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MECHANICAL ENGINEERING/32

MCCHANISM ANALYSIS SIMPLIFIED GRAPHICAL AND ANALYTICAL TECHNIQUES

LXNDON O. BARTON

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DATE DUE

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https://archive.org/details/mechanismanalysiOOOObart

MECHANISM ANALYSIS

MECHANICAL ENGINEERING A Series of Textbooks and Reference Books EDITORS

L. L. FAULKNER

S. B. MENKES

Department of Mechanical Engineering The Ohio State University Columbus, Ohio

Department of Mechanical Engineering The City College of the City University of New York New York, New York

1.

Spring Designer’s Handbook, by Harold Carlson

2.

Computer-Aided Graphics and Design, by Daniel L. Ryan

3.

Lubrication Fundamentals, by J. George Wills

4.

Solar Engineering for Domestic Buildings, by William

A. Himmelman 5.

Applied Engineering Mechanics: Statics and Dynamics,

6.

by G. Boothroyd and C. Poli Centrifugal Pump Clinic, by Igor J. Karassik

1.

Computer-Aided Kinetics for Machine Design, by

Daniel L. Ryan 8.

Plastics Products Design Handbook, Part A: Materials and Components; Part B: Processes and Design for Processes, edited by Edward Miller

9.

Turbomachinery: Basic Theory and Applications, by

Earl Logan, Jr. 10. 11.

Vibrations of Shells and Plates, by Werner Soedel Flat and Corrugated Diaphragm Design Handbook, by Mario

Di Giovanni 12.

Practical Stress Analysis in Engineering Design, by Alexander Blake

13.

An Introduction to the Design and Behavior of Bolted Joints, by

John H. Bickford 14.

Optimal Engineering Design: Principles and Applications,

by James N. Siddall 15. 16.

Spring Manufacturing Handbook, by Harold Carlson Industrial Noise Control: Fundamentals and Applications,

edited by Lewis H. Bell

17.

Gears and Their Vibration: A Basic Approach to Understanding

18.

Chains for Power Transmission and Material Handling: Design

19.

Corrosion and Corrosion Protection Handbook, edited by

Gear Noise, by J. Derek Smith and Applications Handbook, by the American Chain Association

Philip A. Schweitzer 20.

Gear Drive Systems: Design and Application, by Peter Lynwander

21.

Controlling In-Plant Airborne Contaminants: Systems Design and

22.

Calculations, by John D. Constance CAD/CAM Systems Planning and Implementation, by Charles S. Knox

23.

Probabilistic Engineering Design: Principles and Applications,

by James N. Siddall 24.

Traction Drives: Selection and Application, by Frederick W. Heilich III

and Eugene E. Shube 25.

Finite Element Methods: An Introduction, by Ronald L. Huston

and Chris E. Passerello 26.

Mechanical Fastening of Plastics: An Engineering Handbook, by

29.

Bray ton Lincoln, Kenneth J. Gomes, and James F. Braden Lubrication in Practice, Second Edition, edited by W. S. Robertson Principles of Automated drafting, by Daniel L. Ryan Practical Seal Design, edited by Leonard J. Martini

30.

Engineering Documentation for CAD/CAM Applications,

27. 28.

by Charles S. Knox 31.

Design Dimensioning with Computer Graphics Applications, by Jerome C.

Lange 32.

Mechanism Analysis: Simplified Graphical and Analytical Techniques,

by Lyndon O. Barton

OTHER VOLUMES IN PREPARATION

MECHANISM ANALYSIS Simplified Graphical and Analytical Techniques

Lyndon O. Barton Project Engineer E.I. du Pont de Nemours & Co. Wilmington, Delaware

MARCEL DEKKER, INC.

New York and Basel

Library of Congress Cataloging in Publication Data

Barton, Lyndon O., [date] Mechanism analysis. (Mechanical engineering ; 32) Bibliography: p. Includes index. 1. Machinery, Kinematics of. II. Series. TJ175.B29 1984 ISBN 0-8247-7086-2

I. Title.

621.8'll

COPYRIGHT © 1984 by MARCEL DEKKER, INC.

ALL RIGHTS RESERVED

Neither this book nor any part may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, microfilming, and recording, or by any information storage and retrieval system, without permission in writing from the publisher. MARCEL DEKKER, INC. 270 Madison Avenue, New York, New York 10016 Current printing (last digit): 10

98765432

PRINTED IN THE UNITED STATES OF AMERICA

To my wife Olive, my children Rhonda, Loren, Carol, and Leon, and my mother Clarice.

Preface

This book is written primarily for mechanical design engineers and mechan¬ ical design engineering students who are concerned with the design of ma¬ chines, in general, and in particular with mechanism analysis, a subject which forms a principal part of the study of kinematics of mechanisms. The principal aim of this volume is to place at the disposal of the reader a prac¬ tical book that will serve (1) as a handy reference for simplified approaches to problems typically encountered in the analysis and synthesis of a mech¬ anism, and (2) as a supplementary textbook for independent study or quick review of both principles and applications in mechanism analysis. The book presents a wide assortment of graphical and analytical techniques, as well as complete listings in FORTRAN of computer programs and programmable calculator programs for the Hewlett Packard HP-41C for analysis of basic classes of mechanisms. Special emphasis has been given to relatively simple kinematic chains such as slider-crank, four-bar, quick-return, and sliding coupler mechanisms. These mechanisms have been selected because they form the basic elements of most machines and because they are easily adaptable to the teaching of fundamental kinematic principles. Once these principles are fully understood, it is comparatively easy to apply this knowledge to the analysis of more complex mechanisms. Several novel approaches for simplification of the analytic process are presented. These include the rectilinear and angular motion diagrams presented in Chapter 2, the link extension concept for velocity analysis by instant centers in Chapter 5, the generalized procedure for constructing the acceleration polygon in Chapter 9, the Parallelogram Method for slidercrank analysis in Chapter 12, and the Simplified Vector Method and modified version of same in Chapters 14-18. One important feature is that the Simplified Vector Method, unlike conventional methods which rely on calculus and other forms of sophisticated mathematics, relies mainly on basic algebra and tiigonometiy to obtain an analytical solution. This simplified mathematical approach has made it

v

vi

Preface

possible to include several analytical problem solutions rarely found in kinematic textbooks. Hopefully, this approach will not only make this mate¬ rial accessible to a wide body of readers, but will also help to provide the quick insight often needed by designers in the analysis of a linkage. The book is written for easy readability and comprehension, without reliance on any other source. Needed background material on topics such as Uniformly Accelerated Motion (Chapter 2), Properties of Vectors (Chap¬ ter 3), Complex Algebra (Chapter 13), and Trigonometry (Appendix B) is provided for review. Concepts are presented as concisely as possible, employing numerous illustrative examples and graphical aids, as well as step-by-step procedures for most graphical constructions. In addition, the topics are arranged in a logical sequence corresponding to that ordinarily followed in teaching a course in kinematics. Some of the material in this book is based on several technical papers which the author has previously published (see References). Much of the material has been drawn from class notes which have been developed and used over several years of teaching kinematics of mechanisms as part of an Engineering Technology college curriculum. The author gratefully acknowledges his indebtedness to the E. I. DuPont de Nemours and Company Engineering Department and the Delaware Technical Community College Mechanical Engineering Department for pro¬ viding the engineering and teaching opportunities, respectively, that have enabled him to pursue and accumulate the knowledge and experience that form the basis of this book. Grateful acknowledgments and appreciation are also extended to •

Penton Publishing Company, publishers of Machine Design magazine, for permission to reprint portions of previously published articles (including illustrations);

•

American Society for Engineering Education, publishers of Engineering Design Graphics journal for permission to reprint portions of previously published articles (including illustrations);

•

Mr. Albert A. Stewart, for the valuable assistance he has rendered in the calculator program development;

•

Teachers, relatives, and friends who have been a source of inspiration and encouragement in the author's career; and

•

A devoted family, for their love, understanding, and support, always. Lyndon O. Barton

Contents

PREFACE Part I.

v

INTRODUCTORY CONCEPTS 1.

KINEMATIC TERMINOLOGY

1 3

Mechanical Concepts Motion Classification Motion Characteristics

3 9 13

UNIFORMLY ACCELERATED MOTION

16

Rectilinear Motion Angular Motion Conversion Between Angular and Rectilinear Motion Velocity-Time Graph Solutions Summary of Motion Formulas

16 21

1.1 1.2 1.3

2.1 2.2 2.3 2.4 2.5

41

VECTORS 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8

27 33 40

Properties of Vectors Vector Addition Vector Subtraction The Vector Polygon Vector Resolution Orthogonal Components Translational and Rotational Components Effective Components

vii

41 41 45 47 48 51 52 53

viii

Contents

Part H.

GRAPHICAL TECHNIQUES II. A.

4.

GRAPHICAL TECHNIQUES: VELOCITY ANALYSIS EFFECTIVE COMPONENT OF VELOCITY METHOD 4.1 4.2 4.3 4.4 4.5

Introduction The Rigid Body Principle Velocities of End Points on a Link Velocities of Points on a Rotating Body Velocity of Any Point on a Link

4.6 4.7 4.8 4.9

Velocity Analysis of a Simple Mechanism Velocities of Sliding Contact Mechanisms Velocity Analysis of a Compound Mechanism Summary

INSTANT CENTER METHOD 5.1 5.2 5.3 5.4 5.5 5. 6 5.7 5.8 5.9 5.10 5.11

6.7 6.8 6.9 6.10 6.11 6.12 6.13

59 60 60 60 60 61 63 65 68 72 74 75

Introduction Pure Rotation of a Rigid Body Combined Motion of a Rigid Body

75 76 81

Velocity of a Body with Rolling Contact Types of Instant Centers Locating Instant Centers

83 84 85 92 92 95 97 99

Velocity Properties of the Instant Axis Velocity Analysis by Instant Centers Velocity Analysis of a Simple Mechanism Velocity Analysis of a Compound Mechanism Summary

RELATIVE VELOCITY METHOD 6.1 6.2 6.3 6.4 6.5 6.6

57

Introduction Relative Motion Concept The Velocity Polygon The Velocity Polygon Convention Velocity Polygon: Linkage Application Relative Velocity of Two Points on a Rigid Body

101 101 101 103 104 104

Velocities of End Points on a Floating Link Velocity of Any Point on a Link

105 106 108

Velocity of Any Point on an Expanded Link Velocity Analysis of a Simple Mechanism Velocities of Sliding Contact Mechanisms Velocities of a Body with Rolling Contact Velocity Analysis of a Compound Mechanism

109 111 113 117 119

Contents

H.B.

7.

ix

GRAPHICAL TECHNIQUES: ACCELERATION ANALYSIS LINEAR ACCELERATION ALONG CURVED PATHS 7.1 7.2 7.3 7.4 7.5 7. 6 7.7 7.8

8.

EFFECTIVE COMPONENT OF ACCELERATION METHOD 8.1 8.2 8.3 8.4

9.

11.

Introduction Acceleration of End Points on a Link Slider-Crank Analysis Four-Bar Linkage Analysis

RELATIVE ACCELERATION METHOD 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8

10.

Introduction Normal Acceleration Tangential Acceleration Resultant Acceleration Proportionality of Accelerations Relative Acceleration of Two Points on a Rigid Body Acceleration of Any Point in a Floating Link Coriolis Acceleration

Introduction The Acceleration Polygon Acceleration Polygon Convention Generalized Procedure Mechanism with Expanded Floating Link Compound Mechanism Cam-Follower Mechanism Summary

VELOCITY DIFFERENCE METHOD

123 124 124 124 126 127 133 134 138 14 6

151 151 151 154 157 161 161 162 162 163 172 175 179 181 183

10.1 10.2

Introduction Slider-Crank Mechanism Analysis

183 184

10.3 10.4

Quick-Return Mechanism Analysis Four-Bar Mechanism Analysis

187 189

GRAPHICAL CALCULUS METHOD 11.1 11.2

Graphical Differentiation Graphical Integration

194 194 204

Contents

X

12.

