Mechanics - of - Machines - Samuel - Doughty, Manuais, Projetos, Pesquisas de Mecatrônica

Baixe Mechanics - of - Machines - Samuel - Doughty e outras Manuais, Projetos, Pesquisas em PDF para Mecatrônica, somente na Docsity! Mechanics of É Machines Samuel Doughty EE | Professor of Mechanical Engineering ) University of Wisconsin—Platteville WILEY John Wiley & Sons, inc. New York - Chichester » Brisbane + Toronto + Singapore Copyright 6 1988, by John Wiley & Sons, Inc, All rights reserved. Published simultaneousky in Canada. Reproduction or translation of any part of this work beyond that permitted by Sections 107 and 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlaw(ul. Requests for permission or further information should be addressed to the Permissions Department, John Wiley & Sons. Library of Congress Cataloging-in-Publication Data Doughty, Samuel. Mechanics of machines. Bibliography: p. 1. Mechanical engineering. 1. Title, TJ170.D68 1987 621.8 87-23042 ISEN 0-471-84276-1 Printed in the United States of America 1098765432 To Ann Elizabeth and Susan Beth  viii Preface up in Chapter 9, comparing static, Kinetostatic, and dynamic approaches to the problem: The appendices provide supplementary and review material on matrices, numerical methods, and other topics in mechanics and mathe- matics. From the very beginning, the book is thoroughiy computer-oriented. The student is encouraged to look at all problems — whether examples in the text or homework problems-—from the standpoint of preparing them for a computer solution. In some cases, the student will actually carry the solution all.the way through to a computer solution; in other cases, the student will only formulate the problem in a manner appropriate for com- puter solution. (There are also many problems to be solved completely in closed form for practice without the need to have a computer available.) Although computers have been around for many years, the last decade has seen a real revolution in computing with the development of microcompu- ters. It is reasonable to think that every engineer has or shortly will have free access to a computer of some sort, and thatto be computer is most likely a microcomputer. Most microcomputers today have the capacity to solve the kinds of problems encountered in the mechanics of machines. It goes without saying that any modern mainframe computer is sufficient as well, sa ihe choice of whether to use a mainframe or a microcomputer is mainly dependent on what is available, In terms of programming languages, virtu- ally ali microcomputers support some form of BASIC, and some support other languages such as FORTRAN and Pascal as well. The universality of BASIC makes it a useful language for the examples in this text, but there is nothing about the methods of analysis that is tied to BASIC. Any of the methods presented here can be performed satisfactorily using any high- level programming language. Most of the programs presented in BASIC in the body of the text are also given in FORTRAN in Appendix A10. This textbook takes the position vector loop approach to position anal- ysis, in which the configuration of the mechanism is described in terms of closed position vector loops. Scalar components of these vector equations are used because these are solved more casily for the required information using numerical techniques. The derivatives of these same position vector loops provide the velocity and acceleration descriptions for the mecha- nism. The numerical techniques required to solve these systems of equa- tions are discussed where required, as well as in detailed presentations in Appendices 1 and 2. Position-dependent velocity coefficient functions are defined, relating the secondary velocities to the primary velocity. Both the velocity coefficients and the velocity coefficient derivatives are required to express the accelerations. The velocity coefficients and the velocity coeff- cient derivatives are used to connect the kincmatics to the later work in kinetics, In the area of kinetics, a major objective of this book has been to provide an energy-based approach, appropriate to the course level. In Chapter 8, multidegree of freedom dynamics is presented using the well- known Lagrange equation as the basis. The presentation is directed toward Preface ix application of the Lagrange equation to machine systems, so the derivation is deferred to Appendix A9. For single degree of fizedom mechanisms (Chapter 7), the less well-known Eksergian's equation is presented. E pro- vides a powerful tool that offers useful insight into the way various param- eters affect the motion. Chapter 6 presents statícs from the viewpoint of the Principle of Virtual Work. This method is valuable in its own right and aiso as preparation: for the dynamics presentations of Chapters 7 aná 8. The discussion on cams is approached first from the standpoint oí design, by looking at some length into the problem of analytical design of a cam profile to generate a prescribed follower response. The required do- sign decisions are identified, and their effects on size, siate of stress, and other maiters are considered. The later part of Chapter 4 deais with the analysis of prescribed cam and follower systems, The typical machine design textbook contains quite a bit of discussion on the design of gears, often including material on involute kinematics, gear tooth strength, and train ratio calculations. An important kinematic design question that is missing from most textbooks on kinematics or chine design is the matter of. how to determine workable combinations of gear tooth numbers to achieve a specified train ratio. Design approaches to this problem are included in the discussion on gears and gear traine i Chapter 5. There are several ways to organize a course using this book. There is certainly more material included than can be covered in a one-semester course, At the University of Wisconsin— Platteville the course is usually taught using Chapters 1 through 3 for kinematics, part of Chapter 4 on cams, Chapter 5 on gears, Chapter ó for virtual work, and dynamics using Chapter 7. If cams and gears are covered elsewhere, a very satisfaciory course can be constructed using Chapters 1 through 3 for kinematics, and Chapters 6 through 9 for kineties and analysis of forces. Another variation is to focus an single degree of freedom systems using Chapters 1 and 2 for kinematics, Chapter 4 dealing with cams, gears in Chapter 5 (excluding planetary systems), and Chapters 6 and 7 for kinetics. The mais: consides- ations in structuring a course are to be sure to get the necessary rmaterial regarding position analysis, velocity coefficients, and velocity coefficient derivatives— presented in Chapters 2 and 3—-that are required for all of the work ir kinetics, Chapters 6 through 9. Most of the problems at the end of each chapter ask the student to prepare the description for an eventual computer solution. One of the early steps on such preparation is the choice of appropriate variabies to describe the problem. In most cases, there are several possible choices. For some problems all correct choices are equally useful, whereas for other problems one choice is definitely better than the others. The ability to choose a workable set of variables, preferably the best set, is a skill that is only learned by practice, Consequently, the need to avoid guiding the choice of variables makes it virtually impossible to tabulate answers to the problems. Complete solutions are, however, available in the Instructor's Manual, Jtis x Preface also important for students to develop an ability to validat: their own mathematical descriptions; this is rarely ever done when a student is work- ing toward an answer tabulated in the back of a textbook. The mathematical level of the book is appropriate to junior-level engi- neering students who are well grounded in caleulus. A firm grasp of differ- ential calculus is absolutely essential to understanding and applying the methods developed in the text, and integral calculus is essential for under- standing many of the derivations. Matrix notation is used extensively, but only a very modest level of knowledge is reguired in this area. The review of matrix methods provided in Appendix Al coyers all of the topics re- quired for use here. Many of the examples and homework problems are drawn from my own industrial experience and consulting assignments. From a teaching viewpoint, my experience has been that students are attracted to problem descriptions that resemble real machines. I believe that anyone teaching this material and wishing to draw on personal industrial experience will find this book complementary to that effort, Comments and suggestions from those using the book will be welcome so that it may be improved in the future. Samuel Doughty Contents CHAPTER 1 Introduction 1.1 General Introductory Comments 12 Degrees of Freedom 1.3 Use of Matrix Notation 1.4 Computer-Aided Problem Solving 1.5 Computer Languages 1.6 Conclusion Problem Set CHAPTER 2 Single Degree of Freedom Linkages 2.1 An Overview of the Process 2.2 Kinematics of the Slider--Crank Mechanism 23 Kinematics of the Four-Bar Linkage 2.4 Constraints ' 2.5 Multiloop Single Degree of Freedom Mechanisms” . 2.6 General Kinematic Analysis for Single Degree of Freedom Mechanisms 2.7 Conclusion Problem Set CHAPTER3 Multidegree of Freedom Linkages 3.1 Kinematic Analysis in Closed Form 3.2 Kinematic Analysis with Numerical Solution NO 14 14 23 3 4 48 55 58 59 85 85 90  xii Contents 3.3 Numerical Solution for Multiloop Mechanisms 3,4 General Analysis for Multidegree of Freedom Mechanisms 35 Conclusion Problem Set CHAPTER 4 Cam Systems 4.1 Introduction PastT Cam Design 4.2 Displacement Functions and Graphical Cam Designs 43 A Kinematic Theorem for Rigid Bodies 44 Analytical Design of a Cam with a FlatFaced, Translating Follower 45 Analytical Design of a Cam with an Offset, Translating Roller Follower 4.6 Aralytical Design of a Cam with a Pivoted, Flat-Faced Follower 4.7 Analytical Design of a Cam with a Pivoted, Roller Follower Partil Cam Mechanism Kinematics 4.8 Eecentric Circle Cam 4.9 Angle Between Radius and Normal 410 Follower Response for Arbitrary Cam Profile 411 Condusion Problem Set CHAPTER 5 Gears 51 Introduction 5.2 Velocity Ratio 53 Conjugate Profiles 5.4 Properties of the Involute 5.5 Involute as a Gear Tooth 5.6 AGMA Standards and Tooth Proportions 5.7' Contact Ratio, Interference; and Undercutting 5.8 Simple and Compound Gear Trains 5.9 Planetary Gear Trains 5.10 Gear Train Design 511 Concusion Problem Set 95 102 105 * 106 117 17 120 127 128 143 146 150 151 157 159 164 165 174 174 175 179 180 183 187 188 191 195 200 209 210 Contents CHAPTER 6 Statics and the Prínciple of Virtual Work 6,1.. General Comments 62 Principle of Virtual Work 6.3 “Applications of the Principle of Virtual Work 64 Another Look at Virtual Work 65 Condusior Problem Set CHAPTER7 Dynamics of Single Degree of Freedom Machines: Eksergian's Equation of Motion 7.1 Kinetic Bnergy of a System of Rigid Bodies 7.2 Generalized Forces 7.3 Eksergian's Equation of Moon, 74 Potential Energy Representation of Conservative Forces 7.5 Mechanism Simulation 7.6 Mechanism Simulation Examples 7.7 Conclusion Problem Set CHAPTER 8 Dynamics of Multidegree of Freedom Machines: The Lagrange Equation of Motion 8.1 Kinetic Energy for a Multidegree of Freedom Machine 8.2 First General Form for the Lagrange Equation 83 Second General Form for the Lagrange Equation 84 Applications of the Lagrange Equation 8.5 Multidegre of Rreedom Simulation Examples 8.6 An Alternate Approach to First-Order Equations of Motion for Multidegree of Freedom Systems 8.7 Concusion Problem Set CHAPTERO9 Reactions and Intemai Forces 9,1 General Comments 9.2 Three Approaches to the Problem 9.3 Example Slider-Crank Force Analysis 9.4 Calculation of Reactions Alone 9.5 Conclusion Problem Set 216 216 27 231 234 236 237 252 252 255 256 257 258 262 278 278 281 221 293 296 296 302 319 322 323 336 336 337 338 348 350 351 2 Introduction Bether, and implementation of the resulting descriptions via the digital computer, is a relatively recent synthesis. In using this book, it is well to be prepared to recognize these four concepts whenever they appear. For this purpose, they are introduced here. Kinematics is the study of motion without regard for the forces in- volved in the motion. In the area of kinematics, the first of the four major concepts is the use of Position Vector Loop Eguations, In typical machinery systems, the components form closed loops that change shape as the com- ponents move, but remain closed; each loop can be described as a vector sum identically zero at all times. If enough information has been specified to determine the mechanism configuration uniquely, these vector equa- tions, or their scalar equivalents, can be solved for all the remaining posi- tion variables. The details of this process will be discussed at length later, but for the present, it is important to remember that the position vector loop equations provide a means for determining all required position infor- mation. The use of Velocity Coejficients and Velocity Coefficient Derivatives is the second major concept, again originating in the general subjebt area of kinematics. Many readers will be Alias with the fact that for two pulleys connected with a taut belt, the rotation rate of the drive pulley and that of the driven pulley have a fixed ratio; this is a conseguence of the invariant geometry of the system. The actyal value of the speed ratio is fixed by the particular geometry of the systenho in this case the pulley radii. In a system where the geometry varies as the system moves, such as a slider-crank assembly, the ratio of the output to input speeds is variable, depending on the instantaneous position of the system, The key point here is that the speed or velocity ratio is dependent on tho position but is independent of the actual speed. Thus, the speeds of all points can be expressed as appropriate position-dependent, velocity coefiicient functions multiplying à common reference speed. The situation regarding accelerations is semewhat more complicated, but it will be shown that for many systems the acceleration is the sum of two terms. One term involves the acceleration of the reference point and the position-dependent velocity coefficient, and another term involves the derivative of the velocity coefficient and the square of the reference speed. Therefore, all speeds and accelerations are rather simply relateq to the speed and acceleratign of a reference point through the use of the position-dependent velocity coelficients and velocity coefficient deriva- tives. “ The third concept, usually associated with the area of statics, is called the Principle of Virtual Work. It is one of the oldest energy principles of mechanics, and describes the corditions for eguilibrium in a manner fully equivalênt to the more familiar statements regarding the vanishing of the force and moment sums on a body. When force and moment sums are computed for a body such as a dam or a tower, there is no difficulty in that application because the position of the body is specified. In many machinery 1.2 Degrees of Freedom 3 problems, the positions of the components under load are not specified in fhe beginning, but must be determined as part of the solution. This is just the situation for which the Principle of Virtual Work is best súited.. Its “application is heavily dependent ori thê Iinematics of thé system"únder consideration and the ideas of functional dependence and independence. These needs are well accommodated through the use of velocity coeffi- cients, The dynamics of a machine system can be described by applying New- ton's Second law, or by using equations of motion derived from energy considerations. Thus, the fourth major concept is the use of Energy-Bascd Eguations of Motion. These equations apply equally well to conservative and nonconservative systems. The wellknown Lagrange equations of motion apply to systems with any number of independent coordinates. For prob- lems invoiving only one independent coordinate, there is an alternate en- exgy formulation available, which is associated with the name Eksergián. This method is often easier to use and gives more insight than the La- grange equation. In using either of these approaches, the expressions for the kinetic and potential energies as well as those for the virtual work of the external forces are central. The Kinetic energy expression is greatiy facilitated through the use of the velocity coefficients, The velocity coeffi- cients and the potential energy are, in tum, dependent on the system position that is determined from the position vector loop equations. Each of the four major concepts has a long history, and they can certainly be applied together in manual calculations. The greatest benefit comes, however, through joint computer implementation of these con- cepts. The application of the positioá vector loop equations often leads to a very complicated system of nonlinear transcendental equations. In such cases, closed-form solutions are virtually impossible! but numerical solu- tions by iterativo techniques are usually readily available, The calculation of the velocity. coefficients involves the simultaneous solution of a system of linear equations. Although in concept this is straightforward, it becomes excessively laborious for more than two equation. When appiying the principle of virtual work, it is often necessary to solve a system of nonlinear equations, and numerical solution by an iterative technique usirig a com- puter is the only easy way to achieve à solution. Itis important to note that the equations of motion for most mechanisms are highly nionlinear differ- ential eguations, irrespective of the manner in which they are obtained. Consequently, solving them through analytic methods is usually not possi- ble; numerical integration is required and this is practical only when per- formed by a computer. : 1,2 DEGREES OF FREEDOM The term “degrees of freedom” has long been used in classical dynamics to refer to the number of independent variables that are required to describe a  & Introduction mechanical system. The concept is introduced at this point, and under- standing of the term will grow with progress through the discussions in the following chapters. Consider a particle that is free to move relative to a rectangular Carte- sian coordinate system in two dimensions. If two numbers are specified that represent the X and Y coordinates of the particle, then the position of the particle relative to the coordinate system is fully described. Alterna- tively, polar coordinates could be specified for the particle, again two val- ves. In either case, two items of information are required, and the particle “15 said'to have two degrees of freedom, often abbreviated ZDOF. Fora rigid body in the same two-dimensional space, there are three degrees of free- dom. One way to consider these degrees of freedom is to associate two of them with the coordinates of one point on the body, and to associate the third, degree of freedom with an angle describinig the orientation of the body. Ée sure to notice that the number of degrees of freedom speaks of the number of coordinates required, not about any particular choices for the coGrdinates. The angle used is only a coordinate in the sense that it partially describes the position of the body, although not in the sense of a Tectangular Cartesiar coordinate. * There are many other ways to associate the degrees of freedom with “eoordinates describing the position of the body, but there are two impor- tant requirements to be satisfed by any such choice. The first is that the entire set of coordinates is suíficient to describe the position completely. The second is that each coordinate is independent. Using the association deseribed above for the rigid body, the coordinate X can be varied without any change required in Y or the angle; the same statement can be made about cach of the other coordinates, so they are independent. A set of coordinates that is (1) complete in that they are sufficient to specify the system position fully and (2) cach independent of the others is called a set of generalized coordinates, Generalized coordinates may include rectangular Cartesian coordinates, angles, arc lengths, or any other measures that contribute to the description of the system configuration, subject to the requirement that they form a complete, independent set. For the present, the number of gencralized coordinates should be considered equal to the númber of degrees of freedom. Exceptions to this statement will be dis- cussed later under the topic of “Constraints.” For machinery kinematics problems, there will usually be more kine- imatic variables than degrees of freedom. After choosing generalized coor- dinates to be associated with the degrees of freedom and considered as primary variables, the remaining kinematic variables will be called secondary vêriables, For purely Kinematic problems, the primary variables are consid- eréd as assigned values (“inputs”); whereas the secondary variables are among the problem unknewns. The position vector loop equations provide the means to determine the secondary variables based on the assigned values for the primary variables (generalized coordinates). In the case of a dynamics problem, the governing differential equations describe the man- ary coordinates, 1.4 Computer-Aided Problem Solving 5 nex in which the generalized coordinates (primary variables) vary with time and, again, the position vector loop equations determine the second- 1.3 USE OF MATRIX NOTATION In solving mechanics ofmachines problerns, there are many physical quam- tities of a vector character to consider. It is often convenient to use vector notation to analyse such a problem, but when it comes time to compute, such notation is not suitable. The digital computer can not deal with veo- tors as such. It can, however, accommodate one-dimensional arrays, and these can represent a vector quite well. If the vector of interest is à position vector in a two-dimensional space, then a (2 X 1) column matrix, repre- sented by a (2 X 1) array in the computer, can store each component oí-the vecior. There are other times when many variables are described simply as Kas Xas o o, Xp where n is a positive integer; this case can also be repre- sented by a column matrix that transiates to an (n X 1) array in the com- puter. In fact, systems of linear algebraic equations of any size can be neatiy represented and manipulated in the matrix form! o Matrix notation is only a notation but it has much'to recommend it. By Itself, it does nothing that cannot be done in other ways, butit provides an orderly, systematic means tó deal with the computationel bookkeeping problem. it adapts so seadily to computer implementstion that if a problem is formulated using matrix notation, the computer code required for the solution is usually quite easy to write. 