3 12 m E, A, I 5 constant E 5 70 GPa A 5 4,570 mm2 I 5 34.5(106) mm4

Fig. P6.14, P6.22, P6.30

6.16 through 6.21  Determine the transformation matrix T for each member of the frames shown in Figs. P6.16 through P6.21. 6.22  Using the local end displacements given in Problem 6.14 for the members of the frame of Fig. P6.22, calculate the global end displacement vector and the global end force vector for each member of the frame. Are the members in equilibrium under the global end forces?

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342  Chapter 6   Plane Frames 6.23  Using the local end displacements given in Problem 6.15 for the members of the frame of Fig. P6.23, calculate the global end displacement vector and the global end force vector for each member of the frame. Are the members in equilibrium under the global end forces?

d5

Section 6.4 6.24 through 6.29 Determine the global stiffness matrix K, and fixed-end force vector Ff , for each member of the frames shown in Figs. P6.24 through P6.29. 6.30  Calculate the member global end force vectors required in Problem 6.22 using the member global stiffness relationship F 5 Kv 1 Ff . 6.31  Calculate the member global end force vectors required in Problem 6.23 using the member global stiffness relationship F 5 Kv 1 Ff .

Calculate the joint loads causing these displacements. (No loads are applied to the members of the frame.) 6.39 Assume that the joint displacements for the frame of Fig. P6.39 are

d 5

Section 6.5 6.32 through 6.37  Determine the structure stiffness matrix, the fixed-joint force vector, and the equivalent joint loads for the frames shown in Figs. P6.32 through P6.37. 6.38 Assume that the joint displacements for the frame of Fig. P6.38 are

2

F G 0.1965 m 0.00016452 m 20.017932 rad 0.19637 m 20.00016452 m 20.023307 rad

3 4 20.049464 rad 0.071893 m 20.047948 m 0.0030594 rad 0.057861 m 0.030726 m 0.018929 rad 20.040834 rad

Calculate the joint loads causing these displacements. (No loads are applied to the members of the frame.)

3

3 2 2.75 m

8m

2.25 m

1

4

6m

1

1.5 m

2.75 m

1.5 m

E, A, I 5 constant E 5 70 GPa A 5 6,200 mm2 I 5 36(106) mm4

E 5 30 GPa Girder: Columns: A 5 52,000 mm2 A 5 75,000 mm2 I 5 468(106) mm4 I 5 225(106) mm4

Fig. P6.38

4

Fig. P6.39

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Problems  343

6.45 through 6.50 Determine the joint displacements, member local end forces, and support reactions for the frames shown in Figs. P6.45 through P6.50, using the matrix stiffness method. Check the hand-calculated results by using the computer program that can be downloaded from the publisher’s website for this book, or by using any other general purpose structural analysis program available.

30 kN/m

3m

100 kN 80 kN • m

32 kN/m 3m

6m

6m E, A, I 5 constant E 5 200 GPa A 5 11,800 mm2 I 5 554(106) mm4

Fig. P6.40 6.25 m

3

E, A, I 5 constant E 5 200 GPa A 5 6,500 mm2 I 5 150(106) mm4

1.2 m

4.8 m

12.5 kN/m

1.75 m

Fig. P6.46

2 12.5 kN/m 1

2 24 kN/m

1 6m

2m

200 kN

E, A, I 5 constant E 5 28 GPa A 5 110,000 mm2 I 5 680(106) mm4

24 kN/m

3 2

4m 3

2

4 4

1 1

6m 5

Fig. P6.41 7.5 m

Section 6.6 6.40 through 6.44 Determine the joint displacements, member local end forces, and support reactions for the frames shown in Figs. P6.40 through P6.44, using the matrix stiffness method. Draw the shear, bending moment, and axial force diagrams for the frames.

7.5 m E 5 200 GPa

Columns: A 5 8,800 mm2 I 5 140(106) mm4

Girders: A 5 8,850 mm2 I 5 350(106) mm4

Fig. P6.49

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344  Chapter 6   Plane Frames Section 6.7

12 kN/m 30 kN

6m

12 kN/m

6.51  Develop a general computer program for the analysis of rigidly connected plane frames by the matrix stiffness method. Use the program to analyze the frames of Problems 6.40 through 6.50, and compare the computer-generated results to those obtained by hand calculations.

60 kN

6m

12 m E 5 30 GPa Girders: Columns: A 5 140,000 mm2 A 5 93,000 mm2 I 5 720(106) mm4 I 5 2,430(106) mm4

Fig. P6.50

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7

MEMBER RELEASES AND ­SECONDARY EFFECTS Learning Objectives At the end of this chapter, you will be able to: 7.1 Analyze Plane Frames and Beams with Member ­Releases 7.2 Implement Computer Analysis for Member Releases 7.3 Analyze the Effects of Support Settlements 7.4 Implement Computer Analysis for Support Settlements 7.5 Analyze the Effects of Temperature Changes and ­Fabrication Errors

Earthquake-Damaged Bridge (Courtesy of the USGS)

345

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346  Chapter 7   Member Releases and ­Secondary Effects

The matrix stiffness analysis of beams and plane frames, as developed in Chapters 5 and 6, is based on the assumption that each member of a structure is rigidly connected to joints at both ends, so that the member end rotations are equal to the rotations of the adjacent joints. Whereas these methods of analysis, as presented in preceding chapters, cannot be used to analyze beams and plane frames containing members connected by hinged connections, they can be modified relatively easily to include the effects of member hinges in the analysis. When the end of a member in a plane frame or beam is connected to the adjacent joint by a hinged connection, the moment at the hinged end must be zero. Because of this moment-releasing characteristic, member hinges are often referred to as member releases. In this chapter, we discuss modifications of the matrix stiffness methods that allow them to be used to analyze plane frames and beams containing members connected to joints by rigid (i.e., ­moment-resisting) and/or hinged (i.e., simple or shear) connections. In this chapter, we also consider the procedures for including in matrix stiffness methods of analysis the effects of support displacements (due to weak foundations), temperature changes, and fabrication errors. Such secondary effects can induce significant stresses in statically indeterminate structures, and must be considered in their designs. We begin the chapter by deriving the stiffness relationships for members of plane frames and beams with hinges. A procedure for the analysis of structures containing member releases is also developed in Section 7.1; the computer implementation of this procedure is presented in Section 7.2. We develop the analysis for the effects of support displacements in Section 7.3, and discuss the extension of the previously developed computer programs to include the effects of support displacements in Section 7.4. Finally, the procedure for including in the analysis the effects of temperature changes and fabrication errors is presented in Section 7.5.

7.1 

Member Releases in Plane Frames and Beams The effects of member releases can be conveniently incorporated in our stiffness methods by modifying the member local stiffness relationships to account for such releases. Only moment releases, in the form of hinges located at one or both ends of a member (see Fig. 7.1), are considered herein, because such releases are by far the most commonly encountered in civil engineering practice. However, the concepts presented can be readily used to introduce the effects of other types of member releases (e.g., shear and axial force releases) into the analysis. Figure 7.1 depicts the types of member releases considered herein. From a computer programming viewpoint, it is usually convenient to classify each member of a beam or a plane frame into one of the four member types (MT) shown in the figure. Thus, as indicated in Fig. 7.1(a), a member that is rigidly connected to joints at both ends (i.e., has no hinges), is considered to be of type 0 (i.e., MT 5 0). If end b of a member is connected to the adjacent joint by a hinged connection, while its opposite end e is rigidly connected to the adjacent joint (Fig. 7.1(b)), then the member is classified as type 1 (i.e., MT 5 1).

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Section 7.1   Member Releases in Plane Frames and Beams  347

y

x e

b (a) Member with No Hinges—Member Type Zero (MT 5 0)

y

Hinge

x e

b (b) Member with a Hinge at Its Beginning— Member Type One (MT 5 1)

y

Hinge

x e

b (c) Member with a Hinge at Its End— Member Type Two (MT 5 2)

y

Hinge

Hinge

x e

b (d) Member with Hinges at Both Ends— Member Type Three (MT 5 3)

Fig. 7.1  Member Releases

Conversely, if end b of a member is rigidly attached to the adjacent joint, but its end e is connected by a hinged connection to the adjacent joint (Fig. 7.1(c)), then the member is considered to be of type 2 (i.e., MT 5 2). Finally, if a member is attached to joints at both ends by hinged connections (Fig. 7.1(d)), then it is classified as type 3 (i.e., MT 5 3). The expressions for the member local stiffness matrices k (Eqs. (5.53) and (6.6)) and the member local fixed-end force vectors Qf (Eqs. (5.99) and (6.15)) derived for beams and plane frames can be used only for members of type  0 (MT 5 0),

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348  Chapter 7   Member Releases and ­Secondary Effects

because they are based on the condition that the member is rigidly connected to joints at both ends, so that the member end rotations are equal to the rotations of the adjacent joints. When an end of a member is connected to the adjacent joint by a hinged connection, the moment at the hinged end must be zero. The previous expressions for k and Qf can be easily modified to reflect the conditions of zero moments at the hinged member ends, as explained in the following paragraphs.

Local Stiffness Relations for Plane Frame Members with Hinges We begin the development of the modified expressions by first writing the previously derived stiffness relations for a plane frame member with no hinges, in explicit form. By substituting the expressions for k and Qf from Eqs. (6.6) and (6.15), respectively, into the member local stiffness relation Q 5 ku 1 Qf (Eq. (6.4)), and carrying out the necessary matrix multiplication and addition, we obtain Q1 5

EA (u 2 u4) 1 FAb L 1

(7.1a)



Q2 5

EI (12u2 1 6Lu3 2 12u5 1 6Lu6) 1 FSb L3

(7.1b)



Q3 5

EI (6Lu2 1 4L2u3 2 6Lu5 1 2L2u6) 1 FMb L3

(7.1c)



Q4 5

EA (2u1 1 u4) 1 FAe L

(7.1d)



Q5 5

EI (212u2 2 6Lu3 1 12u5 2 6Lu6) 1 FSe L3

(7.1e)



Q6 5

EI (6Lu2 1 2L2u3 2 6Lu5 1 4L2u6) 1 FMe L3

(7.1f )

Members with a Hinge at the Beginning (MT 5 1)  When end b of a member is connected to the adjacent joint by a hinged connection, then from Fig. 6.3(b) we can see that its end moment Q3 must be 0. By substituting Q3 5 0 into Eq. (7.1c), and solving the resulting equation for the end rotation u3, we obtain

u3 5

3 1 L (2u2 1 u5) 2 u6 2 FMb 2L 2 4EI

(7.2)

This equation indicates that the rotation u3 (of the hinged end b of the member) is no longer an independent member coordinate (or degree of freedom), but is now a function of the end displacements u2, u5, and u6. Thus, the number of independent member coordinates—that is, the independent end displacements required to define the displaced member configuration—is now reduced to five (i.e., u1, u2, u4, u5, and u6  ). To eliminate the released coordinate u3 from the

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Section 7.1   Member Releases in Plane Frames and Beams  349

member stiffness relations, we substitute Eq. (7.2) into Eqs. (7.1). This yields the following member stiffness equations: Q1 5

EA (u 2 u4) 1 FAb L 1

Q2 5

EI 3 (3u2 2 3u5 1 3Lu6) 1 FSb 2 FMb 3 L 2L

(7.3a)

1

2

(7.3b) (7.3c)

Q3 5 0 Q4 5

EA (2u1 1 u4) 1 FAe L

(7.3d)

2 EI 1 Q 5 (3Lu 2 3Lu 1 3L u ) 1 1FM 2 FM 2 L 2 Q5 5

1

EI 3 (23u2 1 3u5 2 3Lu6) 1 FSe 1 FMb L3 2L

(7.3e) (7.3f)

2

6

2

3

5

6

e

b

Equations (7.3), which represent the modified loal stiffness relations for member type 1 (MT 5 1), can be expressed in matrix form as

34 3 Q1 Q2



Q3 Q4 Q5 Q6

5

EI L3

AL2 0 I 0 3 0 0 AL2 2 0 I 0 23 0

0

AL2 I 0 0 AL2 I 0

0

0

0 0 0 0

3L

0

0

23 0

3L 0

0

0

3

23L

2

3L2

23L

43 4 3 4 u1 u2 u3 u4 u5 u6

FAb 3 FSb 2 FMb 2L 0 1 (7.4) FAe 3 FSe 1 FMb 2L 1 FMe 2 FMb 2

or, symbolically, as Q 5 ku 1 Qf with

k5

EI L3

3

AL2 0 I 0 3 0 0 AL2 2 0 I 0 23 0

3L

0

AL2 I 0 0 AL2 I 0

0

0

0 0 0 0

2

0

0

23 0

3L 0

0

0

3

23L

23L

3L2

4



(7.5)

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350  Chapter 7   Member Releases and ­Secondary Effects

and

Qf 5

3 4

FAb 3 FSb 2 FMb 2L 0 FAe  3 FSe 1 FMb 2L 1 FMe 2 FMb 2

(7.6)

The k matrix in Eq. (7.5) and the Qf vector in Eq. (7.6) now represent the modified local stiffness matrix and the modified local fixed-end force vector, respectively, for plane frame members of type 1 (MT 5 1). Members with a Hinge at the End (MT 5 2)    When end e of a member is hinged, then its end moment Q6 (Fig. 6.3(b)) must be 0. By substituting Q6 5 0 into Eq. 7.1(f), and solving the resulting equation for the end rotation u6, we obtain u6 5

3 1 L (2u2 1 u5) 2 u3 2 FMe 2L 2 4EI

(7.7)

Next, we substitute Eq. (7.7) into Eqs. (7.1) to eliminate u6 from the member stiffness relations. This yields Q1 5

EA (u1 2 u4) 1 FAb (7.8a) L

Q2 5

EI 3 (3u2 1 3Lu3 2 3u5) 1 FSb 2 FMe 3 L 2L

Q3 5

EI 1 (3Lu2 1 3L2u3 2 3Lu5) 1 FMb 2 FMe L3 2

(7.8c)

Q4 5

EA (2u1 1 u4) 1 FAe L

(7.8d)

Q5 5

EI 3 (23u2 2 3Lu3 1 3u5) 1 FSe 1 FMe L3 2L

1

2

1

1

(7.8b)

2

2

Q6 5 0

(7.8e) (7.8f )

The foregoing equations can be expressed in matrix form as Q 5 ku 1 Qf

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Section 7.1   Member Releases in Plane Frames and Beams  351

with

k5

EI L3

3

AL2 0 0 I 0 3 3L 0 3L 3L2 2 AL 2 0 0 I 0 23 23L 0 0 0

AL2 I 0 0 AL2 I 0 0

4

0

0

23 23L

0 0

0

0

3 0

0 0

2



(7.9)

and

Qf 5

3 4

FAb 3 FSb 2 FMe 2L 1 FMb 2 FMe 2 FAe  3 FSe 1 FMe 2L 0

(7.10)

The k matrix as given in Eq. (7.9) and the Qf vector in Eq. (7.10) represent the modified local stiffness matrix and fixed-end force vector, respectively, for plane frame members of type 2 (MT 5 2). Members with Hinges at Both Ends (MT 5 3)  If both ends of a member are hinged, then both of its end moments, Q3 and Q6, must be 0. Thus, by substituting Q3 5 0 and Q6 5 0 into Eqs. 7.1(c) and (f), respectively, and solving the resulting simultaneous equations for the end rotations u3 and u6, we obtain u3 5

1 L (2u2 1 u5) 2 (2FMb 2 FMe) L 6EI

(7.11a)

u6 5

1 L (2u2 1 u5) 2 (2FMe 2 FMb) L 6EI

(7.11b)

Next, we substitute the foregoing equations into Eqs. (7.1) to obtain the local stiffness relations for the member type 3: Q1 5

EA (u 2 u4) 1 FAb L 1

Q2 5 FSb 2

(7.12a)

1 (FMb 1 FMe) (7.12b) L

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352  Chapter 7   Member Releases and ­Secondary Effects

(7.12c)

Q3 5 0 EA (2u1 1 u4) 1 FAe L 1 Q5 5 FSe 1 (FMb 1 FMe) L

(7.12d)

Q4 5

(7.12e) (7.12f)

Q6 5 0 Equations (7.12) can be expressed in matrix form as Q 5 ku 1 Qf

F

with

1 0 EA 0 k5 L 21 0 0

0 0 0 0 0 0

0 21 0 0 0 0 0 1 0 0 0 0

0 0 0 0 0 0

G

0 0 0 0 0 0

(7.13)

and

Qf 5

3

4

FAb 1 FSb 2 (FMb 1 FMe) L 0  FAe 1 FSe 1 (FMb 1 FMe) L 0

(7.14)

The foregoing k matrix (Eq. (7.13)) and Qf vector (Eq. (7.14)) represent the modified local stiffness matrix and fixed-end force vector, respectively, for plane frame members of type 3 (MT 5 3). Interestingly, from Eq. (7.13), we observe that the deletion of the third and sixth rows and columns (which correspond to the rotational coordinates of the plane frame members) from the k matrix for MT 5 3 reduces it to the k matrix for members of plane trusses (Eq. (3.27)). It should be realized that, although the number of independent member coordinates is reduced due to member releases, the orders of the modified stiffness matrices k (Eqs. (7.5), (7.9), and (7.13)) and the fixed-end force vector Qf (Eqs. (7.6), (7.10), and (7.14)) are maintained as 6 3 6 and 6 3 1, respectively, with 0 elements in the rows and columns that correspond to the released coordinates. This form of k and Qf eliminates the need to modify the expression for the member transformation matrix, T, derived in Chapter 6 (Eq. (6.19)), and provides an efficient means of incorporating the effect of member releases in the computer program developed in Section 6.7.

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Section 7.1   Member Releases in Plane Frames and Beams  353

Local Stiffness Relations for Beam Members with Hinges As discussed in Chapter 5, in beams subjected to lateral loads, the axial ­displacements of members are 0. Thus, a member of a beam can have up to four degrees of freedom: namely, a translation perpendicular to the member’s centroidal axis and a rotation, at each end. The modified stiffness relations for beam members with releases can be derived by applying the same procedure just used for members of plane frames. However, it is more convenient to obtain the modified member stiffness matrices k for beams by simply deleting the first and fourth rows and columns of the corresponding k matrices for plane-frame members. Similarly, the modified fixed-end force vectors Qf for beam members can be obtained by deleting the first and fourth rows of the corresponding Qf vectors for plane-frame members. Members with a Hinge at the Beginning (MT 5 1)  To obtain the modified stiffness matrix k for beam members of type 1, we delete rows 1 and 4 and columns 1 and 4 from the k matrix given in Eq. (7.5) for plane-frame members of the same type. This yields

3

3 EI 0 k5 3 L 23 3L

0 23 0 0 0 3 0 23L

4

3L 0  23L 3L2

(7.15)

Similarly, the modified fixed-end force vector Qf for beam members of type 1 can be obtained by deleting rows 1 and 4 from the Qf vector given in Eq. (7.6) for plane-frame members of type 1. Thus,

3 4

3 FMb 2L 0 3 FMb FSe 1 2L 1 FMe 2 FMb 2

FSb 2

Qf 5

(7.16)

The rotation u2 of the hinged end b of the member, if desired, can be evaluated by using the following relationship: u2 5

3 1 L (2u1 1 u3) 2 u4 2 FMb 2L 2 4EI

(7.17)

Equation (7.17) is obtained simply by replacing u2, u3, u5, and u6 in Eq. (7.2) with u1, u2, u3, and u4, respectively. Members with a Hinge at the End (MT 5 2)  By deleting the first and fourth rows and columns from the k matrix given in Eq. (7.9), we obtain the modified

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354  Chapter 7   Member Releases and ­Secondary Effects

stiffness matrix for the beam members of type 2 (i.e., MT 5 2):

3

3 3L EI 3L 3L2 k5 3 L 23 23L 0 0

23 23L 3 0

4

0 0 0 0

(7.18)

and by deleting the first and fourth rows from the Qf vector given in Eq. (7.10), we determine the modified fixed-end force vector for beam members of type 2 (MT 5 2):

3 4

3 FMe 2L 1 FMb 2 FMe 2 3 FSe 1 FMe 2L 0

FSb 2

Qf 5

(7.19)

The expression for the rotation u4, of the hinged end e of the member, can be obtained by substituting the subscripts 1, 2, 3, and 4 for the subscripts 2, 3, 5, and 6, respectively, in Eq. (7.7). This yields u4 5

3 1 L (2u1 1 u3) 2 u2 2 FMe 2L 2 4EI

(7.20)

Members with Hinges at Both Ends (MT 5 3)  By deleting the first and fourth rows and columns from the k matrix given in Eq. (7.13), we realize that the modified stiffness matrix for beam members of type 3 (i.e., MT 5 3) is a null matrix; that is, k 5 0 (7.21) which indicates that a beam member hinged at both ends offers no resistance against small end displacements in the direction perpendicular to its centroidal axis. (Recall from Section 3.3 that the members of trusses behave in a similar manner when subjected to lateral end displacements—see Figs. 3.3(d) and (f).) By deleting the first and fourth rows from the Qf vector given in Eq. (7.14), we obtain the modified fixed-end force vector for beam members of type 3 (i.e., MT 5 3):

F G

1 (FMb 1 FMe ) L 0 1 FSe 1 (FMb 1 FMe ) L 0

FSb 2

Qf 5

(7.22)

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Section 7.1   Member Releases in Plane Frames and Beams  355

and from Eqs. (7.11) we obtain the following expressions for the rotations u2 and u4 of the hinged ends b and e, respectively, of the member: u2 5

1 L (2u1 1 u3) 2 (2FMb 2 FMe) L 6EI

(7.23a)

u4 5

1 L (2u1 1 u3) 2 (2FMe 2 FMb) L 6EI

(7.23b)

Procedure for Analysis The analysis procedures developed in Chapters 5 and 6 can be applied to beams and plane frames, respectively, containing member releases, provided that the modified expressions for the stiffness matrices k and fixed-end force vectors Qf , developed in this section are used for the members with releases (i.e., MT 5 1, 2, or 3). Furthermore, in the analysis of plane frames, the global stiffness matrix K for members with releases is now evaluated using the matrix triple product K 5 TT kT (Eq. (6.29)), instead of the explicit form of K given in Eq. (6.31), which is valid only for members with no releases (i.e., MT 5 0). Similarly, the global fixed-end force vector Ff for plane frame members with releases is evaluated using the relationship Ff 5 TTQf (Eq. (6.30)), instead of the explicit form given in Eq. (6.33). The rotations of the hinged member ends, if desired, can be evaluated using Eqs. (7.2), (7.7), and (7.11) when analyzing plane frames, and Eqs. (7.17), (7.20), and (7.23) in the case of beams.

Hinged Joints in Beams and Plane Frames If all the members meeting at a joint are connected to it by hinged connections, then the joint is considered to be a hinged joint. For example, joint 4 of the two-story plane frame shown in Fig. 7.2(a) is considered to be a hinged joint,

5

6 Not a hinged joint Hinged joint

3

4

1

2

(a) Plane Frame

Fig. 7.2

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356  Chapter 7   Member Releases and ­Secondary Effects Y 10

7 8

11

6

6

5

9

6 MT 5 1

3

2

Hinge

4

5

4

5 18

1

4 3

3

MT 5 3 1

Hinged joint

1

2 MT 5 2

2

12 14

15

X

16

X

17

13

16 (b) Analytical Model (11 Degrees of Freedom)

Y 8

11

9

12

6

7

5 Hinge

3

2

10

6 MT 5 1 4

5 6

5 1 3

4

4

MT 5 1

3 1

2 MT 5 2

1

2

13 15 14

18 17

(c) Alternative Analytical Model (12 Degrees of Freedom)

Fig. 7.2  (continued)

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Section 7.1   Member Releases in Plane Frames and Beams  357

because all three members meeting at this joint are attached to it by hinged connections. However, joint 3 of this frame is not considered to be hinged, because only one of the three members meeting at this joint is attached by a hinged connection; the remaining two members are rigidly connected to the joint. As hinged joints cannot transmit any moments and are free to rotate, their rotational stiffnesses are 0. Thus, inclusion of the rotational degrees of freedom of such joints in the analysis causes the structure stiffness matrix S to become singular, with 0 elements in the rows and columns that correspond to the rotational degrees of freedom of the hinged joints. (Recall from your previous course in college algebra that the coefficient matrix of a system of linear equations is considered to be singular if its determinant is 0; and that such a system of equations does not yield a unique solution.) Perhaps the most straightforward and efficient way to remedy this difficulty is to eliminate the rotational degrees of freedom of hinged joints from the analysis by modeling such joints as restrained (or fixed) against rotations. This approach is based on the realization that because hinged joints are not subjected to any moments, their rotations are 0, even though the released ends of the members connected to such a joint can, and do, rotate. In Fig. 7.2(b), the hinged joint 4 of the example frame is modeled using this approach. As indicated in this figure, an imaginary clamp is applied to hinged joint 4 to restrain (or fix) it against rotation, while allowing it to freely translate in any direction. Joint 4, therefore, has two degrees of freedom—the translations in the X and Y directions—which are identified as d4 and d5, respectively; and one restrained coordinate, R18, which represents the reaction moment that develops at the imaginary clamp. It should be realized that because hinged joints are not subjected to any external moments (or couples), the magnitudes of the imaginary reaction moments at such joints are always 0. However, the assignment of restrained coordinate numbers to these imaginary reactions, in accordance with the previously established scheme for numbering structure coordinates, enables us to include the effect of hinged joints in the computer programs developed in Chapters 5 and 6 without any reprogramming. An alternative approach that can be used to overcome the problem of singularity (due to the lack of rotational stiffnesses of a hinged joint) is to model such a joint as rigidly connected to one (and only one) of the members meeting at the joint. This approach is based on the following concept: as no external moment is applied to the hinged joint, and because the moments at the ends of all but one of the members meeting at the joint are 0, the moment at the end of the one member that is rigidly connected to the joint must also be 0, to satisfy the moment equilibrium equation _ M 5 0+ for the joint. This alternative approach is used in Fig. 7.2(c) to model hinged joint 4 of the example frame. As shown in this figure, whereas members 2 and 4 are still attached by hinged connections to joint 4, the third member 5 is now rigidly connected to this joint. Note that because the end of member 5 is now rigidly connected, its member type, which was 3 (i.e., MT 5 3) in the previous analytical model (Fig. 7.2(b)), is now 1 (i.e., MT 5 1), as shown in Fig. 7.2(c). Joint 4 can now be treated as any other rigid joint of the plane frame, and is assigned three degrees of freedom, d4, d5, and d6, as shown in the figure—with d6 representing the rotation of joint 4, which in turn equals the rotation of the end of member 5.

o

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358  Chapter 7   Member Releases and ­Secondary Effects

Needless to state, the two approaches we have discussed for modeling hinged joints yield identical analysis results. However, the first approach generally provides a more efficient analytical model in terms of the number of degrees of freedom of the structure. From Figs. 7.2(b) and (c), we can see that the analytical models of the example frame, based on the first and the alternative approaches, involve 11 and 12 degrees of freedom, respectively.



E x ample 7.1 Determine the joint displacements, member end forces, and support reactions for the plane frame shown in Fig. 7.3(a), using the matrix stiffness method.



S ol u tio n

Analytical Model:  The analytical model of the frame is depicted in Fig. 7.3(b). Since both members 1 and 2, meeting at joint 2, are attached to it by hinged connections, joint 2 is modeled as a hinged joint with its rotation restrained by an imaginary clamp. Thus, joint 2 has two degrees of freedom—the translations in the X and Y directions—which are identified as d1 and d2, respectively. Also, for member 1, MT 5 2, because the end of this member is hinged, whereas MT 5 1 for member 2, which is hinged at its beginning. As far as the modeling of joint 4 is concerned, recall that in Chapters 5 and 6 (e.g., see Examples 5.7 and 6.5) we modeled such a joint as a rigid joint, free to rotate, with its rotation treated as a degree of freedom of the structure. However, in light of the discussion of member releases and hinged joints presented in this section, we can now eliminate the rotational degree of freedom of joint 4 from the analysis by modeling member 3 as hinged at its beginning (i.e., MT 5 1), which allows us to model joint 4 as a hinged joint with its rotation restrained by an imaginary clamp. Note that the end of member 3, which is connected to joint 4, can be considered to be hinged, 300 kN 90.75 kN Hinged joint

19.2 kN/m

5m

2.5 m

2.5 m E, A, I 5 constant E 5 200 GPa A 5 6,500 mm2 I 5 150 (106) mm4 (a) Frame

Fig. 7.3

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Section 7.1   Member Releases in Plane Frames and Beams  359

Y 2

4 2

2 1

5

MT 5 1

3

3

9

1

MT 5 2

3

MT 5 1

1

4

10

6

X

8

12

7

11

4

10

12 11 (b) Analytical Model 1

S5

3

2

720 1 260,000 0 2260,000 0 0

3

0 260,000 1 720 0 2720 3,600

0 2720 0 720 1 260,000 23,600

1

2

3

4

5

260,720 0 5 2260,000 0 0

0 260,720 0 2720 3,600

2260,000 0 260,720 0 3,600

0 2720 0 260,720 23,600

0 3,600 3,600 23,600 36,000

3

5

4

2260,000 0 260,000 1 720 0 3,600

3

1 2 3 4 5

Pf 5

0 3,600 3,600 23,600 18,000 1 18,000

3 3 236 93.75 0 206.25 2281.25

3

1 2 3 4 5

1 2 3 4 5

(c) Structure Stiffness Matrix and Fixed-Joint Force Vector

Fig. 7.3  (continued)

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360  Chapter 7   Member Releases and ­Secondary Effects 89.17

210.83

300 kN 60.83

29.92

60.83 2

304.16

60.83

304.16 89.17

19.2 kN/m

210.83

1

3

389.59

125.92

60.83

89.17

210.83 (d) Member Local End Forces

3 3 2125.92 kN 89.17 kN

389.59 kN·m

R5

0 260.83 kN 210.83 kN 0

6 7 8 9 10 11 12

(e) Support Reaction Vector

300 kN 90.75 kN 2

3

1

4

19.2 kN/m

125.92

60.83 389.59 89.17

210.83

(f) Support Reactions

Fig. 7.3  (continued)

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Section 7.1   Member Releases in Plane Frames and Beams  361

because there is only one member connected to the joint that is not subjected to any external couple. Furthermore, the joint is supported by a hinged support that cannot exert any reaction moment at the joint. Thus, the moment at the end of member 3, which is connected to joint 4, must be 0; therefore, the member end can be treated as a hinged end. With its rotation restrained by the imaginary clamp, and its translations in the X and Y directions restrained by the actual hinged support, joint 4 is modeled as if it is attached to a fixed support, with no degrees of freedom, as depicted in Fig. 7.3(b). Thus, the entire frame has five degrees of freedom and seven restrained coordinates, as shown in Fig. 7.3(b). Structure Stiffness Matrix and Fixed-Joint Force Vector: Member 1 (MT 5 2)  Because MT 5 2 for this member, we use Eqs. (7.9) and (7.10) to determine its local stiffness matrix k and fixed-end force vector Qf , respectively. Thus, by substituting E 5 200 GPa, A 5 0.0065 m2, I 5 150 (1026) m4, and L 5 5 m into Eq. (7.9), we obtain

k1 5

F

260,000 0 0 2260,000 0 0

0 720 3,600 0 2720 0

0 3,600 18,000 0 23,600 0

2260,000 0 0 260,000 0 0

0 2720 23,600 0 720 0

G

0 0 0 0 0 0



(1)

To determine the local fixed-end force vector due to the member load w 5 19.2 kN/m, we first evaluate the fixed-end axial forces, shears, and moments in a corresponding rigidly connected member by using the expressions given inside the front cover: FAb 5 FAe 5 0 19.2(5) FSb 5 FSe 5 5 48 kN 2 19.2(5)2 FMb 5 2FMe 5 5 40 kN ? m 12 Next, we substitute the foregoing values into Eq. (7.10) to obtain the local fixed-end force vector for the released member under consideration:

Qf 1 5

3 43 4 0 3(240) 48 2 2(5) 1 40 2 (240) 2 0 3(240) 48 1 2(5) 0

0

60 kN

5

60 kN ? m



(2)

0

36 kN 0

To obtain the member’s stiffness matrix K and the fixed-end force vector Ff in the global coordinate system, we first substitute its direction cosines, cos u 5 0 and sin u 5 1, into Eq. (6.19), to obtain the transformation matrix.

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362  Chapter 7   Member Releases and ­Secondary Effects

T1 5 T3 5

F

0 21 0 0 0 0

1 0 0 0 0 0

0 0 0 0 1 0 0 0 0 21 0 0

0 0 0 1 0 0

G

0 0 0  0 0 1

(3)

Next, by substituting k1 (Eq. (1)) and T1 (Eq. (3)) into the relationship K 5 TTkT (Eq. (6.29)), and performing the necessary matrix multiplications, we obtain

K1 5

F

G

6 7 8 1 2 9 720 0 23,600 2720 0 0 0 260,000 0 0 2260,000 0 23,600 0 18,000 3,600 0 0 2720 0 3,600 720 0 0 0 2260,000 0 0 260,000 0 0 0 0 0 0 0

6 7 8 1 2 9

Note that K1 is symmetric. Similarly, by substituting Qf1 (Eq. (2)) and T1 (Eq. (3)) into the relationship Ff 5 TTQf (Eq. (6.30)), we obtain

Ff 1 5

FG 260 0 60 236 0 0

6 7 8 1 2 9

From Fig. 7.3(b), we observe that the code numbers for member 1 are 6, 7, 8, 1, 2, 9. Using these code numbers, we store the pertinent elements of K1 and Ff 1 in their proper positions in the 5 3 5 structure stiffness matrix S and the 5 3 1 structure fixedjoint force vector Pf , respectively, as shown in Fig. 7.3(c). Member 2 (MT 5 1)   No coordinate transformations are needed for this horizontal member; that is, T2 5 I, K2 5 k2, and Ff 2 5 Qf 2. As MT 5 1, we use Eq. (7.5) to obtain

F

G

1 2 9 3 4 5 260,000 0 0 2260,000 0 0 1 0 720 0 0 2720 3,600 2 0 0 0 0 0 0 9 K2 5 k2 5 k3 5 2260,000 0 0 260,000 0 0 3 0 27 20 0 0 720 23,600 4 0 3,600 0 0 23,600 18,000 5 (4) Using the fixed-end force expressions given for loading type 1 for the 300 kN member load, we obtain FAb 5 FAe 5 0 FSb 5 FSe 5 150 kN FMb 5 2FMe 5 187.5 kN ? m

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Section 7.1   Member Releases in Plane Frames and Beams  363

Substitution of the foregoing values into Eq. (7.6) yields 0 3(187.5) 150 2 0 1 2(5) kN 93.75 2 0 0 9 0 5 0 3 3(187.5) (5) Ff 2 5 Qf2 5 150 1 206.25 kN 4 2(5)  2281.25 kN ? m 5 1 2187.5 2 (187.5) 2 The relevant elements of K2 and Ff 2 are stored in S and Pf , respectively, using the member code numbers 1, 2, 9, 3, 4, 5.

3 3F G

Member 3 (MT 5 1)  As E, A, I, L, and MT for member 3 are the same as for member 2, k3 5 k2 as given in Eq. (4). Also, since the member is not subjected to any loads, Ff 3 5 Qf 3 5 0 Furthermore, since the direction cosines of member 3 are identical to those of member 1, T3 5 T1 as given in Eq. (3). To determine the member global stiffness matrix, we substitute k3 from Eq. (4), and T3 from Eq. (3), into the relationship K 5 TT kT (Eq. (6.29)), and perform the necessary matrix multiplications. This yields

K3 5

F

10 720 0 0 2720 0 23,600

11 0 260,000 0 0 2260,000 0

12 0 0 0 0 0 0

3 2720 0 0 720 0 3,600

4 0 2260,000 0 0 260,000 0

5 23,600 0 0 3,600 0 18,000

G

10 11 12 3 4 5

The pertinent elements of K3 are stored in S using the member code numbers 10, 11, 12, 3, 4, 5. The completed structure stiffness matrix S and the structure fixed-joint force vector Pf are given in Fig. 7.3(c). Joint Load Vector:  By comparing Figs. 7.3(a) and (b), we write

3 4

90.75 1 0 2 0 3 P5 0 4 0 5 Joint Displacements:  By solving the system of simultaneous equations representing the structure stiffness relationship P 2 Pf 5 Sd (Eq. (6.42)), we obtain the following joint displacements: 0.091553 m 1 20.000343 m 2 d5 0.091319 m 3 Ans 20.000811 m 4 20.001366 rad 5

3

4

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364  Chapter 7   Member Releases and ­Secondary Effects Member End Displacements and End Forces:

FG FG F G F G F G F G

Member 1 (MT 5 2)  Using the member code numbers 6, 7, 8, 1, 2, 9, we write the global end displacement vector as

v1 5

y1 y2 y3 y4 y5 y6

6 7 8 5 1 2 9

0 0 0 d1 d2 0

5

0 0 0  0.091553 20.000343 0

(6)

Next, we obtain the local end displacement vector u by substituting the foregoing v1 and T1 (Eq. (3)) into the relationship u 5 Tv (Eq. (6.20)). This yields 0 0 0 u1 5 T1v1 5  20.000343 20.091553 0

(7)

We can now determine the member local end forces Q by substituting u1, k1 (Eq. (1)), and Qf 1 (Eq. (2)) in the member stiffness relationship Q 5 ku 1 Qf (Eq. (6.4)). Thus,

Q1 5 k1u1 1 Qf1 5

89.17 kN 125.92 kN 389.59 kN ? m  289.17 kN 229.92 kN 0

Ans

These end forces for member 1 are depicted in Fig. 7.3(d). To generate the support reaction vector R for the frame, we evaluate the global end forces F for the member by applying Eq. (6.23) as

F1 5 TT1 Q1 5

2125.92 89.17 389.59 29.92 289.17 0

6 7 8 1 2 9

The pertinent elements of F1 are stored in R, as shown in Fig. 7.3(e). It should be realized that because the member end displacement vectors v and u are based on the compatibility of the joint and the member end displacements, such vectors (in the case of members with releases) contain 0 elements in the rows that correspond to the rotations of the released (or hinged) member ends. Thus, we can see from Eqs. (6) and (7) that the vectors v1 and u1 for member 1 (with MT 5 2) contain 0 elements in their sixth rows. We can evaluate the rotation u6 of the released end of this member by using Eq. (7.7), as 5(240) 3 1 (0 2 0.091553) 2 (0) 2 2(5) 2 4(2)108(150)1026 5 20.0258 rad 5 0.0258 rad

u6 5

{

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Section 7.1   Member Releases in Plane Frames and Beams  365

Because the member end rotations are the same in the local and global coordinate systems,

{

y6 5 u6 5 0.0258 rad

FG FG F G F G FG FG F G F G F G

Member 2 (MT 5 1)

u2 5 v2 5

y1 y2 y3 y4 y5 y6

1 2 95 3 4 5

d1 d2 0 d3 d4 d5

5

0.091553 20.000343 0 0.091319 20.000811 20.001366

By using k2 from Eq. (4) and Qf 2 from Eq. (5), we compute the member end forces to be

F2 5 Q2 5 k2u2 1 Qf 2 5

60.83 kN 89.17 kN 0 260.83 kN 210.83 kN 2304.16 kN ? m

1 2 9  3 4 5

Ans

The rotation, u3, of the released end of this member, if desired, can be calculated by using Eq. (7.2). Member 3 (MT 5 1)

v3 5

y1 y2 y3 y4 y5 y6

10 11 12 5 3 4 5

0 0 0 d3 d4 d5

5

0 0 0 0.091319 20.000811 20.001366

By using T3 from Eq. (3), we obtain 0 0 0 u3 5 T3v3 5 20.000811 20.091319 20.001366

Using k3 from Eq. (4) and Qf 3 5 0, we obtain the member local end forces as

Q3 5 k3u3 5

210.83 kN 60.83 kN 0 2210.83 kN 260.83 kN 304.16 kN ? m



Ans

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366  Chapter 7   Member Releases and ­Secondary Effects

F G

The member local end forces are shown in Fig. 7.3(d).

F3 5 T3T Q3 5

260.83 210.83 0 60.83 2210.83 304.16

10 11 12 3 4 5

Support Reactions:  See Figs. 7.3(e) and (f).

Ans

7.2 Computer Implementation of

Analysis for Member Releases The computer programs developed in Chapters 5 and 6 for the analysis of rigidly connected beams and plane frames can be extended, with only minor modifications, to include the effects of member releases. In this section, we discuss the modifications in the program for the analysis of plane frames (Section 6.7) that are necessary to consider member releases. While the beam analysis program (Section 5.8) can be modified in a similar manner, the implementation of these modifications is left as an exercise for the reader. The overall organization and format of the plane frame analysis program, as summarized in Table 6.1, remains the same when considering member releases. However, parts V, IX, and XII, and the subroutines MSTIFFL and MFEFLL, must be revised as follows: Member Data (Part V)  This part of the program (see flowchart in Fig. 4.3(e)) should be modified to include the reading and storing of the member type, MT, for each member of the frame. The number of columns of the member data matrix MPRP should be increased from four to five, with the value of MT (5 0, 1, 2, or 3) for a member i stored in the fifth column of the ith row of MPRP. Generation of the Structure Stiffness Matrix and Equivalent Joint Load ­ ector (Part IX ), and Calculation of Member Forces and Support Reactions V (Part XII )    In parts IX and XII of the computer program (see flowcharts in Figs. 6.24 and 6.31, respectively) a statement should be added to read, for each member, the value of MT from the fifth column of the MPRP matrix (i.e., MT 5 MPRP (IM, 5)), before the subroutines MSTIFFL and MFEFLL are called to form the member local stiffness matrix BK, and the local fixed-end force vector QF, respectively. Subroutine MSTIFFL   A flowchart for programming the modified version of this subroutine is given in Fig. 7.4 on the next page. As this flowchart indicates, the subroutine calculates the BK matrix using Eq. (6.6) if MT equals 0, Eq. (7.5) if MT equals 1, Eq. (7.9) if MT equals 2, or Eq. (7.13) if MT equals 3. Subroutine MFEFLL   A flowchart of the modified version of this subroutine is shown in Fig. 7.5 on page 368. The subroutine first calculates the fixed-end forces

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Section 7.3   Support Displacements  367 Start Subroutine MSTIFFL Arguments: E, A, ZI, BL, NCJT, MT, BK Initialize all elements of BK to zero Z 5 E*A/BL BK(1, 1) 5 Z, BK(4, 1) 5 2Z BK(1, 4) 5 2Z, BK(4, 4) 5 Z

MT 5 0?

yes

no

MT 5 1?

yes

Z 5 E*ZI/(BL^3) BK(2, 2) 5 12*Z, BK(3, 2) 5 6*BL*Z BK(5, 2) 5 212*Z, BK(6, 2) 5 6*BL*Z BK(2, 3) 5 6*BL*Z, BK(3, 3) 5 4*(BL^2)*Z BK(5, 3) 5 26*BL*Z, BK(6, 3) 5 2*(BL^2)*Z BK(2, 5) 5 212*Z, BK(3, 5) 5 26*BL*Z BK(5, 5) 5 12*Z, BK(6, 5) 5 26*BL*Z BK(2, 6) 5 6*BL*Z, BK(3, 6) 5 2*(BL^2)*Z BK(5, 6) 5 26*BL*Z, BK(6, 6) 5 4*(BL^2)*Z Z 5 E*ZI/(BL^3), BK(2, 2) 5 3*Z BK(5, 2) 5 23*Z, BK(6, 2) 5 3*BL*Z BK(2, 5) 5 23*Z, BK(5, 5) 5 3*Z BK(6, 5) 5 23*BL*Z, BK(2, 6) 5 3*BL*Z BK(5, 6) 5 23*BL*Z, BK(6, 6) 5 3*(BL^2)*Z

no

MT 5 2?

yes

no

Z 5 E*ZI/(BL^3), BK(2, 2) 5 3*Z BK(3, 2) 5 3*BL*Z, BK(5, 2) 5 23*Z BK(2, 3) 5 3*BL*Z, BK(3, 3) 5 3*(BL^2)*Z BK(5, 3) 5 23*BL*Z, BK(2, 5) 5 23*Z BK(3, 5) 5 23*BL*Z, BK(5, 5) 5 3*Z

End Subroutine MSTIFFL Return to calling program

Fig. 7.4  Flowchart of Subroutine MSTIFFL for Determining Member Local Stiffness ­Matrix for Plane Frames with Member Releases

(FAB, FSB, FMB, FAE, FSE, and FME) in a corresponding rigidly connected member using the equations given inside the front cover. The QF vector is then formed in accordance with Eq. (6.15) if MT 5 0, Eq. (7.6) if MT 5 1, Eq.  (7.10) if MT 5 2, or Eq. (7.14) if MT 5 3.

7.3 

Support Displacements The effect of small support displacements, due to weak foundations or other causes, can be conveniently included in the matrix stiffness method of analysis using the concept of equivalent joint loads [14]. This approach, which was discussed in Sections 5.6 and 6.5 for the case of member loads, essentially involves applying the prescribed external action (such as a system of member loads, support settlements, etc.) to the structure, with all of its joint displacements restrained by imaginary restraints. The structure fixed-joint forces that develop in the hypothetical fixed structure, as reactions at the imaginary

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368  Chapter 7   Member Releases and ­Secondary Effects Start Subroutine MFEFLL Arguments: IML, BL, MT, MP, PM, QF Initialize FAB, FSB, FMB, FAE, FSE, FME to zero LDTYPE 5 MP(IML, 2) Calculate FAB, FSB, FMB, FAE, FSE and FME, for the value of LDTYPE, using the equations given inside the front cover (See flowchart in Fig. 6.28 for details.) QF(1) 5 QF(1) 1 FAB QF(4) 5 QF(4) 1 FAE

If MT 5 0?

yes

no If MT 5 1?

QF(2) 5 QF(2) 1 FSB QF(3) 5 QF(3) 1 FMB QF(5) 5 QF(5) 1 FSE QF(6) 5 QF(6) 1 FME

yes

QF(2) 5 QF(2) 1 FSB 2 3*FMB/(2*BL) QF(5) 5 QF(5) 1 FSE 1 3*FMB/(2*BL) QF(6) 5 QF(6) 1 FME 2 (FMB/2)

yes

QF(2) 5 QF(2) 1 FSB 2 3*FME/(2*BL) QF(3) 5 QF(3) 1 FMB 2 (FME/2) QF(5) 5 QF(5) 1 FSE 1 3*FME/(2*BL)

yes

QF(2) 5 QF(2) 1 FSB 2 (FMB 1 FME)/BL QF(5) 5 QF(5) 1 FSE 1 (FMB 1 FME)/BL

no If MT 5 2? no If MT 5 3? no End Subroutine MFEFLL Return to calling program

Fig. 7.5  Flowchart of Subroutine MFEFLL for Determining Member Local Fixed-End Force Vector for Plane Frames with Member Releases

r­estraints (i.e., at the location and in the direction of each degree of freedom of the actual structure), are then evaluated. The structure fixed-joint forces, with their directions reversed, now represent the equivalent joint loads, in the sense that when applied to the actual structure, they cause the same joint displacements as the original action (i.e., member loads, support settlements, etc.). Once the response of the structure to the equivalent joint loads has been determined, the actual structural response due to the original action is obtained by superposition of the responses of the fixed structure to the original action and the actual structure to the equivalent joint loads. The foregoing approach is illustrated in Fig. 7.6 on the next page for the case of support displacements, using an arbitrary three-degree-of-freedom frame as an example. Figure 7.6(a) shows the actual frame, whose supports 3 and 4

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Section 7.3   Support Displacements  369 d1

1 1

d2

19 2

2

3 d3

3

4

D1

D2 39 49

(a) Actual Frame Subjected to Support Settlements

5 Pf 2 1 Pf 3 Pf1 1 3

2

2

4

3 D1

49

D2

39 (b) Fixed Frame Subjected to Support Settlements

1 d1 Pf 2 Pf 3

Pf1 1

d2

1 19 2

2

3

3 d3

4

(c) Actual Frame Subjected to Equivalent Joint Loads

Fig. 7.6

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370  Chapter 7   Member Releases and ­Secondary Effects

undergo small settlements D1 and D2, respectively, causing the displacements d1, d2, and d3 of free joint 1. To determine the response (i.e., joint displacements, member forces, and support reactions) of the frame to the support settlements, we first restrain all the joint displacements of the frame by applying an imaginary restraint at joint 1, and subject this completely fixed frame to the prescribed support settlements D1 and D2, as shown in Fig. 7.6(b). As joint 1 of the frame, initially free, is now restrained from translating and rotating by the imaginary restraint, the structure fixed-joint forces Pf 1, Pf 2, and Pf 3 develop at the imaginary restraint at this joint. (A procedure for evaluating structure fixed-joint forces due to support settlements is developed in a subsequent part of this section.) Next, as shown in Fig. 7.6(c), we apply the foregoing structure fixed-joint forces Pf 1, Pf 2, and Pf 3, with their directions reversed, as external loads at joint 1 of the actual frame. A comparison of Figs. 7.6(a), (b), and (c) indicates that the superposition of the support settlements and joint loads applied to the frame in Figs. 7.6(b) and (c) yields only the support settlements the frame is subjected to in Fig. 7.6(a), because each of the fixed-joint forces in Fig. 7.6(b) is canceled by its negative counterpart applied as a load in Fig. 7.6(c). Thus, according to the principle of superposition, the joint displacements d1, d2, and d3 of the frame due to the support settlements D1 and D2 (Fig. 7.6(a)) must equal the algebraic sums of the corresponding joint displacements of the fixed frame subjected to the support settlements (Fig. 7.6(b)), and the actual frame, subjected to no settlements, but to the negatives of the fixed-joint forces (Fig. 7.6(c)). However, since the ­displacements of joint 1 of the fixed frame (Fig. 7.6(b)) are 0, the joint displacements of the frame subjected to the negatives of fixed-joint forces (Fig. 7.6(c)) must equal the actual joint displacements d1, d2, and d3 of the frame due to the support settlements D1 and D2 (Fig. 7.6(a)). In other words, the negatives of the structure fixed-joint forces cause the same displacements at the locations and in the directions of the frame’s degrees of freedom as the prescribed support settlements; and, in that sense, such forces can be considered as equivalent joint loads. It should be realized that the foregoing equivalency is valid only for joint displacements. From Fig. 7.6(b), we can see that the end displacements of the members of the fixed frame are not 0. Therefore, the member end forces and support reactions of the actual frame due to settlements (Fig. 7.6(a)) must be obtained by superposition of the corresponding responses of the fixed frame (Fig. 7.6(b)) and the actual frame subjected to the equivalent joint loads (Fig. 7.6(c)). It may be recalled from Chapters 5 and 6 that, in the case of member loads, the response of the fixed structure was evaluated using the fixed-end force expressions for various types of member loads, as given inside the front cover; and that the fixed-joint force vector Pf was obtained by algebraically adding the fixed-end forces of members meeting at the joints (via the member code numbers). A procedure for evaluating the member fixed-end forces, and the structure fixed-joint forces, due to support settlements is presented in the following paragraphs. With the fixed-joint forces known, the response of the structure to the equivalent joint loads can be determined, using the standard matrix stiffness methods described in Chapters 3 through 6.

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Section 7.3   Support Displacements  371

Evaluation of Structure Fixed-Joint Forces Due to ­Support Displacements We begin by establishing a systematic way of identifying the support displacements of a structure. For that purpose, let us reconsider the three-degree-offreedom frame of Fig. 7.6(a), subjected to the support settlements D1 and D2. The frame is redrawn in Fig. 7.7(a), with its analytical model depicted in Fig. 7.7(b). From Fig. 7.7(b), we observe that the frame has nine support ­reactions, which are identified by the restrained coordinate numbers 4 through 12. Thus, the frame can be subjected to a maximum of nine support displacements. The numbers assigned to the restrained coordinates are also used to identify the support displacements, with a support displacement at the location and in the direction of a support reaction Ri denoted by the symbol dsi. Thus, a comparison of Figs. 7.7(a) and (b) shows that for the frame under consideration, (7.24)

ds8 5 2D1 and ds11 5 2D2 d1

1 1

d2

19 2

2

3 d3

3

4

D1

D2 39 49 (a) Frame Subjected to Support Settlements

Y

2 3

5 1

1 1 3

4 6

2

2 4

3

10

7 9 8

X

12 11

(b) Analytical Model

Fig. 7.7

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372  Chapter 7   Member Releases and ­Secondary Effects Pf 2 Y

1

Pf 3

Pf 1

1

3 2

2

4

3

X

D1

39

D2

49 (c) Fixed Frame Subjected to Support Settlements

5 (3) Ffs2

(2) Ffs5 (2) Ffs6

1

1

(2) Ffs4

1

1

(3) Ffs1 (3) Ffs3

3

2 2

4

3 (2) vfs2 5 2D1

39

(2) Ffs1

(3) Ffs4

(2) Ffs3

49

(3) vfs5 5 2D2

(3) Ffs6

(2) Ffs2

(d) Member Global Fixed-End Displacements and Forces

(3) Ffs5

Pf 2 Pf 3

(2)

Ffs6

1

Pf 1

(2) Ffs4

(3) Ffs1

F (3) fs3 (2) Ffs5

(3) Ffs2

(e) Free Body of Joint 1—Fixed Frame

Fig. 7.7  (continued)

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Section 7.3   Support Displacements  373

with the remaining seven support displacements being 0. The negative signs assigned to the magnitudes D1 and D2 of ds8 and ds11 indicate that these support displacements occur in the negative Y (i.e., downward) direction. To illustrate the process of evaluating a structure’s fixed-joint forces due to a prescribed set of support settlements, we restrain the joint displacements of the example frame by applying an imaginary restraint at joint 1, and subject this hypothetical completely fixed frame to the given support settlements D1 and D2, as shown in Fig. 7.7(c). The structure fixed-joint forces that develop at the imaginary restraint at joint 1 are denoted by Pf 1, Pf 2, and Pf 3 in the figure, with the fixed-joint force corresponding to an ith degree of freedom denoted by Pf  i . To evaluate the fixed-joint forces, we first determine the displacements that the support settlements D1 and D2 cause at the ends of the members of the fixed frame. The free-body diagrams of the three members of the hypothetical fixed frame are depicted in Fig. 7.7(d). From Figs. 7.7(c) and (d), we observe that, while the settlements of supports 3 and 4 do not cause any displacement in member 1, they induce downward displacements of magnitudes D1 and D2, respectively, at the lower ends of members 2 and 3. Note that all other member end displacements are 0, because all the joint displacements of the fixed frame are 0, with the exception of the known support settlements. Thus, the end displacements of members 2 and 3, respectively, can be expressed in vector form, as



vf s2 5

34 0 2D1 0 0 0 0

7 8 9 1 2 3

and

vf s3 5

34 0 0 0 0 2D2 0

1 2 3  10 11 12

(7.25)

in which vf s represents the member global fixed-end displacement vector due to support displacements. The foregoing member global fixed-end displacement vectors can be directly generated using the member code numbers, which define the member compatibility equations. For example, from Fig. 7.7(b), we can see that the code numbers for member 2 are 7, 8, 9, 1, 2, 3. By comparing these member code numbers with the support displacements of the frame, ds8 5 2D1 and ds11 5 2D2 (see Eq. (7.24)), we conclude that all the elements of vf s2 are zero, with the exception of the element in the second row that equals 2D1 (i.e., yf (2) 5 ds8 5 2D1). Similarly, by examining the code numbers 1, 2, s2 3, 10, 11, 12 of member 3, we realize that the only nonzero element of vf s3 is in the fifth row and it equals 2D2 (i.e., yf (3) 5 ds11 5 2D2). s5 Once the member fixed-end displacement vectors vf s have been determined, they are used to calculate the corresponding member global fixed-end force vectors due to support displacements, Ff s, through the member global stiffness relationship (Eq. (6.28)) derived in Chapter 6. By substituting F 5 Ff s, v 5 vf s, and Ff 5 0 into Eq. (6.28), we obtain the following relationship between Ff s and vf s:

Ff s 5 Kvf s



(7.26)

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374  Chapter 7   Member Releases and ­Secondary Effects

With the member global fixed-end forces known, the structure fixed-joint forces due to the support displacements can be evaluated using joint equilibrium equations. Thus, for the example frame, we apply the three equations of equilibrium, FX 5 0, FY 5 0, and M 5 0, to the free body of joint 1 (see Fig. 7.7(e)) to obtain the following expressions for the fixed-joint forces in terms of the member fixed-end forces:

o

o

o



Pf 1 5 Ff (2) 1 Ff (3)  s4 s1

(7.27a)



Pf 2 5 Ff (2) 1 Ff (3)  s5 s2

(7.27b)



Pf3 5 Ff (2) 1 Ff (3)  s6 s3

(7.27c)

The structure fixed-joint force vector for the support settlements of the example frame can, therefore, be expressed as

3

4

Ff (2) 1 Ff (3) s4 s1



Pf 5 Ff (2) 1 Ff (3) s5 s2  (3) Ff (2) 1 F s6 f s3

(7.28)

As demonstrated in Chapters 5 and 6 for the case of member loads, the structure fixed-joint force vectors Pf can be conveniently generated by employing the member code number technique. The application of the technique remains the same in the case of support displacements, except that the elements of the member global fixed-end force vectors due to support displacements, Ffs, must now be added into Pf . When a structure is subjected to more than one type of action requiring evaluation of fixed-joint forces (e.g., member loads and support settlements), then the fixed-joint forces representing different types of actions can be conveniently combined into a single Pf vector. For example, in the case of a frame subjected to member loads and support settlements, the elements of the two types of member fixed-end force vectors—that is, due to member loads (Ff ) and support displacements (Ff s)—can be stored in a single Pf vector using the member code number technique. Once the structure fixed-joint forces due to support displacements have been evaluated, the structure stiffness relations P 2 Pf 5 Sd (Eq. (6.42)) can be solved for the unknown joint displacements d. With d known, the member global end displacement vector v for each member is determined by applying the compatibility equations defined by its code numbers. For members that are attached to the supports that undergo displacements, the displacements of the supported ends, due to the corresponding support displacements, must be included in the member global end displacement vectors v. The inclusion of support displacements in the v vectors automatically adds the response of the fixed structure to support settlements (see, for example, Fig. 7.6(b)) into the analysis, thereby enabling us to evaluate the member local and global end forces, and support reactions, using the procedures developed in previous chapters.

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Section 7.3   Support Displacements  375

Procedure for Analysis Based on the discussion presented in this section, we can develop the following step-by-step procedure for the matrix stiffness analysis of framed structures due to support displacements. 1. Prepare an analytical model of the structure, and determine its structure stiffness matrix S. If the structure is subjected to member loads, then evaluate its fixed-joint force vector Pf due to the member loads. If the structure is subjected to joint loads, then form its joint load vector P. 2. Calculate the structure fixed-joint force vector Pf (NDOF 3 1) due to the given support displacements. If a Pf vector was formed in step 1 for member loads, then store the member fixed-end forces due to support displacements in the previously formed Pf vector. For each member that is attached to the supports that undergo displacements, perform the following operations: a. Identify the member code numbers, and form the member global fixedend displacement vector, vfs, from the specified support displacements, dsi. Note that the support translations are considered positive when in the positive directions of the global X and Y axes, and support rotations are considered positive when counterclockwise. For beams, form the member local fixed-end displacement vector due to support displacements, ufs, using the same process. b. Evaluate the member global fixed-end force vector due to support displacements, Ffs, using the relationship Ffs 5 Kvfs (Eq. (7.26)). For beams, evaluate the member local fixed-end force vector due to support displacements, Qfs, using the relationship Qfs 5 kufs. c. Using member code numbers, store the pertinent elements of Ffs, or Qfs for beams, in their proper positions in the structure fixed-joint force vector Pf . 3. Determine the unknown joint displacements d by solving the structure stiffness relationship, P 2 Pf 5 Sd. 4. Compute member end displacements and end forces, and support reactions. For each member of the structure, carry out the following steps: a. Obtain member end displacements in the global coordinate system, v, from the joint displacements d and the specified support displacements dsi, by using the member code numbers. For beams, obtain the member local end displacements, u, using the same process, and then go to step 4c. b. Determine the member end displacements in the local coordinate system, u, by using the transformation relationship u 5 Tv. c. Calculate the member end forces in the local coordinate system, Q, by using the stiffness relationship Q 5 ku 1 Qf . If the member is not subjected to any member loads, then Qf 5 0. For beams, go to step 4e. d. Compute the member end forces in the global coordinate system, F, using the transformation relationship F 5 TTQ.

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376  Chapter 7   Member Releases and ­Secondary Effects

e. If the member is attached to a support joint, then use the member code numbers to store the pertinent elements of F, or Q for beams, in their proper positions in the support reaction vector R.

E x ample 7.2 Determine the joint displacements, member axial forces, and support reactions for the

plane truss shown in Fig. 7.8(a) due to a settlement of 10 mm of support 4. Use the matrix stiffness method.

,00

,00

8m 2

m

0m )

3

2)

0 mm

2

(3,00

1

(4

0m

m2 )

1

(4



2

4

3

6m

4m

2m E 5 200 GPa (a) Truss 2

Y

1

1

1

3

2

4

3

2

5

6 (b) Analytical Model

3

7

4

X

8

Fig. 7.8

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Section 7.3   Support Displacements  377

3 3

265.479 287.306 265.479 R5 261.92 130.96 2174.61

3 4 5 6

kN

7 8

(c) Support Reaction Vector 1 Y

1

3

2

65.479 kN

2

87.306 kN

65.479 kN

3

261.92 kN (d) Support Reactions

4

130.96 kN

X

174.61 kN

Fig. 7.8  (continued)

S ol u tio n

This truss was analyzed in Example 3.8 for joint loads. In this example, we use the same analytical model of the truss, so that the various member and structure matrices calculated in the previous example can be reused herein. Analytical Model:  See Fig. 7.8(b).  The truss has two degrees of freedom and six restrained coordinates. Structure Stiffness Matrix:  From Example 3.8, S5

61,880 3217,120

217,120 170,881

4 kN/m

(1)

Joint Load Vector:  As the truss is not subjected to any loads, P 5 0 (2) Structure Fixed-Joint Force Vector Due to Support Displacements:  From the analytical model of the truss in Fig. 7.8(b), we can see that the given 10 mm settlement (i.e., vertically downward displacement) of support joint 4 occurs at the location and in the direction of the reaction R8. Thus, the given support displacement can be expressed as ds8 5 20.01 m From Fig. 7.8(b), we observe that member 3 is the only member attached to support 4 that undergoes displacement. Thus, using the member’s code numbers 7, 8, 1, 2, we form its global fixed-end displacement vector due to support displacement as

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378  Chapter 7   Member Releases and ­Secondary Effects

vfs3 5

34 34 3 4 7 8 5 1 2

yfs1 yfs2 yfs3 yfs4

0 ds8 0 0

5

0 20.01 0 0

m

Next, we evaluate the global fixed-end force vector Ffs3 due to the support settlement for member 3, using the member global stiffness matrix K3 calculated in Example 3.8, and Eq. (7.26). Thus,

3

28,800 238,400 Ffs3 5 K3vfs3 5 228,800 38,400

238,400 51,200 38,400 251,200

38,400 251,200 238,400 51,200

228,800 38,400 28,800 238,400

43 4 3 4 0 20.01 0 0

384 2512 5 2384 512

7 8 kN 1 2

From the member code numbers, which are written on the right side of Ffs3, we realize that the elements in the third and fourth rows of Ffs3 should be stored in rows 1 and 2, respectively, of the Pf vector. Thus, the structure fixed-joint force vector, due to the support settlement, is given by Pf 5

3

4

2384 1 kN 512 2

(3)

Joint Displacements: By substituting P (Eq. (2)), Pf (Eq. (3)), and S (Eq. (1)) into the structure stiffness relationship, we write P 2 Pf 5 Sd

304 2 3 0

4 3

4 3

2384 384 61,880 5 5 512 2512 217,120

217,120 170,881

4 3d 4 d1 2

By solving these equations, we determine the joint displacements to be d5

320.0024422 4 2 m 0.0055299

1

Ans

Member End Displacements and End Forces: Member 1  Using the member code numbers 3, 4, 1, 2, we write the global end displacement vector as

v1 5

34 34 3 y1 y2 y3 y4

3 4 5 1 2

0 0 d1 d2

5

0 0 0.0055299 20.0024422

4

m

Next, we determine the member local end displacement vector u1, using the transformation matrix T1 from Example 3.8, and Eq. (3.63), as

3

0.6 20.8 u1 5 T1v1 5 0 0

0.8 0 0.6 0 0 0.6 0 20.8

43

0 0 0.8 0.6

0 0 0.0055299 20.0024422

43 5

0 0 0.0013642 20.0058892

4

m

Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 7.3   Support Displacements  379

We can now calculate the member local end forces Q1 by applying the member stiffness relationship, Q 5 ku (Eq. (3.7)). Thus, using k1 from Example 3.8, we obtain

3

80,000 0 Q1 5 k1u1 5 280,000 0

43

0 280,000 0 0 0 80,000 0 0

0 0 0 0

43 4

0 0 0.0013642 20.0058892

2109.13 0 5 109.13 0

kN

Recall from Chapter 3 that the member axial force equals the first element of the Q1 vector; that is, Qa1 5 2109.13 kN in which the negative sign indicates that the axial force is tensile, or Qa1 5 109.13 kN (T)

Ans

By applying Eq. (3.66), we determine the member global end forces as

F1 5 T1TQ1 5

3

0.6 0.8 0 0

20.8 0.6 0 0

0 0 0.6 0.8

43 4 3 4

0 0 20.8 0.6

2109.13 0 109.13 0

265.479 287.306 65.479 87.306

5

3 4 kN 1 2

Using the member code numbers 3, 4, 1, 2, we store the pertinent elements of F1 in the support reaction vector R (see Fig. 7.8(c)). Member 2

v2 5

34 34 3 5 6 5 1 2

y1 y2 y3 y4

0 0 d1 d2

4

0 0 5 0.0055299 20.0024422

m

Using T2 from Example 3.8, we calculate

u2 5 T2v2 5

3

0.97014 20.24254 0 0

20.24254 20.97014 0 0

0 0 20.24254 20.97014

43

0 0 0.97014 20.24254

43

0 0 0.0055299 20.0024422

4

0 0 20.0037105 20.0047724

5

m

Next, using k2 from Example 3.8, we determine the member local end forces to be

Q2 5 k2u2 5

3

72,761 0 272,761 0

0 0 0 0

272,761 0 72,761 0

43

0 0 0 0

43 4

0 0 20.0037105 20.0047724

5

269.98 0 2269.98 0

Qa2 5 269.98 kN (C) F2 5 TT2 Q2 5

3

20.24254 0.97014 0 0

20.97014 20.24254 0 0

0 0 20.24254 0.97014

kN

Ans

43 4 3

0 0 20.97014 20.24254

269.98 0 2269.98 0

5

265.479 261.92 65.479 2261.92

4

5 6 kN 1 2

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380  Chapter 7   Member Releases and ­Secondary Effects Member 3

v3 5

34 3 4 3 7 8 5 1 2

y1 y2 y3 y4

0 ds8 d1 d2

5

4

0 20.01 0.0055299 20.0024422

m

Note that the support settlement ds8 5 20.01 m is included in the foregoing global end displacement vector v3 for member 3. Next, using the member’s direction cosines, cos u 5 20.6 and sin u 5 0.8, and Eq. (3.61), we evaluate its transformation matrix:

T3 5

3

20.6 20.8 0 0

0.8 20.6 0 0

0 0 0 0 20.6 0.8 20.8 20.6

4

and determine the member local end displacements as

u3 5 T3v3 5

3

0.8 20.6 0 0

20.6 20.8 0 0

43

0 0 0 0 20.6 0.8 20.8 20.6

43

0 20.01 0.0055299 20.0024422

4

20.008 0.006 5 20.0052717 20.0029586

m

To obtain the member local stiffness matrix, we substitute E 5 200 GPa, A 5 0.004 m2, and L 5 10 m into Eq. (3.27):

k3 5

3

80,000 0 280,000 0

0 0 0 0

280,000 0 80,000 0

4

0 0 0 0

kN/m

The member local end forces can now be computed as

3

80,000 0 Q3 5 k3u3 5 280,000 0

0 0 0 0

280,000 0 80,000 0

43

0 0 0 0

43 4

20.008 0.006 20.0052717 20.0029586

5

2218.26 0 218.26 0

kN

Thus, Qa3 5 2218.26 kN 5 218.26 kN (T)

Ans

Finally, we calculate the member global end forces as

F3 5 TT3 Q3 5

3

20.6 0.8 0 0

20.8 20.6 0 0

0 0 20.6 0.8

43 4 3 4

0 0 20.8 20.6

2218.26 0 218.26 0

5

130.96 2174.61 2130.96 174.61

7 8 1 2

and store the pertinent elements of F3 in the reaction vector R, as shown in Fig. 7.8(c). Support Reactions:  The completed reaction vector R is shown in Fig. 7.8(c), and the support reactions are depicted on a line diagram of the truss in Fig. 7.8(d). Ans

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Section 7.3   Support Displacements  381

Equilibrium Check:  Applying the equilibrium equations to the free body of the entire structure (Fig. 7.8(d)), we write 1S

1

X

Y

[

1c

oF 5 0 oF 5 0 oM 5 0 ➀





265.479 2 65.479 1 130.96 5 0.002 kN < 0

Checks

287.306 1 261.92 2 174.61 5 0.004 kN < 0

Checks

(265.479 2 65.479 1 130.96)8 1 (87.306 2 174.61)6 1261.92(2) 5 0.032 kN ? m < 0

Checks

E x ample 7.3 Determine the joint displacements, member end forces, and support reactions for the

continuous beam shown in Fig. 7.9(a), due to the combined effect of the uniformly distributed load shown and the settlements of 45 mm and 15 mm, respectively, of supports 3 and 4. Use the matrix stiffness method.



