SolidWorks Motion Tutorials

PR Do E no RE t c LE op AS y E or D di RA st F rib T ut e ® SolidWorks

2011

SolidWorks Motion

Dassault Systèmes SolidWorks Corporation 300 Baker Avenue Concord, Massachusetts 01742 USA

COMMERCIAL COMPUTER SOFTWARE PROPRIETARY U.S. Government Restricted Rights. Use, duplication, or disclosure by the government is subject to restrictions as set forth in FAR 52.227-19 (Commercial Computer Software Restricted Rights), DFARS 227.7202 (Commercial Computer Software and Commercial Computer Software Documentation), and in the license agreement, as applicable. Contractor/Manufacturer: Dassault Systèmes SolidWorks Corporation, 300 Baker Avenue, Concord, Massachusetts 01742 USA

PR Do E no RE t c LE op AS y E or D di RA st F rib T ut e

© 1995-2010, Dassault Systèmes SolidWorks Corporation, a Dassault Systèmes S.A. company, 300 Baker Avenue, Concord, Mass. 01742 USA. All Rights Reserved. The information and the software discussed in this document are subject to change without notice and are not commitments by Dassault Systèmes SolidWorks Corporation (DS SolidWorks). No material may be reproduced or transmitted in any form or by any means, electronically or manually, for any purpose without the express written permission of DS SolidWorks. The software discussed in this document is furnished under a license and may be used or copied only in accordance with the terms of the license. All warranties given by DS SolidWorks as to the software and documentation are set forth in the license agreement, and nothing stated in, or implied by, this document or its contents shall be considered or deemed a modification or amendment of any terms, including warranties, in the license agreement. Patent Notices SolidWorks® 3D mechanical CAD software is protected by U.S. Patents 5,815,154; 6,219,049; 6,219,055; 6,611,725; 6,844,877; 6,898,560; 6,906,712; 7,079,990; 7,477,262; 7,558,705; 7,571,079; 7,590,497; 7,643,027; 7,672,822; 7,688,318; 7,694,238; and foreign patents, (e.g., EP 1,116,190 and JP 3,517,643). eDrawings® software is protected by U.S. Patent 7,184,044; U.S. Patent 7,502,027; and Canadian Patent 2,318,706. U.S. and foreign patents pending.

Trademarks and Product Names for SolidWorks Products and Services SolidWorks, 3D PartStream.NET, 3D ContentCentral, eDrawings, and the eDrawings logo are registered trademarks and FeatureManager is a jointly owned registered trademark of DS SolidWorks. CircuitWorks, Feature Palette, FloXpress, PhotoWorks, TolAnalyst, and XchangeWorks are trademarks of DS SolidWorks. FeatureWorks is a registered trademark of Geometric Ltd. SolidWorks 2011, SolidWorks Enterprise PDM, SolidWorks Simulation, SolidWorks Flow Simulation, and eDrawings Professional are product names of DS SolidWorks. Other brand or product names are trademarks or registered trademarks of their respective holders.

Copyright Notices for SolidWorks Standard, Premium, Professional, and Education Products Portions of this software © 1986-2010 Siemens Product Lifecycle Management Software Inc. All rights reserved. Portions of this software © 1986-2010 Siemens Industry Software Limited. All rights reserved. Portions of this software © 1998-2010 Geometric Ltd. Portions of this software © 1996-2010 Microsoft Corporation. All rights reserved. Portions of this software incorporate PhysX™ by NVIDIA 2006-2010. Portions of this software © 2001 - 2010 Luxology, Inc. All rights reserved, Patents Pending. Portions of this software © 2007 - 2010 DriveWorks Ltd. Copyright 1984-2010 Adobe Systems Inc. and its licensors. All rights reserved. Protected by U.S. Patents 5,929,866; 5,943,063; 6,289,364; 6,563,502; 6,639,593; 6,754,382; Patents Pending. Adobe, the Adobe logo, Acrobat, the Adobe PDF logo, Distiller and Reader are registered trademarks or trademarks of Adobe Systems Inc. in the U.S. and other countries. For more copyright information, in SolidWorks see Help > About SolidWorks. Copyright Notices for SolidWorks Simulation Products Portions of this software © 2008 Solversoft Corporation. PCGLSS © 1992-2007 Computational Applications and System Integration, Inc. All rights reserved.

Copyright Notices for Enterprise PDM Product Outside In® Viewer Technology, © Copyright 1992-2010, Oracle © Copyright 1995-2010, Oracle. All rights reserved. Portions of this software © 1996-2010 Microsoft Corporation. All rights reserved.

Copyright Notices for eDrawings Products Portions of this software © 2000-2010 Tech Soft 3D. Portions of this software © 1995-1998 Jean-Loup Gailly and Mark Adler. Portions of this software © 1998-2001 3Dconnexion. Portions of this software © 1998-2010 Open Design Alliance. All rights reserved. Portions of this software © 1995-2009 Spatial Corporation. This software is based in part on the work of the Independent JPEG Group.

Document Number: PMT1142-ENG_DRAFT

PR Do E no RE t c LE op AS y E or D di RA st F rib T ut e Contents

Introduction:

About This Course . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Prerequisites . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Course Design Philosophy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Using this Book . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Laboratory Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Training Files . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Windows® 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Conventions Used in this Book . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Use of Color . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 What is SolidWorks Motion? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 What is Motion Simulation? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Understanding Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Mass and Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Degrees-of- Freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Constraining Degrees-of- Freedom . . . . . . . . . . . . . . . . . . . . . . . . 6 Motion analysis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 How is motion analyzed on the computer?. . . . . . . . . . . . . . . . . . . 6 Basics of Mechanism Setup in SolidWorks Motion . . . . . . . . . . . . . . . 7 Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Fixed Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 Floating Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Mates. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Motors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Constraint Mapping Concept . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 i

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Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 Lesson 1: Introduction to Motion Simulation and Forces

Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Basic Motion Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Case Study: Car Jack Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Problem Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Stages in the Process. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Driving Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Understanding Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 Applied Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Force Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Force Direction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Case 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Case 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Case 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Results. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Plot Categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Sub-Categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Resizing Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 Exercise 1: 3D Fourbar Linkage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

Lesson 2: Building a Motion Model and Post-processing Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 Creating Local Mates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 Case Study: Crank Slider Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 Problem Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 Stages in the Process. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 Mates. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 Concentric Mate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 Hinge Mate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 Point-to-Point Coincident Mate . . . . . . . . . . . . . . . . . . . . . . . . . . 37 Lock Mate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 Two Face-to-Face Coincident Mates . . . . . . . . . . . . . . . . . . . . . . 37 Universal Mate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 Screw Mate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 Point-on-Axis Coincident Mate . . . . . . . . . . . . . . . . . . . . . . . . . . 39 Parallel Mate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

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Perpendicular Mate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 Local Mates. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 Function Builder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 Importing Data Points. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 Alternative Units. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 Plotting Kinematic Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 Absolute vs. Relative values. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 Output coordinate system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 Angular Displacement Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 Angular Velocity and Acceleration Plots . . . . . . . . . . . . . . . . . . . 62 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 Exercise 2: Piston . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 Exercise 3: Trace Path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

Lesson 3: Introduction to Contacts, Springs and Dampers Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 Contact and Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 Case Study: Catapult. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 Problem Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 Stages in the Process. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77 Interference Detection. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 Contact . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 Contact groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82 Contact Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 Translational Spring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 Magnitude of Spring Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 Translational Damper . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 Post-processing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 Analysis with Friction (Optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 Exercise 4: The Bug. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 Exercise 5: Door Closer. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 Lesson 4: Advanced Contact Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99 Contact Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 Case Study: Latching Assembly . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 Problem Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 Fixing Motion with Motors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 Motor Input and Force Input Types . . . . . . . . . . . . . . . . . . . . . . 103 Functional Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

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Force Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 STEP Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 Contact: Solid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 Poisson Model (Restitution Coefficient) . . . . . . . . . . . . . . . . . . 110 Impact Force Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 Closing Remarks. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 Geometrical Description of Contacts . . . . . . . . . . . . . . . . . . . . . . . . 115 Instability Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 Modifying Result Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 Closing Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 Precise Contact . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 Integrators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 GSTIFF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 WSTIFF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 SI2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 Discussion: References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 Exercise 6: Hatchback . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 Exercise 7: Conveyor Belt (No Friction). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 Path Mate Motor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 Exercise 8: Conveyor Belt (With Friction) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

Lesson 5: Curve to Curve Contact Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 Contact Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 Case Study: Geneva Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 Problem Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 Curve to Curve Contact . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 Solid bodies vs. curve to curve contact. . . . . . . . . . . . . . . . . . . . . . . 159 Solid Bodies Contact Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 Exercise 9: Conveyor Belt (Curve to curve contact with friction) . . . . . . . . . . . 161 Lesson 6: CAM Synthesis Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 CAMs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 Case Study: CAM Synthesis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 Problem Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 Stages in the Process. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 Generating a CAM Profile . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 Trace Path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169 Exporting Trace Path Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170

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Cycle based motion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173 Exercise 10: Desmodromic CAM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 Exercise 11: Rocker CAM Profile. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

Lesson 7: Flexible Joints

Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193 Flexible Joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 Case Study: System with Rigid Joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 Problem Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 Stages in the Process. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 Calculation of Wheel Input Motion . . . . . . . . . . . . . . . . . . . . . . 198 Understanding Toe Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 System with Flexible Joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

Lesson 8: Redundancies

Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209 Redundancies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210 What are redundancies? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213 Effects of Redundancies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214 How are redundancies removed in the solver? . . . . . . . . . . . . . . 215 Case Study: Door Hinges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 Problem Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 Degrees of Freedom Calculation . . . . . . . . . . . . . . . . . . . . . . . . 219 Total Actual and Estimated DOF . . . . . . . . . . . . . . . . . . . . . . . . 219 Using Flexible Joints Option to Remove Redundancies . . . . . . 222 Limitations of Flexible Mates. . . . . . . . . . . . . . . . . . . . . . . . . . . 223 Bushing Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224 How to Check For Redundancies . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 Typical Redundant Mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226 Dual Actuators Driving a Part . . . . . . . . . . . . . . . . . . . . . . . . . . 226 Parallel Linkages. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227 Exercise 12: Dynamic Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229 Exercise 13: Dynamic Systems 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231 Exercise 14: Kinematic Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233 Exercise 15: Zero Redundancy Model-Part 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238

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Exercise 16: Zero Redundancy Model-Part 2 (Optional) . . . . . . . . . . . . . . . . . . . 243 Exercise 17: Removing Redundancies with Bushings . . . . . . . . . . . . . . . . . . . . . 244 Exercise 18: Catapult. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251

Lesson 9: Export to FEA

Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257 Exporting Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258 Case Study: Drive Shaft . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258 Project Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258 Stages in the Process. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 FEA Export . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262 Load Bearing Faces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 Mate location . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 Export of Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264 SolidWorks Simulation Users Only . . . . . . . . . . . . . . . . . . . . . . 266 Direct Solution in SolidWorks Motion . . . . . . . . . . . . . . . . . . . . . . . 274 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278 Exercise 19: Export to FEA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279

Lesson 10: Event Based Simulation Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289 Event Based Simulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290 Case Study: Sorting Device . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290 Problem Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290 Servo motors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290 Sensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291 Task . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298 Lesson 11: Design Project (Optional) Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299 Design Project. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300 Case Study: Surgical Shear - Part 1 . . . . . . . . . . . . . . . . . . . . . . . . . 300 Problem Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300 Force to Cut the Catheter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301 Self Guided Problem - Part 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303 Stages in the Process. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303 Self Guided Problem - Part 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304 Stages in the Process. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304 Problem Solution - Part 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305 Creating the Force Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308 Force to Cut the Catheter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308

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Creating the Force Expression . . . . . . . . . . . . . . . . . . . . . . . . . . 311 Force Expression. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313 IF Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313 Developing the Expression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313 Case Study: Surgical Shear - Part 2 . . . . . . . . . . . . . . . . . . . . . . . . . 320 Stages in the Process. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330

Appendix A: Motion Study Convergence Solutions and Advanced Options Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334 Accuracy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335 Integrator Type . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336 GSTIFF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336 WSTIFF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337 Stabilized Index Two (SI2). . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337 Integrator Settings. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337 Maximum Iterations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337 Initial Integrator Step Size . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337 Minimum Integrator Step Size . . . . . . . . . . . . . . . . . . . . . . . . . . 337 Maximum Integrator Step Size . . . . . . . . . . . . . . . . . . . . . . . . . . 338 Jacobian Re-evaluation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339 Appendix B: Mate Friction Mate Friction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342 Concentric (Spherical) Mate Friction Model . . . . . . . . . . . . . . . 343 Coincident Translational Mate Friction Model . . . . . . . . . . . . . 344 Concentric Mate Friction Model. . . . . . . . . . . . . . . . . . . . . . . . . 344 Coincident Mate (Planar) Friction Model. . . . . . . . . . . . . . . . . . 344 Universal Joint Friction Model . . . . . . . . . . . . . . . . . . . . . . . . . . 345 Friction Results Reported . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345

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PR Do E no RE t c LE op AS y E or D di RA st F rib T ut e Contents

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SolidWorks 2011

PR Do E no RE t c LE op AS y E or D di RA st F rib T ut e Introduction

1

Introduction

The goal of this course is to teach you the basics of how to use the SolidWorks Motion simulation software to help you analyze the kinematic or dynamic behavior of your SolidWorks assembly model.


About This Course

SolidWorks 2011

The focus of this course is on the fundamental skills and concepts central to the successful use of SolidWorks Motion 2011. You should view the training course manual as a supplement to, and not a replacement for, the system documentation and on-line help. Once you have developed a good foundation in basic skills, you can refer to the on-line help for information on less frequently used command options.

Prerequisites

Students attending this course are expected to have the following:

I I I

Mechanical design experience. Experience with the Windows™ operating system. Completed the on-line SolidWorks tutorials that are available under Help. You can access the on-line tutorials by clicking Help, Online Tutorial.

Course Design Philosophy

This course is designed around a process- or task-based approach to training. Rather than focusing on individual features and functions, a process-based training course emphasizes processes and procedures you should follow to complete a particular task. By utilizing case studies to illustrate these processes, you learn the necessary commands, options and menus in the context of completing a design task.

Recommended Length

The recommended minimum length of this course is two days.

Using this Book

This training manual is intended to be used in a classroom environment under the guidance of an experienced SolidWorks Motion instructor. It is not intended to be a self-paced tutorial. The examples and case studies are designed to be demonstrated “live” by the instructor.

Please note, there may be slight differences in results for certain lessons due to service pack upgrades, etc.

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SolidWorks 2011

Laboratory exercises give you the opportunity to apply, practice and expand the material covered during the lecture/demonstration portion of the course. They are designed to represent typical design, and analysis situations while being modest enough to be completed during class time. You should note that many students work at different paces. Therefore, we have included more lab exercises than you can reasonably expect to complete during the course. This ensures that even the fastest student will not run out of exercise.


Laboratory Exercises

Introduction

Training Files

A complete set of the various files used throughout this course can be downloaded from the SolidWorks website, www.solidworks.com. Click on the link for Support, then Training, then Training Files, then SolidWorks Simulation Training Files. Select the link for the desired file set. There may be more than one version of each file set available. Direct URL:

www.solidworks.com/trainingfilessimulation

The files are supplied in signed, self-extracting executable packages.

The files are organized by lesson number. The Case Studies folder within each lesson contains the files your instructor uses while presenting the lessons. The Exercises folder contains any files that are required for doing the laboratory exercises.

Feature Names

Throughout this course, feature names may be different from those you obtain when doing the case studies and exercises. SolidWorks and SolidWorks Motion name features sequentially, (Hinge1, Hinge2, etc.) so if you apply mates in a different order, or you delete and then recreate a mate, your names will be different. In most cases, mates are referred to by their type (lock, hinge, coincident, etc.) and the components that are mated (link, support, etc.). In addition, images are also included to help avoid ambiguity, but you must always check the instructions carefully to make sure you are selecting the correct feature.

3

Introduction

The screen shots in this manual were made using SolidWorks 2010 and SolidWorks Motion 2010 running on Windows® 7. If you are running on a different version of Windows, you may notice differences in the appearance of the menus and windows. These differences do not affect the performance of the software.


Windows® 7

SolidWorks 2011

Conventions Used in this Book

This manual uses the following typographic conventions: Convention

Bold Sans Serif

SolidWorks Motion commands and options appear in this style. For example, Motion, Delete Results means choose the Delete Results option from the Motion menu.

Typewriter

Feature names and file names appear in this style. For example, Concentric.

17 Do this step

Use of Color

4

Meaning

Double lines precede and follow sections of the procedures. This provides separation between the steps of the procedure and large blocks of explanatory text. The steps themselves are numbered in sans serif bold.

The SolidWorks and SolidWorks Motion user interface make extensive use of color to highlight selected geometry and to provide you with visual feedback. This greatly increases the intuitiveness and ease of use of the SolidWorks Motion software. To take maximum advantage of this, the training manuals are printed in full color.

SolidWorks 2011

Introduction

SolidWorks Motion is a virtual prototyping tool for engineers and designers interested in understanding the performance of their assemblies. Powered by ADAMS® technology, the industry leader for over 25 years, SolidWorks Motion helps you ensure that your designs will work and perform as expected prior to building them. By learning how to effectively utilize the features of the user interface, you will have the key to unlocking a solution to the most complex mechanisms.

What is Motion Simulation?

A mechanism is a mechanical device that has the purpose of transferring motion and/or force from a source to an output. Motion simulation is simply the study of such moving systems or mechanisms. The motion of any system is determined by the following:


What is SolidWorks Motion?

I I I I I

Mates connecting the parts The mass and inertia properties of the components Applied forces to the system Driving motions (Motors or Actuators) Time

Understanding Basics Mass and Inertia

The principle of inertia is one of the fundamental laws of classical physics which are used to describe the motion of matter and how it is affected by applied forces. Today, the concept of inertia is most commonly defined using Isaac Newton's First Law of Motion, which states: Every body perseveres in its state of being at rest or of moving uniformly straight ahead, except insofar as it is compelled to change its state by forces impressed.

Mass and inertia play a very important role in the simulation of dynamic systems and are also important in kinematics. Realistic values of mass and inertia are needed in nearly all simulations.

Degrees-ofFreedom

An unconstrained rigid body in space has six degrees-of-freedom: three translational and three rotational. It can translate along its X, Y, and Z axes and rotate about its X, Y, and Z axes as shown in the figure to the right.

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Introduction

SolidWorks 2011

Constraints are the restrictions placed on a part’s movement in specific degrees-offreedom. Mates are connections that restrict the movement of one part with respect to another.

Motion analysis

The two equations governing three dimensional motion of a rigid body are known as Euler’s equations.


Constraining Degrees-ofFreedom

The first equation is Newton’s second law of motion which states that the sum of externally applied forces on a body is equal to the rate of dP change of linear momentum P, ΣF = -------- . dt

For bodies where mass does not change, the right hand side of the equation simplifies to more commonly known mass times acceleration, ΣF = ma . The second equation is based on the sum of the moments about the center of mass of a rigid body due to external forces, and couples should equal the rate of change of angular momentum H of the body. dH ΣM = -------dt

How is motion analyzed on the computer?

Τhe program uses the modified Newton Raphson iteration method in each time step.

By taking very small time steps, the software can predict the position of parts at the next time step based on initial conditions or the previous time step. The solution must satisfy: I I I

6

Velocity of parts Mates connecting parts Forces and accelerations

SolidWorks 2011

Introduction


The answer is iterated until certain accuracy is reached for that time step for force and acceleration values.

Basics of Mechanism Setup in SolidWorks Motion

The following paragraphs outline how SolidWorks Motion treats parts and sub-assemblies, and how the mates directly define the motion of the mechanism when loaded by external forces (such as gravity or isolated forces) or prescribed motions (motors).

Rigid Body

In SolidWorks Motion, all parts are treated as infinitely rigid. This means that there is no internal deflection within a part and the parts do not deform or change shape during the simulation. A rigid body can be a single SolidWorks part or a sub-assembly. There are two states of the sub-assembly in SolidWorks: Rigid or Flexible. A rigid sub-assembly means that the individual components that make up the sub-assembly are assumed to be rigidly attached (welded) to each other as if they were one single part.

If a sub-assembly status is set to flexible in SolidWorks, it does not mean that the sub-assembly parts become flexible. This setting means that the root level parts of the sub-assembly are treated independently of each other by SolidWorks Motion. The constraints (SolidWorks mates at the sub-assembly level) between these parts are automatically mapped into SolidWorks Motion.

Fixed Parts

A rigid body can be treated as a fixed part or a floating (moving) part. Fixed parts are, by definition, at absolute rest. Each fixed rigid body has zero degrees-of-freedom. A fixed part serves as the reference frame for the remaining rigid bodies that are moving.

In SolidWorks, any component that is fixed in your assembly is automatically treated as a fixed part when you begin a new mechanism and map the assembly constraints.

7

Introduction

Components that move in the mechanism are considered moving parts. Each moving part has six degrees-of-freedom.


Floating Parts

SolidWorks 2011

In SolidWorks, any component that is floating in your assembly is automatically treated as moving when you begin a new mechanism and map the assembly constraints.

Mates

SolidWorks mates fully define how rigid bodies are attached and how they move relative to each other. Mates remove degrees-of-freedom from the parts to which they are attached.

When you add a mate, such as a concentric mate, between two rigid bodies, you remove degrees-of-freedom, causing them to remain positioned with respect to each other regardless of any motion or force in the mechanism.