1.4 COMPUTER-AIDED PROELEM SOLVING Engineering analysis always revolves around problem solving of one type or another, and the computer can be a great aid in many cases. However, the computer can never do the entire job, and the místaken attitude that it can often leads to much lost time! This misconception becomes rather obvious when the various steps to problem solving are considered, so a quick reviewr of the process may be beneficial. - The typical steps in solving a mechanics of machines problem are as follows: 1. Problem Definition. This is the task of determining what problem is to be solved and what information is available to begin that solution. Many extors are made at this point, and these errors usually Icad to major time loss. 1 Throughout this book, matrix notation is used extenstvely. The reader who is nol formiliar veith matrices should review Appendix Al. This appendix provides a summary of the neces sary aspects of matrix notation and manipulation. For further details, a text on matrix algebra should be consulted. 6 Introduction 2, Conceptual Solution. Based on the information gathered in step 1, an overall plan of attack must be established, This plan is formulated with the assumption that no unexpected difficulties will arise; for this reason, it may require modification. Nevertheless, there is no point to begin- ning calculation until a tentative route to the solution is determined. If a computer solution is indicated, a roúgh flow chart should be drafted at this point. 3. Detailed Analysis. The conceptual solution process often presumes that the governing kinematic equations can be written; this is the point at which they are actually written. AlLof the governing equations for the problem are assembled and manipulated to set the stage for the compu- tational work that follows. Note that this stép often involves the substi- tution of one equation into another, the use of calculus, and the recast- ing of equations into matrix fôrm. These are all operations, however, that cannot be done readily in the computer. At the end of this step, detailed equations for use ih a computer program should be available. 4. Computer Implementation. This is the time for writing and executing computer programs: If adequate preparation has been made, this pro cess will move quite zapidly. Note that it is often a good idea to test any program after the first major calculation has been programmed. As an example, in a problem involving a complete kinematic analysis for a linkage, it is usually worthwhile to check the position solution and debug as necessary before programming the velocity and acceleration solutions. 5. Interpretation of the Results. This is the most essential part of any analy- sis. Engineering analysis is always done for a purpose—-to answer some type of question. Typical acceptable answers are (a) “The maximum velogity is 187 in,/s,” (b) “The minimum bearing force occurs when the crank position is 1.277 radians,” or (c) “The arm will fail due to exces- sive dynamic loading.” It is never sufficient to submit a stack of com- puter output as an engineering report. Itis, however, always necessary to Graw a conclusion from the arialysis, and to provide the analysis and the computer results that support the conclusion. The foregoing steps are typical and, as such, one may need to deviate from them during the course of solving some problems. It should be noted that the computer is actually used only in step 4. The availability of the computêr is assumed in steps 2 and 3 and this availability influences the overall planning (step 2) and the final form sought in the analysis (step 3). In days before the general availability of high-speed digital computers, the computational step was one of thelongest; now it should not require much more time than any other step. 1,5 COMPUTER LANGUAGES There are a variety of computer languages used for engineering computa- tions. For many years, FORTRAN was the standard engineering language, Problem Set 7 but more recently others such as BASIC and PASCAL have come into wide use, , In developing an engineering textbook, it is often useful to include sample programs that detail the logic required to implement varicús.soli- tion techniques. The BASIC language has been used for all programs in the body of this text. Similar programs in FORTRAN are provided in Appendix AJ of the text, IÉ the user prefers to work in yet another language, it is a fairly simple matter to transfer the logic of the program from one high-level language to another. There is nothing about the computer techniques that is tied to any particular computer language. As mentioned previously, matrix notation and the associated matrix operations will play a major role in the solution of mechanies of miachines problems. Some versions of BASIC have included matrix operations such as “MATA = B*C”, which will form the matrix product of the arrays Band, C and store the result in the array A. If available, these operations should certainly be used; they are usually written in machine lariguage and will execute very quickly. Unfortunately, most versions of BASIC for micro- computers do not include these matrix operations. To facilitate their use, however, we provide a set of matrix operations subroutines in BASIC in Appendix A1.6; corresponding subroutinés in FORTRAN are given in Ap- pendix A10. These subroutines may be keyed in and debugged a single time with the final results stored on diskette. Development of computer programs that require matrix operations begins with copying these subrou- tines. Note that the BASIC routines begin at line 5000, so there is adequate room for most programs to be stored ahead of the matrix operations rou- times. The FORTRAN matrix operations subroutines should be compiled and called exactly as any other FORTRAN subroutines. 1.6 CONCLUSION The discussions about the use of matrix notation, the steps in computer- aided problem solving, and the matter of computer languagés indicate the general flavor of the material in the remainder of this book. The material will be presented at a moderately sophisticated mathematical level, with a clear orientation toward computer implementation of the analytical results. The four major concepts underlying much of the presentation have been discussed; it will help to be alert to these as they appear throughout the book, and to look for the interplay between these ideas. The discússion on degrees of freedom and generalized variables has laid the foundation of an understanding that will be built upon throughout the remainder of the book. PROBLEM SET Determine the number of degrees of freedom for each of the following systems, and fully document the reasoning in each case. In making this determination, make the following assumptions:  12 Introduction Stationary Pivot Problem Set 13 CHAPTER 2 Single Degree of Freedom Linkages 2.1 AN OVERVIEW OF THE PROCESS This chapter will address kinematic analysis of linkages that have onty one degree of freedom. Recall from Section 1.2 that one degree é freedom indicates that only one primary variable needs to be specified to fully define the position of all parts of the mechanism. For a kimematic analysis, the one degree af freedom is associated with some suitable coordinate and considered as an-assigned value (an “input” to the problem. In some cases, a problem statement may ask for kinematic solutions throughout some specified range of the primary variable, such as over one crank revo- lution, but this, in fact, asks for the solutions for a sequence of problems — one for each crank position considered. The value of using a computer to carry out these repetitive solutions is obvious. The kinematie analysis of a mechanism is usually understood to mean development of equations describing the position, velocity, and accelera- tion at all points of interest'in thg mechanism for chosen values of the primary variable, its speed, and its acceleration. In many cases, the posi- tion, velocity coefficients, and velocity coefficient derivatives (concepts not yet defined) will actually be preferable, particularly if the kinematic solu- tion is for use in a static or dynamic analysis. ' This. section .will present the process in'some; detail by means ofa simple example. Later sections in this chapter will present two important common cases, the slider-crank mechanism and the four-bar linkage. The introduction of constraints deseribing sliding and rolhng are considered in Section 2.4. This is followed by a section on single degree of freedom mechanisms involving multiple loops. The final section presents a general- 14 : 2.1 An Overvieio of the Process 15 ized development of the whole kinematic analysis process for single degree of freedom systems. For the first example; consider the Crank-Lever Reciprocatinig Drive shown pictorially in Figure 2.1 and in “kinematic skeleton” form in Figure 2.2, The crank pivoted at P; rotates throtgh a full circle causing the arm pivoted at P; to have a reciprocating rotational motion, Figure 2.1 shows the crank tip sliding in a slot in the arm for this case, whereas the kinematic skeleton, Figure 2.2, shows the arm as a rod passing through a block at the crank tip. The device could actually be built cither way and the two con- structions are kinematically completely equivalent. This simple mechanism is found in many feeder devices and other machines. The center to center distance C and the crank radius R, labeled in Figure 2.2, are known dimen- sions. With Cand R known, if a value is given for the crank position angle q, the mechanism configuration is fully specified, so this is evidentiy a single degree of freedom (SDOP) system. The position variables that re- main unknown are the arm angle, 4, and the location of the crank-arm contact, length B. The first step in the analysis will be to establish the position vector loop equations, Position Analysis. Consider the three position vectors B, C, and R shown in Figure 2.3. From the closure of the triangle, it is evident that B-R-C=0 or, in scalar form, BcosÃA-— Rcosg— C=0 horizontal component BsinA-Rsing-0=0 vertical component These are the (nonlincar) position vector loop equations for this mecha- nism. The input q and thé two dimensions C and Rare known, whereas the * variables A and B are unknown, The next step is to solve the equations for A and B. Often this step will require numerical solution, but not always. The possibility of an analytic solution should always be considered, and in this case this can be done: = FIGURE 2.1 Pictorial Representation of CrankSlot Recipro- cating Drive q |  16 Single Degree of Freedom Linkages FIGURE 2.2 Kinematic Skeleton Representation of Crank-Slót Reciprocating Drive Eliminating B between the two equations gives : Resina inÃA = Rcosq from which A can be determined nsing the principal value of the are- tangent function. With A known, the distance B can be readily computed by solving either equation for B; po CtRcso po Rsing cos À sin 4 Note that there will be times when sin A is zero and, for some proportions of the linkage, cos 4 can also go to zero. Even if both of these events can cecur, they will hot happen simultaneously, so that B can always be evalu- ated using one of the above expressions. For computer implementation, an appropriate test (LF ABS (COS(A))<0.1 THEN ++. +++) should be in- cluded to choose the second expression for B in the event that the first is approaching an indeterminant condition. JÉ the calculations are to be done numerically, this is quite far enough. K further analysis is to be done with these results, then it may be of value to eliminate À algebraicaly. Velocity Analysis. In performing the velocity analysis, it is assumed that the results of the position analysis are available as well as the original FIGURE 2.3 Position Veclor Loop for Crank-Slot Recipro- cating Drive 2.1 An Queroiew of the Process 17 known data. For this problem, this means that the known values now include q, C,R, À, and B, while the new unknowns to be addressed are À and B. The velocity loop equations are obtained by differemtiation of the position loop equations, thus: B cos A -BAsin A+ Rjsing=0 ÊsinA+BÁcosA-Rjcosq=0 in view of the quantities already known, these equations are actually a pair of simultaneous linear algebraic equations in the two unknowns, À and B. This is more cleaily evident if the equationis ave cast in matrix form cos A -Bsin AB I-sing) . Rd! sina BeosAlA t cosg To solve these equations analytically, it is necessary to premultiply by the inverse of the coefficient matrix on the left.! Note first that the determinant of the coefficient matrix is simply B. Making the premultiplication by the inverse of the coefficient matrix gives Bj Rg[BcosA Bsin Ajf-sing] al ÉB l-snA cos all cos al — fRsm(4-9) | “umcsta-a) Note that this shows that each of the unknown velocíties, À and É, is given by the product of q with 2 factor dependent on the position. The velocity coefficients for this case are BG = Kg) = Rsin(A — q) Alj = Kg) = (R/B) costa — 9) For a specific value of 9, the corresponding values of À and B axe readily calculated from the solution above; without specifying the value of 5. the velocity analysis may be made in generalized forma by evaluating the posi- tion dependent velocity coefficients, Ko(9) and Ki(g). Acceleration Anatysis. To develop the acceleration analysis, the results of boih the position and velocity analyses are presuimed to be known. Thus, the list of known information now includes q, C,R, A,B,q, A, and B. The unknowns to be determined in this step are À and É. There ave two approaches to the acceleration analysis; both will be described here because cach gives a different insight. First, differentiation à For the case of imo eguations and the associated (2 X 2) coefficient matrix, the Appendix ALS provides the system solution in terms of a closed-form inverse. » 22 Single Degree of Freedom Linkages many machinery components do not move about fixed centers. A method applicable to those situations will be demonstrated in Sections 2.2 and 2.3. Summary of Observations from the Example Problem The kinematic analysis of this SDOF system involved the following steps: 1, Developing the position equations and solving those equations for the secondary variables; 2. Differentiating the position equations to obtain the velocity equations that are solved for the secondary velocities, or for the velocity coeffi- cients; 3. Determining secondary accelerations by solving the equations resulting from differentiation of the velocity equations. This may be done by the straightforward differentiation indicated, or by using the velocity coeffi- cients and velocity.coefficient derivatives, obtained by differentiation from step-2; , 4. Defining body coordinates to specify the location of a particular point of interest; 5. Determining base coordinates for the point of interest in terms of the body coordinates, secondary variables, and the primary variable; 6. Determining velocity components, acceleration components, velocity coefficients, and velocity coefficient derivatives as required, beginning with the base coordinate expressions. The position loop equations were written in a scalar form that could be expressed as fila, A, B)=0 horizontal component fXg A, B)=0 — vertical component When the velocity equations were cast in matrix form, the coefficient ma- trix for the unknown velocities was 3h dh 9B 04 [E A -Bsin 4 9] |snA4: BcosA Sá 3h 3B This is calted the Jacobian matrix for this system, This same Jacobian matrix is seen again as tlie coefficient of the unknowi-aecelerations for the acceler- ation equations written in matrix form. If a numerical solution had been required for the position equations, the Jacobian matrix would have been used in the numerical solution process (see Section 2.3 and Appendix A3). The occurrence of the Jacobian matrix in these three different relations is not accidental; it should be expected, and its failure to appear at these 2.2 Kinematics of the Slider-Crauk Mechanism 23 points is an indication of an error in the analysis. Whenever the Jacobian is to be evaluated numerically the same code, in the form of a subroutine, should be used for each evaluation. In each case, the system of eguations (position loop equations, velocity loop equations, “or acteleration loop equations) is solvable only so long as the determinant of the Jacobian matrix is non-zero. Positions for which the Jacobian determinant is zero are known as singular points; these positions require special treatment. 2.2 KINEMATICS OF THE SLIDER-CRANK MECHANISM One of the most important common mechanisms is the slider-crank mech- anism. It is found in pumps, compressors, steam engincs, feeders, crush- ers, punches, and injectors. The slider--crank mechanism is central to the diesel and gasoline internal combustion engines that are so much a part of modem life. In most: cases the crank rotates continuously in the same direction, although in some cases the crank motion may be oscillatory. The analysis presented here is in a general form so that the results apply to any slider-crank device. The kinematic skeleton for a typical slider-crank mechanism is shown in Figure 2,6, The crank turns about a fixed pivot at the origin of coordi- nates, and the slider has reciprocating motion along a line parallel to the X— axis. As shown, the path of the slider pivot is displaced a distance C above the X-axis. For many applications the offset C will be zero; it may also be negative. The link joining the crank and the slider is called the connecting rod, and its length is denoted as L. Number of Degrees of Freedom. Before beginning the kinematic analysis, itis important to determine the number of degrees of freedom associated with the mechanism. For the slider-crank, if a value is specified for the crank angle q, then the configuration of the entire assembly is determined and the mechanism has one degree of freedom. To see this, consider a graphical construction process shown in Figure 2.7. All dimensional data is Comecting Rod Ls FIGURE 2.6 Typical Slider-Crank Mechanism  24 Single Degree of Freedom Linkages x A (o) Locating Crank Throw tb) Path of the Slider Pivot e) Actual Slider Pivot Located 1 from Crank Tip “4 a td) Completed Slider Crank Configuration FIGURE 2.7 Construction Demonstrating that the Slider-Crank is a SDOF Mechanism (a) Locating Crank Throw (b) Path of the Slider Pivot (6) Actual Slider Pivot Located a Distance [ from : Crank Tip x :: “-a (9) Completed Slider Crank Configuration 2.2 Kinematics of ie Slider-Crank Mechanism 25 assumed lnown before the construction begins. The important question is “Wlhat is the minimum number of additional values that must be specified to determine the configuration of the mechanism?” The steps for the con- struction are as follows: à. With 4 given, the location of the crank throw is determined by the polar coordinates (R, q). (The crank throw is the outer end of the crank to which one end of the connecting rod is attached.) b. The path of the slider pivot is a line parallel to the X-axis and is dis- placed from it by a distance C. e. The present location of the slider pivot is the point along the stider locus that is a distance L from the crank throw. à. The figure is now complete. Im this problem, it is evident immediately that the crank rotation will be a useful primary variable (generalized coordinate), so thé construction be- gins with a value assigned for that variable. Becaúse we could complete the construction by assigaing only one variable, the answer to the initial ques- tion is, “One additional value is suíficient.” Therefore the slider-crank mechanism has a single degree of freedom. The graphical thought process described has broad applicability; ir cem be adapted to determine the number of degrees of freedom for à varieiy of mechanisms. Ia each case, it is necessary to answer the question, “How many variables must be specified (in addition to the dimensional data) to graphically construct the mechanism?” The word consfruci mesns thai known lengths or angles can be measured and laid out accordingly. Other points must be located and lines defined strictly in the sense of Euclidean geometric construction, performed with a-compass and a straightedge When this approach is applicd to systeros with multiple degrees of free- dom, the need to specify values for other variables becomes apparent as the construction proceeds. Unlêss the system is very complicated, it is not necessary to actually draw the mechanism; instead, a mental constru is usually sufficient. Position Analysis. As just discussed, the slider-crank mechanism posi- tion is fully determined by specitying a single variable, here assume to be the crank angle. The secondary Kinematic variables of interest are the con- necting rod obliquity angle, 4, and the slider position, X (see Figure 2.6). There is a single position vector loop that runs from the crank pivot to the crank throws, along the conriecting rod to the slider pivot, and back te ihe crank pivot. The scalar position loop equations are ReosgiLeosA-X=0 Rsing-LsinA-C=0 These equations must be soived for 4 and X, with the following results: 26 Single Degree of Freedom Linkages A = Aresinl(R sin q — C)/L] Xx=Reosg+Lceos 4 Because A will always He in the first or fourth quadrants, the principal value of the inverse sine function will.be correct. The solutions may be checked by substitution in the position loop equations. Velocity Analysis, The object of the analysis in this step is to obtain the position-dependent velocity coefficients, K, and K,. Differentiation of the position loop equations gives the velocity loop equations -Rjsing- LÂsinA -X=0 Rá cosq — LÁ cos A =0 Prior to solving for the unknown velocities, these equations are recast in the matrix form -LsinA -D[Ã). ai -Leos A. 0X 4 =R cos q The (2 X 2) coefficient matrix on the left side of the equation is the Jacobian matrix for the slider-crank mechanism, This is a system of linear, simulta- neous algebraic equations that must be solved. (The solution for the case of two equations in two unknowns is given in detail in Appendix AL.5.) For the present system of equations, the solution is À -g [0 t ai) x) IcsAlLcsA -LsinAJ-R cosq Rcosg =qiLcos A -Rsing — Rcosg tan À The expression for X may be simplified further by using the second of the position eguations to replace (R sin q) with (L sin 4 + C). After some algebra, the final result is ' X=-HC+Xtan4) To express the velocity coefficients, the ratios Àlã and X/á are formed: io Rcosg Ei = É! = 4Lcos A . Hj) dl (cs x tan 4) Acceleration Analysis. As mentioned in Section 2.1, there are two ways to approach the secondary accelerations. The following discussion will present both approaches, beginning with the direct differentiation of thc 2.2 Kinematics of the Slider-Crank Mecheriism 27 velocity loop equations with respect to time, Performing the indicated differentiations gives —Rj sing — Rj cos q — LÁsinA — LÃicosA- K=0 Rjcosg — R$? sin q — Là cos À + LÃ2sinA = 0 When cast in matrix form, the Jacobian matrix is evident again in the (2 x 2) coefficient matrix on the left side of the equation: -LsinA -Ià Rsing , IReosq [| LosA = +42 + À2 -L cos À oHX —R cos q, Rsing -Lsin A This system of equations is solved by the established procedure to express the accelerations, À and X. From the second approach to the acceleration calculations, the accclerations will take the form ld) where K, and K, are the velocity coefficients, and L, and L, are the deriva- tives of K, and K, with respect to q. By direct differentiation of K, and Ks with respect to q, the velocity coefficient derivatives are 1, = BEBE A Ko tanA L.