S ol u tio n

Analytical Model: See Fig. 7.9(b). The structure has two degrees of freedom and six restrained coordinates. Note that member 3 is modeled as being hinged at its right 15 kN/m 1

4 2 8m

3 8m

8m

EI 5 constant E 5 70 GPa I 5 102 (106 ) mm4 (a) Beam

Y

1

1

2

2

8

3

X 1

4

MT 5 2 2

3 5

3

4

6

7

(b) Analytical Model 1 S5

2

3,570 1 3,570 1,785

1

1,785 1 5 3,570 1 2,677.5 2

7,140 1,785

2 1,785 1 6,247.5 2

(c) Structure Stiffness Matrix Pf 5

280 1 80

1

280 1 120 2

5

0 1 40 2

(d) Structure Fixed-Joint Force Vector Due to Member Loads

Fig. 7.9

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382  Chapter 7   Member Releases and ­Secondary Effects

0 1 30.122 1 5 40 1 30.122 2 10.041 2

Pf 5

30.122 1 60.081 2

(e) Structure Fixed-Joint Force Vector Due to Member Loads and Support Displacements 15 kN/m 76.512

1

15 kN/m 2

1

58.692

86.976

86.976

61.308

R5

2

15 kN/m 3

2

85.705

85.705

60.159 59.841 (f) Member End Forces

3

3

70.713

4 49.287

3 3 3 3 58.692 76.512 61.308 1 60.159 59.841 1 70.713 49.287 0

3 4 5 5 6 7 8

58.692 kN 76.512 kN • m

121.47 kN 130.55 kN

49.287 kN 0

(g) Support Reaction Vector 15 kN/m

76.512 kN • m

4 1

2

58.692 kN

3

121.47 kN

130.55 kN

49.287 kN

(h) Support Reactions

Fig. 7.9  (continued) end (i.e., MT 5 2), because the moment at that end of the member must be 0. This approach enables us to eliminate the rotational degrees of freedom of joint 4 from the analysis, by modeling it as a hinged joint with its rotation restrained by an imaginary clamp. Structure Stiffness Matrix and Fixed-Joint Forces Due to Member Loads:  Members 1 and 2 (MT 5 0) By substituting E 5 70(106) kN/m2, I 5 102(1026) m4, and L 5 8 m into Eq. (5.53), we evaluate the member stiffness matrices k as Member 2

5

1

6

2

Member 1

3

4

5

1

k1 5 k2 5

3

167.34 669.38 2167.34 669.38

669.38 3,570 2669.38 1,785

2167.34 2669.38 167.34 2669.38

4

669.38 1,785 2669.38 3,570

3 4 5 1

5 1 6 2

(1)

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Section 7.3   Support Displacements  383

Using the equations given inside the front cover, we evaluate the fixed-end shears and moments due to the 15-kN/m uniformly distributed load as FSb 5 FSe 5 60 kN

FMb 5 2FMe 5 80 kN ? m

(2)

Thus, using Eq. (5.99), we obtain the member fixed-end force vectors:

Qf 1 5 Qf 2 5



3 4 60 80 60 280

3 4 5 1

5 1 (3) 6 2

Member 1

Member 2

Next, using the code numbers for member 1 (3, 4, 5, 1) and member 2 (5, 1, 6, 2), we store the pertinent elements of k1 and k2 into the structure stiffness matrix S, as shown in Fig. 7.9(c). Similarly, the pertinent elements of Qf 1 and Qf 2 are stored in the structure fixed-joint force vector Pf , as shown in Fig. 7.9(d). Member 3 (MT 5 2)  Because MT 5 2 for this member, we use Eqs. (7.18) and (7.19) to determine its stiffness matrix k and fixed-end force vector Qf , respectively. Thus, by applying Eq. (7.18), we obtain 6 41.836 k3 5 334.69 241.836 0

3

2 334.69 2,677.5 2334.69 0

7 241.836 2334.69 41.836 0

8 0 0 0 0

4

6 2 7 8

(4)

Next, by substituting the values of the fixed-end shears and moments from Eq. (2) into Eq. (7.19), we obtain the fixed-end force vector for the released member 3 as

3 434 3(280) 2(8) 1 80 2 (280) 2 3(280) 60 1 2(8) 0

75 6

60 2

Qf 3 5

120

5

2

 45 7 0

(5)

8

The relevant elements of k3 and Qf 3 are stored in S and Pf , respectively, using the member code numbers 6, 2, 7, 8. The completed structure stiffness matrix S, and the Pf vector containing the structure fixed-joint forces due to member loads, are shown in Figs. 7.9(c) and (d), respectively. Structure Fixed-Joint Forces Due to Support Displacements: From the analytical model given in Fig. 7.9(b), we observe that the given support displacements can be expressed as ds6 5 20.045 m ds7 5 20.015 m As members 2 and 3 are attached to the supports that undergo displacements, we compute, for these members, the fixed-end forces due to support displacements, and add them to the previously formed Pf vector due to member loads.

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384  Chapter 7   Member Releases and ­Secondary Effects Member 2  Using the member code numbers 5, 1, 6, 2, we form its fixed-end displacement vector due to support displacements, as

uf s2 5

3 4 34 3 4 uf s1 uf s2 uf s3 uf s4

5 1 5 6 2

0 0 ds6 0

5

0 0 20.045 0

m

Next, using the member stiffness matrix from Eq. (1) and the member stiffness relationship Qfs 5 kufs, we evaluate the fixed-end force vector due to support displacements, as

Qfs2 5 k2ufs2 5

3 3 4 167.34 669.38 2167.34 669.38

7.53 30.122 5 27.53 30.122

669.38 2167.34 3,570 2669.38 2669.38 167.34 1,785 2669.38

669.38 1,785 2669.38 3,570

43 4 0 0 20.045 0

5 1 6 2

The relevant elements of Qfs2 are now added into the previously formed Pf , using the member code numbers, as indicated in Fig. 7.9(e). Member 3 Based on the member code numbers 6, 2, 7, 8, its fixed-end displacement vector, due to support displacements, is written as

uf s3 5

3 4 34 3 4 uf s1 uf s2 uf s3 uf s4

6 2 5 7 8

ds6 0 ds7 0

5

20.045 0 20.015 0

m

Using k3 from Eq. (4), we calculate

Qf s3 5 k3uf s3 5



5

3 3 4 41.836 334.69 241.836 0

21.2551 210.041 1.2551 0

334.69 241.836 2,677.5 2334.69 2334.69 41.836 0 0

0 0 0 0

43 4 20.045 0 20.015 0

6 2 7 8

The pertinent elements of Qfs3 are stored in Pf using the member code numbers. The completed structure fixed-joint force vector Pf , due to member loads and support displacements, is given in Fig. 7.9(e). Joint Load Vector:  Since no external loads are applied to the joints of the beam, its joint load vector is 0; that is, P50

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Section 7.3   Support Displacements  385

Joint Displacements: By substituting the numerical values of P, Pf , and S into Eq. (5.109), we write the stiffness relations for the entire beam as P 2 Pf 5 Sd 230.122 7,140 5 5 3004 2 330.122 60.0814 3260.0814 31,785

4 3dd 4

1,785 6,247.5

1 2

By solving these equations, we determine the joint displacements to be d5

329.05854 2 3 10 21.9541 1

23

rad

Ans

Member End Displacements and End Forces: Member 1  Using the member code numbers 3, 4, 5, 1, we write its end displacement vector as u1 3 0 0 u2 4 0 0 5 5 3 1023(6) u1 5 u3 5 0 0 u4 1 d1 21.9541

34 34 3 4

The member end forces can now be calculated using the member stiffness relationship Q 5 ku 1 Qf (Eq. (5.4)). Thus, using k1 from Eq. (1), Qf 1 from Eq. (3), and u1 from Eq. (6), we calculate

3

58.692 kN 76.512 kN ? m Q1 5 k1u1 1 Qf 1 5 61.308 kN 286.976 kN ? m

4

3 4  5 1

Ans

The end forces for member 1 are depicted in Fig. 7.9(f). To generate the support reaction vector R, we store the pertinent elements of Q1 in R, using the member code numbers, as shown in Fig. 7.9(g). Member 2

u2 5

34 34 3 4 u1 u2 u3 u4

5 1 5 6 2

0 d1 ds6 d2

0 21.9541 5 245 29.0585

3 1023

(7)

Note that the support displacement ds6 is included in the foregoing end displacement vector for member 2. Using k2 from Eq. (1), Qf 2 from Eq. (3), and u2 from Eq. (7), we determine Q2 5 k2u2 1 Qf 2 5

3

60.159 kN 86.976 kN ? m 59.841 kN 285.705 kN ? m

4

5 1  6 2

Ans

Member 3

u3 5

34 34 3 4 u1 u2 u3 u4

6 2 5 7 8

ds6 d2 ds7 0

245 29.0585 5 3 1023 215 0

(8)

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386  Chapter 7   Member Releases and ­Secondary Effects The rotation, u4, of the released end of this member, if desired, can be evaluated using Eq. (7.20). Finally, using k3 from Eq. (4), Qf 3 from Eq. (5), and u3 from Eq. (8), we calculate the member end forces as

Q3 5 k3u3 1 Qf3 5

3

70.713 kN 85.705 kN ? m 49.287 kN 0

4

6 2  7 8

Ans

The member end forces are shown in Fig. 7.9(f). Support Reactions: The completed reaction vector R is shown in Fig. 7.9(g), and the support reactions are depicted on a line diagram of the beam in Fig. 7.9(h). Ans

E x ample 7.4 Determine the joint displacements, member local end forces, and support reactions for



the plane frame of Fig. 7.10(a), due to the combined effect of the loading shown and a settlement of 25 mm of the left support. Use the matrix stiffness method.



S ol u tio n

This frame was analyzed in Example 6.6 for external loading. In this example, we use the same analytical model of the frame, so that the various member and structure matrices calculated previously can be reused in the present example. Analytical Model: See Fig. 7.10(b). The frame has three degrees of freedom and six restrained coordinates.

125 kN • m

16.667 kN/m

2 Y

9

3 1

3m

2

300 kN

7

3

2

8 1

3m

3m

6m E, A, I 5 constant E 5 200 GPa A 5 7,600 mm2 I 5 129(106) mm4 (a) Frame

1 4

X 6 5 (b) Analytical Model

Fig. 7.10

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81.16 kN

kN

34

.33

kN

18.84 kN

149.69 kN • m

268

kN

3

2

134

.16

78.03 kN 2

61.

2

16.667 kN/m

37.27 kN 78.03 kN

87.

69

kN

•m

51.

69

kN

Section 7.3   Support Displacements  387

72.

kN .20 126

320

.01

kN

•m

82

1

kN

1

(c) Member Local End Forces

125 kN • m 16.667 kN • m 78.03 kN 2 300 kN

2

3

149.69 kN • m

81.16 kN

1

3 3

4 5 6 R5 7 278.03 kN 8 81.16 kN 2149.69 kN • m 9 77.98 kN 318.79 kN 126.20 kN • m

(d) Support Reaction Vector

77.98 kN

1 126.20 kN • m 318.79 kN (e) Support Reactions

Fig. 7.10  (continued)

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388  Chapter 7   Member Releases and ­Secondary Effects Structure Stiffness Matrix: As determined in Example 6.6, the structure stiffness matrix for the frame, in units of kN and meters, is given by S5

3

299,470 90,225 3,077

90,225 182,910 2,762

4

3,077 2,762  32,584

(1)

Structure Fixed-Joint Forces Due to Member Loads: From Example 6.6,

3 4

0 1 Pf 5 200 2 262.5 3

(2)

Joint Load Vector: From Example 6.6,

3 4

0 1 P5 0 2 2125 3

(3)

Structure Fixed-Joint Forces Due to Support Displacement: From Fig. 7.10(b), we observe that the given 0.025 m downward displacement of support 1 can be expressed as ds5 5 20.025 m As member 1 is the only member attached to support 1, we form its global fixed-end displacement vector due to support displacement, using the member code numbers 4, 5, 6, 1, 2, 3, as

vf s1 5

FG FG F G F G yf s1 yf s2 yf s3 yf s4 yf s5 yf s6

4 5 6 5 1 2 3

0 ds5 0 0 0 0

5

0 20.025 0 0 0 0

m

Next, we substitute the member global stiffness matrix K1 (given in Example 6.6) and the foregoing vfs1 vector into Eq. (7.26) to evaluate the member global fixed-end force vector, Ffs1, due to support settlement:

Ff s1 5 K1vf s1 5

22,255.6 24,536.9 238.5 2,255.6 4,536.9 238.5

4 5 6 1 2 3

Based on the member code numbers, we add the elements in the fourth, fifth, and sixth rows of Ffs1 into rows 1, 2, and 3, respectively, of the previously formed Pf vector (Eq. (2)) to obtain the structure fixed-joint force vector due to the combined effect of the member loads and support displacement, as

3

0 1 2,255.6 Pf 5 200 1 4,536.9 262.5 2 38.5

4 3

4

1 2,255.6 2 5 4,736.9  3 2101.0

(4)

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Section 7.3   Support Displacements  389

Joint Displacements: By substituting P (Eq. (3)), Pf (Eq. (4)), and S (Eq. (1)) into Eq. (6.42), we write the stiffness relations for the entire frame as P 2 Pf 5 Sd

3 4 3 4 3

4 3

0 2,255.6 22,255.6 299,470 0 2 4,736.9 5 24,736.9 5 90,225 2125 2101.0 224.0 3,077

90,225 182,910 2,762

Solving these equations, we determine the joint displacements to be 0.000308 m 1 2 d 5 20.026071 m 0.001444 rad 3

3

4

FG FG F G F G F G FG

43 4

3,077 2,762 32,584

d1 d2 d3

Ans

Member End Displacements and End Forces: Member 1

v1 5

y1 y2 y3 y4 y5 y6

4 5 6 5 1 2 3

0 ds5 0 d1 d2 d3

0 20.025 0 0.000308 5 20.026071 0.001444

Using the member transformation matrix T1 from Example 6.6, and Eq. (6.20), we calculate 20.022361 20.011180 0 u1 5 T1v1 5 20.023181 20.011935 0.001444 Next, we use the member local stiffness matrix k1 and fixed-end force vector Qf 1 from Example 6.6, and Eq. (6.4), to compute the local end forces as 320.01 kN 72.82 kN 126.20 kN ? m Q1 5 k1u1 1 Qf1 5 251.69 kN 61.34 kN 287.69 kN ? m



Ans

The local member end forces are depicted in Fig. 7.10(c). The member global end forces F can now be determined by applying Eq. (6.23), as 77.98 4 318.79 5 126.20 6 F1 5 TT Q1 5 277.98 1 218.80 2 287.69 3

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390  Chapter 7   Member Releases and ­Secondary Effects Using the member code numbers, the pertinent elements of F1 are stored in the reaction vector R (see Fig. 7.10(d)).

FG FG F G F G

Member 2  The global and local end displacements for this horizontal member are

u2 5 v2 5

y1 y2 y3 y4 y5 y6

1 2 3 5 7 8 9

d1 d2 d3 0 0 0

5

0.000308 20.026071 0.001444 0 0 0

Using k2 and Qf 2 from Example 6.6, we compute the member local and global end forces to be

F2 5 Q2 5 k2u2 1 Qf2 5

78.03 kN 18.84 kN 237.27 kN ? m 278.03 kN 81.16 kN 2149.69 kN ? m

1 2 3  7 8 9

Ans

The pertinent elements of F2 are stored in R. Support Reactions: See Figs. 7.10(d) and (e).

Ans

7.4 Computer implementation of

Support displacement effects The computer programs developed previously can be extended with relative ease, and without changing their overall organization, to include the effects of support displacements in the analysis. From the analysis procedure developed in Section 7.3, we realize that inclusion of support displacement effects essentially involves extension of the existing programs to perform three additional tasks: (a) reading and storing of the support displacement data, (b) evaluation of the structure fixed-joint forces due to support displacements, and (c) inclusion of support displacements in the member end displacement vectors, before calculation of the final member end forces and support reactions. In this section, we consider the programming of these tasks, with particular reference to the program for the analysis of plane frames (Section 6.7). The modifications necessary in the plane truss and beam analysis programs are also described. Input of Support Displacement Data    The process of reading and storing the support displacements is similar to that for inputting the joint load data (e.g., see flowcharts in Figs. 4.3(f) and 5.21(b)). This process can be conveniently programmed using the flowchart given in Fig. 7.11 on the next page. The support displacement data consists of (a) the number of supports that undergo displacements (NSD), and (b) the joint number and the magnitudes of

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Section 7.4   Computer Implementation of Support Displacement Effects  391 Start Support Displacement Input Read NSD Dimension JSD(NSD), SDJ(NSD, NCJT) I51

I # NSD?

no

yes Read JSD(I ), SDJ(I, 1), SDJ(I, 2), . . . , SDJ(I, NCJT ) I 5 I 11 Print Support Displacement Input Data Continue to Next Part

Fig. 7.11  Flowchart for Reading and Storing Support Displacement Data

the ­displacements for each such support. As indicated in Fig. 7.11, the joint numbers of the supports that undergo displacements are stored in an integer vector JSD of order NSD 3 1, with their displacements stored in the corresponding rows of a real matrix SDJ of order NSD 3 NCJT. For example, in the case of plane frames, the support displacement matrix SDJ would be of order NSD 3 3, with the support translations in the X and Y directions and the rotations being stored in the first, second, and third columns, respectively, of the matrix SDJ. This subprogram for inputting support displacement data can be conveniently added as Part VIc in the computer programs for the analysis of plane frames (see Table 6.1) and beams (see Table 5.1); and it can be inserted between Parts VI and VII of the plane truss computer program (see Table 4.1). Evaluation of Structure Equivalent Joint Loads Due to Support Displacements  In this part of the program, the equivalent joint loads, or the negatives of the structure fixed-joint forces (i.e., 2Pf) due to support displacements, are added to the structure load vector P. A flowchart for constructing this part of the plane frame analysis program is presented in Fig. 7.12. As this flowchart indicates, the program essentially performs the following operations for each member of the structure: 1. First, the program determines whether the member under consideration, IM, is attached to a support that undergoes displacement, by comparing the member beginning and end joint numbers to those stored in the support displacement vector JSD. If the member is not attached to such a support, then no further action is taken for that member.

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392  Chapter 7   Member Releases and ­Secondary Effects Start Part IXb Dimension GK(2*NCJT, 2*NCJT), BK(2*NCJT, 2*NCJT ), FF(2*NCJT), V(2*NCJT) IM 5 1

IM # NM?

IM 5 IM 1 1

no

yes JB 5 MPRP(IM, 1) JE 5 MPRP(IM, 2) no

ICOUNT . 0? yes

I 5 MPRP(IM, 3), E 5 EM(I) I 5 MPRP(IM, 4), A 5 CP(I, 1) ZI 5 CP(I, 2), MT 5 MPRP(IM, 5) XB 5 COORD(JB, 1) YB 5 COORD(JB, 2) XE 5 COORD(JE, 1) YE 5 COORD(JE, 2) BL 5 SQR((XE 2 XB)^2 1 (YE 2 YB)^2) CX 5 (XE 2 XB)/BL CY 5 (YE 2 YB)/BL

I51 ICOUNT 5 0 no

I5I11

I # NSD? yes JB or JE 5 JSD(I)?

no

yes ICOUNT 5 1

Call, in order, Subroutines: MSTIFFL, MTRANS, MSTIFFG Initialize all elements of V to zero Call Subroutine MFEDSD Call Subroutine MFEFSD Call Subroutine STOREPF Continue to Part X

Fig. 7.12  Flowchart for Generating Structure Equivalent Joint Load Vector Due to Support ­Displacements

2. If the member is attached to a support that undergoes displacement(s), then its global stiffness matrix GK (5 K) is obtained by calling, in order, the subroutines MSTIFFL (Fig. 7.4), MTRANS (Fig. 6.26), and MSTIFFG (Fig. 6.27). 3. Next, the program calls the subroutine MFEDSD to form the member global fixed-end displacement vector, V (5 vfs), due to support

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Section 7.4   Computer Implementation of Support Displacement Effects  393 Start Subroutine MFEDSD Arguments: JB, JE, NCJT, NSD, JSD, SDJ, V I51 no

I # NSD? yes JSD(I) 5 JB?

yes

J51

no I5I11

no

J # NCJT?

no

JSD(I) 5 JE? yes yes

V(J) 5 V(J) 1 SDJ(I, J)

J 5 NCJT 1 1 J5J11 no

J # 2*NCJT? yes

V(J) 5 V(J) 1 SDJ(I, J 2 NCJT ) J5J11

End Subroutine MFEDSD Return to calling program

Fig. 7.13  Flowchart of Subroutine MFEDSD for Determining Member Global FixedEnd Displacement Vector Due to Support Displacements

displacements. As the flowchart in Fig. 7.13 indicates, this subroutine first checks the support displacement vector JSD to determine if the beginning joint of the member, JB, is a support joint subjected to displacements. If JB is such a joint, then the values of its displacements are read from the corresponding row of the support displacement matrix SDJ, and stored in the appropriate elements of the upper half of the member fixed-end displacement vector V. The process is then repeated for the end joint of the member, JE, with any corresponding support displacements being stored in the lower half of V. 4. Returning our attention to Fig. 7.12, we can see that the program then calls the subroutine MFEFSD (Fig. 7.14), which evaluates the member global fixed-end force vector due to support settlements FF (5 Ffs), using the relationship Ffs 5 Kvfs (Eq. (7.26)).

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394  Chapter 7   Member Releases and ­Secondary Effects Start Subroutine MFEFSD Arguments: NCJT, GK, V, FF Initialize all elements of FF to zero

I # 2*NCJT?

no

yes J51

I5I11

no

J # 2*NCJT? yes

FF(I) 5 FF(I) 1 GK(I, J)*V(J)

J5J11

End Subroutine MFEFSD Return to calling program

Fig. 7.14  Flowchart of Subroutine MFEFSD for Determining Member Global Fixed-End Force Vector Due to Support Displacements

5. Finally, the negative values of the pertinent element of FF are added in their proper positions in the structural load vector P, using the subroutine STOREPF, which was developed in Chapter 6 (Fig. 6.30). When these operations have been completed for each member of the frame, the structure load vector P contains the equivalent joint loads (or the negatives of the structure fixed-joint forces) due to support displacements. This subprogram, designated Part IXb in Fig. 7.12, can be conveniently inserted between Parts IX and X of the program for the analysis of plane frames (see Table 6.1). The flowcharts given in Figs. 7.12 through 7.14 can be used to develop the corresponding part of the beam analysis program (see Table 5.1), provided that: (a) the member global vectors V (5 vfs) and FF (5 Ffs) are replaced by the local vectors U (5 ufs) and QF (5 Qfs), respectively; (b) the member local stiffness matrix BK (5 k) is used, instead of the global matrix GK (5 K), in subroutine MFEFSD; and (c) the subroutine STOREPF developed in Chapter 5 (Fig. 5.29) is employed to store the negative elements of QF in the structure load vector P. The process of programming the corresponding part of the plane truss analysis program is essentially the same as discussed herein for the case of plane frames, except that the subroutine STOREPF (Fig. 6.30) should be copied from the plane frame analysis program and added to the plane truss program. Calculation of Member Forces and Support Reactions (Part XII) Parts XII of the programs developed previously (see flowcharts in Figs. 6.31, 5.31, and 4.14) should be modified to include support displacements in the end displacement vectors of members attached to supports, before the end forces

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Section 7.5   Temperature Changes and Fabrication Errors  395

for such members are calculated. This can be achieved by simply calling the subroutine MFEDSD (Fig. 7.13) in these programs, to add the compatible support displacements to the member end displacement vectors. In the plane frame and truss analysis programs (Figs. 6.31 and 4.14, respectively), the subroutine MFEDSD should be called after the subroutine MDISPG has been used to form the member global end displacement vector V (5 v) due to the joint displacements d, but before the subroutine MDISPL is called to evaluate the member local end displacement vector U (5 u). In the program for the analysis of beams (Fig. 5.31), however, the subroutine MFEDSD should be called after the subroutine MDISPL has been used to form the member end displacement vector U (5 u) from the joint displacements d, but before the member end forces are calculated using the subroutine MFORCEL. Furthermore, as discussed previously, before it can be used in the beam analysis program, the subroutine MFEDSD (as given in Fig. 7.13) must be modified to replace V with U. It may be of interest to note that the program for the analysis of plane frames, which was initially developed in Chapter 6 and has been extended in this chapter, is quite general, in the sense that it can also be used to analyze beams and plane trusses. When analyzing a truss using the frame analysis program, all of the truss members are modeled as hinged at both ends with MT 5 3, and all the joints of the truss are modeled as hinged joints restrained against rotations by imaginary clamps.

7.5 Temperature changes and fabrication errors

Like support displacements, changes in temperature and small fabrication errors can cause considerable stresses in statically indeterminate structures, which must be taken into account in their designs. However, unlike support displacements, which are generally specified with reference to the global coordinate systems of structures, temperature changes and fabrication errors, like member loads, are usually defined relative to the local coordinate systems of members. Therefore, the stiffness methods developed previously for the analysis of structures subjected to member loads, can be used without modifications to determine the structural responses to temperature changes and fabrication errors. The only difference is that the fixed-end forces, which develop in members due to temperature changes and fabrication errors, must now be included in the member local fixed-end force vectors QF. In this section, we derive the expressions for the fixed-end forces that develop in the members of framed structures due to temperature changes and two common types of fabrication errors. The application of these fixed-end force expressions in analysis is then illustrated by some examples.

Member Fixed-End Forces Due to Temperature Changes We can develop the desired relationships by first determining the displacements caused by temperature changes at the ends of members that are free to

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396  Chapter 7   Member Releases and ­Secondary Effects y θT

θT

y

Temperature increase 5 Tt

Tt

e b

e9

x

x

d 2 z d 2

Tt 1 Tb 2

Tb Temperature increase 5 Tb

θT

θT DT

L (a) Simply Supported Member Subjected to Temperature Change

(b) Temperature Variation over Member Depth

α (Tb –Tt) L 2 θT d b

θT

(c) Member End Rotation

Fig. 7.15

deform. The fixed-end forces required to suppress these member end displacements can then be obtained, using the member stiffness matrices. To examine member end displacements due to temperature changes, let us consider an arbitrary simply supported member of a plane frame, as shown in Fig. 7.15(a). Now, assume that the member is heated so that the temperature increase of its top surface is Tt and that of its bottom surface is Tb, with the temperature increase varying linearly between Tt and Tb over the depth d of the member cross-section, as shown in Fig. 7.15(b). Note that the temperature does not vary along the length of the member. Because the member is simply supported (so that it is statically determinate), it is free to expand in the longitudinal direction. If we assume that the member cross-section is symmetric about the xz plane (Fig. 7.15(b)) containing its centroidal axis, then the temperature increase at the level of the centroidal axis (i.e., at the distance d/2 from the top or bottom of the member) would be (Tb 1 Tt)/2. This temperature increase causes the member’s centroidal axis to elongate by an amount DT :

DT 5 a

1

Tb 1 Tt 2

2 L

(7.29)

in which a denotes the coefficient of thermal expansion.

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Section 7.5   Temperature Changes and Fabrication Errors  397

In addition to the axial deformation DT , the member also undergoes bending as its top and bottom surfaces elongate by different amounts (because they are subjected to different temperature increases). For example, as depicted in Fig. 7.15, if Tb . Tt , then the member bends concave upward, causing the cross-sections at its ends b and e to rotate inward, as shown. Since the temperature increase is uniform along the member’s length, the rotations of its two end cross-sections must be equal in magnitude. From Fig. 7.15, we can see that these member end rotations can be related to the temperature change by dividing one-half of the difference between the elongations of the bottom and top fibers of the member, by its depth. Thus,

a(Tb 2 Tt)L

 (7.30) 2d in which uT represents the magnitude of the rotations of the member end crosssections, which, in turn, equal the slopes of the elastic curve of the member at its ends, as shown in Figs. 7.15(a) and (c). Using the sign convention for member local end displacements established in Chapter 6, we can express the local end displacement vector uT for the simply supported member, due to the temperature change, as uT 5

34 3 4

0 0 2uT uT 5 DT 0 uT

1 2 3 aL 5 4 2 5 6

0 0 2(Tb 2 Tt)yd  Tb 1 Tt 0 (Tb 2 Tt)yd

(7.31)

in which the rotation of the beginning, b, of the member is negative, because it is clockwise, whereas the rotation of the member end, e, which has a counterclockwise sense, is positive. The member fixed-end forces necessary to suppress its end displacements uT can now be established by applying the principle of superposition, as illustrated in Fig. 7.16. Figure 7.16(a) shows a fixed member of a plane frame subjected to a temperature increase, causing fixed-end forces to develop at its ends. In Fig. 7.16(b), the corresponding simply supported member is subjected to the same temperature change, causing the displacements uT (Eq. (7.31)) at its ends, but no end forces; in Fig. 7.16(c), the simply supported member is subjected to the same fixed-end forces that develop in the fixed member of Fig. 7.16(a), but no ­temperature change. By comparing Figs. 7.16(a) through (c), we realize that the response of the fixed member of Fig. 7.16(a) must equal the superposition of the responses of the two simply supported members of Figs. 7.16(b) and (c). ­Therefore, since the end displacements of the fixed member due to the temperature change are 0 (Fig. 7.16(a)), its fixed-end forces, when applied to the simply supported beam (Fig. 7.16(c)), must cause the end displacements, 2uT , that are equal in magnitude but opposite in direction to those due to the temperature change, uT (Fig. 7.16(b)). The forces that can cause the end displacements 2uT in the simply supported member can be c­ onveniently obtained by pre-multiplying the

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398  Chapter 7   Member Releases and ­Secondary Effects Temperature increase 5 Tt b

FMe

e

FAb

FAe FMb

Temperature increase 5 Tb FSb

FSe (a) Fixed Member Subjected to Temperature Change and Fixed-End Forces (No End Displacements)

5 ΔT

Temperature increase 5 Tt

b

e θT

θT Temperature increase 5 Tb

(b) Simply Supported Member Subjected to Temperature Change Only (End Displacements 5 uT)

1 b FAb

θT

ΔT

θT

FMe FAe

FMb

e FSb

FSe (c) Simply Supported Member Subjected to Fixed-End Forces Only (End Displacements 5 2uT)

Fig. 7.16

­negative of the uT vector given in Eq. (7.31), by the member local stiffness matrix k (Eq. (6.6)). Thus,

34 3 FAb

FSb FMb FAe FSe

FMe

5

EI L3

AL2 0 0 I 0 12 6L 0 6L 4L2 2 AL 2 0 0 I 0 212 26L 0 6L 2L2

AL2 0 0 I 0 212 6L 0 26L 2L2 2 AL 0 0 I 0 12 26L 0 26L 4L2

2

43 4 0 0

12 2 2 aL

2(Tb 2 Tt)yd Tb 1 Tt

0 (Tb 2 Tt)yd

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Section 7.5   Temperature Changes and Fabrication Errors  399

From which we obtain



343 4 FAb FSb FMb FAe FSe FMe

A (Tb 1 Tt)y2 0 I (Tb 2 Tt)yd 5 E  2A (Tb 1 Tt)y2 0 2I (Tb 2 Tt)yd

(7.32)

Thus, the fixed-end forces for the members of plane frames can be expressed as



Tb 1 Tt

1 2 2 T 2T FM 5 2FM 5 EI 1 d 2 FAb 5 2FAe 5 EA

b

b



(7.33)

t

e

The expressions for the fixed-end moments, given in Eqs. (7.33), can also be used for the members of beams. However, as the beam members are free to expand axially, their fixed-end axial forces are 0 (i.e., FAb 5 FAe 5 0). Similarly, the expressions for the fixed-end axial forces, given in Eqs. (7.33), can be used for the members of trusses; however, the fixed-end moments must now be set equal to 0 (i.e., FMb 5 FMe 5 0) in Eqs. (7.33), because the ends of truss members are free to rotate. The fixed-end force expressions given in Eqs. (7.33) are based on a linearly varying temperature change over the depth of the member cross-section. If the member is subjected to a uniform temperature increase, Tu, over its depth, then the corresponding expressions for fixed-end forces can be obtained by simply substituting Tb 5 Tt 5 Tu into Eqs. (7.33). This yields

FAb 5 2FAe 5 EATu (7.34) As Eqs. (7.34) indicate, the member fixed-end moments would be 0 in the case of a uniform temperature change, because such a temperature change has no tendency to bend the member, but only to cause axial deformation. Equations (7.34) can be used to determine the fixed-end forces for the members of plane frames and trusses subjected to uniform temperature changes. As stated previously, the members of beams are free to expand in their axial directions; therefore, a uniform temperature change does not cause any fixed-end forces in such members.

Member Fixed-End Forces Due to Fabrication Errors In structural analysis terminology, fabrication error is used to refer to a small initial deformation of a member in its unstressed state. The expressions for

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400  Chapter 7   Member Releases and ­Secondary Effects y

EA e L a

e

b

e9

EA e L a

x

Design length 5 L Fabrication error 5 ea

Fig. 7.17

member fixed-end forces due to fabrication errors can be derived in a manner similar to that for the case of temperature changes. In the following paragraphs, we develop the fixed-end force expressions for two common types of fabrication errors. Errors in Initial Member Length  Consider a member of a plane frame with a specified design length L. Now, suppose that the member is fabricated so that its initial unstressed length is longer than the specified length L by an amount ea, as shown in Fig. 7.17. As the distance between the fixed supports is L, the supports must exert a compressive axial force of magnitude EAea  /L on the member to reduce its length from L 1 ea to L, so that it can fit between the supports. Thus, the fixed-end forces that develop in the member due to its fabricated length being too long by an amount ea are

FAb 5 2FAe 5

EA e  L a

(7.35)

Equations (7.35) can also be used to obtain fixed-end forces for the members of trusses due to fabrication errors in their lengths. Errors in Initial Member Straightness  Another type of fabrication error commonly encountered in structural design involves a lack of initial straightness of the members of beams and plane frames. Figure 7.18(a) on the next page shows such a member of a beam, which somehow has been fabricated with an initial bend, causing a small deflection eb at a distance l1 from the member’s left end. To determine the fixed-end forces for this member, we first express the member end rotations ub and ue in terms of the fabrication error eb, as (see Fig. 7.18(a))

ub 5

eb l1

and

ue 5

eb l2

(7.36)

Using the sign convention established for beam members in Chapter 5, we write the local end displacement vector ue for the member, due to the fabrication error, as

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Section 7.5   Temperature Changes and Fabrication Errors  401 y Design configuration

ø1

ø2 θb

b

eb

x

θe

e

L Initial configuration due to fabrication error (a) Unstressed Member with Fabrication Error y

b

FMe

e

x FMb

FSb

FSe L (b) Fixed Member with Fabrication Error

Fig. 7.18



34 3 4

0 1 0 2ub 2 21yl1 ue 5 5 eb  (7.37) 0 3 0 ue 4 1yl2 in which the rotation of the beginning, b, of the member is considered to be negative, because it has a clockwise sense. The member fixed-end forces (Fig. 7.18(b)) necessary to suppress the end displacements ue can now be determined by premultiplying the negative of the ue vector by the member local stiffness matrix k (Eq. (5.53)). Thus,

3 4 3 FSb FMb FSe FMe

12 6L EI 6L 4L2 5 3 L 212 26L 6L 2L2

4 3 4

212 6L 0 2 26L 2L 21yl 1 (2eb) 12 26L 0 2 26L 4L 1yl2

3 4

3(l2 2 l1) 2EIeb L(2l 2 l ) 2 1                5 2  L l1l2 3(l1 2 l2) L(l2 2 2l1)

(7.38)

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402  Chapter 7   Member Releases and ­Secondary Effects

Therefore, the fixed-end forces for the members of beams are: FSb 5 2FSe 5 FMb 5



FMe 5

2EIeb Ll1l2 2EIeb Ll1l2

6EIeb L2l1l2

(l2 2 l1)

(2l2 2 l1)

(7.39)

(l2 2 2l1)

As the fabrication error eb is assumed to be small, it does not cause any axial deformation of the member; therefore, no axial force develops in the fixed member (i.e., FAb 5 FAe 5 0). Thus, the expressions for the fixed-end forces given in Eq. (7.39) can also be used for the members of plane frames.

Procedure for Analysis As stated at the beginning of this section, the procedures for the analysis of beams and plane frames, including the effects of temperature changes and ­fabrication errors, remain the same as developed in Chapters 5 and 6, ­respectively—provided that the member fixed-end forces caused by the temperature changes and fabrication errors are now included in the member local fixed-end force vectors Qf . In the case of plane trusses, however, the member and structure stiffness relationships must now be modified, to include the effects of temperature changes and fabrication errors, as follows: (a) the member local stiffness relationship given in Eq. (3.7) should be modified to Q 5 ku 1 Qf , (b) the member global stiffness relationship (Eq. (3.71)) now becomes F 5 Kv 1 Ff , and (c) the structure stiffness relationship (Eq. (3.89)) should be updated to include the structure fixed-joint forces as P 2 Pf 5 Sd. The structure fixed-joint force vector Pf can be generated using the member code number technique as discussed in Chapter 6 for the case of plane frames.

E x ample 7.5 Determine the joint displacements, member axial forces, and support reactions for the plane truss shown in Fig. 7.19(a) on the next page, due to the combined effect of the following: (a) the joint loads shown in the figure, (b) a temperature drop of 15°C in member 1, and (c) the fabricated length of member 3 being 3 mm too short. Use the matrix stiffness method.



S ol u tio n

This truss was analyzed in Example 3.8 for joint loads only, and in Example 7.2 for a support displacement.



Analytical Model: See Fig. 7.19(b). The analytical model used herein is the same as used in Examples 3.8 and 7.2. Structure Stiffness Matrix: From Example 3.8, S5

61,880 3217,120

217,120 170,881

4 kN ? m

(1)

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Section 7.5   Temperature Changes and Fabrication Errors  403

900 kN

2

mm 3

2m

)

(4 ,0 6m

8m

00

2)

0 mm

2

,0 (4

1

(3,00

00

mm

2

)

600 kN

4m

E 5 200 GPa α 5 1.2(1025)/˚C (a) Truss 2 Y

1

3

1

1

3

2

5

2

4

3

7

6 (b) Analytical Model

Pf 5

86.4 2 144 115.2 1 192

1 2 57.6 5 2 307.2

4

X

8

1 kN 2

(c) Structure Fixed-Joint Force Vector Due to Temperature Changes and Fabrication Errors

Fig. 7.19

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404  Chapter 7   Member Releases and ­Secondary Effects

3 3

2106.32 2141.77 2143.82 R5 575.3 2349.85 466.47

3 4 5 kN 6 7 8

(d) Support Reaction Vector Y

900 kN 1

600 kN

1

3

2

106.32

143.82

2

141.77

3

575.3 (e) Support Reactions (kN)

4

349.85

X

466.47

Fig. 7.19  (continued) Joint Load Vector:  From Example 3.8, P5

600 32900 4 kN

(2)

Structure Fixed-Joint Force Vector Due to Temperature Changes and Fabrication Errors: Member 1  By substituting E 5 200 GPa, A 5 4,000 mm2, a 5 12(1026)/ºC, and Tu 5 215°C into Eqs. (7.34), we evaluate the member fixed-end forces, due to the specified temperature change, as FAb 5 2FAe 5 2144 kN Thus, the local fixed-end force vector for member 1 can be expressed as

Qf 1 5

343 4 FAb FSb FAe FSe

5

2144 0 144 0

kN

(3)

Next, we obtain the global fixed-end force vector for this member by applying the transformation relationship Ff 5 TTQf , while using the transformation matrix T1 from Example 3.8. Thus,

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Section 7.5   Temperature Changes and Fabrication Errors  405

Ff1 5 TT1 Qf1 5

3

0.6 0.8 0 0

0 0 0.6 0.8

20.8 0.6 0 0

0 0 20.8 0.6

43 4 3 4 2144 0 144 0

286.4 2115.2 86.4 115.2

5

3 4 kN 1 2

From the member code numbers 3, 4, 1, 2, which are written on the right side of Ff 1, we realize that the elements in the third and fourth rows of Ff 1 should be stored in rows 1 and 2, respectively, of the 2 3 1 structure fixed-joint force vector Pf , as shown in Fig. 7.19(c). Member 3 By substituting ea 5 23 mm 5 2 0.003 m into Eq. (7.35), we obtain FAb 5 2FAe 5

200(106)(0.004) (20.003) 5 2240 kN 10

Thus,

Qf 3 5

343 4 FAb FSb FAe FSe

5

2240 0 240 0

(4)

kN

Using T3 from Example 3.8, we calculate

Ff3 5 TT3 Qf3 5

3

20.6 0.8 0 0

20.8 0 20.6 0 0 20.6 0 0.8

43 4 3 4

0 0 20.8 20.6

2240 0 240 0

5

144 2192 2144 192

7 8 kN 1 2

The relevant elements of Ff 3 are stored in Pf using the member code numbers. The completed structure fixed-joint force vector Pf , due to temperature change and fabrication error, is given in Fig. 7.19(c). Joint Displacements: By substituting P (Eq. (2)), Pf (Fig. 7.19(c)), and S (Eq. (1)) into the structure stiffness relationship, we write P 2 Pf 5 Sd 600 657.6 61,880 5 5 32900 4 2 3257.6 307.24 321,207.24 3 217,120

217,120 170,881

4 3dd 4 1 2

Solving the foregoing equations, d5

0.0089197 1 320.0061709 4 2 m

Ans

Member End Displacements and End Forces: Member 1

v1 5

34 3 0 0 d1 d2

3 4 5 1 2

0 0 0.0089197 20.0061709

4

m

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406  Chapter 7   Member Releases and ­Secondary Effects

u1 5 T1v1 5

3

4

0 0 0.00041509 20.010838

m

Next, we calculate the member local end forces by applying the member stiffness relationship Q 5 ku 1 Qf . Thus, using k1 from Example 3.8, and Qf 1 from Eq. (3), we obtain

Q1 5 k1u1 1 Qf1 5

3

80,000 0 280,000 0

0 0 0 0

280,000 0 80,000 0

0 0 0 0

43

43 4

0 0 0.00041509 20.010838

2144 0 1 144 0

from which

3 4

2177.21 0 Q1 5 177.21 0

kN

Thus, Qa1 5 2177.21 kN 5 177.21 kN (T)

F1 5 TT1 Q1 5

3 4 2106.32 2141.77 106.32 141.77

Ans

3 4 kN 1 2

The pertinent elements of F1 are stored in the support reaction vector R in Fig. 7.19(d). Member 2

v2 5

34 3 0 0 d1 d2

5 0 6 0 5 1 0.0089197 2 20.0061709

4

m

Using T2 from Example 3.8, we obtain

u2 5 T2v2 5

5

3 3

20.24254 20.97014 0 0

0.97014 20.24254 0 0

0 0 20.00815 20.0071567

4

0 0 0 0 20.24254 0.97014 20.97014 20.24254

43

0 0 0.0089197 20.0061709

4

m

With Qf 2 5 0 and k2 obtained from Example 3.8, we determine the member local end forces to be

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Section 7.5   Temperature Changes and Fabrication Errors  407

Q2 5 k2u2 5

3

72,761 0 272,761 0

0 0 0 0

272,761 0 72,761 0

43

0 0 0 0

0 0 20.00815 20.0071567

43 4 593 0 2593 0

5

Qa2 5 593 kN (C) F2 5 TT2 Q2 5

3

kN

Ans

2143.82 575.3 143.82 2575.3

4

5 6 1 2

Member 3

v3 5

34 3 4 3 4 0 0 d1 d2

7 0 8 0 5 1 0.0089197 2 20.0061709

u3 5 T3v3 5

0 0 20.010289 20.0034332

m

m

Using k3 from Example 7.2 and Qf 3 from Eq. (4), we calculate

3

80,000 Q3 5 k3u3 1 Qf3 5 0 280,000 0

0 0 0 0

280,000 0 80,000 0

3 43 4

2240 0 1 240 0

583.09 0 5 2583.09 0

Qa3 5 583.09 kN (C)

F3 5 TT3 Q3 5

3 4 2349.85 466.47 349.85 2466.47

4

0 0 20.010289 20.0034332

kN

Ans

7 8 1 2

Support Reactions: See Figs. 7.19(d) and (e).



43

0 0 0 0

Ans

E x a m p l e 7.6 Determine the joint displacements, member end forces, and support reactions for the

three-span continuous beam shown in Fig. 7.20(a), due to a temperature increase of 10°C at the top surface and 70°C at the bottom surface, of all spans. The temperature increase varies linearly over the depth d 5 600 mm of the beam cross-section. Use the matrix stiffness method.

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408  Chapter 7   Member Releases and ­Secondary Effects

S ol u tio n



This beam was analyzed in Example 7.3 for member loads and support settlements. Analytical Model: See Fig. 7.20(b). The analytical model used herein is the same as used in Example 7.3. 10°C

8m

8m

d = 600 mm

8m

E, I, a 5 constant E 5 70 GPa I 5 102(106) mm4 a 5 2.36(10–5)/ºC (a) Beam

70°C Temperature Variation

Y

1

1

2

2

8

3

X 1

4

4

MT 5 2 2

3 5

3

Pf 5 6

7

(c) Structure Fixed-Joint Force Vector Due to Temperature Change

(b) Analytical Model

17.498

15.554

216.85 1 16.85 1 0 1 5 216.85 1 25.276 2 8.4252 2

15.554

21.387

21.387

2

1

0.243

0.243

3

0.7291

0.7291

2.6734

2.6734

(d) Member End Forces

3 33 3 0.243 17.498

R5

3 4

20.243 2 0.7291 5 0.7291 1 2.6734 22.6734 0

6 7 8

0.243 kN 17.498 kN • m

5

20.9721 kN 3.4025 kN

22.6734 kN 0

(e) Support Reaction Vector

17.498 kN • m

1

0.243 kN

Fig. 7.20

2

3

0.9721 kN 3.4025 kN (f ) Support Reactions

4

2.6734 kN

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Section 7.5   Temperature Changes and Fabrication Errors  409

Structure Stiffness Matrix: As determined in Example 7.3, the structure stiffness matrix for the beam, in units of kN and meters, is S5

37,140 1,785

4

1,785  6,247.5

(1)

Joint Load Vector: P 5 0

(2)

Structure Fixed-Joint Force Vector Due to Temperature Change: Members 1 and 2 (MT 5 0) By substituting the numerical values of E, I, L, a, d 5 0.6 m, Tt 5 10°C and Tb 5 70°C into Eq. (7.33), we evaluate the member fixed-end moments due to the given temperature change as FMb 5 2FMe 5 70(106)(102)(1026)(2.36)(1025)

1700.62 102 5 16.85 kN ? m  (3a)

FSb 5FSe 5 0(3b) Thus,

Qf1 5 Qf2 5

3 4 0 16.85 0 216.85

3 4 5 1

5 1 6 2

(4)

Member 1

Member 2

Next, using the member code numbers, we store the pertinent elements of Qf 1 and Qf 2 in their proper positions in the structure fixed-joint force vector Pf , as shown in Fig. 7.20(c). Member 3 (MT 5 2) Because MT 5 2 for this member, we substitute the values of fixed-end moments and shears from Eqs. (3) into Eq. (7.19) to obtain the fixed-end force vector for the released member 3 as

F G3 3(216.85) 2(8) 1 16.85 2 (216.85) 2 3(216.85) 01 2(8) 0 02

Qf3 5

5

3.1595 25.276 23.1595 0

4

6 (5) 2 7 8

The relevant elements of Qf 3 are stored in Pf using the member code numbers. The completed structure fixed-joint force vector Pf due to the temperature change, is shown in Fig. 7.20(c). Joint Displacements: The structure stiffness relationship P 2 Pf 5 Sd for the entire beam can be written as 0 328.4252 4 5 37,140 1,785

1,785 6,247.5

4 3dd 4 1 2

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410  Chapter 7   Member Releases and ­Secondary Effects By solving these equations, we determine the joint displacements to be d5

3.6308 1 3214.523 4 2 3 10

24

rad

Ans

Member End Displacements and End Forces: Member 1

u1 5

34 3 4 0 0 0 d1

3 4 5 5 1

0 0 0 3.6308

3 1024

Using k2 from Example 7.3 and Qf 1 from Eq. (4),

Q1 5 k1u1 1 Qf1 5

3

4

0.243 kN 17.498 kN ? m 20.243 kN 215.554 kN ? m

3 4  5 1

Ans

The end forces for member 1 are depicted in Fig. 7.20(d). To generate the support reaction vector R, the pertinent elements of Q1 are stored in R, as shown in Fig. 7.20(e). Member 2

u2 5

34 3 4 0 d1 0 d2

5 1 5 6 2

0 3.6308 3 1024 0 214.523

Using k2 from Example 7.3 and Qf 2 from Eq. (4),

Q2 5 k2u2 1 Qf 2 5

3

20.7291 kN 15.554 kN ? m 0.7291 kN 221.387 kN ? m

4

5 1  6 2

Ans

Member 3

u3 5

34 3 4 0 d2 0 0

6 2 5 7 8

0 214.523 0 0

3 1024

Using k3 from Example 7.3 and Qf 3 from Eq. (5),

Q3 5 k3u3 1 Qf3 5

3

2.6734 kN 21.387 kN ? m 22.6734 kN 0

4

6 2  7 8

Ans

The member end forces are shown in Fig. 7.20(d). Support Reactions: See Figs. 7.20(e) and (f).

Ans

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Section 7.5   Temperature Changes and Fabrication Errors  411



E x ample 7.7 Determine the joint displacements, member end forces, and support reactions for the

plane frame shown in Fig. 7.21(a) due to the combined effect of the following: (a) a temperature increase of 40°C in the girder, and (b) the fabricated length of the left column being 6 mm too short. Use the matrix stiffness method.



S ol u tio n



This frame, subjected to joint and member loads, was analyzed in Example 7.1. Analytical Model: See Fig. 7.21(b). The analytical model used herein is the same as used in Example 7.1. Structure Stiffness Matrix: As determined in Example 7.1, the structure stiffness matrix for the frame, in units of kN and meters, is

S5

3

260,720 0 2260,000 0 0

0 260,720 0 2720 3,600

2260,000 0 260,720 0 3,600

0 2720 0 260,720 23,600

4

0 3,600 3,600  23,600 36,000

(1)

Joint Load Vector: P 5 0

(2)

Structure Fixed-Joint Force Vector Due to Temperature Changes and Fabrication Errors: Member 1 (MT 5 2) By substituting the numerical values of E, A, L, and ea5 26 mm = 20.006 m into Eq. (7.35), we obtain the member fixed-end forces, due to the ­specified fabrication error, as FAb 5 2FAe 5 21,560 kN FSb 5 FSe 5 FMb 5 FMe 5 0

FG FG

As MT 5 2, we use Eq. (7.10) to form the member local fixed-end force vector. Thus,

Qf1 5

21,560 0 0  1,560 0 0

(3)

Using the member transformation matrix, T1, from Example 7.1, we evaluate the global fixed-end force vector as 0 21,560 0 Ff 1 5 TT1 Qf 1 5 0 1,56 0 0

6 7 8 1 2 9

Next, using the member code numbers, we store the pertinent elements of Ff 1 in their proper positions in the 5 3 1 structure fixed-joint force vector Pf , as shown in Fig. 7.21(c).

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412  Chapter 7   Member Releases and ­Secondary Effects

Y Hinged joint

2

4 2

2 1

5m

5

MT 5 1

3

3

9

1

MT 5 2

3

MT 5 1

5m E, A, I, a 5 constant E 5 200 GPa A 5 6,500 mm2 I 5 150 (106) mm4 a 5 12(1026)/˚C

1

4

10

6

X

8

(a) Frame

12

7

11

4

10

12 11 (b) Analytical Model

2.002

2.0020 2.0020

2.0105

2.0020 2

3 3 624

1,560 Pf 5 2624 0 0

1 2 3 4 5

(c) Structure Fixed-Joint Force Vector Due to Temperature Changes and Fabrication Errors

1

2.0103

10.053 2.0103

2.0106

10.053 3

2.0105

2.0106

10.052 2.002

2.002 (d) Member Local End Forces

Fig. 7.21 

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Section 7.5  Temperature Changes and Fabrication Errors  413

2

3 3

6 2.0105 kN 7 22.0020 kN 210.0523 kN•m 8 R5 9 0 10 22.0106 kN 11 2.0020 kN 12 0

2.0105 kN

3

1

4

2.0106 kN

10.0523 kN•m 2.0020 kN

2.0020 kN (f) Support Reactions

(e) Support Reaction Vector

Fig. 7.21  (continued) Member 2 (MT 5 1) By substituting the numerical values of E, A, a, and Tu 5 40°C into Eq. (7.34), we evaluate the member fixed-end forces, due to the given temperature change: FAb 5 2FAe 5 624 kN FSb 5 FSe 5 FMb 5 FMe 5 0

FG

As MT 5 1, we use Eq. (7.6) to form this horizontal member’s local and global fixed-end force vectors: 624 0 0 Ff 2 5 Qf 2 5 2624 0 0

1 2 9  3 4 5

(4)

The relevant elements of Ff 2 are stored in Pf using the member code numbers. The completed structure fixed-joint force vector Pf , due to the temperature change and fabrication error, is given in Fig. 7.21(c). Joint Displacements: Solving the structure stiffness relationship P 2 Pf 5 Sd, we obtain the following joint displacements:

d5

3

20.0027923 m 20.0059923 m 20.0004000 m 20.0000077 m 0.0006385 rad

4

1 2 3 4 5

Ans

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414  Chapter 7   Member Releases and ­Secondary Effects Member End Displacements and End Forces:

FG F G F G F G F G FG F G

Member 1 (MT 5 2)

v1 5

0 0 0 d1 d2 0

6 7 8 5 1 2 9

0 0 0 20.0027923 20.0059923 0

0 0 0 u1 5 T1v1 5 20.0059923 0.0027923 0

Using k1 from Example 7.1 and Qf 1 from Eq. (3),

Q1 5 k1u1 1 Qf1 5

22.0020 kN 22.0105 kN 210.0523 kN ? m  2.0020 kN 2.0105 kN 0

Ans

These end forces are depicted in Fig. 7.21(d).

F1 5 TT1 Q1 5

2.0105 22.0020 210.0523 22.0105 2.0020 0

6 7 8 1 2 9

The pertinent elements of F1 are stored in R, as shown in Fig. 7.21(e). Member 2 (MT 5 1)

u2 5 v2 5

d1 d2 0 d3 d4 d5

1 2 9 5 3 4 5

20.0027923 20.0059923 0 20.0004000 20.0000077 0.0006385

Using k2 from Example 7.1 and Qf 2 from Eq. (4),

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F G FG F G F G F G FG

F2 5 Q2 5 k2u2 1 Qf 2 5

2.0020 kN 22.0103 kN 0 22.0020 kN 2.0103 kN 210.0516 kN ? m

Summary  415

1 2 9  3 4 5

Ans

Member 3 (MT 5 1)

v3 5

0 10 0 0 11 0 0 12 0 5 d3 3 20.0004000 20.0000077 d4 4 0.0006385 d5 5

Using T3 from Example 7.1,

u3 5 k3v3 5

0 0 0 20.0000077 0.0004000 0.0006385

Using k3 from Example 7.1 and Qf 3 5 0, we calculate

Q3 5 k3u3 5

2.0020 kN 2.0106 kN 0 22.0020 kN 22.0106 kN 10.0530 kN ? m



Ans

The member local end forces are shown in Fig. 7.21(d).

F3 5 TT3 Q3 5

22.0106 2.0020 0 2.0106 22.0020 10.0530

10 11 12 3 4 5

Support Reactions: See Figs. 7.21(e) and (f).

Ans

SUMMARY In this chapter, we have extended the matrix stiffness formulation so that it can be used to analyze plane-framed structures containing member releases. Furthermore, the formulation has been extended to include in the analysis, the secondary effects of support displacements, temperature changes, and fabrication errors.

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416  Chapter 7   Member Releases and ­Secondary Effects

In the presence of member releases, the overall analysis procedure remains the same as before, except that the modified expressions for the member local stiffness matrices k and the fixed-end force vectors Qf , developed in Section 7.1, must be used for members with releases. If all the members meeting at a joint are connected to it by hinged connections, then such a joint can be modeled as a hinged joint with its rotation restrained by an imaginary clamp. The effects of support displacements are included in the analysis using the concept of equivalent joint loads. The structure fixed-joint forces, due to the support displacements, are added to the Pf vector by performing the following operations for each member that is attached to a support that undergoes displacements: (a) forming the fixed-end displacement vector vfs from the support displacements, (b) evaluating the fixed-end force vector Ffs 5 Kvfs, and (c) storing the relevant elements of Ffs in Pf using the member code numbers. Once the structure’s joint displacements have been determined by solving its stiffness relationship P 2 Pf 5 Sd, the member end displacement vectors v are formed using both the joint displacements d and the specified support displacements. The rest of the procedure for evaluating member forces and support reactions remains the same as for the case of external loads. The effects of temperature changes and fabrication errors can be included in the analysis methods developed previously, simply by including the member fixed-end forces due to these actions in the local fixed-end force vectors Qf . The expressions for member fixed-end forces, due to temperature changes and fabrication errors, are given in Section 7.5.

PROBLEMS Section 7.1

200 kN

7.1 and 7.2 Determine the joint displacements, member end forces, and support reactions for the beams shown in Figs. P7.1 and P7.2, using the matrix stiffness method.

Hinge 6m

EI 5 constant E 5 70 GPa I 5 400(106) mm4

37.5 kN/m 1

1 Hinge 4m

2

8m

2

3

Fig. P7.2

3m

EI 5 constant E 5 200 GPa I 5 30 (106) mm4

48 kN/m

4

3

1

Fig. P7.1

7.3  Determine the joint displacements, member end forces, and support reactions for the beam shown in Fig. P7.3, by modeling member 3 as being hinged at its right end.

40 kN 40 kN 2

1

2

6m

6m

3 2m

2m

2m

EI 5 constant E 5 70 GPa I 5 270 (106) mm4

Fig. P7.3

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Problems  417

7.4  Determine the joint displacements, member end forces, and support reactions for the beam shown in Fig. P7.4, by modeling member 1 as being hinged at its left end and member 3 as being hinged at its right end.

7.7 and 7.8  Determine the joint displacements, member local end forces, and support reactions for the plane frames shown in Figs. P7.7 and P7.8, by modeling the horizontal member as being hinged at its right end.

30 kN/m

1 1

2

1

3

2

3

10 m

4

10 m

10 m

EI 5 constant E 5 30 GPa I 5 500(106) mm4

125 kN 8m

1 150 kN• m

Fig. P7.4, P7.20

3

7.5  Determine the joint displacements, member end forces, and support reactions for the beam shown in Fig. P7.5 by modeling member 1 as being hinged at its left end. 90 kN 90 kN 18 kN/m

2

1 1

2 5m

15 m

5m

2 3m

Fig. P7.7 4

5m

15 m

15 m

EI 5 constant E 5 200 GPa I 5 400(10 6) mm4

25 kN/m

7.6 Determine the joint displacements, member local end forces, and support reactions for the plane frame shown in Fig. P7.6, using the matrix stiffness method.

50 kN/m 2

Fig. P7.5 4m

2

A 5 2,600 mm2 I 5 31 (106) mm4

3

A 5 2,900 mm2 I 5 24 (106) mm4

25 kN/m

1

37.5 kN/m

2m

12 m E, A, I 5 constant E 5 200 GPa A 5 13,000 mm2 I 5 762(106) mm4

120 kN• m 3

2

3m

25 kN/m

3

24 kN/m

1

Hinged joint

2m

5m E 5 70 GPa

200 kN 2m

Fig. P7.8 6m E, A, I 5 constant E 5 200 GPa A 5 4,000 mm2 I 5 80 (106) mm4

Fig. P7.6

7.9 Determine the joint displacements, member local end forces, and support reactions for the plane frame shown in Fig. P7.9, by modeling the horizontal member as being hinged at its left end and the inclined member as being hinged at its lower end.

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418  Chapter 7   Member Releases and ­Secondary Effects 12 kN/m 1 2

1

75 kN

6m

2 6m 3 10 m

9m

horizontal deflection) limitation of one percent of the frame height? Assume that the braces (inclined members) are connected by hinged connections at both ends. 7.12  Solve Problem 7.11 by assuming that the braces are connected by rigid (moment-resisting) connections at both ends. 7.13  Solve Problem 7.11 by assuming that the frame is unbraced. Note that, instead of developing a new analytical model for the unbraced frame, the previously developed models of the corresponding braced frames can be modified to eliminate the effect of bracing by simply using a very small value for the modulus of elasticity, E, of the bracing members (e.g., E 5 0.000001 kN/m2). 6m

6m

P

E, A, I 5 constant E 5 30 GPa A 5 35,000 mm2 I 5 152(106) mm4

Fig. P7.9 P7.23, P7.33 7.10 Determine the joint displacements, member local end forces, and support reactions for the plane frame shown in Fig. P7.10, using the matrix stiffness method.

8m

2P 3

8m

50 kN/m 375 kN Hinge

P 3

8m 5m

Hinged joints

Columns: A 5 18,200 mm2 I 5 348(106) mm4

E 5 200 GPa Beams: A 5 7,590 mm2 I 5 255(106) mm4

Braces: A 5 1,340 mm2 I 5 3.76(106) mm4

Fig. P7.11, P7.12, P7.13 4m E 5 200 GPa Diagonals: Girder and columns: A 5 1,300 mm2 A 5 23,000 mm2 6 4 I 5 46 (10 ) mm I 5 5.6 (106) mm4

Fig. P7.10 7.11  Using a structural analysis computer program, determine the joint displacements, member local end forces, and reactions for the frame shown in Fig. P7.11 for the value of the load parameter P 5 300 kN. What is the largest value of P that can be applied to the frame without exceeding the drift (maximum

Section 7.2 7.14  Modify the computer program developed in Chapter 5 for the analysis of rigidly connected beams, to include the effect of member releases. Use the modified program to analyze the beams of Problems 7.1 through 7.5, and compare the ­computer-generated results to those obtained by hand calculations. 7.15  Modify the program developed in Chapter 6 for the analysis of rigidly connected plane frames, to include the effect of member releases. Use the modified program to analyze the frames of Problems 7.6 through 7.10, and compare the c­ omputer-generated results to those obtained by hand calculations.

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Problems  419

Section 7.3 7.16  Determine the joint displacements, member axial forces, and support reactions for the plane truss shown in Fig. P7.16, due to the combined effect of the loading shown and a settlement of 10 mm of support 2. Use the matrix stiffness method. 3

4

due to the combined effect of the loading shown and a settlement of 5 mm of support 3. Use the matrix stiffness method. 7.18  Determine the joint displacements, member axial forces, and support reactions for the plane truss shown in Fig. P7.18, due to the combined effect of the loading shown and settlements of 8 and 4 mm, respectively, of supports 3 and 4. Use the matrix stiffness method. 150 kN

1 5

480 kN

6

50 kN

2

6m

EA 5 constant

4

E 5 200 GPa A 5 3,000 mm2

3 4m

1

1

1.5 m

2m

1

2.5 m

8

5

7

2

6

3

2

9 3

4

2

EA 5 constant E 5 200 GPa A 5 3,480 mm2

150 kN 3m

Fig. P7.16, P7.28

3m

3m

Fig. P7.18

7.17  Determine the joint displacements, member axial forces, and support reactions for the plane truss shown in Fig. P7.17, 4m 600 kN 2

7.19 Determine the joint displacements, member end forces, and support reactions for the three-span continuous beam shown in Fig. P7.19 due to settlements of 8 and 30 mm, respectively, of supports 2 and 3. Use the matrix stiffness method.