Motors

Motors can be defined for part to control its movement over a period of time. A motor dictates the displacement, velocity, or acceleration of a part as a function of time.

Gravity

Gravity is an important quantity when the weight of a part has an influence on its simulated motion, such as a body in free fall. In SolidWorks Motion, gravity consists of two components:

I I

Direction of the gravitational vector Magnitude of gravitational acceleration

The Gravity Properties dialog allows you to specify the direction and magnitude of the gravitational vector. You can specify the gravitational vector by entering the x, y and z values in the appropriate text box, or by specifying a reference plane. The magnitude must be entered separately. The default value for the gravitational vector is (0, -1, 0), and the magnitude is 9.81 m/s2 (or the equivalent in the currently active units).

Constraint Mapping Concept

One of the reasons SolidWorks Motion is such a time saving tool is that it automatically maps the SolidWorks assembly mates (constraints) to SolidWorks Motion. There are more than 100 ways to mate or constrain parts in SolidWorks.

Forces

When defining various Force objects in SolidWorks Motion, a location and/or direction has to be specified. These directions and locations are derived from selected SolidWorks entities. The entities can be sketch points, vertices, edges or surfaces.

Summary

This short review of motion simulation using SolidWorks Motion is not, of course, all inclusive. It is only intended to get us started with the hands-on lessons. As we progress through the lessons presented in the following chapters, we will occasionally digress from software-specific issues in order to discuss relevant motion simulation fundamentals.

8

PR Do E no RE t c LE op AS y E or D di RA st F rib T ut e Lesson 1 Introduction to Motion Simulation and Forces

Objectives

Upon successful completion of this lesson, you will be able to: I

Use Assembly Motion to animate the motion of a car jack assembly.

I

Use SolidWorks Motion to simulate physical behavior of the car jack and determine the torque required to lift a vehicle.

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Lesson 1

SolidWorks 2011

Introduction to Motion Simulation and Forces

In this lesson, we will perform a basic motion analysis using SolidWorks Motion to simulate the weight of a vehicle on the jack and determine the torque required to lift it. Engineers can then use this information to choose the required electric motor to drive the car jack.

Case Study: Car Jack Analysis

A mechanical jack is a device that lifts heavy equipment. The most common form is a car jack, floor jack, or garage jack which lifts vehicles so that maintenance can be performed. Car jacks usually use mechanical advantage to allow a human to lift a vehicle. More powerful jacks use hydraulic power to provide more lift over greater distances. Mechanical jacks are usually rated for a maximum lifting capacity (e.g., 1.5 tons or 3 tons).


Basic Motion Analysis

Because this is our first motion analysis, no contact is used and the tilting motion of the jack is prevented with the help of the mates.

Problem Description

The car jack will be driven at a rate of 100 RPM and will be loaded with a force of 8,900 N., representing the weight of a vehicle. Determine the torque and power required to lift the load through the range of motion of the jack.

Stages in the Process

I

Create a Motion Study.

This will be a new motion study.

I

Add a rotary motor.

The rotary motor will drive the jack.

I

Add gravity.

Normal gravity will be added so that the weight of the car jack components are considered in the calculations.

I

Add the weight of the car.

The weight of the car will be added as a downward force on the Support.

I

10

Calculate the motion.

SolidWorks 2011

Lesson 1 Introduction to Motion Simulation and Forces


The default analysis will run for five seconds but we will increase it to allow the jack to extend fully. I

Plot the results.

We will create various plots to show the torque and power required.

1

Ensure that SolidWorks Motion is added in. Under Tools, Add-ins, make sure SolidWorks Motion is checked.

2

Open an assembly file. Open Car_Jack from the Lesson01\Case Studies folder.

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Lesson 1

SolidWorks 2011


3

Set the document units.


SolidWorks Motion uses the document units set in the SolidWorks document. Click Tools, Options, Document Properties, Units.

Select MMGS (millimeter, gram, second) for the Unit system. This will set our length units to millimeters and force to Newtons.

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SolidWorks 2011


Change to the Motion Study. Click on the Motion Study 1 tab that appears at the bottom left-hand corner of the window. If this tab is not visible, select Motion Manager on the View menu.


4

13

Lesson 1

SolidWorks 2011


Motion can be driven by gravity, springs, forces or motors. Each has different characteristics that can be controlled.


Driving Motion Introducing: Motors

Motors can create either linear, rotary or path dependent motion or to prevent motion. This motion can be defined in a number of different ways. I

Constant Speed

The motor will drive at a constant velocity.

I

Distance

The motor will move for a fixed distance or degrees.

I

Oscillating

Oscillating motion is a back and forth motion at a specific distance at a specified frequency.

I

Segments

Motion profile is constructed from segments of the most commonly used functions such as linear, polynomial, half-sine and others.

I

Data Points

Interpolated motion is driven by a tabular set of values.

I

Expression

The motor can be driven by a function created from existing variables and constants.

I

Servo Motor

The motor used to implement control actions for the event-based triggered motion.

Where to Find It

14

I

On the MotionManager toolbar, click Motor

.

SolidWorks 2011


Create a Motor that drives the Screw_rod at 100 RPM. Click Motor on the Motion Manager toolbar.


5

Under Motor Type, select Rotary Motor.

Under Component Direction, select the cylindrical face of the Screw_rod part as shown in the figure. The Motion Direction field will automatically populate the same face to specify the direction.

Use the Reverse Direction button to orient the motor (see the figure). Leave the Component to move relative to field empty. This ensures that the motor direction is specified with respect to the global coordinate system. Under Motion, select the Constant speed and enter a value of 100 RPM. Click OK.

Important!

Make sure that the motor is oriented as shown in the figure.

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Lesson 1

SolidWorks 2011



Click the graph in the PropertyManager to view the enlarged plot.

Close the graph plot and click OK to close the Motor PropertyManager.

6

Type of Study.

Make sure that the Motion Type of Study field shows Motion Analysis.

Gravity

Gravity is an important quantity when the weight of a part has an influence on its simulated motion, such as a body in free fall. In SolidWorks Motion, gravity consists of two components: I I

Direction of the gravitational vector Magnitude of the gravitational acceleration

The Gravity Properties allows you to specify the direction and magnitude of the gravitational vector. You can specify the gravitational vector by selecting the X, Y and Z direction or by specifying a reference plane. The magnitude must be entered separately. The default value for the gravitational vector is Y and the magnitude is 9806.55 mm/sec2 or the equivalent in the currently active units.

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SolidWorks 2011


Apply Gravity to the assembly. Click Gravity on the Motion Manager toolbar.


7

For Gravity Parameters, Direction Reference, select the Y direction.

Under Numeric gravity value, type in a value of 9806.65 mm/sec^2. Click OK.

Forces

Force entities (including both forces and moments) are used to effect the dynamic behavior of parts and sub assemblies of a motion model and are usually a representation of some external effect acting on the analyzed assembly. Forces may resist or induce motion, and are defined using similar functions that are used to define motors (constant, step, function, expression or interpolated).

Forces in SolidWorks Motion can be divided into two basic groups: I

Action Forces

A single applied force or moment representing the effect of the external objects and loadings on the part or subassembly. The weight of the vehicle applied on the car jack or an aerodynamic force on the car body are examples of action forces.

I

Action and Reaction Forces

A pair of forces or moments, both action and corresponding reaction, are applied on the parts or subassemblies.

A spring force could be understood as action and reaction force because both are acting on the same line of action and acting on the assembly at the spring mount points. Another example would be a person pushing with his/her arms on the two opposing parts of an assembly. Such a person can then be represented in the motion analysis by a pair of two opposing forces of equal magnitude on the same line of action, i.e. action and reaction forces.

Understanding Forces

A force can define load or compliance on a part. SolidWorks Motion provides the following type of forces:

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Applied forces are forces that define loads at specific locations on a part. You must provide you own description of the force behavior by specifying a constant force value or a function expression. The applied forces available in SolidWorks Motion are the applied force, applied torque, action/reaction force and action/reaction torque.


Applied Forces

The orientation of action-only forces can be fixed or at relative to the orientation of any part in the mechanism. Applied forces are used to model inputs such as actuators, rockets, aerodynamic loads and many more.

Force Definition

To define a force the following information must be specified:

I I I

Part or parts on which the forces act. Point of the force application. Magnitude and direction of the force.

On the MotionManager toolbar, click Force Only in the PropertyManager.

. Select Action-

Where to Find It

I

Force Direction

The force direction is based on the reference part you select in the Force Direction box. An illustration below gives you the three cases on how the force direction changes based on the selected reference parts.

Case 1

Direction of force is based on a fixed component.

If fixed component is the assembly origin then the initial orientation of the force will be held constant throughout the simulation. Reference Fixed Component

F1

18

F1

F1

SolidWorks 2011


Direction of force is based on the selected moving component, which is also the component on which you want to apply the force.


Case 2

If the part to which the force is applied is used as the reference datum, then the force will remain locked in its relative orientation to the body over the entire simulation time (i.e. it will stay in alignment with the geometry on the body used to define the direction). Reference Rotating Component

F1

F1

F1

Fixed Component

Case 3

Direction of force is based on the selected moving component which is different from the component on which you want to apply the force.

If another moving part is used as the reference datum, the direction of the force will change based on the relative orientation of the reference body to the moving body. This is hard to visualize easily, but if you apply the force on a body that is held locked in position, and use a rotating part as the reference datum, you should see the force rotates in concert with the reference body. Reference Rotating Component

F1

Note

F1

F1

Make sure that the gravity symbol shows the orientation in the negative Y direction.

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Create a force of 8900 N to simulate the weight of the car on the car jack. Click Force on the Motion Manager.


8

For Type, select Force.

Under Direction, select Action Only.

Under Action part and point of application of action, select the circular edge on component Support-1 (see image below). For Force Direction, select the vertical edge on the Base-1 component.

The default force direction is defined by the circular edge selected in the Action part and point of application of action field, i.e. perpendicular to the plane of the edge. Because the default direction is correct in this case, the edge selected in the Force Direction field is not required and is done solely for the educational purpose.

Note

Under Force Function, select Constant. Enter a force value of 8900 N.

Make sure that the force is directed downwards.

Note

Click OK to close the Force/Torque PropertyManager.

9

20

Run the Simulation. Click Calculate . The simulation will calculate for 5 seconds.

SolidWorks 2011


10 Run the Simulation for 8 seconds.


Drag the end time key to 8 seconds on the timeline and recalculate.

Results

The primary output from a motion study is a plot of one parameter versus another, usually time. Once the motion is calculated plots can be created for a variety of parameters. All existing plots will be listed at the bottom of the MotionManager tree.

Plot Categories

Plots of the following categories can be created: I I I I

Sub-Categories

Displacement Acceleration Momentum Power

I I I I

Displacement Forces Energy Other quantities

Within each of the categories, plots can be created for: I I I I I I I I I I I I I I

Trace Path Linear Displacement Linear Acceleration Angular Velocity Applied Force Reaction Force Friction Force Contact Force Angular Momentum Angular Kinetic Energy Potential Energy Delta Pitch Roll Bryant Angles

I I I I I I I I I I I I I I

XYZ Position Linear Velocity Angular Displacement Angular Acceleration Applied Torque Reaction Moment Friction Moment Translational Momentum Translational Kinetic Energy Total Kinetic Energy Power Consumption Yaw Rodriguez Parameters Projection Angles

Resizing Plots

Plots can be resized by dragging any border or corner.

Where to Find It

I

Click Results and Plots

on the MotionManager toolbar.

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11 Plot the torque required to lift the weight of the car. Click Results and Plots in the Motion manager.

Under Result, select the category as Forces. Under Sub-category, select Motor Torque.

Under Result component, select Magnitude.

Under Select rotational motor object to create result, select the motor that we created (see image below). Click OK.

The plot of torque required appears in the graphics area.

The required torque is about 7244 N-mm.

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SolidWorks 2011


Once the Rotary Motor1 is selected, a triad is displayed in the graphics area. This triad indicates the local X, Y and Z axes of the motor in which the output quantities may be displayed. In the present case we require the plot of the magnitude which is independent of the coordinate system. The post-processing is described in greater detail in the next lesson.


Note

12 Plot the power consumed to lift a weight of 8900 N. We will add this plot into an existing graph. Click Results and Plots in

the Motion Manager toolbar.

Under Result, select the category as Momentum/Energy/Power. Under Sub-category, select Power Consumption.

Under Select motor object to create result, select the same motor that you selected in the previous step.

Under Plot Results, select Add to existing plot and select Plot1 from the pull down menu. Click OK.

The power consumption is 76 Watts. Based on the torque and the power information, we can select an electric motor and use it to drive the Screw_rod instead of a human hand. You can click Play to see the animation. The vertical time bar in both the MotionManager and the graph indicates the time.

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13 Plot the vertical position of the Support. Click on the Results and Plots icon in the Motion Manager.

Under Result, select the category as Displacement/Velocity/ Acceleration. Under Sub-category, select Linear Displacement. For Result Component, select Y-component.

For Select two points/faces, select the top face of the support. If no second item is selected, the ground serves as the default second component, or the reference. Leave the Component to define XYZ directions field empty. This indicates that the displacement is reported in the default global coordinate system.

Note

The displacement is measured at the origin of the Support part file, indicated as the small blue sphere in the above figure, with respect to the origin of the Car_Jack assembly file. The result is reported in the default global coordinate system. Click OK.

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Lesson 1



The above graph indicates change of the global Y coordinate of the origin of the Support part file. The displacement is therefore 51mm (212-161mm) in the positive global Y axis.

14 Modify the graph.

Modify the ordinate of the graph to show the angular displacement of the motor.

In the MotionManager tree, expand the Results folder. Right-click Plot2 and click Edit Feature. Under Plot Results, Plot Results verus: select New Result.

For Define new result, select Displacement/ Velocity/Acceleration.

Select Angular Displacement under sub-category.

Select Magnitude for result component.

Select RotaryMotor1 for the simulation element. Click OK.

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15 Examine the graph.


The result plot is a little coarse and the data ordinate does not cover the full range of -180 to 180 degrees. To generate a graph with finer detail, more data must be saved to disk.

Introducing: Study Properties

SolidWorks Motion has its own set of properties to control the way the study is calculated and displayed. Study properties will be discussed throughout the book. Click Motion Study Properties


Where to Find It

I

Introducing: Frames per Second

Frames per second controls how often the data is saved on the disk. The higher the frames per second, the more dense the data recorded.

Where to Find It

I

In the Motion Study Properties, expand Motion Analysis and either type the number, use the spinbox arrows or adjust the slider.

16 Modify Motion Study properties. Click Motion Study Properties in the

MotionManager toolbar.

Change the Frames per second to 100.

Click OK.

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SolidWorks 2011


17 Calculate the study.


Notice that we have more detail and the angular displacement is nearly from -180 to 180 degrees.

18 Save and close the file.

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SolidWorks 2011

Exercise 1 3D Fourbar Linkage

Exercise 1: 3D Fourbar Linkage


This assembly is a simple mechanism called 3D Fourbar linkage. There are only four parts in the mechanism. The Support part is grounded, and the rotation of the Lever part will cause a sliding motion of the SliderBlock part.

LeverArm

linkage

Support

SliderBlock

This exercise reinforces the following skills: I I

Project Description

Basic Motion Analysis on page 10. Results on page 21.

The LeverArm will be simply rotated with a constant 360 deg/sec angular velocity. Determine the amount of torque required to drive this mechanism and plot it from the motion simulation.

1

Open an assembly file.

Open 3D Fourbar linkage from the Lesson01\ Exercises folder.

2

Verify fixed and moving components.

Make sure that support is fixed while the other components can move.

3

Motion study.

In the MotionManager, select Motion Analysis.

The default Motion Study 1 will be used for the analysis.

4

Add gravity.

Apply gravity in the negative Z direction.

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3D Fourbar Linkage

Define motion of the Lever Arm. Define a Rotary Motor at 360 deg/sec.


5

You can enter 360 deg/sec directly into the PropertyManager and it will automatically be converted to RPM.

Tip

6

Motion study properties.

Set the Frames per second to 100 and drag the time key to 4 seconds.

7 8

Calculate the simulation.

Determine the torque and power required to drive the mechanism.

Define a graph showing the moment torque and the required power as a function of time. Define both quantities in a single graph.

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SolidWorks 2011

Exercise 1 3D Fourbar Linkage

9

Linear velocity of the SliderBlock.


Plot a graph showing the linear velocity of the SliderBlock as a function of time.

10 Modify the graph.

Modify the ordinate of the graph to show the angular displacement of the Rotary Motor. This way the graph will show the variation of the SliderBlock velocity relative to the angular displacement of the LeverArm.


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3D Fourbar Linkage

32

PR Do E no RE t c LE op AS y E or D di RA st F rib T ut e Lesson 2 Building a Motion Model and Post-processing

Objectives


Build proper SolidWorks Motion models for kinematic simulation.

I

Create local mates for a SolidWorks Motion study.

I

Create and modify plots for post-processing.

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Building a Motion Model and Post-processing

In the previous lesson, the mates created in SolidWorks were used as joints in SolidWorks Motion. If the components are not mated in SolidWorks, or if we wish to examine different connection types in SolidWorks Motion, mates can be added or modified in the Motion Analysis.

Case Study: Crank Slider Analysis

In this lesson, we will setup the mechanism for the crank slider model. We will use SolidWorks mates that most closely represent the real mechanical connections. The crank slider model is used in a variety of engineering applications, such as a steam engine or the cylinder of an internal combustion engine. Therefore, we will apply a motor on the crank part, run the simulation, and then postprocess some results to estimate the required torque.


Creating Local Mates

Crank

Arm Mount

Link2

Link1

Arm

Crank Housing

Collar Shaft Collar

Problem Description

The crank is driven at a constant angular velocity of 60 RPM. Determine the torque required to rotate the crank part.


I

Create a motion study.

I

Preprocessing.

Add local mates to the assembly with the motion study active.

I

Run the simulation.

Calculate the motion.

I

Post-processing.

Plot and analyze the results.

1


Open 3dcrankslider from the Lesson02\Case Studies folder.

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SolidWorks 2011

Lesson 2 Building a Motion Model and Post-processing

2

Examine the assembly.


SolidWorks Motion assumes that all components that are fixed in SolidWorks are considered to be grounded parts, and all components that are floating are assumed to be moving parts. However, the movement of these parts is constrained by the SolidWorks mates.

There are no mates in this assembly, but three parts are fixed. The collar_shaft, arm_mount and crank_housing are fixed as these are parts that would be connect to ground and will have no motion in the assembly.

Fixed

No Mates

The remaining parts will need mates to constrain their motion to that expected of the mechanical system.

Mates

Mates are used to constrain the relative motion of a pair of rigid bodies by physically connecting them.

Note

A rigid body acts and moves as a single unit. SolidWorks components situated at the root level are considered rigid bodies. This means that SolidWorks and SolidWorks Motion treat subassemblies as single rigid bodies. Mates can be classified into two main types: I

Mates used to constrain the relative motion of a pair of rigid bodies by physically connecting them. Examples: Hinge, Concentric, Coincident, Fixed, Screw, Cam, etc.

I

Mates used to enforce standard geometric constraints. Examples: Distance, Angle, Parallel, etc.

Below are some descriptions of some of the most commonly used mate types. For a comprehensive understanding of all the other mates, please refer to the SolidWorks help.

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The concentric mate allows both relative rotation as well as relative translation of one rigid body with respect to another rigid body. The concentric mate origin can be located anywhere along the axis about which the rigid bodies can rotate or slide with respect to each other. Example: Piston sliding and rotating inside a cylinder.


Concentric Mate

Hinge Mate

Hinge mate is essentially concentric mate with the restricted translation between the two components.

In SolidWorks Motion, the hinge mate is used rather than a combination of concentric and coincident because the mechanical joint is a hinge. Hinge mates are found in the Mechanical Mates tab of the Mate PropertyManager.

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SolidWorks 2011


This type of mate permits free rotation about a common point of one rigid body with respect to another rigid body. The origin location of this mate determines the point about which the rigid bodies can pivot freely with respect to each other. Example: Ball and Socket joint.


Point-to-Point Coincident Mate

Lock Mate

The lock mate locks two rigid bodies together so they may not move with respect to each other. For a lock mate, the origin location and orientation does not affect the outcome of the simulation. A real world example of a lock mate is a weld that holds two parts together.

Two Face-to-Face Coincident Mates

This mate allows one rigid body to translate along a vector with respect to a second rigid body. The rigid bodies may only translate, not rotate, with respect to each other.

The location of the origin of a translational joint with respect to its rigid bodies does not affect the motion of the two bodies but does affect the reaction or the bearing loads.

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A universal mate permits the transfer of rotation from one rigid body to another rigid body. This mate is particularly useful to transfer rotational motion around corners, or to transfer rotational motion between two connected shafts that are permitted to bend at the connection point (such as the drive shaft on an automobile).


Universal Mate

The origin location of the universal mate represents the connection point of the two rigid bodies. The two shaft axes identify the center lines of the two rigid bodies connected by the universal joint. Note that SolidWorks Motion uses rotational axes parallel to the rotational axes you identify but passing through the origin of the universal mate.

Screw Mate

The screw mate constrains one rigid body to rotate as it translates with respect to another rigid body.