=-Rcosg-— K?LcosA — LiLsin A This result can be confirmed by completing the solution to the system of simultaneous equations in the first approach to the acceleration caléula- tion. As stated before, the complete accelerations are given by a sum in- volving these velocity coefficient derivatives and the velocity coefficients, multiplied by 9? and à, respectively. Numerical Values. Consider a slider-crank mechanism defined by the following dimensions: R=35in. L=103in. Í C=0.3in. The position values A and X, velocity coefficients, and velocity coefficient derivatives are to be evaluated for the crank angle q = «7/3 radians. The preceding analysis has been implemented in a short computer program, which follows. Notice that the problem data is embedded ín the program from lines 1130 to 1160. The program asks for the crank angle as input at line 1240. The input value used is 1.0472 radians, which is the decimal equivalent to 7/3. . | | |  32 Single Degree of Freedom Linkages Coupler Link Stationany Link FIGURE 2.9 Typical Four-Bar Linkage ated with the four-bar linkage include approximate straight lines (Watt and Scott Russel linkages), closed curves, and even circles (Galloway mecha- nism). This range of adaptability has attracted the interest of designers over the years through the present day. The four-bar linkage consists of four bars or links of constant length, as the nâmé implies, and can be visualized as a closed assembly of four, pin- jointed links. A typical four-bar linkage is shown in Figure 2.9. One of the four links shown is completely stationary, and physically does not look like alink atail. Nevertheless, it functions as a link, maintaining the separation of the two pivot points where the remainder of the linkage is supported. The two links joined to the stationary link are referred to as cranks because their motions are purely circular about their stationary pivots. The final link joining the two cranks is called the coupler link. For most purposes, the four-bar linkage is úsed in one of two ways: (1) the mechanism may be used to-transfer power from one crank (the input crank) to a second crank (the output crank) with a particular motion or (2) a point on the coupler link may be driven in a desired motion by the input crank rotation. It is also possible, but relatively uncommon, for the input link to be the coupler. The four-bar linkage is a single degree of freedom mechanism. This is readily seen by mentally applying the “graphical construction” test de- scribed in Section 2.2. It will be convenient to associate the one degree of freedom with the angular position of the input crank. The following analy- sis will first develop the position, velocity, and acceleration relations for the secondary Kkinematic variables in terms of the motion of the input crank, Closed-form position solutions are again possible, but in this case they are unwieldy. Instead, this opportunity is taken to introduce a general riumérical-techítique for solving' the: position equations (the Newton> Raphson methos). Position Analysis. The link lengths, Cr, Cr, C3, and Cy, are understood to be given for the four-bar linkage shown in Figure 2.10. The configuration of 33 FIGURE 2.10 Kinematic Skeleton for Typical Four-Bar Linkage : the four-bar linkage is then fully determined when the input crank angle 9 is specified, The remaining secondary kinematic variables are the angular positions for the coupler and the second crank, 4 and As. These will be determined from the following position loop equations: fi(g,42,45) = Ci cos q + Co cos Az + Cacos As = C=0 falg.Ao,Ão) = Gosing + Csin A+ CasinÃy = 0 As previously mentioned, a numerical solution procedure called the Newton-Raphson method, will be used to solve the position loop equa- tions. The details of the Newton-Raphson solution are given in Appendix A? and should be reviewed at this point if not familiar. In the notation of the appendix, the vector of unknowns, (5), is here called (A), where 15) = (4) = Col (As, As) and the residual vector is 1F), where TF) = Colfilg, A2As), fa(g, Aa, As)) The Jacobian matrix for this system is determined by partial differentiation, with the result Sh dh 34 BA: [ê sin 44 —Casin “ df df 24: DAS Using the residual vector (Fy and the Jacobian matrix just defined, the position loop solution may be computed iteratively to any sequired degree of precision. “This solution method is demonstrated in an example problem following the velocity and acceleration analyses. One requirement for using the Newton-Raphson solution is an initial estimate for cach of the unknowns. These estimates can be determined by Cocos A) Cacos Às  34 Single Degree of Freedom Linkages a very roughcalculation, by a sketched graphical solution, or by pure guess; any reasonably close estimates will suffice because the solution converges for a wide range of initial estimates. On the other hand, if the initial estimates are too far from the solution, the process will either not converge, or will converge to a solution representing a different configura- tion. In any event, the better the estimates, the quicker the process will converge to an acceptable solution. Velocity Analysis. Velocity loop eguations are determined by differenti- ating the position loop equations. The resulting system of linear equations can be cast in matrix form as —Casin 42 —Cs sin As][Ão [ Cisin q . Cicos ão Cos As lÁs $ —C, cos q Notice that the coefficient matrix on the left side is again the Jacobian matrix. For given values of q and 9, this system of linear equations is readily solved (see Appendix 41.4), Alternatively, the velocity coefficients, Ky and Ky, may be determined by dividing both sides of the equation by q: Kz Adlã —C sin dy —G sin er G sin 4 Ka, Ástá, Cicos A CGeosAs] (-Cicosg The velocity coefficient solution is implemented in the example program that follows the acceleration analysis. Acceleration Analysis. Time differentiation of the velocity loop eguations provides the acceleration loop equations. With the results cast in matrix form and after some rearrangement, the system of linear equations to be solved for the secondary accelerations is Co sin 4 —Cs sim As]fÃo “1[ CG sin q CGeosà Ccos AsilÃs La cosg +a E cos ' 1? [e cos a +ã? E cos a Gy sin q LCo sin 42 Cs sin As, Notice that the coefficient matrix on the left side is again the Jacobian matrix. The right side involves the squares of the secondary velocities, as well as j and qí. These may be expressed in terms of j2? by using the velocity coefficients found: previously. With these substitutions, the secondary ac- celerations are . Às —C sin 4 —G sin tal C sin à Às) | GrosA Cacos As é -—€ cos q, 2.3 Kinematics of the Four-Bar Linhage * 35 1 E cos q + KZC, cos 42 + KiiC cos As Ci sin q + KG, sin A, + K2C; sin As =: aee La Fla, Note that the inverse of the Jacobian matrix multiplied into the coefficient of does, in fact, give the velocity coefficients, K: and Ky. The inverse of the Jacobian matrix multiplied into the coefficient of q? gives the velocity coefficient derivatives, L> and Ly. This last step can be verified by (1) completing the solution for the velocity coefficients (sec Appendix A1.5), (2) differentiating the expressions for K» and K; with respect to q to produce LI and Ls, and (3) completing the solution for the coefficient of q?, above. When Ly and Ls are compared to the coefficients of 4%, they are found to be the same. Numerical Values. For a four-bar linkage defined by the link lengths G=s5in. G=8in. G=7im. Cs, = 10 in. as shown in Figure 2.11, the secondary positions, velocity coefficients, and velocity coefficient derivatives are to be evaluated when the assigned crank angle is q = /3 radians. cy=10 FIGURE 2.11 Four-Bar Linkage for Numerical Example of Section 2.3  36 Single Degree of Freedom Linkages igie Deg; 8 The previous analysis applies directly to this problem, and it has been implemented in the computer program that follows shortly. Notice the organization of the program: . lines 1000 to 1490 perform all initializations . liries 1500 to 1990 perfórm the position solution . lines 2000 to 2150 are the velocity coefficient evaluations « lines 2500 to 2640 perform the velocity coefficient derivative evaluations . lines 2800 to 2880 are the output . lines 3000 to 3110 specify the problem Sm gn AJ the problem data and govemning equations are embedded in the final segment of the program, which is called as a subroutine from lines 1520 and 2030, The Matrix Operations Package is also required (see Appendix AL6). 1000 REM FOUR BAR LINKAGE ANALYSIS tOLO REM USING NENTON-RAPHSON SOLUTION 1020 REM FOR POSITION ANALYSIS 1030 REM logo REM F VECTOR DF FUNCTION GALUES 1050 REM A VECTOR OF ANGLE VALUES 1050 REM D VECTOR OF ANCLE CHANGES 1070 REM FA SQUARE MATRIX DF PARTIAL 1080 REM DERIVATIVES OF F 4RT A 1090 REM : 1200 DIM FtZ) ALZ) DIZ) FA(ZIZ) 1Z10 DIM DI(2),02(2,2),0342,2) DA(Z HE) iz20 REM FUNCTION ERROR LIMIT, EF 1280 EF=0,0001 izdo REM MIN CORRECTION: EA º 1250 EA=0,0091 1260 REM MAX NUMBER OF ITERATINNS, 19 1270 19-20 1260 REM 1400 REM INITIALIZATION 1410 PRINT"ENTER INPUT CRANK ANCLE+ 0” 1420 INPUT Q 1430 PRINT “ENTER INITIAL ESTIMATES FOR AZ & AB” 4440 PRINT"AZ= 1450 INPUT AZ i4B0 PRINT'A3 1470 INPUT Ad 146Q. Atid=Az 1490 At2)=A3 1500 REM BEGIN POSITION SOLUTIQN 1510 FOR l8=1 TD 19 1520 COSUB 3000 1330 E=0 2.3 Kinematics of the Fowr-Bar Linkage 37 1540 FOR I=1 TO 2 1550 1F EXFt(1)*Z THEN E=FtI)CZ todo TF ECE EN GOTO 1880 EF TH isso RER IF EtEF “THEN .THE SOLUTION HAS CONVERGED «4 , 1600 REM NENTON RAPHSON DETAILS 1640 FOR 1=1 TO 2 1650 FOR J=1 TO 2 1860 O2(1sD=FACIod) 1870 NEXT 3 1680 NEXT 1 . 1690 FOR TD 2 izoo 02€1 FtIs 1710 NEXT 1720 Zi=2 1722 Z2=1 1723 231 1730 GOSUB 5400 1740 FOR 1=1 TO 2 1750 D(I)=02(1+2) EXT 1 1580 REM SEARCH OUT & TEST LARGEST CHANGE 1780 E=0 1800 FOR I=1 TO £ 1810 IF ExDtI)“2 THEN E=D(I)"Z 1820 NEXT 1 1030 IF EXEA THEN GOTO 1980 1840 REM IF EXEA THEN THE MIN STEP 1850 REM CRITERION HAS BEEN SATISFIED 1DG0 REM ; 1870 REM MAKE THE CHANGE 1880 FOR I=1 TO Z 1890 AtLI)=A(D+D(T) 1900 NEXT É 1910 NEXT 18 1020 REM IF THE PROCESS GETS TO THIS 1930 REM POINT: IT HAS FAILED TO CONVERGE 1940 PRINT"POSITION SOLUTION FAILED! 1950 END 1980 AZ=A(I) 1090 A3=A(Z) , 2000 REM VELOCITY SOLUTION 2010 REM 2020 REM RE-EVALUATE THE JACOBIAN O GOSUB 3000 Zouo REM SET UP LINEAR EONS FOR VEL COEF 2050 DZ(1 +3)=CLeSINtR) 1 2060 0242,3)=-CISLOS(N) 2070 FOR I=1 TO 2 2080 FOR TD 2 2090 0261) J)=FA(I:d)  42 Single Degree of Freedom Linkages described. IÉ the system motion is to be described using Newton's Second Jaw and the appropriate kinematical relations, discussion of constraints implies that there are force terms required in the sum of forces that canmot be determined until after the equations of motion have been solved! What can be done? The idea of a constraint relation is that these unknown and anknowable forces can be included in the motion description in terms of the effecis they produce. Often this takes the form of one or more additional equations that describe the result of the unknown forces. An example wiil help establish this concept. Consider a two-dimensional problem involving a wheel moving in contact with a horizontal surface. Initially, it appears that the system has two degrees of freedom, and it is convenient to associate one with the position of the center of the wheel, X, and the second with an angular coordinate, 4, describing the rotation of the wheel. Without friction be- túrcen the wheel and the supporting surface, the wheel motion may beany combination of rolling'and sliding, so this is completely correct. If, on the other hand, the surface is perfectly rough, there will be no sliding. The phrase ”perfectly rough” implies the existence of whatever force is needed to assúre that-no sliding occurs, but-the exact magnitude of that force cannot be determined until the motion is known. For the perfectly xough surface, the effect of the friction force is to cause a direct relation between the horizontal displacement, X, and the rotation, A. This relation is dX — RA = 0 or X-— RA + Constant = 0 Either of these may be called the equation of constraint, which has the effect in this example of reducing the number of degrees of freedom from two to one. This reduction is the usual case; but not the only case; read on. Constraint equations axe classified in two types: holomomic constraint or nonholonomic constraint. These are sometimes considered roughly equiva- lent to “less difficult” and “more difficult,” although these are not neces- sarily correct interpretations. Conptráint relations take one of Elec forms: (1) finite equalities (non-differential equátions), (2) finite inequalities, and (3) differential relations (equations). The first forma, the finite equality, cam always be used to reduce the number of degrees of freedom of the system. This form and its effect were illustrated earlier. This is the constraint type of greatest interest in the mechanics of machines; it is called a holonomic constraint. K will be discussed in detail after the other two types have been described, Thesécond'constraint-type, the-inequality relation, is always a nonholonormic constraint, put it usually causes little difficulty. The third type, the differential relation; may be either a holonomic or nonholonomic constraint, depending on the exact nature of the relation, The nonholonomic differen- tial constraint is the source of the “less difficult/more difficult” perception. As an example of the ineguality constraint, consider the motion of a Ê 2,4 Constraints 43 projectile toward a spherical target. The equations of motion formulated for this motion will apply only while the projectile is outside the target. After the projectile strikes the target and begins to penetrate, different equations of motion will apply -In.texms of a rectangular-coordinate system with the target sphere of radius R centered on the origin, if the projectile position is given by (X,Y,Z), the appropriate constraint is simply R+AN+Z>R For the motion problem, the differential equations describing the flight of the projectile toward the target are solved by ignoring the effect of the constraint. Rather, the constraint limits the range of applicability of the solution; the solution is valid only for positions satisfying the constraint. IF the solution is to be done numerically, after each time step the validity of the constraint is tested. Whenever the constraint is no longer satisfied, the a Pandino most be changed for the penetration problem. This is e typical handling of inegualit: ints— i e Price a dir E otia quality constraints-—a relatively simple type of In the case of the differential relation, the constraint condition that it describes may or may not be holonomie. Supposc that the differential relation is in the form CxdX+CydY +C dz =0 TF this is the exact differential of a function F(X,Y,Z), then the differential constraint relation is equivalent to ar = dx + ddr + DS dz =CdX+C,dy + Ci dz =0 which can be integrated to obtain F(X,Y,2)=0 Because this last relation is a finite equality, a holonomic constraint condi- tionis described by either the integrated finite form or th= original differen- tial form. The conditions under which a differential form can be integrated are described in texts on the theory of differential equations.? In the event that no function F exists whose differential is the-given differential con- straint equation, then the constraint is nonholonomic. dm mechanical systems, nonholonomic constraints are most often asso- ciated with rolling contacts. There are three classical problems common: used to illustrate nonholonomic constraints: (1) a sphere rolling on a Tough plane, (2) a coin rolling on a rough plane, and (3) two wheels that have a common axe rolling on a rough plane. A variation on the third problem is 2 See E. L. Ince, Ordinary Difjerential Equations, pp. 52-55, Dover, New York, 1944.  pino 4 Single Degree of Freedom Linkages the motion on ice of two skate blades rigidly joined by a rod that is perpen- dicular to them. Only the sphere rolling on a rough plane will be discussed here. 2.4.1 Sphere Rolling on à Rough Plane Consider a solid sphere of radius R, rolling without slipping on a rough plane. For any rigid Body moving in three dimensions, there are six degrees of freedom requiring six generalized coordinates. It is common to associate three of these generalized coordinates with the location of the center of mass, while the other three are assóciated with angular coordinates that describe the orien- tation of the body. For the sphere under discussion, the six generalized coor- dinates are reduced to five because the sphere is to remain in contact with the rough plane; one of the translational coordinates for the center of mass is fixed by contact with the plane. How many degrees of freedom does this system have? Consider a stationary rectangular cartesian coordinate system with X- and Y-axes in the plane and the Z-axis normal to the plane; the associated unit vectors are 1, j, and k. The angular velocity of the sphere will be o=wdi+taj+ak where w;, wy, and o, are cach functions of À, Àz, and Às, the time derivatives of the angular coordinates, If X and Y denote the center of mass coordinates and R is the radius of the sphere, the velocity of the point on the surface of the sphere in contact with the plane 'is Vp= Xi+Yj + ox(-RI) -—s The sphere remains in contact with the plane, so that Z-component of this velocity equation must be identically zero. The constraint expressing roll- ing without slipping requires that the other two components also be zero, resulting in two differential equations of constraint: fUX Àr Ào, À) =0 fuY, Àr, Àa, Às) =0 These two equations of constraint are, im fact, nonholonomic. Thus, there are five generalized coordinates required to describe the system configura- fon (X, Y, Ay, Az, As), but the system has only three degrees of freedom. In general, for systems involving nonholonomic constraints, the number of coordinates required are more than the number of degrees of freedom. The presentation of the rolling sphere problem has, of necessity, been somewhat sketchy so as to avoid the need for Euler angles and, other “advalhced concépts. For those who wish to pursue the matter further, these three classical examples are discussed in a number of other texts.”7 2H. Goldstein, Classica) Mechanics, p. 13, Addison-Wesley, Reading, MA, 1959. *D. T. Greenwood, Principles of Dynamics, pp. 234.236, Prentice-Hall, Englewood Cliffs, NJ, 1965, 2.4 Constramts 45 Two-Dimensional Rolling Constraint One of the common constraints encountered in the mechanics of machines is the two-dimensional rolling constraint. This refers to the two-dimensional problem of one.body.folling over another, such as à wheel rolling along the ground or one gear rolling on another. The problémi of the wheel zoliing along the ground was briefly discussed at-the beginning of this section. In that case, the equation of constraint was first written in differential foi and then in the equivalent finite (non-differential) form. For two-dimer sional rolling problems such as this, the equation of constraint can always be written in ânite form, and thus, the two-dimensional rolling constraint is always holonomic. This means that the constraint equation car. be used to eliminate one of the coordinates as well as a degree of freedom, keeping the number of degrees of freedom and the number of generalized coordi- nates equal. Im the carlier discussion, to assure rolling motion, certain surfaces were described as perfectly rough. The term perfectly roúgh implies suffi- cient friction to assure no sliding under all conditions. In reality, there are no perfectly rough plane surfaces. Problems in elementary mechârdos are often solved by first assuming that there is no sliding and, after the motion has been determined on the basis of this assumption, calculating the mag- nitude of the required friction force. If this required friction force exceeds that available, then the system must slip, and the problem must be re- worked with slpping included. One device often used in machinery to assure that slip does not occur, and thus to approximate the perfectly rough surface, is to cut mating teeth on the two bodies where they are in contact. If the two bodies are wheels, the resultis a pair of gears. If one body is a straight surface and the other is a wheel, with the addition of teeth they become a rack and pinion. For properly formed teeth, the teeth do not affect the motion but only serve to assure that therc is no sliding. The matter of proper toath formas will be considered in Chapter 5, For the present, it will be sufficient to consider a gear as a perfectly rough wheel. The effective radius of this wheel is knoven in gear terminology as the pitch rdius. The introduction of a rolling constraint modifies the established dure for kinematic analysis. When constraints are involved, the comi set of position loop equations and constraint equations must be solved simultaneousiy for the secondary variables. For velocity and acceleration analyses, the constraint cquation(s) must be differentiated along with the 5 C. W. Kilmister and]. E. Reeve, Rational Mechanics, p. 200, Elsevier, Amsterdam/New York, 1966. 5 E. J. Totby, Advanced Dynantics for Engineers, pp. 261-262, Holt, Rinehayt and Winston, New York, 1984. 7].L. Synge, “Classical Dynamics,” pp. 40-41, in Encyclopedia of Physics, S. Flugge, ed., Vol n1/1, Springer-Verlag, New York/Berlin, 1960.  