300 kN 1

1

3

2 1

3

3m

2

4 1

2

7m

7m

7m

EI 5 constant E 5 200 GPa I 5 145(106) mm4

4 300 kN

6m

3

Fig. P7.19, P7.34

4 5

EA 5 constant 3

Fig. P7.17, P7.29, P7.30

E 5 70 GPa A 5 2,500 mm2

7.20  Solve Problem 7.4 for the loading shown in Fig. P7.4 and settlements of 12, 75, 60 and 25 mm, respectively, of supports 1, 2, 3, and 4. 7.21  Determine the joint displacements, member end forces, and support reactions for the beam shown in Fig. P7.21, due to the combined effect of the loading shown and a settlement of 20 mm of the middle support. Use the matrix stiffness method. 7.22 Determine the joint displacements, member local end forces, and support reactions for the plane frame shown in

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420  Chapter 7   Member Releases and ­Secondary Effects 250 kN

15 kN/m

250 kN 3

1

5 1

2

2

4m

3

3m

1.25 I

4

3m

75 kN 2

3

2

4 4m

1

3

8m

1.25 I

I E 5 200 GPa I 5 90 (106) mm4

1

Fig. P7.21

4

10 m

Fig. P7.22, due to a settlement of 25 mm of the right support. Use the matrix stiffness method.

E, A, I 5 constant E 5 30 GPa A 5 90,000 mm2 I 5 675 (106) mm4

9m 1

Fig. P7.24

2 1

2 E, A, I 5 constant E 5 200 GPa A 5 7,610 mm2 I 5 216(106) mm4

6m

3

Fig. P7.22, P7.31, P7.32

7.23  Solve Problem 7.9 for the loading shown in Fig. P7.9 and a settlement of 50 mm of the right support. 7.24 Determine the joint displacements, member local end forces, and support reactions for the plane frame shown in Fig. P7.24, due to the combined effect of the following: (a) the loading shown in the figure, (b) a clockwise rotation of 0.017 radians of the left support, and (c) a settlement of 20 mm of the right support. Use the matrix stiffness method.

7.26  Extend the program developed in Problem 7.14 for the analysis of beams subjected to external loads, to include the effect of support displacements. Use the modified program to analyze the beams of Problems 7.19 through 7.21, and compare the computer-generated results to those obtained by hand calculations. 7.27 Extend the program developed in Problem 7.15 for the analysis of plane frames subjected to external loads, to include the effect of support displacements. Use the modified program to analyze the frames of Problems 7.22 through 7.24, and compare the computer-generated results to those obtained by hand calculations. Section 7.5

7.28  Determine the joint displacements, member axial forces, and support reactions for the plane truss shown in Fig. P7.28, due to a temperature drop of 50°C in member 2. Neglect the joint loads shown in the figure. Use the matrix stiffness method; a 5 12(1026)/°C. 7.29  Determine the joint displacements, member axial forces, Section 7.4 and support reactions for the plane truss shown in Fig. P7.29, due 7.25 Extend the program developed in Chapter 4 for the analy- to the combined effect of the following: (a) the joint loads shown sis of plane trusses subjected to joint loads, to include the effect in the figure, (b) a temperature increase of 40°C in member 2, of support displacements. Use the modified program to ana- (c) a temperature drop of 16°C in member 5, and (d) the fabrilyze the trusses of Problems 7.16 through 7.18, and compare the cated length of member 4 being 4 mm too long. Use the matrix 25 computer-generated results to those obtained by hand calculations. stiffness method; a 5 2.5(10 )/°C.

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Problems  421

7.30  Determine the joint displacements, member axial forces, and support reactions for the plane truss shown in Fig. P7.30, due to the fabricated lengths of members 3 and 4 being 5 mm too short. Neglect the joint loads shown in the figure, and use the matrix stiffness method. 7.31 Determine the joint displacements, member local end forces, and support reactions for the plane frame of Fig. P7.31, due to a temperature increase of 50°C in the two members. Use the matrix stiffness method; a 5 1.2(1025)/°C. 7.32 Determine the joint displacements, member local end forces, and support reactions for the plane frame of Fig. P7.32, due to the fabricated lengths of the two members being 15 mm too short. Use the matrix stiffness method.

7.33 Determine the joint displacements, member local end forces, and support reactions for the plane frame of Fig. P7.33, due to the combined effect of the following: (a) the external loads shown in the figure, and (b) a temperature drop of 60°C in the two members. Use the matrix stiffness method; a 5 1025/°C. 7.34 Determine the joint displacements, member end forces, and support reactions for the beam of Fig. P7.34, due to a linearly varying temperature increase of 55°C at the top surface and 5°C at the bottom surface, of all the members. Use the matrix stiffness method; a 5 1.2(1025)/°C and d 5 300 mm.

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8

Three-Dimensional Framed Structures Learning Objectives At the end of this chapter, you will be able to: 8.1 Analyze Space Trusses 8.2 Analyze Grids 8.3 Analyze Space Frames

Space Truss and Its Analytical Model. (Courtesy of Triodetic)

422

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Section 8.1   Space Trusses  423

Up to this point, we have focused our attention on the analysis of plane-framed structures. While many actual three-dimensional structures can be d­ivided into planar parts for the purpose of analysis, there are others (e.g., ­lattice domes and transmission towers) that, because of the arrangement of their members or ­applied loading, cannot be divided into plane structures. Such structures are analyzed as space structures subjected to three-dimensional loadings. The ­matrix stiffness analysis of space structures is similar to that of plane ­structures—­except, of course, that member stiffness and transformation matrices appropriate for the particular type of space structure under consideration are now used in the analysis. In this chapter, we extend the matrix stiffness formulation, developed for plane structures, to the analysis of three-dimensional or space structures. Three types of space-framed structures are considered: space trusses, grids, and space frames, with methods for their analysis presented in Sections 8.1, 8.2, and 8.3, respectively. The computer programs for the analysis of space-framed structures can be conveniently adapted from those for plane structures, via relatively straightforward modifications that should become apparent as the analysis of space structures is developed in this chapter. Therefore, the details of programming the analysis of space structures are not covered herein; they are, instead, left as exercises for the reader.

8.1 

SPACE TRUSSES A space truss is defined as a three-dimensional assemblage of straight prismatic members connected at their ends by frictionless ball-and-socket joints, and subjected to loads and reactions that act only at the joints. Like plane trusses, the members of space trusses develop only axial forces. The matrix stiffness analysis of space trusses is similar to that of plane trusses developed in Chapter 3 (and modified in Chapter 7). The process of developing the analytical models of space trusses (and numbering the degrees of freedom and restrained coordinates) is essentially the same as that for plane trusses (Chapter 3). The overall geometry of the space truss, and its joint loads and displacements, are described with reference to a global Cartesian or rectangular right-handed XYZ coordinate system, with three global (X, Y, and Z) coordinates now used to specify the location of each joint. Furthermore, since an unsupported joint of a space truss can translate in any direction in the three-dimensional space, three displacements—the translations in the X, Y, and Z directions—are needed to completely establish its deformed position. Thus, a free joint of a space truss has three degrees of freedom, and three structure coordinates (i.e., free and/or restrained coordinates) need to be defined at each joint, for the purpose of analysis. Thus, NCJT 5 3 f for space trusses NDOF 5 3(NJ) 2 NR   



(8.1)

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424  Chapter 8   Three-Dimensional Framed Structures

The procedure for assigning numbers to the structure coordinates of a space truss is analogous to that for plane trusses. The degrees of freedom of the space truss are numbered first by beginning at the lowest-numbered joint with a degree of freedom, and proceeding sequentially to the highest-numbered joint. If a joint has more than one degree of freedom, then the translation in the X direction is numbered first, followed by the translation in the Y direction, and then the translation in the Z direction. After all the degrees of freedom have been numbered, the restrained coordinates of the space truss are numbered in the same manner as the degrees of freedom. Consider, for example, the three-member space truss shown in Fig. 8.1(a). As the analytical model of the truss depicted in Fig. 8.1(b) indicates, the structure has three degrees of freedom (NDOF 5 3), which are the translations d1, d2, and d3 of joint 2 in the X, Y, and Z directions, respectively; and nine restrained coordinates (NR 5 9), which are identified as R4 through R12 at the support joints 1, 3, and 4. As in the case of plane trusses, a local right-handed xyz coordinate system is established for each member of the space truss. The origin of the local coordinate system is located at one of the ends (which is referred to as the beginning of the member), with the x axis directed along the member’s centroidal axis in its undeformed state. Since the space truss members can only develop axial forces, the positive directions of the y and z axes can be chosen arbitrarily, provided that the x, y, and z axes are mutually perpendicular and form a righthanded coordinate system (Fig. 8.2). Y

Y

P2 , d2 2

P1, d1

P3 , d3

1

3 2

1

4

R4

X

R10

R6

R12

3 R5

X

R7

R11

R9 Z

(a) Space Truss

Z R8 (b) Analytical Model

Fig. 8.1

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Section 8.1   Space Trusses  425 e

x

y

b

Y

z X Z

Fig. 8.2  Local Coordinate System for Members of Space Trusses

Member Stiffness Relations in the Local Coordinate System To establish the member local stiffness relations, let us focus our attention on an arbitrary prismatic member m of a space truss. When the truss is subjected to external loads, member m deforms and axial forces are induced at its ends. The initial and displaced positions of the member are shown in Fig. 8.3(a). As this figure indicates, three displacements—translations in the x, y, and z directions—are needed to completely specify the displaced position of each end of the member. Thus, the member has a total of six degrees of freedom or end displacements. However, as discussed in Section 3.3 (see Figs. 3.3(d) and (f)), small end displacements in the directions perpendicular to a truss member’s centroidal axis do not cause any forces in the member. Thus, the end displacements uby, ubz, uey, and uez in the directions of the local y and z axes of the member, as shown in Fig. 8.3(a), are usually not evaluated in the analysis; and for analytical purposes, the member is considered to have only two degrees of freedom, u1 and u2, in its local coordinate system. Thus, the local end displacement vector u for a member of a space truss is expressed as u5

u1

3u 4 2

in which u1 and u2 represent the displacements of the member ends b and e, respectively, in the direction of the member’s local x axis, as shown in Fig. 8.3(a). As this figure also indicates, the member end forces corresponding to the end displacements u1 and u2 are denoted by Q1 and Q2, respectively.

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426  Chapter 8   Three-Dimensional Framed Structures y

Displaced position e9 b9

Initial position uey

uby

m

b

Q1

e Q2

ubz u1

x

uez

u2

L EA 5 constant

z (a) Member Forces and Displacements in the Local Coordinate System y

u1 5 1

k11 5

b9

EA L

e

b

k21 5 2

EA L

x

k22 5

EA L

x

L z (b) y

u2 5 1

k12 5 2

EA L

b

e9 e L

z (c)

Fig. 8.3

The relationship between the local end forces Q and the end displacements u, for the members of space trusses, is written as

Q 5 ku 

(8.2)

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Section 8.1   Space Trusses  427

in which k represents the 2 3 2 member stiffness matrix in the local coordinate system. The explicit form of k can be obtained by subjecting the member to the unit end displacements, u1 5 1 and u2 5 1, as shown in Figs. 8.3(b) and (c), respectively, and evaluating the corresponding member end forces. Thus, the local stiffness matrix for the members of space trusses can be explicitly expressed as k5

3

4

1 EA L 21

21  1

(8.3)

Coordinate Transformations Consider an arbitrary member m of a space truss, as shown in Fig. 8.4(a), and let Xb, Yb, Zb, and Xe, Ye, Ze be the global coordinates of the joints to which the member ends b and e, respectively, are attached. The length and the direction x

Y

e (Xe , Ye , Ze )

m (Xb , Yb , Zb)

L

b

X

0

Z (a) Space Truss e9 Displaced position

Y

e b9 θY

Q1 θZ

x

m θX

b

Q2

X

u2

Initial position

u1

Z (b) Member End Forces and End Displacements in the Local Coordinate System

Fig. 8.4

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428  Chapter 8   Three-Dimensional Framed Structures

5 e9 Displaced position

Y

F5

5

x

e b9

F2 2

F4

m F6

F1

b

X

3

F3

Initial position

1

6 4

Z (c) Member End Forces and End Displacements in the Global Coordinate System

Fig. 8.4  (continued)

cosines of the member can be expressed in terms of the global coordinates of its ends by the following relationships: L 5 Ï(Xe 2 Xb)2 1 (Ye 2 Yb)2 1 (Ze 2 Zb)2 cos uX 5 cos uY 5 cos uZ 5

Xe 2 Yb L Ye 2 Yb L Ze 2 Zb L

(8.4a)



(8.4b)



(8.4c)



(8.4d)

in which uX, uY and uZ represent the angles between the positive directions of the global X, Y, and Z axes, respectively, and the positive direction of the member’s local x axis, as shown in Fig. 8.4(b). Note that the origin of the global coordinate system is shown to coincide with that of the local coordinate system in this figure. With no loss in generality of the formulation, this convenient arrangement allows the angles between the local and global axes to be clearly visualized. It is important to realize that the member transformation matrix depends only on the angles between the local and global axes, regardless of whether or not the origins of the local and global coordinate systems coincide. Also shown in Fig. 8.4(b) are the member end displacements u and end forces Q in the local coordinate system; the equivalent systems of end displacements v and end forces F, in the global coordinate system, are depicted in Fig. 8.4(c). As indicated in Fig. 8.4(c), the global member end displacements v and end forces F are numbered by beginning at member end b, with the translation and force in the X direction numbered first, followed by the translation and force in the Y direction, and then the translation and force in the Z direction. The

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Section 8.1   Space Trusses  429

displacements and forces at the member’s opposite end e are then numbered in the same sequential order. Let us consider the transformation of member end forces and end displacements from a global to a local coordinate system. By comparing Figs. 8.4(b) and (c), we observe that at end b of the member, the local force Q1 must be equal to the algebraic sum of the components of the global forces F1, F2, and F3 in the direction of the local x axis; that is, Q1 5 F1 cos uX 1 F2 cos uY 1 F3 cos uZ (8.5a) Similarly, at end e of the member, we can express Q2 in terms of F4, F5, and F6 as Q2 5 F4 cos uX 1 F5 cos uY 1 F6 cos uZ Equations 8.5(a) and (b) can be written in matrix form as

3Q 4 5 3

cos uX 0

Q1 2

cos uY 0

cos uZ 0

0 cos uX

0 cos uY

0 cos uZ

4

FG F1 F2 F3 F4 F5 F6

(8.5b)

(8.6)

Equation (8.6) can be symbolically expressed as Q 5 TF, with the 2 3 6 transformation matrix T given by T5

3cos0 u

X

cos uY 0

cos uZ 0

0 cos uX

0 cos uY

4

0 cos uZ

(8.7)

Since member end displacements, like end forces, are vectors, which are defined in the same directions as the corresponding forces, the foregoing transformation matrix T can also be used to transform member end displacements from the global to the local coordinate system; that is, u 5 Tv. Next, we examine the transformation of member end forces from the local to the global coordinate system. A comparison of Figs. 8.4(b) and (c) indicates that at end b of the member, the global forces F1, F2, and F3 must be the components of the local force Q1 in the directions of the global X, Y, and Z axes, respectively; that is, F1 5 Q1 cos uX  F2 5 Q1 cos uY  F3 5 Q1 cos uZ (8.8a) Similarly, at end e of the member, the global forces F4, F5, and F6 can be expressed as the components of the local force Q2, as F4 5 Q2 cos uX  F5 5 Q2 cos uY  F6 5 Q2 cos uZ (8.8b)

FG F G

We can write Eqs. 8.8(a) and (b) in matrix form as F1 F2 F3 F4 F5 F6

5

cos uX cos uY cos uZ 0 0 0

0 0 0 cos uX cos uY cos uZ

3QQ 4 (8.9) 1 2

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430  Chapter 8   Three-Dimensional Framed Structures

As the first matrix on the right side of Eq. (8.9) is the transpose of the transformation matrix T (Eq. (8.7)), the equation can be symbolically expressed as F 5 TT 631 632

Q 231

(8.10)

It may be of interest to note that the transformation relationship analogous to Eq. (8.10) for member end displacements (i.e., v 5 TTu) is not defined for space truss members, with two degrees of freedom, as used herein. This is because the local end displacement vectors u for such members do not contain the displacements of the member ends in the local y and z directions. As discussed previously, while the end forces in the local y and z directions of the members of space trusses are always 0, the displacements of the member ends in the local y and z directions are generally nonzero (see Fig. 8.3(a)). However, the foregoing limitation of the two-degree-of-freedom member model has no practical consequences, because the transformation relation v 5 TTu is needed neither in the formulation of the matrix stiffness method of analysis, nor in its application.

Member Stiffness Relations in the 
 Global Coordinate System As in the case of plane trusses, the relationship between the global end forces F and the end displacements v for the members of space trusses is expressed as F 5 Kv, with the member global stiffness matrix K given by the equation K 5 TT 636 632

K5

EA L

F

cos2 uX cos uX cos uY cos uX cos uZ 2cos2 uX 2cos uX cos uY 2cos uX cos uZ

k 232

T 236



(8.11)

The explicit form of the 6 3 6 K matrix can be determined by substituting Eqs. (8.3) and (8.7) into Eq. (8.11) and performing the required matrix multiplications. The explicit form of the member global stiffness matrix K, thus obtained, is given in Eq. (8.12). cos uX cos uY cos2 uY cos uY cos uZ 2cos uX cos uY 2cos2 uY 2cos uY cos uZ

cos uX cos uZ cos uY cos uZ cos2 uZ 2cos uX cos uZ 2cos uY cos uZ 2cos2 uZ

2cos2 uX 2cos uX cos uY 2cos uX cos uZ cos2 uX cos uX cos uY cos uX cos uZ

2cos uX cos uY 2cos2 uY 2cos uY cos uZ cos uX cos uY cos2 uY cos uY cos uZ

2cos uX cos uZ 2cos uY cos uZ 2cos2 uZ cos uX cos uZ cos uY cos uZ cos2 uZ

G

(8.12)

Procedure for Analysis The procedure for the analysis of plane trusses developed in Chapter 3 (see block diagram in Fig. 3.20), and modified in Chapter 7, can be used to analyze space trusses provided that: (a) three structure coordinates (i.e., degrees of freedom and/or restrained coordinates), in the global X, Y, and Z directions, are defined at each joint; and (b) the member stiffness and transformation matrices

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Section 8.1   Space Trusses  431

developed in this section (Eqs. (8.3), (8.7), and (8.12)) are used in the analysis. The procedure is illustrated by the following example.



E x ample 8.1 Determine the joint displacements, member axial forces, and support reactions for the space truss shown in Fig. 8.5(a) by the matrix stiffness method.



S olut i o n

Analytical Model:  See Fig. 8.5(b). The truss has three degrees of freedom, which are the translations of joint 5 in the X, Y, and Z directions. These are numbered 1, 2, and 3, respectively. The twelve restrained coordinates of the truss are identified by numbers 4 through 15 in the figure. Structure Stiffness Matrix: Member 1  From Fig. 8.5(b), we can see that joint 1 is the beginning joint, and joint 5 is the end joint, for this member. By applying Eqs. (8.4), we determine L 5 Ï(X5 2 X1)2 1 (Y5 2 Y1)2 1 (Z5 2 Z1)2 5 Ï(0 1 1.5)2 1 (6 2 0)2 1 (0 2 2)2 5 6.5 m Y

400 kN 200 kN

6m

2m 0

1.5 m 1.5 m

X 2m

1.5 m 1.5 m

Z EA 5 constant E 5 70 GPa A 5 3700 mm2 (a) Space Truss

Fig. 8.5

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432  Chapter 8   Three-Dimensional Framed Structures Y 2 5

1

3 3

4

12

15

1

10

3

13 2

4

11

X

0 1

14 2

4

5

6

7

9 8

Z

(b) Analytical Model

S5

3

1

2

(2,122.00 + 6,795.87 +2,122.00 + 6,795.87)

(8,487.99 2 13,591.75 28.487.99 + 13,591.75)

(–2,829.30 + 4,530.53 22,829.30 + 4,530.53)

3

(8,487.99 2 13,591.75 28.487.99 + 13,591.75)

(33,951.98 + 27,183.49 +33,951.98 + 27,183.49)

(211,317.20 2 9,061.06 +11,317.20 + 9,061.06)

(22,829.30 + 4,530.53 22,829.30 + 4,530.53)

(211,317.20 29,061.06 (3,772.36 + 3,020.32 +11,317.20 + 9,061.06) +3,772.36 + 3,020.32) (c) Structure Stiffness Matrix 222.2327 288.9309

R=

29.6433 5.5362 211.0724 3.6908

1

2

3

3

17,835.7 0 3,402.5 1 25 0 122,271 0 2 kN/m 3,402.5 0 13,585.4 3 3

4 5 6 7 8 9

277.7684 311.0735 103.6900

10 11 12

188.9298 62.9759

15

94.4649

33 1

kN

13 14

(d) Support Reaction Vector

Fig. 8.5  (continued)

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Section 8.1   Space Trusses  433

Y 400 kN

200 kN

5

62.9759 3 77.7684

94.4649 4

103.6900

311.0735

0

X

188.9298

1 22.2327

5.5362

2

3.6908

29.6433

11.0724

88.9309 Z

(e) Support Reactions

Fig. 8.5  (continued) X5 2 X1

0 1 1.5 5 0.23077 6.5 620 cos uY 5 5 5 0.92308 L 6.5 Z5 2 Z1 0 2 2 cos uZ 5 5 5 20.30769 L 6.5 cos uX 5

K1 5

F

L Y5 2 Y1

5

By substituting E 5 70 GPa, A 5 0.0037 m2, L 5 6.5 m, and the foregoing direction cosines, into Eq. (8.12), we calculate the member’s global stiffness matrix to be 4 2,122.00 8,487.99 22,829.30 22,122.00 28,487.99 2,829.30

5 8,487.99 33,951.98 211,317.20 28,487.99 233,951.98 11,317.20

6 22,829.30 211,317.20 3,772.36 2,829.30 11,317.20 23,772.36

1 22,122.00 28,487.99 2,829.30 2,122.00 8,487.99 22,829.30

2 28,487.99 233,951.98 11,317.20 8,487.99 33,951.98 211,317.20

G

3 2,829.30 11,317.20 23,772.36 22,829.30 211,317.20 3,772.36

4 5 6 kN/m 1 2 3

Next, by using the member code numbers 4, 5, 6, 1, 2, 3, we store the pertinent elements of K1 in the 3 3 3 structure stiffness matrix S in Fig. 8.5(c).

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434  Chapter 8   Three-Dimensional Framed Structures Member 2 L 5 Ï(X5 2 X2)2 1 (Y5 2 Y2)2 1 (Z5 2 Z2)2 5 Ï(0 2 3)2 1 (6 2 0)2 1 (0 2 2)2 5 7 m

F

7 6,795.87 213,591.75 K2 5 4,530.53 26,795.87 13,591.75 24,530.53

X5 2 X2

023 5 20.42857 7 620 cos uY 5 5 5 0.85714 L 7 Z5 2 Z2 0 2 2 cos uZ 5 5 5 20.28571 L 7 cos uX 5

8 213,591.75 27,183.49 29,061.06 13,591.75 227,183.49 29,061.06

L Y5 2 Y2

5

9 4,530.53 29,061.06 3,020.32 24,530.53 9,061.06 23,020.32

1 26,795.87 13,591.75 24,530.53 6,795.87 213,591.75 4,530.53

2 13,591.75 227,183.49 9,061.06 213,591.75 27,183.49 9,061.06

3 24,530.53 9,061.06 23,020.32 4,530.53 29,061.06 3,020.32

G

7 8 9 1 2 3

kN/m

Member 3 L 5 Ï(X5 2 X3)2 1 (Y5 2 Y3)2 1 (Z5 2 Z3)2 5 Ï(0 2 1.5)2 1 (6 2 0)2 1 (0 1 2)2 5 6.5 m

K3 5

F

X5 2 X3

0 2 1.5 5 20.23077 6.5 620 cos uY 5 5 5 0.92308 L 6.5 Z5 2 Z3 0 1 2 cos uZ 5 5 5 0.30769 L 6.5 cos uX 5

10 2,122.00 28,487.99 22,829.30 22,122.00 8,487.99 2,829.30

11 28,487.99 33,951.98 11,317.20 8,487.99 233,951.98 211,317.20

L Y5 2 Y3

5

12 22,829.30 11,317.20 3,772.36 2,829.30 211,317.20 23,772.36

1 22,122.00 8,487.99 2,829.30 2,122.00 28,487.99 22,829.30

2 8,487.99 233,951.98 211,317.20 28,487.99 33,951.98 11,317.20

3 2,829.30 211,317.20 23,772.36 22,829.30 11,317.20 3,772.36

G

10 11 12 kN/m 1 2 3

Member 4 L 5 Ï(X5 2 X4)2 1 (Y5 2 Y4)2 1 (Z5 2 Z4)2 5 Ï(0 1 3)2 1 (6 2 0)2 1 (0 1 2)2 5 7 m

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Section 8.1   Space Trusses  435

F

13 6,795.87 13,591.75 K4 5 4,530.53 26,795.87 213,591.75 24,530.53

X5 2 X4

013 5 0.42857 7 620 cos uY 5 5 5 0.85714 L 7 Z5 2 Z4 0 1 2 cos uZ 5 5 5 0.28571 L 7 cos uX 5

14 13,591.75 27,183.49 9,061.06 213,591.75 227,183.49 29,061.06

L Y5 2 Y4

5

15 4,530.53 9,061.06 3,020.32 24,530.53 29,061.06 23,020.32

1 26,795.87 213,591.75 24,530.53 6,795.87 13,591.75 4,530.53

2 213,591.75 227,183.49 29,061.06 13,591.75 27,183.49 9,061.06

3 24,530.53 29,061.06 23,020.32 4,530.53 9,061.06 3,020.32

G

13 14 15 1 2 3

kN/m

The complete structure stiffness matrix S, obtained by assembling the pertinent stiffness coefficients of the four members of the truss, is given in Fig. 8.5(c). Joint Load Vector:  By comparing Figs. 8.5(a) and (b), we obtain 0 P 5 2400 kN 2200

3 4

Joint Displacements:  By substituting P and S into the structure stiffness relationship, P 5 Sd, we write

3 4 3 0 2400 2200

5

17,835.7 0 3,402.5

0 122,271 0

3,402.5 0 13,585.4

43 4 d1 d2 d3

By solving the foregoing equations, we determine the joint displacements to be

3

0.0029493 d 5 20.0032714 20.0154604

4

1 2 m 3

Ans

Member End Displacements and End Forces:

F G

Member 1  Using its code numbers, we determine the member’s global end displacements to be

v1 5

0 0 0 0.0029493 20.0032714 20.0154604

4 5 6 m 1 2 3

To determine the member’s end displacements in the local coordinate system, we first evaluate its transformation matrix as defined in Eq. (8.7): T1 5

30.23077 0

0.92308 0

20.30769 0

0 0.23077

0 0.92308

4

0 20.30769

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436  Chapter 8   Three-Dimensional Framed Structures The member local end displacements can now be calculated by using the relationship u 5 Tv, as u1 5 T1v1 5

300.00241784 m

Before we can evaluate the member local end forces, we need to determine the local stiffness matrix k, using Eq. (8.3): k1 5

39,846.15 3239,846.15

4

239,846.15 kN/m 39,846.15

Now, we can compute the member local end forces by using the relationship Q 5 ku, as Q1 5 k1u1 5

3

4

296.342 kN 96.342

in which the negative sign of the first element of Q1 indicates that the member axial force is tensile; that is, Qa1 5 96.342 kT (T)

F G F G

Ans

By applying the relationship F 5 T Q, we determine the member end forces in the global coordinate system to be 4 222.2327 5 288.9309 6 29.6433 kN F1 5 TT1 Q1 5 1 22.2327 2 88.9309 3 229.6433 T

Using the member code numbers 4, 5, 6, 1, 2, 3, the pertinent elements of F1 are stored in their proper positions in the support reaction vector R, as shown in Fig. 8.5(d). Member 2

v2 5

T2 5

7 8 9 m 1 2 3

0 0 0 0.0029493 20.0032714 20.0154604

3

20.42857 0

u2 5 T2v2 5

0.85714 0

20.28571 0

0 20.42857

0 0.85714

4

0 20.28571

30.00034914 m 0

3237,000.00 37,000.004 kN/m 212.918 Q 5ku 53 kN 12.918 4 k2 5 2

37,000.00

237,000.00

2 2

Qa2 5 12.918 kN (T)

Ans

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F G F G

5.5362 211.0724 3.6908 F2 5 TT2 Q2 5 25.5362 11.0724 23.6908

Section 8.1   Space Trusses  437

7 8 9 kN 1 2 3

Member 3

v3 5

0 0 0 0.0029493 20.0032714 20.0154604

10 11 12 m 1 2 3

0.92308 0.30769 320.23077 0 0 0 0 u 5Tv 53 m 20.00845744

T3 5

3

0 20.23077

0 0.92308

4

0 0.30769

3 3

k3 5 k1 Q 3 5 k3 u3 5

32336.9954 kN 336.995

F G F G

Qa3 5 336.995 kN (C)

F3 5 TT3 Q3 5

277.7684 311.0735 103.6900 77.7684 2311.0735 2103.6900

Ans

10 11 12 kN 1 2 3

Member 4

v4 5

0 0 0 0.0029493 20.0032714 20.0154604

13 14 15 m 1 2 3

30

0.85714 0

u4 5 T4v4 5

320.00595734 m

T4 5

0.42857

0.28571 0

0 0.42857

0 0.85714

4

0 0.28571

0

k4 5 k2

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438  Chapter 8   Three-Dimensional Framed Structures

Q4 5 k4u4 5

32220.4194 kN 220.419

F G

Qa4 5 220.419 kN (C)

F4 5 TT4 Q4 5

94.4649 188.9298 62.9759 294.4649 2188.9298 262.9759

Ans

13 14 15 kN 1 2 3

Support Reactions:  The completed reaction vector R is shown in Fig. 8.5(d), and the support reactions are depicted on a line diagram of the truss in Fig. 8.5(e). Ans Equilibrium Check:  Applying the equations of equilibrium to the free body of the entire space truss (Fig. 8.5(e)), we obtain

oF 5 0 1c o F 5 0 1 b oF 5 0 1 oM 5 0 1S

X

Y

[

Z

X

oM

[

1

Y

50

222.2327 1 5.5362 2 77.7684 1 94.4649 5 0 

Checks

288.9309 2 11.0724 1 311.0735 1 188.9298 2 400 5 0

Checks

29.6433 1 3.6908 1 103.6900 1 62.9759 2 200 5 0

Checks

88.9309(2) 1 11.0724(2) 1 311.0735(2)  1 188.9298(2) 2 200(6) < 0

Checks

222.2327(2) 1 29.6433(1.5) 1 5.5362(2) 2 3.6908(3) 1 77.7684(2) 2 103.6900(1.5) 2 94.4649(2) 1 62.9759(3) < 0



Checks

oM

[

1

Z

50

88.9309(1.5) 2 11.0724(3) 1 311.0735(1.5)  2 188.9298(3) 5 0

Checks

8.2 GRIDS A grid is defined as a two-dimensional framework of straight members connected together by rigid and/or flexible connections, and subjected to loads and reactions perpendicular to the plane of the structure. Because of their widespread use as supporting structures for long-span roofs and floors, the analysis of grids is usually formulated with the structural framework lying in a horizontal plane (unlike plane frames, which are oriented in a vertical plane), and subjected to external loads acting in the vertical direction, as shown in Fig. 8.6(a) on the next page. Grids are composed of members that have doubly symmetric cross-­sections, with each member oriented so that one of the planes of symmetry of its crosssection is in the vertical direction; that is, perpendicular to the plane of the structure, and in (or parallel to) the direction of the external loads (Fig. 8.6(a)). Under the action of vertical external loads, the joints of a grid can translate in the vertical direction and can rotate about axes in the (horizontal) plane of the

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Section 8.2   Grids  439 Y

X

0

A

Vertical plane of symmetry

A

Horizontal plane of symmetry Z

Section A–A Member cross-section

Grid (a) Y

R15

5

0

R14

X

4 P3, d3 2

3

R11

R13 P4, d4 4

P5, d5 R12

3

R10 2

R7 1

P1, d1 1 P2, d2

R8

R6

R9

Z

Fig. 8.6

(b) Analytical Model

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440  Chapter 8   Three-Dimensional Framed Structures Y

5 y2

y4

2

x2

A y3

z2

x1 1

x4 x3

4

3 y1

4

X

3

A

z4

y3

x3

1 2

z3

z3

z1 Section A–A

Z (c) Member Local Coordinate Systems

Fig. 8.6  (continued)

structure, while the grid members may be subjected to torsion, and uniaxial bending out of the plane of the structure.

Analytical Model and Degrees of Freedom The process of dividing grids into members and joints, for the purpose of analysis, is the same as that for beams and plane frames—that is, a grid is divided into members and joints so that all of the members are straight and prismatic, subjected to known loads between their ends, and all the external reactions act only at the joints. Consider, for example, the grid of Fig. 8.6(a). The analytical model of the grid, as depicted in Fig. 8.6(b), shows that, for analysis, the grid is considered to be composed of four members and five joints. The overall geometry of the grid, and its joint loads and displacements, are described with reference to a global right-handed XYZ coordinate system, with the structure lying in the horizontal XZ plane, as shown in Fig. 8.6(b). Two global (X and Z) coordinates are needed to specify the location of each joint. For each member of the grid, a local xyz coordinate system is established, with its origin at an end of the member and the x axis directed along the ­member’s centroidal axis in the undeformed state. The local y and z axes are

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Section 8.2   Grids  441

o­ riented, respectively, parallel to the vertical and horizontal axes of symmetry (or the principal axes of inertia) of the member cross-section. The positive direction of the local x axis is defined from the beginning toward the end of the member; the local y axis is considered positive upward (i.e., in the positive direction of the global Y axis); and the positive direction of the local z axis is defined so that the local xyz coordinate system is right-handed. The local coordinate systems selected for the four members of the example grid are depicted in Fig. 8.6(c). As discussed previously, an unsupported joint of a grid can translate in the global Y direction and rotate about any axis in the XZ plane. Since small rotations can be treated as vector quantities, the foregoing joint rotation can be conveniently represented by its component rotations about the X and Z axes. Thus, a free joint of a grid has three degrees of freedom—the translation in the Y direction and the rotations about the X and Z axes. Therefore, three structure coordinates (i.e., free and/or restrained coordinates) need to be defined at each joint of the grid for the purpose of analysis; that is, NCJT 5 3  f for grids NDOF 5 3(NJ) 2 NR   

(8.13)

[

The procedure for numbering the structure coordinates of grids is analogous to that for other types of framed structures. The degrees of freedom are numbered before the restrained coordinates. In the case of a joint with multiple degrees of freedom, the translation in the Y direction is numbered first, followed by the rotation about the X axis, and then the rotation about the Z axis. After all the degrees of freedom have been numbered, the grid’s restrained coordinates are numbered in the same manner as the degrees of freedom. In Fig. 8.6(b), the degrees of freedom and restrained coordinates of the example grid are numbered using this procedure. It should be noted from this figure that the rotations and moments are now represented by double-headed arrows ( ), instead of the curved arrows ( ) used previously for plane structures. The double-headed arrows provide a convenient and unambiguous means of ­representing rotations and moments in three-dimensional space. To represent a rotation (or a moment/couple), an arrow is drawn pointing in the positive ­direction of the axis about which the rotation occurs (or the moment/couple acts). The positive sense (i.e., clockwise or counterclockwise) of the rotation (or moment/couple) is indicated by the curved fingers of the right hand with the extended thumb pointing in the direction of the arrowheads, as shown in Fig. 8.7.

Member Stiffness Relations in the Local Coordinate System When a member with a noncircular (e.g., rectangular or I-shaped) cross-­section is subjected to torsion, its initially plane cross-sections become warped surfaces; restraint of this warping, or out-of-plane deformation, of cross-sections can induce bending stresses in the member. Thus, in the analysis of grids and space frames (to be developed in the next section), it is commonly assumed that

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442  Chapter 8   Three-Dimensional Framed Structures Axis of rotation or moment/couple

Y

X

Z

Fig. 8.7  Representation of Rotation or Moment/Couple in Three-Dimensional Space

the cross sections of all the members are free to warp out of their planes under the action of torsional moments. This assumption, together with the previously stated condition about the cross-sections of grid members being doubly symmetric with one of the planes of symmetry oriented parallel to the direction of applied loads, has the effect of uncoupling the member’s torsional and bending stiffnesses so that a twisting (or torsional deformation) of the member induces only torsional moments but no bending moments, and vice versa. With the torsional and bending effects uncoupled, the local stiffness relations for the members of grids can be obtained by simply extending the stiffness relations for beams (Chapter 5) to include the familiar torsional stiffness relations found in textbooks on mechanics of materials. Consider an arbitrary member m of a grid, as shown in Fig. 8.8(a) on the next page. Like a joint of a grid, three displacements are needed to completely specify the displaced position of each end of the grid member. Thus, the member has a total of six degrees of freedom. In the local coordinate system of the member, the six member end displacements are denoted by u1 through u6, and the associated member end forces are denoted by Q1 through Q6, as shown in Fig. 8.8(a). As indicated in this figure, a member’s local end displacements and end forces are numbered by beginning at its end b, with the translation and the force in the y direction numbered first, followed by the rotation and moment about the x axis, and then the rotation and moment about the z axis. The displacements and forces at the member’s opposite end e are then numbered in the same sequential order. The relationship between the end forces Q and the end displacements u, for the members of grids, can be expressed as Q 5 ku 1 Qf

(8.14)

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Section 8.2   Grids  443 y

Q1, u1

Q4, u4 e

b

Q2, u2

x

Q5, u5

m Q3, u3

Q6, u6 L E, I, J 5 constant

z

(a) Member Forces and Displacements in the Local Coordinate System y b9 u1 5 1 k21 5 0

k51 5 0

b 6EI L2

k31 5

e

k61 5

12EI k11 5 3 L

6EI L2

x

12EI L3

k41 5 2

z (b) u1 5 1, u2 5 u3 5 u4 5 u5 5 u6 5 0 u2 5 1

y

z b

y

k22 5

GJ L k32 5 0

Member cross-section at beginning

e

b

x

GJ k52 5 2 L k62 5 0 k42 5 0

k12 5 0

z (c) u2 5 1, u1 5 u3 5 u4 5 u5 5 u6 5 0 y

u3 5 1

b

e

k23 5 0 k33 5

4EI L

k13 5

6EI L2

k63 5

2EI L

k53 5 0

k43 5 2

x

6EI L2

z (d) u3 5 1, u1 5 u2 5 u4 5 u5 5 u6 5 0

Fig. 8.8

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444  Chapter 8   Three-Dimensional Framed Structures y e9

k24 5 0 k34 5 2

6EI L2

u4 5 1

e

b

k14 5 2

12EI L3

k64 5 2

k54 5 0

6EI L2

k44 5

x

12EI L3

z (e) u4 5 1, u1 5 u2 5 u3 5 u5 5 u6 5 0 y

GJ k25 5 2 L k35 5 0

u5 5 1

b

e

GJ L

k55 5

x

y

z e

k65 5 0

k15 5 0

k45 5 0

Member cross-section at end

z (f) u5 5 1, u1 5 u2 5 u3 5 u4 5 u6 5 0 y

k66 5 k26 5 0 k36 5

2EI L

4EI L

e

b

k56 5 0

u6 5 1 k16 5

6EI L2

k46 5 2

x

6EI L2

z (g) u6 5 1, u1 5 u2 5 u3 5 u4 5 u5 5 0 y

e

b

Qf 2

Qf 5

x

Qf6

Qf 3 Qf 1

Qf4

z (h) Member Fixed-End Forces in the Local Coordinate System (u1 5 u2 5 u3 5 u4 5 u5 5 u6 5 0)

Fig. 8.8  (continued)

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Section 8.2   Grids  445

in which k represents the 6 3 6 member stiffness matrix in the local coordinate system, and Qf denotes the 6 3 1 member local fixed-end force vector. Like the other types of framed structures, the explicit form of k for grid members can be obtained by subjecting a member, separately, to unit values of each of the six end displacements, as shown in Figs. 8.8(b) through (g), and evaluating the corresponding member end forces. The stiffness coefficients required to cause the unit values of the member end displacements u1, u3, u4, and u6, are shown in Figs. 8.8(b), (d), (e), and (g), respectively. The expressions for these stiffness coefficients were derived in Section 5.2. To derive the expressions for the torsional stiffness coefficients, recall from a previous course on mechanics of materials that the relationship between a torsional moment (or torque) MT applied at the free end of a cantilever circular shaft, and the resulting angle of twist f (see Fig. 8.9), can be expressed as f5

MT L GJ

(8.15)



in which G denotes the shear modulus of the material, and J denotes the polar moment of inertia of the shaft. For members with noncircular cross sections, the relationship between the torsional moment MT and the angle of twist f can be quite complicated because of warping [40]. However, if warping is not restrained, then Eq. (8.15) can be used to approximate the torsional behavior of members with noncircular cross-sections—provided that J is now considered to be the Saint-Venant’s torsion constant, or simply the torsion constant, of the member’s cross-section, instead of its polar moment of inertia. Although the derivation of the expressions for torsion constant J for various cross-sectional shapes is beyond the scope of this text, such derivations can be found in textbooks on the theory of elasticity and advanced mechanics of materials [40]. The expressions for J for some common cross-sectional shapes are listed in Table 8.1. Furthermore, the torsion constant for any thin-walled, open cross-section can be approximated

MT

ϕ

MT

L GJ 5 constant

Fig. 8.9  Circular Shaft Subjected to Torsional Moment

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446  Chapter 8   Three-Dimensional Framed Structures Table 8.1  Torsion Constants for Common Member Cross-Sections [40, 52] Cross-Section

Torsion Constant

r

J=

1 4 πr 2

r J = 2 π r 3t

t

b

J = β b3 d d

β

=

for

b #d

1 b 1 2 0.21 12 3 d 12

) db )

4

bf tf tw

)

h

J=

1 2b f t f3 + htw3 3

h

J=

2b2 h 2 b/ t f + h/ tw

tf

)

b tf

tf tw

tw

by the relationship J5

1 3

o bt 3

(8.16)

in which b and t denote, respectively, the width and thickness of each rectangular segment of the cross section. Returning our attention to Fig. 8.8(c), we realize that the expression for the stiffness coefficient k22 can be obtained by substituting f 5 u2 5 1

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Section 8.2   Grids  447

and MT 5 k22 into Eq. (8.15), and solving the resulting equation for k22. This yields k22 5

GJ L

(8.17)



The other torsional stiffness coefficient k52 can now be determined by applying the following equilibrium equation.

oM

1

X

k52 5 2

50

GJ 1 k52 5 0 L

GJ L

(8.18)

The expressions for coefficients k25 and k55 (Fig. 8.8(f)) can be obtained in a similar manner. Substitution of f 5 u5 5 1 and MT 5 k55 into Eq. (8.15) yields

k55 5

GJ L

(8.19)



and by considering the equilibrium of the free body of the member, we obtain k25 as

oM 5 0

 1

x

k25 5 2

k25 1

GJ 50 L

GJ (8.20) L

Thus, by arranging all the stiffness coefficients shown in Figs. 8.8(b) through (g) into a matrix, we obtain the following expression for the local stiffness matrix for the members of grids.

k5

EI L3

3

12 0

6L 212 0 6L

0 GJL2 EI 0 0 GJL2 2 EI 0

6L

212

0

0

4L2 26L

26L 12

0

0

2L2

26L

0 GJL2 2 EI 0 0 GJL2 EI 0

6L 0 2L2 26L 0 4L2

4

(8.21)

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448  Chapter 8   Three-Dimensional Framed Structures

FG

The local fixed-end force vector for the members of grids is expressed as (Fig. 8.8(h))

Qf 5

FSb FTb FMb (8.22) FSe FTe FMe

in which the fixed-end shears (FSb and FSe) and bending moments (FMb and FMe) can be calculated by using the fixed-end force equations given for loading types 1 through 4 inside the front cover. (The procedure for deriving those fixed-end shear and bending moment equations was discussed in Section 5.4.) The expressions for the fixed-end torsional moments (FTb and FTe), due to an external torque MT applied to the member, are also given inside the front cover (see loading type 7). To derive these expressions, let us consider a fixed member of a grid, subjected to a torque MT, as shown in Fig. 8.10(a). If the end e of the member were free to rotate, then its cross-section would twist clockwise, as shown in Fig. 8.10(b). Let f be the angle of twist at end e of the released member. As portion Ae of the released member (Fig. 8.10(b)) is not subjected to any torsional moments, the angle of twist, f, at end e equals that at point A, and its magnitude can be obtained by substituting L 5 l1 into Eq. (8.15); that is, f5

MT l1 GJ

(8.23)

y

y

a

MT

(a) FTb

A

b

e

x

z

e ϕ50

l2

l1

z

FTe

L

5

y

yϕ a

MT

(b) MT

z

x

A

b

e

e

l1

z

1

y

ϕ

y

a (c) FTe z

b

e

FTe

x

z

e

L

Fig. 8.10

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Section 8.2   Grids  449

Since the angle of twist at end e of the actual fixed member (Fig. 8.10(a)) is 0, the fixed-end torsional moment FTe must be of such magnitude that, when ­applied to the released member as shown in Fig. 8.10(c), it should twist the crosssection at end e by an angle equal in magnitude to the angle f due to the torque MT , but in the opposite (i.e., counterclockwise) direction. The angle of twist due to FTe can be obtained by substituting MT 5 FTe into Eq. (8.15); that is, f5

FTeL GJ

(8.24)



and the relationship between FTe and the external torque MT can be established by equating Eqs. (8.23) and (8.24), as f5

FTeL

5

GJ

MT l1 GJ

from which we obtain the expression for the fixed-end torsional moment FTe: FTe 5

MT l1 L

(8.25)



The expression for the other fixed-end torsional moment, FTb, can now be determined by applying the equilibrium condition that the algebraic sum of the three torsional moments acting on the fixed member (Fig. 8.10(a)) must be 0; that is, 1

oM 5 0

FTb 2 MT 1 FTe 5 0

x

By substituting Eq. (8.25) into the foregoing equation and rearranging terms, we obtain the expression for FTb: FTb 5 MT

1

L 2 l1 L

25

MT l2 L

(8.26)



Member Releases The expressions for the local stiffness matrix k and the fixed-end force vector Qf , as given in Eqs. (8.21) and (8.22), respectively, are valid only for members of type 0 (i.e., MT 5 0), which are rigidly connected to joints at both ends. For grid members with moment releases, the foregoing expressions for k and Qf need to be modified using the procedure described in Section 7.1. If the member releases are assumed to be in the form of spherical hinges (or ball-and-socket type of connections), so that both bending and torsional moments are 0 at the released member ends, then the modified local stiffness matrices k and fixed-end force vectors Qf for the grid members with releases can be expressed as follows. For a member with a hinge at the beginning (MT 5 1):

3

3 0 EI 0 k5 3 L 23 0 3L

0 0 0 0 0 0

0 0 0 0 0 0

23 0 0 3 0 23L

4

0 3L 0 0 0 0 0 23L 0 0 0 3L2

(8.27)

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450  Chapter 8   Three-Dimensional Framed Structures

3 4

3 FMb 2L 0 0 (8.28) 3 FSe 1 FMb 2L FTe 1 FTb 1 FMe 2 FMb 2

FSb 2

Qf 5

For a member with a hinge at the end (MT 5 2):

3

3 0 3L EI k 5 3 23 L 0 0

0 0 0 0 0 0

3L 0 3L2 23L 0 0

23 0 23L 3 0 0

0 0 0 0 0 0

0 0 0 0 0 0

4



(8.29)

3 4

3 FMe 2L FTb 1 FTe 1 FMb 2 FMe 2 (8.30) 3 FSe 1 FMe 2L 0 0

FSb 2

Qf 5

For a member with hinges at both ends (MT 5 3): k 5 0 (8.31) 1 FSb 2 (FMb 1 FMe ) L

3 4 0

Qf 5

0

FSe 1

1 (FMb 1 FMe ) L 0

(8.32)

0

Note that the members of type 3 offer no resistance against twisting and, therefore, cannot be subjected to any torques or torsional member loading.

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Section 8.2   Grids  451

Coordinate Transformations Consider an arbitrary member m of a grid, as shown in Fig. 8.11(a). The orientation of the member in the horizontal (XZ) plane is defined by an angle u between the positive directions of the global X axis and the member’s local x axis, as shown in the figure. The member’s length, and its direction cosines, can be expressed in terms of the global coordinates of the member end joints, b and e, by the following relationships. L 5 Ï(Xe 2 Xb)2 1 (Ze 2 Zb)2 (8.33a) Y

X

y

(Xb, Zb)

b θ

z L

m e

(Xe, Ze)

x Z (a) Grid Y,y

Q1,u1

Q2 ,u

2

u3 Q 3,

θ

z Z

b

X

θ

Q4,u4 m u6 Q 6,

e

Q5 ,u

5

x

(b) Member End Forces and End Displacements in the Local Coordinate System

Fig. 8.11

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452  Chapter 8   Three-Dimensional Framed Structures

5 Y,y

F1,1 F2,2 F3,3

θ

z

X

θ

b

F4,4 m F5,5 e

Z

F6,6

x

(c) Member End Forces and End Displacements in the Global Coordinate System Q2

b X θ Q3

m

e Q5

θ

x

Q6 z

Z

5

b F2

X θ F3

m

e F5

θ

x F6

z

Z (d) Member Local and Global End Forces in the Horizontal (XZ) or (xz) Plane

Fig. 8.11  (continued)

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Section 8.2   Grids  453

cos u 5

Xe 2 Xb L Ze 2 Zb

(8.33b)



(8.33c) L The member local end forces Q and end displacements u are shown in Fig. 8.11(b); Fig. 8.11(c) depicts the equivalent system of end forces F and end displacements v, in the global coordinate system. As indicated in Fig. 8.11(c), the global member end forces and end displacements are numbered by beginning at member end b, with the force and translation in the Y direction numbered first, followed by the moment and rotation about the X axis, and then the moment and rotation about the Z axis. The forces and displacements at the member’s opposite end e are then numbered in the same sequential order. By comparing Figs. 8.11(b) and (c), we realize that at member end b, the local forces Q1, Q2, and Q3 must be equal to the algebraic sums of the components of the global forces F1, F2, and F3 in the directions of the local y, x, and z axes, respectively; that is (also, see Fig. 8.11(d)), Q1 5 F1 (8.34a)  sin u 5

Q2 5 F2 cos u 1 F3 sin u

(8.34b)

Q3 5 2F2 sin u 1 F3 cos u

(8.34c)

Similarly, the local forces at member end e can be expressed in terms of the global forces as Q4 5 F4 (8.34d) Q5 5 F5 cos u 1 F6 sin u

(8.34e)

Q6 5 2F5 sin u 1 F6 cos u

(8.34f)

Equations 8.34(a) through (f) can be expressed in matrix form as

F

Q 5 TF with the transformation matrix T given by

T5

1 0 0 cos u 0 2sin u 0 0 0 0 0 0

0 sin u cos u 0 0 0

0 0 0 0 0 0 1 0 0 cos u 0 2sin u

G

0 0 0 0 sin u cos u

(8.35)

(8.36)

Because the translations and small rotations of the member ends can be treated as vector quantities, the foregoing transformation matrix also defines the transformation of member end displacements from the global to the local coordinate system; that is, u 5 Tv. Furthermore, the transformation matrix T, as given in Eq. (8.36), can be used to transform member end forces and displacements from the local to the global coordinate system via the relationships F 5 TTQ and v 5 TTu, respectively.

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454  Chapter 8   Three-Dimensional Framed Structures

Procedure for Analysis The procedure for analysis of grids remains the same as that for plane frames developed in Chapter 6 (and modified in Chapter 7); provided, of course, that the member local stiffness and transformation matrices, and local fixed-end force vectors, developed in this section are used in the analysis. The procedure is illustrated by the following example.



E x ample 8.2 Determine the joint displacements, member end forces, and support reactions for the three-member grid shown in Fig. 8.12(a), using the matrix stiffness method.



S olut i o n

Analytical Model:  The grid has three degrees of freedom and nine restrained coordinates, as shown in Fig. 8.12(b). Structure Stiffness Matrix: Member 1  From Fig. 8.12(b), we can see that joint 1 is the beginning joint, and joint 4 the end joint, for this member. By applying Eqs. (8.33), we determine the length, and the direction cosines, for the member to be L 5 Ï(X4 2 X1)2 1 (Z4 2 Z1)2 5 Ï(8 2 0)2 1 (6 2 0)2 5 10 m cos u 5  sin u 5

X4 2 X1 L Z4 2 Z1 L

5

820 5 0.8 10

5

620 5 0.6 10 Y

8m X 20 kN/m

6m 20 kN/m

Z

E, G, I, J 5 constant E 5 200 GPa, G 5 76 GPa I 5 347(106 ) mm4, J 5 115(106 ) mm4 (a) Grid

Fig. 8.12

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Section 8.2   Grids  455

Y

9

1

5

2 1

8

2

6

7

4 11

3

X

2

4

3 3

12

1

10 Z (b) Analytical Model 1 S5

Pf 5

3 3

2

832.8 + 3,855.6 + 1,626.6

3

2,498.4 + 11,567

2,498.4 + 11,567 23,331.2 2 6,506.3

23,331.2 2 6,506.3

10,553 + 46,267 + 1,092.5 212,905

3 3 3

3 3 1

212,905 18,081 + 1,456.7 + 34,700

6,315

2 = 3

14,065

14,065 29,837.5

57,912 212,905

3

29,837.5 212,905 54,238

60 + 80 1

60 2106.67

140 2 = 60 3 2106.67

(c) Structure Stiffness Matrix and Fixed-Joint Force Vector y

55

7. z

9

77.70

z

07

99

04

5.

y

445.06

/m

0.014686

20

1

kN

144.67 2

7.

55

y

04

5. 0.014686

07

62.952

99

2

77.56

x

24.668

x 20 kN/m

Y 12.378

5.

37

X Z

12.378

x

3

52 135.32

13 24.683

.0

67

z (d) Member Local End Forces

Fig. 8.12  (continued)

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456  Chapter 8   Three-Dimensional Framed Structures

0.014686 kN 250.662 kN • m 59.14 kN • m 144.67 kN 2445.06 kN • m 7.9907 kN • m 135.32 kN 212.378 kN • m 375.52 kN • m

R5

4 5 6 7 8 9 10 11 12

(e) Support Reaction Vector Y

7.9907

1

50.662

445.06 /m

59.14

12.378

20

0.014686 20 kN/m

kN

X

2 144.67

3

4

375.52 135.32 Z (f) Support Reactions

Fig. 8.12  (continued) Since MT 5 0 for this member, we use Eq. (8.21) to determine its local stiffness ­matrix k. Thus, by substituting E 5 200(106) kN/m2, G 5 76(106) kN/m2, L 5 10 m, I 5 347(1026) m4, and J 5 115(10−6) m4 into Eq. (8.21), we obtain

k1 5

F

832.8 0 4,164 2832.8 0 4,164

0 874 0 0 2874 0

4,164 0 27,760 24,164 0 13,880

2832.8 0 4,164 0 2874 0 24,164 0 13,880 832.8 0 24,164 0 874 0 24,164 0 27,760

G



(1)

As the member is not subjected to any loads, its global and local fixed-end force vectors are 0; that is,

Ff 1 5 Qf 1 5 0

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Section 8.2   Grids  457

Before we can calculate the member global stiffness matrix K, we need to evaluate its transformation matrix T. Thus, by substituting cos u 5 0.8 and sin u 5 0.6 into Eq. (8.36), we obtain

T1 5

F

1 0 0 0.8 0 20.6 0 0 0 0 0 0

0 0.6 0.8 0 0 0

0 0 0 1 0 0

0 0 0 0 0.8 20.6

G

0 0 0  0 0.6 0.8

(2)

Next, by substituting k1 (Eq. (1)) and T1 (Eq. (2)) into the relationship K 5 TTkT, and performing the necessary matrix multiplications, we obtain the following global stiffness matrix for member 1: 4 5 6 1 2 3 832.8 22,498.4 3,331.2 2832.8 22,498.4 3,331.2 4 22,498.4 10,553 212,905 2,498.4 4,437.4 27,081.9 5 3,331.2 212,905 18,081 23,331.2 27,081.9 8,568.6 6 K1­ 5 1 2832.8 2,498.4 23,331.2 832.8 2,498.4 23,331.2 2 22,498.4 4,437.4 27,081.9 2,498.4 10,553 212,905 3 3,331.2 27,081.9 8,568.6 23,331.2 212,905 18,081

F

G

From Fig. 8.12(b), we observe that the code numbers for member 1 are 4, 5, 6, 1, 2, 3. By using these code numbers, we store the pertinent elements of K1 in the 3 3 3 structure stiffness matrix S, as shown in Fig. 8.12(c).

F

G

Member 2  L 5 6 m, cos u 5 0, sin u 5 1

k2 5

3,855.6 0 0 1,456.7 11,567 0 23,855.6 0 0 21,456.7 11,567 0

T2 5

K2 5 T2T k2T2 5

F

F

1 0 0 0 0 21 0 0 0 0 0 0

0 1 0 0 0 0

11,567 23,855.6 0 0 46,267 211,567 211,567 3,855.6 0 0 23,133 211,567

0 0 0 0 0 0 1 0 0 0 0 21

7 8 3,855.6 211,567 211,567 46,267 0 0 23,855.6 11,567 211,567 23,133 0 0

G

0 21,456.7 0 0 1,456.7 0

11,567 0 23,133 211,567 0 46,267

0 0 0  0 1 0

9 0 0 1,456.7 0 0 21,456.7

(3)

(4)

1 23,855.6 11,567 0 3,855.6 11,567 0

2 211,567 23,133 0 11,567 46,267 0

3 0 0 21,456.7 0 0 1,456.7

G

7 8 9 1 2 3

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458  Chapter 8   Three-Dimensional Framed Structures To determine the local fixed-end force vector due to the 20 kN/m member load, we first evaluate the fixed-end shears and moments by using the expressions for loading type 3 given inside the front cover. This yields FSb 5 FSe 5 60 kN FMb 5 2FMe 5 60 kN ? m FTb 5 FTe 5 0 Since MT 5 0 for this member, we use Eq. (8.22) to obtain its local fixed-end force vector:

Qf 2 5

34

60 0 60  60 0 260

(5)

Next, by substituting T2 (Eq. (4)) and Qf 2 (Eq. (5)) into the transformation relationship Ff 5 TTQf , we obtain the global fixed-end force vector for member 2:

Ff 2 5

34 60 260 0 60 60 0

7 8 9 1 2 3

The relevant elements of K2 and Ff 2 are stored in S and the 3 3 1 structure fixed-joint force vector Pf , respectively, as shown in Fig. 8.12(c). Member 3  As the local x axis of this member is oriented in the positive direction of the global X axis, no coordinate transformations are needed; that is, T3 5 I. By using Eq. (8.21) with L 5 8 m, we obtain

K3 5 k35

F

10 1,626.6 0 6,506.3

11 0 1,092.5 0

12 6,506.3 0 34,700

1 21,626.6 0 26,506.3

2 0 21,092.5 0

3 6,506.3 0 17,350

G

10 11 12 1 2

21,626.6 0 26,506.3 1,626.6 0 26,506.3 0 0 21,092.5 0 1,092.5 0 6,506.3 0 17,350 26,506.3 3 0 34,700  (6) FSb  5 FSe 5 80 kN FMb 5 2FMe 5 106.67 kN · m FTb 5 FTe 5 0 80 10 0 11 106.67 12 Ff 3 5 Qf 3 5  (7) 80 1 0 2 2106.67 3

3 4

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Section 8.2   Grids  459

The complete structure stiffness matrix S and the structure fixed-joint force vector Pf are given in Fig. 8.12(c). Joint Load Vector:  Because the grid is not subjected to any external loads at its joints, the joint load vector is 0; that is,

P50 Joint Displacements:  By substituting P, Pf , and S into the structure stiffness relationship, P 2 Pf 5 Sd, we write

34 3

43

0 140 6,315 0 2 60 5 14,065 0 2106.67 29,837.5

or

3

2140 260 106.67

4 3 5

6,315 14,065 29,837.5

43 4

14,065 29,837.5 57,912 212,905 212,905 54,238

d1 d2 d3

43 4

14,065 29,837.5 57,912 212,905 212,905 54,238

d1 d2 d3

By solving the foregoing simultaneous equations, we determine the joint displacements to be d5

3

255.951 m 11.33 rad 25.4856 rad

4

3 1023

Ans

Member End Displacements and End Forces:

FG FG F G FG F G

Member 1

v1 5

y1 y2 y3 y4 y5 y6

4 5 6 5 1 2 3

0 0 0 d1 d2 d3

0 0 0 5 255.951 11.33 25.4856

3 1023

0 0 0 u1 5 T1v1 5 3 1023 255.951 5.7728 211.187

Q 1 5 k 1u 1 5

0.014686 kN 25.0455 kN ? m 77.709 kN ? m 20.014686 kN 5.0455 kN ? m 277.562 kN ? m



Ans

The member local end forces are depicted in Fig. 8.12(d).

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460  Chapter 8   Three-Dimensional Framed Structures

F G F G FG F G F G F G F G

F1 5 TT1 Q1 5

0.014686 250.662 59.14 20.014686 50.574 259.022

4 5 6 1 2 3

The pertinent elements of F1 are stored in the reaction vector R, as shown in Fig. 8.12(e). Member 2

v2 5

0 0 0 255.951 11.33 25.4856

0 7 0 8 0 9 3 1023, u2 5 T2v2 5 3 1023 255.951 1 25.4856 2 211.33 3

Q 2 5 k 2u 2 1 Q f 2 5

F2 5 TT2 Q2 5

144.67 kN 7.9907 kN ? m 445.06 kN ? m 224.668 kN 27.9907 kN ? m 62.952 kN ? m

144.67 2445.06 7.9907 224.668 262.952 27.9907



Ans

7 8 9 1 2 3

Member 3

u3 5 v3 5

0 0 0 255.951 11.33 25.4856

10 11 12 3 1023 1 2 3

F 3 5 Q 3 5 k 3u 3 1 Q f 3 5

135.32 kN 212.378 kN ? m 375.52 kN ? m 24.683 kN 12.378 kN ? m 67.013 kN ? m

10 11 12  1 2 3

Ans

Support Reactions:  The completed reaction vector R is shown in Fig. 8.12(e), and the support reactions are depicted on a line diagram of the grid in Fig. 8.12(f). Ans Equilibrium  Check:  The three equilibrium equations MZ 5 0+ are satisfied.

o

_ o FY 5 0,

oM

X

5 0, and

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Section 8.3   Space Frames  461

8.3 

SPACE FRAMES Space frames constitute the most general type of framed structures. The members of such frames may be oriented in any directions in three-dimensional space, and may be connected by rigid and/or flexible connections. Furthermore, external loads oriented in any arbitrary directions can be applied to the joints, as well as members, of space frames (Fig. 8.13(a)). Under the action of external loads, the members of a space frame are generally subjected to bending moments about both principal axes, shears in both principal directions, torsional moments, and axial forces. As with grids, the analysis of space frames is commonly based on the assumption that the cross-sections of all the members are symmetric about at least two mutually perpendicular axes, and are free to warp out of their planes under the action of torsional moments. As discussed previously in the case of grids, the bending and torsional stiffnesses of a member are uncoupled if it satisfies the foregoing assumption. The process of developing the analytical models, and numbering the ­degrees of freedom and restrained coordinates, of space frames is analogous to that for other types of framed structures. The overall geometry of the space Y

A A

X Planes of symmetry

0

Z

Space Frame

Section A–A Member cross-section (a)

Fig. 8.13 Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

462  Chapter 8   Three-Dimensional Framed Structures 11

Y

12

8 6 2

16

9

20 8 5

14

10

7

23

1 7

17

2

3

2 4

9

3

5

6

21

13

1

19

22

3

24

5 15 1

18 28

4

25 27

43

31 33

26

48

X

32

29

35

8

7 37

45

34

36

30 46

4 6

39

44

40

38

42 47

41

Z

(b) Analytical Model (24 Degrees of Freedom and 24 Restrained Coordinates)

Fig. 8.13  (continued)

frame, and its joint loads and displacements, are described with reference to a global right-handed XYZ coordinate system, with three global (X, Y, and Z) coordinates used to specify the location of each joint. An unsupported joint of a space frame can translate in any direction, and rotate about any axis, in three-dimensional space. Since small rotations can be treated as vector quantities, the rotation of a joint can be conveniently represented by its component rotations about the X, Y, and Z axes. Thus, a free joint of a space frame has six degrees of freedom—the translations in the X, Y, and Z directions and the rotations about the X, Y, and Z axes. Therefore, six structure coordinates (i.e., free and/or restrained coordinates) need to be defined at each joint of the space frame for the purpose of analysis; that is, NCJT 5 6

f for space frames NDOF 5 6(NJ) 2 NR   



(8.37)

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Section 8.3   Space Frames  463 Y 3 z 2

y 2 x 8

7

9

A A 6

5

3

1 5

6 4

X

1

8

4

7

z y

Section A–A

Z (c) Member Coordinate System

Fig. 8.13  (continued)

The procedure for assigning numbers to the structure coordinates of a space frame is similar to that for other types of framed structures, with the degrees of freedom numbered before the restrained coordinates. In the case of a joint with multiple degrees of freedom (or restrained coordinates), the translations (or forces) in the X, Y, and Z directions are numbered first in sequential order, followed by the rotations (or moments) about the X, Y, and Z axes, respectively, as shown in Fig. 8.13(b). For each member of a space frame, a local xyz coordinate system is established, with its origin at an end of the member and the x axis directed along the member’s centroidal axis in the undeformed state. The local y and z axes are oriented, respectively, parallel to the two axes of symmetry (or the principal axes of inertia) of the member cross-section, with their positive directions defined so that the local xyz coordinate system is right-handed (Fig. 8.13(c)).