SPECIAL METHODS 12.1 12.2 12.3

Part m. 13.

14.

223 229

COMPLEX ALGEBRA

231

13.1 13.2 13.3 13.4 13.5

Introduction Complex Vector Operations Geometric Representation of a Complex Vector Complex Forms The Unit Vector

13.6

Linkage Application

235 237

FOUR--BAR MECHANISM ANALYSIS: SIMPLIFIED VECTOR METHOD

242

Introduction Scope and Assumptions Geometric Relationships Velocity Analysis Acceleration Analysis

SLIDER-CRANK MECHANISM ANALYSIS: SIMPLIFIED VECTOR METHOD 15.1 15.2 15.3 15.4 15.5

16.

212 218

ANALYTIC TECHNIQUES

14.1 14.2 14.3 14.4 14.5 15.

Complete Graphical Analysis Method Equivalent Linkage Method Slider-Crank Acceleration: Parallelogram Method

212

231 231 232 234

242 245 245 246 249

260

Introduction Scope and Assumptions Geometric Relationships Velocity Analysis Acceleration Analysis

260 263 263 265 267

QUICK-RETURN MECHANISM ANALYSIS: SIMPLIFIED VECTOR METHOD

274

16.1 16.2 16.3 16.4 16.5

Introduction

274

Scope and Assumptions Geometric Relationships Velocity Analysis

276 277 277 279

Acceleration Analysis

Contents

17.

xi

SLIDING COUPLER MECHANISM ANALYSIS: SIMPLIFIED VECTOR METHOD 17.1 17.2 17.3 17.4 17.5

18.

Introduction Scope and Assumptions Geometric Relationships Velocity Analysis Acceleration Analysis

289 290 295 295 297

SLIDER-CRANK MECHANISM ANALYSIS: MODIFIED VECTOR METHOD

306

18.1 18.2 18.3 18.4 18.5 19.

289

Introduction Scope and Assumptions Geometric Considerations Velocity Analysis Acceleration Analysis

SLIDER-CRANK MECHANISM ANALYSIS: CALCULUS METHOD 19.1 19.2 19.3

Introduction Scope and Assumptions Displacement, Velocity, and Acceleration Analysis

PROBLEMS

314 314 314 314 319

Kinematic Terminology Uniformly Accelerated Motion Vectors Graphical Techniques: Velocity Analysis Graphical Techniques: Acceleration Analysis Graphical Techniques: Miscellaneous Analytical Techniques: Velocity and Acceleration APPENDIX A A.l A. 2

306 306 306 307 309

319 320 321 327 339 342 353 355

Introduction Calculator Operating Procedure

FORTRAN Programs Four-Bar: Simplified Vector Method Slider-Crank: Simplified Vector Method Quick-Return: Simplified Vector Method Sliding Coupler: Simplified Vector Method Slider-Crank: Modified Vector Method

355 355 358 364 371 377 383

xii

Contents

Calculator Programs Four-Bar: Simplified Vector Method Slider-Crank: Simplified Vector Method Quick-Return: Simplified Vector Method Sliding Coupler: Simplified Vector Method Slider-Crank: Modified Vector Method APPENDIX B B. 1 B.2 B.3

388 396 404 413 421 428

Nomenclature Trigonometry Review Table of Trigonometric Functions

428 431 437

SELECTED REFERENCES

438

INDEX

443

MECHANISM ANALYSIS

I INTRODUCTORY CONCEPTS

Mechanism analysis (or kinematics of machines) is inherently a vital part in the design of a new machine or in studying the design of an existing machine. For this reason, the subject has always been of considerable importance to the mechanical engineer. Moreover, considering the tremen¬ dous advances that have been made within recent years in the design of high-speed machines, computers, complex instruments, automatic controls, and mechanical robots, it is not surprising that the study of mechanisms has continued to attract greater attention and emphasis than ever before. Mechanism analysis may be defined as a systematic analysis of a mechanism based on principles of kinematics, or the study of motion of machine components without regard to the forces that cause the motion. To better appreciate the role of mechanism analysis in the overall design proc¬ ess, consider the following. Typically, the design of a new machine begins when there is a need for a mechanical device to perform a specific function. To fulfill this need, a conceptual or inventive phase of the design process is required to establish the general form of the device. Having arrived at a concept, the designer usually prepares a preliminary geometric layout of the machine or mechanism for a complete kinematic analysis. Here the designer is concerned not only that all components of the machine are prop¬ erly proportioned so that the desired motions can be achieved (synthesis phase), but also with the analysis of the components themselves to determine such characteristics as displacements, velocities, and accelerations (analy¬ sis phase). At the completion of this analysis, the designer is ready to pro¬ ceed to the next logical step in the design process: kinetic analysis, where individual machine members are analyzed further to determine the forces resulting from the motion. Mechanism analysis therefore serves as a necessary prerequisite for the proper sizing of machine members, so that they can withstand the

1

2

Introductory Concepts i-IDEAL FLUIDS

rFLUIDS- -VISCOUS FLUIDS

COMPRESSIBLE FLUIDS

rSTRENGTH OF MATERIALS MECHANICSDEFORMABLE BODIES-

•THEORY OF ELASTICITY

lTHEORY OF PLASTICITY

lSOLIDS-

[-STATICS lRIGID BODIES-

rKINETICS LDYNAMICS—

_MECHANISM 'synthesis lKINEMATICSMECHANISM ANALYSIS

Figure 1.1

Mechanism analysis and other branches of mechanics

loads and stresses to which they will be subjected. Figure 1.1 shows the relationship of mechanism analysis to other branches of mechanics.

1

Kinematic Terminology

1.1

MECHANICAL CONCEPTS

A mechanism is a combination of rigid bodies so connected that the motion of one will produce a definite and predictable motion of others, according to a physical law. Alternatively, a mechanism is considered to be a kine¬ matic chain in which one of the rigid bodies is fixed. An example of a mech¬ anism is the slider crank shown in Figure 1.1. Instruments, watches, and governors provide other examples of mechanisms. A machine is a mechanism or group of mechanisms used to perform useful work. A machine transmits forces. An example of a machine is the internal combustion engine shown in Figure 1.1. The term "machine" should not be confused with "mechanism" even though in actuality, both may refer to the same device. The difference in terminology is related primarily to function. Whereas the function of a machine is to transmit energy, that of a mechanism is to transmit motion. Stated in other words, the term "mechanism" applies to the geometric arrangement that imparts definite motions to parts of a machine.

B

3

4

Introductory Concepts

Figure 1.2

Four-bar linkage mechanism.

A pair is a joint between two bodies that permits relative motion. 1.

A lower pair has surface contact, such as a hinge or pivot. Surface contact is a characteristic of sleeve bearings, piston rings, screwed joints, and ball-and-socket joints. As an example, joints A, B, C, and D of Figure 1.2 are lower pairs. These joints are called revolute or turning pairs.

2.

A higher pair has line or point contact between the surface ele¬ ments. Point contact is usually found in ball bearings. Line contact is characteristic of cams, roller bearings, and most gears.

A link is a rigid body that serves to transmit force from one body to another or to cause or control motion, such as the connecting rod or crank arm in Figure 1.1. Alternatively, a link is defined as a rigid body having two or more pairing elements. A kinematic chain is a group of links connected by means of pairs to transmit motion or force. There are three types of chains: locked chain constrained chain, and unconstrained chain.

Figure 1.3

Locked chain.

Kinematic Terminology

12.

3.

5

A locked chain has no relative motion between the links. An example of this is the three-link chain shown in Figure 1.3. A constrained chain is one in which there is definite relative motion between the links. For example, if one link is fixed and another link is put in motion, the points on all the other links will move in definite paths and will always move in the same paths regardless of the number of lines the motion is repeated. An example is the four-bar mechanism shown in Figure 1.4. An unconstrained chain is one in which, with one link fixed and another link put in motion, the points on the remaining links will not follow definite paths. An example of an unconstrained chain is the five-link mechanism shown in Figure 1.5.

A linkage is a mechanism in which all connections are lower pairs. The four-bar and slider-crank linkages are typical examples. A simple mechanism is one that consists of three or four links, whereas a compound mechanism consists of a combination of simple mech¬ anisms, and is usually made up of more than four links.

Figure 1.5

Unconstrained chain.

6

Introductory Concepts

A structure is a combination of rigid bodies capable of transmitting forces or carrying loads, but having no relative motion in them. Alterna¬ tively, a structure may be thought of as a locked kinematic chain. A frame is a stationary structure that supports a machine or mechanism. Normally, it is the fixed link of a mechanism (e.g., link 1 in Figure 1.2). A driver is that part of a mechanism which causes motion, such as the crank in Figure 1.2. A follower is that part of a mechanism whose motion is affected by the motion of the driver, such as the slider in Figure 1.2.

Figure 1.7

Cam-follower mechanism.

Kinematic Terminology

7

8

Introductory Concepts

Modes of Transmission Motion can be transmitted from driver to follower by: 1.

Direct contact a. Sliding b. Rolling

2.

Intermediate connectors a. Rigid: links b. Flexible: belts, fluids

3.

Nonmaterial: magnetic forces

A gear is a machine member, generally circular, whose active sur¬ face is provided with teeth to engage a similar member to impart rotation from one shaft to another. An example is the spur gear shown in Figure 1.6. A cam is a rotating or sliding machine member whose function is to impart a predetermined motion to another part that rolls or slides along the surface of the member. An example is shown in Figure 1.7.

Inversion Inversion is the process of fixing different links in a chain to create different mechanisms. Many useful mechanisms may be obtained by the inversion of various kinematic chains. An example of such inversion can be seen in the slider crank chain shown in Figure 1.8: 1.

By making link 1 of the chain fixed, we obtain a steam engine mechanism (Figure 1.8a).

2.

By fixing the crank, link 2, we obtain the Whitworth quickreturn mechanism, used in various types of metal shapers (Figure 1.8b).

3.

By fixing the connecting rod, link 3, we obtain the oscillating cylinder engine, once used as a type of marine engine (Figure

4.

Finally, by fixing the slider, link 4, we obtain the mechanism shown in Figure 1.8d. This mechanism has found very little practical application. However, by rotating the figure 90° clockwise the mechanism can be recognized as part of a garden pump.

R is important to keep in mind that the inversion of a mechanism does not change the relative motion between the links, but does change their abso¬ lute motions.

Kinematic Terminology

1.2

9

MOTION CLASSIFICATION

Definitions Motion is the act of changing position. The change of position can be with respect to some other body which is either at rest or moving. Rest is a state in which the body has no motion. Absolute motion is the change of position of a body with respect to another body at absolute rest. Relative motion is the change of position of a body with respect to another body that is also moving with respect to a fixed frame of reference.

Types of Motion Plane Motion In plane motion all points on a body in motion move in the same plane or parallel planes. All points of the body or system of bodies remain at a con¬ stant distance from a reference plane throughout the motion. Typical ex¬ amples are the connecting rod on a slider-crank mechanism (Figure 1.1) and the side rod on a locomotive (Figure 1.9). There are three classes of plane motion: (1) rotation, (2) translation, and (3) combined translation and rotation. 1.

2.