When defining a screw mate, you can define the pitch. The pitch is the amount of relative translational displacement between the rigid bodies for each full rotation of the first rigid body. The displacement of the first rigid body relative to the second rigid body is a function of the rotation of the first rigid body about the axis of rotation. For every full rotation, the displacement of the first rigid body along the translation axis with respect to the second rigid body is equal to the value of the pitch.

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SolidWorks 2011


This type of mate permits one translational and three rotational motions of one part with respect to another. The translational motion between the parts is confined to the orientation axis. The point defines the initial pivot location on the axis.


Point-on-Axis Coincident Mate

Parallel Mate

A parallel mate permits only translational motion of one part with respect to another. No rotation is allowed.

In the picture below, the blue x part can move relative to the ground in the X direction. The red y part can move relative to the x part in the Y direction. The z part can move relative to the y part in the Z direction. Finally, the red/yellow/blue cube on the z part has a curvilinear motion relative to the ground but always stays parallel.

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Perpendicular Mate


The perpendicular mate allows both translational and rotational motion of one part with respect to another. It imposes a single rotational constraint on the components so that the component axes remain perpendicular. This allows relative rotations about either z-axis, but does not allow any relative rotation in the direction perpendicular to both z-axes.

It is recommended that the mates are representing the real mechanical connections as closely as possible, i.e. mechanical hinge should be modeled using the hinge mate and not using a combination of coincident and concentric mates.

Local Mates

Mates created in SolidWorks can be transferred to the Motion Analysis and used as mechanical joints. If there are no mates in the SolidWorks assembly or if we wish to define the connections differently than the SolidWorks mates, we can add local mates directly to the motion study. Local mates only apply to the study to which they were added. To add local mates, make a motion study active and add the mates. With a motion study active, any mate added is only applied in that motion study.

3

Verify the document units.

Verify that the document units are set to MMGS (millimeter, gram, second).

4


Right-click the Motion Study 1 tab and click Create New Motion Study.

Make sure that the Motion Analysis is selected as the Type of Study in the MotionManager.

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SolidWorks 2011


5

Move components.


Move the components that are not fixed to separate the assembly. We are doing this only to make it easier to select faces and to keep track of what components are mated.

6

Create a local mate.

Add a mate and select Hinge from the Mechanical Mates section. For Concentric Selections, select the two cylindrical faces of the shaft and hole shown with red arrows. For Coincident Selections, select the end face of the shaft and crank housing shown with the blue arrows.

Click OK.

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7

Warning.


Because the timeline is active, the mate changes the position of the crank at the starting position of the animation. This is OK for what we are doing.

Click Yes.

8

Examine the mate.

Notice that this mate is only located in the MotionManager and not in the FeatureManager design tree.

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SolidWorks 2011


Add additional mates. Add a concentric mate between the two

Link1


9

spherical surfaces shown on the parts Link1 and crank.

crank

Concentric

10 Mate arm to arm_mount. Add a hinge mate to connect the arm to the arm_mount.

11 Mate Link1 to the arm.

This connection requires two hinge mates, one between Link1 and cardian, and a second hinge mate between cardian and arm.

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12 Mate Link2. Link2 will use a hinge mate to connect the arm. As there is no pin


Hinge

going through the two holes, the coincident selections will be the two touching faces.

Mate the other end of Link2 to the pin on the collar with just a concentric mate.

Concentric

13 Mate collar to collar_shaft.

Add a concentric mate between a cylindrical surface on each part.

14 Test the assembly.

Rotate the crank and make sure the components move as expected.

Check the FeatureManager design tree and the Motion Study tree. All the mates should just be in the Motion Study tree.

15 Add gravity.

Add gravity in the negative Y direction.

16 Calculate.

Adjust the assembly key to 5 seconds. Click Calculate Simulation

.

17 Play the simulation.

Play the simulation at 25% speed.

The crank will rock back and forth as gravity affects the components and potential and kinetic energy are exchanged. As there is no friction, the parts will continue to move without end.

18 Set the timebar to 0s.

To add a motor at time 0s, the timebar needs to be set to 0s.

19 Add a motor.

Create a Motor that drives the crank. Click Motor

on the Motion Manager.


Under Motor Location, select the cylindrical face of the crank part (as shown in the figure). The default selection for the Motor Direction is correct for this analysis. Make sure that the motor is oriented as shown in the figure.

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SolidWorks 2011



Under Motion, select the Motor Type as Data Points. The command invokes the Function Builder window. Make sure that Value (y) and Independent variable (x) are set to Displacements (deg) and Time (s).

Function Builder

Function Builder can be used to construct functional expression for motors and forces.

Introducing: Function Builder

Function Builder can build functional expressions using predefined Segments, imported set of discrete Data Points or mathematical Expressions. The figure below shows the segment view of the Function Builder window.

I

Segments

In Segment view, user select both the independent (typically time) and dependent variable (displacements, velocity or acceleration). For each specified interval, the transition from the initial to final value is controlled using one of the predefined profiles curves. The following profile curves have been implemented: Linear, Cubic,

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Quarter-Sine, Half-Sine, 3-4-5 Polynomial and others. As the function is constructed, the graph windows show the corresponding variation of displacement, velocity, acceleration and the jerk (time derivative of acceleration). Note that it is possible to save and retrieve function from stored location.

I

Data Points

The discrete set of data points can be imported from a *.csv file or entered manually. The functionality as well as the options are similar to the Interpolated option of the input and are explained in this lesson.

I

Expression

Expression enables the construction of functions with the help of predefined mathematical functions, variables and constants, and existing motion study results. As in both previous cases, the function can be saved at a specific location. This procedure will be used in this lesson.

Where to Find It

I

In the Motor or Force/Torque PropertyManagers, under Motor Type or Force Function dialog select Segments, Data Points or Expression.

20 Import data points.

Rather than type the individual values into the table, we can load them from a file. In this case, we have an Excel file. Locate the file crank rotation.csv in the Case Studies folder and examine the file. It is just two columns of numbers representing the time and displacement.

Click the Import Data button. Navigate to and select the crank rotation.csv file and click Open. The values from the file are now inserted into the Time and Value columns. Select Akima as the Interpolation type.

Note

46

The Function Builder graph windows automatically updates the plots for displacement, velocity, acceleration and jerk. The data points describe linear increase of the angular displacement in time, a harmonic motion

SolidWorks 2011



.

Click OK to complete the definition of the profile and close the Function Builder. Click OK to complete the definition of the Motor feature.

Rename this motor feature to Motor - crank.

Importing Data Points

Using imported data points, you can use your own motion data to control the displacement, velocity, or acceleration of the motion. The data points that can be imported into SolidWorks Motion must be in a text file (*.txt) or comma separated file (*.csv) format. The file should contain one data point per line. The data point consists of two values, the time and the value at that time. Commas or spaces can be used as separators between the values. The file is essentially free format aside from these restriction. SolidWorks Motion allows for unlimited number of data points to be used. The minimum number of data points to be defined is four. The first column, Independent variable (x), in the data point template is typically time, but other parameters such as cycle angle, angular displacement and others can be used as well. The second column, Value (y), is the displacement, velocity, or acceleration. These values can be manually entered or imported.

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Besides Linear interpolation, two spline-fitting options are available to smoothen out the data: the Akima spline (AKISPL) and cubic curve spline (CUBSPL). It is recommended that you use a cubic curve because it will work well even if you data points are not evenly spaced. An Akima curve is fast, but will not work as well if you points are not evenly spaced.

21 Run the simulation. Click Calculate to run the simulation for 5 seconds. 22 Plot the torque.

Create a plot for the torque required to turn the mechanism. Define the plot by Forces, Motor Torque and Magnitude. Select the Motor-crank for the Simulation element. Click OK.

23 Examine the plot.

The plot may be improved by recording more data points by increasing the Frames per second option in the Motion Study Properties.

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24 Plot the power.


Create a plot for the power required to turn the mechanism.

Define the plot by Momentum/Energy/Power, and Power Consumption. Select the Motor-crank for the Simulation element.

Click OK.

Note

Knowing the operating RPM, torque and/or power we can select the appropriate motor to drive our system.

Power

Power is the rate at which work is performed, or the amount of work conducted in one second. Forces conduct work on distances, moments then on the angular displacements. For rotating motors the following relationship therefore holds: Power [W]

= Torque [N-m] × Angular velocity [rad/sec]

The power plot in the previous figure can be easily verified. The maximum torque is 10 N-mm = 0.01 N-m Angular velocity

= 360 deg/sec = 2π rad/sec

Students can easily verify this by creating the plot of the angular velocity. The resulting maximum power is then: Power

= 0.01 × 2 π = 0.063W

The graph of the power indicates 0.06 W because two significant digits precision is used by default.

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Lesson 2

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Often times the rating of the electric motors is expressed in maximum power and torque. Alternative units are used frequently.


Alternative Units

If rpm is used for the angular velocity, then:

Torque [N-m] × 2π × Angular velocity [rpm] Power [W] = ------------------------------------------------------------------------------------------------------------------------60

If horsepower is used instead, the following conversion can be used: Mechanical horsepower = 33,000 lb-ft/min = 745.7W

A useful formula when computing power using mechanical horsepower in the English system of units is the following:

Power [hp]

Torque[lb-ft] × RPM Torque [lb-ft] × 2π × Angular velocity [rpm] = ------------------------------------------------------------------------------------------------------ = ----------------------------------------------33,000

5252.1

While mechanical horsepower is common in some industries in the United States (automotive industry, for example), similar measure called metric horsepower is used in Europe and Asia. Metric horsepower is then defined as: Metric horsepower

= 735.5W

Because of this ambiguity in the definition of horsepower, its use today is not recommended.

25 Add a mate.

When we added mates to this motion study, we only added mates essential to describing the motion. Depending on how the assembly is built, a mate preventing the collar from rotating around the collar shaft could be defined. This mate would represent the mechanical function of the keyway. Add a coincident mate between one side face of the key and the corresponding face on the keyway.

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SolidWorks 2011



26 Run the study. Change the Frames per Second to 100, then re-calculate the study. 27 Review results.

The torque values are essentially the same as when we did not have the coincident mate. The plot is now smoother as we have four times more data points.

Following the recommendations that all mates should represent the real mechanical connections for the kinematic analyses, this mate defining the keyway could be defined, even if it is not required for the actual motion analysis.

28 Plot reaction force.

Create a new plot to show the reaction force on the motor.

Define the plot by Forces, Reaction Force, and Magnitude.

In this assembly, the first hinge we defined was between the crank and crank_housing. As the crank_housing is fixed, the mate must transmit the reaction force.

Select the first hinge mate as the Simulation element.

Because the selected mate connects two parts, there are two equal and opposite forces acting in the mate. One of the two parts must be selected for the plot of this force.

Select any face on the crank-1 part as the second component in the Simulation Element field. Click the Show vector in the graphics window checkbox.

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This mate must be selected from the Mate Group 1 folder in the Motion Manager tree because this mate is local and is not listed in the FeatureManager design tree.


Note

Click OK.

29 Warning.

We will receive a warning about redundant constraints. Redundant constraints may have significant impact on the mate forces (forces in the mechanical connections, defined by the mates) and will be discussed later in the course. The resulting force obtained for this mechanism is, however, correct as the redundancies present in this assembly do not have any effect on the force shown in the figure below.

Click No.

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SolidWorks 2011


30 Review the plot.


Play the simulation and observe the reaction force vector.

When viewed from the Right view:

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The Results PropertyManager provides access to various output quantities reviewed in Lesson 1. They can be requested in absolute or relative values, and with respect to another component of the assembly. While in most situations the default output is in the global coordinate system of the top level assembly, it is very easy to transform the values to any other selected local coordinate system.

Absolute vs. Relative values

To request the plot of absolute values, select the component (mate, motor, part etc.) in the Simulation Element field.


Plotting Kinematic Results

To plot values relative to the second component, add this component to the Simulation Element field.

Reference component

Note

The reference component must be selected as the second in the list.

Output coordinate system

Typically, the results are output in the global coordinate system of the assembly. For some simulation components (mates and motors, for example) the default output is, however, in the local system of the selected component. To plot results in other than the default coordinate system, select the desired component in the Component to define XYZ directions field. The values will then be transformed into the coordinate system of the selected part.

Note

54

The requested output coordinate system is indicated by the triad shown in the graphics area.

SolidWorks 2011



In the next section four plots we will demonstrate the results in their absolute and relative magnitudes, evaluated in both the global and local coordinate systems.

31 Absolute result for component in global system.

Create a plot for the X component of the linear displacement of arm.

Define the plot by Displacement/Velocity/Acceleration, Linear Displacement and X Component.

Select any face of the arm component for the Simulation element.

Click OK.

Note

If we select a face, the plot will be of the linear displacement of the part’s origin, indicated by a small blue sphere, with respect to the origin of the assembly in the global assembly system.

Because the input was harmonic motion, the output is an oscillatory motion.

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32 Absolute result for component transformed in local system. Create a plot for the X component of the linear displacement of arm

transformed in its own local coordinate system.

Edit the definition of the previous plot and select arm as the Component to define XYZ directions. Click OK.

Local coordinate system of arm part

Global coordinate system

Note

Note that the triad on the part arm now indicates the output local coordinate system which is misaligned with the global coordinate system. Further more, note that this local output system translates and rotates with respect to the global coordinate system as you play the motion.

Note

The above figure shows the linear displacement of the part’s origin, with respect to the origin of the assembly, transformed in the part’s coordinate system. Alternatively, we can view the above graph as the values from step 31, transformed in the coordinate system of the arm.

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33 Relative result for component in global system.


Create a plot for the X component of the linear displacement of the arm relative to the displacement of the collar. Edit the previous plot.

Clear the Component to define XYZ directions field.

Select the collar part as the second component in the Simulation element field.

We can see that the displacement has somewhat different oscillatory characteristic as the displacement of the arm in the global coordinate system (step 31). Relative result for component transformed in local system.

Note

The above figure shows the linear displacement of the arm’s origin, with respect to the origin of the collar part in the global coordinate system.

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34 Relative result for component in local system.


Create a plot for the X component of the linear displacement of the arm relative to the displacement of the collar. Transform the results in the local coordinate system of Link1. Edit the definition of the previous plot and select Link1 as the Component to define XYZ directions.

Local coordinate system of Link1 part

Global coordinate system

Note

Note that the triad on the part Link1 now indicates the output local coordinate system which is misaligned with the global coordinate system.

The above plot shows the values plotted in step 33, transformed in the coordinate system of the collar.

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Angular displacement plots can be created to measure the angular displacement of a motor, mate, three points or one component relative to another component. Because the angular displacement is not a vector, only the magnitude can be plotted.


Angular Displacement Plots

The previous section introduced generation of kinematic results plots for a component. In the next steps various post-processing plots for other simulation elements (mates, motors etc.) will be generated. For most of these simulation elements, the default output coordinate system is the local coordinate system of the element.

35 Angular displacement of mate.

Create a plot for the angular displacement of the local hinge mate between the part Link1 and the cardian.

Define the plot by Displacement/Velocity/Acceleration, Angular Displacement and Magnitude.

Select the local hinge mate between the Link1 and cardian for the Simulation element.

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Click OK.


Notice the triad at the location of the hinge. It indicates that the output coordinate system is the local system of the hinge mate. Only the magnitude can be requested.

This plot shows the vertical rotation of the Link2 part which is approximately 2 degrees.

36 Angular displacement of motor.

To explore the other options of the angular displacement plot, we will modify our existing plot rather than create a new plot. In the Results folder, right-click the last plot and click Edit Feature.

Delete the hinge mate and select the motion component Motor -

crank for the Simulation element.

Click OK.

The plot shows harmonic motion of the motor. The angular displacement goes from zero to +180 degrees then returns from -180 degrees. The slope of the graph is constant.

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37 Angular displacement of two lines defined by three points.


This time, we will create a plot to show the angular displacement between two lines defined by three points. Create a new plot.

Define the plot by Displacement/Velocity/ Acceleration, Angular Displacement and Magnitude.

For the Simulation element, first select the two vertices shown, then select the edge.

Select Show vector in graphics window. This will show lines between the three selected points.

Vertex 1

Vertex 2

Edge (defining Vertex 3)

Click OK.

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The plot shows the angle between the line defined by Vertex1 and Vertex3 and the line defined by Vertex2 and Vertex3 (Vertex3 therefore defines the center point).

Notice that in the present case the limits of the angular motion is 84 degrees and 121 degrees giving the range of 37 degrees.

Angular Velocity and Acceleration Plots

Similarly to the angular displacement, Angular velocity plot can be generated for a motor, mate and a component relative to another component. Magnitude as well as all three coordinate components are available.

39 Plot angular velocity and acceleration.

Generate a couple of velocity and acceleration plots on your own. Try to plot both the absolute and relative magnitudes in both the global and local coordinate systems.


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In this lesson we analyzed a 3dcrankslider assembly. While both SolidWorks and SolidWorks Motion Simulation assemblies can be built in many ways with various mates, the main objective of this lesson was to show the suggested assembly building procedure for the motion analysis where only kinematic results (displacements, velocities, accelerations etc.) are of interest. We call this type of the analysis “kinematic analysis”. “Dynamic analysis” is then a simulation where mate forces and their distribution throughout the assembly is required. These later types of analyses can be more intricate since the redundancies need to be understood and addressed (redundancies are subject of the later lessons in this course).


Summary

It was suggested that the most suitable approach to obtain kinematic results while investing reasonable about of time in the motion assembly building is to model the mates as closely to the real mechanical connections as possible, i.e. all real mechanical hinges will be modeled as hinge mates. This lesson also introduced the most common mate types and the subject of local mates. Local mates are designed within the SolidWorks Motion Simulation tab and do not affect the original SolidWorks assembly and the design intent in any way. This way, each Motion Simulation study may feature its own independent mates. While motors and forces input may be defined in many ways, this lesson shows the procedure to control the magnitudes using the imported data from the table. The second half of this lesson introduces various available result quantities and shows their definitions in detail.

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Exercise 2 Piston

Exercise 2: Piston


In this exercise, we will manually create local mates and run a motion simulation on a simple engine under the effects of gravity only. We will plot the results and check the assembly for interference. engineblock piston

conrod

crankshaft

bearing


1

Creating Local Mates on page 34. Angular Displacement Plots on page 59.


Open Piston from the Lesson02\Exercises folder.

2

Type of Study.

Select the Motion Study 1 tab and set the Type of Study to Motion Analysis.

3


Verify that the document units are MMGS (millimeter, gram, second).

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Piston

4

Verify fixed and floating states of components.


Examine the assembly. Three parts are fixed and three parts are not fixed and not mated. MateGroup1 is empty. The engineblock and the two bearings are fixed.

The piston, crankshaft and conrod are floating. Fixed Components

5

Floating Components

Move components.

Move the floating components away from their final positions. We are doing this just to make it easier to select faces as we create local mates.

6

Add local mates.

Add the following local mates: I

Note

66

Hinge between the crankshaft and bearing<2>.

The second hinge mate between the crankshaft and the bearing<1> components could have been defined as well. However, it would have no effect on the kinematic results of this simulation.

SolidWorks 2011

Exercise 2 Piston

Hinge between the conrod and crankshaft.


I

I

Concentric between the piston

and the cylindrical face of the engineblock (piston bore).

I

Concentric between the upper hole in the conrod and one of the wrist pin holes in the piston. We do not have the wrist pin

modeled so we are using the concentric mate in its place.

7

Add gravity.

Under Gravity Parameters, Direction Reference, select the Y direction.

Under Numeric gravity value, type in a value of 9806.65 mm/sec^2.

8

Motion Study properties.

Set the study properties to record 100 frames per second.

9

Run the simulation for 2.32 seconds.

Make sure that the study type is set to Motion Analysis.

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Piston

10 Examine the motion.


Playback the study at one-quarter speed. The weight of the piston and conrod will cause the piston to try to move to bottom dead center. As there is no friction, the model will just oscillate as the total energy of the system is conserved.

While the assembly moves freely, we cannot tell if there is interference between the different components. In Lesson 3, interference detection in SolidWorks Motion will be demonstrated.

11 Plot results.

Create a plot of the angular displacement of the crankshaft.

Initially, the plot may look odd, however if you examine it closely you can see that the component is just rocking back and forth.

12 Angular displacement of hinge mate.

Create another plot for angular displacement of the hinge mate between the crankshaft and bearing.

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Exercise 2 Piston


The plot should look the same as the previous plot for the crankshaft, except that the values are of opposite sign and the graph begins at 0 degrees. This is because the displacement plot for the mate, motors and spring features are plotted at the local coordinate system by default.

13 Plot linear displacement.

Create a plot for the linear displacement of the piston in the global coordinate system. Plot the Y-component as this is the direction along the axis of the piston bore. The plot shows normal harmonic motion.

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Piston

14 Transform linear displacement plot.


Transform the above linear displacement plot into the local coordinate system of the crankshaft.

As the local coordinate system of the crankshaft rotates, the values in the plot are changing from positive to negative.


Summary

70

In this exercise you analyzed a small piston assembly. The main objective was to practice the assembly building procedure when kinematic results are of interest only and to plot various result quantities.

SolidWorks 2011

Exercise 3 Trace Path

Exercise 3: Trace Path


In this exercise we will use motors that are driven by tabular data to have a stylus trace a path like a pen plotter.

cross beam

chassis

pointer


Procedure

Local Mates on page 40. Importing Data Points on page 47.

Open the existing assembly from the Exercises folder.