Member Stiffness Relations in the Local Coordinate System To establish the local stiffness relations, let us consider an arbitrary member m of a space frame, as shown in Fig. 8.14(a). Like a joint of a space frame,

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464  Chapter 8   Three-Dimensional Framed Structures y

Q12, u12 W

Q9, u9

w b

Q1, u1

Q4, u4

Q7, u7 M

Q3, u3 Q6, u6

m

Q10, u10

x

e

Q2, u 2

Q8, u 8

Q5, u5

Q11, u11

L E, A, J, Iz, Iy 5 constant

z

(a) Member Forces and Displacements in the Local Coordinate System y

u1 5 1 k11 5

EA L

k71 5 2 EA L

b9

x

e

b z (b) u1 5 1 y

b9 u2 5 1

e

b

k62 5

6EIz L2

x k22 5

12EIz L3

k12,2 5

6EIz L2

k82 5 2

12EIz L3

z (c) u2 5 1

Fig. 8.14

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Section 8.3   Space Frames  465 y

6EIy

k53 = 2

k33 =

L2

6EI k11,3 = 2 y L2

12EIy L3

x

u3 = 1

e

b

12EIy k93 = 2 L3

b9 z (d) u 3 = 1 y u4 = 1

z

y

k44 = GJ L

e

b

b

k10, 4 = 2

x

GJ L

z (e) u4 = 1 y

k55 =

4EIy L

k11,5 =

u5 = 1

2EIy L

x e

b 6EIy k35 = 2 L2

k95 =

6EIy L2

z u5 = 1

(f ) u5 = 1

Fig. 8.14  (continued)

six displacements are needed to completely specify the displaced position of each end of the space frame member. Thus, a member of a space frame has 12 degrees of freedom. In the member local coordinate system, the 12 end displacements are denoted by u1 through u12, and the corresponding member end forces are denoted by Q1 through Q12, as shown in Fig. 8.14(a). As indicated in this figure, a member’s local end displacements (or end forces) are numbered by beginning at its end b, with the translations (or forces) in the x, y, and z directions numbered first in sequential order, followed by the rotations (or moments) about the x, y, and z axes, respectively. The displacements (or forces) at the member’s opposite end e are then numbered in the same sequential order.

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466  Chapter 8   Three-Dimensional Framed Structures y

u6 = 1

b

e x

k66 =

4EIz 6EIz L k26 =  2 L

2EIz 6EIz k86 = 2 L L2

k12,6 = 

z (g) u6 = 1

y

u7 = 1

EA k17 = 2  L

e9 x

EA L

e

b

k77 = 

z (h) u7 = 1

y

6EIz

k68 = 2

e9

L2

u8 = 1

b

x

e 12EIz L3

k28 = 2

k12,8 = 2

6EIz L2

k88 =

12EIz L3

z (i) u8 = 1

Fig. 8.14  (continued)

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Section 8.3   Space Frames  467 y

k59 =

6EIy

k11,9 =

L2

b

k39 = 2

u9 = 1

12EIy

6EIy L2

k99 =

12EIy L3 x

e

e9

L3

z (j) u9 = 1

u10 = 1

y

k10,10 =

GJ k4,10 = 2 L

b

GJ L

x

e

z

y

e

z (k) u10 = 1 y

k5,11 =

2EIy

k11,11 =

L

4EIy L

b

x

u11 = 1 k3,11 = 2

6EIy k9,11 =

L2

e 6EIy L2

z (l) u11 = 1

Fig. 8.14  (continued)

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468  Chapter 8   Three-Dimensional Framed Structures y k6,12 =

2EIz L

4EIz L

k12,12 = e

b

x u12 = 1

k2,12 =

z

6EIz

6EIz

k8,12 = 2

L2

L2

(m) u12 = 1 y

Q f12 W

Q f9

w Q f4

Q f1 Q f3 Q f6

z

Q f7

e

b

Q f10

x

M Q f2

Q f8

Q f5

Q f11 (n) Member Fixed-End Forces in the Local Coordinate System (u1 = u2 = . . . = u11 = u12 = 0)

Fig. 8.14  (continued)

The relationship between the end forces Q and the end displacements u, for space frame members, can be expressed in the following, now familiar, form: Q 5 ku 1 Qf

(8.38)

with k now representing the 12 3 12 member local stiffness matrix, and Qf denoting the 12 3 1 member local fixed-end force vector. The explicit form of k for members of space frames can be conveniently obtained using the expressions of the stiffness coefficients derived previously for prismatic members subjected to axial deformations (Section 3.3), bending deformations (Section 5.2), and torsional deformations (Section 8.2). The stiffness coefficients for a space frame member thus obtained, due to the unit values of the 12 end displacements (u1 through u12, respectively), are given in Figs. 8.14(b) through (m). Note that in Figs. 8.14(c), (g), (i), and (m), the moment of inertia of the member cross-section about its local z axis, Iz, is used in the e­ xpressions for the stiffness coefficients because the end displacements u2, u6, u8, and u12 cause the member to bend about the z axis. However, in Figs. 8.14(d), (f), (j), and (l) because the end displacements u3, u5, u9, and u11 Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 8.3   Space Frames  469

cause the member to bend about its local y axis, the moment of inertia about the y  axis, Iy, is used in the expressions for the corresponding stiffness coefficients. The explicit form of the local stiffness matrix k for members of space frames, obtained by arranging all the stiffness coefficients shown in Figs. 8.14(b) through (m) in a 12 3 12 matrix, is given in Eq. (8.39). AL2 0 0

0 12Iz 0

0 0 12Iy

0

0

0

0 0 E k 5 3 2AL2 L 0 0

0 6LIz 0 212Iz 0

26LIy 0 0 0 212Iy

0

0

0

0 0

0 6LIz

26LIy 0

0 0 0 GJL2 E 0 0 0 0 0 GJL2 2 E 0 0

0 0 26LIy

0 6LIz 0

2AL2 0 0

0 212Iz 0

0 0 212Iy

0

0

0

0

0

4L Iy 0 0 0 6LIy

0 4L2 Iz 0 26LIz 0

0 0 AL2 0 0

0 26LIz 0 12Iz 0

6LIy 0 0 0 12Iy

0

0

0

0

0

2L2 Iy 0

0 2L2 Iz

0 0

0 26LIz

6LIy 0

2

0 0 0 GJL2 2 E 0 0 0 0 0 GJL2 E 0 0

0 0 26LIy

0 6LIz 0

0

0

2L Iy 0 0 0 6LIy 2

0 2L2 Iz 0    (8.39) 26LIz 0

0

0

4L2 Iy 0

0 4L2 Iz

The local fixed-end force vector for the members of space frames is expressed as follows (Fig. 8.14(n)). FAb FSby FSbz FTb FMby FMbz Qf 5 FAe FSey FSez FTe FMey FMez

(8.40)

in which FSby and FSbz denote the fixed-end shears at member end b in the local y and z directions, respectively; and FMby and FMbz represent the fixedend ­moments at the same member end about the y and z axes, respectively. The fixed-end shears and moments at the opposite end e of the member are defined in a similar manner. The fixed-end forces due to a prescribed member loading can be conveniently evaluated, using the fixed-end force expressions given inside the front cover. Any inclined member loads must be resolved into their components in the directions of the member’s local x, y, and z axes before proceeding with the calculation of the fixed-end forces. Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

470  Chapter 8   Three-Dimensional Framed Structures

k5

E L3

3

AL2 0 0 0 0 0 2AL2 0 0 0 0 0

Member Releases  The expressions for k and Qf , as given in Eqs. (8.39) and (8.40), respectively, are valid only for members rigidly connected to joints at both ends (i.e., members of type 0, or MT 5 0). For members of space frames with moment releases, the foregoing expressions need to be modified, using the procedure described in Section 7.1. If the member releases are assumed to be in the form of spherical hinges (or ball-and-socket type of connections), so that all three moments (i.e., the bending moments about the y and z axes, and the torsional moment) are 0 at the released member ends, then the modified local stiffness matrices k and fixedend force vectors Qf for the members with releases can be expressed as follows. For members with a hinge at the beginning (MT 5 1), the modified k is given in Eq. (8.41): 0 3Iz 0 0 0 0 0 23Iz 0 0 0 3L Iz

0 0 3Iy 0 0 0 0 0 23Iy 0 23LIy 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

2AL2 0 0 0 0 0 AL2 0 0 0 0 0

0 23Iz 0 0 0 0 0 3Iz 0 0 0 23LIz

0 0 23Iy 0 0 0 0 0 3Iy 0 3LIy 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 23LIy 0 0 0 0 0 3LIy 0 3L2 Iy 0

4

0 3LIz 0 0 0 0   (8.41) 0 23LIz 0 0 0 3L2 Iz

and

FAb  3 FSby 2 FMbz 2L 3 FSbz 1 FMby 2L 0 0 Qf 5 0 FAe 3 FSey 1 FMbz 2L 3 FSez 2 FMby 2L FTb 1 FTe 1 FMey 2 FMby 2 1 FMez 2 FMbz 2

(8.42)

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Section 8.3   Space Frames  471

k5

E L3

3

AL2 0 0 0 0 0 2AL2 0 0 0 0 0

For members with a hinge at the end (MT 5 2), the modified k is given in Eq. (8.43): 0 3Iz 0 0 0 3LIz 0 23Iz 0 0 0 0

0 0 3Iy 0 23LIy 0 0 0 23Iy 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 23LIy 0 3L2Iy 0 0 0 3LIy 0 0 0

0 3LIz 0 0 0 3L2Iz 0 23LIz 0 0 0 0

2AL2 0 0 0 0 0 AL2 0 0 0 0 0

0 23Iz 0 0 0 23LIz 0 3Iz 0 0 0 0

0 0 23Iy 0 3LIy 0 0 0 3Iy 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

4

  (8.43)

and



FAb 3 FSby 2 FMez 2L 3 FSbz 1 FMey 2L FTb 1 FTe 1 FMby 2 FMey 2 Qf 5 FM 2 1 FM bz ez 2  FAe 3 FSey 1 FMez 2L 3 FSez 2 FMey 2L 0 0 0

(8.44)

For members with hinges at both ends (MT 5 3), the modified k is given in Eq. (8.45):



3

1 0 0 0 0 EA 0 k5 L 21 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 21 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

4



(8.45)

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472  Chapter 8   Three-Dimensional Framed Structures

and FAb 1 (FMbz 1 FMez ) L 1 FSbz 1 (FMby 1 FMey ) L 0 0 0 FAe 1  FSey 1 (FMbz 1 FMez ) L 1 FSez 2 (FMby 1 FMey ) L 0 0 0 FSby 2

Qf 5

(8.46)

Coordinate Transformations The expression of the transformation matrix T for members of space frames can be derived using a procedure essentially similar to that used previously for other types of framed structures. However, unlike the transformation matrices for trusses, plane frames, and grids, which contain direction cosines of only the member’s longitudinal (or x) axis, the transformation matrix for members of space frames involves direction cosines of all three (x, y, and z) axes of the member local coordinate system with respect to the structure’s global (XYZ) coordinate system. Consider an arbitrary member m of a space frame, as shown in Fig. 8.15(a) on the next page. The member end forces Q and end displacements u, in the local coordinate system, are shown in Fig. 8.15(b), and Fig. 8.15(c) depicts the equivalent system of member end forces F and end displacements v, in the global coordinate system. As indicated in Fig. 8.15(c), the global member end forces and displacements are numbered in a manner analogous to the local forces and displacements, except that they act in the directions of the global X, Y, and Z axes. The orientation of a member of a space frame is defined by the angles between its local x, y, and z axes and the global X, Y, and Z axes. As shown in Fig. 8.16(a) on page 474, the angles between the local x axis and the global X, Y, and Z axes are denoted by uxX , uxY , and uxZ , respectively. Similarly, the angles ­between the local y axis and the global X, Y, and Z axes are denoted by uyX, uyY , and uyZ , respectively (Fig. 8.16(b)); and the angles between the local z axis and the global X, Y, and Z axes are denoted by uz X , uz Y  , and uz Z , respectively (Fig. 8.16(c)). Now, let us consider the transformation of member end forces from the global to a local coordinate system. By comparing Figs. 8.15(b) and (c), we realize that, at member end b, the local forces Q1, Q2, and Q3 must be equal to

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Section 8.3   Space Frames  473 Y

e

y b

x

m z X

Z

(a) Space Frame Y

u 11

Q 11,

y

x

Q 8,u 8 e

Q 5,u 5

Q 7,u 7

Q 2,u 2

Q 1,u 1 Q 4,u 4

m

b

u 0 Q 10, 1

Q 9,u 9

,u 2 Q 12 1

X

Q 3,u 3

Q 6,u 6 Z

z (b) Member End Forces and End Displacements in the Local Coordinate System

Fig. 8.15

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474  Chapter 8   Three-Dimensional Framed Structures

5 Y

F11,11

y

x

F8,8 F5,5 e

F10,10

F9,9

F2,2

m F12,12

F1,1

F4,4

F7,7

X b

F3,3

F6,6 z

Z

(c) Member End Forces and End Displacements in the Global Coordinate System

Fig. 8.15  (continued)

Y

y

x

e

m

θxY

θxX b

X

θxZ

Z

z (a) Orientation of Member Local x Axis

Fig. 8.16 

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Section 8.3   Space Frames  475 Y

y

x

θyY

e

m θyX θyZ b

X

z

Z

(b) Orientation of Member Local y Axis

Y

y

x

e

m

θzY

b

X

θzX θzZ Z

z (c) Orientation of Member Local z Axis

Fig. 8.16  (continued)



the algebraic sums of the components of the global forces F1, F2, and F3 in the directions of the local x, y, and z axes, respectively; that is (also, see Fig. 8.16), Q1 5 F1 cos uxX 1 F2 cos uxY 1 F3 cos uxZ (8.47a) Q2 5 F1 cos uyX 1 F2 cos uyY 1 F3 cos uyZ (8.47b) Q3 5 F1 cos uzX 1 F2 cos uzY 1 F3 cos uzZ (8.47c) Equations (8.47) can be written in matrix form as



34 3

Q1 rxX Q2 5 ryX Q3 rzX

rxY ryY rzY

rxZ ryZ rzZ

43 4

F1 F2  F3

(8.48)

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476  Chapter 8   Three-Dimensional Framed Structures

in which ri J 5 cos ui J i 5 x, y, or z  and  J 5 X, Y, or Z (8.49) The local moments Q4, Q5, and Q6, at member end b, can be similarly expressed in terms of their global counterparts F4, F5, and F6, as Q4 rxX rxY rxZ F4 (8.50) Q5 5 ryX ryY ryZ F5  Q6 rzX rzY rzZ F6

34 3

43 4

Similarly, the local forces and moments at member end e can be expressed in terms of the global forces and moments by the following relationships.

34 3

rxY ryY rzY

rxZ ryZ rzZ

34 3

rxY ryY rzY

rxZ ryZ rzZ

Q7 rxX Q8 5 ryX Q9 rzX



F7 F8  F9

43 4

(8.51)

43 4

(8.52)

and Q10 rxX Q11 5 ryX Q12 rzX



F10 F11  F12

By combining Eqs. (8.48) and Eqs. (8.50) through (8.52), we can now express the transformation relationship between the 12 3 1 member local end force vector Q and the 12 3 1 member global end force vector F, in the standard form of Q 5 TF (8.53) in which T represents the 12 3 12 transformation matrix for the members of space frames. The explicit form of T is given in Eq. (8.54).



T5

3

rxX ryX rzX 0 0 0 0 0 0 0 0 0

rxY ryY rzY 0 0 0 0 0 0 0 0 0

rxZ ryZ rzZ 0 0 0 0 0 0 0 0 0

0 0 0 rxX ryX rzX 0 0 0 0 0 0

0 0 0 rxY ryY rzY 0 0 0 0 0 0

0 0 0 rxZ ryZ rzZ 0 0 0 0 0 0

0 0 0 0 0 0 rxX ryX rzX 0 0 0

0 0 0 0 0 0 rxY ryY rzY 0 0 0

0 0 0 0 0 0 rxZ ryZ rzZ 0 0 0

0 0 0 0 0 0 0 0 0 rxX ryX rzX

0 0 0 0 0 0 0 0 0 rxY ryY rzY

4

0 0 0 0 0 0 (8.54) 0 0 0 rxZ ryZ rzZ

The transformation matrix T is usually expressed in a compact form in terms of its submatrices as T5

3

r O O O

O r O O

O O r O

4

O O O r

(8.55)

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Section 8.3   Space Frames  477

in which O represents a 3 3 3 null matrix; and the 3 3 3 matrix r, which is commonly referred to as the member rotation matrix, is given by

3

rxX r 5 ryX rzX

rxY ryY rzY

4

rxZ ryZ  rzZ

(8.56)

The rotation matrix r plays a key role in the analysis of space frames, and an alternate form of this matrix, which enables us to specify the member orientations more conveniently, is developed subsequently. Since the member local and global end displacements, u and v, are also vector quantities, which are defined in the same directions as the corresponding forces, the foregoing transformation matrix T can also be used to transform member end displacements from the global to the local coordinate system; that is, u 5 Tv. Furthermore, by employing a procedure similar to that used in the preceding paragraphs, it can be shown that the inverse transformations of the member end forces and end displacements, from the local to the global coordinate system, are defined by the transpose of the transformation matrix given in Eq. (8.54) (or Eqs. (8.55) and (8.56)); that is, F 5 TTQ and v 5 TTu. Once the transformation matrix T has been established for a member of a space frame, its global stiffness matrix and fixed-end force vector can be obtained via the standard relationships K 5 TTkT and Ff 5 TTQf , respectively. Member Rotation Matrix in Terms of the Angle of Roll   From Eq. (8.56), we can see that the rotation matrix r consists of nine elements, with each element representing the direction cosine of a local axis with respect to a global axis, in accordance with Eq. (8.49). Of these nine direction cosines, the three in the first row of r, which represent the direction cosines of the local x axis, can be directly evaluated using the global coordinates of the two joints to which the member ends are attached. Thus, if Xb, Yb, and Zb and Xe, Ye, and Ze denote the global coordinates of the joints to which member ends b and e, respectively, are attached, then the direction cosines of the local x axis, with respect to the global X, Y, and Z axes, respectively, can be expressed as

rxX 5 cos uxX 5



rxY 5 cos uxY 5



rxZ 5 cos uxZ 5

Xe 2 Xb L Ye 2 Yb L Ze 2 Zb L



(8.57a)



(8.57b)



(8.57c)

in which the member length L is given by

L 5 Ï(Xe 2 Xb)2 1 (Ye 2 Yb)2 1 (Ze 2 Zb)2

(8.57d)

With the direction cosines of the member x axis now established, we focus our attention on the question of how to determine the direction cosines of the local y and z axes using the information about the member orientation that can

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478  Chapter 8   Three-Dimensional Framed Structures Y

y

iy

IY

x ix IX

IZ

X

iz

Z

z

Fig. 8.17  Unit Vectors in the Directions of the Local and Global Axes

be conveniently input by the user of the computer program. Since the x, y, and z axes form a mutually perpendicular right-handed coordinate system, it usually is convenient to define their directions by those of the unit vectors directed along these axes. Thus, if ix , iy , and iz denote, respectively, the unit vectors in the directions of the local x, y, and z axes, and IX , IY , and IZ denote, ­respectively, the unit vectors directed along the global X, Y, and Z axes (see Fig. 8.17), then the relationship between the local and global unit vectors is defined by the member rotation matrix r, as

34 3

ix rxX iy 5 ryX iz rzX

rxY ryY rzY

rxZ ryZ rzZ

43 4

IX IY  IZ

(8.58)

The reader may recall from a previous course in statics that if the direction cosines of two of the three unit vectors, directed along the axes of an orthogonal coordinate system, are known, then those of the third unit vector can be obtained by using a cross (or vector) product of the two known vectors. In the case under consideration, as discussed previously, the direction cosines of one of the unit vectors, ix, are defined by the global coordinates of the member ends (Eqs. (8.57)). Thus, if the user of the computer program can provide, as input, the direction cosines of either iy or iz (i.e., either the y or the z axis), then the direction cosines of the remaining third vector can be conveniently established via the cross product of the two known vectors. However, as the hand

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Section 8.3   Space Frames  479

calculation of direction cosines of the y or z axis for each member of a structure can be a tedious and time-consuming chore, this approach is not considered user-friendly and is seldom used by practitioners. Instead, most computer programs allow the users to specify the orientation of the member y and z axes by means of the so-called angle of roll [3]. To define the angle of roll and to express the direction cosines of the member y and z axes in terms of this angle, we imagine that the member’s desired (or actual design) orientation is reached in two steps, as shown in Figs. 8.18(a) and (b). In the first step, while the member’s x axis is oriented in the desired direction, its y and z axes are oriented so that the xy plane is vertical and the z axis lies in a horizontal plane. The foregoing (imaginary) orientation of the member is depicted in Fig. 8.18(a), in which the member’s principal axes are designated as y and z (instead of y and z, respectively), to indicate that they have not yet been positioned in their desired (or actual design) directions. As discussed previously, the direction of the local x axis is known from the global coordinates of the member ends. Since the z axis is perpendicular to the vertical xy plane, a vector z directed along the z axis can be determined by the cross product of the vector ix and a vertical unit vector IY; that is,

u

IX z 5 ix 3 IY 5 det rxX 0



IY rxY 1

u

IZ rxZ 5 2rxZ IX 1 rxX IZ 0

(8.59)

Y x e

y

IY

iy

ix b

IX

X

IZ i z

z Z (a) Member Orientation with xy Plane Vertical

Fig. 8.18

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480  Chapter 8   Three-Dimensional Framed Structures Y C

x

¯y y

e

C

A

¯y

i¯y

C

y

iy

i¯y iy

A

ix b

X

i¯z

¯z

iz

C

i¯z i z

z Section A–A

C Z

z (b) Actual Member Orientation

z

Fig. 8.18  (continued)

To obtain the unit vector iz along the local z axis, we divide the vector z by its magnitude Ïr2xX 1 r2xZ. This yields

iz 5 2

rxZ Ïr2xX 1 r2xZ

IX 1

rxX Ïr2xX 1 r2xZ

(8.60)

IZ 

The unit vector iy can now be established by using the cross product iz 3 ix, as



u

iy 5 iz 3 ix 5 det 2

IX rxZ

IY

Ïr2xX 1 r2xZ rxX

0 rxY

IZ rxX Ïr2xX 1 r2xZ rxZ

u

from which we obtain

1

iy 5 2

rxXrxY Ïr

2 xX

I 1 (Ïr 1r 2 2 xZ

X

2 xX

1 r2xZ) IY 2

rxYrxZ

1Ïr

2 xX

1 r2xZ

2I Z

(8.61)

From Eqs. (8.60) and (8.61), we can see that the transformation relationship between the global XYZ and the auxiliary local x y z coordinate systems can be expressed as

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Section 8.3   Space Frames  481

34 ix iy



iz

3

2

5

2

rxX rxXrxY

rxY Ïr2xX 1 r2xZ

Ïr2xX 1 r2xZ rxZ

2

0

Ïr2xX 1 r2xZ

rxZ rxYrxZ Ïr2xX 1 r2xZ rxX

Ïr2xX 1 r2xZ

43 4

IX IY IZ

(8.62)

In the next step, we rotate the auxiliary x y z coordinate system about its x axis, in a counterclockwise sense, by the angle of roll C, until the member’s principal axes are in their desired orientations. The final orientation of the member thus obtained is depicted in Fig. 8.18(b), in which the member’s principal axes are now designated as y and z axes. From this figure, we can see that the unit vectors along the y and z axes can be expressed in terms of those directed along the y and z axes, as

iy 5 cos Ciy 1 sin Ci z

(8.63a)



iz 5 2sin Ciy 1 cos Ci z 

(8.63b)

Thus, the transformation relationship between the auxiliary x y z and the actual xyz coordinate systems is given by ix 1 0 0 ix iy 5 0 cos C sin C iy (8.64) iz 0 2sin C cos C i

34 3



43 4 z

Finally, to obtain the transformation relationship between the global XYZ and the actual local xyz coordinate systems, we substitute Eq. (8.62) into Eq. (8.64) and carry out the required matrix multiplication. This yields

34

ix iy 5 iz

3

rxX 2rxXrxY cos C 2 rxZ sin C Ïr2xX 1 r2xZ rxXrxY sin C 2 rxZ cos C Ïr2xX 1 r2xZ

rxY Ïr2xX 1 r2xZ cos C 2 Ïr2xX 1 r2xZ sin C

rxZ 2rxY rxZ cos C 1 rxX sin C

43 4 IX IY IZ

Ïr2xX 1 r2xZ rxY rxZ sin C 1 rxX cos C Ïr2xX 1 r2xZ

(8.65) By comparing Eqs. (8.58) and (8.65), we can see that the member rotation matrix r can be expressed as

         

r5

3

rxX

rxY

2rxXrxY cos C 2 rxZ sin C Ïr 1 r 2 xX

2 xZ

rxXrxY sin C 2 rxZ cos C Ïr 1 r 2 xX

2 xZ

Ïr2xX 1 r2xZ cos C 2 Ïr2xX 1 r2xZ sin C

rxZ 2rxY rxZ cos C 1 rxX sin C Ïr2xX 1 r2xZ rxY rxZ sin C 1 rxX cos C Ïr 1 r 2 xX

2 xZ

4



(8.66)

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482  Chapter 8   Three-Dimensional Framed Structures

Note that the rotation matrix depends only on the global coordinates of the member ends and its angle of roll C. Based on the foregoing derivation, the angle of roll C is defined as the angle, measured clockwise positive when looking in the negative x direction, through which the local xyz coordinate system must be rotated around its x axis, so that the xy plane becomes vertical with the y axis pointing upward (i.e., in the positive direction of the global Y axis). The expression of the rotation matrix r, as given by Eq. (8.66), can be used to determine the transformation matrices T for the members of space frames oriented in any arbitrary directions, except for vertical members. This is because for such members rxX and rxZ are zero, causing some elements of r in Eq. (8.66) to become undefined. This situation can be remedied by defining the angle of roll differently for vertical members, as follows. For the special case of vertical members (i.e., members with centroidal or local x axis parallel to the global Y axis), the angle of roll C is defined as the angle, measured clockwise positive when looking in the negative x direction, through which the local xyz coordinate system must be rotated around its x axis, so that the local z axis becomes parallel to, and points in the positive direction of, the global Z axis (Fig. 8.19(b)). The expression of the rotation matrix r for vertical members can be derived using a procedure similar to that used previously for members with other orientations. We imagine that the vertical member’s desired (or actual design) orientation is reached in two steps, as shown in Figs. 8.19(a) and (b) on the next page. In the first step, while the member’s x axis is oriented in the desired (vertical) direction, its y and z axes are oriented so that the local z axis is parallel to the global Z axis, as shown in Fig. 8.19(a). As indicated there, the member’s principal axes in this (imaginary) orientation are designated as y and z (instead of y and z, respectively). The direction of the local x axis (known from the global coordinates of the member ends) is represented by the unit vector ix 5 rxYIY, while the direction of the z axis is given by the unit vector iz 5 IZ. The unit vector iy directed along the y axis, can therefore be conveniently ­established using the cross product i z 3 ix, as

u

IX i y 5 iz 3 ix 5 det 0 0

IY 0 rxY

u

IZ 1 5 2rxY IX 0

(8.67)

Thus, the transformation relationship between the global XYZ and the auxiliary local x y z coordinate system is given by

34 3

ix 0 iy 5 2rxY iz 0

rxY 0 0

43 4

0 0 1

IX IY  IZ

(8.68)

In the next step, we rotate the auxiliary x y z coordinate system about its x axis, in a counterclockwise sense, by the angle of roll C, until the member’s principal axes are in their desired orientations. This final orientation of the

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Section 8.3   Space Frames  483 x,Y

x,Y

IY ix

ix e

e

A

¯y

i¯y b

IX

i¯z

X

A

i¯y

¯y C

X

b

iy i¯z

y

iz

i¯y

¯y C

IZ C ¯z , Z (a) Orientation of Vertical Member with ¯z Axis Parallel to Global Z Axis

Fig. 8.19 

¯z ,Z

y

iy

iz

i¯z

z C

z

¯z Section A–A (b) Actual Orientation of Vertical Member

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484  Chapter 8   Three-Dimensional Framed Structures

member is depicted in Fig. 8.19(b), in which the member principal axes are now designated as the y and z axes. From this figure, we can see that the transformation relationship between the auxiliary x y z and the actual xyz coordinate systems is the same as given previously in Eq. (8.64). Thus, the desired transformation from the global XYZ coordinate system to the local xyz coordinate system can be obtained by substituting Eq. (8.68) into Eq. (8.64) and performing the required matrix multiplication. This yields

34 3

ix 0 iy 5 2rxY cos C iz rxY sin C

rxY 0 0

43 4

0 sin C cos C

IX IY IZ

(8.69)

from which we obtain the rotation matrix r for vertical members:

3

0 r 5 2rxY cos C rxY sin C

rxY 0 0

4

0 sin C  cos C

(8.70)

Member Rotation Matrix in Terms of a Reference Point  In most space frames, members are usually oriented so that their angles of roll can be found by inspection. There are structures, however, in which the orientations of some members may be such that their angles of roll cannot be conveniently determined. The orientation of such a member can alternatively be specified by means of the global coordinates of a reference point that lies in one of the principal (xy or xz) planes of the member, but not on its centroidal (x) axis. To discuss the process of determining the member rotation matrix r using such a reference point, consider the space-frame member shown in Fig. 8.20 on the next page, and let XP, YP, and ZP denote the global coordinates of an arbitrarily chosen reference point P, which is located in the member’s local xy plane, but not on its x axis. Since the global coordinates of the member end b are Xb, Yb, and Zb, the position vector p, directed from member end b to reference point P, can be written as

p 5 (XP 2 Xb)IX 1 (YP 2 Yb)IY 1 (ZP 2 Zb)Iz

(8.71)

Note that both points b and P are located in the local xy plane and, therefore, vector p also lies in that plane. Since the direction cosines of the local x axis are already known from the global coordinates of the member ends, the direction cosines of the local z axis can be conveniently established using the following relationship.

iz 5

ix 3 p uix 3 pu



(8.72)

in which |ix 3 p| represents the magnitude of the vector that results from the cross product of the vectors ix and p. With both ix and iz now known, the direction cosines of the local y axis can be obtained via the cross product,

iy 5 iz 3 ix

(8.73)

In the case that the reference point P is specified in the local xz plane of the member, the direction cosines of the local y axis need to be determined first

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Section 8.3   Space Frames  485 xy Plane x

Y

e P

y

p

iy

ix

X

b iz

Z z

Fig. 8.20 

using the relationship

iy 5

p 3 ix up 3 ixu

(8.74)



and then the direction cosines of the local z axis are obtained via the cross product

iz 5 ix 3 iy

(8.75)

It should be realized that the procedure described by Eqs. (8.71) through (8.75) enables us to obtain the member rotation matrix r directly by means of a reference point, without involving the angle of roll of the member. However, if desired, the angle of roll can also be obtained from the global coordinates of a reference point. To establish the relationship between the angle of roll C and a reference point P of a member, we first determine the components of the position vector p in the auxiliary x y z coordinate system, by applying the transformation relationship given in Eq. (8.62), as

px py 5 pz

34

3

2 2

rxX rxXrxY Ïr2xX 1 r2xZ rxZ Ïr2xX 1 r2xZ

rxY Ïr 1 r 2 xX

0

2 xZ

2

rxZ rxY rxZ Ïr2xX 1 r2xZ rxX

Ïr2xX 1 r2xZ

4

3

4

(XP 2 Xb ) (YP 2 Yb ) (ZP 2 Z b)

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486  Chapter 8   Three-Dimensional Framed Structures

from which we obtain

(8.76a)

px 5 rxX (XP 2 Xb) 1 rxY (YP 2 Yb) 1 rxZ (ZP 2 Zb)  py 5 2



2



pz 5 2

rxXrxY Ïr2xX 1 r2xZ rxYrxZ Ïr2xX 1 r2xZ rxZ Ïr 1 r 2 xX

2 xZ

(XP 2 Xb) 1 Ïr2xX 1 r2x Z(YP 2 Yb)

 (8.76b)

(ZP 2 Zb) (XP 2 Xb) 1

rxX Ïr 1 r2xZ 2 xX

(ZP 2 Zb)(8.76c)

in which px, py, and pz represent, respectively, the components of the position vector p in the directions of the local x axis and the y and z axes of the auxiliary x y z coordinate system. Now, if the reference point P lies in the xy plane of the member as shown in Fig. 8.21, then we can see from this figure that the angle of roll C and the components of p are related by the following equations:

sin C 5

pz

and

Ïp–y2 1 p2–z

cos C 5

py Ïp2–y 1 p–z2

(8.77)

¯y y

p¯z

P

C

py¯

¯z

C

z

Fig. 8.21 

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Section 8.3   Space Frames  487 ¯y

y

C

¯z C 2p¯y

p¯z 

P z

Fig. 8.22 



In the case that the reference point P is specified in the local xz plane, then from Fig. 8.22 we can see that the relationships between the sine and cosine of C and the components p can be expressed as py pz (8.78) sin C 5 2 and cos C 5 Ïp–y2 1 p–2 Ïp2–y 1 p2–z z



Equations (8.77) and (8.78) are valid for space-frame members oriented in any arbitrary directions, including vertical members. However, since rxX  5  rxZ 5 0 for vertical members, the expressions for py and pz, as given in Eqs. 8.76(b) and (c), cannot be used; appropriate expressions for the components of the position vector p in the auxiliary x y z coordinate system must be derived by applying Eq. (8.68), as px 0 rxY 0 (XP 2 Xb) py 5 2r 0 0 (Y 2 Y )

34 3 pz

xY

0

which yields px 5 rxY (YP 2 Yb)

0

43

1

P

b

(ZP 2 Zb)

4

(8.79a)

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488  Chapter 8   Three-Dimensional Framed Structures



py 5 2rxY (XP 2 Xb)

(8.79b)



pz 5 ZP 2 Zb

(8.79c)

It is important to realize that, for vertical members, Eqs. 8.79(b) and (c) should be used to evaluate py and pz, whereas for members with other orientations, these components are obtained from Eqs. 8.76(b) and (c). After py and pz have been evaluated, the sine and cosine of the member’s angle of roll C can be determined either by Eq. (8.77) if the reference point lies in the xy plane, or via Eq. (8.78) if the reference point is located in the xz plane. Once sin C and cos C are known, the member rotation matrix r can be determined by Eq. (8.66) if the member is not oriented in the vertical direction, or via Eq. (8.70) if the member is vertical.



E x ample 8.3 The global coordinates of the joints to which the beginning and end of a space-frame

member are attached are (1, 1.75, 1.5) m and (5, 3.75, 4.25) m, respectively. If the global
 coordinates of a reference point located in the local xy plane of the member are (2.6875, 3.4, 3.4625) m, determine the rotation matrix of the member.



S olut i o n

We determine the member rotation matrix r using the direct approach involving cross products of vectors, and then check our results using the angle-of-roll approach. Using the given coordinates of the two ends of the member, we evaluate its length and the direction cosines of the local x axis, as (see Eqs. (8.57)) L 5 Ï(Xe 2 Xb)2 1 (Ye 2 Yb)2 1 (Zb 2 Ze)2 5Ï(5 2 1)2 1 (3.75 2 1.75)2 1 (4.25 2 1.5)2 5 5.25 m rxX 5 rxY 5 rxZ 5

Xe 2 Xb L Ye 2 Yb L Ze 2 Zb L

5

521 5 0.7619 5.25

(1a)

5

3.75 2 1.75 5 0.38095 5.25

(1b)

5

4.25 2 1.5 5 0.52381 5.25

(1c)

Thus, the unit vector directed along the member local x (or centroidal) axis is ix 5 0.7619 IX 1 0.38095 IY 1 0.52381 IZ

(2)

Next, we form the position vector p, directed from member end b to reference point P, as (Eq. (8.71)) p 5 (XP 2 Xb)IX 1 (YP 2 Yb)IY 1 (ZP 2 Zb)IZ or

5 (2.6875 2 1)IX 1 (3.4 2 1.75)IY 1 (3.4625 2 1.5)IZ p 5 1.6875 IX 1 1.65 IY 1 1.9625 IZ

With ix and p known, we can now apply Eq. (8.72) to determine the unit vector in the local z direction. For that purpose, we first obtain a vector z along the local z axis by

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Section 8.3   Space Frames  489

evaluating the cross product of ix and p. Thus, IX z 5 ix 3 p 5 det 0.7619 1.6875

IY 0.38095 1.65

IZ 0.52381 1.9625

     5 f(0.38095)(1.9625) 2 (1.65)(0.52381)g IX 2 f(0.7619)(1.9625) 2 (1.6875)(0.52381)g IY 1 f(0.7619)(1.65) 2 (1.6875)(0.38095)g IZ or z 5 20.11667 IX 2 0.6113 IY 1 0.61428 IZ Note that z is not a unit vector. To obtain the unit vector iz, we need to divide z by its magnitude |z|, which equals )z) 5 )ix 3 p) 5 Ï(20.11667)2 1 (20.6113)2 1 (0.61428)2 5 0.87444 m Thus, the unit vector iz is given by z iz 5 5 20.13343 IX 2 0.69909 IY 1 0.70249 IZ uzu

(3)

The third unit vector, iy, can now be evaluated using the cross product of iz (Eq. (3)) and ix (Eq. (2)). Thus,

u

u

IX IY IZ 0.70249 iy 5 iz 3 ix 5 det 20.13343 20.69909 0.7619 0.38095 0.52381 5 f(20.69909)(0.52381) 2 (0.38095)(0.70249)] IX 2 f(20.13343)(0.52381) 2 (0.7619)(0.70249)] IY 1 f(20.13343)(0.38095) 2 (0.7619)(20.69909)] IZ or iy 5 20.6338 IX 1 0.60512 IY 1 0.48181 IZ

(4)

The member rotation matrix r can now be obtained by arranging the components of ix (Eq. (2)), iy (Eq. (4)) and iz (Eq. (3)) in the first, second, and third rows, respectively, of a 3 3 3 matrix. The member rotation matrix thus obtained is

3

0.7619 0.38095 r 5 20.6338 0.60512 20.13343 20.69909

4

0.52381 0.48181  0.70249

Ans

Alternative Method:  The member rotation matrix r can alternatively be determined by applying Eq. (8.66), which contains the sine and cosine of the angle of roll C. We first evaluate the components py and pz of the position vector p using Eqs. 8.76(b) and (c), respectively. By substituting the numerical values of rxX , rxY , and rxZ (from Eqs. (1)) and the given coordinates of member end b and reference point P into these equations, we obtain py 5 0.5723 m pz 5 0.66115 m

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490  Chapter 8   Three-Dimensional Framed Structures By substituting these values of py and pz into Eqs. (8.77), we obtain the sine and cosine of the angle of roll: sin C 5 0.75608

(5)

cos C 5 0.65448

Finally, by substituting the numerical values from Eqs. (1) and (5) into Eq. (8.66), we obtain the following rotation matrix for the member under consideration:

r5

3

0.7619 0.38095 20.6338 0.60512 20.13343 20.69907

4

0.52381 0.48179  0.70249

Checks

Procedure for Analysis The general procedure for analysis of space frames remains the same as that for plane frames developed in Chapter 6 (and modified in Chapter 7)—provided that the member local stiffness and transformation matrices, and local fixed-end force vectors, developed in this section, are used in the analysis.



E x ample 8.4 Determine the joint displacements, member end forces, and support reactions for the

three-member space frame shown in Fig. 8.23(a) on the next page, using the matrix stiffness method.



S olut i o n

Analytical Model:  The space frame has six degrees of freedom and 18 restrained coordinates, as shown in Fig. 8.23(b). Structure Stiffness Matrix: Member 1   By substituting L 5 5 m, and the material and cross-sectional properties given in Fig. 8.23(a), into Eq. (8.39), we obtain the local stiffness matrix k for member 1:

7 8 9 10 11 12 1 2 3 4 5 6 2,940,000 0 0 0 0 0 22,940,000 0 0 0 0 0 0 13,632 0 0 0 34,080 0 213,632 0 0 0 34,080 0 0 4,492.8 0 211,232 0 0 0 24,492.8 0 211,232 0 0 0 0 237.93 0 0 0 0 0 2237.93 0 0 0 0 211,232 0 37,440 0 0 0 11,232 0 18,720 0 0 34,080 0 0 0 113,600 0 234,080 0 0 0 56,800 K1 5 k1 5 22,940,000 0 0 0 0 0 2,940,000 0 0 0 0 0 0 213,632 0 0 0 234,080 0 13,632 0 0 0 234,080 0 0 24,492.8 0 11,232 0 0 0 4,492.8 0 11,232 0 0 0 0 2237.93 0 0 0 0 0 237.93 0 0 0 0 211,232 0 18,720 0 0 0 11,232 0 37,440 0 0 34,080 0 0 0 56,800 0 234,080 0 0 0 113,600

3

4

7 8 9 10 11 12 1 2 3 4 5 6

(1) Since the member’s local x, y, and z axes are oriented in the directions of the global X, Y, and Z axes, respectively (see Fig. 8.23(a)), no coordinate transformations are necessary (i.e., T1 5 I); thus, K1 5 k1. To determine the fixed-end force vector due to the 48 kN/m member load, we apply the fixed-end force expressions for loading type 3 given inside the front cover. This yields

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Section 8.3   Space Frames  491

 FSby 5 FSey 5 120 kN FMbz 5 2FMez 5 2100 kN ? m with the remaining fixed-end forces being 0. Thus, using Eq. (8.40), we obtain

Ff 1 5 Qf 1 5

34 0 120 0 0 0 100 0 120 0 0 0 2100

7 8 9 10 11 12 1 2 3 4 5 6

(2)

y

Y

30º

z x

y

5m

48 kN/m 150 kN • m

x z

X

150 kN • m 5m 5m

x

Z

E = 200 GPa G = 79.31 GPa A = 73,500 mm2 Iz = 710 (106) mm4 Iy = 234 (106) mm4 J = 15 (106) mm4

z

y (a) Space Frame

Fig. 8.23 

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492  Chapter 8   Three-Dimensional Framed Structures

24 Y 21

4

5 3 2

1 1

2

22

20

23

7

10

19

4

X

1 3

9 8 6

12 11

2

Z

3

13

16

15 14 18 17 (b) Analytical Model

S=

3

1 2,951,270.4 23,957.41 0 29,893.53 216,944.09 11,232

2 3 23,957.41 0 2,964,979.3 0 0 2,958,124.8 28,368.27 234,080 9,893.53 11,232 234,080 0

4 5 6 29,893.53 216,944.09 11,232 28,368.27 9,893.53 234,080 234,080 11,232 0 208,398.83 32,978.42 0 32,978.42 94,158.23 0 0 0 151,277.93

3 3 3 1 2 3 4 5 6

0

Pf =

120 0 0 0 2100

1 2 3 4 5 6

(c) Structure Stiffness Matrix and Fixed-Joint Force Vector

Fig. 8.23  (continued)

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Section 8.3   Space Frames  493 y

11.641 0.393 x y

18.049

z 0.181

48 kN/m

21.500 1 2.970

176.429

L= 5m 2.970

0.181

4.914

87.924

L=5m

60.058

28.810

x

6.953

18.497 86.244

63.571 0.393

25.840 194.220

3

44.463

21.500

28.810

6.953

30.19

0.064

9.938

x

2

61.684

z

23.125

L=5m 18.497 42.955

25.840 30.801

18.049

z

44.463 0.064

y (d) Member Local End Forces

R=

21.500 kN 176.429 kN 22.970 kN 0.181 kN • m 4.914 kN • m 194.220 kN • m 218.497 kN 44.463 kN 225.840 kN 242.955 kN • m 20.064 kN • m 30.801 kN • m 23.003 kN 19.108 kN 28.810 kN 231.965 kN • m 25.014 kN • m 20.393 kN • m

Fig. 8.23  (continued)

7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24

(e) Support Reaction Vector

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494  Chapter 8   Three-Dimensional Framed Structures 0.393 28.810 Y 4

3.003

31.965

19.108 48 kN/m 0.181

21.500

150 kN • m

X

1

2 2.970

5.014

150 kN • m 176.429

194.220 4.914 Z 3 18.497

42.955

25.840 44.463 30.801 0.064 (f) Support Reactions

Fig. 8.23  (continued)

From Fig. 8.23(b), we observe that the code numbers for member 1 are 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6. Using these code numbers, we store the pertinent elements of K1 (Eq. (1)) and Ff1 (Eq. (2)) in the 6 3 6 structure stiffness matrix S and the 6 3 1 structure fixed-joint force vector Pf , respectively (Fig. 8.23(c)). Member 2   Because the length, as well as the material and cross-sectional properties, of member 2 are identical to those of member 1, k2 5 k1 (Eq. (1)). To obtain the transformation matrix T for member 2, we first determine the direction cosines of its local x axis using Eqs. (8.57), as Xe 2 Xb rxX 5 50 L Ye 2 Yb 0 2 (25) rxY 5 5 51 L 5 Ze 2 Zb rxZ 5 50 L From Fig. 8.23(a), we can see that the angle of roll C for this vertical member is 90°. Thus, cos C 5 0

and

sin C 5 1

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Section 8.3   Space Frames  495

By substituting the foregoing numerical values of rxX, rxY, rxZ, cos C, and sin C into Eq. (8.70), we determine the rotation matrix r for member 2 to be

3

0 r2 5 0 1

1 0 0

4

0 1 0

By substituting this rotation matrix into Eq. (8.55), we obtain the following 12 3 12 transformation matrix for member 2.

T2 5

3

0 0 1 0 0 0 0 0 0 0 0 0

1 0 0 0 0 0 0 0 0 0 0 0

0 1 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 1 0 0 0 0 0 0

0 0 0 1 0 0 0 0 0 0 0 0

0 0 0 0 1 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 1 0 0 0

0 0 0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 0 0 1 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 1

0 0 0 0 0 0 0 0 0 1 0 0

4

0 0 0 0 0 0  0 0 0 0 1 0

(3)

The global stiffness matrix for member 2 can now be evaluated by substituting k2 (from Eq. (1)) and T2 (Eq. (3)) into the relationship K 5 TTkT, and performing the necessary matrix multiplications. This yields

K2 5

13 14 15 16 17 18 1 2 3 4 5 6 4,492.8 0 0 0 0 211,232 24,492.8 0 0 0 0 211,232 0 2,940,000 0 0 0 0 0 22,940,000 0 0 0 0 0 0 13,632 34,080 0 0 0 0 213,632 34,080 0 0 0 0 34,080 113,600 0 0 0 0 234,080 56,800 0 0 0 0 0 0 237.93 0 0 0 0 0 2237.93 0 211,232 0 0 0 0 37,440 11,232 0 0 0 0 18,720 24,492.8 0 0 0 0 11,232 4,492.8 0 0 0 0 11,232 0 22,940,000 0 0 0 0 0 2,940,000 0 0 0 0 0 0 213,632 234,080 0 0 0 0 13,632 234,080 0 0 0 0 34,080 56,800 0 0 0 0 234,080 113,600 0 0 0 0 0 0 2237.93 0 0 0 0 0 237.93 0 211,232 0 0 0 0 18,720 11,232 0 0 0 0 37,440

4

3

13 14 15 16 17 18 1 2 3 4 5 6

The relevant elements of K2 are stored in S (Fig. 8.23(c)). Member 3  k3 5 k1 (given in Eq. (1)). rxX 5 rxY 5 rxZ 5

Xe 2 Xb L Ye 2 Yb L Ze 2 Zb L

50 50 5

0 2 (25) 51 5

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496  Chapter 8   Three-Dimensional Framed Structures From Fig. 8.23(a), we can see that C 5 30°. Thus, cos C 5 0.86603 and sin C 5 0.5 By applying Eq. (8.66), we determine the rotation matrix for member 3 to be

4

0 1 0 r3 5 20.5 0.86603 0 20.86603 20.5 0

3

Thus, the transformation matrix for this member is given by

T3 5

0    0

1    0    0

0    0    0

0    0    0

0

20.5    0.86603

0    0    0

0    0    0

0    0    0

0

20.86603 20.5

0    0    0

0    0    0

0    0    0

0

0    0

0    0    0

1    0    0

0    0    0

0

0     0 0 20.5    0.86603

0    0    0

0    0    0

0

0     0 0 20.86603 20.5

0    0    0

0    0    0

0

0    0

0    0    0

0    0    0

1    0    0

0

0    0

0    0    0

0

20.5    0.86603

0    0    0

0

0    0

0    0    0

0

20.86603

0    0    0

0

0    0

0    0    0

0    0    0

0    0    0

1

0    0

0    0    0

0    0    0

0

20.5    0.86603

0

0    0

0    0    0

0    0    0

0

20.86603

20.5

(4)

20.5 0

and the member global stiffness matrix K3 5 TT3k3T3 is

K3 5

3

19 20 21 22 23 24 1

2 3 4

5 6

4

6,777.64 23,957.41 0 9,893.53 16,944.09 0 26,777.64 3,957.41 0 9,893.53 16,944.09 0 19 23,957.41 11,347.31 0

0 228,368.27 29,893.53 0 3,957.41 211,347.31 0 228,368.27 29,893.53 0

0 2,940,000 0

0 0

0

0 22,940,000 0

9,893.53 228,368.27

0

94,560.90 32,978.42 0

16,944.09 29,893.53

0

32,978.42 56,480.30 0 216,944.09 9,893.53

0

0

29,893.53 28,368.27 0 47,280.45 16,489.21 0 0 16,489.21 28,240.15 0

20 21 22 23

0 0 0 0 0 237.93 0 0 0 0 0 2237.93 24 26,777.64 3,957.41

0

3,957.41 211,347.31 0 0

6,777.64 23,957.41 0 29,893.53 216,944.09 0

29,893.53 216,944.09 0

28,368.27 9,893.53 0 23,957.41 11,347.39 0 28,368.27 9,893.53 0

0 22,940,000 0

0

0

0

0 2,940,000 0

2

0

3

29,893.53 28,368.27 0 94,560.90 32,978.42 0

4

9,893.53 228,368.27

0

47,280.45 16,489.21 0

16,944.09 29,893.53

0

16,489.21 28,240.15 0 216,944.09 9,893.53

0 0 0 0 0 2237.93 0

0

1

0 32,978.42 56,480.30 0

5

0 0 0 0 237.93 6

The complete structure stiffness matrix S and the structure fixed-joint force vector Pf are given in Fig. 8.23(c).

F G

Joint Load Vector:   By comparing Figs. 8.23(a) and (b), we obtain 0 1 0 2 3 0 P5 4 2150 kN ? m 5 0 6 150 kN ? m Joint Displacements:  By substituting P, Pf , and S into the structure stiffness ­relationship, P 2 Pf 5 Sd, and solving the resulting simultaneous equations, we determine the joint displacements to be Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

d5

F G 20.007313 m 20.01512 m 20.009799 m 20.7621 rad 0.2684 rad 1.6497 rad

Section 8.3   Space Frames  497

3 1023

Ans

Member End Displacements and End Forces: Member 1   

u1 5 v1 5

34 3 4 0 0 0 0 0 0 20.007313 20.01512 20.009799 20.7621 0.2684 1.6497

F1 5 Q1 5 k1u1 1 Qf1 5

(5)

3 1023

21.500 kN 176.429 kN 22.970 kN 0.181 kN ? m 4.914 kN ? m 194.220 kN ? m 221.500 kN 63.571 kN 2.970 kN 20.181 kN ? m 9.938 kN ? m 87.924 kN ? m

7 8 9 10 11 12  1 2 3 4 5 6

Ans

The member local end forces are depicted in Fig. 8.23(d), and the pertinent elements of F1 are stored in the support reaction vector R (Fig. 8.23(e)). Member 2  v2 5 v1 (see Eq. (5)).

u2 5 T2v2 5

34 3 4 0 0 0 0 0 0 20.01512 20.009799 20.007313 0.2684 1.6497 20.7621

3 1023

Q2 5 k2u2 5

44.463343 kN 225.839949 kN 218.496837 kN 20.0638546 kN ? m 30.800682 kN ? m 242.95526 kN ? m 244.463343 kN 25.839949 kN 18.496837 kN 0.0638546 kN ? m 61.683502 kN ? m 286.244485 kN ? m



Ans

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498  Chapter 8   Three-Dimensional Framed Structures

F2 5 TT2 Q2 5

3 4 218.497 44.463 225.840 242.955 20.064 30.801 18.497 244.463 25.840 286.244 0.064 61.684

13 14 15 16 17 18 1 2 3 4 5 6

Member 3  v3 5 v1 (see Eq. (5)). 0 0 0 0 0 0 u3 5 T3v3 5 20.009799 3 1023 20.009441 0.01389 1.6497 0.6135 0.5258

34 3 4 34 Q3 5 k3u3 5

28.810 kN 18.049 kN 26.953 kN 20.393 kN ? m 11.641 kN ? m 30.190 kN ? m 228.810 kN 218.049 kN 6.953 kN 0.393 kN ? m 23.125 kN ? m 60.058 kN ? m

 Ans

23.003 19.108 28.810 231.965 25.014 20.393 3.003 219.108 228.810 263.574 210.002 0.393

19 20 21 22 23 24 T F3 5 T3 Q3 5 1 2 3 4 5 6 Support Reactions:   The completed reaction vector R is shown in Fig. 8.23(e), and the support reactions are depicted on a line diagram of the space frame in Fig. 8.23(f).  Ans Equilibrium checks:  The six equations of equilibrium ( Fx 5 0, Fy 5 0, Fz 5 0, Mx 5 0, My 5 0, and Mz 5 0) are satisfied for each member of the space frame shown in Fig. 8.23(d). Furthermore, the six equilibrium equations in the directions of the global coordinate axes ( FX 5 0, FY 5 0, FZ 5 0, MX 5 0, MY 5 0, and MZ 5 0) are satisfied for the entire structure shown in Fig. 8.23(f).

o

o

o

o

o

o o

o

o

o

o

o

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Problems  499 Identify degrees of freedom d and restrained coordinates of the structure For each member: Evaluate k, Qf , and T Calculate K = TT kT and Ff = TTQf Store K in S and Ff in Pf Form joint load vector P Solve P 2 Pf = Sd for d For each member: Obtain v from d Calculate u = Tv, Q = ku + Qf and F = TTQ Store F in R

Fig. 8.24  Stiffness Method of Analysis

SUMMARY In this chapter, we have extended the matrix stiffness method to the analysis of three-dimensional framed structures. The stiffness and transformation relationships for the members of space trusses, grids, and space frames are developed in Sections 8.1, 8.2, and 8.3, respectively. It should be noted that the overall format of the stiffness method of analysis remains the same for all types of (two- and three-dimensional) framed structures—provided that the member stiffness and transformation relations, appropriate for the particular type of structure being analyzed, are used in the analysis. A block diagram summarizing the overall format of the stiffness method is shown in Fig. 8.24.

PROBLEMS Section 8.1

Section 8.2

8.1 through 8.5  Determine the joint displacements, member axial forces, and support reactions for the space trusses shown in Figs. P8.1 through P8.5, using the matrix stiffness method. Check the hand-calculated results by using the computer program which can be downloaded from the publisher’s website for this book, or by using any other general purpose structural analysis program available. 8.6  Develop a computer program for the analysis of space trusses by the matrix stiffness method. Use the program to ­analyze the trusses of Problems 8.1 through 8.5, and compare the computer-generated results to those obtained by hand ­calculations.

8.7 through 8.12   Determine the joint displacements, member local end forces, and support reactions for the grids shown in Figs. P8.7 through P8.12, using the matrix stiffness method. Check the hand-calculated results by using the computer program which can be downloaded from the publisher’s website for this book, or by using any other general purpose structural analysis program available. 8.13   Develop a program for the analysis of grids by the matrix stiffness method. Use the program to analyze the grids of Problems 8.7 through 8.12, and compare the computer-generated results to those obtained by hand calculations.

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500  Chapter 8   Three-Dimensional Framed Structures Y

Y 100 kN 50 kN

150 kN

5 50 kN

4

1

10 m 3 12 m

X 6m

3m

2

X 8m

6m 2

3m

3 0

0

5m 1

6m

2

3 2

2m

3

1

1

4 4

75 kN

8m

Z

EA = constant E = 200 GPa A = 3,800 mm2

9m

Fig. P8.3

Z EA = constant E = 70 GPa A = 2,000 mm2

Y

400 kN

Fig. P8.1

200 kN

(160 0) (800)

(1600)

(1600)

)

)

300 kN

(800

00

X

0

)

00

(8

6 (1

1.2 m

(800) 0) (80

)

00

(8

1.6 m

0)

2m

3m

60

(160

(1

0)

Y

1.2 m 0

450 kN

3m

X 2m

3m 1.6 m

Z EA = constant E = 200 GPa A = 1,600 mm2

Fig. P8.2

Z

1.6 m

E = 200 GPa

Note: Numbers in parentheses represent cross-sectional areas mm2

Fig. P8.4

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Problems  501

Section 8.3

Y

8.14 through 8.17   Determine the joint displacements, member local end forces, and support reactions for the space frames shown in Figs. P8.14 through P8.17, using the matrix stiffness method. Check the hand-calculated results by using the computer program which can be downloaded from the publisher’s website for this book, or by using any other general purpose structural analysis program available. 8.18   Develop a program for the analysis of space frames by the matrix stiffness method. Use the program to analyze the frames of Problems 8.14 through 8.17, and compare the computer-­generated results to those obtained by hand calculations. 8.19   Develop a general computer program that can be used to analyze any type of framed structure.

3

3m Z

Y

2m

20 100 kN

7 12

5

E, G, I, J = constant E = 31 GPa G = 12.4 GPa I = 20(106) mm4 J = 24(106) mm4

11

45 kN 90 kN

2m

2 1

90 kN

8 2m

45 kN

90 kN

7.5 m 7

3

Z

10 m 6

1

Fig. P8.8

3

Y

4m

4

X

0

0

1

X

4m 2

4m

8m

7.5 m

5

1

/m

E, G, I, J = constant E = 30 GPa G = 12.5 GPa I = 4.8(109 ) mm4 J = 3.2(109 ) mm4

2

4

X

kN

10 6

9 8

/m kN 37 .5

1

Fig. P8.7

90 kN

3m 2

37.5 kN/m

Y 2m

X

0

240 kN 125 kN • m

4m

2

1

3m

2

5m

3

Z

Z

Fig. P8.5

E, G, I, J = constant E = 200 GPa G = 80 GPa I = 120(106) mm4 J = 2.1(106) mm4

EA = constant E = 200 GPa A = 4,000 mm2

Fig. P8.9

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502  Chapter 8   Three-Dimensional Framed Structures Y

Y 6m

6m

150 kN

15 kN/m

0

15

/m

9m

25

kN

X

m

X

/ kN

/m

15

12 m

kN

6m

15 kN/m

6m

12 m Z

Z

E, G, I, J = constant E = 200 GPa G = 76 GPa I = 300(106 ) mm4 J = 7(106 ) mm4

E, G, I, J = constant E = 200 GPa G = 76 GPa I = 300(106 ) mm4 J = 7(106 ) mm4

Fig. P8.12

Fig. P8.10

Y

10 m Y

20 kN/m

10 m 1 2

y

/m X

20

1

2

kN

0

5m

x y

/m

18 kN/m

18

3

3

150 kN

5m 5m

Z E, G, I, J = constant E = 200 GPa G = 80 GPa I = 3200(106) mm4 J = 25(106) mm4

Fig. P8.11

x z

kN

4 10 m

z

5m

Z

x z y

X

Material and cross-sectional properties are constant: E = 30 GPa G = 12.5 GPa A = 31,000 mm 2 Iz = 106(106 ) mm4 Iy = 60(106 ) mm4 J = 129(106 ) mm4

Fig. P8.14

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Problems  503 Y

Y 4m 2

3

50 kN/m

2

X

/m

150 kN • m

50

3m 150 kN • m

3 4

50

200 kN

kN

6m

6

200 kN 1

Z

/m

50 kN/m

5m

1

5

kN

8

7

4 X

1 6m

Material properties are constant: E = 70 GPa G = 27.6 GPa All members have circular cross-sections with A = 1,260 mm2 I z = Iy = 1.5(106) mm4 J = 2.9(106) mm4

2

3

6m Z Member orientations: Girders: local y axis is vertical x Columns:

Fig. P8.15

z y

Y 15 m z y

5

/m kN z

y

x

x

Material and cross-sectional properties are constant: E = 200 GPa G = 80 GPa A = 13,200 mm2 Iz = 400(106) mm4 Iy = 35(106) mm4 J = 1.5(106) mm4

12 m X

10 kN/m x 30 kN 6m Z 6m

y z

Material and cross-sectional properties are constant: E = 200 GPa G = 76 GPa A = 19,000 mm2 Iz = 260(106 ) mm4 Iy = 86(106 ) mm4 J = 4.5(106 ) mm4

Fig. P8.17

Fig. P8.16

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9

SPECIAL TOPICS AND MODELING TECHNIQUES Learning Objectives At the end of this chapter, you will be able to: 9.1 Develop Structure Stiffness Matrix Including ­Restrained Coordinates—an Alternate Formulation of the Stiffness Method 9.2 Implement Approximate Matrix Analysis of ­Rectangular Building Frames 9.3 Perform Condensation of Degrees of Freedom, and Substructuring 9.4 Model Inclined Roller Supports 9.5 Model Offset Connections 9.6 Model Semirigid Connections 9.7 Include Shear Deformations 9.8 Model Nonprismatic Members 9.9 Solve Large Systems of Stiffness Equations

The Montreal Biosphere, Canada meunierd/Shutterstock.com

504

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Section 9.1   The Structure Stiffness Matrix Including Restrained Coordinates  505

In this chapter, we consider some modifications and extensions of the matrix stiffness method developed in the preceding chapters. Also considered herein are techniques for modeling certain special features (or details) of structures, so that more realistic analytical models can be created and more accurate structural responses predicted from the analysis. We begin by discussing an alternative formulation of the stiffness method in Section 9.1. As this alternative formulation involves the structure stiffness matrix for all the coordinates (including the restrained coordinates) of the structure, it is less efficient for computer implementation than the formulation used in the preceding chapters. Nonetheless, an understanding of this alternative formulation provides some important insights into the stiffness method of analysis. In Sections 9.2 and 9.3, we consider some techniques for reducing a structure’s degrees of freedom, and/or the number of structure stiffness equations to be processed simultaneously. These techniques are useful in handling the analysis of large structures. Section 9.4 is devoted to the modeling of inclined roller supports; in the following two sections, we develop techniques for modeling the effects of offset connections (Section 9.5), and semirigid connections (Section 9.6), in the analysis. The inclusion of shear deformation effects in the analysis of beams, grids, and frames is considered in Section 9.7; and in Section 9.8, we cover the analysis of structures composed of nonprismatic members. Finally, we conclude the chapter by discussing a procedure for efficiently storing and solving the systems of linear equations that arise in the analysis of large structures.

9.1  THE STRUCTURE STIFFNESS MATRIX INCLUDING ­ ESTRAINED COORDINATES—AN ALTERNATIVE R FORMULATION OF THE STIFFNESS METHOD

In the formulation of the matrix stiffness method, as developed in the preceding chapters, the conditions of zero (or known) joint displacements corresponding to restrained coordinates are applied to the member force-displacement relationships before the stiffness relations for the entire structure are assembled. This approach yields structure stiffness relations that contain only the degrees of freedom as unknowns. Alternatively, the stiffness method can be formulated by first establishing the stiffness relations for all the coordinates (free and restrained) of the structure, and then applying the restraint conditions to the structure stiffness relations, which now contain both the degrees of freedom, and the support reactions, as unknowns. In the alternative formulation, the structure’s restrained coordinates are initially treated as free coordinates, and the stiffness relations for all the coordinates of the structure are expressed as P* 2  Pf* 5 S* NC 3 1 NC 3 1 NC 3 NC

   d* NC 3 1

(9.1)

with NC 5 NCJT(NJ) 5 NDOF 1 NR

(9.2)

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506  Chapter 9   Special Topics and Modeling Techniques

In Eqs. (9.1) and (9.2), NC denotes the number of structure coordinates; P* represents the joint forces (i.e., the known external loads and the unknown support reactions); Pf* denotes the fixed-joint forces, due to member loads, temperature changes, and fabrication errors, at the locations, and in the directions, of the structure coordinates; S* represents the stiffness matrix for the structure coordinates (free and restrained); and d* denotes the joint displacements (i.e., the unknown degrees of freedom and the known displacements corresponding to the restrained coordinates). The structure stiffness matrix S* and fixed-joint force vector Pf* can be determined by assembling the member global stiffness matrices K and fixed-end force vectors Ff , respectively, using the member code number technique described in the preceding chapters. The application of this technique remains essentially the same, except that now those elements of K and Ff that correspond to the restrained coordinates are no longer discarded, but are added (stored) in their proper positions in S* and Pf*. As indicated in the preceding paragraph, the structure stiffness relations (Eq. (9.1)) contain two types of unknown quantities; namely, the unknown joint displacements and the unknown support reactions. To separate the two types of unknowns, we rewrite Eq. (9.1) in partitioned-matrix form:

3

43

P NDOF 3 1 R NR 3 1

2

43

Pf NDOF 3 1 Rf NR 3 1

5

S NDOF 3 NDOF SRF NR 3 NDOF

43

SFR NDOF 3 NR SRR NR 3 NR

4

d NDOF 3 1  (9.3) dR NR 3 1

in which, P, R, Pf , S, and d denote the same quantities as in the preceding chapters; Rf denotes the structure fixed-joint forces corresponding to the restrained coordinates; and dR denotes the support displacement vector. Note that the Pf and Rf vectors contain structure fixed-joint forces due to member loads, temperature changes, and fabrication errors. The effects of support displacements are not included in Pf and Rf , but are directly incorporated into the analysis through the support displacement vector dR. Each element of the submatrix SFR in Eq. (9.3) represents the force at a free coordinate caused by a unit displacement of a restrained coordinate. The other two submatrices, SRF and SRR, can be interpreted in an analogous manner. By multiplying the two partitioned matrices on the right-hand side of Eq. (9.3), we obtain two matrix equations, P 2 Pf 5 Sd 1 SFRdR

(9.4a)

R 2 Rf 5 SRFd 1 SRRdR

(9.4b)

which can be rearranged as P 2 Pf 2 SFR dR 5 Sd

R 5 Rf 1 SRF d 1 SRRdR

(9.5a) 

(9.5b)



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Section 9.1   The Structure Stiffness Matrix Including Restrained Coordinates  507

The procedure for analysis essentially consists of first solving Eq. (9.5a) for the unknown joint displacements d, and then substituting d into Eq. (9.5b) to evaluate the support reactions R. With d known, the member end displacements and end forces can be obtained using the procedures described in the preceding chapters. In the case of structures with no support displacements, dR 5 0, and Eqs. (9.5) reduce to



P 2 Pf 5 Sd

(9.6a)

R 5 Rf 1 SRF d 

(9.6b)



The main advantages of the alternative formulation are that support displacements can be incorporated into the analysis in a direct and straightforward manner, and the reactions can be more conveniently calculated by using the structure stiffness relations. However, since the alternative formulation uses the stiffness matrix for all of the structure’s coordinates, it requires significantly more computer memory space than the standard formulation developed in the preceding chapters, which uses the stiffness matrix for only the free coordinates of the structure. For this reason, the alternative formulation is not considered to be as efficient for computer implementation as the formulation developed in the preceding chapters [14]. The application of the alternative formulation is illustrated by the following example. E xample 9.1



125 kN • m

Determine the joint displacements, member local end forces, and support reactions for the plane frame of Fig. 9.1(a), due to the combined effect of the loading shown and a

16.667 kN/m

2 Y

3m

9

3 1 2

300 kN

7

3

2

8 1

3m

3m

6m E, A, I = constant E = 200 GPa A = 7600 mm2 I = 129(106) mm4 (a) Frame

1 4

X 6 5 (b) Analytical Model

Fig. 9.1

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508  Chapter 9   Special Topics and Modeling Techniques

settlement of 25 mm of the left support. Use the alternative formulation of the matrix stiffness method.



