Sample Kinematics and dynamics of mechanical systems implementation in Matlab and Simmechanics solutions PDF

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Authors: Kevin Russell , Qiong Shen | Raj S. Sodhi
Published: CRC Press 2018
Edition: 2nd
Pages: 438
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FOLFNKHUHWRGRZQORDG

SOLUTIONS MANUAL FOR

KINEMATICS and DYNAMICS of MECHANICAL SYSTEMS ®

Implementation in MATLAB and SimMechanics®

KEVIN RUSSELL QIONG SHEN RAJ S. SODHI

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FOLFNKHUHWRGRZQORDG

SOLUTIONS MANUAL FOR

KINEMATICS and DYNAMICS of MECHANICAL SYSTEMS ®

Implementation in MATLAB and SimMechanics®

KEVIN RUSSELL QIONG SHEN RAJ S. SODHI

Boca Raton London New York

CRC Press is an imprint of the Taylor & Francis Group, an informa business

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CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2016 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed on acid-free paper Version Date: 20150716 International Standard Book Number-13: 978-1-4987-2499-9 (Ancillary) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

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CONTENTS Preface

…………………………………………...………………………………………..

1

Chapter 2 Mathematical Concepts in Kinematics

………………………………………..

2

Chapter 3 Fundamental Concepts in Kinematics

………………………………………..

8

Chapter 4 Kinematic Analysis of Planar Mechanisms ……………………………………

19

Chapter 5 Dimensional Synthesis

………………………………………………………...

81

Chapter 6 Static Force Analysis of Planar Mechanisms ………………………………….

159

Chapter 7 Dynamic Force Analysis of Planar Mechanisms ……………………………… 210 Chapter 8 Design & Kinematic Analysis of Gears ………………………………………..

288

Chapter 9 Design & Kinematic Analysis of Disk Cams ………………………………….

327

Chapter 10 Kinematic Analysis of Spatial Mechanisms …………………………………..

364

Chapter 11 Introduction to Robotic Manipulators …………………………………………

409

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PREFACE All of the computed solutions in this manual were produced using the 2013 version of MATLAB® (specifically version R2013b, win32). This is because the development of this textbook began in 2013 and the authors preferred to use a single version of MATLAB® throughout the development of this textbook. Any version of MATLAB® post 2013 is also suitable to run the MATLAB® and SimMechanics® files associated with this textbook (provided all of the required toolkits listed in Appendix A.1 are installed). As presented in Chapter 5 (in the textbook), there is an infinite number of solutions for a given dimensional synthesis problem. Because the solution values for a dimensional synthesis problem depend on the dyad angles specified, the solutions given here for Chapter 5 (which were calculated using arbitrary dyad displacement angles) are intended to serve as a guide to the solution calculation process and not as a benchmark for evaluating student solutions. We encourage and look forward to any feedback you may have. For e-mail correspondence, we can be reached at [email protected]. Thank you.

K. Russell Q. Shen R.S. Sodhi

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FOLFNKHUHWRGRZQORDG CHAPTER 2 Problem 2.1 Statement: Formulate an equation for the vector loop illustrated in Figure P.2.1. Consider that vector Vj always lies along the real axis.

Figure P.2.1 Vector loop (3 vectors where V j changes length) in 2-D complex space Problem 2.1 Solution: Taking the clockwise sum of the vector loop in Figure P.2.1 produces the equation

V1e i1  V2e i2  V j  0 . When expanded and separated into real and imaginary terms, the vector loop equation becomes V1 cos 1  V2 cos 2  V j  0 V1 sin 1  V2 sin 2  0

.

Problem 2.2 Statement: Formulate an equation for the vector loop illustrated in Figure P.2.2. Consider that vector Vj always lies along the real axis and vector V3 is always perpendicular to the real axis.

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Figure P.2.2 Vector loop (4 vectors where V j changes length) in 2-D complex space Problem 2.2 Solution: Taking the clockwise sum of the vector loop in Figure P.2.2 produces the equation

V1e i1  V2 e i 2  V3  V j  0 . When expanded and separated into real and imaginary terms, the vector loop equation becomes V1 cos 1  V2 cos 2  V j  0 V1 sin 1  V2 sin 2  V3  0

.

Problem 2.3 Statement: Calculate the first derivative of the vector loop equation solution from Problem 2.2. Consider only angles 1 , 2 and vector V j from Problem 2 to be time-dependent. Problem 2.3 Solution: Differentiating the vector loop equation solution from Problem 2.2 produces the equation

i

.

When expanded and separated into real and imaginary terms, the vector loop equation becomes

 .



0

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Problem 2.4 Statement:

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Calculate the second derivative of the vector loop equation solution from problem 2.2. Consider only angles 1 , 2 and vector V j from Problem 2 to be time-dependent. Problem 2.4 Solution: Differentiating the vector loop equation solution from Problem 2.3 produces the equation .



When expanded and separated into real and imaginary terms, the vector loop equation becomes



0



V

.

Problem 2.5 Statement: Formulate an equation for the vector loop illustrated in Figure P.2.3.

Figure P.2.3 Vector loop (4 vectors) in 2-D complex space Problem 2.5 Solution: Taking the clockwise sum of the vector loop in Figure P.2.3 produces the equation

V1e i1 V2e i2 V3 ei3 V0 ei0  0 . When expanded and separated into real and imaginary terms, the vector loop equation becomes

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V1 cos 1FOLFNKHUHWRGRZQORDG  V2 cos  2  V3 cos 3  V0 cos  0  0 V1 sin 1  V2 sin 2  V3 sin 3 V0 sin 0  0

.