1


Open pant1 from the Lesson02\Exercises folder.

2

Set the document units. Click Tools, Options, Document Properties, Units.

Select MMGS (millimeter, gram, second) for the Unit system.

3

New study.

Crate a new motion study. Make sure you select Motion Analysis.

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Trace Path

4



Existing mates allow the cross beam to move along the rails of the chassis and the pointer to move along the cross beam.

One mate missing is something to keep the pointer from rotating around the cross beam.

5

Add rotary motor.

To prevent the rotation, we will use a rotary motor. Select Axis1 in the Pointer as the Component.

Set the Motion to Distance and make the distance 0 degrees from 0 to 20 seconds.

6

Add linear motor.

The first linear motor will drive the cross beam along the chassis.

Two csv files are provided in the Exercises folder, movx.csv and movy.csv. These files have number pairs with the first number indicating the time and the second number representing position.

Notice that in each set of numbers, the time points are evenly spaced. This will allow us to use the Akima interpolation type. Add a Linear Motor.

Select the face shown in the image below.

Select Data Points to open the Function Builder window. Import Data for the Displacement (movy.csv file).

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Exercise 3 Trace Path

Look at the triad to see that the selected face will move in the Y direction, so we need the movy.csv file instead of the movx.csv.


Note

Click OK.

7

Add another motor.

Add another linear motor to move the pointer across the cross beam using the movx.csv file. Orient the motor in the direction of the negative X axis.

8

Run the study.

Run the study for 20 seconds.

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Trace Path

9

Create a path trace.


Create a new result. Select Displacement/Velocity/Acceleration and Trace Path. Select the vertex at the end of the Pointer.

Check the Show vector in the graphics window checkbox to see the star shape.

Note

The Trace Path plot will be discussed in more in detail in Lesson 6, where it will be used to generate the profile of a CAM.


Summary

74

In this exercise you analyzed a pen assembly. The main objective of this exercise was to define local mate definitions and to import tabulated data to control the motor magnitude.

PR Do E no RE t c LE op AS y E or D di RA st F rib T ut e Lesson 3 Introduction to Contacts, Springs and Dampers

Objectives


Check interference of components.

I

Apply contact to components.

I

Specify solid bodies contact friction.

I

Add a spring with damper to the assembly.

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Introduction to Contacts, Springs and Dampers

In this lesson we will examine the motion of a catapult as it is loaded and throws a projectile. Some of the components in this lesson are not connected to the others through mates or joints but are restricted based on their contact with other components. We will place these dynamic components into our system by defining contact conditions and also include friction between components.

Case Study: Catapult

The crank will rotate the catapult arm, through a belt and pulley, to a position where a projectile can be loaded. The crank motion will also be transmitted through a gear assembly to a trigger mechanism that will release the projectile and allow the spring to push the projectile onto the projectile holder.


Contact and Friction

When released, the counterweight will cause the arm to rotate and throw the projectile.

Catapult-Arm

Gear assembly

Projectile

Counterweight

Trigger mechanism

Hand crank

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SolidWorks 2011

Lesson 3 Introduction to Contacts, Springs and Dampers

Problem Description


The crank will rotate 2.75 turns to load the catapult. The motion of the rack will cause the trigger to release the projectile onto the projectile holder. The mechanism will release the arm and the counterweight will cause the arm to throw the projectile.

Determine the torque required to rotate the crank and load the catapult. Determine the displacement and velocity and force of the loading spring.


I


This will be a new motion study.

I

Apply friction.

Friction will be added to the existing SolidWorks mates.

I

Apply contact.

Contact will be added to the dynamic components.

I

Add a spring.

We don’t use a spring model in the motion simulation. Instead we create a motion element that mathematically represents the spring.

I

Apply gravity.

The catapult operates under conditions of normal gravity.

I

Calculate the simulation.

I

Plot the results.

We will create various plots to show the torque and power required.

1


Open Catapult-assembly from the Lesson03\Case Studies folder.

2


The crank rotation does two things, it rotates the arm through the belt and pulley and it triggers the release of the projectile through a gear train.

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The gear train consists of three gear mates and a rack and pinion mate. When the rack moves, it will come in contact with the release mechanism and lift the projectile holder door.

Rotate the crank to see how the mates work.

3


Verify that the document units are set to MMGS (millimeter, gram, second).

4


Right-click the Motion Study 1 tab and click Create New Motion Study.


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5

Add a motor.


To revolve the crank we need to apply a motor to the end of the shaft. We want to rotate the crank 2.75 turns in 3 seconds. Click Motor in the MotionManager toolbar.

Select the edge of the crank shaft for both the Motor Direction and Motor Location fields.

Select Rotary Motor for the Motor Type and Distance for the Motion

Type.

Type 990 deg (2.75 turns x 360 deg) for the Displacement and 3 seconds for the Duration. Click OK.

6

Disable the motor.

After the motor turns for 3 seconds, we want it to hold the catapult in the loaded position while the projectile moves into the projectile tray. We then want the motor to disengage to allow the counterweight to drive the catapult.

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The entire simulation will run for 5 seconds, so to make it easier to select on the timeline, click Zoom In on the lower right corner of the MotionManager until a little more than 5 seconds fills the time line MotionManager.

Select the RotaryMotor1 in the MotionManager. Right-click in the timeline at 3.4 seconds and click Off. This creates a key that suppresses the motor at 3.4 seconds so that it will have no effect after this time. If you place the key at the wrong time, just drag it to 3.4 seconds.

Tip

7

Motion Study Properties.

Set the Frames per second to 50.

8

Calculate.

Click Calculate and observe the motion.

Notice that as specified, motor rotates the crank by 2.75 turns in 3 seconds. From 3 to 3.4 seconds the motor keeps the crank, as well as the arm, stationary and ready to launch. Finally it disengages at 3.4 seconds when the mechanism begins to move in not specifically defined motion. A few key elements must still be added to the motion model, however.

9

Analyze the motion.

In the MotionManager, right-click Orientation and Camera Views and click Disable Playback of View Keys.

Change to the Front view and zoom in on the left end of the assembly.

Play the simulation in slow motion again and notice that the two triggers move through each other.

To stop this, we must add contact between them. Before defining contact, we will however introduce a feature which can be used to detect the interference automatically.

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The Interference Detection tool in SolidWorks will detect interference between components. However, it will only detect interference for a single static position of the components.


Interference Detection

In SolidWorks Motion, interference can be detected for the motion path of each component.

Where to Find It

I

Right-click on the top level component in the MotionManager and select Check Interference.

10 Check Interference.

In the MotionManager, right-click the Catapult-assembly and click Check Interference. Select the two triggers and click Find Now.

11 Examine the results.

The two triggers interfere starting at Frame 132 at time 2.620 seconds and remain that way until the last frame.

Select the first interference and click Details. We can now see the location and amount of interference.

Close the dialog Find Interferences Over Time.

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Contacts can be defined between multiple bodies or curves to prevent penetration. In this lesson we will only learn how to define the contact between solid bodies and discuss friction. A more detailed discussion on the definition of contacts and its parameters will be presented in the next lesson.


Contact

Introducing: Contact

Contact is used to define the way bodies react with each other. Within the contact definition, we can control the friction and the elastic properties between the bodies.

Where to Find It

I

Click Contact


12 Add contact.

In the MotionManager toolbar click Contact . Select the two Projectile holder trigger parts. For Contact Type select Solid Bodies.

We will keep all contact parameters except friction at their default values - they are the subject of Lesson 4. Make sure that Material PropertyManager is checked and select Steel (Dry) for both materials.

We will run the simulation without considering the friction between these two parts. Clear Friction.

Click OK.

13 Calculate.

14 Examine the trigger.

When the simulation runs, the trigger on the rack mechanism will now contact the trigger on the projectile holder door and lower it.

Contact groups

82

Contacts between the bodies can be defined in a multiple separate definitions (each for two bodies only), or in one (or a few) definition with all bodies included in a single (or a few) definition only. The later one will consider contact between all selected bodies, thus automatically generating multiple contact pairs. While this procedure is easy to define, considering contact for all pairs can be computationally demanding.

SolidWorks 2011



Contact definition with contact groups ignores contact between parts in the group, but considers contact between all combinations of pairs of bodies across the groups. It is possible to define a maximum of two contact groups.

Introducing: Use contact groups

Contact groups enable to place contacting bodies in two separate groups. All contact combinations across the two groups are considered only.

Where to Find It

I

Click Contact on the MotionManager toolbar. Under Selections, check the Use contact groups check box.

15 Additional contacts.

The following additional contacts have to be defined: I I I

projectile - projectile holder door projectile holder - projectile projectile holder pusher - projectile

Using the procedure outlined in the previous steps, three separate definitions would have to be created. Instead, using the contact groups one single definition will suffice in this case.

Create contact definition. Under Contact Type select Solid Bodies, for Material select Steel (Dry) and clear Friction.

Under Selections check Use contact groups.

Select Projectile in Group1 and projectile holder, projectile holder door and projectile holder pusher in Group2. Click OK.

Note

The PropertyManager indicates that three contact pairs will be considered for the calculation.

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When defining contact, there are three friction options which can be used depending on the model.


Contact Friction

I I I

Static Kinematic None

Once you decide what friction types to include in your contact, you must evaluate the static and/or kinematic velocity and friction constants. Coulomb friction forces are calculated based on two different coefficients - static and kinematic.

Static Coefficient

The static coefficient is the constant used to calculate the force necessary to overcome friction when a body is at rest.

Kinematic Coefficient

The kinematic coefficient is the constant used to calculate friction forces once the body is no longer at rest.

In reality, the static friction transition velocity is zero, but numerical solvers, such as SolidWorks Motion, require that a non-zero value be specified to avoid singularity at the origin. More specifically, when a part is in transition from a negative to positive velocity, and when the velocity is zero, the force magnitude cannot instantaneously transition from a positive to negative value. Therefore, the graph Static shows how Kinematic SolidWorks Motion resolves this issue— Force (N) the user specifies a static and kinematic transition velocity where the friction 0.102 mm/sec. 10.16 mm/sec. coefficients are used. Velocity (mm/sec.) From there, SolidWorks Motion fits a smooth curve to solve for the friction force. In the graph above, the default friction parameters for dry steel in contact are used.

I

Static Friction Transition Velocity: vs = 0.102 mm/s. Kinematic Friction Transition Velocity: vt = 10.16 mm/s. Static Friction Coefficient: 0.30. Kinematic Friction Coefficient: 0.25.

I

In the PropertyManager for Contact, select Friction.

I I I

Where to Find It

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16 Additional contact set.

Create a contact set between: I I

projectile Catapult-Arm

Under Material select Steel (Dry).

Make sure that Friction is checked.

Uncheck Material and change the values for the Dynamic Friction Coefficient and Static Friction Coefficient to 0.15 and 0.2, respectively. Click OK.

Note

Unchecking Material opens up the Friction dialog fields for editing. The contact characteristics in Elastic Properties, determined by the selection in the Material dialog, remain unchanged. Elastic Properties are discussed in Lesson 4.

Translational Spring

A translational spring represents the displacement dependent force acting between two parts over a distance and along a particular direction.

When defining a spring, you can readily change the force-displacement dependency from linear to another predefined relationship by selecting the function type from a list. This allows you to select the relationship between the force and displacement. The following force-displacement relationships are supported in SolidWorks Motion: X, X2, X3, X4, 1/x, 1/x2, 1/x3

You specify the location of the spring on two parts.

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SolidWorks Motion calculates the spring force based on the relative displacement between the two parts, the stiffness of the spring and the fabrication or free length.

When the spring force is negative, the spring is in a stretched position relative to the free length.

Note

Spring forces become ill-defined if the end points become coincident because of undefined direction.

Magnitude of Spring Force

The magnitude of the spring force is based on the stiffness and initial force. The spring relationship can be written as: F = -K (X - X0)n + F0

Where:

X= Distance between the two locations that define the spring K= Spring stiffness coefficient (always > 0) F0 = Reference force of the spring (preload)

n = Exponent. For example, if spring force = KX2, then n = 2. Valid values for the exponent n are: -4,-3,-2,-1,1,2,3,4. X0 = Reference length (at preload, always > 0)

Note

86

I

Positive force repels the two parts.

I

Negative force attracts the two parts.

To create a spring that exhibits non-linear force properties not supported in the spring definition, you must use an action-reaction force where you can enter a non-linear force equation.

SolidWorks 2011


Both linear and torsional springs can be added between components. Both the Exponent of the spring force expression (linear to ± 4) and Spring Constant can be specified.


Introducing: Spring

Click Spring


Where to Find It

I

Translational Damper

A translational damper is considered a resistive element used to “smoothen” out oscillations encountered due to outside forces. Typically, dampers are used in conjunction with springs to “dampen” out any oscillations or vibrations created by the spring. In the real world, bodies and even springs have built in structural damping, and the damper element can be used to represent this. The force created by a damper is dependent on the instantaneous velocity vectors between the two defined endpoints.

Note

To create a damper that exhibits non-linear force properties not supported in the Damper definition, you must use an action-reaction force where you can enter a non-linear force equation based on the velocity between the two points of the force entity.

For the translational damper element, the force equation is pre-defined as F = c × v n where c is the user defined damping coefficient, v is the relative velocity between two end points and n is the exponent. For example, if damper force = -c*v2, then n = 2 (valid options are -4,-3,2,-1,1,2,3,4).

Introducing: Damper

The Damper can be added between components in a mechanism. Additionally, both linear and torsional springs can have damping properties that act as the combination of spring and damper together.

Like springs, both the Exponent of the damper force expression (linear to ± 4) and Damping Constant can be specified.

Where to Find It

I

Click Damper


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17 Add a spring.

To move the projectile into position on the Catapult-Arm, we must add a spring. The spring will have a pre-load to hold the projectile against the back of the projectile holder door. When the door drops, the projectile is pushed into position. Click Spring


Select the two faces shown below.

Set the Spring Parameters as shown to create a linear spring with a Spring Constant of 0.15 N/mm and the Free Length of 13mm.

Select Damper and add a Damping Constant of 0.01 N/(mm/s).

For Display set the Coil Diameter to 4.00mm, 5 turns and a Wire Diameter of 0.5mm.

Note

The values entered in the Display area are only used as graphics parameters. Click OK.

18 Calculate.

When the simulation solves, the projectile flies off into space and arm does not release and the counterweight does not stay level. This is because we are still missing a key element, gravity.

19 Add gravity.

Add gravity to the assembly.

20 Calculate.

This time the arm is cranked down to the loading position and is held there by the motor while the trigger releases the door and the projectile is pushed onto the arm by the spring. At 3.4 seconds, the motor turns off and the gravity on the counterweight swings the arm and launches the projectile.

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Now that the simulation is calculated, we can create plots for the different parameters we are interested in.


Postprocessing

21 Motor torque.

Create a new plot.

Define the plot using Forces, Motor Torque and Magnitude.

Select the RotaryMotor as the rotational element.

We observe that the top torque magnitude reaches approximate 7 Nmm.

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22 Spring displacement.


Create a new plot. Define the plot using Displacement/Velocity/Acceleration, Linear Displacement and Magnitude. Select the Linear Spring as the simulation element.

The spring expands from 6 to 13 mm. In the setup of the problem, we specified the length of the uncompressed spring as 13 mm.

23 Spring velocity.

Create a new plot.

Define the plot using Displacement/Velocity/Acceleration, Linear Velocity and Magnitude.

Select the Linear Spring as the simulation element.

From the plot, we can see that the spring reaches a top speed of 91 mm/ sec.

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24 Spring force.


Create a new plot. Define the plot using Forces, Reaction Force and Magnitude.

Select the Linear Spring as the simulation element. Click OK.

25 Warning.

Similarly to Lesson 2, we will receive a warning about redundant constraints. Redundant constraints may have significant impact on the mate forces (forces in the mechanical connections, mates, defined by the mates) and will be discussed later in the course. The resulting force obtained for this mechanism is, however, correct.

Click No.

26 Review the plot.

From the plot we can see that the maximum spring force is 1 N. We can see that the spring only pushes the projectile for about 0.1 seconds.

Analysis with Friction (Optional)

In this part we will study the effect of contact friction on the motion of the projectile. We will use the study we have just done and add friction between the projectile and the projectile holder.

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27 Duplicate the study.


Duplicate the existing motion study and name it Larger Friction. 28 Add friction.

Edit the group contact set containing the projectile and projectile holder. Activate Friction with the default values for Steel (Dry).

29 Motion study properties. Under Motion study Properties set Number of Frames to 100.

Click the Advanced Options button and change the Integrator Type to WSTIFF.

Note

Integrators are discussed in detail in Lesson 4.

30 Run the simulation. 31 Animate results.

Animate the results and notice that the projectile would not slide onto the arm due to the added friction.


Summary

92

In this lesson we analyzed a catapult assembly. The main objective was to rotate the arm to the position where a projectile can be loaded, then release the arm and eject the projectile. The following features were used and explained in detail: interference check through the computed time steps, definition of the spring and damper and the specification of the solid body contact with the contact groups. Because the parameters of the contact setup are subject of Lesson 4, this lesson only introduced the procedure to define the contacts with friction. Both static and kinematic friction types were introduced and shown. This assembly also features multiple gear mates.

SolidWorks 2011

Exercise 4 The Bug

Exercise 4: The Bug


In this exercise, we will use a mechanical bug with an oscillating motor to demonstrate the effects of friction on the movement of parts. We will run the study twice, first without friction and then with friction.

This exercise reinforces the following skills:

I

1

Contact Friction on page 84.


Open Bug Assembly file from the Lesson03\Exercises folder.

The assembly consists of a flat plate and a two piece mechanical bug. The intent is to have the movement of the leg move the bug along the plate. There is a Coincident mate between central planes on the Base and Plate to keep the Bug moving down the middle of the Plate.

2

Verify the document units. Click Tools, Options, Document Properties, Units.

Verify that MMGS (millimeter, gram, second) is selected for the Unit system.

3

New study.


4

Add gravity.


5

Add contact.

Using contact groups, add solid body contact between the Plane and the two parts of the bug (Leg and Base). Select Rubber (Dry) for the material.

Clear Friction.

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The Bug

6

Add a motor.


Add an oscillating Rotary Motor to the Leg. Attach the motor to the edge shown and set the motor to move 30 degrees at 5 Hz.

7

Calculate.

Calculate the analysis for 2 seconds.

While the motor oscillates properly, without friction, the bug does not move.

8

Add friction.

Edit the two contacts and select Friction. The dynamic friction coefficient will be set to that of the specified material (Rubber (Dry)). Select static friction and use the default values.

9

Re-calculate.

Run the analysis for 20 seconds.

With friction added, the bug will move along the plate.


Summary

94

In this exercise you analyzed a small assembly called bug. The main objective was to see the effect of the friction model in the contact specification. While in the model without friction the bug assembly does not move, addition of the friction come close to reality – the bugs moves along the base plane.

SolidWorks 2011

Exercise 5 Door Closer

In public buildings such as schools or offices, door closers are often added to non-motorized swing doors to ensure that the doors automatically close after use. To ensure that the doors do not close too quickly and slam, a spring damper is added to the interior of the door closer.

Door Closer Analysis

In this exercise, we will use the Motion Manager to add an internal spring and damper to the door closer. We will then use SolidWorks Motion to plot the effect of the spring and damper on the door's behavior and adjust the parameters to achieve the desired result.


Exercise 5: Door Closer


Procedure

Translational Spring on page 85. Translational Damper on page 87.

Open the existing assembly from the Exercises folder.

1 2

Open an assembly file. Open door from the Lesson03\Exercises folder. Verify the document units.

Click Tools, Options, Document Properties, Units.

Verify that MMGS (millimeter, gram, second) is selected for the Unit system.

3

New study.


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Create a linear spring. Define a Linear Spring between the gas-piston and gas-cylinder.


4

Use the circular edges as indicated in the figure. You must select the edges and not the faces or else the software does not use the center. The spring must be aligned with the cylinder. Use 1 N/mm and 180 mm for the Spring Constant and the Free Length, respectively.

Use 5 N/(mm/s) for the Damping constant. Input appropriate values in the Display PropertyManager. Edges for the spring definition

It may be necessary to change the transparency of the door closer's gas cylinder in order to select the interior parts necessary to define the linear spring.

Note

Click OK.

The damper is used to prevent doors from slamming shut due to the force of the spring.

Note

5

Run the Motion Analysis.

Run the analysis for 40 seconds.

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Exercise 5 Door Closer

6

Plot door velocity.


Create a plot of the door (center of mass) velocity magnitude.

Notice that the door closes too quickly (within approximately 24 seconds) and passes through the door frame before coming to a complete stop.

We do not wish to close the door so quickly. Furthermore, we do not want the door to actually pass through the door frame and open on the opposite side. To solve this, we need to redefine the spring and damper constants.

7

Duplicate the study.

It is possible to simply change the constants in the Motion Study we just created. However, we want to be able to compare results from the two constant settings. Therefore, we will duplicate the initial Motion Study and make modifications to the duplicate study.

Note

8

Redefine the spring with damper.

Increase the Spring Constant value from 1.00 N/mm to 2.00 N/mm. Increase the Damping Constant value from 5.00 N/(mm/s) to 10.00 N/(mm/s).

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9

Calculate the Motion Analysis.


Calculate and plot the door velocity.