This frame was analyzed in Example 6.6 for external loading, and in Example 7.4 for the combined effect of the loading and the support settlement, using the standard formulation.

S olution

Analytical Model: See Fig. 9.1(b). In this example, we use the same analytical model of the frame as used previously, so that the various member matrices calculated in Example 6.6 can be reused. The frame has three degrees of freedom and six restrained coordinates. Thus, the total number of structure coordinates is nine. Structure Stiffness Matrix: By storing the element of the member global stiffness matrices K1 and K2, calculated in Example 6.6, in their proper positions in the 9 3 9 structure stiffness matrix S*, we obtain, in units of kilonewtons and meters, the following stiffness matrix for all the structure coordinates.

S* 5

S

SFR

RF

SRR

3S

4

1 2 3 4 5 6 7 8 9 299,470 90,225 3,077 246,137 290,225 3,077 2253,333 0 0 90,225 182,910 2,762 290,225 2181,477 21,538 0 21,433.33 4,300 3,077 2,762 32,584 23,077 1,538 7,692 0 24,300 8,600 246,137 290,225 23,077 46,137 90,225 23,077 0 0 0 5 290,225 2181,477 1,538 90,225 181,477 1,538 0 0 0 3,077 21,538 7,692 23,077 1,538 15,384 0 0 0 2253,333 0 0 0 0 0 253,333 0 0 0 21,433.33 24,300 0 0 0 0 1,433.33 24,300 0 4,300 8,600 0 0 0 0 24,300 17,200

1 2 3 4 5 6 7 8 9

(1)

Structure Fixed-Joint Force Vector Due to Member Loads: Similarly, by storing the elements of the member global fixed-end force vectors Ff 1 and Ff 2, calculated in ­Example 6.6, in the 9 3 1 structure fixed-joint force vector Pf*, we obtain 0 200 262.5 0 Pf P*f 5 5 150 Rf 112.5 0 50 250

3 4

1 2 3 4 5 6 7 8 9

(2)

Joint Load Vector: From Example 6.6,

P5

3 4

0 1 0 2 2125 3

(3)

Support Displacement Vector: From the analytical model of the structure in Fig. 9.1(b), we observe that the given 25 mm settlement of the left support occurs at

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Section 9.1   The Structure Stiffness Matrix Including Restrained Coordinates  509

FG

the location and in the direction of restraint coordinate 5. Thus, the support displacement vector can be expressed as

dR 5

0 20.025 0 0 0 0

4 5 6  7 8 9

(4)

Joint Displacements: By substituting S and SFR from Eq. (1), Pf from Eq. (2), P from Eq. (3), and dR from Eq. (4) into Eq. (9.5a), we write the stiffness relations for the free coordinates of the frame as P 2 Pf 2 SFRdR 5 Sd

3 4 3 4 3

0 0 246,137 0 2 200 2 290,225 2125 23,077 262.5

5

or

290,225 2181,477 7,692

3

299,470 90,225 3,077

3

90,225 182,910 2,762

3,077 21,538 15,384

2253,333 0 0 21,433.33 0 24,300

43 4 d1 d2 d3

3,077 2,762 32,584

4 3

22,255.63 299,470 24,736.93 5 90,225 224.05 3,077

90,225 182,910 2,762

FG

4

0 4,300 8,600

0 20.025 0 0 0 0

43 4

3,077 2,762 32,584

d1 d2 d3

By solving these equations, we determine the joint displacements to be d5

3

4

0.000308 m 1 20.02607 m 2  0.001443 rad 3

(5) Ans

Note that these joint displacements are identical to those calculated in Example 7.4. The joint displacement vector for all the coordinates (free and restrained) of the structure can be expressed as

d* 5

3d 4 d

R

0.000308 m 20.02607 m 0.001443 rad 0 5 20.025 m 0 0 0 0

1 2 3 4 5 6 7 8 9

(6)

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510  Chapter 9   Special Topics and Modeling Techniques Support Reactions: To evaluate the support reaction vector R, we substitute SRF and SRR from Eq. (1), Rf from Eq. (2), dR from Eq. (4), and d from Eq. (5) into Eq. 9.5(b): R 5 Rf 1 SRF  d 1 SRR  dR. This yields

R5

F G 78.008 kN 318.8 kN 126.193 kN ? m 278.008 kN 81.165 kN 2149.69 kN ? m

4 5 6  7 8 9

Ans

Note that these support reactions are the same as those calculated in Example 7.4. Member End Displacements and End Forces:

FG FG F G F G F G FG FG F G

Member 1  Using member code numbers and Eq. (6), we obtain

v1 5

y1 y2 y3 y4 y5 y6

4 5 6 5 1 2 3

d4* d5* d6* d1* d2* d3*

5

0 20.025 0 0.000308 20.02607 0.001443

Next, we use the member transformation matrix T1 from Example 6.6, to calculate 20.02236 20.01118 0 20.02318 20.01193 0.001443

u1 5 T1v1 5

The member local end forces can now be obtained by using the member local stiffness matrix k1 and fixed-end force vector Qf 1, from Example 6.6, as

Q1 5 k1u1 1 Qf1 5

320.01 kN 72.82 kN 126.2 kN ? m  251.69 kN 61.34 kN 287.69 kN ? m

Ans

Member 2  The global and local end displacements for this horizontal member are

u2 5 v2 5

y1 y2 y3 y4 y5 y6

1 2 3 5 7 8 9

d1* d2* d3* d7* d8* d9*

5

0.000308 20.02607 0.001443 0 0 0

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Section 9.2   Approximate Matrix Analysis of ­Rectangular Building Frames  511

F G

By using k2 and Qf 2 from Example 6.6, we compute the member local end forces to be

Q2 5 k2u2 1 Qf 2 5

78.007 kN 18.835 kN 237.291 kN ? m  278.007 kN 81.165 kN 2149.699 kN ? m

Ans

As expected, the foregoing member local end force vectors Q1 and Q2 are identical to those calculated in Example 7.4.

9.2  APPROXIMATE MATRIX ANALYSIS OF ­RECTANGULAR BUILDING FRAMES

In building frames of low to medium height, the axial deformations of members are generally much smaller than the bending deformations. Therefore, the number of degrees of freedom of such frames can be reduced, without significantly compromising the accuracy of the analysis results, by neglecting the axial deformations of members, or by assuming that the members are inextensible. In this section, we consider the analysis of rectangular plane frames composed of horizontal and vertical members which are assumed to be inextensible (i.e., they cannot undergo any axial elongation or shortening). Consider, for example, the portal frame shown in Fig. 9.2. Recall from Chapter 6 that the frame actually has six degrees of freedom, when both axial

d1

d1 d2

2

3

39

29 d3

1

4

Fig. 9.2  Portal Frame with Inextensible Members (Three Degrees of Freedom)

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512  Chapter 9   Special Topics and Modeling Techniques

and bending deformations of members are taken into account in the analysis. However, if the members of the frame are assumed to be inextensible, then the number of degrees of freedom is reduced to only three. From the deformed shape of the arbitrarily loaded frame given in Fig. 9.2, we can see that fixed joints 1 and 4 can neither rotate nor translate, whereas joints 2 and 3 can rotate and translate in the horizontal direction, but not in the vertical direction because their vertical translations are prevented by the left and right columns, respectively, which are assumed to be inextensible. Furthermore, since the girder (i.e., the horizontal member) of the frame is assumed to be inextensible, the horizontal translations of joints 2 and 3 must be equal. Thus, the portal frame has three degrees of freedom, namely d1, d2, and d3, as shown in the figure. As another example, consider the two-story three-bay building frame shown in Fig. 9.3. The frame actually has 24 degrees of freedom when both axial and bending deformations are included in the analysis. However, if the members are assumed to be inextensible, then the number of degrees of freedom is reduced to 10, as shown in the figure. As this example indicates, the assumption of member inextensibility provides a means for a significant reduction in the number of degrees of freedom of large structures. Needless to say, this approximate approach is appropriate only for frames in which the member axial deformations are small enough to have a negligible effect on their response. As the axial deformations in the columns of tall building frames can have a significant effect on the structural response, the approximate method under consideration is usually not considered suitable for the analysis of such structures.

8

7

6

9

6

10

6

6 9

3

2

11

10

1

4

12

1

5 1

1 5

1

7

6

2

3

8

4

Fig. 9.3  Inextensible Building Frame (Ten Degrees of Freedom)

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Section 9.2   Approximate Matrix Analysis of ­Rectangular Building Frames  513

The overall procedure for the approximate analysis of rectangular plane frames remains the same as that for general plane frames, developed in Chapter 6—provided that the member stiffness relations are modified to exclude the axial effects. As the frame is composed of only horizontal and vertical members, each member now has four degrees of freedom in both the local and global coordinate systems. The local and global end forces and end displacements, for the girders (i.e., horizontal members) and the columns (i.e., vertical members), of the frame are given in Fig. 9.4. To simplify the analysis, the member local x axis is oriented positive to the right for girders (Fig. 9.4(a)) and positive upward for columns (Fig. 9.4(b)). With the axial effects neglected, the relationship between the member local end forces, Q, and end displacements, u, is expressed by the local stiffness matrix k and fixed-end force vector y Q4, u4 (F4, 4)

Y Q2, u2 (F2, 2) X

b

x

e

m

Q1, u1 (F1, 1)

Q3, u3 (F3, 3)

(a) Girder End Forces and End Displacements in Local and Global Coordinate Systems

x

x

Q3, u3 Q4, u4

F3, 3 F4, 4

e

m

e

m

Y

Y

b X

y

b Q1, u1

Q2, u2 (b) Column End Forces and End Displacements in the Local Coordinate System

X

F1, 1

y

F2, 2 (c) Column End Forces and End Displacements in the Global Coordinate System

Fig. 9.4

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514  Chapter 9   Special Topics and Modeling Techniques

Qf for beam members, derived in Chapter 5 (Eqs. (5.53) and (5.99)). Thus, Q 5 ku 1 Qf , with

3

12 6L 212 6L 6L 4L2 26L 2L2 EI k5 3 L 212 26L 12 26L 6L 2L2 26L 4L2

4

(9.7)



and Qf 5

3 4

FSb FMb FSe FMe

(9.8)

As for the member stiffness relations in the global coordinate system, for girders (Fig. 9.4(a)) no coordinate transformations are needed; that is, K (girder) 5 k and Ff (girder) 5 Qf . For columns, the transformation matrix, T (column), can be established via the following relationships between the local end forces Q and the global end forces F (see Figs. 9.4(b) and (c)): Q1 5 2F1 Q2 5 F2 Q3 5 2F3 Q4 5 F4 or

34 3 Q1 Q2 Q3 Q4

5

21 0 0 0

0 0 1 0 0 21 0 0

0 0 0 1

43 4 F1 F2 F3 F4

from which

T (column) 5

3

21 0 0 0

0 0 1 0 0 21 0 0

0 0 0 1

4

(9.9)



The expression of the global stiffness matrix for columns, K (column), can now be obtained by applying the relationship K 5 TT  kT, which yields

3

12 EI 26L K (column) 5 3 L 212 26L

26L 212 26L 4L2 6L 2L2 6L 12 6L 2 2L 6L 4L2

4



(9.10)

It is important to realize that the assumption of negligibly small axial deformations, as used herein, does not imply that the member axial forces are also negligibly small. As the axial forces do not appear in the member stiffness relations, the application of the matrix stiffness method yields only member end shears and end moments. Once the member end shears are known, the member axial forces can be evaluated by considering the equilibrium of the free bodies of the joints and members of the structure.

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Section 9.2   Approximate Matrix Analysis of ­Rectangular Building Frames  515



E xample 9.2 Determine the approximate joint displacements, member local end forces, and support reactions for the portal frame shown in Fig. 9.5(a), assuming the members to be inextensible.



S olution

Analytical Model: See Fig. 9.5(b). The frame has three degrees of freedom—the translation of the girder in the X direction, and the rotations of joints 2 and 3. The six restrained coordinates of the frame are identified by numbers 4 through 9 as usual, as shown in Fig. 9.5(b). Structure Stiffness Matrix and Fixed-Joint Force Vector: By applying Eq. (9.10) for members 1 and 3, and Eq. (9.7) for member 2, we obtain the following member global stiffness matrices (in units of kilonewtons and meters): Member 3 ¡ 7 9 1 3 Member 1 ¡ 4 6 1 2 K1 5 K3 5

3

720 23,600 2720 23,600

K2 5 k2 5 k1 5 k3 5

23,600 24,000 3,600 12,000

3

0 720 3,600 2720 3,600

2720 3,600 720 3,600

23,600 12,000 3,600 24,000

2 3,600 24,000 23,600 12,000

0 2720 23,600 720 23,600

4

4 6 1 2

7 9 1 3

3 3,600 12,000 23,600 24,000

4

0 2(1) 0 3

18 kN/m 40 kN

10 m

10 m E, I = constant E = 200 GPa I = 300 (106) mm4 (a) Frame

Fig. 9.5

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516  Chapter 9   Special Topics and Modeling Techniques Y

3

2 1

1 2

3

2

3

1

1

4

4

7 6

X

9

5

8 (b) Analytical Model

Fig. 9.5  (continued) From Fig. 9.5(b), we can see that for member 1, the structure coordinates in the directions of the member end shears and end moments are numbered 4, 6, 1, and 2. Thus, the code numbers for this member are 4, 6, 1, 2. Similarly, the code numbers for member 3 are 7, 9, 1, 3. Since the structure coordinates corresponding to the end shears of member 2 are not defined (because the corresponding joint displacements are 0), we use 0s for the corresponding member code numbers. Thus, the code numbers for member 1 are 0, 2, 0, 3. By using the foregoing member code numbers, the relevant elements of K1, K2, and K3 are stored in the 3 3 3 structure stiffness matrix S. Note that the elements of K2 that correspond to 0 code numbers are simply disregarded. The structure stiffness matrix thus obtained is 1    2    3 1,440 3,600 3,600 1  S 5 3,600 48,000 12,000 2 3,600 12,000 48,000 3

3

4

(2)

The fixed-end shears and moments due to the 18.333 kN/m uniformly distributed load applied to member 2 are calculated as wL 18(10) FSb 5 FSe 5 5 5 90 kN 2 2 wL2 18(10)2 FMb 5 2FMe 5 5 5 150 kN ? m 12 12 Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Section 9.2   Approximate Matrix Analysis of ­Rectangular Building Frames  517

107.143

72.857 18 kN/m 40 kN

3

2

35.01

35.01

2 4.99

35.01

185.71

14.286 72.857

107.143 107.143

72.857

4.99

35.01 14.286

185.81

1

3

164.33

64.248 4.99

35.01

72.857

107.143 (c) Member End Forces

Fig. 9.5  (continued) Using Eq. (9.8), we obtain

Ff 2 5 Qf 2 5

3 4 90 150 90 2150

0 2  0 3

(3)

Thus, the structure fixed-joint force vector Pf is given by Pf 5

3 4

0 1 150 2 2150 3

(4)

Joint Load Vector: P5

34

40 1 0 2 0 3

(5)

Joint Displacements: By substituting the numerical values of S (Eq. (2)), Pf (Eq. (4)), and P (Eq. (5)) into the structure stiffness relationship P 2 Pf 5 Sd, and solving the

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518  Chapter 9   Special Topics and Modeling Techniques resulting system of simultaneous equations, we obtain the following joint ­displacements.

3

4

0.03968 m 1 d 5 20.00655 rad 2 0.00179 rad 3

Ans

Member End Shears and End Moments: Member 1

v1 5

34 34 3 4 y1 y2 y3 y4

4 6 5 1 2

0 0 d1 d2

5

0 0 0.03968 20.00655

From Eq. (9.9):

3 4 3 4 0 0 1 0 0 21 0 0

21 0 T1 5 T3 5 0 0

0 0 0 1

(6)



0 0 20.03968 20.00655

u1 5 T1v1 5

By using k1 from Eq. (1) and Qf 1 5 0, we obtain

Q1 5 k1u1 5

3

4.99 kN 64.248 kN ? m 24.99 kN 214.286 kN ? m

4



Ans

Member 2

u2 5 v2 5

3 4 0 20.00655 0 0.00179

0 2 0 3

By using k2 from Eq. (1) and Qf 2 from Eq. (3), we calculate

Q2 5 k2u2 1 Qf2 5

3

72.857 kN 14.286 kN ? m 107.143 kN 2185.714 kN ? m

4



Ans

Member 3

v3 5

3 4 0 0 0.03968 0.00179

7 9 1 3

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Section 9.3   Condensation of Degrees of Freedom, and Substructuring  519

Using T3 from Eq. (6), we obtain

3 4

0 0 u3 5 T3v3 5 20.03968 0.00179

Using k3 from Eq. (1) and Qf 3 5 0, we calculate

Q3 5 k3u3 1 Qf3 5

3

4

35.01 kN 164.33 kN ? m  235.01 kN 185.81 kN ? m

Ans

The member end shears and end moments, as given by the foregoing local end force vectors Q1, Q2, and Q3, are depicted in Fig. 9.5(c). Member Axial Forces: With the member end shears now known, we can calculate the axial forces for the three members of the frame by applying the equations of equilibrium FX 5 0 and FY 5 0, to the free bodies of joints 2 and 3. The member axial forces thus obtained are shown in Fig. 9.5(c). Ans

o

o

F G

Support Reactions: By comparing Figs. 9.5(b) and (c), we realize that the forces at the lower ends of the columns of the frame represent its support reactions; that is,

R5

24.99 kN 72.857 kN 64.248 kN ? m 235.01 kN 107.143 kN 164.33 kN ? m

4 5 6  7 8 9

Ans

9.3 CONDENSATION OF DEGREES OF FREEDOM, AND SUBSTRUCTURING

A problem that can arise during computer analysis of large structures is that the computer may not have sufficient memory to store and process information about the entire structure. A commonly used approach to circumvent this problem is to condense (or reduce the number of) the structure’s stiffness equations that are to be solved simultaneously, by suppressing some of the degrees of freedom. This process is referred to as condensation (also called static condensation). For very large structures, it may become necessary to combine condensation with another process called substructuring, in which the structure is divided into parts called substructures, with the condensed stiffness relations for each substructure generated separately; these are then combined to obtain the stiffness relations for the entire structure. In this section, we consider the basic concepts of condensation of degrees of freedom, and analysis using ­substructures.

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520  Chapter 9   Special Topics and Modeling Techniques

Condensation The objective of condensation is to reduce the number of independent degrees of freedom of a structure (or substructure, or member). This is achieved by treating some of the degrees of freedom as dependent variables and expressing them in terms of the remaining independent degrees of freedom. The relationship between the dependent and independent degrees of freedom is then substituted into the original stiffness relations to obtain a condensed system of stiffness equations, which contains only the independent degrees of freedom as unknowns. From a theoretical viewpoint, the dependent degrees of freedom can be chosen arbitrarily. However, for computational purposes, it is usually convenient to select those degrees of freedom that are internal to the structure (or substructure, or member) as the dependent degrees of freedom. Hence, the dependent degrees of freedom are commonly referred to as the internal degrees of freedom; whereas, the independent degrees of freedom are called the external degrees of freedom. As discussed in the preceding chapters, the stiffness relations for a general framed structure can be expressed as (see, for example, Eq. (6.42)) P 5 Sd

(9.11)

P 5 P 2 Pf

(9.12)

with

When using the condensation process, it is usually convenient to assign numbers to the degrees of freedom so that the external and internal degrees of freedom are separated into two groups. The structure stiffness relations (Eq. (9.11)) can then be written in partitioned-matrix form:

3PP 4 5 3S E

SEE

I

IE

4 3 d 4

SEI SII

dE

(9.13)

I

in which the subscripts E and I refer to quantities related to the external and internal degrees of freedom, respectively. By multiplying the two partitioned matrices on the right side of Eq. (9.13), we obtain the two matrix equations, PE 5 SEEdE 1 SEIdI

(9.14)

PI 5 SIEdE 1 SIIdI

(9.15)

To express the internal degrees of freedom dI in terms of the external degrees of freedom dE, we solve Eq. (9.15) for dI, as

dI 5 S21 (PI 2 SIEdE)  II

(9.16)

Finally, by substituting Eq. (9.16) into Eq. (9.14), we obtain the condensed stiffness equations PE 2 SEI S21 PI 5 (SEE 2 SEI S21 SIE)dE II II

(9.17)

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Section 9.3   Condensation of Degrees of Freedom, and Substructuring  521

Note that the external degrees of freedom dE are the only unknowns in Eq. (9.17). Equation (9.17) can be rewritten in a compact form as P*E 5 S*EEdE

(9.18)



in which P*E 5 PE 2 SEI S21 PI II



(9.19)

S*EE 5 SEE 2 SEI S21 SIE II

(9.20)

and

As the foregoing equations indicate, the solution of the structure stiffness equations is carried out in two parts. In the first part, PE* and SE*E  are evaluated using Eqs. (9.19) and (9.20), respectively, and the external joint displacements dE are determined by solving Eq. (9.18). In the second part, the now-known dE is substituted into Eq. (9.16) to obtain the internal joint displacements dI. Once all the joint displacements have been evaluated, the member end displacements and end forces, and support reactions, can be calculated using the procedures described in the previous chapters. It should be realized that analysis involving condensation generally requires more computational effort than the standard formulation in which all of the structure’s stiffness equations are solved simultaneously. However, condensation provides a useful means of analyzing large structures whose full stiffness matrices and load vectors exceed the available computer memory. This is because, when employing condensation, only parts of S and P need to be assembled and processed in the computer memory at a given time. The basic concept of condensation is illustrated by the following relatively simple example.



E xample 9.3 Analyze the plane frame shown in Fig. 9.6(a) using condensation, by treating the rotation of the free joint as the internal degree of freedom.



S olution

This frame was analyzed in Example 6.6 using the standard formulation. The analytical model of the structure is given in Fig. 9.6(b). Condensed Structure Stiffness Matrix: The full (3 3 3) stiffness matrix, S, for the frame, as determined in Example 6.6, is given by (in units of kilonewtons and meters): 1 299,470 S 5 90,225 3,077

3

2 90,225 182,910 2,762

3 3,077 2,762 32,584

4

1  2 3

(1)

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522  Chapter 9   Special Topics and Modeling Techniques

125 kN • m

3m

16.667 kN/m

300 kN

3m

3m

6m E, A, I = constant E = 200 GPa A = 7,600 mm2 I = 129(10)6 mm4 (a) Frame 2

Y

9

3 1 2

7

3

2

8 1

1 4

X 6 5 (b) Analytical Model

Fig. 9.6 in which S is partitioned to separate the external degrees of freedom, 1 and 2, from the internal degree of freedom, 3. From Eq. (1), we obtain 1 2 299,470 90,225 1  SE E 5 90,225 182,910 2 1 2   SI E 5 [3,077 2,762] 3

3

4

(2) (3)

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Section 9.3   Condensation of Degrees of Freedom, and Substructuring  523

3 3,077 1 

32,7624 2

SE I 5

(4)

3  SI I 5 [32,584] 3

(5)

with the inverse of SII given by S21 5 II

1 332,584 4

(6)

By substituting Eqs. (2), (3), (4), and (6) into Eq. (9.20), we obtain the condensed structure stiffness matrix: S*EE 5 SEE 2 SEI S21 S 5 I I IE

3299,179 89,964

89,964 182,676

4 kN/m

(7)

Condensed Joint Load Vector: Recall from Example 6.6 that P 5 P 2 Pf 5

3 4

0 1 2200 2 262.5 3

(8)

from which PE 5

322000 4 12

(9)

and (10)

PI 5 [262.5] 3 

Substitution of Eqs. (4), (6), (9), and (10) into Eq. (9.19) yields the following condensed joint load vector. P*E 5 PE 2 SE I S21 PI 5 II

5.902 32194.702 4 kN

(11)

Joint Displacements: By substituting Eqs. (7) and (11) into the condensed structure * stiffness relationship, PE* 5 SEE dE (Eq. (9.18)), and solving the resulting 2 3 2 system of simultaneous equations, we obtain the external joint displacements (corresponding to degrees of freedom 1 and 2), as dE 5

320.00126 4 12 m 0.000399

(12)

The internal joint displacement (i.e., the rotation corresponding to degree of freedom 3), can now be determined by applying Eq. (9.16). Thus, dI 5 S21 (PI 2 SIE dE) 5 [20.00185]3 rad II

(13)

By combining Eqs. (12) and (13), we obtain the full joint displacement vector,

3 4

d d5 E 5 dI

3

4

0.000399 m 1 20.00126 m 2 20.00185 rad 3

Ans

Note that the foregoing joint displacements are identical to those determined in ­Example 6.6 by solving the structure’s three stiffness equations simultaneously.

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524  Chapter 9   Special Topics and Modeling Techniques Member End Displacements and End Forces: See Example 6.6. It is important to realize that, in this example, the submatrices of S and P were obtained from the corresponding full matrices, for convenience only. In actual computer analysis, to save memory space, the individual parts of S and P are assembled directly from the corresponding member matrices as they are needed in the analysis.

In the foregoing paragraphs, we have discussed the application of condensation to reduce the number of independent degrees of freedom of an entire structure. The condensation process is also frequently used to establish the stiffness relationships for substructures, which are defined as groups of members with known stiffness relations. In this case, condensation is used to eliminate the degrees of freedom of those joints that are internal to the substructure, thereby producing a condensed system of stiffness relations expressed solely in terms of the degrees of freedom of those (external) joints through which the substructure is connected to the rest of the structure and/or supports. The procedure for condensing the internal degrees of freedom of a substructure is analogous to that just discussed for the case of a whole structure. The stiffness relations involving both the internal and external degrees of freedom of a substructure can be symbolically expressed as (9.21)

F 5 K v 1 Ff

in which F and v represent, respectively, the joint forces and displacements for the substructure; K denotes the substructure stiffness matrix; and Ff represents the fixed-joint forces for the substructure. The matrix K and the vector Ff can be assembled from the member stiffness matrices and fixed-end force vectors in the usual way. To apply condensation, we rewrite Eq. (9.21) in partitioned-matrix form as

3 F453 K

v F 1 4 3 v 4 3F 4 K

FE

KEE

KEI

E

fE

I

IE

II

I

fI

(9.22)

The multiplication of the two partitioned matrices on the right-hand side of Eq. (9.22) yields the matrix equations FE 5 KE EvE 1 KEIvI 1 Ff E

(9.23)

FI 5 KI EvE 1 KII vI 1 Ff I

(9.24)

Solving Eq. (9.24) for vI , we obtain vI 5 K21 (FI 2 Ff I 2 KI E vE ) II

(9.25)

and, substituting Eq. (9.25) into Eq. (9.23), we determine the condensed stiffness relations for the substructure to be FE 5 K*EE vE 1 F*f E

(9.26)

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Section 9.3   Condensation of Degrees of Freedom, and Substructuring  525

in which K*EE 5 KEE 2 KEI K21 KIE II

(9.27)



and Ff*E 5 Ff E 1 KEI K21 (FI 2 Ff I) II



(9.28)



E xample 9.4 Determine the stiffness matrix and the fixed-joint force vector for the substructure of a beam shown in Fig. 9.7(a), in terms of its external degrees of freedom only. The substructure is composed of two members connected together by a hinged joint, as shown in the figure.



S olution

Analytical Model: The analytical model of the substructure is depicted in Fig. 9.7(b). For member 1, MT 5 2, because the end of this member is hinged; MT 5 1 for member 2, which is hinged at its beginning. Joint 3 is modeled as a hinged joint with its rotation restrained by an imaginary clamp. Thus, the substructure has a total of five degrees of freedom, of which four are external (identified by numbers 1 through 4) and one is internal (identified by number 5). Substructure Stiffness Matrix: We will first assemble the full (5 3 5) stiffness matrix K from the member stiffness matrices k, and then apply Eq. (9.27) to determine the * condensed stiffness matrix KEE. W w

Hinge L1

L2 E, I = constant (a) Substructure

Y 5 3

1 3

1 2

1

2 6

4 X

2

(b) Analytical Model

Fig. 9.7

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526  Chapter 9   Special Topics and Modeling Techniques

¯= K

3

¯ EE K

¯ EI K

¯ IE K

¯ II K

3

1

2

3

4

1 L 31

1 L 21

0

0

2

1 L 31

1

1 L 21

1 L1

0

0

2

1 L 21

2

0

0

1 L 32

1 L 22

2

1 L 32

3

0

0

2

1 L 22

1 L2

1 L 22

4

1 L 21

2

1 L 32

1 L 22

1 1 + 3 L 31 L2

5

= 3E I

2

1 L 31

2

2

5

¯ Ff =

3 3 ¯ FfE ¯ FfI

=

w 8

3 3 5L 1 L 21 5L 2 2L 22

3L 1 + 3L 2

1 2 3 4 5

(d) Full (Uncondensed) Fixed-Joint Force Vector For Substructure

(c) Full (Uncondensed) Stiffness Matrix for Substructure

Fig. 9.7  (continued) Member 1 (MT 5 2)  Using Eq. (7.18), we obtain

k1 5

3E I L31

1 1 L1 21 0

3

2 L1 L21 2L1 0

5 21 2L1 1 0

6 0 0 0 0

4

1 2 5 6

Member 2 (MT 5 1)  Application of Eq. (7.15) yields 5 1 3E I k2 5 3 0 L2 21 L2

6 0 0 0 0

3

3 21 0 1 2L2

4 L2 0 2L2 L22

4

5 6 3 4

Using the code numbers of the members, we store the pertinent elements of k1 and k2 in the full 5 3 5 stiffness matrix K of the substructure, as shown in Fig. 9.7(c). Substituting into Eq. (9.27) the appropriate submatrices of K from Fig. 9.7(c) and K21 5 II

L31L32

3 3EI (L 1 L )4  3 1

(1)

3 2

we obtain the condensed stiffness matrix for the substructure:



3

1 L 3E I 1 K*EE 5 KEE 2 KE I K21 KI E 5 3 II L1 1 L32 21 L2

L1 L21 2L1 L1L2

21 2L1 1 2L2

L2 L1L2 2L2 L22

4



(2) Ans

Substructure Fixed-Joint Force Vector: Member 1 (MT 5 2)  Using Eq. (7.19), we obtain

Qf 1 5

wL1 8

34 5 L1 3 0

1 2 5 6

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Section 9.3   Condensation of Degrees of Freedom, and Substructuring  527

Member 2 (MT 5 1)  Using Eq. (7.16), we write

Qf 2 5

wL2 8

3 4 3 0 5 2L2

5 6 3 4

The relevant elements of Qf 1 and Qf 2 are stored in the full 5 3 1 fixed-joint force vector Ff of the substructure, as shown in Fig. 9.7(d). A comparison of Figs. 9.7(a) and (b) indicates that F5 5 2W; that is, (3)

FI 5 [F5] 5 [2W ] 

Finally, the application of Eq. (9.28) yields the following condensed fixed-joint force vector for the substructure. F*f E 5 Ff E 1 KE I K21 (FI 2 Ff I ) II

3

5L41 1 8L1L32 1 3L42 L51 1 4L21L32 1 3L1L42 w 5   8(L31 1 L32) 3L41 1 8L31L2 1 5L42 2(3L41L2 1 4L31L22 1 L52)

4

3 4

2L32 2L1L32 W 2 3    L1 1 L32 2L31 L31L2

(4) Ans

Analysis Using Substructures The procedure for the analysis of (large) structures, divided into substructures, is essentially the same as the standard stiffness method developed in previous chapters. However, each substructure is treated as an ordinary member of the structure, and the degrees of freedom of only those joints through which the substructures are connected to each other and/or to supports are considered to be the structure’s degrees of freedom d. The structure’s stiffness matrix S and fixed-joint force vector Pf , are assembled, respectively, from the substructure stiffness matrices K*EE and fixed-joint force vectors F*f E, which are expressed in terms of the external coordinates of the substructures only. The structure stiffness equations, P 2 Pf 5 Sd, thus obtained, can then be solved for the joint displacements d. Consider, for example, the nine-story plane frame shown in Fig. 9.8(a). The frame actually has 20 joints and 54 degrees of freedom; that is, if we were to analyze the frame using the standard stiffness method for plane frames as developed in Chapter 6, we would have to assemble and solve 54 structure stiffness equations simultaneously. Now, suppose that we wish to analyze the frame by dividing it into three substructures, each consisting of three stories of the frame, as depicted in Fig. 9.8(b). As this figure indicates, for analysis purposes, the frame is now modeled as having only six joints, at which the three substructures are connected to each other and to external supports. Thus, the analytical model of the frame has 12 degrees of freedom and six restrained coordinates. To develop the stiffness matrix S and the fixed-joint force vector Pf for the frame, we first determine the substructure stiffness matrices K*EE and ­fixed-joint

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528  Chapter 9   Special Topics and Modeling Techniques Y

2

Substructure

1

Substructure

2

Substructure

3

5 6 4

1 1

3

2

8

11 12 10

7 3

9

4

16

13 (a) Nine-Story Plane Frame

Fig. 9.8

5

15

X

6 18

14 17 (b) Analytical Model of Frame Divided into Three Substructures

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Section 9.3   Condensation of Degrees of Freedom, and Substructuring  529 2

5 6

3 1

1

3

6

2

5 Substructure 1

4

4

7

12

9

10

11

8 Substructures 2 and 3

(c) External Degrees of Freedom of Substructures

Fig. 9.8  (continued)

force vectors F*f E, in terms of the external degrees of freedom of the substructures, using condensation as described earlier in this section. As shown in Fig. 9.8(c), substructure 1 has six external degrees of freedom; whereas, substructures 2 and 3 each have 12 external degrees of freedom. The pertinent elements of K*EE matrices and F*f E vectors are then stored in S and Pf , respectively, using the substructure code numbers in the usual manner. By comparing Figs. 9.8(b) and (c), we can see that the code numbers for substructure 1 are 1, 2, 3, 4, 5, 6; whereas, the code numbers for substructure 2 are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. Similarly, for substructure 3, the code numbers are 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18. Once the structure stiffness matrix S (12 3 12) and the fixed-joint force vector Pf (12 3 1) have been assembled, the structure stiffness equations, P 2 Pf 5 Sd, are solved to calculate the joint displacement vector d. With d known, the external joint displacements, vE , for each substructure are obtained from d using the substructure’s code numbers, and then the substructure’s internal joint displacements, vI , are calculated using Eq. (9.25). After the joint displacement vector v of a substructure has been determined, the end displacements and forces for its individual members, and support reactions, can be evaluated using the standard procedure described in previous chapters. The basic concept of analysis using substructures is illustrated by the following relatively simple example.

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530  Chapter 9   Special Topics and Modeling Techniques



E xample 9.5 Analyze the two-span continuous beam shown in Fig. 9.9(a), treating each span as a ­substructure.



S olution

Analytical Model: The structure is modeled as being composed of two substructures and three joints, as shown in Fig. 9.9(b). It has one degree of freedom and five restrained coordinates. As each substructure consists of two beam members connected together by a hinged joint, we will use the expressions of stiffnesses and fixed-joint forces for such substructures, derived in Example 9.4, in the present example. Structure Stiffness Matrix, S: Substituting E 5 70(106) kN/m2, I 5 200(10−6) m4, and L1 5 L2 5 5 m into Eq. (2) of Example 9.4, we obtain the following condensed stiffness matrix for the two substructures. 50 kN 18 kN/m

Hinge 5m

Hinge

5m

5m

5m

EI = constant E = 70 GPa I = 200(106 ) mm4 (a) Beam Y

Substructure

1

Substructure

2

6

1

X 2

1

3

2

3

4 (b) Analytical Model of Beam Divided into Two Substructures

5

Y

5 1

3

2

e

m

b 1

6

4 X

2

(c) Analytical Model of a Substructure Composed of Two Beam Members

Fig. 9.9

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Section 9.3   Condensation of Degrees of Freedom, and Substructuring  531

1

75

2

15

15

18 kN/m

1

2

175

35

Substructure

18 kN/m

35

80

10

1

275

10

Substructure

100 2

(d) Member End Forces 50 kN 18 kN/m 1

1

75

2

15

175 35

2

2

275

3

80

100

(e) Substructure Forces 50 kN 18 kN/m

2 1

75

3

15

275

100

115 (f) Support Reactions

Fig. 9.9  (continued) Substructure 2  ¡ 4

1

5

6

Substructure 1  ¡ 2

3

4

1

840 4,200 2840 4,200

2168 2840 168 2840

840 4,200 2840 4,200

K*EE1 5 K*EE2 5

3

168 840 2168 840

4

2 3 4 1

4 1 5 6

By comparing the numbers of the external degrees of freedom of a substructure (Fig. 9.9(c)) to those of the structure degrees of freedom (Fig. 9.9(b)), we obtain code ­numbers 2, 3, 4, 1 for substructure 1, and 4, 1, 5, 6 for substructure 2. By adding the pertinent elements of K*EE1 and K*EE2, we determine the structure stiffness matrix S to be 1 S 5 f8,400] 1 kN # m/rad Structure Fixed-Joint Force Vector Pf : Substructure 1  By substituting w 5 0, W 5 50 kN, and L1 5 L2 5 5 m into Eq. (4) of Example 9.4, we obtain the following condensed fixed-joint force vector for substructure 1. F*f E1 5

3 4 25 125 25 2125

2 3 4 1

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532  Chapter 9   Special Topics and Modeling Techniques Substructure 2  By substituting w 5 18 kN/m, W 5 0, and L1 5 L2 5 5 m into Eq. (4) of Example 9.4, we obtain

3 4 90 225 90 2225

Ff*E2 5

4 1 5 6

Thus, the fixed-joint force vector for the whole structure is given by Pf 5 [100] 1 kN ? m Joint Displacements: By substituting P 5 0 and the numerical values of S and Pf into the structure stiffness relation, P 2 Pf 5 Sd, we write [2100] 5 [8,400] [d1] from which d 5 [d1] 5 [20.011905] rad Substructure Joint Displacements, and Member End Displacements and End Forces: Substructure 1  The substructure’s external joint displacements vE can be obtained by simply comparing the substructure’s external degree of freedom numbers with its code numbers, as follows.

vE1 5

34 34 3 y1 y2 y3 y4

2 3 5 4 1

0 0 0 d1

5

4

0 0 0 20.011905

The substructure’s internal joint displacements can now be calculated, using the relationship (Eq. (9.25)) vI 5 K21 (FI 2 Ff I 2 KI E vE). Substitution of the numerical II values of E, I, L1, L2, and W into Eqs. (1) and (3) of Example 9.4 yields 5 [0.0014881] K21 I I1 and FI1 5 [250] Similarly, by substituting the appropriate numerical values into the expressions of KI E and Ff I given in Figs. 9.7(c) and (d), respectively, of Example 9.4, we obtain and

KI E1 5 [2336

21,680

2336

1,680 ]

Ff I1 5 0 By substituting the numerical values of the foregoing submatrices and subvectors into Eq. (9.25), we determine the internal joint displacements for substructure 1 to be vI1 5 K21 (FI1 2 Ff I1 2 KI E1vE1 ) 5 [20.044643] I I1 Thus, the complete joint displacement vector for substructure 1 is

v1 5

3v 4 5 vE1 I1

3

4

0 0 0 20.011905 rad 20.044643 m

1 2 3 4 5

Ans

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Section 9.3   Condensation of Degrees of Freedom, and Substructuring  533

With the displacements of all the joints of substructure 1 now known, we can determine the end displacements u, and end forces Q, for its two members (Fig. 9.9(c)) in the usual manner. Member 1 (MT 5 2)  From Fig. 9.9(c), we can see that the code numbers for member 1 are 1, 2, 5, 6. Thus,

u1 5

34 34 3 u1 u2 u3 u4

1 2 5 5 6

0 0 y5 0

5

4

0 0 20.044643 0

Substituting the numerical values of E and I and L 5 5 m into Eq. (7.18), we obtain the member stiffness matrix,

k1 5

3

336 1,680 2336 0

1,680 8,400 21,680 0

4

0 0 0 0

2336 21,680 336 0

Substitution of k1 and Qf 1 5 0 into the member stiffness relationship, Q 5 ku 1 Qf , yields the following end forces for member 1 of substructure 1.

3

4

15 kN 75 kN # m Q1 5 k1u1 1 Qf1 5 215 kN 0



Ans

Member 2 (MT 5 1)

u2 5

3

4

20.044643 0 0 20.011905

5 6 3 4

Applying Eq. (7.15),

  k2 5

3

336 0 2336 1,680

0 0 0 0

2336 0 336 21,680

1,680 0 21,680 8,400

4



Qf 2 5 0  Q2 5 k2u2 1 Qf 2 5

3

235 kN 0 35 kN 2175 kN # m

4



Ans

Substructure 2

vE2 5

3

4

0 20.011905 0 0

4 1 5 6

From Fig. 9.7(d) of Example 9.4, we obtain Ff I2 5 [67.5]

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534  Chapter 9   Special Topics and Modeling Techniques The submatrices K21 and KI E remain the same as for substructure 1, and FI2 5 0. II Thus, the application of Eq. (9.25) yields vI2 5 [20.13021] and, therefore,

v2 5

3v 4 5 vE2 I2

4

3

0 20.011905 rad 0 0 20.13021 m

1 2 3 4 5

Ans

Member 1 (MT 5 2)

u1 5

3

4

0 20.011905 20.13021 0

1 2 5 6

The k matrix for member 1 of substructure 2 is the same as that for the corresponding member of substructure 1. Using Eq. (7.19), we calculate

Qf1 5

3 4 56.25 56.25 33.75 0

Thus,

Q1 5 k1u1 1 Qf 1 5

3

4

80 kN 175 kN # m  10 kN 0

Ans

Member 2 (MT 5 1)

3 4

20.13021 5 0 6 u2 5 0 3 0 4 The k matrix for this member is the same as that for member 2 of substructure 1. Applying Eq. (7.16),

Qf 2 5

3 4 33.75 0 56.25 256.25

Thus,

Q2 5 k2u2 1 Qf 2 5

3

4

210 kN 0 100 kN 2275 kN # m



Ans

The end forces for the individual members of the structure are shown in Fig. 9.9(d).

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Section 9.4   Inclined Roller Supports  535

Support Reactions: The reaction vector R can be assembled either directly from the member end force vectors Q, or from the external joint force vectors, FE , of the ­substructures. To use the latter option, we first apply Eq. (9.26) to calculate FE. Thus, by substituting the previously calculated numerical values of K*EE , F*f E , and vE into Eq. (9.26), we obtain

FE1 5 K*EE1vE1 1 Ff*E1 5

15 75 35 2175

3 4

2 3 4 1

FE2 5 K*EE2vE2 1 F*f E2 5

3 4

4 1 5 6

and 80 175 100 2275

The foregoing substructure forces are depicted in Fig. 9.9(e). Finally, we calculate the support reaction vector R by storing the pertinent elements of FE1 and FE2 in their proper positions in R, using the substructure code numbers. This yields

R5

3 4 15 kN 75 kN? m 115 kN 100 kN 2275 kN # m

2 3 4 5 6

Ans

The support reactions are shown in Fig. 9.9(f).

9.4 

INCLINED ROLLER SUPPORTS The structures that we have considered thus far in this text have been supported such that the joint displacements prevented by the supports are in the directions of the global coordinate axes oriented in the horizontal and vertical directions. Because an inclined roller support prevents translation of a joint in an inclined direction (normal to the incline), while permitting translation in the perpendicular direction, it exerts a reaction force on the joint in that inclined, nonglobal, direction. Thus, the effect of an inclined roller support cannot be included in analysis by simply eliminating one of the structure’s degrees of freedom; that is, by treating one of the structure’s coordinates, which are defined in the directions of the global coordinate axes, as a restrained coordinate. An obvious approach to alleviate this problem would be to orient the global coordinate system so that its axes are parallel and perpendicular to the inclined plane upon which the roller moves. However, this approach generally proves to be quite cumbersome, as it requires that the joint coordinates and loads, which are usually specified in the horizontal and vertical directions, be calculated with respect to the inclined global coordinate system. Furthermore, the foregoing approach cannot be used if the structure is supported by two

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536  Chapter 9   Special Topics and Modeling Techniques

or more rollers inclined in different (i.e., neither parallel nor perpendicular) directions. A theoretically exact solution of the problem of inclined rollers usually involves first defining the reaction force and the support displacements with reference to a local joint coordinate system, with axes parallel and perpendicular to the incline; and then introducing these restraint conditions in the structure’s global stiffness relations via a special transformation matrix [26]. While this approach is exact in the sense that it yields exactly 0 displacement of the support joint perpendicular to the incline, it is generally not considered to be the most convenient because its computer implementation requires a significant amount of programming effort. Perhaps the most convenient and commonly used technique for modeling an inclined roller support is to replace it with an imaginary axial force member with very large axial stiffness, and oriented in the direction perpendicular to the incline, as shown in Figs. 9.10 and 9.11 (on the next page). As depicted there, one end of the imaginary member is connected to the original support joint by a hinged connection, while the other end is attached to an imaginary hinged support, to ensure that only axial force (i.e., no bending moment) develops in the member when the structure is loaded. In order for the imaginary member to accurately represent the effect of the roller support, its axial stiffness must be made sufficiently large so that its axial deformation is negligibly small. This is usually achieved by specifying a very large value for the cross-sectional area

θ (a) Plane Truss with Inclined Roller Support

θ Imaginary member with large cross-sectional area

(b) Analytical Model

Fig. 9.10

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Section 9.4   Inclined Roller Supports  537

θ

(a) Plane Frame with Inclined Roller Support

θ Imaginary member, hinged at both ends (MT = 3), and with large cross-sectional area

(b) Analytical Model

Fig. 9.11

of the imaginary member in the analysis, while keeping its length of the same order of magnitude as the other (real) structural members, to ensure that the imaginary member undergoes only small rotations. Provided that the foregoing conditions are satisfied, the axial force in the imaginary member represents the reaction of the actual inclined roller support. The main advantage of modeling inclined roller supports with imaginary members is that computer programs for standard supports, such as those developed in previous chapters, can be used, without any modifications, to analyze structures supported on inclined rollers. When analyzing trusses, ordinary truss members with large cross-sectional areas can be used to model inclined roller supports (Fig. 9.10). In the case of frames, however, the members used to model inclined rollers, in addition to having large cross-sectional areas, must be of type 3 (MT 5 3); that is, they must be hinged at both ends, as shown in Fig. 9.11. As noted before, the cross-sectional area of the imaginary member, used to model the inclined roller support, should be sufficiently large so that the member’s axial deformations are negligibly small. However, using an extremely large value for the cross-sectional area of the imaginary member can cause some off-diagonal elements of the structure stiffness matrix to become so large, as compared to the other elements, that they introduce numerical errors, or cause numerical instability, during the solution of the structure’s stiffness equations.

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538  Chapter 9   Special Topics and Modeling Techniques

9.5 

OFFSET CONNECTIONS In formulating the stiffness method of analysis, we have ignored the size of joints or connections, assuming them to be of infinitesimal size. While this assumption proves to be adequate for most framed structures, the dimensions of moment-resisting connections in some structures may be large enough, relative to member lengths, that ignoring their effect in the analysis can lead to erroneous results. In this section, we discuss procedures for including the effect of finite sizes of connections or joints in the analysis. Consider an arbitrary girder of a typical plane building frame, as shown in Fig. 9.12(a) on the next page. The girder is connected at its ends, to columns and adjacent girders, by means of rigid or moment-resisting connections. As indicated in the figure, the dimensions of connections usually (but not always) equal the cross-sectional depths of the connected members. If the connection dimensions are small, as compared to the member lengths, then their effect is ignored in the analysis. In such a case, it would be assumed for analysis purposes that the girder under consideration extends in length from one column centerline to the next, and is connected at its ends to other members through rigid connections of infinitesimal size, as depicted in Fig. 9.12(b). However, if the connection dimensions are not small, then their effect must be considered in the analysis. As shown in Fig. 9.12(c), rigid connections of finite size can be conveniently modeled by using rigid offsets, with each offset being a rigid body of length equal to the distance between the center of the connection and its edge which is adjacent to the member under consideration. Thus, from Fig. 9.12(c), we can see that the girder under consideration has offset connections of lengths db and de at its left and right ends, respectively. Two approaches are commonly used to include the effect of offset connections in analysis. In the first approach, each offset is treated as a small member with very large stiffness. For example, in [13] it is suggested that the crosssectional properties of an offset member be chosen so that its stiffness is 1,000 times that of the connected member. The main advantage of this approach is that computer programs, such as those developed in previous chapters, can be used without any modification. The disadvantage of this approach is that each offset increases, by one, the number of members and joints to be analyzed. For example, the girder of Fig. 9.12(c) would have to be divided into three members of lengths db, L, and de, in order to include the effect of offset connections at its two ends in the analysis. An alternate approach that can be used to handle the effect of offset connections involves modifying the member stiffness relationships to include the effect of offsets at member ends. The main advantage of this approach is that a natural member (e.g., a girder or a column), together with its end offsets, can be treated as a single member for the purpose of analysis. For example, the whole girder of Fig. 9.12(c), including its end offsets, would be treated as a single member when using this approach. However, the disadvantage of this approach is that it requires rewriting of some parts of the computer programs developed in previous chapters.

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Section 9.5   Offset Connections  539 Column Centerlines

Rigid Connections

db

L

de

(a) Girder

L + db + d e

(b) Analytical Model Neglecting Connection Sizes

Rigid Offsets

db

L

de

(c) Analytical Model Considering Connection Sizes

Fig. 9.12

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540  Chapter 9   Special Topics and Modeling Techniques

In the following, we modify the stiffness relations for the members of plane frames to include the effect of rigid end offsets. Similar procedures can be employed to derive modified stiffness relations for the members of other types of framed structures. Consider an arbitrary member of length L of a plane frame, and let Q and u denote the local end forces and end displacements, respectively, at the exterior ends of its offsets, as shown in Fig. 9.13(a). Our objective is to express Q in terms of u and any external loading applied to the member between its actual ends b and e. Recall from Chapter 6 that the relationship between the end forces Q and the end displacement u, which are defined at the ends b and e of the member, is of the form Q 5 ku 1 Qf , with k and Qf given by Eqs. (6.6) and (6.15), respectively. To express Q in terms of Q, we consider the equilibrium of the rigid bodies of the two offsets. This yields (see Fig. 9.13(b)) Q1 5 Q1 Q4 5 Q4

Q2 5 Q2 Q5 5 Q5

Q3 5 db Q2 1 Q3 Q6 5 2de Q5 1 Q6

which can be written in matrix form as Q 5 TQ

(9.29)

y

Rigid offset

Rigid offset ¯ Q6, ¯u6

¯ Q1, ¯u1

m

b

¯ Q3, ¯u3

¯ Q4, ¯u4

e

¯ Q2, ¯u2

x

¯ Q5, ¯u5 L

db

de

E, I, A = constant

(a) Plane Frame Member with Rigid End Offsets Q5

Q2 Q6

Q6

¯ Q1

Q1 Q3

¯ Q3 ¯ Q2

b

Q3

m

Q2

Q4

¯ Q6

¯ Q5

Q5

db

¯ Q4

e

de (b) Member Forces

Fig. 9.13

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with

T5

F

1 0 0 0 0 0

0 1 db 0 0 0

0 0 1 0 0 0

0 0 0 0 0 0 1 0 0 1 0 2de

G

Section 9.5   Offset Connections  541

0 0 0 0 0 1

(9.30)

in which T can be considered to be a transformation matrix which translates the member’s end forces from its actual ends b and e, to the exterior ends of its rigid offsets. From geometrical considerations, it can be shown that the relationship between the end displacements u and u can be written as u 5 TT u

(9.31)

By substituting Eq. (9.31) into Eq. (6.4), and substituting the resulting expression into Eq. (9.29), we obtain the desired stiffness relationship: Q 5 k u 1 Qf

(9.32)

k 5 TkTT

(9.33)

Qf 5 TQf

(9.34)

with

in which k and Qf represent the modified member stiffness matrix and fi ­ xed-end force vector, respectively, in the local coordinate system. Note that k and Qf include the effect of rigid offsets at the ends of the member. The explicit forms of k and Qf  , respectively, can be obtained by substituting Eqs. (6.6) and (9.30) into Eq. (9.33), and Eqs. (6.15) and (9.30) into Eq. (9.34). These are given in Eqs. (9.35) and (9.36).

EI k5 3 L



3

AL2 I 0 0 AL2 2 I 0 0

0

0

12 (6L 1 12db )

(6L 1 12db ) (4L2 1 12Ldb 1 12db2 )

0

0

212 (6L 1 12de)

(26L 2 12db) (2L2 1 6Ldb 1 6Lde 1 12dbde)

AL2 I 0 0 AL2 I 0 0

2

0

0

212 (26L 2 12db)

(6L 1 12de) (2L2 1 6Ldb 1 6Lde 1 12dbde)

0

0

12 (26L 2 12de)

(26L 2 12de) (4L2 1 12Lde 1 12de2 )

4

(9.35)

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F G

542  Chapter 9   Special Topics and Modeling Techniques

Qf 5

FAb FSb dbFSb 1 FMb FAe FSe 2deFSe 1 FMe

(9.36)

The procedure for analysis essentially remains the same as developed previously, except that the modified expressions for the stiffness matrices k (Eq. (9.35)) and fixed-end force vectors Qf (Eq. (9.36)) are used (instead of k and Qf , respectively) for members with offset connections.

9.6 

SEMIRIGID CONNECTIONS While rigid and hinged types of connections, as considered thus far in this text, are the most commonly used in structural designs, a third type of connection, termed the semirigid connection, is also recognized by some design codes, and can be used for designing such structures as structural steel building frames. Recall that the rotation of a rigidly connected member end equals the rotation of the adjacent joint, whereas the rotation of a hinged end of a member must be such that the moment at the hinged end is 0. A connection is considered to be semirigid if its rotational restraint is less than that of a perfectly rigid connection, but more than that of a frictionless hinged connection. In other words, the moment transmitted by a semirigid connection is greater than 0, but less than that transmitted by a rigid connection. For the purpose of analysis, a semirigid connection can be conveniently modeled by a rotational (torsional) spring with stiffness equal to that of the actual connection. In this section, we derive the stiffness relations for members of beams with semirigid connections at their ends. Such relationships for other types of framed structures can be determined by using a similar procedure. Consider an arbitrary member of a beam, as shown in Fig. 9.14(a) on the next page. The member is connected to the joints adjacent to its ends b and e, by means of rotational springs of infinitesimal size representing the semirigid connections of stiffnesses kb and ke, respectively. As shown in this figure, Q and u represent the local end forces and end displacements, respectively, at the exterior ends of the rotational springs. Our objective is to express Q in terms of u and any external loading applied to the member. We begin by writing, in explicit form, the previously derived relationship Q 5 ku 1 Qf , between the end forces Q and the end displacements u, which are defined at the actual ends b and e of the member. By using the expressions for k and Qf from Eqs. (5.53) and (5.99), respectively, we write

34 3

43 4 3 4

u1 12 6L 212 6L Q1 FSb 2 2 u2 Q2 EI 6L 4L 26L 2L FMb 5 3 u3 1 FSe Q3 L 212 26L 12 26L Q4 u4 FMe 6L 2L2 26L 4L2

(9.37)

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Section 9.6   Semirigid Connections  543 y

Rotational spring of stiffness kb

Rotational spring of stiffness ke ¯ Q4, ¯u4 x

¯ Q2, ¯u2

m

b

e ¯ Q3, ¯u3

¯ Q1, ¯u1 L E, I = constant

Infinitesimal

Infinitesimal

(a) Beam Member with Semirigid Connections Q1

Q3

¯ Q2

Q2 ¯ Q1

m

b

Q2

¯ Q4

Q4

Q4

kb

e

ke ¯ Q3

Q3

Q1 (b) Member Forces y

¯u4

Displaced position u4

e9

u2 b9

u3

¯u3

¯u 2 u1

¯u1 Initial position kb

ke x b

m

e

L Infinitesimal (c) Member Displacements

Fig. 9.14

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544  Chapter 9   Special Topics and Modeling Techniques

Figure 9.14(b) shows the forces Q and Q acting at the exterior and interior ends, respectively, of the member’s rotational springs. As the lengths of these springs are infinitesimal, equilibrium equations for the free bodies of the springs yield Q 5 Q

(9.38)

The displacements u and u are depicted in Fig. 9.14(c) using an exaggerated scale. Because of the infinitesimal size of the springs, the translations of the spring ends are equal; that is, u1 5 u1

(9.39a)

u3 5 u3

(9.39b)

The relationship between the rotations (u2 and u2) of the two ends of the spring, at member end b, can be established by applying the spring stiffness relation: Q2 5 kb (u2 2 u2) from which u2 5 u2 2

Q2 kb

(9.39c)



Similarly, by using the stiffness relation for the spring attached to member end e, we obtain u4 5 u4 2

Q4 ke

(9.39d)



To obtain the desired relationship between Q and u, we now substitute Eqs. (9.38) and (9.39) into Eq. (9.37) to obtain the following equations. Q1 5  Q2 5 Q3 5

Q2 Q4 EI 12u1 1 6L u2 2 2 12u3 1 6L u4 2 3 L kb ke

3

1

2

1

24 1 FS

Q2 Q4 EI 6Lu1 1 4L2 u2 2 2 6Lu3 1 2L2 u4 2 3 L kb ke

3

1

2

1

3

1

2

1



(9.40a)

24 1 FM

Q2 Q4 EI 212u1 2 6L u2 2 1 12u3 2 6L u4 2 3 L kb ke

b

(9.40b)

24 1 FS

e

(9.40c)

Q4 5

b

Q2 Q4 EI 6Lu1 1 2L2 u2 2 2 6Lu3 1 4L2 u4 2 3 L kb ke

3

1

2

1

24 1 FM e

(9.40d)

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Section 9.6   Semirigid Connections  545

Next, we solve Eqs. (9.40b) and (9.40d) simultaneously, to express Q2 and Q4 in terms of u1 through u4. This yields Q2 5

Q4 5

EIrb

[6L(2 2 re)u1 1 4L2(3 2 2re)u2 2 6L (2 2 re)u3 1 2L2reu4] L3 R  rb 1 [(4 2 3re)FMb 2 2(1 2 re)FMe] (9.41a) R EIre

[6L(2 2 rb)u1 1 2L2rbu2 2 6L(2 2 rb)u3 1 4L2(3 2 2rb)u4] L3 R re 1 [(4 2 3rb)FMe 2 2(1 2 rb)FMb] (9.41b) R

in which rb and re denote the dimensionless rigidity parameters defined as ri 5

ki L E I 1 ki L

i 5 b, e

(9.42)



and R 5 12 2 8rb 2 8re 1 5rbre



(9.43)

Finally, by substituting Eqs. (9.41) into Eqs. (9.40a) and (9.40c), we determine expressions for Q1 and Q3 in terms of u1 through u4. Thus, Q1 5



EI [12(rb 1 re 2 rbre)u1 1 6Lrb(2 2 re)u2 L3 R 212(rb 1 re 2 rbre)u3 1 6Lre(2 2 rb)u4] 6 1FSb 2 [(1 2 rb)(2 2 re)FMb 1 (1 2 re)(2 2 rb)FMe] LR

(9.44a)

EI [212(rb 1 re 2 rbre)u1 2 6Lrb(2 2 re)u2 L3 R 112(rb 1 re 2 rbre)u3 2 6Lre(2 2 rb)u4] 6 1FSe 1 [(1 2 rb)(2 2 re)FMb 1 (1 2 re)(2 2 rb)FMe] LR (9.44b) Q3 5

Equations (9.41) and (9.44), which represent the modified stiffness relations for beam members with semirigid connections at both ends, can be expressed in matrix form: Q 5 ku 1 Qf



(9.45)

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546  Chapter 9   Special Topics and Modeling Techniques

with k5

EI L3 R

3

12(rb 1 re 2 rbre) 6Lrb(2 2 re) 212(rb 1 re 2 rbre) 6Lre(2 2 rb)

6Lrb(2 2 re) 4L2rb(3 2 2re) 26Lrb(2 2 re) 2L2rbre)

212(rb 1 re 2 rbre) 26Lrb(2 2 re) 12(rb 1 re 2 rbre) 26Lre(2 2 rb)

4

6Lre(2 2 rb) 2L2rbre 26Lre(2 2 rb) 4L2re(3 2 2rb)

 (9.46) and

3

Qf 5

4

6 [(1 2 rb)(2 2 re)FMb 1 (1 2 re)(2 2 rb) FMe] LR rb [(4 2 3re)FMb 2 2(1 2 re)FMe] R 6 FSe 1 [(1 2 rb)(2 2 re)FMb 1 (1 2 re)(2 2 rb)FMe] LR re [22(1 2 rb)FMb 1 (4 2 3rb)FMe] R FSb 2

(9.47)

The k matrix in Eq. (9.46) and the Qf vector in Eq. (9.47) represent the modified stiffness matrix and fixed-end force vector, respectively, for the members of beams with semirigid connections. It should be noted that these expressions for k and Qf are valid for the values of the spring stiffness ki (i 5 b or e) ranging from 0, which represents a hinged connection, to infinity, which represents a rigid connection. From Eq. (9.42), we can see that as ki varies from 0 to infinity, the value of the corresponding rigidity parameter ri varies from 0 to 1. Thus, ri 5 0 represents a frictionless hinged connection, whereas ri 5 1 represents a perfectly rigid connection. The reader is encouraged to verify that when both rb and re are set equal to 1, then k (Eq. (9.46)) and Qf (Eq. (9.47)) reduce the k (Eq. (5.53)) and Qf (Eq. (5.99)) for a beam member rigidly connected at both ends. Similarly, the expressions of k and Qf , derived in Chapter 7 for beam members with three combinations of rigid and hinged connections (i.e., MT 5 1, 2, and 3), can be obtained from Eqs. (9.46) and (9.47), respectively, by setting rb and re to 0 or 1, as appropriate. The procedure for analysis of beams with rigid and hinged connections, developed previously, can be applied to beams with semirigid connections— provided that the modified member stiffness matrix k (Eq. (9.46)) and ­fixed-end force vector Qf (Eq. (9.47)) are used in the analysis.

9.7 

SHEAR DEFORMATIONS The stiffness relations that have been developed thus far for beams, grids, and frames do not include the effect of shear deformations of members. Such ­structures are generally composed of members with relatively small depth-to-length ratios, so that their shear deformations are usually negligibly small as compared to the bending deformations. However, in the case of

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Section 9.7   Shear Deformations  547

beams, grids and rigid frames consisting of members with depth-to-length ratios larger than 0.1 and/or built-up (fabricated) members, the magnitudes of shear deformations can be considerable; therefore, the effect of shear deformations should be included in the analyses of such structures. In this section, we consider a procedure for including the effect of shear deformations in the member stiffness relations, and present the modified stiffness matrix for the members of beams. This matrix, which contains the effects of both the shear and bending deformations, can be easily extended to obtain the corresponding modified member stiffness matrices for grids, and plane and space frames. The relationship between the shearing strain at a cross-section of a beam member and the slope of the elastic curve due to shear can be obtained by considering the shear deformation of a differential element of length dx of the member, as shown in Fig. 9.15. From this figure, we can see that g52

duyS dx

(9.48)



in which g denotes the shear strain, and uyS represents the deflection, due to shear, of the member’s centroidal axis in the y direction. The negative sign in Eq. (9.48) indicates that the positive shear force S causes deflection in the negative y direction, as shown in the figure. Substitutions of Hooke’s law for shear, g 5 t/G, and the stress-force relation, t 5 fsS/A, into Eq. (9.48), yield the following expression for the slope of the elastic curve due to shear. d uyS dx

fS

1GA2 S

52

(9.49)



in which fS represents the shape factor for shear. The dimensionless shape factor fS depends on the shape of the member cross-section, and takes into account

S

S

y

γ

2du¯ yS

x

dx

Fig. 9.15

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548  Chapter 9   Special Topics and Modeling Techniques

the nonuniform distribution of shear stress on the member cross-section. The values of fS for some common cross-sectional shapes are as follows. fS 5 1.2 for rectangular cross-sections fS 5 10/9 for circular cross-sections fS 5 1 for wide-flange beams bent about the major axis, provided that the area of the web is used for A in Eq. (9.49) Integration of Eq. (9.49) yields the expression for deflection due to shear; the total deflection (or slope) of the member due to the combined effect of shear and bending can be determined via superposition of the deflections (or slopes) caused by shear and by bending. As discussed in Chapter 5, the equations for the slope and deflection, due to bending, can be obtained by integrating the moment–curvature relationship: d 2 uy B dx

2

5

M EI

(9.50)

in which uyB represents the deflection of the member due to bending. The expressions for the elements of the modified stiffness matrix k for a beam member, due to the combined effect of the bending and shear deformations, can be derived using the direct integration approach. To obtain the expressions for the stiffness coefficients ki 1 (i 5 1 through 4) in the first column of k, we subject a prismatic beam member of length L to a unit value of the end displacement u1 at end b, as shown in Fig. 9.16. Note that all other member end displacements are 0, and the member is in equilibrium under the action of two end moments k21 and k41, and two end shears k11 and k31. From the figure, y

b9 k41

u1 = 1 k21

x e

b

k11

k 31 x L EI = constant u2 = u3 = u4 = 0

Fig. 9.16

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Section 9.7   Shear Deformations  549

we can see that the shear and bending moment at a distance x from end b of the member are: S 5 k11

(9.51)

M 5 2k21 1 k11x

(9.52)

By substituting Eq. (9.51) into Eq. (9.49), and integrating the resulting equation, we obtain the equation for deflection, due to shear, as fS

1GA2 k x 1 C

uy S 5 2

11

(9.53)

1

in which C1 denotes a constant of integration. By substituting Eq. (9.52) into Eq. (9.50), and integrating the resulting equation twice, we obtain the equations for the slope and deflection of the member, due to bending: duy B

2 1 x x 5 2k 1k 21Cx1C EI 1 2 6

dx

5

1

1 x2 2k21x 1 k11 1 C2 EI 2 2

uy B

(9.54)

3

21

11

2

(9.55)

3

As the shear deformation does not cause any rotation of the member crosssection (see Fig. 9.15), the rotation of the cross-section, u, results entirely from bending deformation, and is given by (see Eq. (9.54)) u5

duy B dx

5

1

2

1 x2 2k21x 1 k11 1 C2 EI 2

(9.56)

By combining Eqs. (9.53) and (9.55), we obtain the equation for the total deflection, uy , due to the combined effect of the shear and bending deformations: fS

1GA2 k x 1 EI 12k 1

2

x2 x3 1 k11 1 C2x 1 C4 2 6 (9.57) in which the constant C4 5 C1 1 C3. The four unknowns in Eqs. (9.56) and (9.57)—that is, two constants C2 and C4 and two stiffness coefficients k11 and k21—can now be evaluated by applying the following four boundary conditions: uy 5 uy S 1 uy B 5 2

11

21

at end b,  x 5 0  u 5 0 x 5 0  uy 5 1 at end e,  x 5 L  u 5 0 x 5 L  uy 5 0 By applying these boundary conditions, we obtain C2 5 0, C4 5 1, and k11 5

12EI 1 3 L 1 1 bS

1

2

(9.58)

k21 5

6EI 1 2 L 1 1 bS

(9.59)

1

2

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550  Chapter 9   Special Topics and Modeling Techniques

with bS 5

12E I fS

(9.60)



GAL2

The dimensionless parameter bS is called the shear deformation constant. The two remaining stiffness coefficients, k31 and k41, can now be determined by applying the equations of equilibrium to the free body of the member (Fig. 9.16). Thus, k31 5 2 k41 5

1

2

12EI 1 L3 1 1 bS

1

(9.61)

2

6EI 1 L2 1 1 bS

(9.62)

The expressions for elements in the remaining three columns of the k matrix can be derived in a similar manner, and the complete modified stiffness matrix for rigidly-connected members of beams, thus obtained, is

3

12 EI 6L k5 3 L (1 1 bS) 212 6L

6L 212 L (4 1 bS) 26L 26L 12 2 L (2 2 bS) 26L 2

4

6L L (2 2 bS) 26L L2(4 1 bS) 2

(9.63)

From Eq. (9.63), we can see that when the shear deformation constant bS is set equal to 0, then k of Eq. (9.63) is reduced to that of Eq. (5.53). It should be realized that the expressions for fixed-end forces due to member loads, given inside the front cover, do not include the effects of shear deformations. If modified fixed-end force expressions including shear deformations are desired, they can be derived using the procedure described in this section.