i | 46 Single Degree of Freedom Linkages posítion loop equations and again included in the solutions. The kinematic analysis for a system involving a rolling constraint will be illustrated in the . following example. For this example, the original assembly configuration plays an important role, as is often the case. 2.4.2 Rolling Constraint Example Consider a system involving a wheel rolling without slip on a plane, as shown. in Figure 2.13, The wheel is driven by pulling on the connecting rod that passes through a-pivoted eye block. The radii, R and 7, arc known. Figure 2.13(a) shows the system as originally assembled. Im the original assembly configuration, the position of the contact point, So, is known; the subscript o is used to denote original assembly values. The connecting zod pivot is initially directly above the point of contact between the wheel and the supporting plane, and the connecting rod extends from that pivot to the eye block pivot. As the connecting rod is drawn through the eye block, the wheel rolls to the left as shown in Figure 2.13(b). Set up all equations required to determine the secondary kinematic variables. First consider the information known for the original assembly posi- tion. There is enough information to determine the connecting rod length, Bo, and the initial inclination angle, Cs, as follows: = VETA Co = Avetan[(R + 1)/59] The original rotation angle is A, = 0. Now look at the displaced configuration. The system has one degree of freedom, here associated with the length of rod drawn through the eye Eye Block Connecting Rod FIGURE 2.33 (a) Initial Configuration for Rolling Constraint Example Dose 2.4 Constraints az Rolls Without Slip FIGURE 2.13 (b) Displaced Position for Rolling Constraint Example block, q. There is one position loop, from which the position loop equa- tions are (B-gcsC+rsnA-S=0 (Bo-a)sinC-rcsA-R=0 The unknowns appearing in the two loop equations are C, 4, and S; it is evident that another equation is required. The remaining equation is the constraint relation expressing the fact that the wheel rolls without shipping. On the plane surface, the point of contact moves by the amount (So — 8) from the original configuration to that shown in Figure 2.13(b). On the surface of the wheel, the length of the arc through «which the contact point moves is RA, and these two lengths must be equal for rolling withoúé slip, so So- 5S-RA=0 This is the holonomic constraint expressing rolling without slipping for this problem. The full set of equations that must be solved for the secondary kinematic variables C, 4, and $ are (Bo-qcsC+rsinA-S=0 (Be-g)sinC-rcsA-R=0 Se—S-RA=0 A closed-form solution may be sought for the three simultaneous, nontin- ear equations, ór a numerícal solution may be determined using the New- ton-Raphson method. Por the Newton-Raphson solution, it is necessary i to consider three unknowns; this is a simple matter when the program is written in matrix form, as in Appendix A2. The analysis of velocities and accelerations proceeds in the usual fashion, with the constraint relation differentiated along with the loop equations.  ! 52 Single Degree of Freedom Linkages 2.5 Multiloop Síngie Degree of Freedom Mechanisms 53 i ; the signs of the velocity coefficients, all of the angles decrease as q in- 1610 NEXT K : creases, while the slider is moving down, Ky > 0. These observations arc Í 1620 REM in accord with an intuitive understanding of the working of the machine. Í 1700 REM TEST FOR TERMINATION . | The program listing follows. The Matrix Operations Package from Ap- 1710 REM FORM nora FOR RESTDUAL AND ADJUSTMENT i pendix AJ must-be added to the listing given. 1720 h (1) NBR (2)AZAR IS) ZePrÁIOE + | i7a0 S (1) B+DS(2)"2ADS(3)CP4DE(A)"Z 1000 REM FOUR BAR / TOGGLE LINKAGE 1780 1 N$>E1 THEN GOTO 1770 iO10O REM GOMERNING EQUATIONS AND 1750 V=1 Í 1020 REM JACODIAN MATRIX ELEMENTS ARE 1760 GOTO 1900 1030 REM PROGRANMED AT LINE 4000 1770 IE N2>E2 THEN GOTO iBoo 1046 REM : ' 1780 V=Z 1050 DIM F(4):5(4)-DS(4) /JM(a sa) 1790 GOTO 1900 1060 DIM Ol(4)DZ(415) DAC4 HA) 1B00 REM UPDATE SOLUTION ESTIMATES : 1070. REM 1810 FOR K=1 TD 4 , 1100 REM SET CONTROLS 1820 S(K)=5(K)+DS(KS) : 1110 REM Ei = MAX ALLOMABLE RESIDUAL 1B3o NEXT K i 1120 REM EZ = MIN ALLOHABLE ADJUSTMENT 1840 NEXT É Há 1130 REM 15 = MAX NUMBER OF ITERATIONS 1850 REM, i 1140 E1=0,0001 . 1560 PRINT"MAX NUMBER OF ITERATIONS" Í 1150 E2=0,0001 1870 REM i 4 ilgo I9=10 1900 REM OUTPUT : A 1170 REM in10 PRINTºSOLUTION VALUES” bl 1200 REM ENTER CRANK POSITIDN AND 1920 PRINT Í ú 1210 REM INITIAL ESTIMATE FOR UNKNDUNS 1930 PRINT'S(1)=AZ="I5(15 | 1230 HOME 1946 PRINT'S(2)=B3="IS(2) Í E t240. PRINT'ENTER O! 1950 PRINT"S(3)=B5="45(5) j 1250 INPUT Q iSG0 PRINTYS(M)=Y ="15(4) m 1260 PRINT“ENTER INITIAL ESTIMATE FOR AZ=S(1)" 1970 PRINT ; 1270 INPUT S(1) 19B0 PRINT"FUNCTION VALUES" , 1280 PRINT “ENTER INITIAL ESTIMATE FOR B3=5(2)" i 1980 PRINTºE(I)D="3FCH) ] 1290 INPUT 5(2) 2000 PRINT"E(Z)="5E (2) o 1300 PRINT"ENTER INITIAL ESTIMATE FOR BS=5(3)" 2010 PRINT"E(B)="4F(3) E 1310 INPUT St3) z020 PRINT"F(U)="F(a) a 1320 PRINT“ENTER INITIAL ESTIMATE FOR v=S(4)" 2090 PRINT!NUMBER DF ITERATIONS ="51 Po Ê 1330 INPUT S(4) 2040 IF U=1 THEN PRINT"TERMINATION BASED ON E a 1340 REM MAGNITUDE DF RESIDUAL E 1400 REM BEGIN THE ITERATION 2050 IF U=2 THEN PRINTºTERMENATION BASED ON º E 1410 FOR f=1 TO 19 MAGNITUDE DF ADJUSTHENT : E 1420 REM EVALUATE THE RESIDUAL AND THE JACOBTAN 2ogo REM pn 1440 COSUB 4000 2160 REM VELOCITY COEFFICIENT ANALYSIS i 1500 REM EVALUATE THE ADJUSTMENT 2110 REM F 1510 FOR K=i TO 4 2120 REM SET UP RIGHT SIDE VECTOR ; á 1520 02(K15)=-F(K) . 2130 W2(145)=RESINCO) 4 is30 FOR L=1 TO 4 2140 02(2,5)=-R*COS(M) is à] 1340 BECKAL)EJMIKOLO - «a : 2150 02(3,5)=0 E À 1550 NEXT L 2160 02(4,5)=0 i 1560 NEXT & * 2170 REM RE-EVALUATE JACOBIAN : E 1370 Zl=4 2160 GOSUB 4000 : is72 Z2=1 : 2180 REM SET COEF MATRIX EQUAL TO JACOBIAN a 1578 z3=1 2200 FUR to 4 À 1580 GOSUB 5400 22i0 FOR to 4 y 1590 FOR K=1 TO 4 ! 100 DS(K)=02(K 15) 54 Single Degree of Freedom Linkages 2220 0261 +J)=IMEI NH 2230 NEXT J 2240 NEXT 1 2250 REM SOLVE FOR VELOCITY COEFFICIENTS 2280 GUSUB 5400 2270 REM PRINT VELOCITY COEFFICIENT RESULTS 2280 PRINT 2290 PRINT"VELOCITY COEFFICIENTS" 2900 PRINT"KZ="102(1,5) 2310 PRINT"K3="402(2:5) 2320 2300 2340 2350 4000 REM PROBLEM GEOMETRY 4010 R=5 4920 Cz=i4 4030 C3=27 4040 cS=4a 4050 X1=13,6 4060 v1=18 4100 REM FUNCTION EVALUATIONS ao Fti)=XL+RACOS(O)-C2ACDSCS(1))-CIASINCSIZI) GiZO F(Z)SVIARASINID) +CZASINCB(L))-C3XCOS(S(2)) 4130 F(3 JASIN(S(2))-CS*SIN(S(3)) gião F(4)=5(4)-CI*COS(S(2))-C54COSCS(3)) 4150 REM JACOBIAN EVALUATION 41B0 JM(1+1)=C2*SINCSC1)) ULTO JM(L+42)=-DINTOS(S(Z)) a180 JM(143)=0 4190 JM(1,4)30 4200 JM(Z11)=024C0S(S(ID) 4210 JMIZ,2)=CO*SIN(S(ZI) 4220 JMIZ:3)=D 4230 JMIZ +4)=0 4200 JMt3:1)=0 az50 JM(312)=C34CDS(S(2Z)) 4ZB0 JM(3+,3)=-C5*COS(S(G) H270 JMIIHa)=O azBo JM(M+1)=0 4280 JM(d)Z)=C3ASINCSIZ)) 4300 JM(A 3)=C5*SINCS(3)) UILO JMIA 451.0, 4320 RETURN 4330 END As illustrated here, the existence of multiple independent loops in- creases the number of equations required in each of the position, velocity, and acceleration solutions. Otherwise, it is exactly like any other single degree of freedom system analysis. . 2.6 General Kinematic Analysis for Single Degree of Freedom Mechanismis 55 2.6 GENERAL KINEMATIC ANALYSIS FOR SINGLE DEGREE OF FREEDOM MECHANISMS The kinematic analysis for a single degree of freedom mechanism-has.been presented by means of examples for several important cases. At this point a general analysis will be presented that will include all of the previous examples as special cases and will apply to all other single degree of free- dom mechanisms as well. Recall that the idea of a primary kinematic vari- able or gencralized coordinate associated with the degree of freedom was introduced in Section 1.2. The remaining kinematic variables were labeled as secondary variables. Conseguently, the notation for this general analysis is q = generalized coordinate & = secondary coordinate, i=1,2,...N Position Analysis. The position vector loop equations, written in scalar form, and the constraint equations, if any are required, form a system of N;, simultaneous, nonlinear equations: filg, sa Sao folg, si Sa. -8m)=0 «smj=0 This system of equations must be solved either in closed form or by numer- ical means. For the numerical solution by the Newton--Raphson method, the solution vector (S) and the residual vector (F] each contain N; ele- ments: (SJ = Col(s,, 52,83,» +, Sn) tFh= Col(fi, fa faro o cofre) The Jacobian matrix is formed by differentiating the position loop func tions, fi, with respect to the secondary coordinates, 5;. Note the locations for the various derivatives in the Jacobian mattix: ês à ds àsm El = dh df oh (df as às é ds àsnz dfne df df dfw ds, ds 053 dsn  56 Single Degree of Freedom Linkages As demonstrated previously, the Jacobian matrix plays a most significant role in the Newton-Raphson solution of the position equations, the veloc- ity solution, and the acceleration solution. From this point on, the position solution is assumed to be complete, so that values are available for all position variables. Velocity Analysis. The velocity loop equations are obtained by differenti- ating the position loop equations with respect to time, including the con- straints if any were required. Expressed in matrix form, this is apoc af 5 [dia llído (Nox) (Nox No) (1x1) (No x 1) The elements of the coefficients matrix will usually be functions of the position variables, but these are known from the current position analysis. This partitioned matrix relation, however, may be separated into a system of linear simultaneous algebraic equations to be solved for the secondary velocities. = E (5 4 ag Alternatively, the system may be rearranged to be solvable for the velocity cóefficients, Eva 8 Acceleration Analysis. A second time differentiation is required for the accelerations. In terms of the velocity coefficients (now known either nu- merically or in functional form), the velocity relation is tó = qtkS which is readily differentiated using the chain rule: dUK, (p= mx + qo ELA = Kd) + a If the velocity coefficients were determined in functional form, they can be differentiated to determine the velocity coefficient derivatives directiy. If the yelacity coefíicients were determined númeérically, this differentiation is not possible. For a numerical approach to evaluate the velocity coeffi- Gtent derivatives, consider first the relation defining the column vector of velocity coefficients: fa ntkd = — a 2.6 General Kinematic Analysis for Sinigle Degree of Freedom Mechanisms 57 where [J] is the Jacobian matrix. That matrix is Enowen explicitly, as is the column vector on the right side. Taking an ordinary derivative of this equation with respect to q gives ey - dEK dá Ei 0 ey + q HS É dn ag The velocity coefficient derivatives appear as a factor in the second term, so the equation is solved for that term, Ur E (81) “go dg log Thus, the vector of velocity coefficient derivatives may be obtained as the solution of a system of linear equations, again with the Jacobian as the coefficient matrix. Correct interpretation of the right side of this equation requires careful observance of the rules for ordinary and partial differen- tiation. [REL] = Position, Velocity, and Acceleration for Oihey Points. Considera poini P on a mechanism member in curvilincar motion. Let (Xo, Yo) be the coordi- nates of a point on the same body whose position is readily expressed ir terms of the generalized coordinate and the secondary coordinates. Using this point as an origin, consider a body coordinate system (U, V) on the body containing P, in which (Up, Vp) are the body coordinates oí P, The base coordinates of P are (Xp, Yo), which are readily expressed as Xp = Xo + Upcos 4 — Vo sinA = +U snAÃ+ Vo cos A > where A is the angle between the X-axis and the U-axis. The angle A is either the primary variable, a secondary coordinate, or can be determined from them. This expression is readily evaluated when the secondary coor- dinates are known. The velocity of point P is determined by differentiating the two posi- tion equations with respect to time, and the accelcration requires yet an- other differentiation. Velocity coeficiente and velocity coefficient derivas tives are defined for the point P in the obvious manner. In this way, ihe position, velocity, and acceleration for any point P are readily expressed in terms of the generalized coordinate, q, and its velocity and acceleration, G and . The purpose of this demonskration is not to provide general formulae for use in problem solving. Rather, the reasons for the general analysis are (1) to show that the same processes, resulting in equations of the same form, apply for all closed-joop, single degree of freedom mechanisms and - (2) to focus on the -analytical process, independent of any particular sys- tem. If the final results for a particular problem do not have the forms shown in these general results, that suggests that an error has been made. | | | 62 Single Degree of Freedom Linkages Drive Pin 5 Slot External Geneva Mechanism Geneva Wheel a. Determine the output position, A(g), during the time the drive pin is engaged with. the slot (in closed form); b. Determine the maxdmum and minimum values of q for engagement; c. Determine the velocity coefficient, Ka, and the velocity coefficient deriv- ative, Ls, during pin-slot engagement; d. Given that Ry'= 4.335 in., Ro = 5.9666 in., and D = 7.3751 in., deter- mine 4, À, and À forg = 0.15 tad, q = 12.4xad/s, and q = —29.0 radis?. 2.6 To avoid impact as the drive pin engages the slot in the Geneva sheel, the pin velocity must be along the centerline of the slot. For a drive pin of zero diameter that is located at a xadius Rs, as shown im view (a), engaging a wheel with N slots, (3) (o) Problem Set 63 a. Determine the appropriate value for Ry; b. Determine the appropriate value for D. The value determined in part a for Ry was based on a drive pin. of zero diameter, which, of course, is not the actual case. lt is necessary to use a slightly larger radius, Ry', as shown in view (b). c. If the drive pin radius is R,, determine the appropriate value of Ra"; d. Ifa drive pin diameter of 0.562 in. is to be used with a five-slot Geneva wheel that has the same dimensions as in part d of the preceding problem, evaluate Ry'. 2.7 A crank-lever drive is to be designed to meet the following cri- teria: 1. The lift at the Hip of the slotted member is to be 7.85 in.; 2. The maximum speed at the tip of the slotted member is to be 120 in./sin absolute value with the crank specd of 10 rad/s; 3. The tip of the slotted member is to extend 1.45 in. beyond the outermost position of the slider pivot center point, Determine the following three essential dimensions: € = distance between fixed pivots, R = crank radius, and L = length of the slotted member from the stationary pivot to the tip. Lift 2.8 The figure shows the mechanism of a double-acting gas compres- sion cylinder. À prime mover (not shown) turns the crank, causing the pos Dp N 7; y ou , ' s Ds * bh N  | 1 i 64 Single Degree of Freedom Lintages crosshead, piston rod, and pistôn assembly to move left and right. The piston rod, piston, and cylinder cavity are all assumed to be right circular cylinders, and the dimensions R, Lr, Lo, X1, Xa, Dp, Dr, and T are all known. a. Determine expressions for the chamber volumes, V; and Va, in terms of the known parameters; b. Determine the derivatives dVy/dg and dVo/dg. 2.8 For the dóuble-acting gas compressor.shown, the following di- mensions apply: , R=5im. Dy=8in. K=Qin DD =223in L,=20m To=2in. X=35jn. X =49in. a. Make a plot of V, and dVi/dg versus q; b. Make a plot of V; and dV,/dg versus q. 2.10 For the slider-crank mechanism shown, consider the slider po- sition as the input variable, q. The dimensions cf C;, C2, and L, and radii K ur ppp IVO UM s Problem Set 65 Rs and Ry are known. À spring of length S stretches from the slider to a point on the crankwheel directiy opposite the connecting rod attachment point. a. In closed form, determine S as a function of q; : b. In dosed form, determine K, = dS/dg; c. Given the following data, evaluate S and K; for q = 105in. CG=10in G=45in L=82in. R=Win. R=30m 2.11; The figure shows a slider-crank mechanism falling under the effett-of gravity. The dimensions Cy, Ca, Ui, Uh, and D, and the angle Co are known. a. Determine A(g) and S(g); b. Determine the velocity coefficients K, and Ks; c. Determine the velocity coefficient derivatives L, and Ls; d. Determine the center of mass velocity coefficienis Kre, Kiys Ka Koy- 2.12, The mechanism shown is used in a small, portable air compres- sor”Nofice that the piston, wrist pin, and connecting rod are all one piece. The piston is spherical in shape, and the cavity is a right circular cylinder «with a hemispherical end. The dimensions R+, Rz, Ra, and D arc known, If the piston is 0.010 inch from the end of the cavity at top-dead center, express the cylinder volume as a function of the crank angle, q.  66 Single Degree of Ercedom Linkages i Problem Set 67 2.14; Some early steam engines used the mechanism shown, and itis ! still Sesi in some working model steam engines today, The cylinder oscil- i lates about a fixed pivot as the piston moves back and forth. The oscillation | ; of the cylinder is used to open and close inlet and exhaust portg'to the ' t cylinder, thus providing the necessary valve action. a. Write the position loop eguations and solve them in closed form; b. Determine analytic expressions for the velocity coefficients and velocity ! coefficient derivatives associated with the secondary variables; c. Determine analytic expressions for the velocity coefficients and velocity coefficient derivatives for the piston center of mass; d. Using the data following part e, determine A, À, and À for q = 0.75 rad, à= 14 rad/s, and g = —27 rad/s2, e. Using the folowing data, determine the magnitude of the velocity and the acceleration vectors for the piston center of mass under the condi- il tions of part d. A pa 13) The mechanism shown is used to engage and advance a work piederghe step for each crank revolution. The dimensions Cy, Ca, Cs, and fadius R are known. The working tip of the device is located at P. Assume that an initial kinematic analysis has been done, so that A, Ka, and Ly are known. R=14in. L=45in. C=30in. : . Determine the base coordinates (Xp, Yp) for point P; . Determine velocity coefficients Kpv, Kpy for point P; | Determine velocity coefficient derivatives Lpx, Lpy for point P; . Given the data that follow part e, evaluate Xp, Yp, Kpu, Kpys Lpxs Lyy for q = 0.8 radians; . Given the data that follow, evaluate Xp, Yp, Xp, Yp fox q = 0.8 rad, = 14.5 rad/s, and j = 28 rad/s2, G=0lim — G=005m G=00m R=0055m A=02857 - K = 0.297 La = —0.2599 Al for q = 0,8 poros 2 2.15 A hand-actuated pump mechanism is shown. The dimensions Ci, Co, Ca, Cs, Cs, Dy, and Do are all known. Assume that the initial kinematic analysis for the secondary variables A and B has been completed and that 4, B, Ka, Kb, La, and Ly are known. The condition of interest is when q = 2.27 in. a. Using the data that follows part d, determine the numeric values for X, , ! and Y,, which are the base coordinates for point P; | b. Using the data that follows part d, determine numeric values for the velocity coefficients Kp;, Kpy for point P; . y c. Using the data that follows part d, determine numeric values for the : velocity coefficient derivatives, Lp Lpy for point P;  72 Single Degree of Freedom Lintages | ot: The mechanism shown is a type of garage door. The dimen- —onBhvélues Ci, Ca, . - . Ca are known data, À tension spring is to be atiached between point E on the door and the stationary point designated as the spring anchor. The length of the spring is denoied as 5) ed, a. Analyze the mechanism for all secondary variables (do not solve); b. Express the velocity coefficients; c. Express the spring length, S; d. Express the spring length derivative dS/dg. Y / pan Spring Anchor PiCro, Cit) Mep EX, o Abtachment Details Ga B &— Linkage Pivot Gs Ed spring Attachment +— Dogr Panei Problem Set 73 (2) For the garage door mechanism of problem 2.23, the following dimbnsiônal data are available: Cy = 7.00 ft C = 3.20% Co = 6.82 ft 4.60 ft 0.90 & 0.30 ft 0.90f * Co=7.50k 0.70 ft Cy = 3.50 ft 0.60 ft Cs In this problem gm = 1.9199 radians. Consider the door in nine positions, at evenly spaced intervals on q, including the q = O and q = mes positions, a. Will the lower edge of the door clear the líntei? b. Plot S versus q. e. On the plot of part b, also plot dS/dg versus q. 2.25 The emergency main steam shut-off valve for « nuclear power plant is a massive, tapered plug that swings into a tapered seat to stop the flow of steam. When released, gravity causes the valve to close, with the impact reduced by the dashpot with coefficient B. The plug and its sup- porting links form a four-bar linkage with link lengihs Li, Lo, La, and Lu, 38 shown in the detail sketch. The center of mass of the plug is located by the body coordinates (Lc, Vo). a. Set up the equations required for numerical evaluation of A, and Aq as functions of q (do not solve); b. Set up the equations required for numerical evaluatior of the velocity coefficients Ka and K,s; €. Set up the equations required for numerical evaluation of the velocity coefficient derivatives La and Lis. 7a Single Degree of Freedom Linkages 2.26 The mechanism shown is used to dress the semicircular edge of a grinding wheel. As the handle is displaced through the angle q, the diamond dressing tool moves in a circular arc about point C. Because the wheel spins about its rotation axis while the dressing too] moves, the result is that the surface of the wheel is reshaped to the desired semicircular contour. The dimensional data D+, Da, Da, Da, and Dy are known. The left and right side linkage components are identical. a. Show that the tool actually moves as described; b. Express the radius R in terms of the adjustable parameter D; and the other fixed dimensions of the mechanism. Edge of Grinding Wheel T Grinder axis af Rotation Diamond Dressing Tool Problem Set 75 2.27 The wheel shown, while under the control of a crank and con- necting rod, rolls without slipping on the horizontal surface. The system was assembled originally in the configuration shown by the broken lines, with the crank and connecting rod colinear and the wheel a distance S, to the right of the crank pivot. The dimensional values C;, C>, 7, R, and So are known. a. Setup the equations required to determine the secondary variables (do not solve); b. Set up the equations to be solved for the velocity coefficients; c. Set up the equations required for numerical determination of the veloc- ity coefficient derivatives. 