10 Compare results.

Clicking either of the motion studies we just completed will enable you to compare the results from both studies. You can observe that in the second study, the door closes slower and comes to a complete stop without actually passing through the frame.

Conclusion

From the data in the two simulations, we can determine the appropriate spring and damper constants for the door to close as desired and without slamming.

Summary

In this exercise you analyzed a door assembly. The main objective was to practice the definition of the spring and damper to model the door closer and to find an optimum spring and damper parameters to close the door slowly without it passing through the frame.

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PR Do E no RE t c LE op AS y E or D di RA st F rib T ut e Lesson 4 Advanced Contact

Objectives


Understand the definition as well as the description of contacts.

I

Use expressions to prescribe the magnitude of forces and motors.

I

Analyze some causes of the incorrect solution or a contact solution failure.

I

Use alternative numerical integrators.

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Contact Forces


The objective of this lesson is to get familiar with the definition of solid body contacts, as well as understanding their limitations and use in SolidWorks Motion. The expressions utilizing various mathematical functions prescribing displacements and other study features will be introduced. Contact force as the latch closes and the force needed to close the latch will be extracted; accuracy of the contact force will be discussed as well.

Case Study: Latching Assembly

In this assembly, an over-center latch is used to hold the Carriage part against a spring.

Problem Description

For the latching mechanism, determine: I I

1

The contact force generated on the Spring Lever and Keeper as the latch closes. The forced needed to close the latch.


Open Full Latch Mechanism. from Lesson04\Case Studies folder.

2


The assembly has several mates however not all components have enough mates to allow the parts to move based on the mechanical motion of the final assembly.

The Carriage part is concentric to the center spindle, but can rotate through the side spindles.

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Three components of the latch, knurled_pin, spring and Series Lever are not restrained laterally.

3

Verify the units.

Verify that the document units are set to MMGS.

4

Create a new Motion study.

Name the study Tessellated geometry and set Type of Study to Motion Analysis.

5

Center the latch.

Add a Coincident mate between the Front planes of the Base and Series Lever.

This is a local mate. If you select the Model tab in the MotionManager, the Series Lever can still move.

We could add another mate to restrict the motion of the J_Spring. In the next few steps we will practice an alternative approach to constrain the motion of free parts.

Fixing Motion with Motors

An alternative approach to additional mates is the addition of a motor. The advantage of such an approach may not be immediately apparent, but we will use it in this motion model.

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One reason for using a motor instead of a mate is that it does not introduce additional constraints to the motion model and helps to reduce the number of the redundant constraints. Redundant constraints will be discussed in Lesson 8: Redundancies.

6

Restrict the linear translation of the latch.

Create a Linear Motor.

Attach the motor to the face shown.

For Motion, select Distance and set it to 0 mm.

Set the Start Time to 0s and the Duration to 3.5s.

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Restrict rotation of the Carriage. Create a Rotary Motor.


7

Attach the motor to the edge shown.

For Motion, select Distance and set it to 0 deg.

Set the Start Time to 0s and the Duration to 3.5s. The simulation will run for 3.5 seconds, so this motor will stop the Carriage from rotating during the entire simulation.

Motor Input and Force Input Types

SolidWorks Motion allows you to set the motor input to a number of different types. We have used Constant Speed, Distance and Data Points in most of our lessons thus far, but Expression, Oscillating and Segments are also available. Expression lets us to define a profile that dictates the motion of the motor with a help of various mathematical functions.

Functional Expressions

You can use functional expressions to define magnitudes of input used in: I

Motors

I

Forces

Functions can depend on time or other system states, such as displacement, velocity, and reaction forces and may be composed of any valid combination of simple constants, operators, parameters, and available supported solver functions such as Step (STEP) and Harmonic (SHF), for example. For a detailed list of functions and its syntax, please refer to the on-line help.

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The following is a list of accepted functions: Function

104

Definition

ABS

Absolute value of (a)

ACOS

Arc cosine of (a)

AINT

Nearest integer whose magnitude is not larger than (a)

ANINT

Nearest whole number to (a)

ASIN

Arc sine of (a)

ATAN

Arc tangent of (a)

ATAN2

Arc tangent of (a1, a2)

COS

Cosine of (a)

COSH

Hyperbolic cosine of (a)

DIM

Positive difference of a1 and a2

EXP

e raised to the power of (a)

LOG

Natural logarithm of (a)

LOG10

Log to base 10 of (a)

MAX

Maximum of a1 and a2

MIN

Minimum of a1 and a2

MOD

Remainder when a1 is divided by a2

SIGN

Transfer sign of a2 to magnitude of a1

SIN

Sine of (a)

SINH

Hyperbolic sine of (a)

SQRT

Square root of a1

STEP

Smoothed step function

TAN

Tangent of (a)

TANH

Hyperbolic tangent of (a)

DTOR

Degrees to radians conversion factor

PI

Ratio of circumference to diameter of a circle

RTOD

Radians to degrees conversion factor

TIME

Current simulation time

IF

Defines a function expression

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There are five types of force functions that can be used to define the force:


Force Functions

I

Constant: Sets a constant value.

I

Step: Defines a step by an Initial Value, Start Time, Final Value,

Final Time.

I

Harmonic: Defines the value by Amplitude, Frequency, Average

and Phase Shift.

I

Segments: Defines the value by combining segments of the most

commonly used functions such as linear, polynomial, half-sine and others.

I

Data Points: Takes the values from a table of data points and

interpolates a spline between the data points.

I

STEP Function

Expression: Defines the value using a formula.

A STEP function prescribes the given quantity (displacement, velocity, acceleration or force magnitude, for example) between two values with a smooth transition. Before and after the transition, the displacement, velocity or acceleration magnitude is constant. For example, consider the illustration at the right where:

d0 = Initial value of displacement d1 = Final value of displacement t0 = Start step time

t1 = Final step time

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Create a rotary motor to drive the latch. Hide the J_Spring.


8

In the Motion Manager, click Motor

.


Under either the Motor Location or Components/Direction fields, select Axis1 of the Series Lever as indicated in the figure. This motor will simulate the action of the hand operating the Series Lever to open and close the latch.

Under Motor Type, in the Motion field, select Expression. The command brings up the Function Builder window.

9

Build motor expression. In the Function Builder, make sure that the Expression button is

selected.

Select Mathematical Functions for the input type and double-click STEP(x,x0,h0,x1,h1) to insert the step function. Modify the functional expression to read STEP(TIME,0,0D,1,90D).

Note

The TIME variable can be typed in or inserted by chancing the input type to Variables and Constants and double-clicking TIME. Complete the expression to its final form:

STEP(TIME,0,0D,1,90D)+STEP(TIME,1.5,0D,3,-90D)

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The Function Builder graph windows will update the plots for displacement, velocity, acceleration and jerk automatically.


Note

Click OK to complete the definition of the expression and close the Function Builder. Click OK to complete the definition of the Motor feature.

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The above expression is a combination of two step functions.

STEP(TIME,0,0D,1,90D)


Note

The first rotates the Series Lever component by 90 degrees between 0 and 1 second and then it keeps the vertical position for 0.5 seconds until the time 1.5 seconds. At time 1.5 seconds, we add the second step function which changes the rotational displacement back to zero between the 1.5 and 3 seconds.

Both functions as well as the combination (the final motion of the Series Lever) are shown in the figures.

1.0 sec

STEP(TIME,1.5,0D,3,-90D)

1.5 sec

Combined

3.0 sec

10 Define Spring and Damper.

We now need to define a spring with a damper which generates tension to keep the latch pulled tight.

In the Motion Manager, click Spring .

Choose a Linear Spring with a spring constant of 10 N/mm, and create the spring at the locations shown in the figure below. Keep the Free Length at its default value. Turn on the linear

Damper and specify a magnitude of 0.10 N/(mm/s).

Notice that the free length of the spring is automatically populated into the Free Length field. Click OK.

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Contacts are defined between two or more bodies or two curves (a contact pair). During the definition of the contact between solid bodies, whatever feature you pick on the parts, the corresponding body will be selected (and used for the contact analysis). During the solve, the software calculates at each frame the bounding boxes of the parts interfere. As soon as it is the case, a finer interference calculation is done between the two bodies and from the center of gravity of the interference volume, an impact force is computed and applied on both bodies.


Contact: Solid Bodies

This procedure is schematically shown in the figure below.

1.

2.

3.

4.

To understand the contact treatment in the SolidWorks Motion, we first need to reiterate the very original assumption of this modulus: all parts participating in the motion simulation are rigid. Contact conditions are used to simulate impact of the two or more colliding parts (which are not rigid in real life). Nearly without exceptions all impacts feature high relative velocity, which result in elasto-plastic deformations with severe localized strains and significant changes in the local geometry (geometry of the contact region). Approximations are therefore necessary. SolidWorks Motion allows for the specification of the contact parameters using two distinct approaches: Impact properties (Impact force model) and Restitution coefficient (Poisson model).

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Restitution coefficient (Poisson model): Poisson model is based on the utilization of the restitution coefficient e is defined in the following relationship:


Poisson Model (Restitution Coefficient)

v′ 2 – v′ 1 = e ( v 1 – v 2 )

Where v1 and v2 are the velocities of the spheres before the impact and v1’ and

v1’

v2’

1 v2’ v2’ are the velocities after the impact. The bounding values of this coefficient are (0;1), where 1 indicates perfectly elastic impact where no energy is lost, while 0 indicates perfectly plastic impact where the parts adhere after the impact and maximum possible energy is lost.

v

The restitution coefficient is geometry dependent and spheres in the above illustration are used for the demonstration purposes only.

Poisson model does not require specification of the damping coefficient (as is the case of the Impact force model, discussed below) and does account correctly for the energy dissipation. The use of this model is therefore recommended if energy dissipation is of the great importance in the simulation. Also, determination of the Poisson model parameters, restitution coefficient e, is more straight forward than in the case of the Impact force model; in many instances, the restitution coefficient can be measured using the standardized methods (see ASTM F1887-98 Standard Test Method for Measuring the Coefficient of Restitution (COR) of Baseballs and Softballs, for example) or found in various tables. This model is not suitable for the persistent impacts (impacts, where contact is developed for a prolonged periods of time); Impact force model should be used instead in these situations.

Impact Force Model

Impact properties (Impact force model): Impact properties in SolidWorks Simulation allow for the calculation of the contact force using the following expression: e

F contact = k ( x 0 – x ) – c • v

where k represent the stiffness of the contact, e is the elastic force exponent, and c is the damping coefficient (cmax) is then the maximum possible damping coefficient). As in the case of the restitution coefficient, these parameters are both material and geometry dependent and can not be apparently found in the material tables. The following sections describe the Impact force model parameters in more detail.

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To correctly determine the stiffness, possible solution is to model the configuration of the contact in SolidWorks Simulation finite element software, apply any force in the direction of the impact and solve for the displacements. Stiffness can then be readily obtained from the force magnitude and the resulting displacements. A figure below demonstrates the impact configuration of two spheres meshed in SolidWorks Simulation software.


Stiffness k

In many instances, the elastic solution can be found in various engineering publications. It is apparent that the computation of the contact stiffness k can be a daunting task and simplifications have to be introduced.

Exponent e

This parameter controls the degree on nonlinearity in the elastic force; e=1 then constitutes a linear elastic force.

Damping Coefficient c and Penetration d

When two objects collide and deform, portion of the kinetic energy is consumed on the plastic deformation, heat and similar phenomena. Approximately, this value can be obtained from the results of the nonlinear dynamic solution (of the above problem of the two spheres, for example) with advanced material models. Utilizing this procedure is, however, unrealistic and simplifications are necessary. It is assumed that the damping coefficient (a measure of the capacity to dissipate energy) increases from zero (at the beginning of the impact) to its maximum value cmax, when certain specified deformation is achieved; we call this deformation value penetration d. For any deformation larger than the penetration d, the damping coefficient is constant and equal to cmax. A typical value for the maximum damping coefficient cmax is 0.1% - 1% of the contact stiffness k.

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It is now apparent that the determination of the above parameters is non-trivial, time consuming and significant simplifications have to be introduced. A corollary of the foregoing is that the solution of the collision characteristics (impact forces, accelerations of the impacting regions and etc.) can only be approximate. Their accurate magnitudes can only be determined by more advanced computational methods, such as nonlinear dynamic solutions using SolidWorks Simulation Premium package, which can be computationally very demanding.


Closing Remarks

It is important to clarify that for the purpose of this section, impact force and the acceleration of the impacting regions terms represent the contact quantities at the onset of the contact where severe deceleration forces are encountered, i.e. impact or collision. The duration of these collisions is typically very short. After a certain time, when the impacting or colliding components are touching and the dynamics aspects of the solution is less important, contact forces are accurate and can be extracted from SolidWorks Motion. This is demonstrated at the end of this lesson. In conclusion, if an important objective of the motion simulation is to obtain the impact quantities (impact force, impact region acceleration etc.), time needs to be invested in the determination of the above parameters, or more advanced analysis type must be carried out. Typically, users are not interested in the accurate impact region results but rather they need to determine the kinematics or dynamics of large systems. Approximate values are then used for the contact characteristics and accurate solution of the system kinematics and dynamics can be carried out efficiently.

To assist users with the impact characteristics, SolidWorks Motion contact library features approximate values for some contact material configurations (note that the geometry is not clearly defined). You may use these values as a base line if the material composition of your parts participating in the contact is similar. However, if more accurate impact solution is needed, correct impact parameters have to be determined.

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11 Define contact between latch and latch keeper. In the Motion Manager, click Contact .

Under Contact Type select Solid Bodies.

Select the latch arm (J_Spring), the latch lever (Lever), and the latch keeper (keeper).

Select Specify Material to allow us to define the impact parameters. Select Steel (Dry) from the list for both materials. Keep the Friction on at its default values.

Here we are trying to make the impact more realistic by simulating two hard metals colliding. As discussed above, the elastic properties of the contact are only approximate. More realistic values would be required for a contact region solution (contact reliable force and acceleration of the contact region).

Click OK.

12 Define gravity.

Define gravity in the negative X direction.

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13 Motion analysis Properties. Verify that Frames per second is set at the default value of 25.

In the Motion Study Properties, set the 3D Contact Resolution slider all the way to the left, to its lowest resolution setting.

Note

The contact resolution parameters are explained in the discussion below.

14 Run the simulation for 3.5 seconds.

Notice that the solution was achieved, but is incorrect. The Spring passes through the other components without developing any of the specified contacts. There can be a few reasons for such behavior:

I I I

The time step of the integrator (solver) is too large, in which case the contact is not even detected. The accuracy setting is too high or too low. The geometrical description of contact is insufficient.

In the present case it is the last one causing the incorrect solution.

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SolidWorks Simulation treats the geometries of the contacting solid bodies in two distinct ways:


Geometrical Description of Contacts

I

Tessellated geometry

The surfaces of the contacting bodies are meshed with the triangular elements to simplify the description. The density of the mesh, i.e. the contact geometry resolution, is controlled with the 3D Contact Resolution parameter in the study properties. Because this description is very efficient, yet typically sufficient to obtain accurate solutions, tessellated geometry is the default choice. Very coarse description may result in inaccurate solution or even failing to develop the contacts. This is also the cause of the solution failure in the present case. I

Precise geometry

If the tessellated geometry description if not sufficient (solution is not sufficient or can not be obtained), Use Precise Contact option can be used instead. Exact description of the bodies’ surfaces is then used. While this is the most accurate description, it can be computationally expensive and should be used with caution. Use this option if your contacting bodies feature complex or point like geometries. Examples of the tessellated geometries at two resolution levels as well as the precise geometry are shown below.

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15 Adjust the Study Properties.

Improve the accuracy of the tessellated data. In the MotionManager click Motion Study Properties .

Move the 3D Contact Resolution slider to the right to a value of 94. Click OK.

16 Run the simulation.

Notice that the computation is noticeably slower.

The simulation will fail and display the following message: The solver failed to converge. Possible causes are:

1. The solver is failing to achieve the specified accuracy. Relax the Accuracy setting in Motion Analysis Properties. 2. If parts in the model are moving quickly, evaluate the Jacobian more often. 3. The mechanism may be getting locked. Start the simulation with a different initial configuration or change you motors to get valid motion. 4. If the failure is happening right at the beginning of the simulation, use a smaller Initial Integrator Step Size. 5. Try to use a stiff solver like ‘WSTIFF’. 6. Try to avoid sharp discontinuities in the model like sudden motion changes, force changes or mate activation/ deactivation. 7. You may have motors with very high speeds. Try to reduce the motor speed. 8. Make sure that only one motor is driving a given component at any time.

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After reviewing the message, there are some possible problems. The first possible problem is item 1, where the solver is failing to achieve the specified accuracy. We will try decreasing the accuracy of the solution.

The second message suggests that if the parts move too quickly, the Jacobian should be evaluated more often. Since the Jacobian setting is already at its maximum value, we will achieve this by also reducing the Maximum Integrator Step Size in the Advanced Options of the Motion Study Properties.

The point where the solution fails is when the latch reaches the over center point because of instability in the solution.

17 Adjust the study properties.

We will reduce the accuracy in order to let the solver handle the over center solution. In the MotionManager click Motion Study Properties .

Reduce the Accuracy to 0.001.

Set the Frames per second to 120 to save more instances of data on the disk. Click Advanced Options and reduce the Maximum Integrator Step Size to 0.001.

Click OK to close the Advanced Motion Analysis Options. Click OK to close the Motion Study Properties.

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Accuracy and the Maximum Integrator Step Size parameters have significant impact on the contact solution and should be used first when contact solution problems occur. To read more about the advanced motion analysis options, integrators and their options, refer to Appendix A: Motion Study Convergence Solutions and Advanced Options.


This time the simulation will run, but it may take several minutes to complete.

19 Animate.

Play the animation and zoom in on the latch mechanism.

Notice that when the latch is closed, there is a small oscillation because all the energy is not being damped. This does not happen in the physical model and is a sign that the damping values used in this simulation can be increased to represent the real situation more closely.

Instability Points

Instability points can be defined as instances where self equilibrated structure does not move, however a small impulse in either direction will result in rapid motion during which the stored elastic energy is rapidly transformed in kinetic energy. Such instances are difficult to overcome numerically. This point is featured in our solution and the solver is expectedly facing difficulties. Also, notice the time required for the solution to complete.

20 Plot contact forces.

Plot the contact forces between the latch and the keeper. Create a new plot.

Define the plot using Forces, Contact Force and Magnitude.

Select the two faces shown.

Tip

118

To make it easier to select the faces, move the timeline to a position where the components are in the position shown.

SolidWorks 2011




We can observe that the graph exhibits significant oscillations with apparently unbounded peaks. This interval (approximately 2.85 to 3.5 seconds) corresponds to the small oscillations observed when the latch mechanism is closed, discussed in Step 19. Each one of those peaks correspond to an impact (or collision) force, magnitude of which depends nearly exclusively on the contact stiffness characteristics. Because these are highly approximate, the peaks of the impacts forces in this interval should be ignored.

Modifying Result Plots

Default plots are created with the X axis showing the duration of the simulation and the Y axis scaled to the maximum value of the variable being plotted. There are times when we want to scale the plots differently.

Introducing: Chart Properties

Almost all aspects of a plot can be modified, from the titles, to background color to axis values and titles.

Where to Find It

I

Right-click on the time in the plot and select Chart Properties.

22 Modify the plot.

Right-click on the X axis of the plot and click Axis Properties. Select the Scale tab.

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Clear End Point and type 3 for the new value of the end point of the X axis.

Use the same procedure to change the Y axis to a maximum value of 50.


We have a very sharp peak at 0.5 seconds (point 1) where the spring hits the carriage. Because the peak is so sharp, and the contact force at this instance qualifies as an impact (or collision) force, we do not know how accurate this data is. We would need accurate contact elasticity parameters and more data points to get better accuracy and understand this impact force.

Just before 2.5 seconds (point 2), the latch reaches the over center point and we see the maximum contact force of about 36 N. This solution is reliable and its dependence on the contact parameters is significantly smaller than at point 1.

2

1

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24 Data points.


Right-click on the curve and click Curve Properties. Select the Marker tab. Select Symbol, then OK.


Move you cursor over the data points and the callout will show that the maximum value is 36 N at 2.42 seconds.

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26 Plot closing torque.


Create a new plot. Define the plot using Forces, Motor Torque and Magnitude.

Select the RotaryMotor that closes the latch as the Simulation

element.

Again, we observe similar peaks after approximately 2.85 seconds. These peaks should be ignored for the reason specified in the previous steps.

27 Modify the plot.

Modify the plot to show the first 3 seconds and a maximum magnitude of 200 N-mm.


We can see a maximum torque of 96 N-mm at about 2.10 seconds.

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Closing Force


Having the torque to rotate the latch, we can determine the force required by dividing the torque by the distance over which the closing force acts.

29 Determine the distance. Click Measure on the Tools

menu.

Measure the distance between the end of the latch and the axis on which the motor acts.

30 Required force.

The required force is:

96 N-mm / 25.04 mm = 3.83 N.

Precise Contact

Using precise contact instead of tessellated geometry should result in a more accurate solution, but with a penalty of additional solution time.

We will now solve the problem again with precise contact and compare the results.

1

Create a new study.

Duplicate the existing study. Right-click the tab for the study Tessellated geometry and click Duplicate. Name the new study Precise geometry.

If we experience sudden changes in forces or motions more accurate solution can be obtained using WSTIFF integrator where the integrator coefficients are adjusted based on the current step size. The discussion below describes all integrator types available in SolidWorks Motion and states when they should be used.