Figure 1.9

Rotation: When one point in a body remains stationary while the body turns about that point, the body is said to be in rotation. That is, all points in the body describe circular paths about a stationary axis that is perpendicular to the plane of rotation. Crank AB is Figure 1.1 has rotary motion. Translation: When a body moves without turning, that body is said to be in translation: or, the distances between particles

Locomotive side rod drive. Link BC undergoes curvilinear

motion; link AD undergoes rectilinear translation.

CD

0)

*CJ

£ be

d

o PS

B

B/C

co

(in the direction of AV)

SECOND CHANGE (Figure 7.16) Consider the tangential velocity of slider S as it moves outward from the center (a change in Vg due to the change in slider distance from A). Here the magnitude of AV is given by AV = VB, - VB = Arco and

148

Graphical Techniques: Acceleration Analysis

AY _ Ar AT AT Therefore,

ab2) = Vc “ or r( 2) Ag =

(in the direction of AV)

SUMMARY Combining the results of the two changes, we obtain the Coriolis acceleration:

AC°r = a£> + A® or rC/Or

A

-

(in the direction of AV)

The Coriolis acceleration vector is always directed perpendicular to the rotating link and pointed as if it had been rotated about its tail through an angle 90° with vector VB/c in the direction of rotating link on which the sliding occurs.

go,

the angular velocity of the

Linear Acceleration Along Curved Paths

PQ U

(a)

(b)

(c)

(d)

Figure 7.17 Coriolis acceleration for the four cases of Example 7.4: (a) case 1; (b) case 2; (c) case 3; (d) case 4.

150

Graphical Techniques: Acceleration Analysis

EXAMPLE 7.4 (Figure 7.17) Consider a block B which slides on a rotation rod AD and has a velocity of 5 ft/sec relative to a point C on the rod (Vg/Q = 5 ft/sec). The angular velocity of the rod is 4 rad/sec. Determine the Coriolis acceleration in terms of both magnitude and direction for the following cases. 1. 2. 3. 4.

Block B moves outward and toward D as AD rotates counter¬ clockwise (Figure 7.17a). Block B moves inward and toward A as AD rotates counter¬ clockwise (Figure 7.17b). Block B moves inward and toward A as AD rotates clockwise (Figure 7.17c). Block B moves outward and toward D as A rotates clockwise (Figure 7.17d).

SOLUTION The magnitude of the Coriolis acceleration in each case is given by Cor 2Vb/c“ad = 2(5)(4) = 40 ft/sec2 Figure 7.17 shows the direction of the Coriolis acceleration in each case. Note that the direction of ACor is obtained from rotating the vector VB/C through an angle 90° in the same direction (clockwise or counterclockwise) as that of rotating link AD.

8 Effective Component of Acceleration Method

8.1

INTRODUCTION

The effective component of acceleration method is similar to that of the effective component of velocity method in that it is also based on the rigid body principle. However, it is somewhat more complex for bodies with combined motion since it involves the concepts of relative acceleration and relative velocity. It will be seen that the effective component of acceleration along a line connecting two points, A and B, on a rigid body is composed of two parts:

8.2

1.

The acceleration component of point A relative to point B due to rotation of A about B

2.

The acceleration component of the reference point B due to translation

ACCELERATION OF END POINTS ON A LINK

Consider link AB shown in Figure 8.1. Both the velocity and acceleration of point A are known in magnitude and direction, whereas only the line of action of the acceleration of point B (or Ag) is known. This line of action is the line "b"-"b". Determine the complete acceleration of point B (Ag). The acceleration of point B is related to that of point A by the vector equation

B

aa + ab/a

and similarly, the effective components of A^ and A related by

151

along link AB are

152

Graphical Techniques: Acceleration Analysis

-AB -AB , -AB A„ = A . + A^ , 4 B A B/A To determine Ab/a> we must consider the motion of point B relative to point A. That is, we assume that point A is fixed while B rotates about it. Therefore, the only effective component of acceleration that B can have along link AB is the normal or radial acceleration (A^ ). In other words,

ab/a

ab/a

where N = B/A

V2 B/A AB

Therefore, AB = AB/A

V2 B/A AB _ AD

This means that we must find Vb/a bi order to find Ab/a- vB/A can determined from a velocity polygon.

I

Figure 8.1

Linkage.

Effective Component of Acceleration Method

153

"b"

Figure 8.2 polygon.

Acceleration of end points of a link: (a) linkage; (b) velocity

Graphical Techniques: Acceleration Analysis

154

PROCEDURE (Figure 8.2a) 1.

At point A, lay out the effective component of the given accel¬ eration Aa, that is, AAB. This is obtained, as in the velocity case, by dropping a perpendicular line from the terminus AA

2.

to the extended line AB. Find A^B on link AB. This is obtained from the relationship B -AB -AB aa + ab/a

3.

where Ag/A is the radial or normal component of acceleration of point B relative to point A. _ Determine the effective component of Ag on link AB (Ag ). a. Determine Vg/A. This may be obtained from a velocity poly¬ gon, as in Figure 8.2b. b. Calculate the magnitude of A^B .

AB AB/A

N AB/A

V2 B/A AB

-AB c. From point B, lay out vector Ag (the summation of vectors AB/A anc* Af> alonS Bnk AB. 4.

8.3

Knowing the path of acceleration Ag to be along line "b"-"b", this vector can now be defined by constructing a perpendicular line from the terminus of A^B to meet line "b"-"b", similar to B the procedure used in the velocity case.

SLIDER-CRANK ANALYSIS

Crank AB of the slider-crank mechanism shown in Figure 8.3 rotates with an angular velocity of 2 rad/sec (counterclockwise) and an angular accelera¬ tion of 2 rad/sec2 (counterclockwise). Determine the acceleration of the slider. The acceleration of slider C is related to that of point B by the vector equation

C

ab + a C/B

and similarly, the effective components of these accelerations along BC are related by

Effective Component of Acceleration Method

155

B

-BC

Ac

a-BC

=ab

-BC

+ac/b

where abc = an ac/b c/b and N = X/B

V2 C/B BC

PROCEDURE (Figure 8.4b) 1.

From point B, lay out acceleration vector Ag to a convenient scale using the vector equation

where A

N = AB B

,

X co2

= 1. 5(2)2 = 6.0 in./sec2 A

T = AB 13

X

a

= 1.5(2) = 3 in. /sec2 2-

3.

-BC Construct the effective component of Ag on BC (Ag ). Drop a perpendicular from the terminus of Ag to extended link BC. -BC Determine the effective component of A^ on link BC (A^, ). a- Determine the magnitude Vg/g from a velocity polygon (Figure 8.4a).

Graphical Techniques: Acceleration Analysis

156

fa)

b

fb)

Figure 8.4 Acceleration of points on a slider-crank mechanism: (a) acceleration diagram; (b) velocity polygon.

V= 2.3 in./sec

b. Lay out A^C^ from point C along link BC. The magnitude of this vector is obtained A^g - ■'

;

= 1.76 in./sec2

c. Lay out vector A®c added to vector A®^ along BC. 4.

5.

Determine the absolute acceleration of point C (Ac) • Since the path of the slider is a straight line, the direction of its accel¬ eration must be along the same path. Therefore, to find the magnitude of this acceleration, drop a perpendicular from the effective component A®c to intersect the line of action of A . The point of intersection defines the acceleration Scale the vector Ac to determine its magnitude. A^ = 7.5 in./sec2

Aq.

Effective Component of Acceleration Method

8.4

157

FOUR-BAR LINKAGE ANALYSIS

Crank AB of the four-bar linkage in Figure 8.5 has an angular velocity of 2 rad/sec (counterclockwise) and is accelerating at the rate of 1 rad/sec2. Determine the acceleration of point C. In the example of Section 8.4, the acceleration of point C was related to point B by the vector equation

ab + a C/B and the effective component of the acceleration of point C along link BC was obtained from

abc

Ac

abc ac/b

In that example, since the direction of acceleration of point C was known, only one effective component of that acceleration was required to define the acceleration completely. In this example, however, the direction of accel¬ eration of point C is not known. Therefore, another effective component is needed to define Ac- This leads us to consider the effective component of point C along link CD, which is given by the equation

C

Figure 8.5

Four-bar mechanism.

Graphical Techniques: Acceleration Analysis

158

-CD AC

-CD _ AD

-CD + ac/d

where -CD AD = °

(s 11106 P0^ D is fixed)

t-CD ac/d

t-N C/D

an ac/d

——

where

vc

(directed from C to D)

Therefore, -CD A =

V2 C

(directed from C to D)

—BC —CD With the two effective components Aq and Ac determined, the required acceleration of point C can readily be found. PROCEDURE 1. From point B (Figure 8.6b), lay out acceleration vector Ag to a convenient scale, applying the relationship

A

•D

= A^ + A 13

13

(vectorial sum)

where A

N 13

= AB

X to2

= 1.5(4) = 6 in./sec2 A

T

= AB x a

ID

=

2. 3.

1.5(1) = 1.5 in. /sec2

Construct the effective component of Ag on BC (Ag ). Drop a perpendicular from the terminus of Ag to extended link BC. Find the effective component of Ac on link BC (A^).

c

^BC = aBC + ABC

c

where

ab

ac/b

159

Effective Component of Acceleration Method

BC AC/B

N = AC/B

V2 . C/B BC

a. Therefore, construct a velocity polygon (Figure 8.6a) to find VC/B. V

C/B

4.15 in./sec

b. Determine BC = 4.152 AC/B ~ 2.5

6.8 in./sec2

Therefore, 6.8 in./sec2

(directed along BC toward B)

b

Figure 8.6

Accelerations of points on a four-bar mechanism: (a) velocity

polygon; (b) acceleration diagram.

Graphical Techniques: Acceleration Analysis

160

— ■pC1

c. From point C, add vector A , to vector Ar, along link BC -BC C/B B to obtain Ag • _ 4. Determine the effective component of Ac along link CD (Ac )•

A^° = A^ L

v_/

(radial acceleration of point C)

where V2 N C Ac = CD 2.152

(VC fr°m velocity polygon) =1.54 in./sec2

Therefore, -CD , , A^ = 1.54 in./sec^

(directed from C to D)

5.

From point C, lay out vector A^ the fixed axis D.

on link CD directed toward

6.

With the two effective components of point C now known, the

7.

required acceleration Ac Is obtained by projecting perpendicu¬ lars from the termini of these two components until they inter¬ sect. This point of intersection defines the magnitude and direc¬ tion of required vector Ac. Scale the magnitude of vector Ac • A^ = 14.5 in./sec2 Therefore, Ac = 14.5 in./sec2

(directed as shown)

9 Relative Acceleration Method

9.1 INTRODUCTION The relative acceleration method is probably the fastest and most common among the graphical methods used for acceleration analysis. This is merely an extension of the relative velocity method used for velocity analysis and utilizes an acceleration polygon that is very similar to the velocity polygon. The method is based on the following principles: 1. 2.

All motions are considered instantaneous. The instantaneous acceleration of a point A relative to another point B on a rigid link is obtained from the vectorial relationship + A

A/B

3. The instantaneous motion of a point may be considered one of pure rotation in which the acceleration can be resolved into two rectangular components: one normal and one tangential to the path of rotation. Thus the expression in step 2 can be repre¬ sented graphically as -N -T -N ,-T -N , -T aa + aa = ab + ab + aa/b + aa/b 4. The absolute as well as relative velocities of various points in the mechanism are known. This requirement makes it most desirable to use the relative velocity or instant center method to determine the velocities involved.