2.28 Under the control of a crank and connecting rod, the wheel rolls without slipping on a semicireular support of radius Cs. The original as- sembly configuration is shown in broken lines, with the crank and connect- ing rod colinear. The dimensions Cr, Ca, Ca, Cu, Cs, +, and R are known. a. Set up the equations governing the secondary position variabies (do not solve); b. Set up the equations to be solved for the velocity coefficients; e. Set up the equations for nimerical determination of the velocity coeffi- cient derivatives. r  76 Single Degree of Freedom Lintages Problem Set 7 2.29 The wheel rolls to the right without slipping on the horizontal surface as shown, À slider on an overhead guide at height D is coupled to the wheel by à connecting rod of length L. The dimensions L, D, R, and » are known. When the process begins with q = 0, the wheel is at a known position Xp = Xv. a. Write equations solvable for the secondary position variables (do not solve); | b. Writo the equations for numerical determination of the velocity coeffi- cient derivatives. . = k Dn e He . | , 2 2.81 A disk rolls without slip on the circular arc, dragging the slider , N and connecting rod. When q = 0, the connecting rod is horizontal, with Ph: + Ve the pin joining the connecting rod to the disk located on the Y-axis, as : shown in broken line. The dimensions L, Ro, R, ari Ry are known values, x a. Setup the equations governing the secondary position variables (do not ; E Y í . “ solve); Ro. b. Set up the equations for the velocity coefficients; 2,36 A right-triangular connecting link with sides L and H joins a c. Setup the equations for mumerical determination of the vclocity coeffi- » cent derivatives. slider to a wheel. As the slider moves to the right, the wheel rolls without . - shpping on the incline. The process begins withg = 0,B =0,andS = 5. Later, when q = 5in., the secondary variables have the values 4 = 0,7094 ad, 5 = 0.3642 rad, S = 14.27in. Wheng = 5in.,q = 16.5in./s, andj = —27.7 in./s?, determine numerical values for . Base coordinates for point G; . Velocity coefficients for point G; . Velocity coefficient derivatives for point G; , . Velocity magnitude and acceleration magnitude for point G L=220in. D=182M4in. Rj=60in. C=0482im. Ra =70im, S, = 10.9495 in. H=40in. 2n gp 82 Single Degree of Freedom Linkagos 2.87 For the clamshell bucket of problem 2.36, the following data apply: Di=5.0in. D; = 6.0in. Do =220in. C = 1,7825 radians a. Solve the position equations numerically and make an overlay plot of the mechanism for one-inch increments on q from q = 18 inches to g = 28 inches; b. Determine numerical values for the velocity coefficients K, and K, at each of the positions determined in part a; c. Plot the velocity coefficients K, and Ky versus q using the data from part b. 2.38 The mechanism shown is called a trammel crank drive. There are two sliders attached to the connecting rod. These sliders move in grooves on the crank plate, The same unbroken connecting rod extends to the external slider. The only known dimensions are L; and L;, the two segments of the connecting rod. a. Set up the equations required for numerical determination of the sec- ondary position variables (do not solve); b. Setup the equations governing the velocity coefficients; e. Set up the equations to be solved for numerica! determination of the velocity coefficient derivatives; d. TÉ the extemnal slider is used in a punch action, how many cutting strokes does the punch make per revolution of the crank plate? Problem Set 83 2.39 The mechanism shown is called a “quick return.” As the crank rotates, as described by q, the two sliders must move in their respective guides with positions described by X and Y. The dimensions R and L are known values, a. Formulate the equations that determine the secondary position vari- ables (do not solve); b. Set up the equations for determining the velocity coefficients; e: Jf the crank rotates with constant velocity j, for what crank position is the absolute value of the velocity of the horizontal slider at its maxi- mum? lis minimum? : 84 Single Degree of Freedom Linkages | 2.40 The mechanism shown is called Peaucellier's straight line mech- amism. As the crank is moved through the angle q, the point P describes à straight line as shown in the short broken line. The lengths Ly, La, and R are all known values. Give an analytic proof that point P moves in a CHAPTERS3 - Multidegree of Freedom Linkages | Linkages and other mechanisms with multiple degrees of freedom require | . multiple generalized coordinates that are equal in number to the number of 1 degrees of freedom. The linkage configuration is not fully determined un ui à ali of the generalized coordinates are specified. For kinematic analysis, this is equivalent to saying that there will be multple, assigned primary vari- : e ables. This has relatively little effect on the position solution, but it compli- : E cates the velocity and acceleration analyses. in Chapter 2, in the context of single degree of freedom mechanisms, most of the ideas and tools required for this chapter were introduced. These include primary and secondary variables, position loop equaions, the Newton-Raphson numerical solution technique, the Jacobian matrix, velocity and acceleration equations, velocity coefficients, velocity coeffi cent derivatives, and body coordinates. With mostly minor modifications, all of these will be studied in this chapter as applied to mullidegree of ; freedom systems. Because there are few new theoretical tools to be devel- opeg, most of the presentation will be by means of examples. 3.1 KINEMATIC ANALYSIS IN CLOSED FORM As seen in the preceding chapter, the kinemátic analysis for some mecha- nisms can be carried out in closed form. Although this is true for only a e limited class of mechanisms, it is useful to consider this case for the insight it gives into the analysis process; sometimes the logic of the analysis is clouded by the use of numerical solutions. In the following two degree of freedom example, the position, velocity, and accelerations for the second- ary variables are determined im closed fórm. The modifications required for 85 86 Multidegree of Freedom Linkages multiple degreés of freedom become evident in the velocity and accelera- tion analyses. The motion of a particular point of interest is also developed. 3.1.1 Sliding Four-Bar Mechanism A slíding four-bar mechanism is shown in Figure 3.1, It differs from the conventional four-bar linkage in that the pivot joining the second crank and the coupler is on a slider, so that the effective coupler length is variable. The system has two degrees of freedom that are associated with the two crank angles, q and g,. The secondary coordinates are the coupler inclination angle, A, and the effective length of the coupler, B. The point P is a point of interest on the coupler link, located by body coordinates (Up, Vo). Determine the positions, velcities, and accelerations for the secondary variables, as well as the motion of lhe point P. Position Analysis. The mechanism involves a single position loop that provides tivo scalar position loop equations: hi=Cicosg + BeosA-Cicosga—C3=0 h=Gsing+BsinA-Csing=0 These equations are to be solved for the two secondary coordinates, A and B. Although numerical methads are certainly applicable, in this case a closed form solution is also avaijable. When B is eliminated, the result is Cesinga = Crsing tan ds C4+ Crcosga- Cicosg Pp Vo A FIGURE 3.1 Sliding Four-Har Linkage o , 3.1 Kinematic Analysis in Closed Form 87 For the linkage proportions indicated in the figure, -m/2 < A < m/2. Consequently, this equation can be properly solved for A using the princi- pal value of the inverse tangent function. This value is then available for the solution for B.., ENE Vo p - Cat Cacos gy Ci cos q cos À Velocity Analysis. The velocity loop equations are the result of time dif- ferentiation of the position loop equations; they are —Cijising, + BcosA — BÁsinA + Cajasinqa=0 Cijicosg + BsinA + BÃÁcosA — Cajacosq =0 Recalling that q1, 92, fi, and > are to be assigned values, and that the values of 4 and B have already been determined, these are à pair of linear, simultaneous equations in the two unknown velocities, À and B. When these are cast in matrix form, the coefficient matrix on the left side is recognized as the Jacobian matrix. -BsinA cosÁ à] | Gsing —Co sim galfãr BcosA4 sinAlB) [-Ccosg Crcosgllãr For a closed-form solution, this equation is premultiplied by the inverse of the Jacobian matrix, and the result is algebraically simplified to give (3 e eo cos (gi — 4) (Co/B) cos (A — “o B G sin (gy — A) Co sin (A — q2) da, This shows that the secondary velocities, À and 6, can be expressed as linear combinations of the two primary velocities, q; and 42. The velocity coefficients for this case form a (2 X 2) matrix, rather than simply a column vector (as in the previous chapter): Ka Ki 9=[ a ] Ko Kia o [ee cos (q — A) (Ca'B) cos (4 — ] Crsin (gy — 4) Cosin (A — q3) Then the relation betweén the secondary velocities, velocity coefficients and primary velocites is 6) ” le rala B Ko Kuilgo In general, for the multidegree of freedom situation, the velocity coeffi- cients will form a rectangular matrix, rather than the column vector associ- ated with the velocity coefficients for a single degré £ freedom. The fact  52 Maultidegree of Freedom Linkages However, the numerical evaluation is obtained beiter as the solution of à system of linear simultaneous equations. With the velocity coefficient de- rivative matrices determined, the secondary accelerations can be evaluated from the expression just developed. For the following example, secondary positions, velocities, and accelerations are determined as well as center of mass positions and velocities. 3.2.1 Four-Bar Mechanism with Translating Crank Pivot The mechanism shown in Figure 3.2 is called'a “four-bar mechanism” with translating crank pivot. In reality, itis not a four-bar mechanism at all because there is no fourth bar of fixed length. The terminology only suggests the association. The two primary variables are the crank pivot location and the crank rotation, go and q1, respectively. The two secondary variables are the ángles 4; and Às. The second crank pivot is stationary at the point (X«, Y;). Position, velocity, and acceleration are to be determined for the secondary variables, and the position and velocity are required for each center of mass. Position Analysis. The position loop equations depend on both of the primary variables, as follows: go + Cicos q + Cacos As + Cicos As - K=0 Csing+CsinA, + CssinAs-Y,=0 For specified values of gg and q1, these two equations are to be solved for 4; and As. A closed-form solution may be possible, but a riumerical solution will certainly be simpler to obtain. Such a solution is assumed to have been completed so as to proceed to the next stage of analysis. FIGURE 3.2 Kinematic Skeleton Four-Bar Mechanism with Translating Crank Pivot 3.2 Kinematic Analysis with Numerical Solution 93 Velocity Analysis. The position loop equations are differentiated to give the velocity relations: 74) - [89 E 7 2. , —Co sin Ay —C sin As o o r É, sim el Cocos Az Cacos Ás)lÃs 0 -cicosgillgr The left side of this equation consists of the Jacobian matrix that premult- plies the columi vector of secondary velocities, The right side hos the coefficient matrix [B] multiplying the column vector óf primary velocities. A mumerical solution: of this system of lincar equations is accomplished readily, with the result expressing the secondary velocities À; and Às as Jinear combinations of gp and: GRI Àa) 4 Ki Kay] fão bi edil EA Acceleration Analysis. Differentiating the velocity equation above gives the secondary accelerations: Ào do ão fo = dolLo] + AL +] | 5 folhal G felt E) lr The necessary velocity coefficient derivative matrices are determined as solutions of the following two equations: O 0] [Cikacos A CsKaucos Ay] UlLal = l; j [e sim 42 CaKoo sin q] xi O Cicosg CoKam cos A Cskam cos q] fe (lil = k Cy str “| t po Sim Az ak sir As, The secondary accelerations are now readily evaluated in terms of the primary velocities and accelerations and the cocfficient matrices [K], [iol, and [Li]. Base Coordinates and Velocities. Consider next the matter of base coordi- nates and base coordinate velocities for the centers oí mass for the three links as shown in Figure 3.3. The centers of mass are located in their respective body coordinate systems by the body coordinate paírs (Us, Vo) (Uco, Voa), and (Ua, Voa). The center of mass position. vectors are written by inspection: Xe] [ão + Us cos q — Va sin " E o oo sing: + Va cos q Ko do + Cy cos q; + Ua cos Az — Va sim “ (3) ” E sin gy + Lo sin Az + Vo cos Ay  94 Multidegree of Freedom Linkages - vao x FIGURE 3.3 Pictorial Representation for a Four-Bar Mechanism with Translating Crank Pivot Showing CM Locations Xe (Xu — Us cos As + Va sin 3) Ya) Lys - Us sim As — Va cos As When these expressions are differentiated with respect to time, in addition to terms involving do and qr, there will also be terms involving 42 and Ás. The elements of the rectangular velocity coefficient matrix will be required to re-express the latter in terms of jo and qj. With the required substitutions made, the center of mass velocity components are E] o | —Ua sin q — Va cos ci Ya 0: Uacosgm— Vasing]lá = [ki] E EA E . e e] fo = [K) tj Ya Kao Kymlljs dt — Kao(Ua sin A, + Va cos 43) = sing — Kin(Ulo sin Ag + Va.cos 42) Kyo = Kan(Uo Sin 42 — Vo sin 42) Ka = Ci cos qu + Kan(Ua cos Ao — Va sin Aa) EEE eg Ya, Ka Koji A 3.3 Numerical Solution for Multiloop Mechanisms 95 where Kog = Kaso(Usa Sin As + Vos cos 43) Ka = Ka(Uo sin 43 + Va cos As) Ko = Kao(-Utg cos Ag + Vis sin Aa) Kay = Kigs(-Ua cos Ag + Voa sin Aa) For some purposes (see Section 8.4 for details), it will be useful to combine the foregoing into a single velocity coefficient matrix, which gives all of the center-of-mass velocity components and the rotation rates for all links, Such a velocity coefficient matrix will be (9 x 2), and may be defined as às [Ki Ya Ka 7 a (o o a Gr E) Ya A 01 À Ko Ka Ay Ko Kas It is a simple matter to evaluate this (9 x 2) velocity coefficient matrix numerically. Note that for this particular problem, it is also possible. to obtain analytic expressions for all of the elements of the velocity coefficient matrix. 3.3 NUMERICAL SOLUTION FOR MULTILOOP MECHANISMS Each of the examples considered to this point have involved only à single position loop, resulting in two scalar loop equations. However, this is by no means the only case. As with single degree of freedom mechanisms, there are multiloop, multiple degree of freedom systems that result in one pair of position loop equations for cach independent loop. This increases the number of equations to be dealt with simultaneously, but through matrix notation this is not significantly more difficult. The following exam- ple will demonstrate the process. 3.3,1 Power Shovel A typical crawler-mounted power shovel is shown pictorially in Figure 3.4. The three hydraulic cylinders control the motion of the boom and the bucket, and immediately suggest that the mechanism has three degrees of freedom.  » Multidegree of Freedom Linkages /) UA FIGURE 3,4 Pictorial Representation of a Power Shovel The operator, by means of the hydraulic control valves in the cab, controls the lengths of these three components, Consequently, these may be considered as the three input variables. The dimensions and Kinematic variables are shown on Figure 3.5, where the three hydraulic cylinder lengths arc identified as 1 9a, and ga. The dimensions Cy, Ca, . - Cas are lengths from a pivot or welded intersection to the next such point, without regard for the fact that a member may continue beyond the end of a specific dimension. The angular dimen- síons Ci and Ciz are fixed and known, The secondary variables are 4,, B Aa, Ba, As, Ba, As, and Ba; E is also an unknown, but it is readily oxpressible Bm FIGURE 3.5 Kinematic Skeleton for the Power Shovel Showing Segment Lengths, Angles, and Loops a 3.3 Numerical Solution jor Muitiloop Mechanisms in terms of Ay. Note that all of the As are measured from either a horizontal or vertical reference to once of the machine members. The Bs, on the other hand, measure the relative angle from a major member to 5 nearby member. Position Amakysis. For the position analysis, for loops must be de- scribed. They are identified on Figure 3.5, Before writing the loop equi tions, note that the angle E is dependent on Ay; E=Cu- A qi!2 The loop equations are Loop 1 Di-qcsB:—CsinE=0 De- q snBi+CicsE=0 Loop 2 Cs cos Ay + Ca sin Ay + Cocos A Cs sin Ay — Cs cos Ay + Co sin Az + Cy cos 42 — q — CysinAo— queos (A+ Bi) =6 n(4+8)=0 Loop 3 Cs cos Ay + 43 sin(Ão + Ba) — Ci cos As — Cio sin A = 0 Cs sin 42 — qa cos(Ãa + Ba) + Cap sin 43 + Cr cos A =0 Loop 4 Cy cos Ay + Css sin(Ão — By) — Cia cos As — Cn sin 42 =0 —Ci sin 43 — Cu cos(Ãa — Ba) + Cu sin As + Cu cos Az =0 One approach to this problem would be to treat this simply as a system of nine nonlinear equations to be solved simultaneously. Altornatively, be- cause the equations above are weakly couple, they can be considered as four pairs of equations, each pair to be solved in tum. If this altornate approach js adopted, the equations from the first loop give values for B; and E; these are used to evaluate 4, and E is not required agair. Thercaf- ter, the equations of the second loop give 47 and Ba, and so forth. Velocity Analysis. “The velocity arialysis is based on time differentiation of the position loop equations. With E elirinated and the remaining eight equations differentiated, the resulting systêm of equations can be written in matrix form as gm ts = [Bl tg (8x8) (8x1) (8x3) (3x1) “a | | | | : | | W2 Multidegree of Frecdom Linkages | ; / / A : dederdecdo dido dd dee a sf] i St, FIGURE 3.6 Computer Generated Configuration Plot for the Power Shovel column? Do the signs of the various elements appear reasonable for the configuration shown? All of these questions should be readily answerable, if the material presented has been understood. ; With the machine in this same position, the velocity coefficient matrix 1L:] is evaluated as described earlier, The result is ' O.iz8812 . 0.0 0.0 —-0.088756 0.0 0.0 | 0,1728812 0.0 0.0 | [El = (003) Jo co -0.178812 0.0 0.0 0.0 0.0 0.0 —0.165373 —0,010753 0,160926 —0.096208 0.076981 —0.073073, At this point, evaluation of the secondary accelerations must wait, pending Í numerical evaluation of the two remaining velocity coefficient partial deriv- ative matrices. 3.4 GENERAL ANALYSIS FOR" . MULTIDEGREE OF FREEDOM MECHANISMS “ As shown by the preceding examples, there are mechanisms for which more than one generalized coordinate is required to specify the system configuration; these are multidegree of freedom mechanisms. Such mecha- nisms may involve one or more position vector loops, as has been demon- 3.4 General Analysis for Multidegree of Freedom Mechanisms 103 strated. In this section, the analyses demonstrated in the preceding exam- ple problems are summarized in general form. The notation is as follows: Ny = number of generalized coordinates N; = number of secondary-coordinates mn 5; = secondary coordinates, i = 1, 1... N; u generalized coordinates, i=1,2,...Ni Position Analysis. The position analysis begins by formulating the posi- tion vector loop equations in scalar form, just as for a single degree of freedom case. Any necessary constraint equations are also included with the position loop equations, This gives rise to a system of simultaneous, nonlinear equations of the form falga qr Qui, Sr, Sar São SN) =O fog gr cc qui Si Sr 83.» Sm) =D With all of the generalized coordinates specified, this is a system of No equations to be solved for the N; unknown secondary coordinates. These may be solved in closed form or numerically using the Newton-Raphson method. The fact that there are multiple generalized coordinates has no effect on the solution for the secondary coordinates because all of the gencralized coordinate values are specifed. The position solution vector, 15), and the residual vector, (F), are defined as before: 18) = Colts, sa, Sa... na) tEP = Colkfi, fr for. - fra) The Jacobian matrix is dh 2h dh dh às ds às 95no Et = los às 38 àsnz às . . . . . afnz dfna Gfw Sfm Os Os 883 sn The significance of the Jacobian matrix in the Newton-Raphson solu- tion of the position equations, the velocity solution, the acceleration solu- tion, and the numerical evaluation of the velocity coefficient derivatives has been discussed previously. From this point on, it will be ássumed that the position solution is complete, and that values are available for all posi- tion variables.  : : i E 104 Multidegree of Freedom Linkages Velocity Analysis. The velocity loop equations are obtained by time dif- ferentiation of the position loop equations (and any necessary constraint equations). Assuming no explicit time dependence, this differentiation may be written in matrix form as poaoioBr Gl = t q 1 ds ] [ é ] = No X N1) (Nox No) (Nx) . (No x 2) When the partitions are separated, the result is a system of linear simulta- neous algebraic equations, to be solved for the secondary velocities; that is, is -- Ella The cosfficient matrix on the left side is the Jacobian matrix, as seen before The formal solution, in terms of the inverse of the Jacobian matrix, shows that the secondary velocities can each be written as a linear combination of the generalized velocítics, º 181=- [PL a os! Lag (65 = [KKg) where the velocity coefficient matrix, [KJ, is --[ rt m-lollãl (Na x Ni) (No X No) (No X Ni) Acceleration Analysis. Matrix notation is awlkward for a completely gen- eral analysis of the secondary accelerations. Matrix notation & [mica to one- and two-dimensional arrays, but a three-dimensional array would be most convenient. Nevertheless, the following approach works well. The secondary velocities are expressed in terms of the velocity coefficients and the primary velocities as À 16] = IKHG) When this relation is differentiated with respect tó time, the chain rule is required to reflect the changes in [K] as the generaliz h paso to refe gesin [X] as the generalized coordinates change ax] (8) = 6 lg + qo O om = [LONG + qollago + + IKHO) o ++ IRIA 3,5 Conclusioy 105 11X] is known in explicit analytical form, the required derivatives can be evaliated. This is a direct approach to evaluating ihe secondary accelera- tions, TF [X] has only been determined numerically, then a numerical evalua- tion of the velocity coefficient derivative matrices is also required. Fer this purpose, consider again the relation defining the velocity coefficient matrix: ab] = [Eb [É tm oq When this equation is differentiated with respect to a typical primary vari- able, q, the result includes the required velocity coefficient derivative matrix as a factor in the second term, as follows: [sil + Bálto = - Laçóm] The velocity coefficient derivative matrix [Ln] is then evaluated by the numerical solution of the system. of equations: i 2£ 2 Gdica = = [ad 68 [ço] | nós Jnôskl — qnd; The coefficient'piatrik on the leít side is the familiar Jacobian. matrix that is known explicitly. The first matrix on the right is the partial derivative of the Jacobian matrix with respect to the primary variables, gn; this can be writ- ten explicitly. The second term on the right is the partial derivative of the previous righthand-side matrix [BL also differentiated with respect to the same primary variable, qn; this is also known explicitly, (Recall that the definition of [B] included a negative sign, here shown explicitiy.) Thus, all of the matrices required for the numerical evaluation of [L,] arc known. As usual, this should be treated as a system of simultaneous linsar equations to be solved. As discussed in Section 2.6 (dealing with the general analysis of single degres of freedom systems), the purpose here is not to provide general formulae for problem solving. Instead, the intent is (1) to show the forma of the eguations to be expected in all multidegree of irecdom kinematic analy- ses and (2) to focus on the analytical process without reference to a specific problem. For problem solving, cach problem should be formulated from fundamentals and carried through to the final results. If the results do not follow the forms indicated here, it is likely that an error has been made. 3.5 CONCLUSION For multidegree of freedom mechanisms, the kinematic analysis is on ex- fension of the ideas introduced previously for the analysis of single degree of freedom mechanisms. Although the number of primary coordinates  106 Multidegres of Freedom Linkages increases in the case of multiple degrees of freedom, the position solution process scarcely chaniges at all. The solution for secondary velocities or velocity coefficients is more complicated because the velocity coefficient matrix for multiple degrees of freedom is rectangular, rather than a column vector. The complication extends further to the velocity cocfficient deriva- tives, where partial derivatives of the rectangular velocity coefficient matrix are required. The use of body coordinates and base coordinates for points of interest remains exactly as it was established previously for a single degree of freedom. The velocity coefficients and velocity coefficient partial deriva- tives are, again, rectangular matrices, as noted previously. Despite the increased complexity usually associated with multidegree of freedom problems, for some cases closed-form solutions are available for every step. In general, however, the required solutions will need to be obtained by numerical procedures. The position solution, velocity coeffi- cients, and velocity coefficient partial derivatives—all useful concepts as presented for kinematic analysis -—will also be very significant in the multi- degree of freedom static and dynamic analyses presented in Chapters 6 and 8. REFERENCES Paui, B., Kinematics and Dynamics of Planar Machinery. Englewood Cliffs, N.].: Pren- tice-Hall, 1979. Svoboda, A., Computing Mechanisnis and Linkages. New York: Dover, 1969. Tutte, S. B., Mechanisms for Engineering Design. New York: John Wiley, 1967. PROBLEM SET ; The mechanism shown may be considered a “floating” slider cran) Mechanism because the crank pivot is not stationary. The dimen- sions R, L, and C arc known. Problem Set 107 a. Setup the necessary position equations and solve in closed form for the secondary variables À and X; b. Obtain a closed-form solution for the velocity coefficient matrix; e. Obtain a closed-form solution for the velocity coefficient derivative ma- trices [Li] and [Lo]; d. Using the dimensions given in part £, evaluate 4 and X for q, = 0.85 radians and q, = 0.095 m; e. For q, and q, as in part d, and qa = 1.85 rad/s, 42 = —2.45 m/s, evaluate the velocity coefficient matrix and the velocities A and X; £ Forgy, qu, dr, and ja as in parts d ande, and dj = 11.66 rad/s2, j; = 4.25 m/sº, evaluate the velocity coefficient derivative matrices [L,] and [Ls], and the accelerations 4 and X. R=0.12m L=0287m C=0.008m 3.2 In the modified slider-crank mechanism shown, the plunger op- erates through a guide bushing that itself moves in a vertical guide. The crank radius R and the connecting rod length L are known. a. Set up the position loop equations and solve in closed form for À and X as functions of q, and 42; b. Obtain a closed-form solution for the velocity coefficient matrix; e. Obtain a closed-form solution for the velocity coefficient derivative ma- trices [L4] and [La]; d. Using the data following part £, evaluate A and X for q, = 1.55 radians and q; = 2.27 in. e. Forg; and g,as in part d, and 9; = 26.83 rad/s and 1; = —14.09in./s, evaluate the velocity coefficient matrix and the velocities 4 and X; £ Forg, q, dr, and 4; às in parts dand e, and j = —57.27 rad/s2and o = 8,25 in./s2, evaluate the velocity coefficient derivative matrices [L1] and [La], and the accelerations À and X. R=738in. L=22.85in. Guide Bushing EN fume Nome x —  112 Multidegree of Freedom Linkages is 19, A disk and linkage assembly is initially à 1 it y in the configuration shovir by the broken lines, with the length D; vertical and the length D; horizorital, The horizontal slider is then displaced a distance q; to the right while the disk rolls through the angle q2, where there is no slip between the disk and the supporting surface. The dimensions Rr, Re, Dr, and D are known. o a. Set up position equations solvable for secondary variables A and B (do riot solve); - b. Setup the matrix equation solvable for the velocity coefficient matrix; e. Set up the equations required for numerical determination of the veloc- ity coefficient derivative matrices [L1] and [La]. Rolling Without Ship + 1 2.11 The motion of the disk is controlled b; j in ) y the slider position, g1, and the elevation of the table, qa, through the comnecting rod et length The disk rolls on the table surface without slipping. When qr = qa = O, the gisk e centeres on the able, X = 4 = 0, The dimensions L and D are NOW. : a. Write the position equations for the system; b, Set up the matrix equation to determine the velocity cocfficient matrix; Problem Set 13 e. Set up the equations to numerically determine the velocity cocfficient derivative matrices [L1) and [Lo]. 3.17 For the mechanism shown, the acm is held in contact with the disk at all times by a guide (not shown), and the disk rolls-without slipping on the arm. The initial state is indicated in broken line, with the slider (Dat the origin and the arm along the X-axis. As q, increases, the slider moves to the right and the arm moves past the underside of the disk, thereby increasing 4. As q2 increases, slider (1D, carsying the disk pivot, moves upward, causing the angle A to decrease. The dimensions €, D, E, R, and Ware known. a. Formulate the position equations (do not solve); b. Develop equations solvable for the velocity coefficient maírix (in matrix form); c. Setup the equations to solve for the velocity coefficient derivatives [14] and [L9]; d. Determine the base coordinates for the point M; e. Determine the velocity coefficients for point M; . Determine the velocity coefficient derivative matrices for poini M. | Disk y 1 Radius R Ay aeee; * 3.14 A small lathe is driven by an electric motor with a gravity-ten- sioning arrangement as shown. At rest, the system configuration js as shown in the broken lines. When power is applied, the motor torque puts tension in the lower side of the belt, causing the motor-to rise and the upper side of the belt to slacken. The rotations of the lathe spíndle and the motor pulley are q, and 92, respectively, and the length of the tight side of the beltis D. The dimensions C;, Ca, Ca, Cs, R1, and Ro are known. a. Setup the position equations to be solved for 4, B, and D (do not solve);  14 Multidegree of Ercedom Linkages b. Set up the equations to be solved for the velocity coefficient matrix (in matrix form); c. Set up the equations to be solved for the velocity coefficient derivative magrices [L1) and [Li]. “8.14: The mechanism shown was used formerly as a mechanical ana- log tomíputing component. The vertical motion R results from the two input motions q1 and q2. The angle between the two arms is a right angle. . Determine R as a function of q: and qo; Determine the velocity coefficient (row) vector (Ka, Ka); . Determine the acceleration É in terms of q1, d1, di, G2, ja, and jo; . Forq =35in, À = 2.75 ins, da = —4.55in./92, q, = 118in, do = —6.29 in./s, and q = 3.86 in./s?, evaluate R, R, and k. po rp Problem Set 115 3.15 The figure shows a variation of the mechanism in Problem 3,14. The difference is that the arms no longer form a right angle; the included angle is now C. a. Determine the response R.as a funcãon of q; anda; b. Determine the velocity coefficient (row) vector (Kr, Ka); c. Express the acceleration É in terms of 91, Gu, da, 92, dz, and da. A BU ———— Lo or. In mechanical analog computation, the mechanism shown was callsd-a Summing linkage. The distances B and C are known, Do not assume that the angle À is small. q a, Determiné the response A and X in terms of q, and qu; b. Determine the velocity coefficient matrix in closed form; e. Determine the velocity coefficient derivative matrices [L1] and [L2] in closed form;  116 Multidegrec of Freedom Linkages d. For the dimensions and state described, evaluate X, X, and X. 0.15 m €C=0.08m n= 0.135 m g2 = 0.029 m f = 0.95 m/s 9 = 0.83 m/s jr=-015mis? jr= 054 m/s 3.17 The mechanism shown provides a variable stroke Tength, de- pending on the position of the control arm, 43, as the input cranlk rotation, 41. moves through a revolution. The stroke is minimum for q; = O and maximum for q; = /2. The dimensions L, Ry, Rs, and S are known. a. Determine the position equations for this mechanism (do not solve); . Set up the matrix equation to determine the velocity coefficient matrig e, Set up the equations required for a numerical soluti i Pp U ution for the vi coefficient derivative matrices [Ls] and [Lo]; “loctty &. Obtain a closed-form solution for the stroke (stroke = — ano 08 pe ano stroke (stroke - Xmas — Kin) AS à Ba ma Stroke Xnax CHAPTER 4 Cam Systems 4,1 INTRODUCTION A cam is a solid body shaped such that its motion imparts a presexibed motion to a second body, called the follower, that is maintained in contact with te cam. The shape of the cam and the physical relation between ihe cam and the follower define a particular functional relaúon between the cam position and the follower position over a range of cam positions. Using cams is one of the simplest ways to generate complex motions rwith high repeatability and reliability, and with reasonable costs. For these rea sons, cams are widely used in many types of modern machinery. There are many types of cam and follower systems, some of which are illustrated in Figure 4.1 and Figure 4.2. The cams shown in Figure 4.1 are of the type knowm as disk or pinte cams; this type of cam will be the principal subject of this chapter. Disk cams are commonly fourid in internal combus- tion engines, timer mechanisms, machine tools, and numerous other de- vices. Other types of cams are shown in Figure 4.2, including the wedge cam, the barrel or cylinder cam, and the face cam. The wedge cam is used ín some vending mechanisms, whereas the barrel cam can be found in vari- ous machine tools. The cam types shown it! Figure 4.2 are less common; they will not be considered further here. .The four cam-follower systems shown in Figure 4.1 all use the same disk cam, but each illustrates a different follower type. These followers ray be classified according to the type of surface that contacts the cam and to the type of motion given ta thie follower, The followers shown in the top row, Figure 4.1(a) and (0) are both flat-faced; ioller followers are showm in Figure 4.1(c) and (d). The folower motions in the left column are transla- 7  122 Cam Systems with respect to time, The acceleration is obtained by another differenti- ation, HA) = À FA) + AFA) JE FA) is very large, this indicates very high acceleration of the cam fol- lower. Discontinuities in the second derivative are likely to occur at the transition from one displacement description to the next; (A) is often zero on one side of the junction and non-zero on the other side. If (A) is discontinuous, then the cam-follower acceleration is also discontinuous. When this happens, the contact force between the cam and follower is discontinuous, which is an impact situation. This leads to surface damage and accelerated wear of the cara and follower, as well as exciting vibration throughout the remainder of the system. Evidently, in choosing the dis- placement function, it is necessary to consider carefully the first and sec- ond derivatives as well as the function itself. Consider the construction of .a very simple rise curve, constructed from parabolic segments: The rise interval extends from zero to 4:, and the lft is LV A single parabola cannot be adjusted to fit smoothly over this entire interval, but using two parabolas, one can be fitted at each end and the two can be joined in the middle. Thus, consider the two parabolas p=kA4 Bb=L-K (4-4)? with derivatives : CD Fy=2MA Ft =2K Ff = SKA A) Bf= —2k Itis readily evident that Fi(0) = Fy/(0) = 0, FA) = L, and F'(As) = O; this assuresa smooth transition from the rise curve to the adjacent dwells. What remains is to make the curve continuous over the rise interval. If the two fúnctions E, and F; are to meet at the center of the interval with a common slope, t P(A) = Fo(Ai/2) Fy(AVD) = Es'(AN2) From these two conditions, the values for K; and K, must be K = K = 21/47 Thus, the complete description “of the rise cúive for this example is as follows; HA)= 2L(AIA 0SA<Ail2 (4 — AR 2 =L-2L A AÍLEASA, 4.2 Displaceiment Functions and Graphical Cam Designs 123 This function and its first derivative are continuous throughout the entire interval, including the end points. The second derivative, f'(A), is continu- ous everywhere except for à finite discontinuity at the ends and the mídcie oÉ the interval where the two parabólas meet. Developing a.similay retuun curve appropriate to the interval 44 5 A < Asisa useful exercise; the process is illustrated in the next example. Consider next the development of a sinusoidal displacement function. The required function will provide a sinusoidal rise over the interval zero to A, à dwell at full lift from A; to Ao, à return over the interval Az to Às, anda dwell at zero from Ag to the end of the cycle. The required sinusoid is o£ the form (1 — cos X), and it must move through one half cyclein the interval zero to 4; and again in the interval Az to Às. To express the return curve, given the form fox a rise curve, two substitutions are required: (1) replace the argument 4 with 43 — À and (2) replace the phase duration 4: with As — Az. Note that the range of the function (1 — cos X) is from zero to two, so the coefficient is L/2 rather than L. The required displacement function is given piecewise as follows: 0=A<A HA) = (LDU — costmAA)] =L ASAS A = (LB — cosim(Ãa — AJA — ADI ASA<As =0 ASA Notice that although the function and the first derivative are continuous over the entire cycle, the second derivative has a finite discontinuity at each of the break points. e The matter of choosing a suitable displacement function for a particu- lar design assignment is an ast, and has been explored extensively for many years. Many types of displacement functions have been studied, including sinusoidal, cycloidal, and polynominal fopioos. Several ele- mentary displacement fânctions are listed in Table 4.1: Of the functions tabulated, only the cycloidal displacement function is not restricted to low- speed operations. The parabolic and sinusoidal funcions have finite dis- continuities in the second derivative at both ends of the rise interval, and the parabolic and cubic curves have second derivative discontinuities at the middle of the rise interval. An extensive presentation of the topic is beyond the scope of this book, but further information can be obtained from the references at the end of this chapter. The volume of published work on this matter attests to the significance of cams as useful machine elements and to their often difficult design requirements. The displacement function fuliy specifies the required motion for a particular cam. As such, it provides all oÉ the Kinematic information reguixed to design.the cam and, in the past, was used directly in graphical cam design techniques. Although such graphical methods are no longer recommended, they are useful in clarify- ing the relationship betwcen the displacement function and the cam pro-  E 124 Cam Systems TABLE 4.) Elementary Rise Curves Name Functional Form Parabolic MA) =2UA/A? 0=A< AZ = LI -2KA — AA) APZA<A, Cubie FA) = AAA 0=A <A? = LI — aa — AVA) AUISASA, o L zA simusoidal sta) = 51 - cos (E8)] 0=A<A DLTEA A (274 Cyeoidal HA E [é à sm (5)] 0<A<A where FLA) = cam-follower displacement É = lift, maximum followes displacement A am rotation . A = angular duration of the rise file, For this reason only, two graphical cam design procedures are dis- cussed in the following sections and several graphical design problems are also included at the end of the chapter. Graphical Design of a Cam with a Flat-Faced Translating Follower The graphical design process for this case is shown in Figure 4.4. The development of the cam profile is done in two parts: 1. The radiai position of the cam follower is determined for each of a sequence of cam angular positions; 2. The cam profile is drawn as a smooth curve, tangent to each of the folower face positions in the sequence. The design begins with a layout of the previously chosen displacement function in the upper right corner of the Figure 4.4 (the displacement diagram). Note that this particular displacement function has no dwell at full hft; the return begins immediately after the rise. The abscissa of this plot is divided into a number of points over the cam rotation cycle (12 points are shown in Figure 4.4). Construction then moves to a point chosen for. the'cam axis of rotation (there is a comment later about the choice of this point) where radial lines are constructed at angles equal to the angular divisions of the cam cycle. These radial lines are numbered corresponding to the angular division points. The base line of the displacement diagram is then, projected to intersect a vertical line through the cam axis of rotation, and the cam follower is outlined in this position, as shown. A circle, cen- B a 4.2 Displacement Functions and Graphical Cam Desigas Displacement Diagram ZA sw 7 8 35101127 2 me ciute 2 Prime Ciscle & Base Circle a FIGURE 4.4 Graphical Design of a Cam wlth Plat-Paced Translating Foliowor tered at the cam axis of rotation and tangent to the foliower face lines, is drawn next; this is the base circle for this camfollower System. “The ordinate corresponding to each of the cycle division points is them laid off radially outside the base circle. This process is performed graphi- caliy by projection; the broken lines show the transfer of the ordinate from point é on the displacement diagram to the vertical line representing zero cam displacement, then by rotation about the cam axis of rotafion to the intersection with the radial line 4. At this point a perpendicular to the radial line is erected, representing the follower face. This process is carried out for each division point, resulting in a sequence of perpendicular that bound the cam profile. The cam profile is then drawn asa smooth curve that is tangent to each of the follower face images. The process just described should be reviewed carefully to make clear the manner in which the cam profile is determined from the displacement function. Noté that the cam follower is also involved in this determinanion because à sequence of images of the flat follower face define the boundaries of the cam profile. However, note that the vertical location of the point representing the cam axis of rotation is actually a design decision because this determines the base circle radius. Finally, the size of the base circle controls the overall size of the cam because the cam profile is developed outside the base circle. Graphical Design of a Cam with a Radial, Translating Roller Foliower “Figure 4.5 describes this graphical design process. The same two-stage development applies here as was used for the cam with flat-faced transiat- ing follower:  126 Cam Systems 1. The radial position of the cam follower is determined for each of a - sequence of câm angular positions; 2. The cam profile is drawn as a smooth curve tangent to each of the follower outlines in the sequence. There is one major difference between this problem and the previous problem; here, the follower is circular instead of flat and, consequentiy, its images will be circular arcs. Although this will change the graphical pro- cess distinctly, the two solutions still have much in common. As before, the design begins with a layout of the previously chosen displacement function in the upper right comer of Figure 4.5. The cycle is divided as before, with 12 divisions also used for this example, This is followed by erection of radial res from the cam axis of rotation at the angular positions represented by the division points; these lines are then numbered to correspond with the division points. The base line of the displacement diagram is then projected to intersect a vertical line through the previously selected cam axis 6f rotation; this defines the trace point (the roller axis) for zero cam rotation. The follower outline is then constructed about this trace point, as shown. À circle is drawn, centered on the cam axis of rotation and passing through the roller axis. This circle is the prime circle. The base circle is also drawn about the same center but is tangent to the roller in the follower outline at zero cam rotation. The trace point positions are then measured on each radial line, outward from the prime circle. As previously, this is performed graphically by projection. The bro- ken lines in this case show the projection of ordinate 3, first horizontally to the vertical line through the cam axds of rotation, then by rotation about ihat center to intersect the radial line 3. This defines the trace point when Lit Displacement Diagram 121274577 837101127 2 uu 2 Base Circle A Circular Arc. With Roller Radius. Piteh Curve. Cam Profile FIGURE 4.5 Graphical Design of a Cam With Radial, Translating Foliower 4.3 A Kinematic Theorem for Rigid Bodies 127 the cam rotation is 7/2. The roller-follower image is then drawn at this position, shown as an arc with short broken lines. The process is per- formed for the ordinate associated with each division point. In this case it results in a sequence of circular áres that bound the cám profile: The cárn profile is then drawn as a smooth curve, tangent to each of the arcs repre- senting the roller-follower images. The process for the roller-follower design problem should be com- pazed to that for the flat-faced follower, particularly to identify steps com- mon to both. The role of the cam-follower shape in each should also be compared. Two design decisions were made in this process, (1) the assign- ment of a follower roller radius and (2) the assignment of the prime circle radius. Both of these choices were made arbitrarily in the present develop- ment, yet they have a significant impact on the size, stress conditions, and function of the final design. There will be more discussion on these matters . in the later sections that deal with analytical design methods. 