Integrators

A set of coupled differential and algebraic equations (DAE) define the equations of motion of a SolidWorks Motion model. A solution to these equations is obtained by integrating the differential equations in such a way that the algebraic constraint equations are also satisfied at every time step. The speed of the solution depends upon the numerical stiffness of these equations; the stiffer the equations the slower the solution.

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A set of ordinary differential equations are characterized as numerically stiff when there is a wide spread between high and low frequency eigenvalues, with the high-frequency eigenvalues being overdamped. Special efficient integration methods are required to solve numerically stiff differential equations because usual methods for solving differential equations perform poorly and are too slow. The SolidWorks Motion solver offers three stiff integration methods for computing motion.

GSTIFF

The GSTIFF integration method developed by C. W. Gear is a variable order, variable step size integration method. It is the default method used by the SolidWorks Motion solver. The GSTIFF method is a fast and accurate method for computing displacements for a wide range of motion analysis problems. For more information on this integrator see Gear (1971a and 1971b).

WSTIFF

WSTIFF is another variable order, variable step size stiff integrator. Both methods are very similar in formulation and behavior. Both of them use a backwards difference formulation. The only difference is that the coefficients used internally by GSTIFF are calculated assuming a constant step size whereas in WSTIFF, these coefficients are a function of the step size. So if the step size changes suddenly during integration, GSTIFF introduces a small error in the solution whereas WSTIFF can handle it without any loss of accuracy. So the problems run more smoothly in WSTIFF. Sudden step size changes occur whenever there are discontinuous forces, discontinuous motions or abrupt events such as 3D contacts in the model. For more information on WSTIFF integrator see Van Bokhoven (1975).

SI2

The Stabilized Index Two (SI2) method offered in SolidWorks Motion is a modification of the GSTIFF integration method. This method provides better error control over velocity and acceleration terms in the equations of motion.

Provided the motion is sufficiently smooth, SI2 velocity and acceleration results are more accurate than those computed with GSTIFF or WSTIFF, even for motions with high frequency oscillations. SI2 is also more accurate with smaller step sizes, but is significantly slower. For more information see Brenan et. al. (1996) and Gear et. al (1985). All references are listed at the end of this lesson. For more information please see Appendix A.

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Change study Properties. Click Motion Study Properties on the MotionManager toolbar.


2

Select Use Precise Contact.

Click Advanced Option. Select WSTIFF integrator and reduce the Maximum Integrator Step Size to 0.0005. Click OK.

3

Run the simulation for 3.5 seconds.

This simulation will take longer to run and depending on your computer may be around 7 minutes.

4

Contact forces.

Create a plot for the contact forces between the latch and the keeper.

5

Examine the plot.

The plot is similar to the plot we obtained with the tessellated geometry except that the area where significant oscillations and peaks are present.

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6

Modify the plot.


Modify the plot to show 3 seconds of time and a maximum force of 50 N.

Within the area of interest, we essentially have the same plot as obtained with the tessellated data. The maximum force is again 36 N at 2.42 seconds.

We can therefore conclude that we did not need precise geometry to get accurate results.

7

Summary

Save and close the file.

In this lesson we analyzed a closing and latching operation of the latching mechanism. The action of a human hand was simulated with a help of a motor which controlled the motion of the J_Spring component. The objective of this lesson was to extract the closing force and obtain the contact force between the Spring Lever and the Keeper. The assembly, initially not fully defined, was completed with a help of additional mates and zero displacement motors. At some occasions it is beneficial to restrict the motion with a help of a zero displacement motor rather than an additional mate because no extra degree of freedom is removed (motor is a force added to the system, mate is a constraint removing certain degrees of freedom). The magnitude of the motor closing the latch (i.e. simulating the action of a human hand) was expressed with a help of an expression containing mathematical functions. List of all accepted functions was presented; the STEP function was discussed in detail.

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This analysis also involved high velocity contact of solid bodies. Both available impact models, Poisson and Impact force models, were discussed in detail. The accuracy of the contact characteristics (parameters and geometrical description) as well as the accuracy of some of the resulting quantities, namely contact forces and accelerations of the impacting regions, was discussed in detail as well. The study was run using both available geometrical description models: tessellated and precise geometry. Several convergence issues were presented and their solution was shown. Precise geometry study also introduces various numerical integrators available in SolidWorks Motion simulation. Alternative WSTIFF integrator was also used to solve this part of the problem.

Discussion: References

Gear, C.W. (1971a). The Simultaneous Solution of Differential Algebraic Systems. IEEE Transactions on Circuit Theory, CT-18, No. 1, 89-95. Gear, C.W. (1971b). Numerical Initial Value Problems in Ordinary Differential Equations. New Jersey, Prentice-Hall. Van Bokhoven, W.M.G. (1975, February). Linear Implicit Differentiation Formulas of Variable Step and Order. IEEE Transactions on Circuits and Systems, 22 (2).

Brenan, K.E., Campbell, S.I. and Perzold, L.R. (1996). Numerical Solution of Initial Value Problems in Differential-Algebraic Equations, Classics in Applied Mathematics. ISBN: 0-89871-353-6 (pkb.).

Gear, C.W., Leimkuhler, B. and Gupta, G.K. Automatic Integration of Euler-Lagrange Equations with Constants. Journal of Computation and Applied Mathematics, 12 & 13, pp. 79-90, North-Holland: 1985.

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PR Do E no RE t c LE op AS y E or D di RA st F rib T ut e Lesson 4

Advanced Contact

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SolidWorks 2011

Exercise 6 Hatchback

Exercise 6: Hatchback


A number of contemporary car models are designed as hatchbacks. Similar to station wagons but smaller in size, hatchback cars allow for cargo to be loaded into the back of the car, and typically the rear seat folds down to increase the cargo area.

This exercise reinforces the following skills: I I I I

Project Description

Contact Forces on page 100. Contact: Solid Bodies on page 109. Motor Input and Force Input Types on page 103. Modifying Result Plots on page 119.

Key to the hatchback car's functionality is the hatchback door itself. These doors are attached to the car via an upward swinging hinge and are both supported and assisted by gas pistons. To achieve the same result in SolidWorks Motion, we will apply a motor to the assembly. Determine the force exerted by the gas pistons on the door.

1


Open hatchback from the Lesson04\Exercises folder.

2

Verify units.


3

Create a new Motion Study. Name the new study Hatchback Steel and set the Type of Study to Motion Analysis.

We will be using reference points. In order to use reference points, make sure that all components are resolved.

Note

4

Apply Gravity to the assembly.

Apply gravity in the negative Y direction.

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Hatchback

5

Apply Force to the assembly.


The pressure in the piston will be simulated using Action only force acting on the piston. (It is therefore assumed that the piston force remains constant as the piston opens). We will begin with the force definition of the Left_Cylinder.

Apply a 420 N Linear, Action only force as shown below. Make sure that the force is applied at the indicated point and its direction is referenced with respect to the cylinder. This way the force will be always directed along the axis of the rotating piston.

Under Force Function, make sure the Constant button is selected and enter 420 N in the F1 field.

Click OK.

Note

130

Make sure that the force is oriented as shown in the figure.

SolidWorks 2011


6

Repeat.


Repeat step 5 for the Right_Cylinder.

Mass Properties

On occasions, mass properties of the SolidWorks parts may be modified. This should, however, be an exception rather than a frequent task as most of the SolidWorks parts reflect the design intent and their mass properties are computed automatically. When mass properties are discreetly assigned, they override the properties associated with the material specifically applied to the component.

7

Adjust Mass Properties of Lid-1. Under Tools, select Mass Properties. The Mass Properties window

will appear.

In the Selected items field, right-click and select Clear Selections.

In the assembly view window, click Lid-1 as shown below. Select Assigned mass properties.

In the Mass field, enter 13000 grams.

Click OK.

8

Adjust duration of the simulation.

Set the study duration to 2 second.

9

Set the study properties. Set the Motion Analysis properties to 100 Frames per second.

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Hatchback

10 Contacts - left side.


Define contact conditions between the Left_Cylinder-1 and Left_Piston-1. For both materials select Steel (Dry).

Keep all other contact options at their default values.

11 Contacts - right side.

Repeat for the opposite side of the assembly, creating contacts for Right_Cylinder-1 and Right_Piston-1.


In the SolidWorks Motion Manager, click Calculate

.

The hatchback assembly will open correctly.

13 Graph the cylinder position. Create a Y Component plot of the Center of Mass Position of the Left_Cylinder-1. 14 Examine the plot.

Notice that because the plot is created by default in the global coordinate system, the initial Y value is -63 mm and the final Y value is 289 mm. We can also observe that the initial collision occurs at approximately 0.83 seconds, while the assembly has completely opened and stopped moving at approximately 1.1 second.

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15 Contact force.


As stated in the lesson discussions, with generic contact parameters the contact force solution will be approximate.

Create a new plot for the magnitude of the contact force between the Piston and the Cylinder (you may use either of the two pairs since the assembly is symmetrical).


The two spikes in the graph indicate the initial and the secondary collision between the piston and the cylinder.

The force magnitudes (22,503 N and 4,210 N at the two peaks) represent the contact forces at the instant of the collision and have to be understood as approximate due to the quality of the contact input characteristics. We can further observe that as the motion ceases, the contact force reaches a constant static value. To determine the contact value the limits of the graph need to be modified.

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Hatchback

17 Modify the plot format.


Modify the plot of the contact force so that the static value can be read conveniently.

We can observe that, when the motion ceases, the static equilibrium is reached and the contact force is at that stage approximately 367 N. The accuracy of the static solution is not affected by the selection of the impact model, nor by the selection of the impact model parameters. We can therefore conclude that the static solution is accurate.

It was mentioned already a couple of times that the contact Elastic properties significantly effect the resulting impact contact forces and accelerations of the impacting region. In most scenarios only approximate characteristics are available and as a consequence the resulting impact forces as well as kinematic characteristics of the impacting objects are approximate. We will now modify the contact elastic properties and study their effect on the solution.

18 Copy study.

Copy the study Hatchback Steel into a new study called Hatchback Aluminum.

19 Change contact material.

Change the contact material for both contacts to Aluminum (Dry).

20 Run the study.

21 The cylinder position.

Create an identical plot of the Y Component of the Center of Matt Position.

While the minimum and maximum positions are identical and the general shape of the graph is very similar, notice that the assembly stops moving at somewhat later time of 1.15 seconds (as opposed to 1.1 seconds when the material specification was Steel (Dry)).

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Because the contact elastic properties have effect on the accelerations of the impacting regions as well as on the amount of the energy dissipated during the collision, the resulting velocities after the initial impact will be different. The assembly will, therefore, cease to move at a different (now later) time.

22 Contact force.

Create an identical plot of the contact force.

The maximums at the two peaks are again different, 13,412 N and 2,727 N, respectively. But, the absolute values can not be relied on.

As expected, however, the static force magnitude after the motion ceases is nearly identical to the solution obtained in the previous study, 367 N.

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Hatchback

23 New study (optional).


Repeat the above procedure and change the contact properties to Rubber (Dry).

Examine the results and notice that this is an unrealistic scenario. You will have to extend the length of the study to 20 seconds to reach a point where the motion ceases. The Lid will bounce many times before eventually coming to rest.

The static value of the contact force, 376 N, is again very close to the previous solutions.


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In this exercise we analyzed the opening of a vehicle hatchback. While in reality the two pistons may generate non-constant and non-linear force, we simplified the simulation and applied a constant piston force only. While the force magnitude and its dependence on the piston position can, of course, be modified in a complex way, it was not the objective of this exercise.


Summary

The final phase of the hatchback opening is when the piston contacts the back side of the cylinder. We used solid body contact and studied the hatchback opening characteristics (such as opening time, contact forces, etc.) as functions of the contact specifications. It was found that with various specifications the hatchback stops moving at different times. The last study went to an extreme when we used unrealistic contact specifications: rubber on rubber. In this situation the hatchback exhibited large repeated oscillations which would be undesirable.

The contact force magnitudes were also analyzed. While the peak magnitudes coinciding with the short duration collisions are not reliable since they require very precise contact characteristics, the static contact force after the motion ceases is accurate. This was demonstrated by a very similar result obtained from all three simulation.

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Exercise 7

SolidWorks 2011

Conveyor Belt (No Friction)

Exercise 7: Conveyor Belt (No Friction)


A conveyor, consisting of segmented panels, is driven around a track.


Project Description

Functional Expressions on page 103. Modifying Result Plots on page 119.

Our goal is to drive the conveyor at a speed of 0.62 m/sec using a force that is controlled by a function. In the first part of the exercise we will move the belt with a controlled force. In the second part the force will be replaced with a motion on a path.

1


Open Conveyor_Belt from the Lesson04\Exercises folder.

2

Review the assembly.

The assembly has all the mates needed for the conveyor belt to move correctly. There are many CAM mates that create the tangency conditions between the wheels and the closed loop conveyor paths.

SolidWorks Motion also supports other SolidWorks Advanced mates like the Gear mates and Limit mates.

Note

3

Verify units.

Verify that the document units are set to MKS (meter, kilogram,

second).

4


Create a new motion study and name it Conveyor.

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Exercise 7 Conveyor Belt (No Friction)

5

Apply a force.


We will start by creating a force on Plate-1 simulating a force applied to push the plates on the conveyor.

Apply a 100 N Constant, Action Only, Linear Force on the Plate-1 indicated in the

following figure. Make sure the force is oriented as shown and its direction is referenced with respect to the same plate (i.e. the direction of the force must change as the plate moves around the guides).

6


Set the Number of Frames to 100 and select the WSTIFF integrator.

This problem can be conveniently solved using the faster GSTIFF integrator as well. The WSTIFF integrator is used here only for practice.

Note

7

Run the simulation.

Run the simulation for 5 seconds.

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Plot the velocity magnitude of the Plate-1.


8

The velocity of the conveyor plates is increasing linearly. We now want to maintain the conveyor plate at a constant speed of 0.62 meter/ second.

What are we going to do next?

We are going to change the definition of the force so that it varies as a function of the difference in the current conveyor velocity from our desired conveyor speed. Based on the speed difference, the magnitude and the direction of the force changes to accelerate or decelerate the conveyor based on the following expression: Force = Gain * (Desired Speed - Current Speed) = Gain * (0.62 Current Speed)

When the current speed is less that the desired speed, a positive force is applied to accelerate the conveyor. If the current speed is greater than the desired speed, then a negative force is applied to decelerate the conveyor. The gain value controls the force applied to accelerate or decelerate the conveyor.

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Modify the force.


9

Edit the force and change it’s magnitude from a constant of 100 N to the following functional expression: 100*(0.62-{Velocity1})

Note

To get the {Velocity1} feature into the Expression field, double-click the Velocity1 feature in the Motion Study Results list.


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The plot shows that the velocity is being held to 0.62 m/sec but it is getting there too slowly. We will increase the gain to shorten the time it takes to reach the target speed.

12 Modify the force.

Edit the force and change the equation to: 500*(0.62-{Velocity1})

13 Run the simulation. 14 Examine the plot.

This time, the conveyor reaches the target speed by 1 second and it then holds there as the force varies. The variation of the speed is, however, significant and not acceptable for the manufacturing operation. We can make it smoother by increasing the gain further.

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Modify the force again so that the gain is 5000. Then re-run the analysis.


This time the plot is much smoother.

17 Plot input force.

It can be seen that the force initial magnitude is very high. To accelerate the conveyor from its initially zero velocity. As the conveyor reaches the desired velocity of 0.62 m/sec, the force magnitude tends to reduce to zero. Alternatively, instead of using the force input the conveyor constant velocity can be ensured by using a path mate motion. This is shown in the next part of this exercise.

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Exercise 7

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Path Mate Motor feature prescribed motion of a point along a path. It is required to create a PathMate in SolidWorks prior to defining the Path Mate Motor in SolidWorks Motion.


Path Mate Motor

The options of the PathMate in SolidWorks controls the Pitch, Yaw and Roll rotational degrees of freedom of the point along the path.

18 PathMate.

Choose one of the wheels on the plate where the driving force is applied. Delete the CamMate.

Delete CamMate

In SolidWorks feature tree, unsuppress the Sketch1 feature.

Sketch1

Define a new PathMate between the center point of the wheel and the path defined by Sketch1. Keep all PathMate constraints at their default values of Free.

Note

144

The Pathmate constraints are set to Free because the mechanism is fully constraints due the presence of the remaining CamMates features.

SolidWorks 2011


19 Path Mate Motor.


Define Path Mate Motor.

For the PathMate field select the PathMate defined in the previous step.

Motor orientation

Make sure that the orientation of the motion is the same as is the orientation of the force used to drive the belt.

Select Constant Speed and enter 0.62m/s. Click OK.

20 Suppress force.

Suppress the force feature. This feature is not needed because the

motion is driven by the motor.

21 Run the simulation. 22 Velocity plot.

This time the plot is much smoother.

Notice the oscillatory variation of the velocity. With the constant velocity of 0.62m/s prescribed in step 19, we would expect the resulting velocity profile of the plate to remain constant as well. Can you explain these oscillations?


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Exercise 8

SolidWorks 2011

Conveyor Belt (With Friction)

Exercise 8: Conveyor Belt (With Friction)


This is the same conveyor used in the previous exercise.

We will run the same study, but this time we will include friction and examine the changes in the forces and velocities.

Project Description

Our goal is to drive the conveyor at a speed of 0.62 m/sec using a force that is controlled by a function. This exercise reinforces the following skills: I I I

1

2

Contact Forces on page 100. Functional Expressions on page 103. Precise Contact on page 123.

Open an assembly file. Open Conveyor_Belt from the Lesson04\Exercises\ Conveyor Belt\with contact folder. Review the assembly.

Examine the mates.

The first coincident mate keeps the top of one of the plate pins in the same plane as the end plate of the conveyor. This prevents the side to side motion of the conveyer plates.

There are groups of concentric and coincident mates that hold adjacent plates together. The remaining mates are the CAM mates that create the tangency conditions between the wheels and the closed loop conveyor paths. Instead of using the CAM mates, we will use solid body contact. Suppress all the CAM mates.

3

Verify units.

Verify that the document units are MKS.

4


Create a new motion study and name it Solid body contact.

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Exercise 8 Conveyor Belt (With Friction)

5

Add Contacts.


Add a solid body contact between each wheel and the side plate on the left side of the model (the same side where the CAM mates were applied). There will be 12 contact sets.

Select Steel (Greasy) for the material and keep the default values for both the static and kinematic friction.

We only create contacts on the left side of the assembly. The contacts could be defined on the opposite side to model the problem more realistically. However, similarly to the previous study with CAM mates (where the mates were defined on one side only to avoid redundancies), we will keep the contacts on one side only. The final resultant contact forces will have to be then divided by two.

Note

Redundancies will be covered in a later lesson.

6

Add Gravity.


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Exercise 8

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7

Add a driving force.


Add a constant force of 5,000 N to plate1<3> just as we did in the previous exercise. We need to add the constant force first so that we can generate a velocity graph that can be used in the functional expression to control the force. We need a relatively large force to get the belt moving. In the previous exercise, any force would move the conveyor as there was no friction.

8

Local mate.

There are two instances of a part called plate_adjust_p1 on the bottom of the conveyor that are used to tension the belt.

Add a Lock mate to keep these two parts in the same position relative to each other.

9


This study will be very sensitive to contact accuracy, so we need to use Precise Contact. Also set the Frames per second to 100 and select the WSTIFF integrator.

10 Run the study.


11 Play the animation.

Play the animation at 25% speed to see how the belt moves.

12 Plot the results.

Plot the velocity magnitude of the plate1.The velocity does not increase linearly as before since the friction forces act against the input force and the motion with the contact is more complex.

Note

148

To speed up the work you may interrupt the computations at any time. This run is only important to enable us to define a velocity plot used in the following expression.

SolidWorks 2011


13 Edit the force.


Change the force to a function of velocity using the equation: 5000*(0.62-{Velocity1})

14 Run the study.

15 Examine the velocity plot.

The velocity approaches 0.62, but the variation is too large.

16 Plot the Force magnitude. Create a new plot using Forces, Reaction Force, Magnitude and then select Force1 as the Simulation element.

We will get a warning message about redundant constraints, click Yes.

Contrary to the previous study, the force does not go to zero because of the friction.

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17 Edit the plot.


Make the maximum Y value 1000 so that we can see the oscillations easier.

18 Increase the force.

Edit the force and increase the gain to 50,000. 50000*(0.62-{Velocity1})

19 Run the study.

20 Examine the plots.

The velocity is now nearly constant at 0.62 m/sec.

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The force variation is similar.


Summary

In this exercise we analyzed the motion of the conveyor belt on the fixed guide plates.

The belt was accelerated by an action only force applied on one of the plates. The magnitude of the force was controlled with the help of an expression which included the velocity of the belt as a variable. This way, the input force was directly dependent on the resulting velocity.

Two approaches were shown: the first study simulated the tangential contact between the wheels and the guides using the CAM mates. To reduce the redundancies and to simplify the solution only mates on one side were included. Therefore, the resulting contact forces would have to be reduced by half. To add more realism to the simulation, the second study replaced the CAM mates with the solid body contact. While this approach allows us to add friction, the computation took longer. When the desired speed of 0.62 m/sec was achieved the input force never came to zero in order to overcome the opposing friction forces.