161

Graphical Techniques: Acceleration Analysis

162

9.2

THE ACCELERATION POLYGON

Let us consider the two points A and B on link AB shown in Figure 9.1. As we saw earlier, the acceleration of point B relative to point A is given by the vector expression

B

= A

+ A. B/A

Just as in the velocity case, an acceleration polygon can be developed to represent the vector expression. Furthermore, the construction procedure and convention employed for this development are the same in both cases. PROCEDURE 1.

2.

3.

Define a point o', called the pole, as the origin for the construc¬ tion of the polygon. This is the point of zero acceleration. All absolute acceleration vectors originate from this point. By "absolute acceleration" we mean the real or true acceleration of a body as observed from a stationary frame of reference such as the earth. Since the accelerations A^ and Ag are absolute accelerations, draw vectors A^ and Ag from point o' and define their respective termini as a' and b'. From point a', the terminus of A^i, draw a third vector to ter¬ minate at point b', the terminus of Agt, thereby closing the polygon. This vector defines the relative acceleration Ag/^.

9.3

ACCELERATION POLYGON CONVENTION

The convention used to develop the acceleration polygon of Figure 9.1 can be summarized as follows:

Figure 9.1 Acceleration polygon.

Relative Acceleration Method

163

o'a' represents the acceleration vector AA. o'b' represents the acceleration vector A3, a'b' represents the acceleration vector AB//A. As in the velocity polygon case (Section 6.4), note that the letter to which the vector is directed indicates the acceleration under consideration. For example, in the vectors o'a' and o'b', the arrows point toward a' and b\ respectively. Hence the vectors represent the accelerations of points A and B, respectively. Similarly, in the vector a'b', the arrow points away from a' toward b'; hence the vector represents the acceleration of point B rela¬ tive to point A (Ab/^). If the arrow were reversed, pointing toward a' and away from b', the vector b'a' would represent the acceleration of A relative to B (Aa/b)-

9.4

GENERALIZED PROCEDURE

The construction procedure for developing the acceleration polygon of a mechanism will now be generalized using the four-bar mechanism as a model. Consider the four-bar mechanism ABCD shown in Figure 9.2. In this mechanism, AB, the driven member, has a clockwise angular accelera¬ tion coAb and a counterclockwise angular acceleration «Ab- It is required to find the acceleration of point C on the follower CD (Aq). PROCEDURE 1.

2.

3.

Determine the velocities of all points on the mechanism, includ¬ ing those with combined motion. This may be done using the instant center method, the effective component method, or as in the present case, the relative velocity method (Figure 9.3). Define a starting point o', called the polar origin. All absolute acceleration vectors originate from the polar origin. By "abso¬ lute acceleration" we mean the real or true acceleration of the point as observed from a fixed frame of reference such as the earth. Lay out the known components of absolute acceleration of the drive member, starting with the normal acceleration, which has the magnitude A^ = AB x co^B and is directed parallel to AB. Then add the tangential acceleration, which has the magnitude A^ = AB x £(,„ and is directed perpendicular to the normal B AB _>p acceleration, or Ab 1 Ab* The summation of these^two compo¬ nents defines the absolute acceleration of point B (Ab = ab + _T Ab) (Figure 9.4).If an angular acceleration 0^3 is not given, _rp _rp _n Ab does not exist (AB = 0), and Ab become_s the absolute accel¬ eration of point B. Define the terminus of Ab as b'.

164

Figure 9.2

Figure 9.4

Graphical Techniques: Acceleration Analysis

Four-bar mechanism.

Acceleration polygon.

Relative Acceleration Method

165

4. Select another point on the mechanism that has absolute motion and is rigidly connected to point B (call it point C) and lay out its normal acceleration component A^, noting that this vector is defined by the magnitude

N = C

5.

V2 , C/D CD

and is directed parallel to link CD. The normal components of acceleration are readily determined from the velocity data and the link orientation. -N Through the terminus of vector A^, construct a perpendicular to represent the tangential acceleration (A^), whose direction v

6.

is known but those magnitude is as yet undefined. This line contains point c', the terminus of vector Ac• Consider next the relationship between point B, whose accelera¬ tion is completely known, and point C, whose acceleration is only partially known. This means that we must consider the ac¬ celeration of point B relative to point C.

an +at ab/c ab/c

^B/C rN where A ,

can be completely determined from the velocity ^

rp

analysis and Ag/^ is known in direction only. Therefore, lay out vector A^c knowing that: a. Its magnitude is given by

B/C

VB/C BC

Its direction is obtained by assuming that point C on link is fixed while point B rotates about it. A^^, therefore lies on the link and is directed toward the center of rotation assumed. Also, in accordance with the polygon convention, relative acceleration vectors normally do not originate at the pole o' but extend between the termini of the absolute acceleration vectors to which they relate. For example, the notation Ag/c means that polygon vector is directed from c' to b', and A^/g means that polygon vector is directed from b' to c', where b' and c' are the termini of absolute vectors on the accelera¬ tion polygon, and B and C are the corresponding points in the linkage.

166

Graphical Techniques: Acceleration Analysis

Figure 9.5

Slider-crank mechanism.

Figure 9.6

Velocity polygon—step 1.

Figure 9.7

Acceleration polygon.

Relative Acceleration Method

7.

.

8

167

Through the tail* of A^,^, construct a perpendicular line of undefined length to represent the acceleration Ab/c, whose direction is known but whose magnitude is yet to be determined. This line also contains point c1, the terminus of Ap Now since c' lies on both AT (step 7) and AT (step 5), it fol¬ lows that the intersection of these two lines will define that point. Therefore, extend Aand A£, until they intersect and

9.

.

10

label the point of intersection c'. Finally, lay out a vector from origin o' to terminus c' to define the required absolute acceleration Aq and a vector from ter¬ minus c' to point b' to define the relative acceleration Ab/c* A quick check of the completed polygon should reveal the bal¬ anced vector equation

ab

ac + ab/c

or -N

-T

-N

-T

-N

-T

ab + ab = Ac + Ac + ab/c + ab/c Special Cases Slider-Crank Mechanism The slider-crank mechanism shown in Figures 9.5 to 9.7 may be considered a special case of the four-bar linkage where the follower link—for example, CD in Figure 9.2—is infinitely long. This means that the velocity of point C must be a straight line. Therefore, in step 4 (Figure 9.7) the normal accel¬ eration of the slider A^ is zero, and

-T In effect, the tangential component of acceleration Ac becomes the absolute acceleration of C and originates at o'.

*Note that had the normal relative acceleration A^g been chosen instead of Ag/c, the perpendicular drawn to represent A^g would have been con¬ structed through the terminus of Ac/g. This is because, by polygon con¬ vention, the normal acceleration vector A^^ must head toward c' and away from b'.

168

Graphical Techniques: Acceleration Analysis

ORIGIN

2V_

Figure 9.10

Acceleration polygon.

Relative Acceleration Method

169

Quick-Return Mechanism Another special case is the quick-return mechanism shown in Figures 9.8 to 9.10. Here point B slides on link CD, which is itself rotating. This means that in addition to the normal acceleration of point B relative to a coincident point C on CD, we must consider an additional component of acceleration: the Coriolis acceleration. Also, since the path of B relative to C is a straight line, the accel¬ eration of B relative to C can only be tangential. That is, B can have no normal component of acceleration relative to C; or

Had the path of B on CD been a curve, there would have existed a normal acceleration component of B relative to C (A^5 , ) as will be seen later (see Section 9. 7). To determine the Coriolis acceleration, we note from earlier dis¬ cussion that the magnitude of this vector is given by

A

Cor

2V

B/C^CD

where Vg/g and wgg are obtained from the velocity analysis. Also, we note that in defining the direction of this vector, we always consider (1) the linear velocity of the sliding body relative to that of the rotating body, and (2) the angular velocity of the same body on which the sliding occurs. In this example, since body B slides on link CD, we consider the velocity of B relative to C or Vg/C (not Vg/g) and the angular velocity of link CD or cogg (not oj^g). Accordingly, in step 6 (Figure 9.10), connect the vector A^01" to terminus b' on the polygon, observing the following rules: 1.

2.

This vector has the orientation of vector Vg/q when rotated 90° about its tail in the direction of cogg (counterclockwise in this case). Since point B relative to point C on the link is being considered, the polygon vector must go from c' to b' (note the reversed letter sequence).

EXAMPLE 9.1 (see Figure 9.11) Let AB = 1.5 in., BC = 1.5 in., CD = 1 in., coAg = 1 rad/sec (clockwise), and «ab = 0*5 rad/sec1 2 (counterclockwise). Determine Ac-

Graphical Techniques: Acceleration Analysis

170

a

Figure 9.11

Four-bar mechanism.

SOLUTION 1.

Determine the velocities. Construct a velocity polygon (Figure 9.12) and obtain V

= AB x co

= 1.5(1) = 1.5 in./sec

Aid

-D

V

= 1.45 in./sec

VB/C = 0.8 in./sec

2. 3.

Define polar origin o' for the acceleration polygon (Figure 9.13). Lay out the acceleration of the drive point, Ag.

T Ag - AB x

o

V

~ 1.5(0. 5) = 0.75 in./sec2

c V. B/C SCALE: 1 in. = 1 in. sec.

Figure 9.12

Velocity polygon.

Relative Acceleration Method

A

171

B

The vector originates at o' and terminates at b'. 4. Lay out the normal acceleration of the driven point, A(j. The magnitude of this vector is given as V2 , N C 1.452 AC “ CD “ 1.0

2.13 in./sec2

The vector is directed parallel to link CD, originating from o'. 5. Add the tangential acceleration A^ (undefined length). -T C 6. Lay out the normal acceleration of the drive point relative to the driven point, A^ . .

-N = AB/C

V2 B/C = 0.82 BC 1.5

0.426 in./sec2

The vector is directed parallel to link BC, pointing to b' (or away from c'). ^ 7. Add the tangential acceleration Ag/C (undefined length).

IT

rN

ab/c 1 ab/c

SCALE: 1 in. = 1 in,, sec.

Figure 9.13

Acceleration polygon.

172

Graphical Techniques: Acceleration Analysis

8.

9.

Determine the magnitude of tangential accelerations Aq and A^ , Extend undefined lines for these acceleration components in steps 6 and 7, until the polygon is closed. Aq intersects A-q/q at c'. Determine the required acceleration Aq. Connect point c' to point o' and measure o'c', or magnitude Aq. A

=3.2 in./sec2

(directed as shown in Figure 9.13)

L-

9.5

MECHANISM WITH EXPANDED FLOATING LINK

It was shown earlier that if the accelerations of two points on a link are known, the acceleration of a third point on that link can be found by propor¬ tion. In the following example it will be seen that the polygon construction procedure can also be used to determine the acceleration of the third point. EXAMPLE 9.2 Consider the mechanism ABCE shown in Figure 9.14a, where the crank AB rotates with an angular velocity of 2 rad/sec (clockwise). It is required to find the acceleration of point E on the expanded floating link BEC. SOLUTION 1.

Develop the velocity polygon for the mechanism as shown in Figure 9.14b. From this polygon the absolute and relative veloc¬ ity magnitudes are obtained as follows: V

=2.5 in./sec ID

Vc =

2.8

in./sec

=

1.0

in./sec

V£/c = 1.5 in./sec

2.

Develop the acceleration polygon for the basic mechanism ABC (Figure 9.14c) (ignoring point E for the time being), following the construction procedure discussed earlier. From this polygon the accelerations of points B and C are obtained. A-g = 5 in. /sec2 = 3.9 in./sec2

(directed as shown) (directed as shown)

Relative Acceleration Method

173

Figure 9.14 Mechanism with expanded floating link: (a) mechanism; (b) velocity polygon; (c) basic acceleration polygon (for ABC); (d) complete acceleration polygon.