4.3 A KINEMATIC THEOREM FOR RIGID BODIES Consider a rigid body rotating about a fixed axis at point O, as shown in Figure 4.6. There are two intersecting lines inscribed on the body, 0-0" and P-P*, The rotation of the body is described by the angle A(t), mea- sured from the stationary X-axis to the line 0-0". At the instant shown, the line P-P” makes an angle B with the X-axis. A: this instant, a unit vector along the line P-P' is Ep, which is defined as E=icosB+jsinB The intersection of lines O-O! and P-P” defines a point denoted by di The position of that point is Rj, where Ri =|Ri](icos A + jsin A) and the position of any other point 2 on the line P-P* is Ra, where R=R; + ED The velocity of point 2 is Va, which is expressed as W=kÃxR =kÁx(R + E D) =Y+kxBEáD where Vj is the velocity of point 1. The component of Va in the direction of the line P-P'is : VW E=E (G+kxEÁD) =B:V+E-(kXEB)JÁD = BY  132 Cam Systems edge contact of two circular disks is given by Budynas.? For the contact stress calculation, consider the following definitions: * P, and P; are the radii of curvature of the two disks at the point of contact; * E; and E; are the Young's moduli for the two disk materials; « tis the thickness of the disks; * Fis the total force transferred between the bodies. Then, the maximum normal stress on the contact surface is ' FEED) To = Vi Pi Pa (Es + Es) If one of the surfaces is flat, then the associated radius of curvature is infinite, and the contact surface maximum normal stress simplifies to BETA TT VrtP(E + E) where P is the radius of curvature of the curved body in contact with the flat surface. The most damaging stress state occurs a short distance below the contact surface; at that point the shear stress is a maximum at a value approximately 0.300. Radius of Curvature. For proper operation, the cam profile must be smooth and have a continuously turning tangent. If this condition is not met, the cam could have a cusp or “corner.” Ata cusp, the tangent line is undefined, and the radius of curvature is zero. This would certainly lead to stress difficulties because the normal contact stress varies with the recipro- cal of the square root of the radius of curvature. For a iypical cam profile, the radius of curvature varies from point to point along the profile, so that for finite rotations, the radius of curvature must clearly be considered a variable, However, for infinitesimal rotations, the radius of curvature may be considered constant and the location of the center of curvature to be fixed on the cam. Thus, for an infinitesimal rotation, the contact zone may be replaced locally with a circular arc. Furthermore, because the rest of the cam profile has no function for such a rotation, it may have any form whatever. This type of replacement is shown in Eigure 4.8, where à short circular arc replaces the actual profile near the point of contact, and an irregular boundary replaces the remainder of the cam profile. This figure . emphasizes the fact that this part of the boundary makes no difference. The center of curvature is onthe line of coritact at point C”, and the distance from C* to the point of contact is P, the radius of curvature. The 2R. G. Budynas, Advanced Strength and Applied Stress Annlysis, pp. 154-158, MeGraw-FRI, New York, 1972. - 4.4 Analytical Design of a Com With a Flat-faced, Transiating Poliower 433 FIGURE 4.8 Determination of the Radius of Curvatare distance from the cam rotation axis to Ct is Re, an unknown. Each of the angles C and A are measured from the common vertical reference line to a line fixed on the cam, Thus, they differ by at-most a constarit, and dc . ga” +1 Consider now the position equations: DA) =ÍMA)=Resnc. “MA)=Ro+HA)=-ResC+P The negative sign in the vertical equation is required because, as shown, € is a second quadrant angle. These equations involve three unknowns: Re. C, and P. To eliminate two of them in a single substitution, the horizontal position equation is differentiated with respect to 4, as follows: aD A) = de ASA) = RecosCi =RecosC. “Ahen this is applied im the vertical position equation, the result may be solved for the radius of curvature; thatis,. P=Ro+ HA + IMA) Presented in this form, the equation provides 4 means to evaluate the radius of curvature at any point on the cam profile in terms of the base radius and the displacement function.  134 Com Systems Base Radius Determination. The expression just developed can be used as a guide to choose the base circle radius. In most cases, the base circle should be as smail as possible to minimize the material and space required for the cam. The previous relation can be solved for the base circle radius, and then the minimum taken for both sides of the equation, as follows: Ronin = (P — f = [Din LÉ the contact force is constant, such as for a gravity-loaded follower in à slowly moving system, the minimum acceptabie radius of curvature can be determined from the allowable contact stress; this fixes the first term on the right side. The second and third terms on the right side are functions of the angle À, so it is evident that the minimum value for Ro is determined by minimizing the sum of these two terms with respect to 4, Note, however, that what is required is the minimum of a sum, not the sum of individual minima! If the contact force is not donstant, as is usually the case, then the minimum value of Ro must be determined iteratively to assure satisfactory contact stress for all cam positions. Milling Cutter Coordinates. In the manufacture of cams, it is often desir- able to cut the cam profile using a milling machine, The milling cutter is a rotating, cylindrical tool with cutting edges evenly spaced around the cir- cumference. To manufacture a specified cam, it is necessary to know the required location of the milling cutter axis with respect to the work piece to generate each point on the cam profile. This information is given in terms of bocly coordinates fixed on the cam blank material. The following discus- sion gescribes the calculation of these body coordinates for manufacture of a cam to function with a flat-faced; translating follower. The cam profile to be manufactured is shown in Figure 4.9, along with the oufline of the milling cutter. The body reference line OM serves as a basis for a full set of body coordinates (U, V), with U along OM, and V normal to LE such that the angle G rotates from LU toward V. The objective is to specify body coordinates for the milling cutter, (Um, Vm) as functions of the cara rotation angle, A. Then, as A sweeps from O to 2m, ali of the required positions for the milling cutter are generated. It should be noted that these calculations are for a specific value of the milling cutter radius, Ra, and that use of a different diameter cutter will necessitate calculation of different milling cutter coordinates. Tn the position shown in Figure 4.9, the milling cutter axis is located at (Xm, Xw) relative to the X>Y coordinates, These values are: Xn= FA) O Ya = H(A) + Ro = Ro + HA) + Rm The transformation from X-Y coordinates to U-V coordinates requires a rotation through the angle (4 — 7/2). When this is done, the milling cutter 4.5 Analytical Design of a Cam With an Offset, Translating Roiler Follower 135 Y FIGURE 4.9 Milling Cutter Coordinate Determination for Cam With a Flat-faced Translating Follower coordinates are: Um = Xo sin 4 — Ya cos A Va = Km COS Á + Ya sin A ' These coordinates describe the required location of the cam cutter axis to generate the point in contact with the foliower at the rotation angle 4. 4.5 ANALYTICAL DESIGN OF À CAM WITH AN OFFSET, TRANSLATING ROLLER FOLLOWER To maintain the translational response motion of the previous system, but with reduced friction and wear, a cam with a translating roller follower can be used. The foliower motion need not be radial with respect to the cam axis of rotation; the follower may be offset as shown in Figure 4.10. As before, the caih axis of rotation is point O and the cam rotation angle is 4. The rotation is measured from a stationazy line parallel to the follower motion, around to the body reference line OM. For zero cam rotation, the cam follower is at its lowest position, as shown in Figure 4.10(b); the vertical position of the trace point in this condition is a function of the radius of the prime circle, Rpo, and the offset, E, that is, Ho = VRpd — E? The prime circle radius, Rpo, and the roller radius, Rr, are design decisions; for the present they are considered to have bcen assigned (more discussion  136 Cam Systems (a) System in a Typical N Position (bj System in the Reference Position, A=O FIGURE 4.10 Cam With Offset, Translating, Roller Follower on that choice follows later). The position of the trace point as a function of cam rotation is º í H(A) = Ho + HA) = VR? — EP + (A) Pressure Angle. The line of contact is the normal to the tangent at the point of contact between the cam and the follower. This lme necessarily also passes through the roller axis, which is the trace point, The angle 4.5 Analytical Design of a Cam With an Offset, Translating Roller Follôwer 157 between the line of contact and the axis of motion for the follower is called the pressure angle, Ap, às shown in Figure £.10(a). It is necessary to expres the pressure angle, which is useful in itself and as a step toward determin- ing the required cam profile: The following analysis relates the pressure angle to the displacement function. The cam follower must neither penetrate the cam profile nor separate from it. As before, this condition is expressed in terms of velocities by requiring that the velocity components along the line of contact for the particles in contact (one in the cam and the other in the follower) must be the same. The velocity of the contact point im the 'cam follower may be considered to conisist of two parts: the rotational velocity term, and the vertical motion term, H, which is superimposed on all points in the rolier body. The rotational velocity term is normal to the line of contact. There- fore, it contributes nothing to the component of velocity along the line of contact, The component of Ff that is along the line of contact is Fi cos Ap, so the velocity along the line of contact for the contaci partície in the follower is Vi= É cosA, For the contact particle in the cam, the velocity along the line of contact is ve= à 60 To determine the length So, refer to Figure 4.11, where it is evident that 6) =HsinA, + Ecos Ap which makes the cam velocity component V.= A(H sin Ap + E cos 4p) HsinAg FIGURE 4.11 Evaluation of 6Ô  142 Cam Systems ered, The smallest roller radius results in the outermost cam profile, smooth and apparentiy Kinematically acceptable as shown. Consider the largest roller shown; the cam surface ÁB is required to support the roller as it approaches the position of rainimum pitch curve radius of curvature. Then, as the roller moves away, having passed that point, the follower must pass through the profile segment ÁB while being supported on the segment ÁC. This is a contradictory situation; each part of the motion requires material for support that must be cut away to allow another part of the motion to occur. This situation is called undercutting, and it results in a cusp on the cam profile. The distinguishing features among these three rollers are their radii compared to the pitch curve minimum radius of curvature. For any roller zadius less than the minimum pitch radius of curveture, a smooth cam profile with a continuously tuning tangent line Sesults, FÉ the roller radius is greater than the pitch curve minimum radius of curvature, undercutting will always result and the trace point will not follow the pitch curve. For the intermediate case, where the roller radius is exactly equal to the pitch curve prima radius of curvature, the trace paint follows the pitch curve but the cam profile has an unacceptable cusp. Milling Cutter Coordinates. As discussed in Section 4.5, it is often desir- able to determine milling cutter coordinates for manufacturing the cam " profile, this time for à cam intended to function with à translating, roller foliower. The required cam profile and the milling cutter outline are shown in Figure 4.15. The milling cutter xadius is Rm, Whereas the radius of the FIGURE 4.15 Milling Cutter Coordinate Determination for Cam with Translating, Roller Foliower 4.6 Analytical Design of a Cam With a Pivoted, Flat-Faced Follower 143 intended roller follower is R$. With respect to the X-Y coordinate system, the milling cutter axis coordinates are Xm = E + (Rio Rm) sin Ap Ya = H(A) — (Re — Rm) cos 4p With the necessary rotation by (A — 7/2) to the body coordinate system, the body coordinates for the milling cutter axis are Uia = Xm Sin A — Ya cos à Vm = Xy Cos Á + YaSinà As the angle A sweeps through the range 0 to 27, these functions deter- mine all locations for the milling cutter axis required to generate the speci- fied cam profile. 4.6 ANALYTICAL DESIGN OF A CAM WITH A PIVOTED, FLAT-FACED FOLLOWER A cam system with a pivoted, flat-faced follower is called for when an angular motion is required for the cam follower response. À typical system of this type is shown in Figure 4.16. The displacement function, f(A), for this type of system describes the variation in the angular position of the follower, and consequently, is measured in radians. In the analysis to follow, the displacement function and the dimensions C;, Co, and C; are assumed to have been chosen previously. The cam rotates about a fixed axis at O, with the rotation measured by the angle A from a stationary reference line to the body reference line OM. When the cam rotation is zero, the angular position of the follower is Bo; the value B, depends on the base circle radius and the previously chosen dimensions C;, C>, and Cs. As the caih rotates, the follower is required to move according to the relation B(A) = Bo + f(A) Contact Location. The line of contact is the normal to the tangent line at the point of contact; in. this case the normal to the follower face at the contact point. On the follower, the point of côntact is located by the dis- tance D, as shown in Figure 4.16. For sustained contact without penetra- tion, the velocity components along the line of contact for the two pointsin contact--one on the cam and one on the follower—muist be the same. For the contactpoint on the follower, the component of velocity along the line of contact is y=BD Applying the theorem presented in Section 4.3, the velocity component along the line of contact for the contact point on the cam is v.=A 60  144 Cam Systems FIGURE 4.16 Cam With Pivoted, Flat-faced Follower Equating these two velocity components leads to o SO Bd, , arm Pet The next steps arc to eliminate the distance Gl) and to solve for the contact position, D. To this end, consider an equation expressing position mea- sured parallel to the follower face: 6 +D+CsinB=CicosB Using these two relations, the length ÓQ is eliminated and the contact position is determined . D = (Les B = CosnB 1+f(A) í Cam Profile, with the distance D known, the rectangular coordinates (X, Y) of the contact point are readily determined, as follows: - X=G-CsnB-DcosB; PYIG-CeosB+DsnB. The cam rotation, 4, the cam profile polar coordinate angle, G, and the angular position of the coniact point are related, such that : A+G-—-2= Atan(Y, X) The polar coordinate angle, G, is determined from the preceding relation, whereas the cam profile radius, R, is determined by the Pythagorean 4.6 Analytical Design of a Cam With a Pivoted, Plat-Faced Follower 345 theorem: no G=mi2- A+ Atam(Y, X) R=evVE+Y , Asin the previous cases, these equations give the polar coordinates of the cam profile in' parametrio form, vrhere the parameter is the cam rotation angle, A Radius of Curvature. Following the now familiar pattern, the actual cam profile at the contact point is replaced Locally with a circular arc, vehile the rest of the profile is of no consequence; this is shown in Figure 4.17, The center of curvature is on the line of contact at point C*, a distance P below the point of contact, where P is the-cam profile radius of curvature. The center of curvatare is fixed on ihe cam at an unknown distance Rc from the cam center of rotation. The angles C and 4 cach measure thc angular position of a line in the cam from a common reference. Thus, they differ by at most a constant, and Consider now the two position loop eguaions: ResmC+PsinB+DcosB+CysinB-C = -RecosC+PcosB-DsinB+CcosB-C,=0 These involve the three unknowns Re, C, and P. To eliminate Rand Cin a single substitution, the derivative of the vertical equation is required to eliminate the term Re sin C in the horizontal equation: ResinCC'-PsinBB'-D'sinB-DcosBB'- CasinBB'=0 x FIGURE 4.17 Radius of Curvature Determination é | | | |  146 . Cam Systems Itis also necessary to differentiate the expression for the contact location to express D': º em o v ms mao 4 t =fMCr sin B + Cocos B) = f'D | RR With these substitutions, and recalling that C' = +1 and B' = fº, the quantity Re sin C is eliminated and the result solved for the radius of curvature, P. After much algebraic manipulation, the cam profile tadius of curvature is p= (Lt2f ACI Sim E + Co cos B) +fD. e . q+f% ú * 4.7 ANALYTICAL DESIGN OF A CAM * WITH A PIVOTED, ROLLER FOLLOWER A cam with a pivoted, roller follower provides angular follower response combined with low fricion and wear. A system of this type is shown in Figure 4.18.'The displacement function is assumed to have been chosen previously, as are the dimensions Cs, Ca, and Ca. The cam rotates about a fixed pivot at O, while the follower rotates about the fixed pivot at E. The cam rotation is measured by the angle À from a fixed reference line to the body reference line OM. The follower position is described by the angle B, measured as shown in the figure. The angle B is given by B(A) = Bo + (8) FIGURE 4.18 Cam WithPivoted Roller Follower 4.7 Analytical Design of a Cam With a Pivoted, Roller Follotwer . 147 where B, is the angular position of the follower arm when the cam rotation angle A is zero. The value'B,, the base radius for the cam, the roller follower radius, and C;, C2, and C3 are related as follows: (Ry+ Rj = (ED És cos Bor + (Cad Ca sin BO Orientation of the Line of Contact. The angle D is introduced to express the orientation of the line of contact. In some respects, this angle is compa- xable to the pressure angle of Section 4.5. As shown in Figure 4.18, the angle D is measured from a horizontal reference through point E, down to the normal from the line of contact through point E (the latter shown as the broken line VE). Note that this angle also appears several other places in the figure, and in particular, it is the deviation of the line of contact from the vertical. For the point of contact in the cam, the velocity along the line of contact is v.=A 60 For the velocity of the contact point in the follower, recall that the rota- tional velocity component is normal to the line of contact, so that the component along the line of contact is v=BYE (this is unclear, review the discussion in Section 4.5 regarding determina- tion of the pressure angle.) As before, the cam and foliower must remain in contact without penetration, which is expressed by equiring that the pre- ceding velocity components be equal. This leads to é relation between and VE: M-s Mp Evidently, it is necessary to express the length YE from the right triangle involving the follower arm and the side VÊ, as follows: : FE=Gos(B+D) |, =CalcosBcosD-sinBsinD) The position equation in terms of lengths measured parallel to the line VEis 60 + VÊ = CG cosD+CsinD This is a system of three equations in the three unknowns Sa, YE, and D. Whem and VE are eliminated, the result can be solved for D: = Ca +f)cosB-—C; D = Arctan [é (+f)smB + é]  152 Com Systems process of “splieing” several circular arcs together, as illustrated in Figure 4.21(c), provides the means to generate rather complex motion based en- tirely on circular arcs. Furthermore, it is relatively inexpensive to manufac- ture cam profiles made up of circular arcs. The following discussions apply to any situation where contact between the foliower and the cam profile oecurs on an eccentric circular arc. Fiat-Faced Translating Follower, A flat-faced translating follower is shown with an eccentric arc cam in Figure 4.22, For conveniente, let the body reference line OM pass through the center of curvature for the con- tact surface, the point CX, The length R is called the eccentricity. The radius ofthe circular arcis P, the radius of curvature, For this system, the folower position is H(A) = -RecosA+P =R(l-cs4)+P-R. ? By inspection, the displacement function is . HA)=Re(d- cos A) from which the first two derivatives are En=f(A)=ResinA In=f(A)=Recos A EIA) FIGURE 4.22 Eccentric Arc Cam with Flat-aced Translating Follower 4.8 Eccentric Circle Com 153 The velocity and acceleration of the follower, expressed in terms ofthe cam rotation angle A and its derivatives, are FA) = ÁSIA) = ÁResin A HA) = ÁfÇA) + ÁPA) ' =ÁR.sinA + Á?R cos A É 7 ircle e i ith am Ofjset, Translating Roiler Follower. A full-eircle eccentric cam with ar Ci maneio roller follower is shown in Figure 4,23. The body refe: ence line OM is again taken through the center of curvature for the contact surface, point C*, The known data include the following: Ro eccentricity P: cam profile radius of curvature Re: follower radius E: follower offset To determine the position of the cam follower, consider the horizontal and vertical loop equations: ResinA-(R+P)sin4p-E=0 -Re cos 4 + (Ry + Pj cos Ap -H=0 FIGURE 4.23 Full Circie Eccentric Cam with Offset, Translating, Roller Follower  154 Cam Systems In writing these equations, it has been necessary to introduce the unknown pressure angle, Ap. These equations can be solved explicitly for the pres- sure angle and the follower position: o . (ResmAÃ-E Ap = Aresin (Eae ) H=(R(+P)cosAÃ,— Recs A When the cam rotation angle A is zero, let the pressure angle be denoted Apa, and the follower position is H(0) = (R$ + P) cos Apo — Re This last amount is the constant difference between H(A) and f(A), so that FA) = (Re + Pjlcos Ap — cos Apo) — Re£os À — 1) To determine the velocity relations, the preceding loop equations are differentiated, with the result À Recos A — Àp (Ri + Pjcos Ap =0 Resina - Ap (R+P)sinAp- H=0 These may be solved in closed form to give the velocities Ap and E: ÁR.cos A (Rr + P) cos Ap E=ÁRc(inA — cos A tan Ap) Àp= From this last result, the velocity coefficient Kp can be identified: Ku= f(4)= Re(sin 4 — cos 4 tan A) Also, the derivative dA /dA is available; that is, dA, R cos 4 dA É (Rr + Pj cos Ap with f!