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Exercise 8

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PR Do E no RE t c LE op AS y E or D di RA st F rib T ut e Lesson 5 Curve to Curve Contact

Objectives


Understand the definition as well as the description of contacts.

I

Use expressions to prescribe the magnitude of forces and motors.

I

Analyze some causes of the incorrect solution or a contact solution failure.

I

Use alternative numerical integrators.

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Lesson 5

SolidWorks 2011

Curve to Curve Contact

Contact Forces


The objective of this lesson is to get familiar with the definition of curve to curve contact. This lesson builds on the knowledge acquired in the previous lesson where solid to solid contact was treated in detail.

Case Study: Geneva Mechanism

The geneva mechanism was traditionally used in the movie projectors where each frame is exposed for a certain fraction of a second. The mechanism allows for the transformation of the continuous rotation of the drive wheel into the intermittent rotation of the driven wheel.

Problem Description

Driven wheel

Driving wheel

For the geneva mechanism, determine: I I

1 2

The contact force generated on the driving wheel. Time variation of the driven wheel rotation.

Open an assembly file. Open stargeneva from Lesson05\Case Studies folder. Examine the assembly.

Both the driving wheel and the driven wheel are connected to the base with two hinge mates. There is no mate relation between the wheels - this interaction will be handled with the help of the curve to curve contact.

3

Verify the units.


4


Name the study curve to curve contact.


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SolidWorks 2011

Lesson 5 Curve to Curve Contact



Curve to curve contact can be defined between two curves, either of which can form a closed loop or remain open. The curve geometry is approximated by a discrete set of points. It is possible to specify whether the contact is persistent, i.e. curves are not allowed to separate, or intermittent, where separation may occur. Curve to curve contact supports friction and two contact models, Restitution coefficient and Impact force, described in detail in the preceding lesson.

Introducing: Curve to Curve Contact

Contact is used to define the way two curves interact. Within the contact definition, we can control the friction and the elastic properties between the bodies.

Where to Find It

I

5

Click Contact on the MotionManager toolbar. Under Contact Type click Curves.

Driven wheel and driving wheel contact #1. Specify an intermittent curve to curve contact between the driven wheel and the left knob of the driving wheel.

In the Contact PropertyManager, select Curves under the Contact Type.

Under Selections click the Selection Manager button and set it to Standard Selection. Select the indicated curve on the driving wheel as Curve 1.

Switch the Selection Manger to Select Group setting.

Select the indicated curve and the click the

Tangent button. The tangent closed loop defining the edge of the driven wheel will be

populated.

Click OK in the Selection Manager to end the selection process. The second curve will be constructed and Closed Group will be shown as Curve 2.

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Under Materials specify Steel (Dry) for both components. Make sure that the Friction with the default values is used.

Make sure that the orientation of the outward normal for the Closed Group in Curve 2 field is as indicated in the figure above. The orientation of the curve can be changed with the Outward Normal . Direction button Click OK to close the Contact PropertyManager.

The Curves always touch button must remain unchecked, because the two curves come into an intermittent contact only.

Note

6

Driven wheel and driving wheel contact #2. Following the same procedure specify an intermittent curve to curve contact between the indicated curves.

Use the same contact parameters as those in the preceding step.

Note

156

Make sure that the curves are property oriented.

SolidWorks 2011


Driven wheel and driving wheel contact #3. Continue with the definition of the intermittent curve to curve contact between the indicated segment of the driven wheel and the closed loop curve of the driving wheel.


7

Use the same contact specifications as those used in step 5.

Make sure that the curves are property oriented.

Note

8

Driven wheel and driving wheel contact #4 to #6. Define the intermittent curve to curve contacts between the remaining three segments of the driven wheel and the closed loop curve of the driving wheel.

The last four contact sets can be defined in various ways, for example in a single definition between two closed loop curves. While this is also a valid contact definition, it is preferable to define contacts with simple curves rather than one very complex curve.

Note

9

Driving motor. Apply a 360 deg/sec driving Rotary Motor to the driving wheel.

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10 Motion analysis Properties. Set the Frames per second to 100.

Important!

The 3D Contact Resolution and Use Precise Contact options are only applicable to the contact between solid bodies.

11 Run the simulation for 4.235 seconds. 12 Plot contact forces.

Plot the contact forces between the driven wheel and the left knob of the driving wheel. Define the plot using Forces, Contact Force and Magnitude.

For the selection field, select the Curve Contact1 item from the Motion FeatureManager. Click OK.

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Similarly to the contact force results in solid bodies, the contact force for the curve to curve contact exhibits sharp peaks due to the contact stiffness approximations and they have to be ignored. Nonlinear dynamics solutions would be required for the accurate collision forces. Also, changing the limits for the graph will not yield meaningful static results for the contact force (as was the case in Lesson 4, where static contact force existed). Try to answer why.

13 Rotation of the driven wheel.

Plot the variation of the rotation of the driven wheel in time.

The above plot indicates that the output rotation rate for the driven wheel is 90 deg/sec, or 360 deg in 4 seconds.

Solid bodies vs. curve to curve contact

Lesson 4 and the current lesson introduced the two contact types available in SolidWorks Motion: solid bodies contact and curve to curve contact. The question may arise as to which contact definition to use when. Mots of the contact situations are best resolved with the solid bodies contact type, especially when the solution of the system depends on external forces acting on the objects (dynamic systems). If the contact path can be described using closed loop or open curves, curve to curve contact type may be used. However, if the curves used in the contact definitions encircle the entire objects, and especially if they are very complex, solid bodies contact may still be favored. Therefore, the above problem of the stargeneva mechanism could be solved with the solid bodies contact definition instead.

In the last part of this lesson we will solve this assembly again with the solid bodies contact.

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Lesson 5

SolidWorks 2011


Solid Bodies Contact Solution


In the second part of this lesson the same assembly will be solved with the solid bodies contact. 1

Solve problem with solid bodies contact.

Solve the simulation again with the solid bodies contact. Specify the appropriate geometry description for this contact solution. When finished, compare the curve to curve and solid bodies contact solutions.

2

Summary


In this lesson we analyzed a stargeneva mechanism. This mechanism was traditionally used in movie projectors where each frame is exposed for a certain fraction of a second. The mechanism allows for the transformation of continuous rotation of the drive wheel into intermittent rotation of the driven wheel. To achieve this transformation, a contact between the two wheels must be specified. The analysis began with the introduction of the curve to curve contact that was subsequently defined between the various parts of the assembly. Curve to curve contact allows for the selection of both open and closed loop curves and features the same contact models as the solid bodies contact: Restitution coefficient and Impact force, both described in detail in Lesson 4.

The solution of the contact force and the time variation of the driven wheel rotation were plotted and discussed. It was demonstrated and discussed that the contact force solution in the curve to curve contacts features sharp peaks corresponding to the collision instances. While the high magnitudes of the collision instances should be ignored, the static (or non-collision) contact forces can always be extracted. Finally, the differences and the proper usage of the solid bodies contact and curve to curve contact was discussed. The problem was solved once more with the solid bodies contact and the solution were compared.

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SolidWorks 2011

Exercise 9 Conveyor Belt (Curve to curve contact with friction)

Exercise 9: Conveyor Belt (Curve to curve contact with friction)


This is the same conveyor used in Exercise 7: Conveyor Belt (No Friction) on page 138 and Exercise 8: Conveyor Belt (With Friction) on page 146 where solid body contact was used.

This exercise reinforces the following skills: I I I

Project Description

Contact Forces on page 100. Functional Expressions on page 103. Precise Contact on page 123.

In this exercise the solid body contact will be replaces with the curve to curve contact and the results will be compared. Our goal is to drive the conveyor at a speed of 0.62 m/sec using a force that is controlled by a function.

1


Open Conveyor_Belt from the Lesson05\Exercises folder.

This assembly contains completed file set from the Exercise 8: Conveyor Belt (With Friction) on page 146, where the solid body contact was used to simulate the CAM tangent conditions.

2

Motion study.

Duplicate the Solid body contact study into a new study named curve to curve contact.

3 4

Delete all solid body contacts.

Curve to curve contacts.

Add a curve to curve contact between the edge curve of each wheel and the edge curve of the conveyor_path on the left side of the model (the same side where the solid body contacts were deleted in the preceding step). Again, there will be 12 contact sets.

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Exercise 9

SolidWorks 2011

Conveyor Belt (Curve to curve contact with friction)


Select Steel (Greasy) for the material and keep the default values for both the static and kinematic friction. Check Curves always touch check box.

The Outward Normal Direction is not shown because Curves always touch check box was activated.

Note

5

Motion Study Properties. Set the Frames per second to 100 and select the GSTIFF integrator from the Advanced Options.

Check the Replace redundant mates with bushings checkbox.

The Replace redundant mates with bushings option is used in this model due to the complex redundancies situation. Both this option as well as redundancies are subject of Lesson 8.

Note

6

Run the study.


7

Play the animation.

Play the animation at 25% speed to see how the belt moves.

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Exercise 9 Conveyor Belt (Curve to curve contact with friction)

8

Plot the results.


Plot the velocity magnitude of the plate1.The velocity does not increase linearly as before since the friction forces act against the input force and the motion with the contact is more complex.

Comparing the above variation of the velocity with the results of Exercise 8: Conveyor Belt (With Friction) on page 146, we can conclude that they are very similar. Both show nearly constant velocity of 0.62 m/sec.

9

Summary


In this exercise we analyzed the motion of the conveyor belt on the fixed guide plates. While in Exercise 8: Conveyor Belt (With Friction) on page 146 this problem was solved with the help of the solid bodies contact, in the this exercise curve to curve contact was used instead. It was shown that both approaches provide similar results.

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Exercise 9

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Conveyor Belt (Curve to curve contact with friction)

164

PR Do E no RE t c LE op AS y E or D di RA st F rib T ut e Lesson 6 CAM Synthesis

Objectives


Use of a spline curve to control the motor.

I

Create a trace path of a point to get the CAM profile.

I

Create a SolidWorks part with this CAM profile.

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Lesson 6

SolidWorks 2011

CAM Synthesis

SolidWorks Motion can be used to create CAM profiles based on tabular data or input function such as STEP function. We can work backward by driving the follower with the desired motion, then use the motion of the follower to create the CAM profile.


CAMs

Case Study: CAM Synthesis

In this case study we will generate a CAM profile based on an input follower displacement from a data set.

Problem Description

Create a CAM that will move the follower based on the following curve.

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Lesson 6 CAM Synthesis


To create the CAM, we will follow the steps below: Define the motion of the follower.


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This can be done from a table of values and drive the follower through a motor.

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Create a Trace Path.

The trace path will be in the exact shape of the CAM surface.

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Export the curve to SolidWorks as a sketch.

The trace path can be imported into SolidWorks as a curve and used in a sketch.

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Extrude the sketch to create the CAM.

Open the assembly file. Cam Synthesis.sldasm. Open Cam Synthesis located in the Lesson06\ Case Studies folder.

The assembly consists of a undefined CAM and a follower.

2


Verify that the units are set to MMGS (millimeter, gram,

second).

3

Generating a CAM Profile


To generate a CAM profile, the follower motion is prescribed to the path profile while the CAM component rotates 360°. Both are specified in the next two steps.

4

Define a motor to drive the CAM.

Add a rotary motor to drive the CAM shaft at a constant speed of 120 deg/sec. This will rotate the CAM once every 3 seconds.

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CAM Synthesis

Examine the profile data. Open the file CAM Input.xls located in the Lesson06\Case Studies\CAM Synthesis folder.


5

Part of the file is shown at right. It consists of X and Y coordinates for the position of the CAM follower. The file also contains a plot of the CAM profile based on the tabular data. Review it, then close the file.

6

Define a motor to drive the Follower. Add a linear motor to the top face of the Follower_Guide. Make sure the direction is as shown in the image.

Select Data Points to open the Function Builder window. Select Displacement for Value (y), Time for Independent variable (x) and Akima for the Interpolation type.

Click Import Data and select the CAM Input.csv file. This file contains just the X and Y data that was in the Excel file.

7

Add gravity.


8

Simulation Study Properties.

Change the study properties to save 100 Frames per second.

9

Run the study.


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SolidWorks Motion allows you to graphically display the path that any point on a moving part follows. This is called a Trace Path and it was already used once in Exercise 3: Trace Path on page 71. In this lesson we will use it to generate a profile of a CAM.


Trace Path

You can select the part that will be used to generate the trace curve by selecting it in the box labelled Select Trace Point Component. This field enables you to select a face, edge or a vertex to define a point generating the trace.

Optionally, you can select a reference component that defines a reference frame for the trace path. The default reference frame is the global reference frame defined by the global coordinate system.

Where to Find It

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Create a new plot and select Displacement/Velocity/ Acceleration, then Trace Path.

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10 Create a trace path which will define the CAM profile. Click Results and Plots on the MotionManager toolbar.

Select Displacement/Velocity/Acceleration, then Trace Path.

Select the vertex on the Follower1 to define the CAM profile and the surface of the cam to define the reference component.

Leave the Component to define XYZ directions empty. Click OK. to show the trace.

Notice how a CAM profile is generated. We will now copy this trace path curve directly onto the SolidWorks part from SolidWorks Motion.

Exporting Trace Path Curves

Now that we have the shape of the CAM, we can use this path in SolidWorks to create the CAM itself. The trace path curve can be exported to a SolidWorks part.

Introducing: Create Curve From Trace Path

The Trace Path curve can be used to create a curve in a SolidWorks part to create geometry. This can be done in two ways: I

Create curve from path in reference part.

A part already exists, so the trace path curve can be imported to the existing part.

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Create curve from path in new part.

If a part has not been created, it can be done directly using this command.

Where to Find It

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In the MotionStudy tree, right-click a Trace Path plot under the Results folder and select Create curve from trace path.

SolidWorks 2011



11 Copy trace path curve to SolidWorks Part. Right-click the Trace Path plot under the Results folder and click Create curve from trace path, then Create curve from path in reference part.

12 Open the CAM part.

Open the CAM part in its own window.

The curve has been inserted into the part as a new feature.

13 Extrude the profile.

Create a new sketch on the Front plane. In the SolidWorks FeatureManager, select Curve1.

Click Convert Entities on the Sketch toolbar to project the curve onto the sketch plane.

Also select the outer cylindrical edge of the CAM profile and use Convert Entities to project this edge into the active sketch.

Extrude the sketch to a mid-plane depth of 50 mm.

Make sure that the Merge results checkbox is unchecked.

14 Save and close the part.

Return to the main assembly.

In the last part of this lesson, we will re-run the simulation with the 3D Contact and verify that the cam profile was generated correctly.

We will need to create solid body contact between the follower and the cam, and drive the motion with the rotary motor on the cam and turn off the linear motor on the follower.

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CAM Synthesis


15 Add solid body contact. Add Solid Bodies Contact between the follower and the cam.

Specify Steel (Greasy) for both materials. Clear Friction.

16 Remove the motion for the follower. Right-click LinearMotor1 and click Suppress. 17 Add gravity.


18 Motion Study properties. In the Motion Study Properties, select Use Precise Contact.

Whenever we have point contact, we should use precise contact.


Notice how the follower traverses vertically based on the CAM profile.

20 Examine the motion. Change to the Back view.

The image is at 1.7 seconds. Notice that the follower is not touching the cam. This separation is the result of the momentum of the follower. Just prior to this time, the follower was driven up by cam. The cam profile requires the follower to change direction rapidly, however the only thing holding the follower in contact is gravity.

Is this a problem? Probably not as the follower will eventually have additional components on top of it to force contact with the cam.

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21 Plot the vertical displacement of the follower-1. Create a plot of the Y Component displacement of the center of mass of the follower and compare it to the plot in the Excel file. For clarity,

the Excel plot has been inverted. Both plots have the same shape.

Cycle based motion

In machine design the independent variable TIME is often not the most convenient choice. It may be more comfortable to design all tasks in terms of one master cycle. Typically, the duration of the master cycle is set to 360 degrees.

Introducing: Cycle Based Motion

Cycle based motion allows user to easily modify the duration of the action, or productivity, in the machine design.

Where to Find It

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In the FunctionBuilder window set the input type to Variables and

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Constants and select CycleAngle.

The duration of the cycle is then specified in the Motion Study Properties.

22 Edit rotary motor. Under Motor Type select Segments to open the Function Builder.

In the Function Builder dialog, make sure that the Segments button is selected. Keep Displacements for Value (y) and set the Independent variable (x) to Cycle Angle. Add a row and enter 0deg and 360deg cycle angle for the Start X and End X columns, respectively. Enter 360deg for the final value of the rotational displacement.

Note

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Make sure that the Initial value for the rotational displacement is 0deg.

SolidWorks 2011

Lesson 6


CAM Synthesis

The four graphs indicate the linear increase of the displacement, constant velocity and zero acceleration and jerk. The 360 degree rotation in 360 degree cycle angle indicates one revolution per output cycle.

Note

The duration of the cycle angle (or output cycle) will be specified in the next step. Click OK to close the Function Builder.

Click OK to save the new definition of the Motor.

23 Study properties. Set the Cycle time to 3s.


Notice that the resulting motion of the follower-1 is the same as in the step 21. This is to be expected as both simulation are identical, the former solved using time as independent variable, the later one then using cycle angle as independent variable.

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25 Analyze results.


Notice that the resulting motion of follower-1 is the same as in the step 21. This is to be expected as both simulations are identical the definition of the independent variable. The former one solved the simulation using time as the independent variable, the later one then used cycle angle.

26 Adjust the cycle time to 1.5s. 27 Run the simulation. 28 Analyze results.

Notice the cam now rotates twice in 3 seconds (study duration).

However, reviewing the trace path we see that follower-1 detaches from the cam this is unacceptable. The cycle time of 1.5 seconds is therefore too small for this mechanism.

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CAM Synthesis

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SolidWorks 2011

Exercise 10 Desmodromic CAM

The mechanisms can be actuated and controlled in various directions using various mechanisms. One conventional solution is using springs to return the mechanism to the original position (i.e. valve springs in engines). An alternative solution may be a system of cams called desmodromic cams.


Exercise 10: Desmodromic CAM

In the following exercise, we will build a simple mechanism using a traditional torsional spring first. Then we will build a second cam replacing the torsional spring in the system. This way the mechanism will be driven using a system of cams only. This exercise reinforces the following skills: I I I

Project Description

see Generating a CAM Profile on page 167. see Trace Path on page 169. see Create Curve From Trace Path on page 170.

In this project, we have already designed a cam that will drive the link in a predictable motion. As the cam rotates, it will push the link counterclockwise through contact. As the cam continues to rotate, some force is required to have the link follower stay in contact with the cam. In the first part of the exercise, we will apply a torsional spring to the link to keep it in contact. Separation if no return force

Spring

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Desmodromic CAM

Open an assembly file. Open Desmodromic CAM from the Lesson06\Exercises folder.


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The first cam (cam1) is already created and mated to the follower roller1 with a cam mate.

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Units.

Confirm that the assembly is set to use MMGS units.

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New study.

Create a new motion study.

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Restrict axial motion.

The shaft is currently free to move in an axial direction. Add a linear motor to prevent any axial movement of the shaft. Set the Duration time to 10 s.

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Add rotary motion.

Add a rotary motor to the shaft to have it rotate 360 degrees in 10 seconds.

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Cam mate.

Examine the mates in SolidWorks and notice that there is a cam mate between cam1 and the cam follower (roller<1>). While this mate is acceptable for animation, it is unrealistic for analysis because it forces the two surfaces to stay together even if they would not in reality.

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Run study.

Set the study length to 10 seconds and run it. The study will run and show the motion we desire.

8

Remove the cam mate.

In the FeatureManager design tree, suppress the cam mate.

You must return the timeline to zero before suppressing the mate.

Note

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Run study.

The cam1 still rotates, but the link does not move because there is no connection between the cam1 and the upper follower roller<1>.

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10 Add a spring.


Explode the assembly to make it easier to select the correct surface on the link. Add a torsional spring to hold cams together.

Use a Spring Constant of 10 N-mm/deg, and 30 degrees for Free Angle. The direction should be clockwise when viewed in the Front view.

11 Add contact.

Apply solid body contact between cam1 and the upper follower roller<1>. For Specify Material select Steel (Greasy) and select Friction.

12 Run the study.

The motion is correct and the design works well at slow speeds.

If we run this system at higher speeds, we could run into a problem where the spring cannot keep the follower in contact with the cam. If we get separation, we could then run into additional problems with the follower bouncing of the cam and getting a motion other than that which we were trying to design.

To force contact, we will design a second cam. When our system is viewed from the Front view, our first cam was able to rotate the link counterclockwise through contact, but clockwise motion depended on the spring. In the next part of this lesson, we will replace the spring with a second cam which will be able to rotate the link in the clockwise direction. The two cams work together to maintain positive contact between the cams and followers.

13 Suppress the torsional spring.

Note

You must return the timeline to zero before suppressing the spring.

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Desmodromic CAM

14 Delete contact and unsuppress the cam mate.


We are going use the Trace Path function to generate our second cam path. As we need to generate a path that maintains contact throughout the full rotation, we will use the cam mate to force the contact. Delete the contact between cam1 and its follower roller<1>. In the FeatureManager design tree, unsuppress the cam mate.

15 Run the study. 16 Trace Plot.

Create a new plot to generate the curve of the second cam.

We need to select the center point of the second follower roller. We can do this by selecting the edge of the second follower roller which will define the center point. Also select the face of cam2.