174

Graphical Techniques: Acceleration Analysis

To determine the acceleration of point E (Ag), complete the following steps (Figure 9.14d). a. Lay out the vector for the normal acceleration of E relative to B • The magnitude of this vector is obtained from

N = AE/B

V2 E/B = (1^ EB 1

= 1 in./sec2 This vector is directed from point E toward point B on the mechanism, but on the polygon, heads from point b' (which is defined) to point e' (which is undefined), still maintaining its basic orientation. _JS[ b. Through the terminus of vector Ag /g, draw a line of unde¬ fined length perpendicular to this vector to indicate the direc¬ tion of AT^. This line contains point e', the terminus of vector Ag. c. From point c' on the polygon, lay out the vector for the nor¬ mal acceleration of E relative to C. The magnitude of this acceleration is obtain from

N

=

E/C

V2 E/C = (1.5)2 EC

1.5

=1.5 in./sec2 This vector is directed from point E to point C on the mech¬ anism but on the polygon heads from point c' (which is de¬ fined) toward point e' (which is undefined), still maintaining its basic orientation. d. Through the terminus of vector Ag/C, draw a line of unde¬ fined length perpendicular to this vector to indicate the direc¬ tion of AT This line contains point e', the terminus of A . rp E e. Since point e' is to be found on both lines Ag/g [step (c)] and AE/C [s*"eP (d)]» it must be located at the intersection of these lines. This point e' defines the terminus of the vector Ag drawn from the pole o'. Therefore, A

=5.6 in./sec2

(directed as shown)

Relative Acceleration Method

175

Acceleration Image As in the relative velocity case, it should be observed from the polygon that the triangle formed by points b', e', c' is similar to link BEC, and for this reason it is often referred to as the acceleration image of link BEC. Conse¬ quently, the acceleration of point E could have been determined more directly by constructing the acceleration image on link b'c' of the polygon for the basic mechanism ABC. (Note that the letters used to designate both the links and the image must run in the same order and direction.)

9.6

COMPOUND MECHANISM

Let ABCDEF represent a compound mechanism in the form of a shaper, where crank AB rotates with a constant angular velocity of 1 rad/sec (counterclockwise) (Figure 9.15a). Determine the acceleration of point E on the slider. PROCEDURE 1

Develop the velocity polygon (Figure 9.15b) and determine the velocities as follows:

VB ~ VC =

AB x co = 1.5(1) = 1.5 in. /sec 1.2 in./sec

1.0 in./sec V. B/C =

V

2. 3.

VD

OCX VC = 4(1'2) = l*8 in-/sec

V

2.1 in./sec

D/E

0.75 in./sec

Define polar origin o' for the acceleration polygon_(Figure9.15c). Lay out the acceleration of the driver at point B (AB). A^ = AB x oj2 = 1.5(1)2 = 1.5 in./sec2 B A

T = AB x a = 0 B

_ A

_at _rp = A + A = 1.5 in./sec2 B B B

(directed as shown)

176

Graphical Techniques: Acceleration Analysis

z u Figure 9.15 Acceleration analysis of a shaper mechanism: (a) shaper mechanism; (b) velocity polygon; (c) acceleration polygon.

Relative Acceleration Method

177

-N 4. Lay out the normal acceleration of point-driven point C (Aq)

J Ac

V2

c oc

N = as point d' on polygon. Determine the acceleration of point E relative to point D. r AD/E

rN rT ad/e + AD/E

where V2 . D/E DE

rN

D/E

(properly directed)

(0.75)2 0.1875 in./sec2

(properly directed)

rN

a. Lay out vector Aj-j/g. This vector must be pointed toward d' on the polygon and be parallel to DE on the mechanism. b. Add vector A^ to vector AN D/E D/E’ rT

,

7n

ad/e 1 ad/e 12. Determine the acceleration of point E (AE). Note that since the sLyer Pa^ a straight line, there is no normal acceleration (ae = 0). so A£ is equal to the tangential acceleration AT,

E

which exists along the same path. Therefore:

179

Relative Acceleration Method

Lay out the direction of vector Ap by drawing a line "e"-"e" through pole o' parallel to the slider path, b. Extend vector A^,^ in step 11(b) until it intersects line "e"-"e"

13. 14.

(direction line). This point of intersection defines the mag¬ nitude of ApDefine the point of intersection in step 12(b) as e'. Measure line o'-e', the vector Ap. A

= 0.625 in./sec2

(directed as shown)

-L

9.7

CAM-FOLLOWER MECHANISM

Consider the cam-follower mechanism shown in Figure 9.16a. The cam (2) rotates counterclockwise at a constant angular velocity of 2 rad/sec. Find the acceleration of the follower (4). At first glance it would appear that to determine the acceleration of 4 it would be necessary first to determine the motion of the roller (3). However, since the path that 3 traces on 2 is generally not easily recogniz¬ able, because of the two curved surfaces in contact, a more direct approach is to do the following: 1. 2. 3. 4.

Assume that the roller does not turn. Then the acceleration of contact point C is the same as that of any point in 4. Construct a pitch line to represent the locus of the center of the roller as it rolls on the cam. Designate the new point of contact between 2 and 4 as P2 or P4. Proceed with the analysis based on the assumption that the fol¬ lower 4 actually slides on the expanded cam defined by the pitch line.

Figure 9.16b shows the velocity polygon for the mechanism. Note that since 4 rides on 2, the relative velocity VP4/p2 is considered for Coriolis accel¬ eration. This velocity has a direction tangent to the curvature of the cam at the contact point. Figure 9.16c shows the acceleration polygon for the mechanism. Here it should be noted that the acceleration of P4 relative to P2 (or AP4/P2) consists of three components: -N ,T + yCor AP4/P2 “ AP4/P2 + AP4/P2 P4/P2 This equation, incidentally, is unlike those of earlier examples where A^ not exist, because the sliding paths in those cases were stiaight

180

Graphical Techniques: Acceleration Analysis

Figure 9.16 Acceleration analysis of a cam-follower mechanism: (a) camfollower mechanism; (b) velocity polygon; (c) acceleration polygon.

Relative Acceleration Method

181

lines. In the present case, since the path of P4 on P2 is a curve (or 4 is forced to move_in a curved path as it rides on 2), there must exist a normal acceleration, -^p4/p2> whose magnitude is given by

N _ VP4/P2 P4/P2 r + 0.5 where r + 0.5 or R is the radius of curvature of the path. For the Coriolis acceleration A(-'or, we consider, as before, the motion of P4 relative to P2, computing its magnitude from Cor _ AP4/P2 “ 2VP4/P2W and defining its direction by rotating the vector Vp4/p2 through an angle of 90° in the same angular direction (counterclockwise) as co2. Following is a summary of calculations and results:

V V V

P2 P4

P4/P2 an AP2

Cor VP4/P2

1.6 in./sec

(from the velocity polygon)

3.35 in./sec

(from the velocity polygon)

1. 5(2)2 = 6 in./sec2

2(3.35)(2) = 13.4 in./sec2

N

= 5.6 in./sec2

P4/P2 T P4/P2 AP4

9.8

1. 5(2) = 3 in./sec

4.0 in./sec2

(from the acceleration polygon)

2.8 in./sec2

(from the acceleration polygon)

SUMMARY

The generalized procedure for constructing the acceleration polygon may be summarized as follows: 1.

Proceed from the "known" to the "unknown." That is: a. Lay out those absolute vectors whose magnitude and direction are known.

Graphical Techniques: Acceleration Analysis

182

2.

3.

4.

b. Lay out those components of absolute and relative vectors that are known (magnitude and direction) or can be deter¬ mined. These include normal accelerations and Coriolis acceleration. c. Add to the components in step (b) their corresponding tangen¬ tial accelerations (directions only), and extend these to close the polygon. All absolute vectors on the acceleration polygon originate from pole o', while relative vectors extend between the termini of the absolute vectors. a. A vector that originates from b' and terminates at c' repre¬ sents the relative acceleration of point C to that of point B on the link. b. The choice between Ag/g and A^/g makes no difference in the polygon configuration or the results, except that these vectors have opposite senses. Two undefined vectors such as A^^ and A^ will contain a common point c' on the polygon, and will define that point where they intersect.

10 Velocity-Difference Method

10.1

INTRODUCTION

The velocity-difference method of determining accelerations of points in a mechanism is probably the most straightforward of all the graphical methods used in kinematic analysis. Applicable to any type of mechanism—pin con¬ nected, rolling contact, or sliding contact—the method does not rely on the use of sophisticated formulas, but instead employs the simple relationship

based on the fundamental definition, which states: The acceleration (A) of a point is the rate of change of velocity (AV) of that point over the time interval (AT) in which the change occurs. In terms of mechanism analysis, if the linear acceleration is required for a point in a mechanism at any given position of the mechanism, this acceleration may be found by first consid¬ ering a small change in position (A0) of the point, then determining the change in velocity (AV) resulting from the change in position, and finally dividing the change of velocity by the time interval (AT) during which the change has taken place. In applying the velocity-difference method, experience has shown that, generally, for reasonably accurate results, a position change of the mechanism based on a crank angular displacement of 1/10 rad is accept¬ able. This displacement is normally measured such that the given crank position is centrally located between its initial and final positions. That is, the initial and final crank positions are each 1/20 rad on either side of the given position. The associated time interval AT depends on the angular velocity of the crank and its angular displacement. If, for example, the crank of a mechanism turns at a constant speed of 5 rad/sec, AT is the time it takes

183

184

Graphical Techniques: Acceleration Analysis

for the crank to move through 1/10 rad. In our earlier discussion it was noted that for constant angular speed, AT is obtained from

co

Ad AT

or AT = -

Ad co

1 /rad \ 5 rad/sec V 10 '

= 1/50 sec If, on the other hand, the crank turns with a uniform acceleration, which means that the angular velocity is changing, then again from our earlier discussion, AT is obtained from

a

Aco AT

or Af,'

AT = —

a

where Ago is the change in angular velocity between the initial and final positions of the mechanism. This uniform or constant acceleration case will be discussed in more detail in Section 10.4. The following two sections illustrate the procedure for applying the velocity-difference technique to mechanisms where the crank arm rotates with constant angular velocity.

10.2

SLIDER-CRANK MECHANISM ANALYSIS

Consider slider-crank mechanism ABC shown in Figure 10.1, where crank arm AB is rotating at a constant angular velocity of 1 rad/sec (counter¬ clockwise). Let it be required to find the acceleration of point C for the position shown. PROCEDURE 1.

Lay out the given mechanism in the position ABjCj (see Figure 10.2), where ABj indicates the initial position of the crank, dis¬ placed 1/20 rad (or 1/2 of 1/10 rad) clockwise from the given

Velocity-Difference Method

Figure 10.1

185

Slider-crank mechanism.

position. Note that the angular displacement 1/10 rad can be accurately laid out by applying the well-known relationship S = R A9 where S is the length of an arc that subtends angle A0 and R is the radius of the arc. Hence the angle 1/20 rad is obtained by laying out a segment of a circle having some convenient radius, say 6 in., in which case the arc length is

S = 6^2o) = 0,3 ln‘

A

Figure 10.2

Slider-crank mechanism.

Graphical Techniques: Acceleration Analysis

186

2.

Determine the linear velocity of point Cj using any of the meth¬ ods studied earlier. The relative velocity method, used in this example, yields V

C.

=1.0 in. /sec

(directed as shown in Figure 10.2)

l

3.

4.