(4) known, the velocity coefficient derivative Lo = f'(4) can be obtained directly by differentiation: Re cost A ) TR; + P) cosÃp The acceleration of the follower is then written in the familiar form H=ÃK + A? In=fMA)= Re (cos A+sinAtanA, Flat-Faced, Oscillating Follower. Figure 4.24 shows a full-cirele, eccentric cam with à flat-faced, oscillating follower. As before, the body reference line OM is chosen to pass through the center of curvature, point €*. The known data include the dimensions Ci, Co; Ca, P, and Re. Horizontal and 4.8 Eccentric Circle Cam 155 FIGURE 4.24 Full Circle Eccentric Cam with Oscillating Flat-faced Follower vertical loop equations-can be written involving the two unknowns B and D. When these are solved for the follower angle B, the result is B = —Arccos[(P + Ca)/W(A)] + E where x Y E = Atam(C, — Resin A, C2 + Recos À) WA) = ViC; — Resin AJ + (Co + Recos A) With B known, the distance D may be determined from the two loop equations with the result D=(C- Resin A) cosB-— (C)+ Recos A)sinB = W(A) sin (E — B) Differentiating the loop equations with respect to time gives the veloc- ity loop equations. With some manipulation, the velocity coefficient for the follower angle is k = Resin(A + B) Further differentiation of the velocity coefficient with respect to À gives the velocity coefficient derivative: Ly = Re (1 + Ki) cos(A + B) Pivoted Roller Follower. An oscillating follower with roller contact is shown with an eccentric circle cam in Figure 4.25. Once again, the body reference line OM has been chosen to pass through the center of curvature, point C*. The dimensions C;, C, Ca; P, Re, and Ryare considered known. The combined effect of the cam disk and the follower disk is to maintain a  156 Cam Systems FIGURE 4.25 Full Circle Eccentric Cam with Pivoted, Roller Fol- lower constant distance between the center of the cam disk and the trace point at the tip of the follower arm. The same result could be obtained with a link of length (P + Rg). Itis evident then, that this cam-follower system is equiva- lent to a four-bar linkage. As suggested previously in Chapter 2, a numeri- cal solution of the position loop equations is recommended. That solution «will provide values for both bf the angles B and D as functions of the input rotation À. When the position loop equations are differentiated with respect to time, the resulting pair of equations is linear in the unknown velocities, B and D. These can be solved explicitly for the unknown velocities and then further rearranged to give the velocity coefficients, as follows: — Resin(A + D) * Ticos(B — D) =Recos (A + B) (P+R9cosB-D) K K= Finally, the velocity coefficient derivative, Ly, can be obtained by differenti- ating K, with respect to 4, The result is lcos(B — Dj cos (A + D) (1 + Ka) BO AC cosXB — D) The foregoing results for the eccentric circle cam suggest an approach that côuld be applied to analyse the foilower motion that results from a cam of arbitrary shape. This is possible because the arbitrary shape can be approx- imated locally at each point as a circular arc. However, it is also an awk- ward approach, so a more direct alternative is used in the following discus- sion for cams of arbitrary shape. pa, 4.9 Angle Between Radius and Normal 157 4.9 ANGLE BETWEEN RADIUS ANO NORMAL Before discussing the follower response for a cam of arbitrary shape, itis useful to establish a means to calculate the angle between a radial line and the outward normal at a point on thecam profile as shown in Figure 4.26. The cam profile is assumed to be known in polar form R=R(C) and the angle of interest is denoted V. The unit vectors E, and E, are along and perpendicular to the radial line, respectively. The vector E. is along the radial line, R=R(G) E The first step is to determine a unit vector along the outward normal aí the point considered. When the position vector is differentiated with respectto G, the result is a vector tangent to the cam profile: dR ON E+RE A unit tangent vector may be constructed simply by normalizing this last vector, g DR E+RE VRET RO Let E, denote the unit normal vector, which can be written as a linear combination of E, and Es, E = 0, + bE, FIGURE 4.26 Angle Between a Radial Line and the Normal to the Cam Profile 162 Cam Systems xequired for Gand H if A is considered as the independent variable. Taking G as independent, these equations may pe solved explicitly for 4 and Has foltows: A H n -G — B— Axcsin(E/D) -Rcos(A + 6) > Rrcos(4 + G + V) n where B= Atam(Resin V, R + Rrcos V) D= VIR FRrcos VY + (Resin V) “ The value D is the distance from the fixed pivot point of the cam to the trace point on the follower. The angle B is between the line defined by D - and the radial line through the point of contact. Because G is considered as the-independent variable, the chain rule is required àgain to evaluate the derivative /'(4) = afidA. The derivative oEA with respect to G is determined by. differentiating the horizontal loop equa- * on with respect'to G, with the following result: dA Ri Sin(A + 6) — R cost + 6) REQ + VI costa + G + V) ao R costà + G) + Recos(ã + G + V) | Ina similar manner, the derivative of H with respect to G is determined from the vertical loop equation: aH SE = Rcos(A +.) + ROL + 49 sin(A + q) ERO+HA AVI sinA + O +V) where 4! is understood to mean the derivative dA/dG. The derivative of H . with respect to À is formed as in the previous case: am am |. dG ZA dA ac The second derivative is obtained by another application of the chain rule. Flat-Faced, Oscillating Follower. For the case of a flat-faced, osciliating followez with a specified cam profile, the geometry is as shown in Figure 4.29. The dimensions Ci, Ca, and Cs are known, as is the cam profile R(G). Ta this case, it is convenient to resolve the position loop equations parallel and perpendicular to the follower face; those components are shown in broken line on the figure. The loop equations are ' RenV4D+GsnB-CosB=0 ResV+GQ-GesB-Gsint=0 4.0 Follower. Response for Arbitrary Cam” rofile 163 G FIGURE 4.29 Cam With Specified Profile and Flat-faced, Oscillating Fol- lower ” As in the two preceding cases, it is convenient to take the angle G as the independent variable, and to determine. Band D. The follower angle B is determined from the second loop equation, with the result . (ReosV+hC = E + Arcsin (e -esin ( VCRS CÊ ) where E = AtaníG, CG) With B known, the first loop equation can be used to evaluate D, The angle 4 is then determined from the geometric relation. ALG+V+B=a As before, the solutions for R, V, B, Dyand A can be tabulated readily às functions of G. When a particular value of A is specified, inverse intérpola- tion can be used to determine the associated value of G as well as the other variables. If the loop equations are differentiated with respect tó time, the result is a pair of equations that are linear in the unknown velocities É and D. The equations can then be rearranged to give the velocity coefficients Ks and Ka, and be further differentiated to-yield Ly. Pivoted, Roller Follower. The final case to be considered is that of a pivoted folower with roller contact, as shown in Figure 4.30. The horizon- tal and vertical position loop equations for that system are às follows: Rsin(A+0)+ResidA+G+V)+CcosB- CG =0 -R costA + G) — Ricos(4 + G + V)- CasinB=- G=0  164 Cam Systems FIGURE 4.30 Cam With Specified Profile and Pivoted, Roller Follower There is no advantage to using G as the independent variable in place of 4 because à numerical solution will be required in any case, Following that numerical position solution, the velocity coefficients and velocity coeffi- cient derivatives will be obtained as the solutions of the systems of linear eguations resulting from successive time differentiations of the position equations, 4,11 CONCLUSION The subject of cams is very large, and the discussion of this chapter has been limited necessarily only to disk cams and their four common follower types. An elementary discussion of displacement function has also béen presented, but for most purposes the displacement function is assumed to have been previously specified. Displacement functions have been studied pencively in fre literature, and the interested reader is encouraged to look elsewhere for more informati i i Ie elnenhere for mor ation on the art of choosing an appropriate The first part of the chapter dealt with the analytical design of a cam profile to generate a specified follower response when employed with a particular follower type. In addition to purely kinematic considerations, the Hertz contact stress analysis applicable to cams was presented also. in the second part of the chapter, the follower motion generated by a speci- fied cam profile was analyzed. The special case of response to eccentric circle cams was developed before going to the general discussion for cams of arbitrary profile. Because of their high reliability, accuracy, and relatively low cost in large volume, cams will continue to be used widely for many years to come. The incorporation of computerized synthesis and analysis methods Problem Set 165 writh computerized manufactuzing will make cams more easily incorpo- rated into various types of machinery. REFERENCES Boresi, A. P., Sidebottom, O. M., Seely, F. B., and Smi of Materials. 8rd ed. New York: Wiley, 1978. Juvinall, R C., Fundamentals of Machine Component Design. New York: Wiley, 1983. Mabie, EL H. and Ocvirk, F. W., Mechanisms and Dynamics cf Machinery, 3rá ed. New York: Wiley, 1975. Poritsky, H., “Stresses and Deflections of Cylinárical Bodies in Contact With Ap- plication to Contact of Gears and Locomotive Wheels,” ]. of Applied Mechanics, “Trans. ASME, June 1950, pp. 191-201. Rothbart, H. A., Cams. New York: Wiley, 1956. Shigley, J. E. and Uicker, J. 1, Theory of Machines and Mechanisms. McGraw-Hill, 1980. 1]. O., Advanced Mechanics PROBLEM SET 41 Considera cam for which the rise extends over 0.82 racdian cf cam zotation for a lift of 2.3 in. The rise is in the form of two identical cubic ares (Y = 2X2), each tangent to the adjoining dwell and tangent to each other at the midpoint of the rotation interval, Determine 11 equally spaced poinis on this displacement curve, including both end points. 4.2 The rise portion of a particular cam displacement curve is to be described as two identical hyperbolic sine-squared arcs. The lower arc is of the form HA) = B sinhiCA) with appropriate values of B and C, and the upper arc is similar. Bach arcis tangent to the adjacent dwell, and the two arcs are tangent to each other ai the midpoint of the rotation interval. Let L denote the lift and T the cam rotation during the rise phase. The value of C is 2.0. a. Determine an expression for the coefficient B as a function of L and T; b. For the'case L = 1.2 im, and T =.0.5 radians, evaluate f(T/4); c. For Land Tas in past (b), evaluate f(57T/8). 4.3 The return phase of à cam cycle spans 0.70 radian of cam rota- tion, beginning at Az = 4.22 radians, and removes a lift of 1.65 in. The displacement function during the return is in the foim of a sine-squared arc, taken over one-quarter period of the sine function. (Remember that the displacement function must be tangent to the dwells occurring both before and after the return.) Determine nine equally spaced points on the dis- placement function over the return phase, including both end poinis. | | À : |  166 Cam Systems, 4.4 The rise phase of the two cam profiles are to be compared. For the first cam, the displacement function is sinusoidal; for the second, the displacement function is cycloidal. For cach cam, the rise extends from zero to 4 = 0.67 with the ft L = 0,45 in. a. Tabulate and plot both displacement functions at 11 evenly spaced points over the rise interval, including both end points; b. Tabulate and plot both velocity, coefficient functions, f(A), at the same 12 points; e. Tabulate and plot both velocity coefficient derivative functions, f'(A), at the same 11 points; a. Which of the two displacement functions will provide the most desir- able operating characteristics? Why? 4.5 A particular cemis intended to provide sinusoidal motion during the retum phase of the cam cycle. The retum occurs in the interval 4, = 1.157 to 43 = 1.787, and a lift of 0.375 in. is removed. Tabulate and plot “the following functions at 11 evenly-spaced points over the retum interval: . Displacement function, f(A); . Velocity coefficient, f'(A); . Velocity coefficient derivative, fA); . Velocity coefficient second derivative, f(A); = What is the largest absolute value for the velocity coefficient second derivative? Where does it occur? po ge 4.6 Use the graphical approach to design a plate cam for a flat-faced, translating follower to meet the following kinematic specification: Lift: = 15 mm Rise: sinuscidal 0=A< 1,35 radians Diwell: 198 <A < 1.75 radians Return: parabolic 175 = A < 4.22 radians Dwell: 4.292 = A < 2x radians Take the base circle radius as 50 mm (this is a design decision). Make a full- sig» engineering drawing that shows the following: a. Displacement function; : ». Follower in contact with the cam profile at A=0; - c. Development of the cam profile, based on 12 divisions of the circle. 4.7 Using the graphical approach, design a plate cam for use with a radial, translating roller follower that meets the kinematic specification Problem Set 167 given in Problem 4.6. Take the roller zadius as 9.5 mm, and the prime circle fadius as 65 mm (these are design decisions). Make a full-size engineering drawing that shows the following: a. Displacement finetion; 7 Ss Eur asi e b. Follower in contact with the cam profile at 4 = 0; c. Development of the cam profile, based on 12 divisions of the circle. 4.8 Consider a cam for use with a flat-faced, translating follower having a rise and immediate return spanning 3.0 radians of cam rotation and described by HA) = L sintam 4/3) 0<A<30 =0 30=A<27 | and where L is the lift, 0.887 in. Both the cam and the follower are made of steel, E = 3(107) 1b/in.? Assume the base radius, Ro is 2.35 in., and í a. Determine the length of the cam foliower wear surface; b. Determine polar coordinates (R, G) for points on the cam profile for A = 0,0, 0.5, 1.0, 1.5, 20, 2.5, and 3,0 radians; c. Make a sketch of the cam profile; d. The minimum radius of curvature for this cam profile occuxs at A=15 radians. If the contact force is 20 Ib and the contact stress is to be limited to 7800 Ib/in.?, determine the minimum thickness for the cam. 49 Considera cam, working with a fat-faced, translating follower, where the follower guide is radial with respect to the cam axis of rotation. The system is designed to meet the following kinematic specification: Lift; L=0.45in. Rise: — half-cycle of sinusoid 0=A<042 Dwell: 0.4722 5 À < 2.3030 Retum: half-cycle of sinusoid 2.3030 = À < 3.7525 Dwell: 3752; = A<2m Base Circle Radius = 3.365 in. « a. What is the minimum follower face length to the left of the guide? To the right? Minimum total face length? b. Determine polar coordinates (R, G) for the profile point in contact with the follower for each of the following positions: A = 0, n/6, n/3, u/2, 2513, 5n/6, 7, and 77/6 radians. 4.10 Consider a cam, operating with a flat-faced, translating fol- lower, and having the following kinematie specification:  172 Cam Systems Spring HA) FA) ) 1 A º A Az As 2x The displacement function is as shown-—a sinusoidal rise, dwell, sinusoi- dal retum, and a dweil at zero. The break points for this curve are A = 2.55 rad A 3.45 rad Ay = 5.65 rad while the lftis L = 0.85 im. The follower is spring loaded against the cam, with a preload of 12 Ib when the follower is in a static condition: at its lowest position. The spring rate is K = 5 Ib/in. The spring, the follower, and the shaft above may all be considered massless, but they drive a block weigh- ing 4.73 lb. The follower, the shaft, and the block are all rigid. There is no gravitational effect to be considered. At the time t = 2.0 s, determine numerical values for the following: , Acceleration of the weight; . Force in the spring; . Contact force between the cam and follower; . Force in the shaft at the section indicated G. por» Problem Set 1% ) . OVAS) A cam with a pivoted, flat-faced follower is connected to E 5 ing linifás shown, The angular displacement of the follower is given B(A) = Bo + HA) where f(A) is the displacement function and Bo is the rihimum rotation. The distance D and the angle C are known constants. a. Obtain an expression for the velocity coefficient, K, = dX/24, in terms of Bs, C, D, f, and f'; b. If the displacement function is CA) = 0.32 (1 — cos(2r ALA] 0=A<A =20 AMSEASIm where Ay = 1.657 radians .21 radians .02 radians  CHAPTER 5 Gears 5.1 INTRODUCTION The transmission of power from one shaftto another, often with a change in speed of rotation, is frequently accomplished with gears. Gears are Guia im à wide variety of everyday items such as alarm clocks, Kitchen mixers, automobiles, office copiers, and electric drills. Gears are also used in industrial machinery such as lathes, winches, tractors, and printing presses. Certainly, gears are among the most commonly used machine ments. . de TE two disks are pressed together, edge-to-edge, itis possible to trans- mit rotation from the first disk to the second disk. Friction between the two contact areas acts to transmit torque from one disk to the other; such disks are called friction twheeis. As long as the disks roll without slipping, there is a definite relation between the angular positions of the two disks. H the torque required excecds the friction capacity of the contact, the dists vt skip. The positional relation of the two disks is then changed an me required torque is not met. To prevent sipping, the contact force can be increased, resulting in an increase in the friction capacity of the contact. This occurs at the expense of increased bearing loads, bearing friction, stress and deflection in the shafting, and stresses within the disks. A better «way to overcome slipping is to increase the coefficient of friction through a more rough surface. The rougher surface involves larger surface irregulari- tes, which are more difficult to shear away and, thus, more cffectively zesist slippage. The extreme extension of this ídea occurs when teeth are cut on the two wheels, in which case they are called gears. o TÉ the teeth are cut-straight across the edge of a disk, in a direction parallel to the axis of rotation, the resulting gear is called a spur gear. 474 5,2 Velocity Ratio 175 Alternatively, teeth may be cut along a helix generated around the axis of the gear. In this latter case, the teeth will cross the edge of the disk at an angle, and these gears are called helical genrs. Spur gears or helical gears are used where power is to be transmitted between parallel shafts. If the two shafts intersect, power may be transmitted from oné to thé other using bevel gears, for which each gear is cut from a cone. There are also situations where the two shaft axes are norrintersecting, non-parallel lines; these situations are served with skew gears, hypoid gears, or worm gents. However, in this chapter only spur gears will be discussed; for more information on other types, see the references listed at the end of the chapter. 5.2 VELOCITY RATIO The velocity ratio is one of the most important aspects ofa pair of gears. Itis the ratio of the angular velocity of one gear to that of a second gear that is in mesh with the first. Itis important that this ratio be constant, not only on the average, but instantaneously. Consider first a situation in which the average velocity ratio is constant but the instantaneous, value is variable. Nonconstant Velocity Ratio. Figure 5.1 shows a gear pair such that the teeth of one gear consist of M; radial pins of uniform width that are equally spaced around the circumference of the disk; this is called a spoked gear. The teeth of the second gear consist of M; circular pins evenly spaced between twro disks, each pin with its axis perpendicular to the gear disks and paral- lel to their axis of rotation. This gear type is called a lantern. Gearing of this sort was used in medieval Europe in the construction of grain mills and other machines. Referring to Figure 5.2, the loop equations describing the position relation are: Ricosg + (Ro + W) snA+DcosÃ-C=0 Resing + (Ra + W)cos4 - DsinÃ=0 FIGURE 5.1 Medieval Lantern and Spoked Gears in Mesh  176 Gears FIGURE 5.2 Schematic Diagram for Lanterm and Spoked Gears These relations apply over an interval q; == q = q2 during the engagement of a single pair of teeth, with the corresponding range 4, = 4 = Ao. The dimensions Rj, R>, W, Ds, and C are presumed known, and the loop equations are solvable for A and D as functions of the lantem rotation q. The velocity loop equations are the result of differentiating the positior: equatiors, and can be used to determine velocity coefficients for the sys- tem. The solutions for À and K, as functions of q for typical proportions are shown in Figure 5.3, The range of values for q shown there spans the duration of one tooth-pair engagement, and thus presents one cycle of the periodic function K,. However, the angle A is not periodic because it is increasing continuously. It is evident that K, is not a constant but, instead; varies with position. This indicates that the velocity ratio, À/ã, is not con- FIGURE 5.3 Position and Velocity Coefficient Curves For Lantern and Spoked Gear Pai 5.2 Velocity Ratio a77 stant, but varies with position through the mesh. If the input angular velocity, 9, is constant, the output angular velocity, À, will vary periodi- cally during the engagement of each tooth pair. It should be noted that the average output shaft angular velocity, taken over a full revolution of the output gear, will be constant for constant input. velocity even though the instantaneous value varies. , What difference does it make if the angular velocity ratio varies? Con- sider again the case where the input speed is constant, perhaps because of the use of a very large flywheel. The output gear is connected through a shaft of finite stiffmess to a rotating machine component with finite inertia. As the angular velocity of the output gear varies, the shaft will be alter- nately twisted and relaxed in an effort to cause the driver component to follow the speed variations of the output gear. This alternating torque ir. the shaft will excite torsional vibrations in the driven machine, with resuli- ing noisê, vibration, and fatigue damage. For crude, low-speed, low- torque machinery this situation may be tolerable, but for modem high speed, high-power, machinery it can not be allowed. Condition for Constant Velocity Ratio. Because it is important to have a constant velocity ratio, under what conditions does it occur? Fr Figure 5.4, two cams are shown in contact, representing teeth on tsvo gears. The two cams rotate about fixed pivois at Cy and CG», separated by the center dis- tance C. The cams have instantaneous angular velocities À; and À;, respec- tively. At the point of contact, point H, the tangént line is B-B,, and Ni Ny is the normal line. The normal is also called the “line of action” or the “line of contact.” The cam on: the left is the driver, and the cam on the right is driven. As discussed previously, the cam contact surfaces should never FIGURE 5.4 Two Cams Representing Teeth on Two Gears | | u | |