Problem 2.6 Statement: Calculate the first derivative of the vector loop equation solution from Problem 2.5. Consider only angles 1 , 2 and  3 from Problem 5 to be time-dependent. Problem 2.6 Solution: Differentiating the vector loop equation solution from Problem 2.5 produces the equation

i

.

When expanded and separated into real and imaginary terms, the vector loop equation becomes

 .

 Problem 2.7 Statement: Calculate the second derivative of the vector loop equation solution from Problem 2.5. Consider only angles 1 , 2 and  3 from Problem 5 to be time-dependent. Problem 2.7 Solution: Differentiating the vector loop equation solution from Problem 2.6 produces the equation .



When expanded and separated into real and imaginary terms, the vector loop equation becomes



0



0

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Problem 2.8 Statement:

Formulate an equation for the vector loop illustrated in Figure P.2.4.

Figure P.2.4 Vector loop (5 vectors) in 2-D complex space Problem 2.8 Solution: Taking the clockwise sum of the vector loop in Figure P.2.4 produces the equation

V1e1  V2 e2  V3 e3  V4 e4  V0 e0  0 . When expanded and separated into real and imaginary terms, the vector loop equation becomes V1 cos 1  V 2 cos 2 V3 cos 3 V 4 cos  4  V 0cos  0  0 V1 sin 1  V 2 sin  2  V3 sin  3  V 4 cos  4  V 0 sin  0  0

.

Problem 2.9 Statement: Calculate the first derivative of the vector loop equation solution from Problem 2.8. Consider only angles 1 , 2 , 3 and  4 from Problem 8 to be time-dependent. Problem 2.9 Solution: Differentiating the vector loop equation solution from Problem 2.8 produces the equation .

i

When expanded and separated into real and imaginary terms, the vector loop equation becomes

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

FOLFNKHUHWRGRZQORDG .

 Problem 2.10 Statement: Calculate the second derivative of the vector loop equation solution from Problem 2.8. Consider only angles 1 , 2 , 3 and  4 from Problem 8 to be time-dependent. Problem 2.10 Solution: Differentiating the vector loop equation solution from Problem 2.9 produces the equation



.

When expanded and separated into real and imaginary terms, the vector loop equation becomes

 

4  0

.

 

4  0

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FOLFNKHUHWRGRZQORDG CHAPTER 3 Problem 3.1 Statement: Planar four-bar linkages have many everyday applications (some are illustrated Figure 3.1). Identify and describe four additional everyday applications for the planar four-bar linkage. Problem 3.1 Solution: As illustrated in Solution 3.1 Figure 1, the (a) elliptical trainer, (b) fan-style exercise bicycle, (c) model car steering linkage and (d) excavator bucket all include the planar four-bar linkage.

Solution 3.1 Figure 1

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Problem 3.2 Statement:

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a) Why is it important to know if a mechanism has a single degree of freedom? b) Why is a crank-rocker mechanism more useful than a double-rocker mechanism? c) Should the transmission angle for the planar four-bar linkage be close to 0? Explain? Problem 3.2 Solution: a) Because the DOF of a mechanism is the number of mechanism parameters required to fully define its motion, a mechanism having a single DOF requires a single mechanism motion parameter to fully define mechanism motion. As a result, a single DOF condition provides maximum mechanism motion control. b) Because the driving joint in a crank-rocker mechanism is grounded, it can be directly affixed to a grounded drive system (to compel mechanism motion). The driving joint in the doublerocker mechanism however is not grounded and cannot be directly affixed to a grounded drive system. When considering a grounded drive system, the crank-rocker mechanism is a more practical design option than the double-rocker mechanism. c) Revolute joints by design are better suited to handle transverse loads (loads perpendicular to a link) than loads acting along the link length. For the follower link, the proportions these two follower load components are determined by the transmission angle. It can be determined from the equations in Figure 2.8b that the optimum transmission angle is 90 because it produces a zero force along the follower length.

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Problem 3.3 Statement:

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For the two linkages illustrated in Figure P.3.1, which (if any) of the links can undergo a complete rotation relative to the other links? How do you know?

Figure P.3.1 Planar four-bar linkages with dimensionless link lengths Problem 3.3 Solution: For the first mechanism, the Grashof condition S  L  P  Q holds true. This makes the first mechanism a Change-Point mechanism and capable of full crank rotation. For the second mechanism, the Grashof condition S  L  P  Q holds true and the shortest link is ground. This makes the second mechanism a Double-Crank (or Drag-Link) mechanism and capable of full crank rotation.

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Problem 3.4 Statement:

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Determine the number of links, joints and the mobility of each of the three planar mechanisms in Figure P.3.2.

Figure P.3.2 Planar mechanisms Problem 3.4 Solution: Mechanism #1 includes 5 links ( L  5 ), 5 1-DOF joints ( J1  5) and 1 2-DOF joint ( J 2  1 ). Using Equation (3.2) mechanism #1 has a mobility of 1. Mechanism #2 includes 6 links ( L  6 ) and 7 1-DOF joints ( J1  7 ). Using Equation (3.2) mechanism #2 has a mobility of 1. Mechanism #3 includes 11 links (L  11 ) and 14 1-DOF joints ( J1  14 ). Using Equation (3.2) mechanism #3 has a mobility of 2.

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