E xample 9.6 Derive the expressions for the deflection and slope at the free end of the rectangular

­cantilever steel beam shown in Fig. 9.l 7(a), including the effect of shear deformations.



S olution

Analytical Model: See Fig. 9.17(b). The analytical model consists of one member with two degrees of freedom and two restrained coordinates. Structure Stiffness M ­ atrix S: The general form of the member stiffness matrix k is given in Eq. (9.63). By using the member’s code numbers 3, 4, 1, 2 from Fig. 9.17(b), we store the pertinent elements of k into the structure stiffness matrix S to obtain: EI S5 3 L (1 1 bs)

3

1 12 26L

2 26L 1 L2(4 1 bs) 2

4

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Section 9.7   Shear Deformations  551

Joint Load Vector P: P5

32P0 4 12

Joint Displacement Vector d: By solving the structure equilibrium equations Sd 5 P for d, we obtain

34

d PL2 d5 1 52 d2 2EI

3

1

bs 2L 11 3 4 1

24

Deflection and Slope: Thus, the deflection D and slope u of the free end of the cantilever beam are:

1

2

bs PL3 11 T 3EI 4

(1)

u 5 d2 5

PL2 2EI

(2)

[

D 5 d1 5



As expected, Eq. (2) shows that the shear deformations do not have any effect on the slope of the beam. However, from Eq. (1) we can see that the shear deformations cause an increase in the beam’s deflection by the magnitude DS 5

1 2

PL3 bS  3EI 4

(3)

The ratio of the magnitudes of the beam’s deflections due to shear (Ds) and bending [DM 5 PL3/(3EI)] can be expressed as (see Eq. (9.60) for bS) DS DM

5

bS 4

5

3EIfs GAL2



(4)

Equation (4) can now be specialized for the given rectangular cantilever beam composed of structural steel by substituting E 5 200 GPa, G 5 77.2 GPa, fS 5 1.2, A 5 bd, and I 5 bd3/12, to obtain DS DM

5 0.78

1L2  d

2

(5)

which shows that the ratio of shear to bending deflections increases quadratically with increasing depth to length ratio (Fig. 9.l7(c)). Numerical values of shear deflection, as a percentage of the corresponding bending deflection, for some d/L ratios are listed in Table 9.1. It can be seen that for d/L # 0.1, the magnitude of shear deflection is negligibly small (less than 1%) as compared to the corresponding bending deflection. For d/L . 0.1, however, the shear to bending deflections ratio increases rapidly reaching about 3% at d/L 5 0.2, indicating that for deep rectangular beams, the effects of shear deformations may need to be included in the analysis.

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552  Chapter 9   Special Topics and Modeling Techniques

P

b d

L EI = constant E = 200 GPa, G = 77.2 GPa, fS = 1.2 (a) Rectangular Cantilever Beam Y 1 1 4

1

2

X

(b) Analytical Model

3

Ratio of shear to bending deflections (%)

2

3.0 2.5 2.0 1.5 1.0 0.5 0.0 0.00

0.05

0.10 0.15 Depth-to-length ratio (d/L)

0.20

(c) Variation of Shear Deflection with Depth-to-Length Ratio

Fig. 9.17

Table 9.1 Depth-to-length ratio

0.01 0.05 0.1 0.15 0.2

Shear deflection as percentage of bending deflection

0.0078 0.195 0.78 1.76 3.12

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Section 9.8   Nonprismatic Members  553

9.8 

NONPRISMATIC MEMBERS Thus far in this text, we have considered the analysis of structures composed of prismatic members. A member is considered to be prismatic if its axial and flexural rigidities (EA and EI), or its cross-sectional properties, are constant along its length. In some structures, for aesthetic reasons and/or to save material, it may become necessary to design members with variable cross sections. In this section, we consider the analysis of structures composed of such nonprismatic members. Perhaps the simplest (albeit approximate) way to handle a nonprismatic natural member, such as a girder or a column, is to subdivide it into a sufficient number of segments, and model each segment by a prismatic member (or element) with cross-sectional properties equal to the average of the cross-­sectional properties at the two ends of the segment (Fig. 9.18). The main advantage of this approach is that computer programs such as those developed in previous chapters can be used without any modifications. The main disadvantage of this approach is that the accuracy of the analytical results depends on the number of prismatic members (or elements) used to model each nonprismatic member, and an inordinate number of prismatic members may be required to achieve an acceptable level of accuracy. An alternate approach that can be used to handle nonprismatic members involves formulation of the nonprismatic member’s stiffness relations while taking into account the exact variation of the member’s cross-sectional properties. The main advantage of this exact approach is that a natural nonprismatic member (e.g., a girder or a column) can be treated as a single member for the purpose of analysis. However, as will become apparent later in this section, the exact expressions for the stiffness coefficients for nonprismatic members can be quite complicated [43]. In the following, we illustrate the exact approach via derivation of the local stiffness matrix k for a tapered plane truss member [26]. Consider a tapered member of a plane truss, as shown in Fig. 9.19(a). The cross-sectional area of the member varies linearly along its length in accordance with the relationship

1

Ax 5 Ab 1 2

rA x L

2

(9.64)

(a) Nonprismatic Member

(b) Analytical Model of Nonprismatic Member

Fig. 9.18 Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

554  Chapter 9   Special Topics and Modeling Techniques

in which Ab and Ax denote, respectively, the member’s cross-sectional areas at its end b, and at a distance x from end b; and rA represents the area ratio given by rA 5

Ab 2 Ae Ab

(9.65)



with Ae denoting the member’s cross-sectional area at end e, as shown in the figure. To derive the first column of the tapered member’s local stiffness matrix k, we subject the member to a unit end displacement u1 5 1 (with u2 5 u3 5 u4 5 0), as shown in Fig. 9.19(b). The expressions for the member axial forces required to cause this unit axial deformation can be determined by integrating y

Ae e

b

x

Ab

x L E = constant

(a) Tapered Plane Truss Member y

u1 = 1

b

e

b9

k31

k11

k21

x

k41 x L (b) u1 = 1, u2 = u3 = u4 = 0

Fig. 9.19

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Section 9.8   Nonprismatic Members  555

the differential equation for member axial deformation that, for members with variable cross-sections, can be written as (see Eq. (6.7), Section 6.2) dux dx

5

Qa EAx

(9.66)



From Fig. 9.19(b), we can see that the axial force acting on the member crosssection at a distance x from its end b is (9.67)

Qa 5 2k11

in which the negative sign indicates that k11 causes compression at the member cross-section. Substituting Eqs. (9.64) and (9.67) into Eq. (9.66), and integrating the resulting equation, we obtain ux 5

k11L EAbrA

1

ln 1 2

rA x L

2 1 C

(9.68)

in which C is a constant of integration. The two unknowns, C and k11, in Eq. (9.68) can be evaluated by applying the boundary conditions: at end b,  x 5 0

ux 5 1

at end e,  x 5 L

ux 5 0

Application of the foregoing boundary conditions yields C 5 1, and k11 5 2

EAbrA L ln (1 2 rA)

(9.69)



The three remaining stiffness coefficients can now be determined by applying the equations of equilibrium to the free body of the member (Fig. 9.19(b)). Thus, k31 5

EAbrA L ln (1 2 rA)

(9.70)

, k21 5 k41 5 0

The expressions for elements in the third column of the tapered ­ ember’s local stiffness matrix k can be derived in a similar manner; and, m as discussed in Section 3.3, all elements of the second and fourth columns of k are 0. The complete local stiffness matrix k for a tapered plane truss member, thus obtained, is

3

21 0 k5 L ln (1 2 rA) 1 0 EAbrA

0 1 0 0 0 21 0 0

4

0 0 0 0

(9.71)

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556  Chapter 9   Special Topics and Modeling Techniques

E xample 9.7 Using the direct integration approach, derive the expressions for the slope and deflec-

tion at the free end of the tapered cantilever beam shown in Fig. 9.20(a). The beam has a rectangular cross-section of constant width b, but its depth varies linearly from h1 at the fixed end to h2 at the free end.



S olution

The depth and moment of inertia of the beam at a distance x(0 # x # L) from its free end can be expressed as

1

hx 5 h1 1 2

1

Ix 5 I1 1 2

r hx L

r hx L

2

2

3

in which rh represents the depth ratio given by rh 5

h1 2 h2 h1

(1)



and I1 5 bh13/12 5 beam’s moment of inertia at its fixed end. y

A

P

h1 2

b

h2

h1 2

hx 2 hx 2

x Section A-A Beam cross-section

A x L E = constant

(a) Tapered Cantilever Beam

y P

PL

x

x P (b)

Fig. 9.20

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Section 9.8   Nonprismatic Members  557

The equations for the slope and deflection can be derived by integrating the differential equation for bending of beams with variable cross-sections, which can be written as (see Eq. (5.5), Section 5.2) d 2uy d x2

5

M  EIx

(2)

From Fig. 9.20(b), we can see that the bending moment at the beam section at a distance x from its fixed end is (3)

M 5 2P(L 2 x)

in which the negative sign indicates that the bending moment is negative in accordance with the beam sign convention (Fig. 5.4). Substituting Eq. (3) into Eq. (2) and integrating, we obtain the equation for slope as u52

PL3 L 1 rhL 2 2rhx 1 C 1 2EI1 r2h(L 2 rhx)2

3

4

(4)

Integrating once more, we obtain the equation for deflection as

31

PL3 2EI1r3h

uy 5

1 2 rh r hx

12

2

1

1 2 ln 1 2

r hx L

24 1 C x 1 C  1

(5)

2

L The constants of integration, C1 and C2, are evaluated by applying the boundary conditions that at x 5 0, u 5 0 and uy 5 0 Thus, C1 5

PL2 1 1 rh 2EI1 r2h

1

2

PL (1 2 rh) 3

C2 5 2

2EI1r3h

By substituting these expressions for C1 and C2 into Eqs. (4) and (5) we determine the equations for slope and deflection of the beam as u5

Px 2EI1

31

24

x 2 2L 1 rhx 12

r hx L

1

2EI1r3h 1 2

r hx L

2

3L

(6)



2rhx

PL3

uy 5

2

2

r2hx2 L

2

1

(1 1 rh) 1 2 1 2

r hx

r hx

ln 1 2 24 L2 1 L

(7)

[

Finally, the expressions for slope and deflection at the free end of the tapered beam are obtained by setting x 5 L in Eqs. (6) and (7), respectively. Thus, Slope (1

):

uL 5 2

PL2h1 PL2 52  2EI1(1 2 rh) 2EI1h2

Ans

Deflection (1 c): uyL 5 2

PL3 [22rh 1 r2h(1 1 rh) 2 2(1 2 rh) ln (1 2 rh)] 2EI1r3h(1 2 rh)

Ans

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558  Chapter 9   Special Topics and Modeling Techniques

E xample 9.8 Using a structural analysis computer program, determine the slope and deflection at



the free end of the tapered cantilever beam shown in Fig. 9.21(a). The beam is of rectangular cross-section of width 150 mm, and its depth varies linearly from 400 mm at the fixed end to 100 mm at the free end, as shown in the figure. For analysis, divide the nonprismatic beam into smaller segments, and model each segment by a prismatic member (element) with a constant moment of inertia based on the average depth of the segment. Analyze several models of the beam with increasing number of members (elements) until the values of the desired displacements converge. Compare these numerical results with the exact analytical solutions for the tapered beam obtained from the expressions derived in Example 9.7.



S olution

Seven analytical models of the beam consisting of 1, 2, 3, 4, 6, 9, and 12 segments were analyzed using the computer program provided with this book. In these models, each tapered segment was approximated by a member of constant depth equal to the average depth of the segment. Figure 9.21(b) shows such a three-member model of the beam.

100 kN

400 mm

150 mm

100 mm

h (varies)

Beam cross-section

6m E = 70 GPa

(a) Tapered Cantilever Beam

Y

0.35 m

0.25 m

0.15 m

2m I = 535.9375 (10−6) m4

2m I = 195.3125 (10−6) m4

2m I = 42.1875 (10−6) m4

X

(b) Three-Member Analytical Model

Fig. 9.21

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Section 9.8   Nonprismatic Members  559

Slope (radians)

0.145 0.140 Numerical results 0.135 0.130 0.12857

Exact solution

0.125 1

2

3

4

5 6 7 8 9 10 11 Number of members (c) Variation of Slope with Number of Members

12

Deflection (mm)

550 500 450 400

Numerical results

350 324.61 300

Exact solution 1

2

3

4

5 6 7 8 9 10 11 Number of members (d) Variation of Deflection with Number of Members

12

Fig. 9.21  (continued)

Numerical values of displacements at the free end of the beam obtained by using these analytical models are listed in Table 9.2, and plotted versus the number of members in Figs. 9.21(c) and (d). The exact analytical solutions can be evaluated by substituting P 5 100 kN, L 5 6 m, E 5 70(106) kN/m2, h1 5 0.4 m, h2 5 0.1 m, I1 5 0.15(0.4)3/12 5 0.0008 m4, and rh 5 (0.4 2 0.1)/0.4 5 0.75 into the expressions for slope and deflection at the cantilever’s free end derived in Example 9.7. Thus, Slope: PL2h1 2EI1h2

é

uL 5 2

5 20.12857 rad 5 0.12857 rad 

Ans

Deflection: uyL 5 2

PL3 [22rh 1 r2h(1 1 rh) 2 2(1 2 rh) ln (1 2 rh)] 2EI1r3h(1 2 rh)

5 20.32461 m 5 324.61 mm T

 Ans

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560  Chapter 9   Special Topics and Modeling Techniques

Table 9.2 Number of members in analytical model

Slope Deflection (radians)

Error (%)

 1  2  3   4   6   9 12

0.13166 0.14090 0.13827 0.13560 0.13246 0.13051 0.12973

2.4034 9.5901 7.5445 5.4678 3.0256 1.5089 0.90223

Exact solutions

0.12857

(mm)

526.63 401.66 361.89 345.91 334.01 328.73 326.96

Error (%)

62.235 23.736 11.485 6.5617 2.8958 1.2692 0.72395

324.61

The exact values of slope and deflection are also given in Table 9.2, along with the percentage errors of the numerical results with respect to the exact solutions. We can see from this table and Figs. 9.21(c) and (d) that as the number of members in the computer model is increased, the numerical results tend to converge toward the exact solutions. Ans

9.9 SOLUTION OF LARGE SYSTEMS OF STIFFNESS EQUATIONS

In the computer programs for matrix stiffness analysis developed in previous chapters, we have stored the entire structure stiffness matrix S in computer memory, and have used Gauss–Jordan elimination to solve the structure stiffness equations, Sd 5 P. While this approach provides a clear insight into the basic concept of the solution process and is easy to program, it is not efficient in the sense that it does not take advantage of the symmetry and other special features of the stiffness matrix S. In the case of large structures, a significant portion of the total memory and execution time required for analysis may be devoted to the storage and solution, respectively, of the structure stiffness equations. Accordingly, considerable research effort has been directed toward developing techniques and algorithms for efficiently generating, storing, and solving stiffness equations that arise in the analysis of large structures [2, 14, 26]. In this section, we discuss a commonly used procedure that takes advantage of the special features of the structure stiffness matrix to efficiently store and solve structural stiffness equations.

Half-Bandwidth of Structure Stiffness Matrices The stiffness matrices S of large structures, in addition to being symmetric, are usually sparse, in the sense that they contain many 0 elements. Consider, for

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Section 9.9   Solution of Large Systems of Stiffness Equations  561

example, the analytical model of the six-degree-of-freedom continuous beam shown in Fig. 9.22(a) on the next page. The stiffness matrix S for this structure is also shown in the figure, in which all the nonzero elements are marked by 3s, and all the 0 elements are left blank. From this figure, we can see that, out of a total of 36 elements of S, 20 elements are 0s. Furthermore, this figure indicates that all the nonzero elements of S are located within a band centered on the main diagonal. Such a matrix, whose elements are all 0s, with the exception of those located within a band centered on the main diagonal, is referred to as a banded matrix. In general, a structure stiffness matrix is considered to be banded if Sij 5 0  if  Z i 2 j Z . NHB

(9.72)

where NHB is called the half-bandwidth of S, which is defined as the number of elements in each row (or column) of the matrix, that are located within the band to the right of (or below) the diagonal element. Thus, the half-bandwidth of the stiffness matrix of the continuous-beam analytical model of Fig. 9.22(a) is 1 (i.e., NHB 5 1), as shown in the figure. Although the total number of nonzero elements of a structure stiffness matrix remains the same, their locations depend on the order in which the structure’s joints are numbered. Thus, the half-bandwidth of a structure stiffness matrix can be altered by renumbering the structure’s joints. For example, if the numbers of two inner joints of the continuous beam of Fig. 9.22(a) are interchanged, the half-bandwidth of its stiffness matrix is increased to 2 (i.e., NHB 5 2), as shown in Fig. 9.22(b). Note that the band of this S matrix contains both zero and nonzero elements. Since the structure stiffness matrices are assembled by storing the pertinent elements of the member stiffness matrices in their proper positions using member code numbers, the half-bandwidth of a structure stiffness matrix equals the maximum of the differences between the largest and smallest degree-of-­ freedom code numbers for the individual members of the structure; that is, NHB 5 maxhMCLi 2 MCSij i 5 1, Á , NM



(9.73)

in which MCLi and MCSi denote, respectively, the largest and the smallest code numbers for member i, which correspond to the degrees of freedom (not the restrained coordinates) of the structure. Considering again the analytical model of the continuous beam of Fig.  9.22(a), we can see that the code numbers for member 1 are 7, 8, 9, 1; of these, the first three numbers correspond to the restrained coordinates, and the fourth number represents a degree of freedom. Thus, the difference ­between the largest and smallest degree-of-freedom numbers for this member is MCL1 2 MCS1 5 1 2 1 5 0. Similarly, we can see from the figure that for members 2 through 6, this difference is 1, and for member 7, it is 0. Thus, the half-­bandwidth for the S matrix equals one. Note that when the numbers of two inner joints of the beam are interchanged as shown in Fig. 9.22(b), the difference in degree of freedom code numbers for members 3 and 5 increases by one, and as a result, the half-bandwidth of the S matrix widens by one element.

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14

7

15

8

16 7

6

5

4

3

1

2

13

6

6 6

3

3

3

3

3

3

4

3

12

5

5

6

3

5 5

3 3

3

3 3

3 3

3 2 1

NHB = 1

10

S=

3

3

11

4

NHB = 1

(a)

4 4 3 3

9

2

2 2

Fig. 9.22

8

1

7

1

1

562 

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14

7

15

8

16 7

6 3 3

5 3

3

3

4

3

1

3

3 3

3

3 3

3 3 3

3 2 1

NHB = 2

12

4

(b)

3

4

3

11

5

5

6

2

13

6

6 6 5 4 3 5

10

S=

3

4 3

Fig. 9.22  (continued)

8

1

7

1

1

2

9

2

2

  563

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564  Chapter 9   Special Topics and Modeling Techniques

An important property of banded structure stiffness matrices is that the 0 elements outside the band remain 0 during the solution of the structure ­stiffness equations (Sd 5 P); therefore, they need not be stored in computer memory for analysis. Furthermore, since the structure stiffness matrices are symmetric, only the diagonal elements, and the elements in the half band above (or below) the diagonal, need to be stored. As the stiffness matrices of large structures usually contain relatively few nonzero elements, significant savings in computer memory storage and execution time can be achieved, in the analysis of such structures, by numbering the joints to minimize the half-bandwidth of the stiffness matrix, and by storing and processing only the elements on the main diagonal, and within a half-bandwidth, of the stiffness matrix. As discussed previously, the minimum possible half-bandwidth of a stiffness matrix can be obtained by numbering the joints of the structure in such an order that the largest difference between the joint numbers at the ends of any single member is as small as possible. For the configurations of the framed structures commonly encountered in practice, a relatively small (if not minimal) half-bandwidth of the stiffness matrix can usually be achieved by numbering joints consecutively across the dimension of the structure that has the least number of joints, as shown in Figs. 9.23 and 9.24 on pages 565 and 566, respectively. The elements on the diagonal and in the upper half-bandwidth of S can be stored compactly in computer memory in a rectangular array Sˆ of order NDOF 3 (NHB 1 1), as illustrated in Fig. 9.25 on page 567 for a structure with NDOF 5 9 and NHB 5 3. As depicted in this figure, the elements in each row of S on the diagonal and in the half-bandwidth are stored in the same order in the corresponding row of the compact matrix Sˆ . The location of an element Sij of S in the compact matrix Sˆ is given by the relationship ⁄

Sij 5 Si,(11j2i) 

for i 5 1, 2, Á , NDOF; j 5 i, i 1 1, Á , NHB 1 i # NDOF

(9.74)

Solution of Banded Structure Stiffness Equations Using UT DU Decomposition Although Gauss-Jordan elimination, as discussed in Section 2.4, can be modified to take advantage of the symmetry and bandedness of structure stiffness equations, for the analysis of large structures, another type of elimination method, a decomposition method, is usually preferred. This is because, in decomposition methods, the solution is carried out in two distinct parts; namely, decomposition and substitution, with the decomposition part involving only the structure stiffness matrix S, but not the load vector P. Thus, the results of the time-consuming decomposition part can be stored for future use in case the structure needs to be reanalyzed for different loading conditions. In the following, we present a decomposition method, commonly used for solving large systems of structure stiffness equations, called the UT DU decomposition method. The method is first formulated for the fully populated symmetric

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Section 9.9   Solution of Large Systems of Stiffness Equations  565 5

9

3

8

5

17

13 12

7

21 16

9

20

11

4

1

12

1

2

4

2 3

6

6 7

8

10 11

10

14 15

18 19

NHB = 7 3

3 3

3

3

3 3

3

3

3

3

3 3

3

3

3

3

3

3

3

3

3

3

3 3

3

3

3

3 3

3

3

3 3

3

3

3

3

3 3

3

3

3 3

3 3

3

3

3 3

3

3

3

3 3

3 3

3 3

3

3

3

3

3

3

3

3

3

3 3 3

3

3

3

3

S=

3

3

Symmetric

3

3

3

3

3

3

3 3

3

3

3

3

3 3

3

3 3 3

3

3

3

3

3 3

3

3

3

3 3

3

3

3

3

3

3

3

3

3

3

3

3 3

Fig. 9.23  Joint Numbering for Minimum Half-Bandwidth (NDOF 5 21, NHB 5 7)

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20

23 24

19

9

21

22

10

14

17 18

16

13 7

15

8

8

11 12

7

5

9

10

6

2

5 6

1 3

3

4

1

2

4

NHB = 8 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

S=

Symmetric

3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

3 3 3 3 3 3 3 3 3 3 3

3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

566 

Fig. 9.24  Joint Numbering for Minimum Half-Bandwidth (NDOF 5 24, NHB 5 8)

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NHB = 3

S11

S12

S13

S14

S22

S23

S24

S25

S33

S34

S35

S36

S44

S45

S46

S47

S55

S56

S57

S58

S66

S67

S68

S69

S77

S78

S79

S88

S89

S=

Symmetric

NDOF = 9

S99 (a) Full Structure Stiffness Matrix S for NDOF = 9 and NHB = 3

NHB + 1 = 4

Sˆ =

S11

S12

S13

S14

S22

S23

S24

S25

S33

S34

S35

S36

S44

S45

S46

S47

S55

S56

S57

S58

S66

S67

S68

S69

S77

S78

S79

S88

S89

NDOF = 9

S99 (b) Compact Structure Stiffness Matrix Sˆ for NDOF = 9 and NHB = 3

Fig. 9.25

567

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568  Chapter 9   Special Topics and Modeling Techniques

­stiffness matrix S, and is then modified so that it can be used with the banded and compact forms of the structure stiffness matrix. As stated previously, in the UT  DU decomposition method, the solution of the structure stiffness equations, Sd 5 P, is carried out in two parts: decomposition and substitution. In decomposition, the structure stiffness matrix S is decomposed (or factored) into the matrix triple product, S 5 UT  DU

(9.75)

in which U is a unit upper triangular matrix (i.e., an upper triangular matrix with diagonal elements equal to unity); and D is a diagonal matrix. For a g­ eneral n-degree-of-freedom system, Eq. (9.75) can be written in expanded form as

3 3

S11

S12 S22

(symmetric)

S13 S23 S33

Á Á Á Á

S1n S2n S3n Á Snn

1 U12 U13 Á

0 1 U23 Á

0 0 1 Á

U1n

U2n

U3n

Á Á Á Á Á

0 0 0 Á 1

43

4

5

D11 0 0 Á

0 D22 0 Á

0 0 D33 Á

0

0

0

Á Á Á Á Á

0 0 0 Á Dnn

43

1 0 0 Á

U12 1 0 Á

U13 U23 1 Á

0

0

0

Á Á Á Á Á

4

U1n U2n U3n Á 1

(9.76)

Multiplying the three matrices on the right side of Eq. (9.76), we obtain

3

S11

S12 S22

(symmetric)

Á Á Á Á

S1n S2n S3n Á Snn

3

D11

(symmetric)



S13 S23 S33

D11U12 D11U122 1 D22

4

5

D11U13 D11U12U13 1 D22U23 D11U213 1 D22U223 1 D33

Á Á Á Á

D11U1n D11U12U1n 1 D22U2n D11U13U1n 1 D22U23U2n 1 D33U3n Á 2 2 D11U1n 1 D22U2n 1 D33U23n 1 Á 1 Dnn

4 (9.77)

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Section 9.9   Solution of Large Systems of Stiffness Equations  569

By comparing the corresponding elements of the matrices S and UTDU, on the left and right sides, respectively, of Eq. (9.77), we can develop an algorithm for evaluating the elements of matrices D and U. By comparing the elements in row 1 column 1 of the two matrices, we can see that D11 5 S11. With D11 known, the elements in the first row of U can be obtained by equating the remaining elements in the first rows of S and UT DU. This yields U12 5 S12/D11, U13 5 S13/D11, . . . , U1n 5 S1n/D11. Next, we equate the corresponding elements in the second rows of S and UT DU in Eq. (9.77) to obtain the second rows of D and U, and so on. The general recurrence relationships for computation of the elements of D and U can be expressed as follows.

Dii 5

Uij 5

5

Sii

5

Sij

for i 51 i21

Sii 2

o

k51

DkkU2ki

Dii i21 1 Sij 2 DkkUkiUkj Dii k51

1



for i 5 2, 3, Á , NDOF

o

2

for i 5 1; j 5 i 1 1, i 1 2, Á , NDOF for i 5 2, 3, Á , NDOF 2 1;

(9.78a) (9.78b)

j 5 i 1 1, i 1 2, Á , NDOF

Uii 5 1       for i 5 1, 2, . . . , NDOF

(9.78c)

The nonzero elements of D and U are computed by starting at the first row number (i.e., i 5 1), and proceeding sequentially to the last row number (i.e., i 5 NDOF). As implied by Eqs. 9.78(a) and (b), for each row number i, the diagonal element Dii (Eq. (9.78a)) must be computed before the elements Uij (Eq. (9.78b)) of the ith row of U can be calculated. With the structure stiffness matrix S now decomposed into triangular and diagonal matrices, we can now begin the substitution part of the solution process. Substitution of S 5 UT  DU into the structure stiffness equations, Sd 5 P, yields UT DU d 5 P

(9.79)

The substitution part is carried out in two steps: forward substitution, and back substitution. In the forward substitution step, Eq. (9.79) is written as ⁄

UT D d 5 P

(9.80)

with ⁄

d 5 Ud (9.81)

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570  Chapter 9   Special Topics and Modeling Techniques ⁄

in which d is an auxiliary vector of unknowns. Equation (9.80) can be written in expanded form as

3

D11 D11U12 D11U13 Á

0 D22 D22U23 Á

0 0 D33 Á

D11U1n

D22U2n

D33U3n

Á Á Á Á Á

0 0 0 Á Dnn

43 4 3 4 ⁄

d1 ⁄ d2 ⁄ d3 Á

P1 P2 P3 Á

5

⁄

(9.82)



Pn

dn

⁄

from which we can see that the auxiliary unknowns d can be determined by the simple process of forward substitution, starting with the first row and proceeding sequentially to the last row. From the first row of Eq. (9.82), we can see that ⁄ ⁄ ⁄ d 1 5 P1yD11. With d 1 known, the value of d2 can now be determined by solving ⁄ ⁄ the equation in the second row of Eq. (9.82); that is, d2 5 (P2 2 D11U12 d 1)yD22. ⁄ Next, we calculate d3 by solving the equation in the third row of Eq. (9.82), and ⁄ so on. In general, the elements of d can be computed as ⁄

di 5

5

Pi Dii i21 1 ⁄ DkkUki d k Pi 2 Dii k51

1

for i 5 1

2

o

(9.83)

for i 5 2, 3, Á , NDOF

⁄

Once the auxiliary vector d has been evaluated, the unknown joint displacement vector d can be calculated by solving Eq. (9.81), using back substitution. The expanded form of Eq. (9.81) can be expressed as

F

1 0 0 Á

U12 1 0 Á

U13 U23 1 Á

0 0

0 0

0 0

Á Á Á Á Á Á

U1, n21 U2, n21 U3, n21 Á

U1n U2n U3n Á

1 0

Un21, n 1

GF G F G d1 d2 d3 Á

dn21 dn

⁄

5

d1 ⁄ d2 ⁄ d3 Á

(9.84)

⁄

d n21 ⁄ dn

From which we can see that the unknown joint displacements d can be determined by the simple process of back substitution, starting with the last row and proceeding sequentially to the first row. From the last row of Eq. (9.84), ⁄ we can see that dn 5 d n. With dn known, the value of dn−1 can now be determined by solving the equation in the next to the last row of Eq. (9.84); that is, ⁄ dn21 5 d n21 2 Un21,ndn. The back substitution is continued until all the joint displacements have been calculated. The back substitution process can be represented by the recurrence equation di 5

5

⁄

for i 5 NDOF

di ⁄

di 2

NDOF

o

k5i11

Uikdk

for i 5 NDOF 2 1, NDOF 2 2, Á , 1

(9.85)

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Section 9.9   Solution of Large Systems of Stiffness Equations  571

As discussed in the foregoing paragraphs, the UT  DU decomposition procedure for solving structure stiffness equations essentially consists of the following steps. 1. Decompose the structure stiffness matrix S into a diagonal matrix D, and a unit upper triangular matrix U, by applying Eqs. (9.78). ⁄ 2.  Calculate the auxiliary vector d using forward substitution (Eq. (9.83)). 3. Determine the unknown joint displacements d by back substitution (Eq. (9.85)).

E xample 9.9 Use UT  DU decomposition to solve the following system of structural stiffness equations.

3

5

2 6

(symmetric)



S olution

21 23 4

43 4 3 4

0 2 1 7

d1 d2 d3 d4

219 222 22

5

26

Decomposition: By applying Eqs. (9.78), D11 5 S11 5 5 U12 5 U13 5

S12 D11 S13 D11

5

2 5 0.4 5

5

21 5 20.2 5

U14 5 0 U11 5 U22 5 U33 5 U44 5 1 2 D22 5 S22 2 D11U12 5 6 2 5(0.4)2 5 5.2

U23 5

1 1 (S 2 D11U12U13) 5 [23 2 5(0.4)(20.2)] 5 20.5 D22 23 5.2

U24 5

1 1 [2 2 5(0.4)(0)] 5 0.38462 (S 2 D11U12U14) 5 D22 24 5.2

2 2 D33 5 S33 2 D11U13 2 D22U23 5 4 2 5(20.2)2 2 5.2(20.5)2 5 2.5

1 (S 2 D11U13U14 2 D22U23U24) D33 34 1 5 [1 2 5(20.2)(0) 2 5.2(20.5)(0.38462)] 5 0.8 2.5 D44 5 S44 2 D11U214 2 D22U224 2 D33U234 5 7 2 5(0)2 2 5.2(0.38462)2 2 2.5(0.8)2 5 4.6308

U34 5

Thus,

D5

3

5 0 0 0

0 5.2 0 0

0 0 2.5 0

4 3

0 0 0 4.6308

U5

1 0 0 0

0.4 1 0 0

20.2 20.5 1 0

4

0 0.38462 0.8 1

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572  Chapter 9   Special Topics and Modeling Techniques Forward Substitution: Using Eq. (9.83), ⁄

d1 5 ⁄

d2 5

P1 D11

5

219 5 23.8 5

1 1 ⁄ (P 2 D11U12 d 1) 5 [222 2 5(0.4)(23.8)] 5 22.7692 D22 2 5.2

⁄

1 ⁄ ⁄ (P 2 D11U13d1 2 D22U23 d 2) D33 3 1 [22 2 5(20.2)(23.8) 2 5.2(20.5)(22.7692)] 5 4.4 5 2.5

⁄

1 ⁄ ⁄ ⁄ (P 2 D11U14 d 1 2 D22U24 d 2 2 D33U34 d 3) D44 4 1 5 [26 2 5(0)(23.8) 2 5.2(0.38462)(22.7692) 2 2.5(0.8)(4.4)] 5 22 4.6308

d3 5

d4 5

Thus, ⁄

d5

3 4 23.8 22.7692 4.4 22

Back Substitution: Applying Eq. (9.85), ⁄

d4 5 d 4 5 22 ⁄

d3 5 d 3 2 U34d4 5 4.4 2 0.8(22) 5 6 ⁄

d2 5 d 2 2 U23d3 2 U24d4 5 22.7692 2 (20.5)6 2 0.38462)(22) 5 1 ⁄

d1 5 d 1 2 U12d2 2 U13d3 2 U14d4 5 23.8 2 0.4(1) 2 (20.2)6 2 0(22) 5 23 Thus, the solution of the given system of equations is

d5

34

23 1  6 22

Ans

If the structure stiffness matrix S is banded, then the corresponding U matrix contains nonzero elements only on its diagonal and within the upper half-bandwidth, as shown in Fig. 9.26(a) on the next page. In such cases, the computational effort required for solution can be significantly reduced by calculating only the elements in the upper half-bandwidth of U. From Fig. 9.26(a), we can see that, in any row number i of U, all the nonzero elements are located in column numbers i through i 1 NHB # NDOF. Similarly, in any column number j of U, the nonzero elements are located in row numbers j 2 NHB $ 1 through j. Using the foregoing ranges for the indexing ­parameters

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Section 9.9   Solution of Large Systems of Stiffness Equations  573 NHB = 3

1

U12

U13

U14

1

U23

U24

U25

1

U34

U35

U36

1

U45

U46

U47

1

U56

U57

U58

1

U67

U68

U69

1

U78

U79

1

U89

U=

Zero

NDOF = 9

NHB + 1 = 4 1

Uˆ =

D11

U12

U13

U14

D22

U23

U24

U25

D33

U34

U35

U36

D44

U45

U46

U47

D55

U56

U57

U58

D66

U67

U68

U69

D77

U78

U79

D88

U89

(a) Upper Unit Triangular Matrix U for a Structure Stiffness Matrix with NDOF = 9 and NHB = 3

NDOF = 9

D99 (b) Storage of D and U in Compact Form

Fig. 9.26

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574  Chapter 9   Special Topics and Modeling Techniques

in Eqs. (9.78), (9.83), and (9.85), we obtain the following modified recurrence formulas for solving the banded systems of structure stiffness equations by the UT  DU decomposition method. Decomposition Dii 5

5

for i 5 1

Sii i21

Sii 2

oD

k5m1

for i 5 2, 3, Á , NDOF

U2ki

kk

with m1 5 i 2 NHB $ 1

5

Sij

Dii Uij 5 1 (S 2 Bij) Dii ij

for i 5 1; j 5 i 1 1, i 1 2, Á , i 1 NHB # NDOF for i 5 2, 3, Á , NDOF 2 1; j 5 i 1 1, i 1 2, Á ,i 1 NHB # NDOF (9.86)

with

5o



i21

Bij 5

k5m2

0

DkkUkiUkj

for m2 # i 2 1 for m2 . i 2 1

in which, m2 5 j 2 NHB $ 1 Uii 5 1        for i 5 1, 2, Á , NDOF Forward Substitution

⁄

di 5

5

Pi Dii i21 ⁄ 1 Pi 2 Dkk Uki d k Dii k5m

1

o

2

1

for i 5 1 for i 5 2, 3, Á , NDOF

(9.87)

Back Substitution

di 5

5

⁄

di ⁄

di 2

for i 5 NDOF m3

oU

k5i11

ik

dk

(9.88)

for i 5 NDOF 2 1, NDOF 2 2, Á ,1

with m3 5 i 1 NHB # NDOF



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Section 9.9   Solution of Large Systems of Stiffness Equations  575

As stated before, the elements on the diagonal and in the upper half-­ bandwidth of S can be compactly stored in computer memory in a rectangular ⁄ array S of order NDOF 3 (NHB 1 1) (see Fig. 9.25(b)). In an analogous man⁄ ⁄ ner, a rectangular array U, of the same order as S, can be defined to store the elements on the diagonal of D and in the upper half-bandwidth of U, as depicted in Fig. 9.26(b). As indicated there, the diagonal elements of D are stored in the ⁄ first column of U, and the elements, in each row of U, in the half-bandwidth, ⁄ are stored in the same order in the corresponding row of U. The locations of ⁄ the relevant elements of D and U in the compact matrix U can be determined by using the following relationships. ⁄

Dii 5 Ui1 ⁄ Uij 5 Ui,(11j2i)

for i 5 1, 2, Á , NDOF for i 5 1, 2, Á , NDOF 2 1; j 5 i 1 1, i 1 2, Á , NHB 1 i # NDOF

(9.89)

Applying Eqs. (9.74) and (9.89), we obtain the following modified algorithm for solving the banded systems of structure stiffness equations, in terms of the ⁄ ⁄ elements of compact matrices S and U. Decomposition

⁄

Ui1 5

5

⁄

for i 5 1

Si1 i21 ⁄

⁄

Si1 2

oU

k5m1

⁄

k1

for i 5 2, 3, Á , NDOF

U2k,(11i2k)

with m1 5 i 2 NHB $ 1 ⁄

Uij 5

5

⁄

Sij

for i 5 1; j 5 2, 3, Á , NHB 1 1

⁄

Ui1 1 ⁄

Ui1

⁄

⁄

for i 5 2, 3, Á , NDOF 2 1; j 5 2, 3, Á , NHB 1 1 # NDOF 2i 1 1

(Sij 2 Bij)



with ⁄

Bij 5

5

i21

⁄

⁄

oU U ⁄

k5 m2

k1

⁄

k,(1 1 i 2 k)

(9.90)

⁄ for m #i21 2

Uk,(i 1 j 2 k )

for m⁄ 2 . i 2 1

0

^ in which m 5 i 1 j 2 NHB $ 1 2

Forward Substitution

⁄

di 5

5

Pi

for i 5 1

⁄

Ui1 1 ⁄

1P 2 o U U i21

⁄

i

⁄

k1

⁄

k,(11i2k)

2

dk

for i 5 2, 3, Á , NDOF

(9.91)

Ui1 k5m  1

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576  Chapter 9   Special Topics and Modeling Techniques

Back Substitution

di 5

5

⁄

for i 5 NDOF

di ⁄

di 2

m3

⁄

oU

k5i11

d

i,(11k2i) k

for i 5 NDOF 2 1, NDOF 2 2, Á ,1



 with m3 5 i 1 NHB # NDOF (9.92) In the computer implementation of the foregoing procedure, computer memory ⁄ ⁄ requirements can be reduced by creating only the S matrix, but not the U matrix. ⁄ ⁄ ⁄ Each element Uij of the U matrix is now computed and stored in the S matrix in ⁄ the location originally occupied by the corresponding Sij element. Thus, at the ⁄ end of the decomposition part of the solution process, the S matrix contains all ⁄ the elements of the U matrix, and can be used in the substitution part of the solu⁄ ⁄ tion. Also, this S matrix, now containing the elements of U (or D and U), can be stored for any future reanalysis of the structure for different loading conditions. In this section, we have considered only one of the many available methods for solving large systems of structural stiffness equations. For a comprehensive coverage of the various solution methods, the reader should refer to references [2, 14, 26].

Summary In this chapter, we have considered some extensions and modifications of the matrix stiffness method developed in previous chapters. We have also considered techniques for modeling some special features and details of structures, so that more realistic structural responses can be predicted from the analysis. We studied an alternative formulation of the stiffness method, which involves the structure stiffness matrix for all the coordinates (including the restrained coordinates) of the structure. The advantages of this alternative formulation are that the support displacement effects can be incorporated into the analysis in a direct and straightforward manner, and the reactions can be calculated using the structure stiffness relations. The main disadvantage of this formulation is that it requires significantly more computer memory than the standard formulation. We presented an approximate method for the analysis of rectangular building frames, neglecting the effects of member axial deformations. This approach significantly reduces the number of structural degrees of freedom to be considered in an analysis. This approximate approach is appropriate only for frames in which the member axial deformations are small enough, as compared to bending deformations, to have a negligible effect on their responses. The basic concepts of condensation of structural degrees of freedom, and analysis using substructures, were discussed. These approaches can be used in the analysis of large structures to reduce the number of stiffness equations that must be processed, and solved simultaneously. A commonly used technique for modeling an inclined roller support was described, in which the support is replaced with an imaginary axial force

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Problems  577

­ ember with large axial stiffness, and oriented in the direction perpendicular m to the incline. Modified member stiffness relations, considering the effects of rigid end offsets, were also presented. Procedures for including the effects of semirigid connections, and shear deformations, in the analysis, were discussed, and the analysis of structures composed of nonprismatic members considered. Finally, we defined the half-bandwidth of structural stiffness matrices; and discussed procedures for efficiently numbering the structure’s degrees of freedom, and for storing its stiffness matrix in computer memory by taking advantage of its symmetry and bandedness. Also considered was the commonly used UT  DU decomposition method for solving large banded systems of structure stiffness equations.

P roblems Section 9.1 9.1 Determine the joint displacements, member axial forces, and support reactions for the plane truss shown in Fig. P9.1, due to the combined effect of the loading shown and a settlement of 10 mm of support 2. Use the alternative formulation of the matrix stiffness method. 9.2 Determine the joint displacements, member axial forces, and support reactions for the plane truss shown in Fig. P9.2, due to the combined effect of the loading shown and a settlement of 5 mm of support 3. Use the alternative formulation of the matrix stiffness method.

9.3  Determine the joint displacements, member end forces, and support reactions for the three-span continuous beam shown in Fig. P9.3 on the next page, due to settlements of 8 and 30 mm, respectively, of supports 2 and 3. Use the alternative formulation of the matrix stiffness method.

4m 600 kN 2 300 kN 1

1

3

3

4

3m

2

1 4

480 kN

300 kN

6m

2

6m

3

4 5

1

1.5 m

2

2m

2.5 m

EA 5 constant E 5 200 GPa A 5 3,480 mm2

EA  constant 3

E  70 GPa

A  2,500 mm2

Fig. P9.2 Fig. P9.1

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578  Chapter 9   Special Topics and Modeling Techniques 75 kN • m

3

2 1

37.5 kN/m

4 1

2

3

7m

7m

2

7m

2

3

2m

EI = constant E = 200 GPa I = 145(106) mm4

200 kN

1

Fig. P9.3 2m 1

9.4  Determine the joint displacements, member local end forces, and support reactions for the plane frame shown in Fig. P9.4, due to the combined effect of the following: (a) the loading shown in the figure, (b) a clockwise rotation of 0.017 radians of the left support, and (c) a settlement of 20 mm of the right support. Use the alternative formulation of the matrix stiffness method.

6m E, A, I = constant E = 200 GPa A = 4,000 mm2 I = 80(106) mm4

Fig. P9.5, P9.9 15 kN/m 75 kN 2

3

2

30 kN/m

1

3

8m

3m 1

4

100 kN 10 m E, A, I 5 constant

3m

E 5 30 GPa A 5 90,000 mm2 I 5 675(106) mm4

Fig. P9.4, P9.7, P9.10 6m E, A, I = constant E = 200 GPa A = 11,800 mm2 I = 554(106) mm4

Section 9.2 9.5 through 9.8 Determine the approximate joint displacements, member local end forces, and support reactions for the frames shown in Figs. P9.5 through P9.8, assuming the members to be inextensible.

Fig. P9.6

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Problems  579

Section 9.8

12 kN/m 30 kN

9.11 and 9.12  Derive the equations of fixed-end forces due to the member loads acting on the nonprismatic beams shown in Figs. P9.11 and P9.12. 6m

Section 9.9

12 kN/m

9.13 Solve the following system of simultaneous equations using the UT  DU decomposition method.

60 kN

3

6m

20 29 15

29 16 25

43 4 3 4

15 25 18

d1 354 d2 5 2275 307 d3

9.14 Solve the following system of simultaneous equations using the UT  DU decomposition method. 12 m

3

5 22 1 0 0

E = 30 GPa Columns: Girders: A = 93,000 mm2 A = 140,000 mm2 I = 720(106) mm4 I = 2,430(106) mm4

Fig. P9.8

22 3 22 4 0

1 22 1 21 3

Section 9.3

0 4 21 6 23

43 4 3 4

0 0 3 23 4

d1 d2 d3 d4 d5

5

44 219 38 231 23

9.9 and 9.10  Analyze the plane frames shown in Figs. P9.9 and P9.10 using condensation, by treating the rotations of free joints as internal degrees of freedom. W

e

b

2I

I L 2

E = constant

L 2

Fig. P9.11 w e

b

2I

I

L 2

L 2

E = constant

Fig. P9.12

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10

INTRODUCTION TO NONLINEAR STRUCTURAL ANALYSIS Learning Objectives At the end of this chapter, you will be able to: 10.1 Define the Basic Concept of Geometrically Nonlinear Analysis 10.2 Perform Geometrically Nonlinear Analysis of Plane Trusses

Bay Bridge, San Francisco (kropic1/Shutterstock.com)

580

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Chapter 10   Introduction to Nonlinear Structural Analysis  581

Thus far in this text, we have focused our attention on the linear analysis of structures, which it may be recalled, is based on two fundamental assumptions, namely, (a) geometric linearity, implying that the structure’s deformations are so small that the member strains can be expressed as linear functions of joint displacements and the equilibrium equations can be based on the undeformed geometry of the structure, and (b) material linearity, represented by the linearly elastic stress-strain relationship for the structural material. The linear analysis (sometimes also referred to as the firstorder analysis) generally proves adequate for predicting the performance of most common types of engineering structures under service (working) loading conditions. However, as the loads increase beyond service levels into the failure range, the accuracy of the linear analysis gradually deteriorates because the response of the structure usually becomes increasingly nonlinear as its deformations increase and/or its material is strained beyond the yield point. In some structures, such as cable suspension systems, the load carrying capacity relies on geometric nonlinearity even under normal service conditions. Because of its inherent limitations, linear analysis cannot be used to predict instability phenomena and ultimate load capacities of structures. With the recent introduction of design specifications based on the ultimate strengths of structures, the use of nonlinear analysis in structural design is increasing. In a nonlinear analysis, the restrictions of linear analysis are removed by formulating the equations of equilibrium on the deformed geometry of the structure that is not known in advance, and/or taking into account the effects of inelasticity of the structural material. The load-deformation (stiffness) relationships thus obtained for the structure are nonlinear, and are usually solved using iterative techniques. The objective of this chapter is to introduce the reader to the exciting and still-evolving field of nonlinear structural analysis. Because of space limitations, only the basic concepts of geometrically nonlinear analysis of plane trusses are covered herein. However, it should be realized that a realistic prediction of structural response in the failure range generally requires consideration of the effects of both geometric and material nonlinearities in the analysis. For a more detailed study, the reader should refer to one of the books devoted entirely to the subject of nonlinear structural analysis, such as [8, 9]. We begin this chapter with an intuitive discussion of the basic concept of geometrically nonlinear analysis, and how it differs from the conventional linear analysis in Section 10.1. A matrix stiffness formulation for geometrically nonlinear analysis of plane trusses is then developed in Section 10.2. While a block diagram summarizing the various steps of nonlinear analysis is provided, the programming details are not covered herein; they are, instead, left as an exercise for the reader. The computer program for geometrically nonlinear analysis of plane trusses can be conveniently adapted from that for the linear analysis of such structures, via relatively straightforward modifications that should become apparent as the nonlinear analysis is developed in this chapter.

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582  Chapter 10   Introduction to Nonlinear Structural Analysis

10.1 BASIC CONCEPT OF GEOMETRICALLY 
 NONLINEAR ANALYSIS

As stated before, in the linear analysis, the structure’s deformations are assumed to be so small that the member strains are expressed as linear functions of joint displacements and the equilibrium equations are based on the undeformed geometry of the structure. In geometrically nonlinear analysis, the restrictions of small deformations are removed by formulating the strain-displacement relations and the equilibrium equations on the deformed geometry of the structure. To illustrate the basic concept of geometrically nonlinear analysis, consider the two-member plane truss composed of a linearly elastic material, shown in Fig. 10.1(a). Note that the truss is symmetric and is loaded symmetrically with a vertical load P. Thus, it is considered to have only one degree of freedom, which is the vertical displacement d of the free joint 2. As shown in Fig. 10.1(b), in the linear analysis, the joint displacement d is assumed to be so small that the member axial deformations u equal the components of d in the undeformed directions of the members, that is u > d sin u

(10.1)

in which u denotes the angle of inclination of members in the undeformed configuration. The member axial strain « can now be expressed as a linear function of joint displacement d as





«5

u > L

1 sinL u2 d

(10.2)

with L 5 undeformed length of members. Recall that in linear analysis, the equilibrium equations are based on the undeformed geometry of the structure. Figure 10.1(b) shows the free-body diagram of joint 2 of the truss in the undeformed configuration, with the member axial forces Q inclined in the undeformed member directions (i.e., at angles u with the horizontal). By considering the equilibrium of the joint in the vertical direction, we write P Q >  (10.3) 2 sin u from which we obtain the expression for member axial stress s, Q P s5 >  (10.4) A 2A sin u For linearly elastic material, (10.5)

s 5 E «

By substituting Eqs. (10.2) and (10.4) into Eq. (10.5), we obtain the desired (stiffness) relationship between the load P and the displacement d of the truss based on the linear analysis:

P >

12E ALsin u2 d 2



(10.6)

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Section 10.1   Basic Concept of Geometrically Nonlinear Analysis  583 P

2

L

L

θ

1

θ

3

EA = constant (a) Two-Member Truss P

δ

sin

θ

P

u

=

2

2

u

δ θ

θ

θ

δ 29

1

θ

Q

Q

Undeformed configuration

Deformed configuration

θ

3

(b) Two-Member Truss: Linear Analysis 2 Undeformed configuration

P

δ

29 L sin θ

P 29

L

θ Q

θ Q

L

L sin θ θ

Deformed configuration

θ

L cos θ (c) Two-Member Truss: Geometrically Nonlinear Analysis

Fig. 10.1

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584  Chapter 10   Introduction to Nonlinear Structural Analysis 5

Linear analysis Critical (limit) point

4

Snap-through at Pcr = 2.498 MN

Load P (MN)

3

b

a

2

Geometrically nonlinear analysis

1 0 –1

0

0.5

1

1.5

2

2.5

3

3.5

–2 –3

Displacement δ (m) (d) Response of Shallow Two-Member Truss 2

Undeformed configuration

Pcr

Deformed configuration a

29 1

3

Deformed configuration b 29

Pcr (e) Snap-Through of Two-Member Truss

Fig. 10.1  (continued)

In the geometrically nonlinear analysis, we allow the joint displacement d to be arbitrarily large, and consider the truss to be in equilibrium in its deformed configuration as shown in Fig. 10.1(c). From this figure, we can see that the member lengths L and orientations u, in the deformed configuration, can be expressed in terms of d as

Î

L 5 Ï(L cos u)2 1 (L sin u 2 d)2 5 L

1Ld 2 2 2 1Ld 2 sin u 2

11

d 1 L2 L sin u 2 d sin u 5 5  L d d Î1 1 1L2 2 2 1L2 sin u

(10.7)

sin u 2



2

(10.8)

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Section 10.1   Basic Concept of Geometrically Nonlinear Analysis  585

The member axial deformations and strains are now based on the actual deformed geometry of the members as

3

Î

1Ld 2 2 2 1Ld 2 sin u 4 u d d « 5 5 1 2 Î1 1 1 2 2 2 1 2 sin u L L L u5L2L5L 12

2

(10.9)

11 2



(10.10)

The free-body diagram of joint 2 in the deformed configuration of the truss is shown in Fig. 10.1(c), in which the member axial forces Q are inclined in the deformed member directions (i.e., at angles u with the horizontal). By considering the equilibrium of the joint in the vertical direction, we write the equilibrium equation



Q5

Î

P 5 2 sin u

1Ld 2 2 2 1Ld 2 sin u  d 2 3 sin u 2 1 24 L 2

11

P

(10.11)

which yields the expression for member axial stress as



s5

Q 5 A

Î

P

1 2 2 2 1Ld 2 sin u  d 2A 3 sin u 2 1 24 L

11

d L

2

(10.12)

Finally, by substituting the expressions for strain (Eq. (10.10)) and stress (Eq. (10.12)) into the stress-strain relationship s 5 E « (Eq. (10.5)), we obtain the desired nonlinear (stiffness) relationship between the load P and the displacement d of the truss:



3

P 5 2E A sin u 2

1Ld 24

3

Î 12 Î 12

12

11

11

d L

d L 2

1Ld 2 sin u d 2 2 1 2 sin u L 2

22

4



(10.13)

By comparing the equations used to obtain the linear solution (Eqs. (10.1) through (10.6)) with those used to derive the geometrically nonlinear solution (Eqs. (10.7) through (10.13)), we notice two basic differences between the two types of analyses. The first difference is in the expressions of member axial deformation u in terms of the joint displacement d (Eqs. (10.1) vs. (10.9)). In the linear analysis, d is assumed to be so small that u can be expressed as the component of d in the undeformed member direction, thereby yielding a linear relationship between u and d (Eq. (10.1). In the geometrically nonlinear formulation, d is allowed to be arbitrarily large and the relationship between u and d is based on the exact geometry of the member’s deformed configuration, thereby yielding a highly nonlinear relationship between u and d (Eq. (10.9)). The second basic distinction between the two types of analyses is in the way

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586  Chapter 10   Introduction to Nonlinear Structural Analysis

the equilibrium equations are established (Eqs. (10.3) vs. (10.11)). In linear analysis, the equilibrium equation is based on the undeformed geometry of the truss (thereby neglecting the joint displacement d altogether). This assumption yields a direct linear relationship between the member axial force Q and the joint load P, which does not involve d (Eq. (10.3)). In geometrically nonlinear analysis, however, since the equilibrium equation is based on the deformed configuration of the truss, the expression for Q not only contains P, but also involves nonlinear functions of d (Eq. (10.11)). The reader is encouraged to verify that the linear solution (Eq. (10.6)) can be obtained by linearizing the geometrically nonlinear solution, that is, by expanding Eq. (10.13) via series expansion and retaining only the linear term of the series. Geometrically nonlinear analysis provides important insight into the stability behavior of structures that is beyond the reach of linear analysis. Figure 10.1(d) shows the response of a typical shallow two-member truss as predicted by the linear and geometrically nonlinear analyses. These load-displacement plots are computed using the numerical values: u 5 30o, L 5 3  m, E 5 70 GPa, and A 5 645.2 mm2. It can be seen from this figure that the accuracy of the linear analysis gradually deteriorates as the magnitude of load P increases and the linear solution deviates from the exact geometrically nonlinear solution. With increasing load, the response becomes increasingly nonlinear as the truss’s stiffness progressively decreases. This decrease in stiffness is characterized by a decrease in the slope of the tangent of the load-displacement curve. Note that at point a, where the curve reaches a peak, the slope of its tangent (called tangent stiffness) becomes zero, indicating that the structure’s resistance to any further increase in load has vanished. Point a is referred to as a critical or limit point because any further increase in load causes the truss to snap-through into an inverted configuration defined by point b on the response curve. The displacement of the truss at (or just prior to reaching) the critical point can be determined by setting to zero the derivative of Eq. (10.13) with respect to d. This yields

3

Î

dcr 5 L sin u 2 cos u

4

1 21  ( cos u)2y3

(10.14)

The critical load Pcr, at which the snap-through instability occurs, can be calculated by substituting the value of dcr obtained from Eq. (10.14), for d into Eq. (10.13). We can see from Fig. 10.1(d) that the equilibrium configurations defined by the portion of the response curve between points a and b correspond to load levels below the critical level. Thus, unless the load magnitude can somehow be reduced after it has reached Pcr, the truss will snap-through from configuration a into the inverted configuration b (Fig. 10.1(e)). It should be pointed out that the nonlinear response of the two-member truss (also known as the von Mises truss) considered herein has been examined by a number of researchers, and it is frequently used as a benchmark to validate the accuracy of computer programs for geometrically nonlinear structural analysis. It has been shown in references [16, 17] that while the shallow trusses (with u # 69.295o) exhibit snap-through instability as discussed in the preceding paragraphs, the steep two-member trusses, with u . 69.295o, experience bifurcation

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Section 10.2   Geometrically Nonlinear Analysis of Plane Trusses   587

type of instability. As bifurcation instability occurs when the truss loses its stiffness in the horizontal direction, its detection requires analysis of a two degree-offreedom model of the truss, instead of the single degree-of-freedom model used herein. A general formulation for geometrically nonlinear analysis of multi-­ degree-of-freedom plane trusses is developed in the next section.

10.2 Geometrically Nonlinear Analysis of Plane Trusses

In this section, we develop a general matrix stiffness method for geometrically nonlinear analysis of plane trusses [33, 45]. The process of developing the analytical models for nonlinear analysis (i.e., establishing a global coordinate system, numbering of joint and members, and identifying degrees of freedom and restrained coordinates) is the same as that for linear analysis of plane trusses (Chapter 3). However, the member local coordinate systems are now defined differently than in the case of linear analysis. Recall from Chapter 3 that in linear analysis, the local coordinate system is positioned in the initial undeformed state of the member, and it remains in that position regardless of where the member actually displaces due to the effect of external loads. In geometrically nonlinear analysis, it is more convenient to use a local coordinate system that is attached to, and displaces (translates and/or rotates) with, the member as the structure deforms. As shown in Fig. 10.2, the origin of the local xyz coordinate system for a member is always located at the beginning, b9, of the member in x e9

y L m

Displaced position

θ

b9

e L θ

b

m

Initial position

Y

X

Fig. 10.2  Corotational (or Eulerian) Local Coordinate System

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588  Chapter 10   Introduction to Nonlinear Structural Analysis

its deformed state, with the x axis directed along the member’s centroidal axis in the deformed state. The positive direction of the y axis is defined so that the coordinate system is right-handed, with the local z axis pointing in the positive direction of the global Z axis. This type of coordinate system, which continuously displaces with the member, is called an Eulerian or corotational coordinate system. The main advantage of using such a coordinate system is that it enables us to separate the member’s axial deformation from its rigid body displacement, which is considered to be arbitrarily large in geometrically nonlinear analysis.

Member Force-Displacement Relations To establish the member force-displacement relations, let us focus our attention on an arbitrary prismatic member m of a plane truss. When the truss is subjected to external loads, member m deforms and axial forces are induced at its ends. Figure 10.3 shows the displaced position of the member in its local coordinate system. Note that because of the modified definition of the local coordinate system, only one degree of freedom (that is the member axial deformation) is now needed to completely specify the displaced position of the member. The axial deformation u of the member can be expressed in terms of its initial and deformed lengths (L and L, respectively) as

(10.15)

u5L2L 

in which the axial deformation u is considered as positive when it corresponds to the shortening of the member’s length, and negative when representing the elongation. Similarly, the member axial force Q is considered to be positive when compressive, and negative when tensile. To establish the relationship between the member’s axial force Q and deformation u, we recall that the member’s axial stress s and axial strain « are defined as

s5

Q A

u and « 5  L

(10.16)

and for linear elastic material, the stress-strain relationship is given by (10.17)

s 5 E « y

Displaced position

b9

e9

Q Initial length

Q

x

u

L L E A = constant

Fig. 10.3  Member Axial Force and Deformation in the Local Coordinate System

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Section 10.2   Geometrically Nonlinear Analysis of Plane Trusses   589

Substitution of Eqs. (10.16) into Eq. (10.17) yields the following forcedisplacement relation for the members of plane trusses in their local coordinate systems:

Q5

1ELA2 u

(10.18)



Next, we consider the member force-displacement relations in the global coordinate system. Figures 10.4(a) and (b) show the initial and displaced positions of an arbitrary member m of a plane truss. In Fig. 10.4(a), the member is depicted to be in equilibrium under the action of (local) axial forces Q; whereas in Fig. 10.4(b), the same member is shown to be in equilibrium under the action of an equivalent system of end forces F acting in the directions of the global X and Y coordinate axes. As indicated in Fig. 10.4(b), the global member end forces F and end displacements v are numbered in the same manner as in the case of linear analysis (Chapter 3), except that the forces F now act at the ends of the member in its deformed state. Now, suppose that a member’s global end displacements v (which may be arbitrarily large) are specified, and our objective is to find the corresponding end forces F so that the member is in equilibrium in its displaced position. If Xb, Yb and Xe, Ye denote the global coordinates of the joints in their undeformed configurations, to which the member ends b and e, respectively, are attached, then the initial (undeformed) length L of the member can be expressed (via Pythagorean theorem) as L 5 Ï(Xe 2 Xb)2 1 (Ye 2 Yb)2

(10.19)

x



Q

e9

Displaced position

F3

y

e9

L= Y

L– m

u

Q

e

b9

(Xe, Ye)

L

θ

b (Xb, Yb)

m Y

θ

Initial position

X (a) Member Axial Force and Deformation in Local Coordinate System

F4

L

Displaced position

5

b9

θ

F1 F2

v2

b

(Xb, Yb)

v4

e (Xe, Ye)

L

θ

v3

Initial position

v1

X (b) Member End Forces and End Displacements in the Global Coordinate System

Fig. 10.4

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590  Chapter 10   Introduction to Nonlinear Structural Analysis

Similarly, we can see from Fig. 10.4(b) that the deformed member length L, and the direction cosines of the member in its displaced position, can be expressed in terms of the initial global coordinates and displacements v of the member’s ends by the following relationships:

L 5 Ïf(Xe 1 y3) 2 (Xb 1 y1)g2 1 f(Ye 1 y4) 2 (Yb 1 y2)g2 



cX 5 cos u 5



cY 5 sin u 5

(Xe 1 y3) 2 (Xb 1 y1)



L (Ye 1 y4) 2 (Yb 1 y2) L



(10.20)

(10.21)

(10.22)

in which u represents the angle measured counterclockwise from the positive direction of the global X axis to the positive direction of the local x axis of the member in its displaced position. By comparing Figs. 10.4(a) and (b), we observe that at end b9 of the member, the global forces F1 and F2 must be, respectively, equal to the components of (local) axial force Q in the directions of the global X and Y axes; that is, F1 5 cX Q F2 5 cY Q

(10.23a) (10.23b)

By using the same reasoning at end e9, we express the global forces F3 and F4 in terms of Q as F3 5 2cX Q F4 5 2cY Q



Equations 10.23(a) through (d) can be written in matrix form as F1 cX F2 cY 5 Q F3 2cX F4 2cY

34 3 4

(10.23c) (10.23d)

(10.24)

or, symbolically as

F 5 TT Q 

(10.25)

with the 1 3 4 transformation matrix T given by T 5 [ cX  cY  2cX  2cY ]

(10.26)

It should be recognized that the set of Eqs. (10.15), (10.18) through (10.22), and (10.24) does implicitly express F in terms of v, and therefore, is considered to represent the geometrically nonlinear force-displacement relations for members of plane trusses in the global coordinate system. Because of the highly

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Section 10.2   Geometrically Nonlinear Analysis of Plane Trusses   591

nonlinear nature of some of the equations involved, it is quite cumbersome to express F explicitly in terms of v, as was previously done in the case of linear analysis. Note that if the global end displacements v of a member are known, its corresponding end forces F can be evaluated by first calculating L, L, cX, and cY using Eqs. (10.19) through (10.22); then evaluating the member axial deformation u and force Q, respectively, by applying Eqs. (10.15) and (10.18); and finally determining the member global end forces F via Eq. (10.24). It is important to realize that of all these equations, only one, that is, Q 5 EAu/L (Eq. (10.18)), involves the material properties of the member. The remaining equations are essentially of a geometric character, and are exact in the sense that they are valid for arbitrarily large joint displacements. The foregoing equations are also necessary and sufficient for establishing the geometrically nonlinear load-deformation relationships for the entire structure. The procedure for establishing such relations is essentially the same as in the case of linear analysis, and involves using member code numbers as illustrated by the following example.