Lay out the given mechanism in position ABfCf (see Figure 10.2), where ABf indicates the final position of the crank, dis¬ placed 1/20 rad counterclockwise from the given position. Determine the linear velocity from point Cf as in step 2. This velocity has been determined to be V

=1.17 in./sec

(directed as shown in Figure 10.2)

°f 5.

Determine the velocity difference AV^, which is given by AV

= V

- V I

(vectorial difference) 1

This equation, which is represented graphically in Figure 10.2, yields AV

= 0.17 in./sec

(directed to the left)

L/

6.

Note that the vectorial difference in this example yields the same results as the algebraic difference since Vq and Vq. both have the same line of action. Determine the acceleration of point C. The magnitude of this vector is given by AV, A„ =

AT

where A0 AT = — co

0.1 rad/1.0 rad/sec = 0.1 sec Therefore, 0.17 . A^ = in./sec2 C 0.1 and the vector A

L/

= 1.70 in./sec2 (directed as shown in

Velocity-Difference Method

187

Figure 10.2). Note that the direction of acceleration is always the same as that for AVq.

10.3

QUICK-RETURN MECHANISM ANALYSIS

Now consider the quick-return mechanism ABCD in Figure 10.3. Crank AB turns with a constant angular speed of 30 rad/sec (counterclockwise) and the linear acceleration of point C on the rod is required. PROCEDURE We follow the same procedure as that given in Section 10.2. 1.

2.

Lay out the mechanism AB^C^D (Figure 10.4), showing the initial position of the mechanism rotated back 1/20 rad from the given position. Determine the linear velocity of point C (Vq.) for this position (see Figure 10.4). VP

= 72 in. /sec

(directed as shown)

'■'i

3.

Lay out the mechanism ABfCfD (Figure 10.4), showing the final position of the mechanism rotated forward l/20 rad from the given position.

Figure 10.3

Quick-return mechanism.

(ON

BLOCK)

(ON

ROD)

188

Graphical Techniques: Acceleration Analysis

4. Determine the linear velocity of point C (Vc ) for this position (see Figure 10.4), Vq

5.

= 69.5 in./sec

Determine the velocity difference AVC from the relationship AV^, = Vc^ = 15.0 in./sec

6.

(directed as shown)

(vectorial difference) (directed as shown in Figure 10.4)

Determine the linear acceleration Ac. The magnitude of this vector is given by

Avc

A

= _—

C

AT AV

— C AO

15.0

'0.1/

4500 in./sec2 Therefore,

Velocity-Difference Method

189

= 4500 in. /sec2

7.

(directed like AV^ as shown)

To obtain the angular acceleration of CD (acE)), aPPty the relationship Aco

a

CD

CD AT

where CD " wCDf " ^CDi VCf

VCi

" CDr “ CD. f l = 69.5 _ 72 1.7 1.52 = 40.9 - 47.4 = -6.5 rad/sec AT = — GO

= -(-) 10^30^

1/300 sec

Therefore, = -6.5 1/300 = -1950 rad/sec2 Note that in this analysis there was no need to determine the Coriolis acceleration, which is ordinarily the case in the relative acceleration method. In this respect, the velocity-difference method offers additional simplifica¬ tion to the solution of a problem that can otherwise be more complicated.

10.4

FOUR-BAR ME CHANISM ANALYSIS

Consider the four-bar mechanism ABCD, where crank AB turns with a constant angular acceleration of 1 rad/sec2 and for the position shown has an angular speed of 2 rad/sec (Figure 10.5). It is required to find the linear acceleration of point C on the follower. The procedure for solving this problem is basically the same as that used in Sections 10.2 and 10.3, with the exception that in this case there is

Graphical Techniques: Acceleration Analysis

190

C

Figure 10.5

Four-bar mechanism.

an angular acceleration, and therefore the magnitudes of the linear velocities of Vj} and Vg^ are not the same. These velocity magnitudes are calculated from V td — ABco. Bi i

(10.1)

VBf = ABco{

(10.2)

and

where

is the initial angular velocity of ABj obtained from

2

co

=

2 + 2a AO

CO

(10.3)

and coj is the final angular velocity of ABf, obtained from

2 =

CJ

2 - 2a AO

GO

(10.4)

Also, since the angular acceleration is constant, the angular speed is changing. AT must be expressed in terms of the angular acceleration a and the changing speed w. That is,

Velocity-Difference Method

191

(10.5) where cof = coj. PROCEDURE 1. 2.

Figure 10.6 shows the mechanisms ABjC^D in the initial position and ABfCfD in the final position. Determine coj. This is found by applying Equation (10.3). co2 = co? + 2a A9 l

22 = “U 2(1>(ij;) co? = 4 - 0.1 = 3.9 l

co. =1.97 rad/sec l

3.

Determine VB . From Equation (10.1), the magnitude of this vector is computed as VB. = 1.5(1.9) = 2.95 in./sec Therefore, VBi = 2 .45 in./sec

4.

(directed as shown)

Determine toj . Since a is constant, cof is easily determined by recognizing that its increment from co at the given crank position is equal to that of the same co from co^, due to the same angular displacement of 1/20 rad in both cases. Therefore, a>

= 2 + (2 - 1.97) =2.03 rad/sec

5.

Determine VBf- The magnitude of this vector is computed as

VBf = 1.5(2.03) =3.04 in./sec

192

Graphical Techniques: Acceleration Analysis

Figure 10.6

Four-bar mechanism.

Therefore, = 3.04 in./sec 6.

(directed as shown)

Determine VCf using the mechanism ABfCfD, as before. vCf ~

7 in./sec

(directed as shown)

Velocity-Difference Method

7.

193

Determine Vc. using the mechanism ABjCjD, as before. Vc. = 1.25 in./sec

8.

Determine AVc from AV^ = Vc

- Vc.

= 0.46 in./sec 9.

(directed as shown)

(vectorial difference) (directed as shown)

Determine Ac- The magnitude of this vector is computed as

where AT, from Equation (10.5), is given by

AT

ACO

a 2.03 - 1.97 1 0.06 sec

Therefore, A_

0.46 0.06 7.7 in./sec2

Hence A„ = 7.7 in./sec2

(directed like AV^ as shown)

Despite the simplicity of the velocity-difference method, it should be understood that because of the need to measure small changes in dis¬ placements as well as velocities, a high degree of drawing precision is required if sufficiently accurate results are to be achieved. In some cases, this requirement can impose practical limitations such that the results obtained may not be reliable.

11 Graphical Calculus Method

11.1

GRAPHICAL DIFFERENTIATION

A common method for obtaining velocity and acceleration curves for a mechanism consists of constructing a time-displacement curve and then by graphical differentiation developing the velocity and acceleration curves. Graphical differentiation consists of obtaining the slopes of various tangents along one curve and plotting these values as ordinates to establish a second curve. The method, as applied to the displacement and velocity curves, is based on the following principles. 1.

The instantaneous velocity of a moving point can be represented graphically as the slope of the displacement curve at that instant, or As v = —

2.

where

At is very small

The instantaneous acceleration can be represented as the slope of the velocity curve at that instant, or Av a = —

where

At is very small

Graphical differentiation can be an effective tool for the designer, particularly when used during the preliminary stages of a mechanism design since it affords an overall picture of the velocity and acceleration throughout the motion cycle. For example, from motion curves such as the s-t, v-t, and a-t curves, one can readily determine: The maximum absolute values of displacement, velocity, and accel¬ eration

194

Graphical Calculus Method

195

Where and when the maximum values of angular displacement occur Whether there is any abrupt change in displacement, velocity, and acceleration during the cycle Also, there are many instances where it is necessary to differentiate a curve for which an equation is difficult to obtain. In such instances, graph¬ ical differentiation is the most convenient approach.

Tangent Method Consider the displacement-time curve shown in Figure 11.1. Suppose that it is required to develop the velocity-time curve from this curve. Since the instantaneous velocity at any point is represented by the slope of the displacement curve at that point, a tangent drawn through any point of the curve defines the velocity at that point. Therefore, at point A, the velocity v^ is obtained from

s

Figure 11.1

Graphical differentiation.

196

Graphical Techniques: Acceleration Analysis

= slope of tangent at A

where kg = displacement scale, in./in. = time scale, sec/in.

Similarly,

v

= slope of tangent at B =

vc = slope of tangent at C =

B (k ) s s

w w

Here the negative sign indicates a negative slope of the curve and hence a negative velocity. Also, when the tangent to the curve is a horizontal line, this simply means that the velocity at the point of tangency is zero. In addition, it should be clear that by making the time intervals A^, B^, Ct, and so on, equal, the velocities v^, vg, v^, and so on, will be pro¬ portional to As, Bg, Cs, and so on. Hence, once the velocity v^ has been found, the velocities Vg, v^, and so on, are readily obtained as follows: B

s

s In this manner, the velocities of other points can be obtained to enable a smooth continuous curve to be drawn. EXAMPLE 11.1 The curve shown in Figure 11.2a represents a typical displacement-time curve for a cam mechanism. It is desired to obtain the velocity-time curve from this displacement curve, and subsequently, the acceleration-time curve from the velocity curve.

Graphical Calculus Method

Figure 11.2

197

Graphical differentiation—tangent method: (a) displacement

curve; (b) velocity curve.

Graphical Techniques: Acceleration Analysis

198

SOLUTION 1. 2.

Draw tangents to various points on the given displacement-time curve: points A, B, C, D, and E. Using the tangents as hypotenuses, construct right triangles at each point, making all the bases equal, that is, A

=

= C

etc.

3. Calculate the velocities at points A, B, C, D, and E as follows: k

s

=2 in. /in.

k^

= 3 sec/in.

k

=1 (in./sec)/in.

v

A k s s Atkt B A C

v.

A D =

A E

v„ =

/'0.5 \ 2 '1.0'3 ^

s s

VA

VO.5'

VA

(2.20\ VO.5 '

s s s s

'sec

/O.SA VA ' Vo. 5'

s

VA

( 0 ), Vo. 5'

Note that the remaining points on the velocity curve can be located by inspection based on the symmetrical shape of the displacement curve. 4. 5. 6. 7.

Draw the velocity-time axis using the same time scale as that used for the displacement-time axis (Figure 11.2b). Plot the velocities calculated in step 3 on the velocity-time axis. Draw a smooth curve connecting the plotted points. To obtain the acceleration curve, repeat steps 1 through 6, re¬ placing the displacement curve with the velocity curve just found (Figure 11.3a). The required acceleration curve is shown in Figure 11.3b.

Graphical Calculus Method

Figure 11.3

199

Graphical differentiation—tangent method: (a) velocity curve;

(b) acceleration curve.

Polar Method An alternative graphical differentiation method commonly used is the polar method. This method has been found to produce greater accuracy than the

200

Graphical Techniques: Acceleration Analysis

I

I

Figure 11.4 Graphical differentiation—polar method: (a) displacement curve; (b) velocity curve.

tangent method mainly because it requires less graphical construction. The method is best described by considering the displacement-time curve given in Figure 11.4a. Let it be required to develop the velocity and acceleration diagrams from the given curve.

Graphical Calculus Method

201

PROCEDURE 1.

2.

Lay out the axes for the velocity-time graph directly below those for the displacement curve, maintaining the same time scale in both cases (see Figure 11.4b). Through the point of steepest slope on the displacement curve (point M), draw a tangent line to define the point of maximum velocity (t^j). Now, since by definition, velocity equals slope, the maximum velocity at point M is given by As vmax = T7 At where As is the change of displacement over any interval At, or

v

max

= (—) ^ me4 5 * 7

where k

s

= displacement scale, in./in.

k^. = time scale, sec/in.

3.