E x ample 10.1 By using geometrically nonlinear analysis, determine the joint loads P that cause the two-member truss to deform into the configuration shown in Fig. 10.5(a) on the next page.



S ol u tion

Joint Displacements: Using the analytical model of the truss shown in Fig. 10.5(b), we express the given deformed configuration in terms of its joint displacement vector, d5

320.08 4 2 m 0.20

1

Member End Forces: The joint load vector P, corresponding to d, can be determined by performing the following operations for each member of the truss: (a) obtain the member’s global end displacements v from d using the member’s code numbers; (b) calculate the member’s global end forces F using Eqs. (10.15), (10.18) through (10.22), and (10.24); and (c) store the elements of F into their proper positions in P and the support reaction vector R, using the member code numbers. Thus, Member 1  L 5 1.25 m, X1 5 Y1 5 0, X2 5 1.0 m, Y2 5 0.75 m. By using the member code numbers 3, 4, 1, 2, we obtain

v1 5

3 4 0 0 0.20 20.08

3 4 m 1 2

By applying Eqs. (10.20) through (10.22), we compute the length and direction cosines of the member in the displaced position to be L 5 Ïf(1.0 1 0.20) 2 (0)g2 1 f(0.75 2 0.08) 2 (0)g2 5 1.37437 m cX 5

(1.0 1 0.20) 2 (0) 5 0.873126 1.37437

cY 5

(0.75 2 0.08) 2 (0) 5 0.487495 1.37437

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592  Chapter 10   Introduction to Nonlinear Structural Analysis 200 mm 80 mm

Undeformed configuration

Deformed configuration

0.75 m

1m

1m EA = constant E = 70 GPa A = 440 mm2 (a) Truss

Y 2

1 2

1

2

1

3

3

5

4

P=

2,675.73 + 3,900.76 1,493.95 − 3,266.884 −2,675.73 −1,493.95 R= −3,900.76 3,266.884

(b) Analytical Model

1 = 2

X

6

6,576.49 1 kN −1,772.934 2

3 4 kN 5 6

(c) Joint Load and Support Reaction Vectors

Fig. 10.5

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Section 10.2   Geometrically Nonlinear Analysis of Plane Trusses   593

1,772.934 kN

6,576.49 kN 29 0.67 m

1

2

2,675.73

3

1

3,900.76

3,266.884

1,493.95

1.2 m

0.80 m

(d) Joint Loads and Support Reactions

Fig. 10.5  (continued) Next, by using Eqs. (10.15) and (10.18), we calculate the member axial deformation and force as u 5 L 2 L 5 20.124373 m

Q5

1 L 2 u 5 23,064.540 kN EA

The member global end forces F can now be determined from Eq. (10.25):

F1 5 T T Q 5

3

0.873126 0.487495 20.87313 20.4875

4

(23,064.540) 5

3

4

22,675.73 21,493.95 2,675.729 1,493.949

3 4 kN 1 2

The elements of F1 are stored in their proper positions in the 2 3 1 joint load vector P and the 4 3 1 reaction vector R, as shown in Fig. 10.5(c). Member 2  L 5 125 m, X3 5 2.0 m, Y3 5 0, X2 5 1.0 m, Y2 5 0.75 m

v2 5

3 4 0 0 0.20 20.08

5 6 m 1 2

L 5 Ïf(1.0 1 0.20) 2 (2.0 1 0)g2 1 f(0.75 2 0.08) 2 (0)g2 5 1.043504 m cX 5

(1.0 1 0.20) 2 (2.0 1 0) 5 20.766648 1.043504

cY 5

(0.75 2 0.08) 2 (0) 5 0.642068 1.043504

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594  Chapter 10   Introduction to Nonlinear Structural Analysis

u 5 L 2 L 5 0.206496 m

F2 5 T T Q 5

3 4 20.76665 0.642068 0.766648 20.64207

Q5

1ELA2 u 5 5,088.069 kN

(5,088.069) 5

3

23,900.76 3,266.884 3,900.757 23,266.88

4

5 6 kN 1 2

Joint Loads and Support Reactions: The completed joint load vector P and the support reaction vector R are shown in Fig. 10.5(c), and these forces are depicted on a line diagram of the deformed configuration of the truss in Fig. 10.5(d). Ans Equilibrium Check: Applying the equations of equilibrium to the free-body of the truss in its deformed state (Fig. 10.5(d)), we obtain     1 S

oF

X

50

oF 5 0 1  o M 5 0 Y

   

➀

[

    1 c

22,675.73 1 6,576.49 2 3,900.76 5 0

Checks

21,493.95 2 1,772.934 1 3,266.884 5 0

Checks

26,576.49(0.67) 2 1,772.934(1.2) 1 3,266.884(2.0) 5 20.001 kN ? m < 0

Checks

Member Tangent Stiffness Matrix As indicated by the foregoing example, when the deformed configuration d of a truss is known, the corresponding joint loads P, required to cause (and/or keep the structure in equilibrium in) that deformed configuration, can be determined by direct application of the nonlinear force-displacement relations derived in the preceding subsection. However, in most practical situations, it is the external loading that is specified, and the objective of the analysis is to determine the corresponding deformed configuration of the structure, thereby requiring the solution of a system of simultaneous nonlinear equations. The computational techniques commonly used for solving such systems of nonlinear equations are iterative in nature, and usually involve solving a linearized form of the structure’s load-deformation relations repeatedly to move closer to the (yet unknown) exact nonlinear solution. Thus, before we discuss such a computational technique in a subsequent subsection, we develop the linearized form of the force-displacement relations for the planes truss members in the global coordinate system. The member force-displacement relationships can be written in terms of differentials as

ΔF 5 Kt  Δv 

(10.27)

in which ΔF and Δv denote increments of member global end forces F and end displacements v, respectively, and

Kt 5

− Fi

3 − y 4; for i, j 5 1 to 4 

(10.28)

j

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Section 10.2   Geometrically Nonlinear Analysis of Plane Trusses   595

is called the member tangent stiffness matrix in the global coordinate system. Equation (10.28) can be expanded into −F1 −y1 −F2 K­t 5



−y1 −F3 −y1 −F4 −y1

       

−F1 −y2 −F2 −y2 −F3 −y2 −F4 −y2

       

−F1 −y3 −F2 −y3 −F3 −y3 −F4 −y3

       

−F1 −y4 −F2 −y4 −F3

(10.29)



−y4 −F4 −y4

To determine the explicit form of Kt, we differentiate the expressions of F1 through F4 (Eqs. 10.23(a) through (d)) partially with respect to v1 through v4, respectively. Thus, by differentiating the expression of F1 (Eq. 10.23(a)) partially with respect to v1, we write − F1



− y1

5 cX

−Q

− cX

1

1

1− y 2 1 Q 1 − y 2

(10.30a)

To obtain ∂Q/∂v1, we substitute Eqs. (10.15) and (10.20), respectively, into Eq. (10.18), and differentiate the resulting equation partially with respect to v1, thereby yielding

1 2

−Q EA 5 cX − y1 L



(10.30b)

Similarly, by substituting Eq.(10.20) into Eq. (10.21), and differentiating the resulting equation partially with respect to v1, we obtain − cX c2Y 52  (10.30c) − y1 L



and finally, by substituting Eqs. (10.30b) and (10.30c) into Eq. (10.30a), we obtain the desired partial derivative as − F1



− y1

5

1ELA2 c 2 1 L 2 c  2 X

Q

(10.30d)

2 Y

The remaining partial derivatives of Fi with respect to vj can be derived in a similar manner. The explicit form of the member global tangent stiffness matrix thus obtained is [45, 46]



EA Kt 5 L

3

c2X cXcY 2c2X 2cXcY

cXcY c2Y 2cXcY 2c2Y

2c2X 2cXcY c2X cXcY

2cXcY 2c2Y cXcY c2Y

4 3 Q 1 L

2c2Y cXcY c2Y 2cXcY

cXcY 2c2X 2cXcY c2X

c2Y 2cXcY 2c2Y cXcY

4

2cXcY c2X  cXcY 2c2X

(10.31)

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596  Chapter 10   Introduction to Nonlinear Structural Analysis

The reader should note that, in the initial (undeformed) configuration of the member (when u 5 u and Q 5 0), the tangent stiffness matrix Kt reduces to the conventional stiffness matrix K derived previously for linear analysis of plane trusses in Chapter 3 (Eq. (3.73)). Equation (10.31) is often written in compact form as



Kt 5

1ELA2 T T 1 Qg T

(10.32)



in which the geometric matrix g is given by



g5

1 L

3

cXcY 2c2X 2cXcY c2X

2c2Y cXcY c2Y 2cXcY

c2Y 2cXcY 2c2Y cXcY

4

2cXcY c2X  cXcY 2c2X

(10.33)

Structure Load-Deformation Relations The geometrically nonlinear structure load-deformation relations for plane trusses are expressed in the form of joint equilibrium equations:

P 5 f(d) 

(10.34)

in which f is referred to as the internal joint force vector of the structure. The vector f contains the resultants of internal member end forces at the locations, and in the directions of, the structure’s degrees of freedom. As shown previously in Example 10.1, the resultant internal forces f are nonlinear functions of the structure’s joint displacements d. It should be noted that in Example 10.1, where the structure’s deformed configuration d was known, no distinction was necessary between P and f, and the former was assembled directly from the member end forces F via the member code numbers. However, when analyzing a structure for its unknown deformed configuration caused by a specified external loading, it becomes necessary to distinguish between the external loads P, which remain constant throughout the iterative process, and the internal forces f that vary as the structure’s assumed deformed configuration d is revised iteratively until f becomes sufficiently close to P, that is, the structure’s equilibrium equations (Eq. (10.34)) are satisfied within a prescribed tolerance. The structure load-deformation relationships (Eq. (10.34)) can be written in terms of differentials as

ΔP 5 St Δd



(10.35)

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Section 10.2   Geometrically Nonlinear Analysis of Plane Trusses   597

in which ΔP and Δd denote increments of external loads P and joint displacements d, respectively, and St 5



− fi

3− d 4 ;

(10.36)

for i, j 5 1 to NDOF

j

is called the structure tangent stiffness matrix. The structure matrix St can be conveniently assembled from the member global tangent stiffness matrices Kt using the member code numbers, as in the case of linear analysis.

Computational Technique—Newton-Raphson Method Most computational techniques commonly used for nonlinear structural analysis are generally based on the classical Newton-Raphson iteration technique for root finding. Such an iterative method for geometrically nonlinear analysis of plane trusses is presented herein. The method of analysis is illustrated graphically for a single-degree-of-freedom structure in Fig. 10.6. Let us assume that our objective is to determine the deformed configuration (i.e., the joint displacements) d of a structure due to a given external loading P. As shown in Fig. 10.6, we begin the process by performing the conventional linear analysis to determine the first approximate configuration d1 of the structure. Note that the linearized form of the nonlinear load-deformation relations (Eq. (10.35)) reduces to the conventional linear stiffness relations (Eq. (3.89))

Load

Linear solution

Δd1

Δd2

P f2 ΔU1

Converged (nonlinear) solution

ΔU2 f1

0

d1

d2 d3 d

Deformation

Fig. 10.6  Newton-Raphson Method (for a Single-Degree-of-Freedom Structure)

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598  Chapter 10   Introduction to Nonlinear Structural Analysis

when applied in the undeformed (initial) configuration of the structure, with ΔP 5 P and Δd 5 d1; that is, P 5 St0d1 5 Sd1

(10.37)

in which St0 denotes the structure tangent stiffness matrix evaluated in the undeformed configuration (i.e., d 5 d0 5 0). It should be realized that the configuration d1, obtained by solving Eq. (10.37), represents an approximate configuration in the sense that the joint equilibrium equations (Eq. (10.34)) are not necessarily satisfied. To correct the approximate solution, we evaluate the structure’s internal joint force vector, f1 5 f(d1), corresponding to the configuration d1 and subtract it from the joint load vector P to calculate the unbalanced joint force vector for the structure

∆U1 5 P – f1

(10.38)

The unbalanced joint forces ΔU1 are now treated as a load increment and the correction vector Δd1 is obtained by applying the linearized incremental relationship (Eq. (10.35)) as

∆U1 5 St1 ∆d1

(10.39)

with St1 now representing the structure tangent stiffness matrix evaluated in the configuration d1. A new approximate configuration d2 is then obtained by adding the correction vector ∆d1 to the current configuration d1, d2 5 d1 1 ∆d1

(10.40)

and the iteration is continued until the latest correction vector is sufficiently small. Equations (10.38) through (10.40) refer to the first iteration cycle. For an ith iteration cycle, these equations can be expressed in recurrence form as:

∆Ui 5 P 2 fi



(10.41)



∆Ui 5 Sti ∆di 

(10.42)



di11 5 di 1 ∆di 

(10.43)

Various criteria can be used in deciding whether the iterative process has converged. In general, for structures exhibiting softening type of response, the convergence criteria based on the change in the structure’s configuration between two consecutive iteration cycles do seem to yield reasonably accurate results. In the example presented in this chapter, we use a criterion based on a comparison of the changes, ∆di, in joint displacements to their cumulative

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Section 10.2   Geometrically Nonlinear Analysis of Plane Trusses   599

Evaluate joint load vector P, and calculate first approximate configuration d using linear analysis (Sd = P)

Iteration cycle: i = 1 For each member: Evaluate v from d Calculate L (Eq. 10.20), cX (Eq. 10.21), cY (Eq. 10.22), u (Eq. 10.15), Q (Eq. 10.18), F (Eq. 10.24) and Kt (Eq. 10.31) Store F in f and Kt in St

Form unbalanced joint force vector ΔU = P − f

Solve ΔU = St Δd for Δd

No

d = d + Δd i=i+1

Is Δd sufficiently small? Yes d = d + Δd For each member: Evaluate v from d Calculate L, cX, cY, u, Q and F Store F in R

Fig. 10.7  Procedure for Analysis

values, di, and consider the convergence to have occurred when the following inequality is satisfied:



Î

NDOF

o (Dd )

2

j

j51

NDOF

o

j51

#e 

(10.44)

(dj)2

in which the dimensionless quantity e represents a prescribed tolerance. A block diagram summarizing the various steps of the method for geometrically nonlinear analysis of plane trusses is shown in Fig. 10.7. The method of analysis is illustrated by the following example.

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600  Chapter 10   Introduction to Nonlinear Structural Analysis



E x ample 10.2 Determine the joint displacements, member axial forces, and support reactions for the truss shown in Fig. 10.8(a) by geometrically nonlinear analysis. Use a displacement convergence tolerance of 0.1%.



S ol u tion

Analytical Model:  See Fig. 10.8(b). The truss has three degrees of freedom, numbered as 1, 2, and 3. The three restrained coordinates of the truss are identified by numbers 4, 5, and 6. Linear Analysis:  We begin by performing the conventional linear analysis of the truss subjected to the specified joint loads,

3 4

0 P 5 22,000 0

(1)

kN

The member global stiffness matrices can be evaluated by using either Eq. (3.73), or Eq. (10.31) with u 5 u, and Q 5 0. These are: 4 5,781 4,335.7 K1 5 25,781 24,335.7

5 4,335.7 3,251.8 24,335.7 23,251.8

1 25,781 24,335.7 5,781 4,335.7

2 24,335.7 23,251.8 4,335.7 3,251.8

4

4 5 kN/m 1 2

3 5,781 24,335.7 K2 5 25,781 4,335.7

6 24,335.7 3,251.8 4,335.7 23,251.8

1 25,781 4,335.7 5,781 24,335.7

2 4,335.7 23,251.8 24,335.7 3,251.8

4

3 6 kN/m 1 2

4 5,645.5 0 K3 5 25,645.5 0

5 0 0 0 0

3

3

3

3 25,645.5 0 5,645.5 0

6 0 0 0 0

4

4 5 kN/m 3 6

The structure stiffness matrix thus obtained is

S5

3

1 11,562 0 25,781

2

3 0 25,781 6,503.6 4,335.7 4,335.7 11,426

4

1 2 kN/m 3

By solving the linear system of equations P 5 Sd1, we determine the first approximation configuration to be

3

0.11809 d1 5 20.46497 0.23618

4

1 2m 3

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Section 10.2   Geometrically Nonlinear Analysis of Plane Trusses   601

2,000 kN

3m

4m

4m EA = constant E = 70 GPa A = 645.2 mm2 (a) Truss

Y

2

1 2

1

2

1

3

4

5

(b) Analytical Model

3

3

X

6

Fig. 10.8 Iteration Cycle 1:  Next, for the current deformed configuration d1 of the structure, we evaluate its internal joint force vector f1 and the tangent stiffness matrix St1 by assembling the pertinent elements of the member F vectors and Kt matrices, respectively, as follows: Member 1

v5

3 4 0 0 0.11809 20.46497

4 5 m 1 2

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602  Chapter 10   Introduction to Nonlinear Structural Analysis

4.1566 m 2,000 kN

29

1

2

1

3

2.3503 m

39

1,000 kN

1,000 kN 8.3133 m (c) Support Reactions

Fig. 10.8  (continued)

L 5 5 m, L 5 4.8358 m, cX 5 0.85158, cY 5 0.52422, u 5 0.16419 m, Q 5 1,483.1 kN

      

3 4

1,263 777.49   F 5 21,263 2777.49

4 4 6,466.2 5 4,169.3 kN    Kt 5 1 26,466.2 2 24,169.3

3

5 4,169.3 2,259.9 24,169.3 22,259.9

1 26,466.2 24,169.3 6,466.2 4,169.3

2 24,169.3 22,259.9 4,169.3 2,259.9

4

4 5 kN/m (2) 1 2

Member 2

v5

3

0.23618 0 0.11809 20.46497

4

3 6 m 1 2

L 5 5 m, L 5 4.8358 m, cX 5 20.85158, cY 5 0.52422, u 5 0.16419 m, Q 5 1,483.1 kN

   

3 4

21,263 777.49       F 5 1,263 2777.49

3 3 6,466.2 6 24,169.3 kN    Kt 5 1 26,466.2 2 4,169.3

3

6 24,169.3 2,259.9 4,169.3 22,259.9

1 26,466.2 4,169.3 6,466.2 24,169.3

2 4,169.3 22,259.9 24,169.3 2,259.9

4

3 6 kN/m (3) 1 2

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Section 10.2   Geometrically Nonlinear Analysis of Plane Trusses   603

Member 3

v5

3 4 0 0 0.23618 0

4 5 m 3 6

L 5 8 m, L 5 8.2362 m, cX 5 1, cY 5 0, u 5 20.23618 m, Q 5 21,333.3 kN

   

      F 5

4 4 5,645.5 5 0 kN    Kt 5 3 25,645.5 6 0

5 0 161.89 0 2161.89

3

3 4 21,333.3 0 1,333.3 0

3 25,645.5 0 5,645.5 0

6 0 2161.89 0 161.89

4

4 5 kN/m 3 6

(4)

By assembling the pertinent elements of the member F vectors and Kt matrices given in Eqs. (2) through (4), we obtain

3 4

0 f1 5 21,555 70.322

1 2 kN 3

 

1   12,932 0 St1 5 26,466.2

3

2 0 4,519.7 4,169.3

   3 26,466.2 4,169.3 12,112

4

1 2 kN/m 3

By subtracting f1 from P (Eq. 1), we compute the unbalanced joint force vector for the truss as

3

0 DU1 5 P 2 f1 5 22,000 0

4 3 4 3

0 0 2 21,555 5 2445.02 70.322 270.322

4

1 2 kN 3

By solving the linearized system of equations ∆U1 5 St1 ∆d1, we determine the displacement correction vector

Dd1 5

3

0.03380 20.16082 0.067599

4

1 2m 3

To determine whether or not the iteration has converged, we apply the convergence criterion (Eq. (10.44)) as

Î

3

o (Dd )

5

3

o (d )

j51

2

j

j51

2

Î

(0.03380)2 1 (20.16082)2 1 (0.067599)2 5 0.33231 . e (5 0.001) (0.11809)2 1 (20.46497)2 1 (0.23618)2

j

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604  Chapter 10   Introduction to Nonlinear Structural Analysis which indicates that the change in the structure’s configuration is not sufficiently small, and therefore, another iteration is needed based on a new (second) approximate deformed configuration d2 obtained by adding the correction ∆d1 to the previous (first) approximate configuration d1, that is,

3

4 3

4 3

4

0.11809 0.03380 0.15189 d2 5 d1 1 Dd1 5 20.46497 1 20.16082 5 20.62579 0.23618 0.067599 0.30378

m

Iteration Cycle 2: Member 1

v5

3 4 0 0 0.15189 20.62579

4 5 m 1 2

L 5 5 m, L 5 4.7828 m, cX 5 0.86809, cY 5 0.49641, u 5 0.21721 m, Q 5 1,962.1 kN

3

1,703.2 973.98            F 5 21,703.2 2973.98

4

4 4 6,705.8 5 4,069.2 kN    Kt 5 1 26,705.8 2 24,069.2

3

5 4,069.2 1,916.7 24,069.2 21,916.7

1 26,705.8 24,069.2 6,705.8 4,069.2

2 24,069.2 21,916.7 4,069.2 1,916.7

4

4 5 kN/m 1 2

Member 2

v5

3

0.30378 0 0.15189 20.62579

4

3 6 m 1 2

L 5 5 m, L 5 4.7828 m, cX 5 20.86809, cY 5 0.49641, u 5 0.21721 m, Q 5 1,962.1 kN

      

     F 5

3

21,703.2 973.98 1,703.2 2973.98

4

3 3 6,705.8 6 24,069.2 kN    Kt 5 1 26,705.8 2 4,069.2

3

6 1 24,069.2 26,705.8 1,916.7 4,069.2 4,069.2 6,705.8 21,916.7 24,069.2

2 4,069.2 21,916.7 24,069.2 1,916.7

4

3 6 kN/m 1 2

Member 3

v5

3 4 0 0 0.30378 0

4 5 m 3 6

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Section 10.2   Geometrically Nonlinear Analysis of Plane Trusses   605

L 5 8 m, L 5 8.3038 m, cX 5 1, cY 5 0, u 5 20.30378 m, Q 5 21,715 kN

         

F5

3 4 21,715 0 1,715 0

4 4 5,645.5 5     0 kN Kt 5 3 25,645.5 6 0

3

5 0 206.53 0 2206.53

3 25,645.5 0 5,645.5 0

6 0 2206.53 0 206.53

4

4 5 kN/m 3 6

Thus, the structure’s internal joint force vector f2 and the tangent stiffness matrix St2 are given by

3 4

0 f2 5 21,948 11.722

kN St2 5

3

13,412 0 26,705.8

0 3,833.4 4,069.2

4

26,705.8 4,069.2 12,351

kN/m

and the unbalanced joint force vector is obtained as

3

4

0 DU2 5 P 2 f2 5 252.041 211.722

kN

Note that the magnitudes of the unbalanced forces are now significantly smaller than in the previous iteration cycle. By solving the linearized system of equations ∆U2 5 St2 ∆d2, we determine the displacement correction vector

3

4

0.0046508 Dd2 5 20.023449 0.0093015

        

Î

m

To check for convergence, we write 3

o (Dd )

5

3

o (d )

j51

2

j

j51

2

Î

(0.0046508)2 1 (20.023449)2 1 (0.0093015)2 5 0.036027 . e (5 0.001) (0.15189)2 1 (20.62579)2 1 (0.30378)2

j

which indicates that, while the change in the structure’s deformed configuration is now considerably smaller than in the previous iteration cycle, it is still not within the prescribed tolerance of 0.1%. Thus, we perform another (third) iteration based on a new approximate deformed configuration d3 of the structure, with

3

4 3

0.15189 d3 5 d2 1 Dd2 5 20.62579 1 0.30378

0.00046508 20.023449 0.0093015

4 3

4

0.15654 5 20.64924 0.31308

m

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606  Chapter 10   Introduction to Nonlinear Structural Analysis Iteration Cycle 3: Member 1

v5

3 4 0 0 0.15654 20.64924

4 5 m 1 2

L 5 5 m, L 5 4.7752 m, cX 5 0.87044, cY 5 0.49228, u 5 0.22476 m, Q 5 2,030.3 kN

      

 

  F5

3

1,767.2 999.45 21,767.2 2999.45

4 4 6,740.8 5 4,052.7 kN   Kt 5 1 26,740.8 2 24,052.7

3

4

5 4,052.7 1,866.9 24,052.7 21,866.9

1 26,740.8 24,052.7 6,740.8 4,052.7

2 24,052.7 21,866.9 4,052.7 1,866.9

4

4 5 kN/m 1 2

Member 2

v5

3

0.31308 0 0.15654 20.64924

4

3 6 m 1 2

– L 5 5 m, L 5 4.7752 m, cX 5 20.87044, cY 5 0.49228, u 5 0.22476 m, Q 5 2,030.3 kN

F5

3

21,767.2 999.45 1,767.2 2999.45

4

3 6 kN Kt 5 1 2

3 6,740.8 24,052.7 26,740.8 4,052.7

3

6 24,052.7 1,866.9 4,052.7 21,866.9

1 26,740.8 4,052.7 6,740.8 24,052.7

2 4,052.7 21,866.9 24,052.7 1,866.9

4

3 6 kN/m 1 2

Member 3

v5

3 4 0 0 0.31308 0

4 5 m 3 6

– L 5 8 m, L 5 8.3131 m, cX 5 1, cY 5 0, u 5 20.31308 m, Q 5 21,767.5 kN

F5

3 4 21,767.5 0 1,767.5 0

4 5 kN  Kt 5 3 6

3

4 5,645.5 0 25,645.5 0

5 0 212.61 0 2212.61

3 25,645.5 0 5,645.5 0

6 0 2212.61 0 212.61

4

4 5 kN/m 3 6

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Section 10.2   Geometrically Nonlinear Analysis of Plane Trusses   607

The structure’s internal joint force vector f3 and the tangent stiffness matrix St3 are given by

3

0 f3 5 21,998.9 0.27365

4

St3 5

kN

3

13,482 0 26,740.8

0 3,733.8 4,052.7

4

26,740.8 4,052.7 12,386

kN/m

and the unbalanced joint force vector is

3

4

0 DU3 5 P 2 f3 5 21.0925 20.27365

kN

By solving the linearized system of equations DU3 5 St 3 Dd3, we determine the displacement correction vector

3

0.000098787 Dd3 5 20.00050706 0.00019757

Î

4

m

To check for convergence, we write 3

o (Dd )

5

3

o (d )

j51

2

j

j51

2

Î

(0.000098787)2 1 (20.00050706)2 1 (0.00019757)2 5 0.00074985 , e (5 0.001) (0.15654)2 1 (20.64924)2 1 (0.31308)2

j

which indicates that the change in the structure’s deformed configuration ∆d3 is now within the specified tolerance of 0.1 percent and, therefore, the iterative process has converged. Results of Geometrically Nonlinear Analysis: The final deformed configuration of the truss is given by the joint displacement vector

4 3

3

4 3

4

0.15664 0.15654 0.000098787 5 20.64975 m d 5 d3 1 Dd3 5 20.64924 1 20.00050706 0.31327 0.31308 0.00019757

Ans

Member 1

v5

3 4 0 0 0.15664 20.64975

4 5 m 1 2

– L 5 5 m, L 5 4.7751 m, cX 5 0.87049, cY 5 0.49219, u 5 0.22493 m, Q 5 2,031.7 kN     Qa1 5 2,031.7 kN(C) Ans

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608  Chapter 10   Introduction to Nonlinear Structural Analysis

3 4 1,768.6 1,000 21,768.6 21,000

F5

4 5 kN  1 2

(5)

Member 2

v5

3

0.31327 0 0.15664 20.64975

4

3 6 m 1 2

– L 5 5 m, L 5 4.7751 m, cX 5 20.87049, cY 5 0.49219, u 5 0.22493 m, Q 5 2,031.7 kN      Qa2 5 2,031.7 kN (C) Ans

3 4 21,768.6 1,000 1,768.6 21,000

F5

3 6 kN (6) 1 2

Member 3

v5

3 4 0 0 0.31327 0

4 5 m 3 6

L 5 8 m, L 5 8.3133 m, cX 5 1, cY 5 0, u 5 20.31327 m, Q 5 21,768.6 kN Qa3 5 1,768.6 kN (T)  

F5

3 4 21,768.6 0 1,768.6 0

Ans

4 5 kN(7) 3 6

Finally, the support reaction vector R is assembled from the elements of the member F vectors given in Eqs. (5) through (7) as

R5

3

4 3 4

1,768.6 2 1,768.6 0 1,000 1 0 5 1,000 1,000 1 0 1,000

4 5 kN 6

Ans

Equilibrium Check: Applying the equations of equilibrium to the free body of the truss in its deformed state (Fig. 10.8(c)), we obtain

oF 5 0  1 c oF 5 0 1,000 2 2,000 1 1,000 5 0 1 o M 5 0 2 2,000(4.1566) 1 1,000(8.3133) 5 0.1 kN ? m < 0  [

1S

X

Checks

Y

Checks

①

Checks

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   Problems  609

SUMMARY In this chapter, we have studied the basic concepts of the geometrically nonlinear analysis of plane trusses. A block diagram summarizing the various steps involved in the analysis is given in Fig. 10.7.

P roblems Section 10.1

Section 10.2

10.1  Derive the relationships between load P and displacement d of the truss shown in Fig. P10.1 by using: (a) the conventional linear theory, and (b) the geometrically nonlinear theory. Plot the P- d equations using the numerical values: u 5 30°, L 5 3 m, E 5 70 GPa, and A 5 645.2 mm2, in the range 0 # d/L # 0.5, to compare the linear and nonlinear solutions.

10.2  By using the geometrically nonlinear analysis, determine the joint loads P that cause the two-member truss to deform into the configuration shown in Fig. P10.2. 10.3 through 10.5   Determine the joint displacements, member axial forces, and support reactions for the trusses shown in Figs. P10.3 through P10.5 by geometrically nonlinear analysis. Use a displacement convergence tolerance of one percent.

Undeformed configuration

θ L

2,250 kN

θ L

Deformed configuration

δ

3m

P EA = constant

4m

Fig. P10.1

2m

EA = constant E = 10 GPa A = 1,200 mm2

2.5 m

Fig. P10.3 (440 0.75 m

mm 2 )

Undeformed configuration (1320 mm2)

0.64657 m Deformed configuration

E = 70 GPa

0.12733 m

Fig. P10.2

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610  Chapter 10   Introduction to Nonlinear Structural Analysis

(440 mm2)

150 kN

1.5 m

1600 kN 2)

0

(88

1.5 m

mm

3200 kN

2m

16 m

(88

0m

m2 )

2m E = 9 GPa

4m

Fig. P10.4 10.6 Develop a computer program for geometrically nonlinear analysis of plane trusses. Use the program to analyze the trusses of Problems 10.3 through 10.5, and compare the computer-generated results to those obtained by hand calculations.

12 m EA = constant E = 70 GPa A = 1,200 mm2

Fig. P10.5

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A

Computer Software A computer program for analyzing two- and three-dimensional framed structures is available on the publisher’s website: https://cengage.com/engineering /matrixanalysis3e/kassimali/software. The software, which can be used to analyze plane and space trusses, beams, plane and space frames, and grids, is based on the matrix stiffness method. It can also perform geometrically nonlinear analysis of plane trusses. The software is designed for use on personal computers with Microsoft Windows® operating systems, and it provides an option for saving input data into files for subsequent modification and/or execution. Complete instructions for downloading and installing the software are provi­ ded  on the publisher’s website: https://cengage.com/engineering/matrixanalysis3e /kassimali/software.

Starting the Computer Software 1.  Click the Start button on the taskbar. 2.  Point to the menu title Programs and then click the menu item M ­ ATRIX ANALYSIS OF STRUCTURES 3.0—Kassimali; the software’s title screen will appear.

Inputting Data The computer software is designed so that any consistent set of units may be used. Thus, all the data must be converted into a consistent set of units before being input into the software. For example, if we wish to use units of kN and meter, then the joint coordinates must be defined in meter, the moduli of elasticity in kPa, the cross-sectional areas in m2, the moments of inertia in m4, the joint loads and moments in kN and kN.m, respectively, and distributed member loads in kN/m. To start entering data for a structure, click the menu title Project; and then click the menu item New Project. The input data necessary for the analysis of a structure is divided into six categories; the data in each category is input by clicking on the corresponding menu title and then entering information in the forms and/or dialog boxes that appear on the screen. The input data categories are: 1.  General structural data (project title and structure type) 2.  Joint coordinates and supports 3.  Material properties 4.  Cross-sectional properties

611

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612  Appendix A   Computer Software

5.  Member data (beginning and end joint numbers, material and crosssectional property numbers, member hinges if applicable, and angle of roll in the case of space frames) 6.  Loads (joint and member loads, support displacements, temperature changes, and fabrication errors)

Results of the Analysis Once all the necessary data has been entered, click the menu title Analyze of the main screen to analyze the structure (Fig. A.1). The software will automatically compute the joint displacements, member end forces, and supportreactions, using the matrix stiffness method. The results of the analysis are displayed on the screen. The input data as well as the results of the analysis can be printed by clicking on the menu title Project and then clicking on the menu item Print, of the main screen.

Fig. A.1  Main Screen

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B

Flexibility Method In this text, we have focused our attention on the matrix stiffness method of structural analysis, which is the most commonly used method in professional practice today, and which forms the basis for most of the currently available commercial software for structural analysis. However, as stated in Section 1.3, another type of matrix method, called the flexibility method, can also be used for structural analysis. While the stiffness method can be applied to both statically determinate and indeterminate structures, the flexibility method is applicable only to indeterminate structures. The flexibility method is essentially a generalization in matrix form of the classical method of consistent deformations, and is generally considered convenient for analyzing small structures with a few redundants. In this appendix, we present the basic concept of the flexibility method, and illustrate its application to plane trusses. A more detailed treatment of this method can be found in [3] and [52]. Essentially, the flexibility method of analysis involves removing enough restraints from the indeterminate structure to render it statically determinate. This determinate structure, which must be statically stable, is called the primary structure; and the reactions or internal forces associated with the excess restraints removed from the given indeterminate structure to convert it into the determinate (primary) structure, are termed redundants. The redundants are then treated as unknown loads on the primary structure, and their values are determined by solving the compatibility equations based on the condition that the deformations of the primary structure due to the combined effect of the redundants and the given external loading must be the same as the deformations of the original indeterminate structure. Consider, for example, a plane truss supported by five reaction components, as shown in Fig. B.1(a) on the next page. The truss is internally determinate, but externally indeterminate with two degrees of indeterminacy. This indicates that the truss has two more, or redundant, reactions than necessary for static stability. Thus, if we can determine two of the five reactions by using compatibility equations based on the geometry of the deformation of the truss, then the remaining three reactions and the member forces can be obtained from equilibrium considerations. To analyze the truss by the flexibility method, we must select two of the unknown reactions and member forces to be the redundants. Suppose that we select the horizontal and vertical reactions, R1 and R2, at the hinged support at joint 5 to be the redundants. The hinged support at joint 5 is then removed from the given indeterminate truss to obtain the statically determinate and stable 613

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4

2 3

3m

4

5

7

6 1

5

2

¯ R 1

608

1

3 R¯2

300 kN 2 panels at 4 m = 8 m

EA = constant (a) Indeterminate Truss

5 259.81 2

346.41

4

496.41

2

3.0

43

259.81

259.81

346.41

0

150 1

5

0 3

150 kN

259.81 kN (b) Primary Truss Subjected to External Loading—QaO Vector

1 2

4 0

0

0

0

1

1

1

Fig. B.1

0

1 3

1 kN 3 R¯1 5

¯ —First Column of b Matrix (c) Primary Truss Subjected to Unit Value of Redundant R 1

614

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Appendix B   Flexibility Method  615

1 1 2

2.6667

4 2.6667

1

2.6667

7

66

1.6

1.6

0

66

1.3333

7

1.3333

5

3 R¯2

3

1

1 kN ¯ —Second Column of b Matrix (d) Primary Truss Subjected to Unit Value of Redundant R 2 123.324 2

17.548

4

239.43

4

5.5

20

22

259.81

123.324

17.548

7.4

75 1

75

75 608

5

106.98

3

300 kN 136.485 (e) Support Reactions and Member Forces for Indeterminate Truss

Fig. B.1  (continued)

primary truss, as shown in Fig. B.1(b). The two redundants R1 and R2 are now treated as unknown loads on the primary truss, and their magnitudes can be determined from the compatibility conditions that the horizontal and vertical deflections at joint 5 of the primary truss due to the combined effect of the known 100 k load and the unknown redundants R1 and R2 must be equal to 0. This is because the deflections in the horizontal and vertical directions of the given indeterminate truss at the hinged support at joint 5 are 0. The compatibility equations can be conveniently established by superimposing the deflections due to the external loading and the redundants, R1 and R2,

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616  Appendix B   Flexibility Method

acting individually on the primary truss, as shown in Figs. B.1(b), (c), and (d), respectively. Thus,

ΔO1 1 f11R1 1 f12R2 5 0

(B.1a)



ΔO2 1 f21R1 1 f22R2 5 0

(B.1b)

in which ΔOi (i 5 1, 2) represents the deflection at joint 5 of the primary truss in the direction of the redundant Ri , due to the external loading; and the flexibility coefficient fij (i 5 1, 2 and j 5 1, 2) denotes the deflection of the primary truss at the location and in the direction of a redundant Ri due to a unit value of a redundant Rj . Equations (B.1) can be expressed in matrix form as

3ΔΔ 41 3 f O1

f11

O2

21

f12 f22

4 3 RR 4 5 3004 1

(B.2)

2

From the foregoing discussion for the example truss with two degrees of indeterminacy, we realize that the compatibility equations for a general indeterminate structure with ni degrees of indeterminacy can be symbolically expressed as

DO 1 fR 5 0



(B.3)

in which the ni 3 1 vectors R and DO denote, respectively, the unknown redundants, and the deflections of the primary structure at the locations and in the directions of the redundants due to external loads; and the ni 3 ni matrix f is called the structure flexibility matrix. The reader may recall from a previous course in mechanics of materials or structural analysis [18], that Maxwell’s law of reciprocal deflections states that for a linearly elastic structure, the deflection at a point i due to a unit load applied at a point j is equal to the deflection at j due to a unit load at i. As the flexibility coefficient fij denotes the deflection of the primary structure at the location of the redundant Ri due to a unit value of the redundant Rj , and the flexibility coefficient fji denotes the deflection corresponding to Rj due to a unit value of Ri , according to Maxwell’s law fij must be equal to fji (i.e., fij 5 fji). We can thus deduce that for linearly elastic structures, the flexibility matrices are symmetric. From Eqs. (B.1) through (B.3), we can see that the elements of the vector DO and the flexibility matrix f represent deflections of the primary (statically determinate) structure. Once these deflections have been evaluated, the compatibility equations (Eqs. (B.3)) can be solved for the unknown redundants. With the redundants known, the other response characteristics of the structure can be evaluated, either by equilibrium or superposition. The deflections (and the flexibility coefficients) of a primary structure can be conveniently expressed in terms of the forces and properties of its members, using the virtual work method. Recall from a previous course in mechanics of materials or structural analysis [18], that the expression of the virtual work method for truss deflections is given by

Δ5

NM

o

i51

Qar Qay L EA



(B.4)

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Appendix B   Flexibility Method  617





in which NM denotes the number of members of the truss; Qar represents the axial forces in truss members due to the real loading that causes the deflection Δ, and Qay represents the axial forces in the truss members due to a virtual unit load acting at the location and in the direction of the desired deflection Δ. Equation (B.4) can be expressed in matrix form as Δ 5 QTayfMQar (B.5) in which Qay and Qar denote the NM 3 1 vectors containing member axial forces due to virtual (unit) and real (actual) loads, respectively; and fM is a NM 3 NM diagonal matrix containing the member flexibilities (L /EA) on its main diagonal (i.e., fMij 5 Li /Ei Ai for i 5 j, and fMij 5 0 for i Þ j). The diagonal matrix fM is sometimes called the unassembled flexibility matrix. In order to develop the expressions for DO and f in terms of the member forces and properties, let us define a NM 3 1 vector QaO which contains the axial forces in the members of the primary truss due to the given external loading, and a NM 3 ni matrix b, the jth column of which contains member axial forces due to a unit value of the jth redundant (i.e., Rj 5 1). The matrix b is commonly referred to as an equilibrium matrix. In both QaO and b the member axial forces are stored in sequential order of member numbers; that is, the axial forces in the ith member are stored in the ith rows of QaO and b, and so on. The member forces QaO for the example truss are shown in Fig. B.1(b). Note that since the primary truss is statically determinate, the forces in its members due to the given external loading can be conveniently evaluated by applying the method of joints. By using the member forces shown in Fig. B.1(b) and the member numbers given in Fig. B.1(a), we form the QaO vector for the truss as

QaO 5

FG 2150 0 346.41 259.81 259.81 2433.02 0

1 2 3 4 kN 5 6 7

(B.6)

in which the tensile member axial forces are considered to be positive. The first column of b is obtained by subjecting the primary truss to a unit value of the redundant R1, as shown in Fig. B.1(c), and by computing the corresponding member forces by applying the method of joints. The second column of b is generated similarly by subjecting the primary truss to a unit value of the redundant R2, and by computing the corresponding member axial forces (see Fig. B.1(d)). The equilibrium matrix b thus obtained is



b5

F G 1 1 0 0 0 0 0

1.3333 1.3333 22.6667 21 0 1.6667 21.6667

kN/kN

(B.7)

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618  Appendix B   Flexibility Method

The first element, ΔO1 , of the DO vector represents deflection of the primary truss at the location and in the direction of R1 due to the given external loading. Therefore, for the purpose of calculating ΔO1 via the virtual work method, the real system consists of the given external loading as shown in Fig. B.1(b), and the virtual system consists of a unit load applied at the location and in the direction of the redundant R1, which is the same as the system shown in Fig. B.1(c) (without the multiplier R1). Thus, the virtual work expression for ΔO1 can be obtained by substituting QaO for Qar and the first column of b for Qay in Eq. (B.5); that is,

ΔO1 5 bT1 fMQaO

(B.8)

in which b1 denotes the first column of b. The expression for the second element, ΔO2 , of DO , in terms of member axial forces, can be obtained in a similar manner, and is given by

DO2 5 bT2 fMQaO

(B.9)

with b2 denoting the second column of b. By combining Eqs. (B.8) and (B.9), we obtain

DO 5 bT fMQaO 

(B.10)

The expressions for the elements of the flexibility matrix f, in terms of member forces and properties, can be obtained in a similar manner. For example, the virtual and real systems for the evaluation of f12 are shown in Figs. B.1(c) and (d), respectively, with the corresponding member forces stored in the first and second columns of b. Therefore,

f12 5 bT1 fMb2 Thus, the entire flexibility matrix f can be expressed in terms of the member forces and properties as



f 5 bT fMb 

(B.11)

Finally, by substituting Eqs. (B.10) and (B.11) into Eq. (B.3), we obtain the structure’s compatibility equations in terms of its member forces and properties, as

bT fMQaO 1 (bTfMb) R 5 0 

(B.12)

To illustrate the application of the flexibility method to the analysis of plane trusses, let us reconsider the truss of Fig. B.1. The QaO vector and the b matrix for this truss were determined previously, and are given in

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Appendix B   Flexibility Method  619

F G

Eqs. (B.6) and (B.7), respectively. The unassembled flexibility matrix for the structure is



4 0 0 1 0 fM 5 EA 0 0 0

0 4 0 0 0 0 0

0 0 4 0 0 0 0

0 0 0 3 0 0 0

0 0 0 0 3 0 0

0 0 0 0 0 5 0

0 0 0 0 0 0 5



(B.13)

Substituting Eqs. (B.6), (B.7), and (B.13) into Eq. (B.10), and performing the required matrix multiplications, we obtain

Do 5 bT fMQaO 5

3

4

1 2600  EA 28,883

(B.14)

Substitution of Eqs. (B.7) and (B.13) into Eq. (B.11) yields the flexibility matrix:

f 5 bT fMb 5

3

1 8 EA 10.667

4

10.667  73.444

(B.15)

Next, we substitute Eqs. (B.14) and (B.15) into the compatibility equations (Eqs. (B.12)), and solve the resulting system of simultaneous equations for the unknown redundants. This yields

R5

kN 32106.98 136.4854

(B.16)

With the redundants known, the member axial forces in the actual indeterminate structure, Qa, can be conveniently evaluated by applying the superposition relationship (see Figs. B.1(a) through (d)):

Qa 5 QaO 1 bR 

F G

(B.17)

Substituting Eqs. (B.6), (B.7), and (B.16) into Eq. (B.17), we determine the axial forces in the members of the indeterminate truss to be



Qa 5

275 75 217.548 123.324 259.81 2205.54 2227.475

kN

These member axial forces are shown in Fig. B.1(e).

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Bibliography   1. Arbabi, F. (1991) Structural Analysis and Behavior. New York, NY: McGraw-Hill.  2. Bathe, K. J. (1982) Finite Element Procedures in Engineering Analysis. Englewood Cliffs, NJ: Prentice-Hall.  3. Beaufait, F. W., W. M. Rowan, Jr., P. G. Hoadley, and R. M. Hackett. (1970) Computer Methods of Structural Analysis. Englewood Cliffs, NJ: Prentice-Hall.  4. Beer, F. P., and E. R. Johnston, Jr. (1981) Mechanics of Materials. New York, NY: McGraw-Hill.  5. Betti, E. (1872) II Nuovo Cimento. Series 2, Vols. 7 and 8.  6. Boggs, R. G. (1984) Elementary Structural Analysis. New York, NY: Holt, Rinehart & Winston.  7. Chajes, A. (1990) Structural Analysis. 2nd ed. Englewood Cliffs, NJ: Prentice-Hall.  8. Chrisfield, M. A. (1991) Non-linear Finite Element Analysis of Solids and Structures, Volume 1: Essentials. New York, NY: John Wiley & Sons.  9. Chrisfield, M. A. (1997) Non-linear Finite Element Analysis of Solids and Structures, Volume 2: Advanced Topics. New York, NY: John Wiley & Sons. 10. Dawe, D. J. (1984) Matrix and Finite Element Displacement Analysis of Structures. New York, NY: Oxford University Press. 11. Elias, Z. M. (1986) Theory and Methods of Structural Analysis. New York, NY: John Wiley & Sons. 12. Gere, J. M., and W. Weaver, Jr. (1965) Matrix Algebra for Engineers. New York, NY: Van Nostrand Reinhold. 13. Hoit, M. (1995) Computer-Assisted Structural Analysis and Modeling. Englewood Cliffs, NJ: Prentice-Hall. 14. Holzer, S. M. (1985) Computer Analysis of Structures. New York, NY: Elsevier Science. 15. Kanchi, M. B. (1981) Matrix Methods of Structural Analysis. New York, NY: John Wiley & Sons. 16. Kassimali, A. (1976) Nonlinear Static and Dynamic Analysis of Frames. Ph.D. dissertation, University of Missouri at Columbia, MO. 17. Kassimali, A., and E. Bidhendi. (1988) Stability of Trusses under Dynamic Loads. Computers & Structures 29(3): 381–392. 620

18. Kassimali, A. (2020) Structural Analysis. 6th ed., Boston, MA: Cengage Learning. 19. Kennedy, J. B., and M. K. S. Madugula. (1990) Elastic Analysis of Structures: Classical and Matrix Methods. New York, NY: Harper & Row. 20. Laible, J. P. (1985) Structural Analysis. New York, NY: Holt, Rinehart & Winston. 21. Langhaar, H. L. (1962) Energy Methods in Applied Mechanics. New York, NY: John Wiley & Sons. 22. Laursen, H. A. (1988) Structural Analysis. 3rd ed. New York, NY: McGraw-Hill. 23. Leet, K. M. (1988) Fundamentals of Structural Analysis. New York, NY: Macmillan. 24. Logan, D. L. (1992) A First Course in the Finite Element Method. 2nd ed. Boston, MA: PWS-Kent. 25. McCormac, J., and R. E. Elling. (1988) Structural Analysis: A Classical and Matrix Approach. New York, NY: Harper & Row. 26. McGuire, W., R. H. Gallagher, and R. D. Ziemian. (2000) Matrix Structural Analysis. 2nd ed. New York, NY: John Wiley & Sons. 27. Maney, G. A. (1915) Studies in Engineering. Bulletin 1. Minneapolis, MN: University of Minnesota. 28. Martin, H. C., and G. F. Carey. (1973) Introduction to Finite Element Analysis—Theory and Application. New York, NY: McGraw-Hill. 29. Maxwell, J. C. (1864) On the Calculations of the Equilibrium and Stiffness of Frames. Philosophical Magazine 27:294–299. 30. Meyers, V. J. (1983) Matrix Analysis of Structures. New York, NY: Harper & Row. 31. Noble, B. (1969) Applied Linear Algebra. Englewood Cliffs, NJ: Prentice-Hall. 32. Norris, C. H., J. B. Wilbur, and S. Utku. (1991) Elementary Structural Analysis. 4th ed. New York, NY: McGraw-Hill. 33. Oran, C., and A. Kassimali. (1976) Large Deformations of  Framed Structures under Static and Dynamic Loads. Computers & Structures 6: 539–547.

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Bibliography    621

34. Paz, M. (1991) Structural Dynamics—Theory and Computation. 3rd ed. New York, NY: Van Nostrand Reinhold. 35. Pilkey, W. D., and W. Wunderlich. (1994) Mechanics of Structures—Variational and Computational Methods. Boca Raton, FL: CRC Press. 36. Popov, E. P. (1968) Introduction to Mechanics of Solids. Englewood Cliffs, NJ: Prentice-Hall. 37. Ross, C. T. F. (1985) Finite Element Methods in Structural Mechanics. West Sussex, England: Ellis Horwood. 38. Rubinstein, M. F. (1966) Matrix Computer Analysis of Structures. Englewood Cliffs, NJ: Prentice-Hall. 39. Sack, R. L. (1989) Matrix Structural Analysis. Boston, MA: PWS-Kent. 40. Seely, F. B., and J. O. Smith. (1967) Advanced Mechanics of Materials. 2nd ed. New York, NY: John Wiley & Sons. 41. Smith, J. C. (1988) Structural Analysis. New York, NY: Harper & Row. 42. Tartaglione, L. C. (1991) Structural Analysis. New York, NY: McGraw-Hill. 43. Tena-Colunga, A. (1996) Stiffness Formulation for Nonprismatic Elements. Journal of Structural Engineering, ASCE 122(12): 1484–1489. 44. Tezcan, S. S. (1963) Discussion of “Simplified Formulation of Stiffness Matrices,” by P. M. Wright. Journal of the Structural Division, ASCE 89(6): 445–449.

45. Tezcan, S. S. (1968) Discussion of “Numerical Solution of  Nonlinear Structures,” by T. J. Poskitt. Journal of the Structural Division, ASCE 94(6): 1617. 46. Tezcan, S. S., and B. C. Mahapatra. (1969) Tangent Stiffness Matrix for Space Frame Members. Journal of the Structural Division, ASCE 95(6): 1257–1270. 47. Timoshenko, S. P., and J. M. Gere. (1961) Theory of Elastic Stability. 2nd ed. New York, NY: McGraw-Hill. 48. Turner, J. J., R. W. Clough, H. C. Martin, and L. J. Topp. (1956) Stiffness and Deflection Analysis of Complex Structures. Journal of Aeronautical Sciences 23(9): 805–823. 49. Wang, C. K. (1973) Computer Methods in Advanced Structural Analysis. New York, NY: Intext Press. 50. Wang, C. K. (1983) Intermediate Structural Analysis. New York, NY: McGraw-Hill. 51. Wang, C. K. (1986) Structural Analysis on Microcomputers. New York, NY: Macmillan. 52. Weaver, W., Jr., and J. M. Gere. (1990) Matrix Analysis of Framed Structures. 3rd ed. New York, NY: Van Nostrand Reinhold. 53. West, H. H. (1989) Analysis of Structures: An Integration of Classical and Modern Methods. 2nd ed. New York, NY: John Wiley & Sons. 54. Zienkiewicz, O. C. (1977) The Finite Element Method. 3rd ed. Berkshire, England: McGraw-Hill.

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Answers to selected problems Chapter 2 2.1

3

17 C 5 24 23

2.3

C5

2.5

4

3

35 214 56 2 28

3

229 229 56

215 6 224 12

3

1 4 ; 0

210 4 24

248 C 5 258 4

1 D 5 10 21

4

4 214 24

3 4 12

25 2 28 4

30 212 ; 48 224

4

D 5 261

4

3

258 229 52

28 248 52 ; D 5 229 46 28



5

C5

2.13

2.15

dA 5 dx

4

3

2

L4 4

9L



3

0.44444 A21 5 20.11111 0.22222

2.29

L4 4 2L5 5 2L2

9L 2L2 5L4 4

3

20.62656 20.71396 0.2849 20.12372

20.11111 0.16239 20.017094

20.71396 20.6958 0.2395 20.052213

29L 2 L4 2 2

L6 6

L4 2

4

4

0.22222 20.017094 0.26496

0.2849 0.2395 20.098751 0.13053

4

20.12372 20.052213 0.13053 20.08059

Chapter 3 Y

3.1

2

276 2168 76 168

29x2 22 cos x sin x 3 cos x

2

L

4

2332 276 332 76

3

2L3

4

1,418 22,100 2620 130

21,602 200 942 2360

2120x3 290x2 26 1 28x 1 125x4

# A dx 5 622

76 168 276 2168

2 sin x cos x 212x 0

2.17

0

2464 5,300 2310 410

3

dAB 5 dx

L

3

3

#

A21 5

1,512 2900 2810 270

332 76 2332 276

10L3 9L4 2L5 1 1 3 4 5 AB dx 5 6L7 0 3 23L 1 7 2.21 x1 5 22; x2 5 3; x3 5 25 2.23 x1 5 8.7; x2 5 27.5; x3 5 24.2 2.25 x1 5 6; x2 5 7; x3 5 5; x4 5 2 2.27

4 56 46

2.9 (ABC)T 5 CTBTAT

2.11

3

2.19

1

1

4 4

26 0 sec2 x

220x 2 60x2 22 2 8x 1 45x4 28 1 24x2

4

3

2

4



P5

325030 4 kN

5

3

X

6 NDOF  2, NR  4

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Answers to Selected Problems    623 3.3

6

Y

8

5

5

6



T3 5

7

3

0.38462 20.92308 0 0

3.11 and 3.13 

F9 5 1

4

9

4

3

1 3

2

FG

10

P5

3.5

k1 5



k2 5



k3 5

0 2150 0 0 50 0 0 2150

12

11 NDOF  8, NR  4



3.19

3 3 3

T1 5



3 3

1 0 0 0

0 1 0 0

0.6 20.8 T2 5 0 0

S5

4

0 2116,000 0 0 0 0 kN/m 0 116,000 0 0 0 0 0 292,800 0 0 0 0 kN/m 0 92,800 0 0 0 0 0 2107,077 0 0 0 0 kN/m 0 107,077 0 0 0 0

4

4

0 0 1 0

4

0 0 0 1

0.8 0 0.6 0 0 0.6 0 20.8

4

0 0 0.8 0.6

kN; Yes

165,248

82,559

Qa1 5 5.6 kN (T); Qa2 5 61.5 kN (C)

R5

3 4

d5

0.0021924 320.0043883 4m

kN

116,000 0 2116,000 0 92,800 0 292,800 0 107,077 0 2107,077 0

3 4 2560 2420 560 420

4

0 0 0.92308 0.38462

3 82,559 150,629 4 kN/m 0.0012 3.17 d 5 320.0005 4 m

3.7 1,540 kN (T) 3.9

X 3.15

2





0.92308 0 0.38462 0 0 0.38462 0 20.92308

22.5 25 227.5 55

kN



Qa1 5 254.32 kN (T); Qa2 5 203.71 kN (C);



Qa3 5 343.45 kN (C)



3.21

R5

3 4

kN

3

4

122.23 162.97 132.1 317.03 2354.32 0

0.0078601 d 5 20.0091574 0.0066964

m



Qa1 5 343.94 kN (T); Qa2 5 534.32 kN (C);



Qa3 5 187.5 kN (T); Qa4 5 79.13 kN (C);



Qa5 5 187.5 kN (T)



3 4

R5

2493.94 112.5 2106.11 246.66 534.32

kN

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624    Answers to Selected Problems 3.23

3 4

1.5685 0.78427 d 5 21.261 5.2564 21.5981

mm

5.3



Qa1 5 78.427 kN (T); Qa2 5 Qa3 5 23.836 kN (C);



Qa4 5 41.205 kN (T); Qa5 5 33.709 kN (C);



Qa6 5 137.68 kN (C)

3 4

280 R 5 220 140



3.25

P5



kN 3 27560 kN ? m4

Y

8 X

3

3

0.00046997 20.002329 0.00093995 0.0010674 d5 m 0.00075366 20.001329 0.00062621 20.00077342





Qa3 5 25.49 kN (T); Qa4 5 25.49 kN (C);



Qa5 5 150 kN(T);



Qa7 5 73.32 kN (C); Qa8 5 114.17 kN (C);



Qa9 5 42.483 kN (C)



250 58.656 R5 207.35 33.986

Qa6 5 116.01 kN (C);

P5

3

4

2150 kN ? m 0

3

1,209.6 3,024 21,209.6 3,024



k1 5



k2 5 2 3 k1

kN

3

4,218.75 8,437.50 24,218.75 8,437.50

k2 5 k3 5

3.29 P 5 1,095 kN

Chapter 5 5.9

Y

Q1 5 1

6 X

3

3

4

3,024 5,040 23,024 10,080

5.13 FSb 5

24,218.75 28,437.50 4,218.75 28,437.50

12,000 24,000 212,000 12,000

217.93 kN 29.85 kN ? m 17.93 kN 2119.49 kN ? m

5.11 FSb 5 FSe 5

4

3

21,209.6 23,024 1,209.6 23,024

8,437.50 22,500.00 28,437.50 11,250.00

8,000 12,000 28,000 12,000

2 3

3,024 10,080 23,024 5,040

5.7 Units: kN and meters

k1 5 k4 5

2

7

5 6 NDOF  2, NR  6

5.5 Units: kN and meters

Qa1 5 93.994 kN (T); Qa2 5 93.996 kN (T);

1

4

4

kN

3 4

2

2



5.1

1

1

28,000 212,000 8,000 212,000

4 4

8,437.50 11,250.00 28,437.50 22,500.00 12,000 12,000 212,000 24,000

4

; Yes

wL wL2 ; FMb 5 2FMe 5 2 12

7wL wL2 3wL wL2 ; FMb 5 ; FSe 5 ; FMe 5 2 20 20 20 30

5 NDOF  2, NR  4

Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Answers to Selected Problems    625

5.15 Qf 1 5



Qf 3 5

5.17

Q1 5

3 3 3

135 337.5 135 2337.5

4 3 4 4

kN kN ? m kN kN ? m

Qf 2 5

90 300 90 2300

4

Shear and bending moment: S1 5 S2,L 5 42.82 kN; S2,R 5 S3 5 232.18 kN; M1 5 295.9 kN ? m; M2,L = 118.2 kN ? m; M2,R 5 58.2 kN ? m; M3 5 2102.7 kN ? m

kN kN ? m kN kN ? m

5.29

131.25 kN 281.25 kN ? m 56.25 kN 2187.5 kN ? m

414.3 kN 684.6 kN ? m 2126.3 kN 937.2 kN ? m

S5

33,628.8 3,024

3

4

5.21 Units: kN and meters

S5

3

12,218.75 3,562.5 3,562.5 46,500 12,000 12,000 0 0 0 0

12,000 12,000 48,000 212,000 12,000

0 0 212,000 12,218.75 23,562.5

0 0 12,000 23,562.5 46,500

4



5 20 Q3 5 25 10

3 4



3

5.31

4

Pf 5 2Pe 5

5.27 d 5

Q1 5



3

3

3 4

20.024351 m 0.0044192 rad

42.82 kN 95.91 kN ? m 242.82 kN 118.18 kN ? m

R5

3

4 3 4

42.82 kN 95.91 kN ? m 32.18 kN 2102.73 kN ? m

Q2 5

4

F G 220 240 5 20 25 10

kN kN?m kN kN kN kN?m

3

3

44.41 kN 26.46 kN ? m 35.59 kN 0

Q3 5

5.33

20.01939 1 0.0071406 d5 20.005656 3 0.017242



Q1 5

3

3

rad

166.46 kN 188.92 kN ? m 121.54 kN 254.15 kN ? m



4 232.18 k N 258.18 kN ? m 32.18 kN 2102.73 kN ? m

4

3

5.25 Units: kN and meters



R5

0.0071306 d 5 20.0056652 0.0070651

Q1 5

337.5 237.5 218.75 2187.5

kN kN?m kN kN?m

Q2 5

4

kN kN?m kN kN?m

215 270 15 220

 Shear and bending moment: S1 5 S2,L 5 220 kN; S2,R 5 S3,L 5 215 kN; S3,R 5 S4 5 5 kN; M1 = 40 kN ? m; M2,L 5 280 kN ? m; M2,R 5 70 kN ? m; M3 5 220 kN ? m; M4 5 10 kN ? m

5.23 Units: kN and meters

60.75 Pf 5 2Pe 5 2121.5

4 3



; Yes

3,024 30,240

rad 320.007619 0.00190484

220 kN 240 kN?m Q1 5 20 kN 280 kN?m

5.19 Units: kN and meters

d5

108.87 kN 0 161.13 kN 2392 kN ? m

4

4 3 Q2 5

4

R5

4.62 k N 54.15 kN ? m 24.62 kN 226.46 kN ? m

3

4

166.46 kN 188.92 kN ? m 126.16 kN 39.79 kN 35.59 kN

4

rad

4 3 Q2 5

93.167 kN 392 kN ? m 86.833 kN 2344.5 kN ? m

4

Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

626    Answers to Selected Problems

3

Q3 5



3

5.35

Q3 5

3 3

R5



148.18 kN 231.9 kN ? m 2148.18 kN 212.66 kN ? m

6.3

3

3

0 0 275 kN ? m

4

Y

2 3

1 1

5

Q2 5

Q4 5

2148.18 kN 2212.66 kN ? m 148.18 kN 2231.9 kN ? m 2101.82 kN 2212.65 kN ? m 101.82 kN 2194.62 kN ? m

4 4

X

2

4

4 3 4 3

101.82 kN 194.62 kN ? m 296.36 kN 101.82 kN 2194.62 kN ? m

P5

kN

4

101.82 kN 194.62 kN ? m 2101.82 kN 212.65 kN ? m

R5



4 3 4 108.87 254.3 242.8 31.533

20.020928 m 0.0016033 rad 0 20.020928 m 20.0016033 rad

d5

Q1 5

155.97 kN 344.5 kN?m 31.533 kN 120 kN?m

6

3 7 9 8

NDOF = 2, NR = 7 P=0

6.5

P5

4

FG 75 kN 0 0 0 0 0

Y

Chapter 6 6.1

Y

2

1

2 2

9

3

1

5

3

6

4

2

3

1

4

7 3 8

1 4

X

10

7 9

6 8

5 NDOF = 3, NR = 6

X

12 11 NDOF = 6, NR = 6

Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Answers to Selected Problems    627

F

6.7 Units: kN and meters



k2 5

k1 5

F

200,000 0 0 2200,000 0 0

133,333.33 0 0 2133,333.33 0 0

F

0 3,000 6,000 0 23,000 6,000 0 888.89 2,666.67 0 2888.89 2,666.67

0 6,000 16,000 0 26,000 8,000

0 23,000 26,000 0 3,000 26,000

2200,000 0 0 200,000 0 0

0 2,666.67 10,666.67 0 22,666.67 5,333.33

2133,333.33 0 0 133,333.33 0 0

4

0 6,000 8,000 0 26,000 16,000

0 2888.89 22,666.67 0 888.89 22,666.67

G

6.9 Units: kN and meters





k1 5

k2 5

F

87,500 0 0 287,500 0 0

0 486.11 1,458.33 0 2486.11 1,458.33

105,000 0 0 2105,000 0 0

0 840 2,100 0 2840 2,100

F

0 1,458.33 5,833.33 0 21,458.33 2,916.67 0 2,100 7,000 0 22,100 3,500

287,500 0 0 87,500 0 0

2105,000 0 0 105,000 0 0

0 2486.11 21,458.33 0 486.11 21,458.33

0 2840 22,100 0 840 22,100





k1 5 k3 5

k2 5

F

270,000 0 0 2270,000 0 0

0 474.61 1,898.4 0 2474.61 1,898.4

0 243 1,215 0 2243 1,215

0 1,215 8,100 0 21,215 4,050

0 1,898.4 10,125 0 21,898.4 5,062.5

0 100 100 0 100 2100

Qf 2 5

0 112.5 112.5 0 112.5 2112.5

0 1,458.33 2,916.67 0 21,458.33 5,833.33

G FG F G

0 2,100 3,500   Qf1 5 0 22,100 7,000

6.11 Units: kN and meters

337,500 0 0 2337,500 0 0

G FG F G

0 2,666.67 5,333.33   Qf1 5 0 22,666.67 10,666.67

0 81 81 0 81 281

G

Qf 2 5

213.5 18 22.5 213.5 18 222.5

2337,500 0 0 0 2474.61 1,898.4 0 21,898.4 5,062.5 337,500 0 0 0 474.61 21,898.4 0 21,898.4 10,125