Define a point P (called the pole) on the time axis of the velocity¬ time curve, at any convenient distance left of the origin, and from it draw line t]yp parallel to tangent line tM to intersect the velocity axis. This line locates a point M' on the velocity axis which repre¬ sents the point of maximum velocity, or V-,., = V

M

4.

5.

max

Plot the velocity of point M at the intersection of the horizontal line drawn from M' on the velocity curve and a vertical line drawn from M on the displacement curve. Label this point M". Use the value obtained for vmax to establish the scale for the velocity axis of the velocity-time graph. To do this, compare the actual measurement on the diagram with the computed maximum velocity and then by ratio determine the velocity value that is represented by 1 in. on the diagram. In this example we find that

Graphical Techniques: Acceleration Analysis

202

where k

s

= 5 in./in. = 2 sec/in.

Therefore,

v

max

5 2

4 in./sec

Figure 11.5 Graphical differentiation—polar method: (a) velocity curve; (b) acceleration curve.

Graphical Calculus Method

203

Hence 2 in. on the velocity curve represents 4 in./sec, or 1 in. represents 2 in./sec, or = 2 (in./sec)/in. 6.

To establish the remaining velocity points at ordinates A, B, C,

(B) PAPER

Figure 11.6

Use of a mirror in constructing tangents: (a) incorrect mirror

position; (b) correct mirror position.

204

Graphical Techniques: Acceleration Analysis

7.

D, and so on, follow the procedure outlined in steps 3 and 4. Figure 11.5a shows the completed velocity diagram. After the velocity diagram has been completed, the acceleration diagram can be developed by following the procedure outlined above. Figure 11.5b shows the required curve.

Note on Accuracy It should be emphasized that accuracy of the graphical differentiation meth¬ ods depends primarily on: 1. 2. 3.

How accurately the tangents can be drawn The number of intervals chosen The ability to fit a smooth curve to a given set of points

For greater accuracy in constructing tangents, any straight edge having a reflective surface, such as the rectangular mirror shown in Figure 11.6, may be used. The straight edge, held vertical to the paper, is placed across the curve and rotated such that the visible curve and its reflection form a continuous curve. In this position, the straight edge is then normal to the curve. Therefore, a line drawn perpendicular to the normal and touching the curve is the required tangent. Accuracy is increased with the number of intervals into which the curve is divided. The greater the number of intervals chosen, the greater the accuracy that is achieved. Finally, the ability to fit a smooth curve to a given set of points, particularly if double differentiation is required, greatly minimizes the risk of compounding errors from one curve to the other.

11.2

GRAPHICAL INTEGRATION

Just as it is possible to go from the displacement curve to the velocity curve and then to the acceleration curve using graphical differentiation, it is also possible to reverse the process, going from acceleration to velocity and then to displacement. The process by which this is achieved is called graphical integration. The process of integration can be thought of as the procedure for obtaining the area under a given curve. As applied to the motion of a mech¬ anism, the integration of the acceleration curve gives the velocity curve, and the integration of the acceleration curve gives the displacement curve. This is based on the motion laws, which state that Av = a At and

or

V]? - vQ = a(tp - tQ)

Graphical Calculus Method

As = v At

or

205

sp - sQ = v(tF - tQ)

where the subscripts F and O denote final and original conditions, and a and v denote average acceleration and velocity. Hence the following rules apply. 1.

To derive the velocity-time curve from the acceleration-time curve, the change in velocity between any two times equals the the area under the acceleration-time curve between the same two times.

2.

To derive the displacement-time curve from the velocity-time curve, the change in displacement between any two times equals the area under the velocity-time curve between the same two times.

Mid-ordinate Method There are several graphical methods available to determine the area under the curve. One of the simplest is the mid-ordinate method. The following example will illustrate the procedure. EXAMPLE 11.2 The curve shown in Figure 11.7a represents a typical acceleration curve for a cam mechanism. It is desired to obtain the velocity-time curve. SOLUTION 1.

2.

3.

Divide the given curve into an equal number of sections, S1; S2, S3, and so on, as shown in Figure 11.7a, and construct mean ordinates alt a2, a3, and so on, for each section to intersect the curve at 1, 2, 3, and so on. Through points 1, 2, 3, and so on, draw horizontal lines extending between the boundaries of alternate sides of the curve. For ex¬ ample, in section S1, triangles A and B are on alternate sides of the curve. From this it is easily seen that, provided that the slope of the curve within the section remains fairly constant, the alter¬ nate triangles are approximately equal. This means that in sec¬ tion Sj, the area of triangle A can be considered approximately equal to that of triangle B; similarly, in section S2, the area of triangle C is equal to that of triangle D; and so on. Therefore, the area under the curve in section Sj can be approximated by the rectangle aj Atx; the area under section S2 can be approximated by the rectangle a2 At2; and so on. Since the velocity equals the area under the curve, the velocities vx, v2, and v3 may be found as follows:

206

Graphical Techniques: Acceleration Analysis

Figure 11.7 Graphical integration—mid-ordinate method: (a) acceleration curve; (b) velocity curve.

Graphical Calculus Method

Vj

= aj

207

Atj

v2 = Vj + a2 At2 v3 = v2 + a3 At3

etc.

For example, in Figure 11.7a, Vj = 1.4(1) = 1.4 in./sec v2 = 1.4 + 2(1) = 3.4 inr/sec v3 = 3.4 + 2.4(1) = 5.8 in./sec

4.

The velocity curve is shown in Figure 11.7b. After the required velocity curve has been completed, repeat steps 1 to 3 to develop the required displacement curve.

Polar Method Another graphical integration method that is commonly used is the polar method, which is relatively fast and easy to apply. To illustrate the method, let us consider the acceleration curve shown in Figure 11.8a. From this curve we will develop the velocity curve and, in turn, use the velocity curve to develop the displacement curve. PROCEDURE 1.

Erect coordinates BB, CC, DD, and so on, to divide the curve into an equal number of line intervals, as shown in Figure 11.8a. In this case, five intervals have been chosen with ordinates a1 to a5 defined at the midpoints of the intervals. Also, since the horizontal axis is the time axis, each interval is defined at Atx, At2, At3, and so on. Note that all time scales for acceleration, velocity, and displacement curves must be the same, as with

2. 3.

4.

graphical differentiation. From points 1, 2, 3, and so on, project horizontals to meet a line parallel to the a axis at points 1', 2', 3', and so on. From a point P', located on the time axis, at any convenient distance left of the origin, draw connecting straight lines to points 1', 2', 3', and so on. From the origin of the velocity axis (Figure 11.8b) draw line ob' parallel to P'l' to meet ordinate BB at b', line b'c' parallel to P'2' to meet ordinate CC at c', line c'd' parallel to P'3' to meet ordinate DD at d, and so on, until the complete velocity curve is obtained. From this velocity curve, the velocities at points 1, 2,3, and so on, are given by

208

Graphical Techniques: Acceleration Analysis

v, = P'Q' (bb')k k 1 cl t v

= P'Q’(cc')k k ^

v

cL X

= P'Q'(dd')k k a t

etc.

where b, c, d, and so on, are the intercepts of the first, second, third, and so on, ordinates with the time axis, and ka [(in./sec2)/ in.] and k^ (sec/in.) are the acceleration and time scales, re¬ spectively. The relationships above can be verified by applying the properties of similar triangles to triangles P'Q'l' and obb',

Graphical Calculus Method

209

Figure 11. 8 Graphical integration—polar method: (a, facing page) acceleration curve; (b, facing page) velocity curve; (c) displacement curve.

where bb' is the vertical leg of the velocity curve segment ob. From these triangles, we obtain Q'l' _ P'Q' bb' - ob

(11.1)

But Q'l’ = ^ and ob = Atj Therefore, Equation (11.1) becomes _*i_ = P'Q' bb' Atj from which follows a.1 Atj = P'Q'(bb') But

(all parameters in inches)

Graphical Techniques: Acceleration Analysis

210

ai Atj(kakt) = Vj - v0 where v0 = 0 Therefore, vx = P'Q'(bb')kakt Similarly, from triangles P'Q'2' and b'c"c', where c"c' is the vertical leg of the velocity curve segment b'c", we obtain Q'2' = P'Q' c"c' b'c" from which follow a2 At2 = P'Q'(c"c') v2 - vx = P'Q'(c"c')kakt v2 = Vj + P'Q'(c"c')kakt v 2 = P'Q' (bb')kak|. + P'Q'(c"c')kakt = P'Q'(bb' + c"c')kak^ = P'Q'(cc')kakt and from triangles P'Q'3' and c'd"d', Q'3' = P'Q' d"d' c'd" from which follow a3 At3 = P'Q'(d"d') v3 - v2 = P'Q'(d"d')kakt v3 = v2 + P'Q'(d"d')kakt v3 = P'Q'(cc')kakt+ P'Q'(d"d')kakt = P'Q'(cc' + d"d')kakt - P'Q' (dd')kakt

Graphical Calculus Method

211

To obtain the displacement diagram, repeat steps 1 to 4, replacing the acceleration curve with the velocity curve. Figure 11.8c shows the required displacement diagram.

Note on Accuracy As in the case of the graphical differentiation methods, accuracy of the graphical integration method increases with the number or size of the inter¬ vals chosen. The smaller the interval chosen, the more likely the slope of the curve within that interval (or section) will remain constant, and hence the closer the approximation of equality between the alternate triangles. It is therefore desirable to make the intervals or sections as small as possible for greater accuracy. Also, if double differentiation is required, accuracy will depend on the ability to fit a smooth curve to a set of given points.

12 Special Methods

12.1

COMPLETE GRAPHICAL ANALYSIS METHOD

In the graphical methods for acceleration analysis presented so far, one of the required calculations to be performed was that used to determine the normal acceleration components for points in the mechanism. In this section we demonstrate a complete graphical method where, by proper choice of link, velocity, and acceleration scales, both the magnitude and direction of all normal acceleration components can be determined without the need for calculations. To illustrate this method, let us consider link AB, shown in Figure 12. la, which rotates at an angular velocity of to rad/sec. First, we obtain the magnitude of the velocity of point B relative to A (VB/A) by multiplying the length of the link by the angular velocity,

vb/a = abx"ab

(12-D

We then lay out the vector perpendicular to the link as shown. Here it is useful to recall that the magnitude of Vg/A is related to that of AN, by the expression r>/A

.N _ B/A

V2 B/A BA

(12.2)

Rearranging this equation, we obtain the relationship

A V

N B/A B/A

Va (12.3)

BA

212

Special Methods

213

c

(c) Figure 12.1

which can be represented as BD _ BC BC BA

(12.4)

Graphical Techniques: Acceleration Analysis

214

using two similar triangles, ACB and CDB, as shown in Figure 12. lb, where

BD and BC = V

B/A

Here is can be seen that if the scales of link AB, velocity Vg/^, and accel¬ eration A^, in Equation (12.3) are properly chosen, in accordance with the relationship expressed in Equation (12.4), the length of line BD will accurately represent the required normal component. Let the scales be defined as follows: kg = space scale, ft/in. k^ = velocity scale, (ft/sec)/in. k^ = acceleration scale, (ft/sec2)/in.

where the inch units in the denominators represent actual measurements of the drawing in inches. For example, if the space scale is given as ks = 5 in./in., a link length of 5 in. will be represented by a line 1 in. long on the drawing. Similarly, if the velocity scale is given as ky = 10 (ft/sec)/in., a velocity of 20 ft/sec will be represented by a line 2 in. long on the drawing; and for an acceleration scale of 50 (ft/sec2)/in. an acceleration of 150 ft/sec2 will be represented by a line 3 in. long. The required relationship of the scales is obtained by considering tiiangle abc in Figure 12.1c, where sides x, y, and z are proportional, respectively, to sides AB, BC, and BD of triangle ABC in Figure 12.1b. In other words, if x represents the link length, y the velocity, and z the acceleration, we can write xk

s

= BA

or

- M

x - k yk or

v

s

= BC

(12-5)

215

Special Methods

BC k v

(12.6)

BD

zk or

BD

(12.7)

and from similarity, z = JL y x

(12.8)

Substitution of Equations (12.5) to (12.7) into Equation (12.8) yields BD/k BC/k _a _ _v BC/k " BA/k

(12.9)

or

A* Vk B/A

VB/A/kv BA/k

a

VB/A/kv

(12.10)

from which

,N B/A

V2 k k B/A a s BA k2 v

(12.11)

Thus the scales must be related by the equation k k a s k2 v

(12.12)

or k k a s

k2 v

(12.13)

This means that any two scales may be chosen arbitrarily, but the third must be chosen from Equation (12.12).