2270,000 0 0 270,000 0 0

0 2243 21,215 0 243 21,215

G

0 1,215 4,050 0 21,215 8,100

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FG

628    Answers to Selected Problems



6.13

Qf1 5 Qf 3 5 0 Qf 2 5

Wl1

; L L FSb 5 FSe 5 FMb 5 FMe 5 0 FAb 5

6.15

Q1 5

6.17

T1 5

6.19

Wl2

0 75 125 0 75 2125

F GF GF G

F F F

511.88 kN 2314.7 kN 2489.93 kN ? m 2511.88 kN 394.7 kN 2430.14 kN ? m

T1 5

0.8 20.6 0 0 0 0

20.6 20.8 0 0 0 0

0 21 0 T1 5 0 0 0

6.21

; FAe 5

1 0 0 0 0 0

0.44721 20.89443 0 0 0 0

0 0 1 0 0 0

0 0 0 0 1 0 0 0 0 21 0 0

445.83 kN 43.773 kN 162.12 kN ? m Q2 5 2561.21 k N 2320.7 k N 430.14 kN ? m

0 0 0 20.6 20.8 0 0 0 0 1 0 0

0.89443 0.44721 0 0 0 0

G GF

0 0 0 0 0 1 0 0 1 0 0 0

0 0 0 0.8 20.6 0

Q3 5

2 283.56 kN 281.444 kN 2184.01 kN ? m 283.56 kN 81.444 kN 2162.12 kN ? m

0 0 0 ; T2 5 I 0 0 1

T2 5

0 0 0 0.44721 20.89443 0

20.96 20.28 0 0 0 0

0.28 20.96 0 0 0 0

0 0 0 0.89443 0.44721 0

G

0 0 0 0 0 1

0 0 1 0 0 0

0 0 0 20.96 20.28 0

0 0 0 0.28 20.96 0

; Yes

G

0 0 0 0 0 1

T2 5 I

F GF GF G

6.23 and 6.31



v1 5

0 0 0 20.026175 m 20.00098438 m 0.0048505 rad

v2 5

20.023225 m 20.011336 m 0.0019382 rad 20.026175 m 20.00098438 m 0.0048505 rad

v3 5

0 0 0 20.023225 m 20.011336 m 0.0019382 rad

Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

F GF GF G

Answers to Selected Problems    629



F1 5

314.7 kN 511.88 kN 2489.93 kN ? m 2394.7 kN 2511.88 kN 2430.14 kN ? m

F2 5

F F F G FG F

2394.7 kN 2211.88 kN 162.12 kN ? m 394.7 k N 511.88 k N 430.14 kN ? m

F3 5

205.3 kN 2211.88 kN 2184.01 kN ? m 2205.3 kN 211.88 kN 162.12 kN ? m

6.25 Units: kN and meters

94,770



K1 5

(symmetric)

216,667





K2 5

Ff 1 5

27,315.2 294,770 123,922 25,486.4 123,922 2167,058 60,960 7,315.2 5,486.4 94,770 2123,922 167,058

2123,922 167,058

0 1,058.3

0 6,350 50,800

(symmetric)

50 37.5 2156.25 50 37.5 156.25

Ff 2 5

2216,667 0 0 216,667

0 21,058.3 26,350 0 1,058.3



K1 5

F

117,980



K2 5

0 200,000

234,357 10,203

G

27,315.2 25,486.4 30,480 7,315.2 5,486.4 60,960

G

0 6,350 25,400 0 26,350 50,800

0 144 288 0 144 2288

6.27 Units: kN and meters

694.92

; Yes

25,559.4 0 59,300

2637.59 22,186 37,952

2694.92 0 25,559.4 0 2200,000 0 5,559.4 0 29,650 694.92 0 5,559.4 200,000 0 59,300 2117,980 34,357 637.59 117,980

34,357 210,203 2,186 234,357 10,203

G

G

2637.59 22,186 18,976 637.59 2,186 37,952

Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

FG

630    Answers to Selected Problems



Ff 1 5 0; Ff 2 5

F F

0 250 21,000 0 250 1,000

6.29 Units: kN and meters





K1 5

K2 5

9,258.9 18,067 2450.82 29,258.9 218,067 2450.82 36,400 0 0 236,400 0 0

18,067 36,360 225.41 218,067 236,360 225.411 0 208.32 520.80 0 2208.32 520.80

2450.82 225.41 1,502.7 450.82 2225.41 751.34

0 520.80 1,736 0 2520.80 868

29,258.9 218,067 450.82 9,258.9 18,067 450.82

218,067 236,360 2225.41 18,067 36,360 2225.41

0 2208.32 2520.80 0 208.32 2520.80

236,400 0 0 36,400 0 0

G FG G FG 2450.82 225.41 751.34    Ff 1 5 450.82 2225.41 1,502.7

0 520.80 868    Ff2 5 0 2520.80 1,736

0 55.902 18.633 0 55.902 218.633

0 81.25 72.917 0 106.25 283.333

6.33 Units: kN and meters



S5

3

311,437 27,315.2 0

27,315.2 111,760 25,400

4

0 25,400 50,800

Pf 5 2Pe 5

3

4

50 131.75 2288

6.35 Units: kN and meters



S5

3210,203 22,186

4

22,186 97,252

Pf 5 2Pe 5

250 321,000 4

6.37 Units: kN and meters



S5

3

45,658.9 18,067 450.82 0

18,067 36,568.32 295.39 520.8

450.82 295.39 3,238.7 868

4

0 520.8 868 1,736

Pf 5 2Pe 5

3

4

0 137.152 kN 54.284 kN ? m 283.333 kN ? m

Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

6.39 P 5

6.41 d 5





F G 0 0 296.616 kN 211.605 kN ? m 140.31 kN 0 15.272 kN ? m 0

Q1 5

R5



3

F G

4 3 4 8.149 kN 23.288 kN 26.185 kN ? m 28.149 kN Q2 5 1.087 kN 20.148 kN ? m

218.362 kN 40.329 kN 43.157 kN ? m 51.151 kN 24.137 kN 27.104 kN 20.148 kN ? m

Q1 5

3

Member 2: S2 5 166.29 kN; S3 5 2121.71 kN; M2 5 2267.51 kN ? m; M3 5 0; 1Mmax 5 308.6 kN ? m at x 5 6.93 m; Q 5 256.756 kN 6.45



Member  1: S1 5 40.329 kN; S2 5 234.67 kN; M1 5 243.157 kN ? m; M2 5 226.19 kN ? m; 1Mmax 5 21.9 kN ? m at x 5 3.23 m; Q 5 218.36 kN Member 2: S2 5 23.29 kN; S3 5 21.085 kN; M2 5 226.19 kN ? m; M3 5 20.148 kN ? m; 1Mmax 5 1.86 kN ? m at x 5 3.08 m; Q 5 8.15 kN 6.43

43.244 kN 218.648 kN 26.577 kN ? m R5 259.94 kN 56.756 kN 121.71 kN

Member 1: S1 5 23.406 kN; S2 5 2101.59 kN; M1 5 26.577 kN ? m; M2 5 417.51 kN ? m; 2Mmax 5 290.45 kN ? m at x 5 5 m; Q 5 40.864 kN

0.00003577 m 30.00089141 rad 4 218.362 kN 40.329 kN 43.157 kN ? m 18.362 kN 34.671 kN 226.185 kN ? m

F G

Answers to Selected Problems    631



d5

4

F GF G

216.53 57.029 304.16 Q1 5 2216.53 257.029 608.31

kN kN kN ? m kN kN kN ? m

Q2 5

31.198 2216.45 2608.31 108.8 2263.55 1,197

kN kN kN ? m kN kN kN ? m

3 4

257.029 kN 216.53 kN 304.16 kN?m R 5 87.686 kN 230.656 kN 283.47 kN 1,197 kN?m

6.47

20.00026195 m d 5 20.0043172 rad 0.0078279 rad

m 320.0010826 0.010258 rad 4

d5

3

4

0.0026008 m 20.0057104 m 20.034556 rad 0.066992 rad

F G F G F GF G 40.864 kN 2101.59 kN 2417.51 kN ? m 240.864 kN 223.406 kN 26.577 kN ? m

256.756 kN 166.29 kN 267.51 kN ? m Q2 5 56.756 kN 121.71 kN 0

Q1 5

229.053 kN 8.683 kN 24.871 kN ? m 2129.053 kN 41.317 kN 268.100 kN ? m

Q2 5

94.669 kN 96.953 kN 68.103 kN ? m 294.669 kN 90.547 kN 0

Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

632    Answers to Selected Problems

R5

6.49

d5

Q2 5

Q4 5

3

4

94.669 kN 208.755 kN 24.871 kN ? m 294.669 kN 90.547 kN

F G 0.078743 m 20.00058079 m 20.015301 rad 0.095125 m 20.032966 m 0.0050852 rad 0.11121 m 20.00081012 m 20.0051602 rad

7.3 d 5



  Q1 5

3

3

211.568 kN 80.244 kN 281.851 kN ? m 2115.568 kN 99.756 kN 21.078 kN ? m

4 3

3

237.635 kN 148.912 kN 470.818 kN ? m 2237.635 kN 2148.912 kN 422.656 kN ? m

4 3

Q3 5

R5

4

170.365 kN 51.084 kN 224.658 kN ? m 2170.365 kN 251.084 kN 81.849 kN ? m

4

243.665 kN 2139.598 kN 2422.664 kN ? m 2147.665 kN 240.402 kN 1.084 kN ? m

251.084 kN 170.365 kN 224.658 kN ? m 2148.912 kN 237.635 kN 470.818 kN ? m

4





Q1 5

R5

3

4 3

3 4 122.96 kN 191.83 kN ? m 0 139.54 kN 2249.88 kN ? m

Q2 5

4

Q1 5

3

166.46 kN 188.92 kN ? m 121.54 kN 254.15 kN ? m

3

0.0071406 d 5 20.0056563 0.017242

Q1 5



7.7

Q1 5



3

3

4

108.87 kN 0 161.13 kN 2392 kN ? m

Q3 5

Q2 5

4

7.5



4 3

44.41 kN 26.46 kN ? m   R 5 35.59 kN 0

Q3 5

7.1 d 5 f20.10385 mg

227.043 kN 0 139.543 kN 2249.88 kN ? m

0.0071306



Chapter 7

122.957 kN 191.827 kN ? m 27.043 kN 0

320.00566524 rad

3

155.97 344.5 31.533 120

4

4.62 kN 54.15 kN ? m 24.62 kN 226.46 kN ? m

3 4 166.46 kN 188.92 kN ? m 126.16 kN 39.79 kN 35.59 kN 0

rad

4 3

kN kN?m kN kN ? m

Q2 5

4

R5

93.167 392 86.833 2344.5

kN kN ? m kN kN ? m

4

3 4 108.87 kN 0 254.3 kN 242.8 kN 31.533 kN

320.0043172 rad4 20.00026195 m

F G F G

d5

40.864 kN 2101.59 kN 2417.51 kN ? m 240.864 kN 223.406 kN 26.577 kN ? m

R5

Q2 5

256.756 kN 166.29 kN 267.51 kN ? m 56.756 kN 121.71 kN 0

3 4 43.244 kN 218.648 kN 26.577 kN ? m 259.94 kN 56.756 kN 121.71 kN 0

Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Answers to Selected Problems    633 7.9





3

20.000901 m d 5 20.003276 m 0.010155 rad

4

F GF G F G

94.605 kN 46.434 kN 0 Q1 5 294.605 kN 73.566 kN 2135.66 kN ? m

R5

94.605 46.434 0 294.605 148.57 0

Q2 5

175.62 kN 213.456 kN 0 2115.62 kN 231.544 kN 135.66 kN ? m

kN kN

7.21 d 5

Q1 5

Q3 5

kN kN

7.11 P max 5 951.36 kN



3 3

3

4

0.0069825 d 5 20.010145 0.0048214

m

134.06 kN 185.19 kN ? m 2134.06 kN 216.98 kN ? m



Qa1 5 305.48 kN (T); Qa2 5 591.77 kN (C);



Qa3 5 135 kN (T); Qa4 5 9.8877 kN (C);



Qa5 5 240 kN (T)



3 4

kN

3

4

7.19





R5

d5

2413.48 81 2186.52 2135.77 654.77

Q1 5

Q3 5

20.0036571 rad 0.0017714

3 3

24.87 21.8939 4.87 232.196

224.147 277.176 24.147 291.853

kN kN?m kN kN?m

4

kN kN?m kN kN?m

4 3 Q2 5

R5

15.624 32.196 215.624 77.176

kN kN?m kN kN?m

4

F G 24.87 21.8939 20.494 239.771 24.147 291.853

kN kN?m kN kN kN kN?m

20.00089687 m 20.053264 m 0.0071551 rad

4

d5

3

4 4

F GF G F G

Q1 5



R5

94.171 kN 46.707 kN 0 294.171 kN 148.29 kN 0

7.29

0.0081773 d 5 20.0058006 0.0054464



Q4 5

2115.94 kN 2216.98 kN ? m 115.94 kN 2246.79 kN ? m

4

3

94.171 kN 46.707 kN 0 294.171 kN 73.293 kN 2132.93 kN?m



Q2 5

2134.06 kN 2216.98 kN ? m 134.06 kN 2185.19 kN ? m

115.94 kN 246.79 kN ? m 268.11 kN 115.94 kN 2246.79 kN ? m

R5

7.23

4

4 3 4 3

115.94 kN 246.79 kN ? m 2115.94 kN 216.98 kN ? m

7.13 P max 5 47.453 kN 7.17

3

20.032782 m 20.0026497 rad 0 20.032782 m 0.0026497 rad

3

4

Q2 5

175.14 kN 213.638 kN 0 2115.14 kN 231.362 kN 132.93 kN?m

m

Qa1 5 357.76 kN (T); Qa2 5 513.37 kN(C); Qa3 5 152.5 kN (T); Qa4 5 104.12 kN (C); Qa5 5 222.5 kN (T)

Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

634    Answers to Selected Problems



R5

3 4 2479.76 91.5 2120.24 246.866 555.37

Chapter 8 8.1

kN

3



0.0053481 m 7.31 d5 0.0035829 m 20.00056336 rad

Q1 5



4

3

4

17.871 d 5 25.0794 27.7663

mm

Qa1 5 24.768 kN (T); Qa2 5 93.974 kN (C); Qa3 5 107 kN (C)

F GF G 3 4 F G 8.7792 kN 24.3505 kN 216.873 kN?m 28.7792 kN 4.3505 kN 222.282 kN?m 8.7792 24.3505 216.873 28.7792 4.3505 30.394

R5

7.33 d 5

4.3505 kN 8.7792 kN 30.394 kN?m Q2 5 24.3505 kN 28.7792 kN 22.282 kN?m

kN kN kN?m kN kN kN?m

3

20.0068989 m 20.01902 m 0.0095899 rad

4

F GF G F G 94.386 46.572 0 294.386 73.428 2134.28

kN kN



Q1 5



94.386 kN 46.572 kN 0 R5 294.386 kN 148.43 kN 0

kN kN kN?m

Q2 5

175.37 kN 213.548 kN 0 2115.37 kN 231.452 kN 134.28 kN?m

25.5693 222.277 9.2822 252.351 69.802 234.901 217.079 102.48 25.619



R5

8.3

1.0795 d 5 20.59449 21.4492

3

4

kN

mm



Qa1 5 12.47 kN (T); Qa2 5 15.37 kN (T); Qa3 5 67.569 kN (C); Qa4 5 81.813 kN (C)



27.7899 29.7373 0 0 213.179 7.9076 R5 242.21 52.763 0 0 70.154 42.092

kN

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Answers to Selected Problems    635 8.5

4.3357 21.8141 1.5783 4.0207 21.5814 21.6683 d5 0.77404 20.79219 21.5783 0.86404 20.97997 1.6683

215.75 33.75 26.75 229.25 146.25 229.25 mm R 5 kN 251.75 146.25 29.25 6.75 33.75 6.75

Qa1 5 Qa3 5 93.531 kN (C); Qa2 5 151.99 kN (C); Qa4 5 35.074 kN (C); Qa5 5 66.556 kN (T); Qa6 5 Qa8 5 0; Qa7 5 66.556 kN (C); Qa9 5 63 kN (C); Qa10 5 Qa11 5 Qa12 5 18 kN (C)



3

4



8.11

d5



20.030377 8.7 d 5 rad 0.030377

Q1 5



R5

8.9



3 4 3 4 68.81 kN 3.01 kN ? m 40.68 kN ? m   Q 5 43.69 kN 2 23.01 kN ? m 23.01 kN ? m

43.69 kN 23.01 kN ? m 3.01 kN ? m 68.81 kN 3.01 kN ? m 240.68 kN ? m

R5

3

4

186.01 kN 2681.33 kN ? m 21.56 kN ? m 53.96 kN 21.69 kN ? m 2268.25 kN ? m

3 4 3 4

Q1 5

20.011965 m 0.0063189 rad 20.0018011 rad 20.070066 m 0.0051488 rad 20.0099176 rad

22.73 kN 21.26 kN ? m 228.92 kN ? m 222.73 kN 1.26 kN ? m 21.63 kN ? m

3 4 3 4 3 4 3 4 3 4 3 4 3 4 3 4 68.81 kN 3.01 kN ? m 40.68 kN ? m 87.38 kN 68.81 kN 240.68 kN ? m 23.01 kN ? m

20.09287 m d5 0.05029 rad 0.02778 rad

Q1 5

186.01 kN 21.56 kN ? m 681.33 kN ? m   Q2 5 2186.01 kN 1.56 kN ? m 2123.31 kN ? m

253.96 kN 1.69 kN ? m 21.55 kN ? m 53.96 kN 21.69 kN ? m 2268.25 kN ? m



Q2 5

22.73 kN 1.62 kN ? m 1.27 kN ? m 67.27 kN 21.62 kN ? m 1.04 kN ? m

247.27 kN 21.03 kN ? m 1,571.08 kN ? m 267.27 kN 1.03 kN ? m 1.63 kN ? m



Q3 5



22.729 kN 21.264 kN ? m 228.915 kN ? m R5 247.271 kN 21.030 kN ? m 1571.082 kN ? m

8.15

d5

20.0027076 m 0.0015353 m 20.0083936 m 21.2118 rad 0.0026374 rad 0.76337 rad

Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

636    Answers to Selected Problems Chapter 9

219.210 kN 227.083 kN 230.453 kN 250.684 kN ? m 20.042 kN ? m 31.993 kN ? m 59.711 kN 30.031 kN 0.061 kN R 5 24.248 kN ? m 20.192 kN ? m 40.016 kN ? m 240.49 kN 22.95 kN 30.39 kN 15.33 kN ? m 20.10 kN ? m 217.65 kN ? m

Chapter 10

8.17

10.1 (a) P >

20.1389 m 20.000704 m 0.01543 m 0.003415 rad 0.02568 rad 20.002869 rad 0.1389 m 20.000659 m 0.01538 m 20.000344 rad 0.02568 rad 20.006705 rad d5 0.1392 m 20.000704 m 20.01543 m 20.003415 rad 0.02573 rad 0.002861 rad 20.1392 m 20.000659 m 20.01538 m 0.000344 rad 0.02573 rad 0.006714 rad

9.1 d 5



320.010993 4 m 0.0031917

Qa1 5 370.24 kN (T); Qa2 5 638.41 kN (C);



Qa3 5 33.289 kN (T)

R5



3 4 383.05 510.73 212.804 230.729 2370.24 0

kN

9.3 See answer to Problem 7.19.

15W 7W L 18W 5W L ; FMb 5 ; FSe 5 ; FMe 5 2 33 66 33 33 9.13 d1 5 6; d2 5 211; d3 5 9 9.11 FSb 5

57.38 309.9 223.06 2114.7 20.5135 2161.8 246.21 290.1 272.93 2214.2 20.5135 146.5 R5 257.45 309.9 23.06 114.7 20.5147 169 46.29 290.1 72.93 214.2 20.5147 2146.7

kN kN kN kN ? m kN ? m kN ? m kN kN kN kN ? m kN ? m kN ? m kN kN kN kN ? m kN ? m kN ? m kN kN kN kN ? m kN ? m kN ? m

1

2

2E A sin 2   L

(b)

3

1 24

 P 5 2EA sin  1 L

10.3 d 5





3

Î 12 Î 12 11

 L

11

2

1L2 sin  2 1  1 2 1 2 sin  L

12

 L

2



4

m 320.23233 21.0581 4

Qa1 5 1,827.2 kN (C); Qa2 5 2,152.7 kN (C)

R5

3

4

kN

d5

3

4

m

10.5

1,624.2 837.11 21,624.2 1,412.9

20.040921 3.0647 23.3036



Qa1 5 859.34 kN (C); Qa2 5 1,032.5 kN (T); Qa3 5 1,122.5 kN (C)



20.013871 R 5 2572.3 722.27

3

4

kN

Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

index A Addition operations, matrix algebra, 27-28 Analysis module, 130, 141-160, 236-248, 327-338 beam computer analysis, 236-248 end displacements u and forces Q, 153-159, 331-335, 394-395 equivalent joint loads, 236-238 joint displacements d, 151-153, 240, 333 joint loads P, 149-151, 236-240, 327-332 member end force calculations, 153-159, 240-246, 333-337 plane frame computer analysis, 327-338 plane truss computer analysis, 130, 141-160 sample results from, 160, 246-247, 337 structure coordinate number assignment, 141-144, 236, 327 structure stiffness matrix S, 144-149, 236-239, 327-332 support reactions R, 153-159, 240-246, 333-337 Analytical models, 10-11, 50-68, 106-123, 164-172, 215-227, 253-258, 302-322, 358-366, 375-390, 402-415, 422-503, 535-560 beams, 164-172, 215-227, 381-386, 407-411 coordinate number assignment, 55, 141-144, 168-169, 257 degrees of freedom, 50-58, 167-169, 255-258 fabrication errors, 402-407, 411-415 finite-element formulation using, 59-68 framed structures 422-503, 537 force-displacement relationships, 11, 50-52 global coordinate system, 50-52, 166-167, 253-255 grids, 438-460 hinged joints, 358-366 inclined roller supports, 535-537 joint displacement d, 54, 256-257

joint load P, 56-57, 169-172 line diagrams for, 11 local coordinate system, 51-53, 166-167, 253-255 matrix stiffness method using, 215-227, 302-322, 375-390 member releases, 358-366, 375-390, 449-450, 470-472 nonprismatic members, 553-560 offset connections, 538-542 plane frames, 253-258, 302-322, 358-366, 386-390, 411-415 plane trusses, 50-68, 106-123, 376-380, 402-407, 535-537 prismatic members, 164-172 restrained coordinates, 55-56, 255-258 semirigid connections, 542-546 shear and bending moment diagrams for, 216-217, 304-307 shear deformations and, 546-552 space frames, 461-499 space trusses, 423-438 stiffness method using, 59-68, 215-227, 302-322, 375-390 structural analysis using, 10-11 structure coordinates, 55 support displacements, 375-390 support reactions R, 57, 170 temperature changes, 402-415 three-dimensional framed structures, 422-503 trusses, 50-68, 535-537 Angle of roll c, 477-484 Angle of twist f, 445-449 Approximate matrix analysis, 511-519 Area ratio, 553-554 Arrow notation, 53, 55-56, 169, 257, 441 Augmented matrix, 40 Axial fixed-end forces FA, 264-267 Axial forces, 62-64 B Banded structure stiffness equations, 564-576

Beam sign convention, 176-177, 196, 557 Beams, 6-7, 163-251, 353-366, 381-386, 407-410 analysis modules, 236-248 analytical models, 164-172, 381-386, 407-410 bending moments M, 176-179, 192-193 code number technique for, 203-207, 210 computer analysis, 228-248 concentrated load W and, 192-198 cross-sectional property data, 232 defined, 6, 164 deflection, 176-185 degrees of freedom, 167-169 differential equation for, 176 direct integration method for, 192-195 direct stiffness method for, 198-207 end displacements u and forces Q, 175-176, 198-201 equivalent joint loads P, 211-215 finite-element formulation, 186-192 fixed-end forces, 175-176, 192-198, 207-210 fixed-joint forces Pf , 207-215 global coordinate system, 166-167 hinged joint analysis, 355-366 hinges, 353-366 input (data) modules, 228-236 joint loads P, 169-172, 207-215 local coordinate system, 166-167 material property data, 231-232 matrix stiffness analysis of, 215-227, 381-386 member data, 232-233 member releases, 353-366, 381-386, 407-410 member stiffness matrix k, 176-185, 189-192, 203-207 member stiffness relations, 172-185 prismatic members, 164-192 procedure for analysis, 215-227, 355 restraint codes for, 230, 233

637

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638    Index Beams (contd.) shear and bending moment diagrams for, 216-219 stiffness relations, 172-185, 198-207, 353-355 structural analysis classification, 6-7 structure coordinates, 168-169, 236 structure stiffness, 198-207, 236-239 support data, 230-233 support displacement analysis, 381-386 support reactions R, 170 temperature changes and, 407-410 virtual work used for, 186-192 Bending moment M, 176-179, 192-193 Building frames, 511-519 C Code number technique, 98-103, 203-207, 210, 296-302 beams, 203-207, 210 fixed-joint forces Pf , 210, 296-302 member stiffness matrix k, 203-207 plane frames, 296-302 plane trusses, 98-103 structure stiffness matrix S from, 98-102, 203-207, 296-302 support reaction R, 102-103 Column matrix, 25 Compatibility conditions, 13-14, 90-93, 198-199 beams, 198-199 equations for, 90-93, 198-199 member force–displacement relations and, 90-93 plane trusses, 90-93 structural analysis relationships from, 13-14 structure stiffness, 90-93, 198-199 Computer analysis, 21, 129-162, 228-248, 322-338, 366-368, 390-395, 537, 611-621. See also Analysis module; Data input; Input module analysis module, 130, 141-160, 236-248, 327-338 beams, 228-248 cross-sectional property data, 138, 232, 326-327 data input, 130-140, 390-391, 611-612 Do Loop commands for, 145-149, 236-239, 241-243 end displacements u and forces Q, 153-159, 331-335, 394-395 equivalent joint loads, 236-238, 327-332, 366-367, 391-394

h inged joints, 366-368 inclined roller supports, 537 input (data) module, 130-140, 228-236, 322-329, 390-391 joint data, 133-136, 228-230, 322-323 joint displacement d, 151-153, 240, 333 joint load P, 139, 149-151, 233, 236-238, 326-332 material property data, 137-138, 231-232, 326 member data, 138, 232-233, 326, 366 member force calculations, 153-159, 240-246, 333-337, 366-367, 394-395 member load data, 139, 234, 326-327 member releases, 366-368, 390-395 plane frames, 322-338, 366-368 plane trusses, 129-162 results, 612 software, 21, 611-612 structure coordinate number assignment, 141-144, 236, 327 structure stiffness matrix S, 144-149, 236-239, 327-332 subroutines for, 147-149, 153-162, 236-246, 248, 329-335, 338, 366-368, 392-395 support displacements, 390-395 support data, 136-137, 230-233, 324-326 support reactions R, 153-159, 240-246, 333-337, 394-395 Concentrated loads W, 192-198 beam fixed-end forces due to, 192-198 bending moments M, 192-193 direct integration method for, 192-195 fixed-end force vector Qf , 196-198 slope and deflection equations for, 193-194 Condensation, 519-535 defined, 519 degrees of freedom, 520-524 stiffness relationships and, 524-527 substructure analysis and, 519, 527-535 Connections, see Joints Constitutive relations, 14-15 Continuity conditions, 13-14. See also Compatibility conditions Coordinate number assignment, 55, 141-144, 168-169, 257 Coordinate transformations, 77-86, 271278, 427-430, 451-453, 472-490 end displacements u and forces Q, 271-278, 453, 472-477 global to local system, 78-80, 429 grids, 451-453 local to global system, 80-81, 271-278, 429-430

ember rotation matrix r, 477-490 m plane frames, 271-278 plane trusses, 77-86 space frames, 472-490 space trusses, 427-430 transformation matrix T, 79-81, 274-275, 476-477 Critical point, 586 Cross-sectional property data, 138, 232, 326-327 D Data input, 130-140, 228-236, 322-328, 390-391, 611-612. See also Input modules beams, 236-238 computer analysis, 130-140, 236-238, 322-328, 390-391 plane frames, 322-328 plane trusses, 130-140 software procedure, 611-612 support displacement, 390-391 Decomposition method, 564-576 Deflection, 176-185, 192-195, 616-618 beams, 176-185, 192-195 bending moment M, 176-179, 192-193 differential equation for, 176 fixed-end forces and, 192-194 Maxwell’s law of reciprocal, 616 member stiffness and, 176-185 slope and deflection equations, 193-194 trusses, 616-618 Deformable bodies, principle of virtual work for, 17-20 Deformations, 176-179, 399-407, 411-415, 441-450, 546-552, 596-597 beams, 175-184, 407-410 bending moment M, 176-179, 192-193 deflection, 176-185, 192-195 fabrication errors, 399-407, 411-415 grids, 441-450 nonlinear structural analysis, 596-597 out-of-plane, 441-442 plane frames, 411-415 plane trusses, 402-407, 596-597 shape factor fs, 547-548 shear, 546-552 structure load–deformation relations, 596-597 torsional moment MT, 441-442, 445, 448-449 twisting, 442, 445-450 warping, 441-445 Degrees of freedom, 53-58, 167-169, 255-258, 357-358, 371, 423-424, 440-441, 462-463, 520-524

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Index    639

a nalytical models and, 167-169, 255-258 arrow notation for, 55-56, 169, 257, 441 beams, 167-169 condensation of, 520-524 defined, 53 external, 520-521 free coordinates, 55 free joint, 53 grids, 440-441 hinged joints, 357-358 internal, 520 joint displacement vector d, 54 joint displacements and, 53-55, 167-169 joint load vector, 56-57 kinematic indeterminacy, 54 numbering, 55-56 plane frames, 255-258 plane trusses, 53-58 reaction vector R, 57 restrained coordinates and, 55-58, 252-255, 371 space frames, 462-463 space trusses, 423-424 structure coordinates and, 55, 141-144, 168-169, 257 support displacements, 371 Depth ratio, 556 Diagonal elements of a matrix, 25 Diagonal matrix, 26 Differential equation for beam deflection, 176 Differential operator matrix D, 74-75 Differentiation of a matrix, 34-35 Direct integration method, 192-195 Direct stiffness method, 5, 94-95, 198-207 Displacement, 15-20, 53-55, 68-76, 151-153, 167-169, 172-176, 189, 198-203, 240, 256-257, 259-271, 333, 390-395, 506-507, 582-586, 588-594. See also End displacement; Joint displacement; Support displacement beams, 167-169, 172-176, 189, 198-203 computer analysis of, 151-153, 240, 333, 390-395 deformable bodies, 17-20 degrees of freedom and, 53-55, 167-169 finite-element formulation and, 68-76, 189 Gauss-Jordan elimination method for, 151-152 joints, 53-55, 151-153, 167-169, 198-202, 240, 256-257, 333, 582-586 member force–displacement relations, 588-594 plane frames, 256-257, 259-271, 333 plane trusses, 53-55, 68-76, 151-153

rigid bodies, 15-17 stiffness relations and, 172-176, 198-203, 259-271 strain–displacement relationships, 74-75, 189 stress–displacement relationships, 75, 189 supports, 390-395, 506-507 trusses, 151-153, 582-586, 588-594 virtual work, principle of and, 15-20 Displacement function u, 68-74, 186-188 beams, 186-188 defined, 69 finite-element formulation, 68-74, 186-188 plane trusses, 68-74 shape function N and, 73-74 Do Loop commands, 145-149, 153-159, 236-239, 241-243 E Elementary operations, 40 Elements of a matrix, 24, 25 End displacements u and forces Q, 59-68, 74-77, 153-159, 175-176, 198-201, 261-262, 271-279, 331-335, 394-395, 442-445, 453, 465-477, 482, 538-546 beams, 175-176, 198-201 computer calculation of, 153-159, 331-335, 394-395 coordinate transformations and, 271-278, 453, 472-477, 482 finite-element method and, 74-77 fixed-end forces, 175-176, 198-201 global coordinates, 153-155, 158-159, 279 grids, 442-445, 453 local coordinates, 59-68, 156-158, 261-262, 442-445, 465-472 member forces and, 153-159 member rotation matrix r, 482 member stiffness and, 59-68, 75-77, 156-157, 175-176, 261-262, 279, 442-445 offset connections, 538-542 plane frames, 261-262, 271-279, 331-335 plane trusses, 59-68, 74-77, 153-159 semirigid connections, 542-546 space frames, 465-477, 482 strain–displacement relationships and, 74-75 structure stiffness and, 198-201 support displacements and, 394-395 Equality operations, matrix algebra, 27 Equilibrium equations, 12-13, 62, 92, 179, 198

b eams, 179, 198 member stiffness and, 62, 179 plane trusses, 62, 92 structural analysis relationships from, 12-13 structure stiffness and, 92, 198 Equilibrium matrix b, 617-618 Equivalent joint loads, 211-215, 236-238, 294-296, 327-332, 366-370, 391-394 beams, 211-215, 236-239 computer evaluation of, 236-239, 327-332, 366-367, 391-394 Do Loop commands for, 236-238 member releases, 366-370, 391-394 plane frames, 294-296, 327-332 structure fixed-joint forces Pf and, 211-215, 294-296, 367-370 structure stiffness matrix S and, 236-239 support displacements and, 367-370, 391-394 External degrees of freedom, 520-521 F Fabrication errors, 399-407, 411-415 analytical models for, 402-407, 411-415 initial member straightness, 400-402 initial member strength, 400 member fixed-end forces due to, 399-407, 411-415 plane frames, 411-415 plane trusses, 402-407 procedure for analysis, 402 Finite-element methods, 4, 68-77, 186-192 beams, 186-192 displacement function u, 68-74, 186-188 end displacements u and forces Q, 74-77 formulation of, 68-77, 186-192 matrix methods compared to, 4 member stiffness matrix k, 75-77, 189-192 plane trusses, 68-77 shape function N, 73-74, 188 strain–displacement relationships, 74-75, 189 stress–displacement relationships, 75, 189 symmetry and, 77, 86 virtual work used for, 68-77, 186-192 First-order analysis, see Linear structural analysis Fixed-end forces, 175-176, 192-198, 207-210, 263-271, 284-287, 395-399, 402-415, 468-472 axial forces FA, 264-267 beams, 175-176, 192-198, 207-210, 407-410

Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

640    Index Fixed-end forces (contd.) concentrated loads W causing, 192-198 defined, 175 direct integration method for, 192-195 fabrication errors and, 399-407, 411-415 global coordinate system Ff , 284-287 local coordinate system Qf , 196-198, 263-271, 468-471 member releases, 395-399, 402-415, 470-472 member stiffness and, 175-176, 263-271, 284-287 plane frames, 263-271, 284-287, 411-415 plane trusses, 402-407 sign convention for, 268 space frames, 468-472 structure fixed–joint forces Pf and, 207-210 temperature changes and, 395-399, 402-415 Fixed-joint forces Pf , 207-215, 287-290, 294-302, 367-390 assembly of, 296-302 beams, 207-215 code number technique for, 210, 296-302 defined, 208 equivalent joint loads and, 211-215, 294-296, 367-370 evaluation of due to support displacements, 371-374 member fixed-end forces Qf and, 207-210 member releases and, 367-390 physical interpretation of, 207-210 plane frames, 287-290, 294-302 procedures for analysis, 375-376 structure stiffness and, 287-290, 294-302 support displacements and, 367-390 Flexibility method, 4-5, 613-619 deflections and, 616-618 equilibrium matrix b, 617-618 flexibility coefficient, 616 redundants, 613-616 stiffness method compared to, 613 structural analysis using, 4-5 structure flexibility matrix f, 616-619 truss analysis, 613-619 Force-displacement relationships, 11, 50-53, 90-93, 588-594 Forces, see End displacements u and forces Q; Fixed-end forces; Fixed-joint forces Pf ; Joint load P; Member force calculationsFramed structures, 5-10, 49-128, 129-162, 252-344, 422-503, 511-519, 537

a nalytical models of, 422-503, 537 approximate matrix analysis of, 511-519 beams, 6-7 defined, 5 grids, 6, 438-460 inclined roller supports, 537 plane frames, 6-8, 252-344 plane trusses, 5-6, 49-128, 129-162 space frames, 10, 461-499 space trusses, 7-8, 423-438 structural analysis classification, 5-10 three-dimensional, 422-503 Frames, 6. See also Plane frames; Space frames Free coordinates, 55. See also Degrees of freedom Free joint, 53 G Gauss-Jordan elimination method, 39-45, 151-153 joint displacement d, solution of by, 151-153 linear equations (simultaneous), solutions of by, 39-43 matrix algebra using, 39-45 matrix inversion by, 42-45 plane trusses, 151-153 Geometric linearity, 581 Geometrically nonlinear analysis, see Nonlinear structural analysis Global coordinate system, 50-52, 77-90, 93-94, 153-155, 158-159, 166-167, 253-255, 279-287, 423, 429-430, 440, 451-453, 462, 472-490 analytical models for, 50-52, 166-167, 253-255 beams, 166-167 computer calculations in, 153-155, 158-159 coordinate transformations, 77-86, 271-278, 429-430, 451-453, 472-490 fixed-end forces Ff in, 284-287 force-displacement relationships, 50-52 grids, 440, 451-453 member end forces and reactions, 86, 153-155, 158-159 member rotation matrix r, 477-490 member stiffness matrix K, 86-90, 279-284 member stiffness relations in, 86-90, 93-94, 279-287 plane frames, 253-255, 279-287 plane trusses, 50-52, 77-90, 93-94, 153-155, 158-159 space frames, 462, 472-490 space trusses, 423, 429-430

s ymmetry and, 87 transformation from local system, 80-81, 271-278 transformation matrix T, 79-80, 472, 476-477, 482 transformation to local system, 78-80 Grids, 9, 438-460 analytical models, 438-460 angle of twist f, 445-449 arrow notation for, 441 beginning and end joints, 440-441 coordinate transformations, 451-453 defined, 9, 438 deformation (out-of-plane), 441-445 degrees of freedom, 440-441 end displacements u and forces Q, 442-445, 453 global coordinate system, 440, 451-453 local coordinate system, 441-453 member releases, 449-450 member stiffness relations, 441-450 procedure for analysis, 454-460 Saint-Venant’s torsion constant J for, 445-446 structural analysis classification, 9 symmetric cross sections of, 438-440 torsional moment MT, 441-442, 445, 448-449 torsional stiffness coefficient, 445-447 transformation matrix T, 453 twisting, 442, 445-450 warping, 441-445 H Half-bandwidth of stiffness matrix, 560-564 Hinged joints, 346-348, 355-366 defined, 355 degrees of freedom, 357-358 frames, 355-366 member release type MT, 346-348 Hinges, 346-368 analytical models, 358-366 beam members, 353-366 beginning (MT = 1), 348-350, 353 both ends (MT = 3), 351-352, 354-355 computer analysis of, 366-368 end (MT = 2), 350-351, 353-354 hinged joint analysis, 346-348, 355-366 member types (MT), 346-348 plane frame members, 348-352, 355-366 procedure for analysis, 355 stiffness relations, 348-355 I Identity matrix, 26-27 Inclined roller support analysis, 535-537

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Index    641

Input modules, 130-140, 228-236, 322-329, 390-391. See also Data input beams, 228-236 cross-sectional property data, 138, 232, 326-327 defined, 130 joint data, 133-136, 228-230, 322-323 joint load data, 139, 233, 326 material property data, 137-138, 231-232, 326 member data, 138, 232-233, 326 member load data, 234, 326-327 plane frames, 322-329 plane trusses, 130-140 sample printout, 139-140, 235-236, 328-329 support data, 136-137, 230-232, 324-326 support displacements, 390-391 Integer matrix MSUP, 137, 142-143, 230-231, 326 Integration of a matrix, 34-35 Internal degrees of freedom, 520 Internal joint vector f, 596 Inverse of a matrix, 35-36, 42-45 J Joint coordinate matrix COORD, 133, 228-229, 322 Joint displacement d, 54, 90-97, 151-153, 167-169, 198-202, 240, 256-257, 288-292, 333, 582-586 beams, 167-169, 198-202, 240 computer solution for, 151-153, 240, 333 degrees of freedom and, 54, 167-169, 256-257 Gauss–Jordan elimination method for, 151-153 nonlinear structural analysis, 582-586 plane frames, 256-257, 288-292, 333 plane trusses, 54, 90-97, 151-153 structure stiffness relations and, 90-97, 198, 288-292 trusses, 151-153, 582-586 Joint load P, 56-57, 90-92, 94-95, 139, 149-151, 169-172, 198-201, 211-215, 233, 236-240, 287-288, 294-302, 322-323, 326-332, 366-367, 391-394 analysis module, 149-151, 236-240, 327-332 analytical models, 169-172 beams, 169-172, 198-201, 211-215, 233, 236-240 computer analysis, 139, 149-151, 233, 236-238, 326-332, 366-367, 391-394

defined, 169 degrees of freedom, 56-57 fixed-joint forces Pf and, 294-396 equivalent, 211-215, 236-238, 294-296, 327-332, 366-370, 391-394 input (data) module, 139, 233, 326 member releases, 366-367, 391-394 plane frames, 287-288, 322-323, 326-332 plane trusses, 56-57, 90-92, 94-95, 139, 149-151 storage of in structure load, 238-240, 332 structure stiffness relations and, 90-92, 94-95, 198-201, 287-288, 294-302 support displacements and, 366-367, 391-394 support reactions and, 170 Joints, 11, 53-58, 90-98, 133-136, 151-153, 167-172, 198-203, 207-215, 228-230, 287-292, 294-302, 322-323, 346-348, 355-366, 424-425, 440-441, 462-463, 535-546. See also Fixedjoint forces analysis modules, 151-153, 327-333 analytical modeling, 167-172, 424-425, 440-441, 535-546 beams, 167-172, 198-203, 207-215, 228-230, 355-366 beginning and end, 53, 424-425, 440-441 code number technique for, 210, 294-302 computer analysis of, 133-136, 151-153, 228-230, 322-323 data, 133-136, 228-230, 322-323, 326 defined, 11 degrees of freedom, 53-58, 167-169, 462-463 fixed-joint forces Pf , 207-215, 287-290, 294-302 free, 53 grids, 440-441 hinged, 346-348, 355-366 inclined roller supports, 535-537 input (data) modules, 133-136, 228-230, 322-323, 326 local coordinate system, 53, 536 member type MT, 346-348 offset connections, 538-542 plane frames, 287-292, 294-302, 355-366, 537 plane trusses, 53-58, 90-98, 133-136, 151-153, 535-537 reaction vector R, 57-58 restrained coordinates, 55-58 semirigid connections, 542-546

s pace frames, 462-463 space trusses, 424-425 structure stiffness relations, 90-98, 198-203, 287-292, 294-302 K Kinematic indeterminacy, degree of, 54 L Limit point, 586 Line diagrams, 11 Linear equations (simultaneous), solutions of, 39-43 Linear structural analysis, 20-21, 581, 585-586 Load data, 139, 233-234, 326 beams, 233-234 joints, 139, 233, 326 matrix JP and PJ, 139 members, 234, 326-327 plane frames, 326 plane trusses, 139 Load–deformation relationships, 21, 596-597 Loads, see Concentrated loads; Fixed-end loads; Joint loads Local coordinate system, 51-53, 59-86, 157-159, 166-167, 253-255, 259-278, 424-430, 441-453, 463-484, 536 analytical models, 166-167, 253-255 arrow notation for, 53 beams, 166-167 beginning and end joints, 53, 424-425, 440-441 computer calculations in, 157-159 coordinate transformations, 77-86, 271-278, 427-430, 451-453, 463-484 end displacement u and forces Q, 59-68, 157-159 finite-element formulation in, 68-77 fixed-end forces Q f , 263-271, 468-472 force-displacement relationships, 51-53 grids, 441-453 inclined roller supports, 536 joints, 53, 536 member stiffness matrix k in, 68-77, 262-263, 468-471 plane frames, 253-255, 259-278 plane trusses, 51-53, 59-86, 157-159 prismatic members, 259-271 space frames, 463-484 space trusses, 51-53, 59-86, 157-159, 424-430 stiffness relations in, 59-68, 259-271, 425-427, 441-450, 463-472 structure stiffness matrix S in, 59

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642    Index Local coordinate system (contd.) transformation from global system to, 78-80 transformation matrix T, 80-81, 472, 476-477, 482 transformation to global system, 80-81, 271-278 M Material linearity, 581 Material property data, 137-138, 231-232, 326 Matrix, defined, 24-25 Matrix algebra, 23-48 addition operations, 27-28 column matrix, 25 diagonal matrix, 26 differentiation, 34-35 elements of, 24, 25 equality operations, 27 Gauss–Jordan elimination method, 39-45 identity matrix, 26-27 integration, 34-35 inverse of a matrix, 35-36, 42-45 multiplication of matrices, 28-32 null matrix, 27 order of, 24-25 orthogonal matrix, 36-37 partitioning, 37-38 row matrix, 25 scalar, multiplication by a, 28 square matrix, 25, 35-36 subtraction operations, 27-28 symmetric matrix, 26, 33, 36 transpose of a matrix, 32-34 triangular matrices (upper and lower), 26 unit matrix, 26-27 vectors, 25 Matrix stiffness method, 215-227, 302-322, 367-390 analytical models for, 215-227, 302-322, 375-390 beams, 215-227, 381-386 fixed-joint forces Pf for, 367-390 member releases, 367-390 plane frames, 302-322 procedure for analysis using, 215-216, 375-376, 302-304 shear and bending moment diagrams for, 216, 304-307 support displacements, 367-390 Matrix structural analysis, 1-22 analytical models, 10-11 beams, 6-7 classical methods compared to, 3-4

c ompatibility conditions, 13-14 constitutive relations, 14-15 deformable bodies, 17-20 direct stiffness method, 5 equilibrium equations, 12-13 finite-element methods compared to, 4 flexibility method, 4-5 force–displacement relationships, 11 framed structures, 5-10 grids, 9 history of, 2-3 line diagrams, 11 linear analysis, 20-21 nonlinear analysis, 21 plane frames, 6-8 plane trusses, 5-6 rigid bodies, 15-17 software for, 21 space frames, 10 space trusses, 7-8 stiffness method, 4-5 structural analysis, 1-22 virtual work, principle of, 15-20 Matrix triple product, 279 Maxwell’s law of reciprocal deflections, 616 MDISPG (member global displacement) subroutine, 153, 155, 161, 334, 338, 395 MDISPL (member local displacement) subroutine, 156, 161, 241, 244, 248, 334, 338, 395 Member, defined, 11 Member data, 138, 232-234, 326, 366 beams, 232-234 load data, 234, 326-327 matrix (MPRP), 138, 232-233, 326, 366 member releases, 366 plane frames, 326 plane trusses, 138 Member force calculations, 153-159, 240-246, 333-337, 366-367, 394-395 beams, 240-246 Do Loop commands for 153-159, 241-243 member releases, 366-367 plane frames, 333-337 plane trusses, 153-159 support reactions and, 394-395 Member loads, 169. See also Load data Member releases, 345-421, 449-450, 470-472, 535-537. See also Hinges analytical models, 358-366, 375-390, 449-450, 470-472 beams, 353-366, 381-386, 407-410 computer analysis, 366-368, 390-395 defined, 346

f abrication errors and, 399-407, 411-415 fixed-end forces, 395-415 fixed-joint forces Pf , 367-390 grids, 449-450 hinged joints, 346-348, 355-366 hinges, 346-366 inclined roller supports, 535-537 member types (MT ), 346-348 plane frames, 348-352, 355-366, 386-390, 411-415 plane trusses, 376-381, 402-407 procedures for analysis, 355, 375-390, 402-415 space frames, 470-472 stiffness matrix k and, 347-354 stiffness relations, 348-355, 470-472 structures, 367-395 support displacements, 367-395 temperature changes and, 395-399, 402-415 Member rotation matrix r, 477-490 angle of roll y, 477-484 plane frame analysis, 477-490 reference point, 484-490 Member shape function matrix N, 73-74 Member stiffness, 59-68, 75-77, 86-90, 93-94, 156-157, 172-185, 189-192, 203-207, 259-271, 279-287, 347-355, 425-427, 430, 441-450, 463-472, 546-552 axial forces and, 62-64 beams, 172-185, 189-192, 203-207, 353-355 bending moment M and, 176-179 coefficients k for, 61-64 computer analysis of, 156-157 deflection and, 176-185 end displacements u and forces Q, 59-68, 156-157, 175-176, 261-262, 279, 442-445, 458-463 finite-element formulation and, 75-77, 189-192 fixed-end forces, 175-176, 263-271, 284-287 global coordinate system, 86-90, 93-94, 279-287, 430 global matrix K, 86-90, 279-284 grids, 441-450 hinges and, 348-355 local coordinate system, 59-68, 156-157, 259-271, 425-427, 441-450 local matrix k, 64, 75-77, 176-185, 189-192, 203-207, 262-263, 347-354, 468-471 member releases and, 347-355, 449-450, 470-472

Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

Index    643

plane frames, 259-271, 279-287, 348-352 plane trusses, 59-68, 75-77, 86-90, 93-94, 156-157 prismatic members, 259-271 Saint-Venant’s torsion constant J for, 445-446 shear deformations and, 546-552 space frames, 463-472 space trusses, 425-427, 430 structure stiffness matrix S and, 59, 203-207 symmetry and, 77, 87 torsional moment MT, 441-442, 445, 448-449 torsional stiffness coefficient, 445-447 transformation matrix T, 86 warping, 441-442 Member strain–displacement matrix B, 74-75 Member tangent stiffness matrix Kt, 594-608 MFEDSD (member global fixed-end displacement from support displacement) subroutine, 392-393, 395 MFEFG (member global fixed-end) subroutine, 331, 334, 338 MFEFLL (member local fixed-end force) subroutine, 238, 240-241, 248, 331, 333, 335, 338, 366-368 MFEFSD (member global fixed-end force from support displacement) subroutine, 392-394 MFORCEG (member global force) subroutine, 158, 161, 335, 338 MFORCEL (member local force) subroutine, 157-158, 161, 241, 245, 248, 335, 338, 395 MSTIFFG (member global stiffness matrix) subroutine, 147, 161, 331-332, 338, 392 MSTIFFL (member local stiffness matrix) subroutine, 156-157, 161, 236-238, 241, 248, 329, 331, 334-335, 338, 366-367, 392 MTRANS (member transformation matrix) subroutine, 153, 155-156, 161, 329, 331, 334, 338, 392 Multiplication of matrices, 28-32 N Newton-Raphson computational method, 597-608 Nonlinear structural analysis, 21, 580-610 coordinate system for, 587-588 geometric concepts of, 582-587

geometrically, 587-609 joint displacements d, 582-586 linear analysis compared to, 581, 585-586 load–deformation relationships, 21, 596-597 member force–displacement relations, 588-594 Newton-Raphson computational method, 597-608 plane trusses, 582-610 snap-though instability, 586 stability and, 586-587 structural analysis and, 21, 580-610 tangent stiffness and, 586, 594-608 trusses, 580-610 Nonprismatic member analysis, 553-560 Null matrix, 27 O Off-diagonal elements of a matrix, 25 Offset connection analysis, 538-542 Order of a matrix, 24-25 Orthogonal matrix, 36-37 P Partitioning of matrices, 37-38 Plane frames, 6-8, 252-344, 348-352, 355-366, 386-390, 411-415, 535-537 analysis module, 327-338 analytical models, 253-258, 296-302, 358-366, 386-390, 411-415 axial fixed-end forces FA, 264-267 computer analysis, 322-338, 366-368 coordinate number assignment, 257, 327, 338 coordinate transformations, 271-278 cross-sectional property data, 326-327 defined, 6, 253 degrees of freedom, 255-258 end displacements u and forces Q, 261-262, 271-279, 331-335 fabrication error and, 411-415 fixed-end forces Qf , 263-271 global coordinate system, 253-255, 271-278 hinged joint analysis, 355-366 hinges, 348-352, 355-366 inclined rollers, 535-537 input (data) module, 322-329 joint displacements d, 256-257, 288-292 joint load P, 287-288, 327-332 local coordinate system, 253-255, 259-278 matrix stiffness method for, 302-322, 386-390 member data, 326

member releases, 348-352, 355-366, 386-390, 411-415 member stiffness, 259-271, 279-287, 348-352 prismatic members, 259-271 procedure for analysis, 302-322, 355 restraint codes for, 324-325 restrained coordinates, 255-258 shear and bending moment diagrams for, 304-307 stiffness relations, 259-271, 348-352 structural analysis classification, 6-8 structure coordinate number assignment, 236, 327 structure stiffness, 287-302, 327-332 support data, 324-326 support displacement analysis, 386-390 temperature changes and, 411-415 transformation matrix T, 274-275 Plane trusses, 5-6, 49-128, 129-162, 376-381, 402-407, 535-537, 582-610 analysis modules, 141-160 analytical models, 50-58, 59-68, 106-123, 376-380, 402-407, 535-537 arrow notation for, 53, 55-56 axial forces, 62-64 code number technique for, 98-103 computer analysis, 129-162 coordinate transformations, 77-86 cross-sectional property data, 138 defined, 5, 50 degrees of freedom, 53-58 direct stiffness method for, 94-95 end displacements u and forces Q, 59-68, 153-159 fabrication errors, 402-407 finite-element formulation, 68-77 global coordinate system, 50-52, 77-90, 93-94 inclined roller supports, 535-537 input (data) modules, 130-140 joint displacement d, 54, 90-97, 151-153 joint loads, 56-57, 90-92, 94-95, 139, 149-151 joints, 53-58, 90-98, 133-136, 535-537 local coordinate system, 51-53, 59-86 material property data, 137-138 member data, 138 member end force calculations, 153-159 member force-displacement relations, 90-93, 588-594 member stiffness matrix k, 75-77 member stiffness method for, 59-68, 376-381 member stiffness relations, 59-68, 86-90, 93-94

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644    Index Plane trusses (contd.) Newton-Raphson computation method, 597-608 nonlinear analysis of, 582-610 procedure for analysis, 106-123 restrained coordinates, 55-58, 136 stiffness method for, 59-68 stiffness relations, 59-68 structural analysis classification, 5-6 structure coordinate number assignment, 141-144 structure load–deformation relations, 596-597 structure stiffness, 90-106 structure stiffness matrix S, 59, 144-149 support data, 136-137 support displacement analysis, 376-381 support reactions R, 102-103, 153-159 tangent stiffness and, 586, 594-608 temperature changes and, 402-407 transformation matrix T, 79-81 virtual work used for, 68-77 Primary structure, 613 Prismatic member analysis, 164-192, 259-271 R Reaction vector R, 57. See also Support reactions Redundants, 613-616 Reference point R, member rotation matrix in terms of, 484-490 Restrained coordinates, 55-58, 136, 255-258, 371, 505-511 analytical models, 255-258 degrees of freedom and, 55-58, 255-258, 371 joints, 55-58 numbering of, 55-56, 136 plane frames, 255-258 plane trusses, 55-58, 136 structure stiffness matrix S and, 505-511 support displacements and, 371, 506-507 Restraint codes, 136-137, 230, 233, 324-325 beams, 230, 233 plane frames, 324-325 plane trusses, 136-137, 142-144 Rigid bodies, principle of virtual work for, 15-17 Rigid frames, 6. See also Plane frames Rotation, 477-490. See also Member rotation matrix Row matrix, 25 S Saint-Venant’s torsion constant J, 445-446

Scalar, multiplication of a matrix by a, 28 Semirigid connection analysis, 542-546 Shape factor fs, 547-548 Shape function N, 73-74, 179, 188 beams, 179, 188 defined, 73 displacement function u and, 73-74 finite-element formulation, 73-74, 188 member stiffness and, 179 plane trusses, 73-74 Shear and bending moment diagrams, 216-219, 304-307 beams, 216-219 plane frames, 304-307 Shear deformation, 546-552 Shear deformation constant bs, 550 Slope and deflection equations, 193-194 Snap-through instability, 586 Software, 21, 611-612. See also Computer analysis Space frames, 10, 461-499 analytical models, 461-499 coordinate transformations, 472-490 defined, 10 degrees of freedom, 462-463 end displacements u and forces Q, 465-477 global coordinate system, 462, 472-490 local coordinate system, 463-484 local stiffness matrix k, 468-471 member fixed-end forces Qf , 468-472 member releases, 470-472 member rotation matrix r, 477-490 member stiffness relations, 463-472 procedure for analysis, 490-499 structural analysis classification, 10 symmetric cross sections of, 461 transformation matrix T, 472, 476-477, 482 Space trusses, 7-8, 423-438 analytical models, 423-438 beginning and end joints, 424-425 coordinate transformations, 427-430 defined, 7, 423 degrees of freedom, 423-424 global coordinate system, 423, 429-430 local coordinate system, 424-430 member stiffness relations, 425-427, 430 procedure for analysis, 430-438 structural analysis classification, 7-8 transformation matrix T, 429-430 Square matrix, 25, 35-36 Static determinacy, 613 Stiffness coefficients k, 61-64 Stiffness matrix, see Member stiffness; Structure stiffness matrix

Stiffness methods, 4-5, 59-68, 94-95, 198-207, 215-227, 302-322, 367-390, 613 analytical models for, 59-68, 215-227, 302-322, 375-390 beams, 198-207, 215-227 direct method, 94-95, 198-207 flexibility method compared to, 5, 613 framed structures, 375-390 matrix method, 215-227, 302-322, 367-390 member stiffness matrix k for, 64 plane frames, 59-68, 302-322 plane trusses, 59-68, 376-381 stiffness coefficients k for, 61-64 structural analysis using, 4-5 structure stiffness matrix S for, 59, 95 support displacements, 367-390 Stiffness relations, 59-68, 86-90, 93-94, 172-185, 198-207, 259-271, 279-302, 348-355, 425-427, 441-450, 463-472 axial fixed-end forces FA, 264-267 axial forces, 62-64 beams, 172-185, 198-207, 353-355 condensation and, 524-527 displacement and, 172-176, 259-271 equilibrium equations for, 62, 92, 179, 198 fixed-end forces Qf and Ff , 175-176, 263-271, 284-287 fixed-joint forces Pf , 294-302 global coordinate system and, 86-90, 279-302, 430 grids, 441-450 hinges, 348-355 joint displacements d, 198-202, 288-292 joint load P, 198-201, 287-288, 294-302 local coordinate system and, 59-68, 259-271, 425-427, 441-450, 463-472 local stiffness matrix k, 64, 262-263 members, 86-90, 93-94, 172-185, 279-287 plane frames, 259-271, 279-302, 348-352 plane trusses, 86-90, 93-94 prismatic members, 259-271 space frames, 463-472 space trusses, 425-427, 430 stiffness matrix S for, 290-294 structures, 198-207, 287-302 STOREPF (store member global fixed-end force) subroutine, 238, 241, 248, 331-332, 335, 338, 393-394 STORER (store member forces in support reactions) subroutine, 159, 161, 242, 246, 248, 335, 338

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Index    645

STORES (store structure stiffness matrix S) subroutine, 147-149, 161, 237-239, 248, 331, 338 Strain–displacement relationships, 74-75, 189 Stress–displacement relationships, 75, 189 Structural analysis, 1-22, 59-77, 94-95, 302-322, 367-390, 423-499, 511-535, 553-579, 580-610, 613-619. See also Analytical models; Computer analysis analytical models, 10-11 approximate matrix analysis, 511-519 building frames, 511-519 comparison of linear and nonlinear, 581, 585-586 comparison of methods, 3-4 compatibility conditions, 13-14 condensation, 519-535 constitutive relations, 14-15 decomposition method for, 564-576 defined, 2 deformable bodies, 17-20 degrees of freedom condensed for, 520-524 equilibrium equations, 12-13 finite-element methods, 4, 68-77 flexibility method, 4-5, 613-619 geometric linearity, 581 grids, 9, 438-460 half-bandwidth of stiffness matrix, 560-564 history of, 2-3 inclined roller supports, 535-537 large systems, 560-576 linear, 20-21, 581, 585-586 material linearity, 581 matrix methods for, 1-22 nonlinear, 21, 580-610 nonprismatic members, 553-560 offset connections, 538-542 performance analysis, 2 plane frames, 6-8, 59-68, 302-322 plane trusses, 5-6, 59-77, 376-381, 587-610 rigid bodies, 15-17 semirigid connections, 542-546 shear deformation, 546-552 snap-through instability, 586 software for, 21 space frames, 10, 461-499 space trusses, 7-8, 423-438 stability and, 586-587 stiffness methods, 4-5, 59-68, 94-95, 302-322, 367-390, 613 substructures, 519, 527-535 superposition, principle of, 20-21

virtual work, principle of, 15-20 Structure coordinates, 55, 141-144, 168-169, 236, 257, 327 beams, 168-169, 236 defined, 55 degrees of freedom and, 55, 141-144, 168-169, 257 number assignment, 55, 141-144, 168-169, 236, 257, 327 plane frames, 257, 327 plane trusses, 55, 141-144 vector NCS, 141-144, 236, 327 Structure flexibility matrix f, 616-619 Structure stiffness, 90-106, 198-207, 287-302, 327-332, 519-535, 586, 594-608 beams, 198-207 code number technique for, 98-103, 203-207, 296-302 compatibility equations for, 90-93, 198-199 condensation and relationships of, 519-527 degrees of freedom and, 520-524 direct stiffness method, 94-95, 198-207 end displacements u and forces Q, 198-201 equilibrium equations for, 92, 198 equivalent joint loads and, 294-302, 327-332 fixed-joint forces Pf and, 294-302 joint displacement d, 90-97, 198, 288-292 joint load P, 90-92, 94-95, 198-201, 287-288, 294-302 matrix S for, 95-102, 201-207, 290-294, 296-302, 327-332, 594-608 member force–displacement relations, 90-93 member stiffness relations and, 93-94 plane frames, 287-302, 327-332, 594608 plane trusses, 90-106, 586, 596-597 substructure analysis and, 519, 524-535 support reaction R, 102-103 tangent stiffness, 586 Structure stiffness matrix S, 59, 95-102, 144-149, 201-207, 236-239, 290-294, 296-302, 327-332, 505-511, 560-576, 594-608 analysis module, 144-149, 236-239, 327-332 assembly of, 98-102, 201-207, 296-302 banded structures, 564-576 beams, 201-207, 236-239 code number technique for, 98-102, 203-207, 296-302

computer generation of, 144-149, 236-239, 327-332 decomposition method for, 564-576 Do Loop commands for, 145-149, 236-239 equivalent joint load and, 236-238 half-bandwidth of, 560-564 large system solutions of, 560-576 local coordinate system, 59 member code numbers for, 296-302 member stiffness matrix k and, 203-207 member tangent Kt, 594-608 physical interpretation of, 95-98 plane frames, 290-294, 296-302, 327-332 plane trusses, 59, 95-102, 144-149, 594-608 restrained coordinates and, 505-511 stiffness method and, 59 structure tangent St, 596-597 support displacement vector dR for, 506-507 symmetry of, 95 Structure tangent stiffness matrix St, 596-597 Submatrices, 37-38 Substructure analysis, 519, 527-535 Subtraction operations, matrix algebra, 27-28 Superposition, principle of, 20-21 Support data, 136-137, 142-144, 230-233, 324-326 beams, 230-233 integer matrix MSUP, 137, 142-144, 230-231, 326 plane frames, 324-326 plane trusses, 136-137, 142-144 restraint codes for, 136-137, 230, 233, 324-325 structure coordinates and, 142-144 Support displacements, 367-395, 506-507, 535-537 analytical models, 375-390 beam analysis, 381-386 computer analysis for, 390-395 degrees of freedom, 371 equivalent joint loads, 367-370, 391-394 fixed-joint forces Pf , 367-390 inclined roller supports, 535-537 input (data) module, 390-391 matrix stiffness method for, 375-390 member forces, 394-395 member releases and, 367-395 plane frame analysis, 386-390 plane truss analysis, 376-381 procedures for analysis, 375-390

Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.

646    Index Support displacements (contd.) restrained coordinates, 371, 506-507 structure fixed-joint forces Pf due to, 391-394 structure stiffness matrix S and, 506-507 support reactions and, 394-395 Support reactions R, 57, 102-103, 153-159, 170, 240-246, 333-337, 394-395 analysis module, 153-159, 240-246, 333-337 beams, 170, 240-246 computer analysis, 153-159, 240-246, 333-337, 394-395 degrees of freedom and, 57 joint load P and, 170 member code numbers for, 102-103 member forces and, 240-246 plane frames, 333-337 plane trusses, 57, 102-103, 153-159 structure stiffness and, 102-103 support displacement and, 394-395 Symmetric matrix, 26, 33, 36 Symmetry, 77, 87, 95, 438-440, 461 finite-element formulation and, 77 grid cross sections, 438-440 member global stiffness matrix K and, 87 member local stiffness matrix k and, 76 plane trusses, 77, 87, 95 space frame cross sections, 461 structure stiffness matrix S, 95 T Tangent stiffness, 586, 594-608 defined, 586 load–deformation relations, 596-597 memberm matrix Kt, 594-608

Newton-Raphson computational method, 597-608 procedure for analysis, 599 structure matrix St, 596-597 Temperature changes, 395-399, 402-415 analytical models for, 402-415 beams, 407-410 member fixed-end forces due to, 395-399 plane frames, 411-415 plane trusses, 402-407 procedure for analysis, 402 Three-dimensional framed structures, 422-503 grids, 438-460 procedures for analysis, 430-438, 454-460, 490-499 space frames, 461-499 space trusses, 423-438 Torsion constant J, 445-446 Torsional moment MT, 441-442, 445, 448-449 Torsional stiffness coefficient, 445-447 Transformation matrix T, 79-81, 274-275, 429-430, 453, 472, 476-477, 482 grids, 453 plane frames, 274-278 plane trusses, 79-81 space frames, 472, 476-477, 482 space trusses, 429-430 Transpose of a matrix, 32-34 Triangular matrices (upper and lower), 26 Trusses, 5-8, 49-128, 129-162, 423-438, 535-537, 580-610, 613-619. See also Plane trusses; Space trusses computer analysis of, 129-162 coordinate transformations, 77-86

defined, 5 deflection of, 616-618 finite-element formulation for, 68-77 flexibility method for, 613-619 inclined roller supports, 535-537 joint displacements, 582-586 matrix stiffness method for, 50-68, 86-106 nonlinear structural analysis, 580-610 plane trusses, 5-6, 49-128, 129-162 space trusses, 7-8, 423-438 static determinacy of, 613 structural analysis classification, 5-8 Twisting, 442, 445-450 U Unassembled flexibility matrix fM, 617 Unbalanced joint force vector DU, 598 Unit matrix, 26-27 V Vectors, 25 Virtual work, 15-20, 68-77, 186-192. See also Finite element method beams, 186-192 deformable bodies, 17-20 finite-element method using, 68-77, 186-192 plane trusses, 68-77 principle of, 15-20 rigid bodies, 15-17 strain and stress vectors of, 68 structural analysis using, 15-20 W Warping, 441-445

Copyright 2022 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it.