216

Graphical Techniques: Acceleration Analysis

Figure 12.2

Four-bar mechanism.

EXAMPLE 12.1 For the four-bar mechanism ABCD shown in Figure 12.2, make a complete graphical acceleration analysis given that crank AB is rotating with an angular velocity of 2 rad/sec (counterclockwise) and an angular acceleration of 1 rad/sec2 (clockwise). SOLUTION 1.

Select a space scale ks = 2 in./in. or 1 in. (space scale) = 2 in. and velocity scale kv = 2 (in./sec)/in. or 1 in. (velocity scale) = 2 in./sec, and from the relationship

v obtain

b

VEL. 1"

Figure 12.3

Velocity polygon.

SCALE

= 2

in /sec

217

Special Methods

k

.

2

3.

4.

5.

a

(2/l)z 2/1

2 (in./sec2)/in.

or 1 in. (acceleration scale) = 2 in./sec2. Construct the velocity polygon (Figure 12.3) using the ky scale. This requires the calculation of the velocity magnitude Vg/^ as a first step, then the layout of the vectors in accordance with the velocity polygon procedure already discussed. Transfer vectors Vg, Vq, and Vg/C from the velocity polygon to points B and C on the linkage, maintaining the same orientation. Vg is drawn from B perpendicular to AB; Vq is drawn from C perpendicular to CD; and Vg/C is drawn from B perpendicular to BC. using the construction outlined Determine A^, A^J, and A^ B/C 13 C -N above. For example, in Figure 12.4, Ag is obtained by constructa right-angle triangle Abx in which z. b is 90°, Ax is the extended link AB, and the line Bb, perpendicular to AB, is the velocity magnitude of point B. The line segment Bx then represents the required value of Ag. Construct the acceleration polygon using the k~ scale (1 in. = 2 in./sec2), starting with the calculation of A^, then following the acceleration polygon procedure already discussed. Figure

Figure 12.4

Normal acceleration construction

Graphical Techniques: Acceleration Analysis

218

SCALE

n 2

ACC.

12.5 shows the acceleration polygon, from which the required acceleration Aq is determined to be A

= 12.2 in./sec2

(as directed)

In the example above, it is to be noted that only two calculations were neces¬ sary to complete the analysis after the scales were determined. These were to determine and A^, the velocity and acceleration magnitudes of the first link. If this link were to rotate with a constant angular velocity, calcu¬ lation of A^ would not have been necessary, since the value of this accelera¬ tion would be zei-o. Thus, for a complete graphical analysis, the maximum number of calculations necessary is two.

12.2

EQUIVALENT LINKAGE METHOD

Determining the acceleration of points on many higher-paired mechanisms, such as those with rolling and sliding contacts, can become rather involved if a point-to-point analysis is attempted. This is because of the need to know the curvature of the path traced by a point on one link relative to the

Special Methods

Figure 12.6

Kinematically equivalent four-bar linkages.

220

Figure 12.7

Graphical Techniques: Acceleration Analysis

Kinematically equivalent slider-crank linkages.

Special Methods

221

other and to apply the Coriolis law. If no easily recognized path is found, the analysis can be difficult. To simplify this problem, the use of equivalent linkages has been found most effective. In application, an equivalent linkage replaces a higherpaired contact with appropriate lower pairs that will produce the correct values of velocities and accelerations for the instantaneous phase under consideration. An equivalent linkage may then be defined as one that pro¬ duces identical motion as the part being analyzed for a given position or phase. Figures 12.6 and 12.7 show several mechanisms with their equivalent linkages depicted by dashed lines. Note that the rolling and sliding surfaces have been replaced by pin joints as part of a more simplified four-bar link¬ age. Note also that in each case, the floating link of the equivalent linkage is drawn along the common normal of the two contacting surfaces and con¬ nects the centers of curvature of the surfaces. Although an equivalent linkage is generally valid only for a given instant or phase and does not ordinarily apply to a complete cycle, there are some instances where the equivalent linkages of some higher-paired mechanisms will duplicate the input/output motion of those mechanisms throughout their motion cycle. Some examples are shown in Figures 12.6 and 12.7. EXAMPLE 12.2 Consider the cam mechanism shown in Figure 12.8a. The cam (2) rotates counterclockwise at a constant angular velocity of 2 rad/sec. Find the acceleration of the follower (4) using the equivalent linkage method. SOLUTION The equivalent mechanism for the cam mechanism given is the simple slider-crank ABC shown in Figure 12.8b, for which the velocity and accel¬ eration diagrams are readily obtained, as shown in Figure 12.8c and d. Applying the velocity polygon construction procedure, we obtain

B

V. V

C/B

0.9(2) = 1.8 in./sec 1.45 in./sec 0. 6 in./sec

(from the velocity polygon) (from the velocity polygon)

Applying the acceleration construction procedure, we obtain

A

13

= 0.9(2)2 =3.6 in./sec2

222

Graphical Techniques: Acceleration Analysis

Fig. 12.8 Acceleration analysis of a cam-follower mechanism: (a) camfollower mechanism; (b) equivalent linkage; (c) velocity polygon; (d) accel¬ eration polygon.

223

Special Methods

*Tb =

0

an ac/b

^

= 0.18 in./sec2

= 3.8 in./sec2 at C/B AC =

2.8 in./sec2

(from the acceleration polygon) (directed as shown)

Note that the resulting acceleration of point C (A^) is exactly the same as that obtained for point P4 (the same point) in Section 9.7, using an alter¬ native method.

12.3

SLIDER-CRANK ACCELERATION: PARALLELOGRAM METHOD

Despite the wide use of the relative acceleration method in solving linkage problems, it is not uncommon for one to experience some confusion in cor¬ rectly applying this method to the slider-crank mechanism. The confusion most often encountered arises from uncertainties or oversight as to the proper location of the relative radial acceleration vector in the acceleration polygon. As a result, many errors are made. To avoid such confusion or oversight, the graphical method presented here makes use of simple parallelogram constructions which serve as guides in laying out the vectors. The method is not only simple to apply, but also saves time.

Scope In a typical slider-crank mechanism, as shown in Figure 12.9a, crank AB has angular rotation to and angular deceleration o about point A. Angle 0q is the instantaneous angular position of the crank with respect to the line of action AC, and dc is the angle between the sli_der arm BC and line AC. It is requii-ed to determine slider acceleration Ac at any angle Qq. Construction Development 1.

Define o' as the point of zero acceleration coincident with pivot B on the mechanism, and draw vector o'b' (see the acceleration diagram in Figure 12.9c) to represent acceleration, AB, as determined from (12.14)

224

Graphical Techniques: Acceleration Analysis

(b)

KEY POINT: vN

,

Ag/C (°r

-rN

. _

_.

if used)

is

always located on the 'reflected' connecting rod.

(c)

Figure 12.9 Graphical construction procedure: (a) slider-crank mecha¬ nism; (b) velocity polygon; (c) acceleration polygon.

225

Special Methods

where the magnitudes of Ag and Ag are given by

OBw2

AT- = OBa Jd

2.

3.

With o'b' as the diagonal and 9q as_the subtended angle, construct a parallelogram o'Cb'c such that o'C is equal and parallel to cb', and b'C is equal and parallel to o'c. (Note the geometric identity between triangular sections o'b'c and o'b'C. Vector o'b' may be considered the axis of asymmetry to the parallelogram.) Determine Vg/c from the vectorial relationship (12.15) where V

B

OB co

(directed perpendicular to OB)

(12.16)

and Vc is known in direction only (along the slider path). Vg/c can therefore be obtained from a velocity diagram, as follows. From a point o (see the velocity diagram in Figure 12.9b), draw line ob scaled to represent Vg perpendicular to the instantaneous position of crank OB and pointing in the direction of the motion. Then draw a line parallel to the direction of the slider motion to represent velocity V^.. Finally, draw a line_from point b perpen¬ dicular to arm BC so that it intersects the Vc line at point c. The velocity magnitude Vg/C is determined by measuring line be and converting to the appropriate velocity according to the chosen scale. 4. Compute the acceleration magnitude A^c from

N = AB/C

V2 , B/C BC

(12.17)

Next, mark a segment of line b'c on the parallelogram to repre¬ sent vector Ag ^ scaled the same as o'b' and heading toward b . (This requirement is in keeping with the polygon convention, in which the acceleration vector relative to a point on a link must be be directed toward the corresponding point on the acceleration polygon. Alternatively, if aN/b is considered, this vector must point from b' to c' on the polygon.)

226

Graphical Techniques: Acceleration Analysis -N From the tail of vector Ag/^, draw a perpendicular line to inter¬ sect line o'c^at point c'. Point c' defines the acceleration, Ac. along line o'c, and the perpendicular drawn from a£J represents rji

B/ C

the tangential acceleration, Ag/C. The value of Ac may be checked with the vectorial relationship

aN4.

ab

TT ab

a-N rT ab/c ~ ab/c

(12.18)

(Note that for a uniform velocity of crank OB, triangle o'b'c is identical to the configuration of the given mechanism OBC, in which case C and C are coincident points. Note also that A^ alB/ C ways lies on the "reflected" slider arm.) As an example, consider a slider-crank mechanism where OB = 1.5 in., BC = 3 in., co = 1 rad/sec, and a - 0. Determine the slider acceleration, Ac, for eQ = 45° (Figure 12.10), 135° (Figure 12.11), 225° (Figure 12.12), and 315°(Figure 12.13). PROCEDURE 1*

From Equation (12.14), acceleration Ag = 1.5 in./sec and line o b is drawn 1.5 in. in length to represent this vector.

Figure 12.10 polygon.

Crank angle at 45°: (a) velocity polygon; (b) acceleration

Special Methods

Figure 12.11 polygon.

227

Crank angle at 135°: (a) velocity polygon; (b) acceleration

225°

b

Figure 12.12 polygon.

(b) Figure 12.13 polygon.

Crank angle at 315°: (a) velocity polygon; (b) acceleration

228

Graphical Techniques: Acceleration Analysis

.

2 3.

4.

Construct the parallelogram o'Cb'c, using o'b' as the diagonal. From Equation (12.16), Vg =1.5 in./sec. With line ob drawn 1.5 in. long to represent vector Vg, a velocity polygon is con¬ structed from which Vg/c = 1.12 in./sec. From Equation (12.17), A^^ = o.42 in. and a 0.42-in. segment

5.

of line b'c is denoted on the parallelogram to represent Ag/^. A perpendicular line is drawn from the tail of vector A^^ to intercept line o'c at point c'. The length of line o'c' is measured to be 1.05 in., so that Aq = 1.05 in./sec2 (to the left). Use of Equation (12.18) confirms this value as being accurate. Similarly, Ac for the crank position