We now have the basic path, but it is too large because we had to trace the center of the second follower roller<2>.

Measure the second follower roller<2>. As it is 52 mm, we will have to reduce the size of the cam2 by half of this, or 26 mm.

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18 Export the curve in the reference part. 19 Open the part. Open the part cam2 in its own window. 20 Extrude the new cam. Create a sketch on the Front plane of

the part.

Use Convert Entity to create a circle in the sketch based on the outer edge of the existing part. Use Convert entity again to create a curve from the trace. Set the type of this converted curve to For construction.

Select the For construction curve from the trace plot and create an Offset Curve, 26 mm to the inside. Extrude the new cam2 a depth of 10 mm so that the two solids coincide. Merge the results.

21 Motion Study.

Return to the assembly window.

We will now run the study using the two cams to drive the motion. Suppress the cam mate.

Add contact between each of the cams and its respective follower.

User Steel (Greasy) for the material and select Friction.

22 Run the study.

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Desmodromic CAM



Both cams stay in contact with their rollers throughout the rotation as one takes care of counterclockwise rotation of the link and the other controls clockwise rotation.

Tip

Use a vertical split screen to be able to watch both the Front and Back views as the shaft rotates.


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Exercise 11 Rocker CAM Profile

In this exercise we will create a multi-piece cam that is used to control the motion of a slider.


Exercise 11: Rocker CAM Profile

The toothed wheel rotates and has attached to it a drive plate and guides for the slider.

The roller will ride in a path between two stationary cam plates. This system uses the inner cam to move the slider radially outward and the outer cam to move the slider radially inward. This exercise reinforces the following skills:

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see Generating a CAM Profile on page 167. see Trace Path on page 169. see Create Curve From Trace Path on page 170.

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Rocker CAM Profile

Project Description


The assembly rotates at 8,000 deg/sec. On each rotation, the rocker will move radially based on a predefined schedule which is provided in an attached file. Create the cams from the existing parts based on a predefined motion path provided in the separate file.

1


Open rocker cam profile exercise from the Lesson06\Exercises folder.

2


If we hide the toothed wheel and drive_plate assembly. We can see that the two cam plates are in place, but the cam paths have not been defined.

3

Units.

Confirm that the assembly is set to use MMGS units.

4

New study.


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5

Define the rocker motion.


Add a linear motor to the bottom face of the rocker. This motion must be specified relative to another component, so select the guide plate (699-0431) shown. Use Data Points, Displacement and load the file Slide Translation Motion.csv. For Interpolation type, select Cubic. Make sure that the direction is radially outward.

You may hide the Plate CAM Assembly for easier definition.

Note

6

Define the rotation.

Add a rotary motor to the drive_plate assembly (or part 699-0414).

Set the motor to rotate at a constant speed of 8,000 deg/sec. The direction should be counterclockwise when viewed from the Top view.

7


As the time of the simulation is very short, we will need a high frame rate to have sufficient points to get a smooth result.

Set the Motion Study Properties to capture 2,500 frames per second.

8

Run the study.

Set the length to 0.045 seconds. This will be one full revolution of the assembly at the speed of 8,000 deg/sec. The assembly should make one revolution.

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Rocker CAM Profile

9

Define a result plot.


Create a trace path of the center of the roller (699-0413) on the rocker.

Note

If the curve does not look smooth, increase the image quality in SolidWorks Tools, Options.

10 Create curve.

With nothing selected, right-click the Trace Path plot and select Create curve from trace path, and then Create curve from path. Because we have nothing selected, this curve will be a feature in the assembly FeatureManager design tree.

11 Model.

We are now going to work on individual components of the assembly, so we do not want to be in the Motion Study. Click the Model tab.

12 Hide components.

We will be creating the cam paths while in the assembly, so it will be easier to see what we are doing if parts that are not affected are hidden. Hide the toothed wheel, Slide Assembly and drive plate

assembly.

13 Edit part.

Select the part 699-0416 in the Plate CAM Assembly and click Edit Part .

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14 Edit sketch. Edit Sketch3 under Base-Extrude.

This is just a circular sketch that defines the outer face of the part. We will replace this sketch with the trace path curve with an offset for half the diameter of the roller. In the FeatureManager design tree, select the curve (it will be above the parts and assemblies). Use Convert Entities to create a curve from the trace path curve created in the previous step and set its property to For construction. Click Offset Entities and type 6 mm for the offset (half of the roller diameter). Make sure the direction of the offset is inside. Click OK to confirm the offset.

Delete the original circle from the sketch.

Exit the sketch and the part edit mode. The profile should look like the image.

15 Outer cam.

Edit the part 699-0417 in the Plate CAM Assembly.

On the face facing the 699-0416 component (closer face when viewed using the Bottom view), create a sketch using the same procedure. Offset the same curve, but this time 6 mm to the outside. Extrude a cut to a depth of 8.8 mm. Exit the part edit mode.

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16 Determine the inner radius.


Measure the distance from the center of the outer cam plate to the vertex shown. This is the same radius that is needed to create the curve of the profile on the keeper.

Highlight the distance and press Ctrl-C to copy this value to the clipboard as we will need it in the next step.

17 Show part.

Return to the Edit Assembly mode and Show the part keeper.

This is the keeper that is used to allow access when assembling the rocker.

18 Edit sketch.

Edit the sketch for Boss-Extrude1.

Double-click the dimension for the radius of the arc and paste the measured distance from the clipboard.

19 Examine the completed cams.

Return to the assembly and examine the cams that have been created. We should now have a smooth cam path.

20 New motion study.

Duplicate the existing motion study into a new one. Name the new study with contacts.

21 Suppress the linear motor.

Duplicate the existing motion study into a new one. Name the new study with contacts and suppress the linear motor feature.

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22 Contacts.


Create solid body contacts between all necessary components. Tip

You can conveniently use the contact groups to minimize the number of definitions.

23 Motion study properties. Activate Use Precise Contact. 24 Calculate the motion study. 25 Analyze the results.

Verify that the designed cam assembly provides the desired motion of the rocker.


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Rocker CAM Profile

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PR Do E no RE t c LE op AS y E or D di RA st F rib T ut e Lesson 7 Flexible Joints

Objectives


Learn about Flexible connectors (Bushings).

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Create Advanced Plots.

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SolidWorks 2011

Flexible Joints

In the physical world, nothing is absolutely rigid as materials have the ability to deform elastically and plastically. To this point in the course, mates were all simulated as rigid, which is not realistic. In this lesson, we will start with rigid mates and then make them flexible to more realistically model them as they would react in the physical world.


Flexible Joints

Case Study: System with Rigid Joints

A vehicle is being driven on a test track, which has rumble strips that are 100 mm in height and spaced 2,100 mm apart. The vehicle is moving at a speed of 60 km/h. A suspension-steering system is set-up and will be tested for these conditions. The model is a geometric representation of a short-long arm (SLA) suspension subsystem with the steering mechanism. Steering

IntermittentShaft

Steering Shaft

Steering Rack

Body Ground Tie Rod

Base Caps

Strut Upper

Upper Arm

Lower Arm

wheel

Strut Lower

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Lesson 7 Flexible Joints

Problem Description


The goal of this study is to inspect the toe angle that the wheel exhibits throughout its vertical travel of 100 mm in jounce and rebound. The toe angle that the wheel exhibits is for the steering wheel angles of 45 degrees, 0 degrees, and -45degrees. We will first run the study at the three angles with rigid joints. Then we will change the joints to flexible and run the study again for comparison.


To analyze the suspension system, we will follow the steps below:

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Create mates.

We will make sure that all the mechanical mates that are required have been included in the assembly.

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Define the motion.

Add a linear motor that is driven at the frequency that is created by the vehicle speed and rumble strip spacing.

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Plot the results.

Create plots of the tire yaw angle versus the vertical displacement.

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Modify the joints.

Change the joints from rigid to flexible.

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Re-run the study.

The results of the study will be compared against the previous study with rigid joints.

1


Open Suspension_Steering_System from the Lesson07\Case Studies folder.

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2



Before we create a motion study, we need to examine the assembly and determine how the linkages are connected. In the Mates folder, there is an Angle mate. Examine the mate. This mate controls the angle of the steering wheel and will be one of our study parameters as we can use this mate to turn the steering wheel to specific angles. Move the tire vertically and rotate it. Notice that the lower arm is not connected to the lower strut. Also notice that the tire can turn, even though the steering wheel doesn’t because of the mate.

Fixed

Not Fixed

Fixed

The five Base_Caps (top of the strut and two per each arm) are fixed and cannot move.

3

Prepare to apply mates.

We are going to add two mates to the assembly, a rack and pinion mate to connect the steering rack to the steering shaft, and a lock mate to connect the bottom of the strut to the lower arm. Before applying these mates, the tire needs to be returned to its zero position.

Important!

Either close the assembly without saving, and then reopen it to return to the starting point. Alternatively, use Reload to copy the assembly on disk back into RAM.

4

Attach Base_Caps<5> to Lower_Arm. In SolidWorks add a Lock mate between the parts Lower_Arm and Base_Caps<5>.

Now the two parts are rigidly connected to each other.

5

Float Base_Caps<5>. Base_Caps<5> is Fixed when we open the assembly. Now that we have created the Lock mate to the Lower_arm, we must remove the Fixed mate to allow the suspension to move.

Right-click Base_Caps<5> and click Float.

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Create a Rack and Pinion mate between Steering_Shaft and

Steering_rack.


6

Add a Rack Pinion mate between Steering_shaft and Steering_rack which would be connected through a worm gear.

When the Steering part (attached to Steering_Shaft) rotates by 7 degrees, the Steering_rack part travels 1.0 mm.

Select Rack travel/revolution and type 51.43 mm [(360°/7°) x 1 mm/ rotation = 51.43mm/rotation]. Select Reverse to make the direction correct.

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Examine Angle mate. There is an Angle mate on the Steering part. With all the mates


7

properly applied, the angle mate will be the equivalent of the driver’s input. To get -45 degrees, input 45 degrees and click Flip dimension. Angle = +45°

Angle = 0°

Angle = -45°

Set the angel value back to 0 degrees before leaving this step.

Calculation of Wheel Input Motion

A simple harmonic function motion will be imposed on the wheel to simulate this condition. To achieve this, some preliminary calculations are done based on the inputs. For a harmonic function, we need to find the frequency (deg/sec) and the amplitude (which in this example is the 100 mm height of the rumble strip). Frequency can be computed from the spacing of the rumble strips (2,100 mm) and the velocity (60 km/h.).

Frequency (Hz) = velocity / spacing = 60 (km/h) / 2,100 (mm) = 16,666.67 (mm/s)/2,100 mm=7.94 Hz.

The peak-to-peak amplitude desired is 100 mm.

8

Create a motion study.


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9

Create input motion.


We will add a motor to drive the vertical motion of the wheel based on the frequency of the vehicle moving over the rumble strip at the desired speed. Add a Linear Motor.

Select the vertex in the center of the wheel hub for the position of the motor. For direction, select the Top plane in the part wheel.

Important!

You must use the Top Plane in the wheel part and not a plane outside of the part.

Select Oscillating for the motion type. The amplitude is 50 mm (half the height of the rumble strip) and the frequency is 7.94 Hz. Keep 0deg for the Phase Shift.

Click OK.

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Flexible Joints


10 Create a spring and damper between Strut_Lower and Strut_Upper. Define a spring that is attached at the Point at the top of the strut and

the edge at the bottom.

Enter 60.0 N/mm for the Spring Constant and 405 mm for the Free Length. Add a linear Damper with Damping Constant of 0.46 N/(mm/s).

For Display, Coil Diameter = 60 mm; Number of Coils = 10; Wire Diameter = 10 mm.

Click OK.

11 Study properties.

Set the study properties to record 500 Frames per second.

12 Run the study.

Run the study for 0.12 second which is one cycle at the input frequency of 7.94 Hz.

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When a pair of wheels is set so that their leading edges are pointed slightly towards each other, the wheel pair is said to have toe-in. If the leading edges point away from each other, the pair is said to have toeout. The amount of toe can be expressed in degrees (from the angle to which the wheels are out of parallel), or more commonly, as the difference between the track widths (as measured at the leading and trailing edges of the tires or wheels). Toe settings affect three major areas of performance: tire wear, straight-line stability and corner entry handling characteristics.


Understanding Toe Angles

The pink arrow denotes the direction of travel of the car.

13 Animate.

Play back the study at slow speed to observe the motion. If you select Loop , it will continue to play.

14 Plot the pitch.

Create a new plot and select Other quantities, Pitch/Yaw/Roll and Pitch.

Select the tire face of the part wheel for the part to create results. This will plot the Pitch at the center of mass for the wheel part. The toe angle can easily be determined from the plot.

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15 Plot Toe angle vs. wheel height (the Y displacement).


The previous plot is not important to use as what we are really interested in is the Toe angle as a function of the vertical displacement of the spindle.

Edit the previous plot. Under Plot Results versus, select New Result, then Displacement/Velocity/Acceleration, Center of Mass Position, Y Component. Select the same face of the wheel.


Because we have rigid joints, we have two lines that fall on top of each other. One line is the wheel moving up, while the other is the wheel moving down. We will now repeat the simulation for two more configurations: steering angles 45 deg and -45 deg (simulating a left and right turn, respectively).

17 Change the steering angle to 45 deg.

Set the timeline to zero.

Important!

If you do not return the timeline to zero, before editing the mate, the mate will still be at zero degrees at time zero and will change to 45° at whichever point the timeline was when the edit was made.

Edit the Angle mate and change it from 0 to 45 deg (to simulate a left turn).

18 Re-run the simulations.

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The curves are still straight, but the values are slightly different.

20 Change the steering angle to -45 deg.

Set the timeline to zero.

Edit the Angle mate and change it from 45 to -45 deg (to simulate a right turn).

Again we have the same shape curves, but different values.

From the three graphs shown, notice how the toe angle changes with the change in the steering angle.

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Flexible Joints

In the physical world, nothing is absolutely rigid as materials have the ability to deform elastically and plastically. In the previous study, the joints were all simulated as rigid, which is not realistic. In the following part of the lesson, we will change the joints to be flexible, which will more realistically model the real world.

Introducing: Bushings

Bushing objects are added to model flexible mates used on physical suspensions. Bushing elements allow deformation in a certain degreeof-freedom that is not accounted for if the attachment is modeled as rigid. In this lesson, notice how the Lower_Arm is connected to the Base-Caps with two concentric mates. These two mates could be replaced with bushings in order to simulate a flexible connection between the Lower_Arm and Base-Caps.


System with Flexible Joints

Typical bushings used in automotive vehicle design consist of steel-onsteel, Urethane, or Nylon. The stiffness and damping characteristics of these bushings are measured by SAE testing methods (see Reference 1) and depend on the type of vehicle (see Reference 2). Orthotropic bushings can greatly affect the kinematics (camber, toe angles) and dynamics (joint, shock forces) results of your model when compared to a rigid connection. In our simulation, we will use isotropic bushings.

21 Review the mates in the model.

Review the mates associated with the Lower_Arm and the Upper_Arm parts. Notice how they are connected to the Base-Caps. The Base-Caps are connected to the automobile frame. There is no slack in the mates. However, in real life there is some slack or play between the arms and base-caps. To incorporate this slack, a flexible connector or, in other words, a bushing will be used.

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22 Create bushings between the Base_Caps and Arms.

We want to edit the global mates locally, so we must edit the mates in the SolidWorks FeatureManager design tree while staying in the motion study tab.

Locate the four Concentric mates between Base_Caps 1 through 4 and the upper and lower arms.

Edit each mate in turn. I I I I I

Select the Analysis tab and make the following changes to each mate: Select Bushing. Select Isotropic for both Translational and Torsional. For Translational, change the Stiffness to 3,500 N/mm, Damping to 2.63 N-s/mm and Force to 0. Leave the Torsional values at their default settings.

Each mate will now have a Bushing symbol next to the mate type.

23 Steering angle.

Set the steering angle back to 0°.


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25 Plot Toe angle vs. wheel height (the Y displacement).


The steering angle should already be plotted on the screen. We can see that there is now some slack in the bushings.

The plot below shows the same plot when joints are used instead of bushings.

Comparing the results, we confirm that the toe angle is different between the two conditions (bushings vs. joints).

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26 Review Simulation.


Zoom into the location of the Lower_Arm where it connects to the Base_Caps. Notice how Lower_Arm interacts with the Base_Caps. There will also be some slack between the moving parts and the Base_Caps.

Time = 0.0

Time = 0.025

Time = 0.05

27 Obtain results for the two additional configurations. Obtain the graph of the Toe angle vs. the wheel height for the two

additional configurations: steering angles 45 deg and -45 deg.

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Flexible Joints


Summary

In this lesson, we learned the use of springs, dampers, and bushings in SolidWorks Motion. We explored several post-processing options to analyze the rotational displacements of the model. We also studied the effect of making joints flexible by introducing bushings.

References

[1] Adams, Herb, “Chassis Engineering”, The Berkley Publishing Group, 1993. [2] Kirschenbaum, Al, “The Official Ford Mustang 5.0”, Bentley Publishers, 1993.

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PR Do E no RE t c LE op AS y E or D di RA st F rib T ut e Lesson 8 Redundancies

Objectives

Upon successful completion of this lesson and exercises, you will be able to: I

Understand Redundancies and how they affect the simulation.

I

Use Flexible mates to automatically remove redundancies in a mechanism.

I

Assign the stiffness to each mate individually.

I

Understand how to build assemblies without redundancies.

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Redundancies

In Lesson 2, 3D Crank Slider Mechanism, we studied the recommended approach to building assemblies for the kinematic simulation, i.e. simulation where our main objective is to obtain displacements, velocities, accelerations, jerks or possibly some reaction forces. We covered the fact that mates are used to connect the assembly components and thus constrain the relative motion of a pair of rigid bodies. Mates therefore determine how the assembly moves and we also reviewed some of the most common mate types. In the last part of this lesson we will discuss this topic in greater detail; we will review how many degrees of freedom do mates constrain and why can this be important for the solution of our motion simulation. Before we move ahead, let us review some of the basic terminology and concepts.


Redundancies

Each unconstrained body in space has six degrees of freedom: three translations and three rotations about X, Y and Z axes. Any rigid body, i.e. SolidWorks part or rigidly attached parts forming sub-assemblies, therefore feature all six degrees of freedom. When we use mates to connect rigid parts or subassemblies together, each mate (or connection type) removes certain number of degrees of freedom from the system. Below, we will review the basic mate types and state how many degrees are removed when two rigid bodies are connected. The table below lists the most common mates representing some of the common mechanical connections. Translational DOF removed

Rotational DOF removed

Total DOF removed

Hinge mate

3

2

5

Concentric (2 cylinders)

2

2

4

Concentric (2 spheres)

3

0

3

Lock mate

3

3

6

Universal mate

3

1

4

Screw mate

2

2 (+1)

5

Point to point coincident

3

0

3 (this mate is identical to the concentric spheres mate)

Mate Type

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The table below list some of the special mates, which do not necessary represent a real mechanical connection, but do impose a geometric constraint on the two connected bodies. Translational DOF removed

Rotational DOF removed

Total DOF removed

Point on axis

2

0

2

Parallel (2 planes)

0

2

2

Parallel (2 axes)

0

2

2

Parallel (axis and plane)

0

1

1

Parallel (2 axes)

0

2

2

Perpendicular (2 axes)

0

1

1

Perpendicular (2 planes)

0

1

1

Perpendicular (axis and plane)

0

2

2

Mate Type

A very large number of mates can be listed in the above tables. As you can see, not only the mate type determines the number of the constrained degrees of freedom, but also the pair of selected entities is important. Lesson 2 recommended that for the models where kinematic quantities (displacements, velocities, accelerations etc.) are required, all mates should, up to the reasonable extent, represent real mechanical connections. In the figure to the right, the door is connected with two hinges. Both hinges should be defined as hinge mates for the kinematic solution.

As you will soon learn, that due to the redundancies, this approach is not sufficient when joint forces are required, or when the part is to be exported to SolidWorks Simulation for the stress analysis.

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Based on the number of degrees of freedom, mechanical systems are divided into two categories: I I

Kinematic System

Kinematic systems Dynamics systems

For a kinematic system, mates and motors fully constrain all the degrees of freedom on the mechanism. So the position, velocity and acceleration of each part are fully defined at every time step based upon the mates and motions applied by motors. Mass and inertia information is not needed to decide the motion. Such mechanism is said to have zero degrees of freedom. For example, consider the Scissor lift model shown to the right. The motion of the scissor lift will always be the same regardless of the mass of the links or platform, or the weight of people (external load) standing on the platform. Only the force required to drive the lift will change, depending upon a change in the mass of any component or the external load. More weight means that more force is needed to get from height A to height B.

Dynamic System

In a dynamic system, the resulting motion of parts depends upon the mass of components and the applied forces. If the mass or applied forces change, then the motion behavior is different. Such a mechanism is said to have more than zero degrees-of-freedom. In the mass string example to the right, depending upon the mass of the balls, the motion will be different. Or, if you swing the ball on the left with a different force, the motion of the balls will be different.

In summary, the primary difference between kinematic and dynamic system is that a kinematic system motion is not influenced by the mass or applied loads, whereas a dynamic system motion is easily influenced by changing mass and applied loads.

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All of the systems that were analyzed in Lesson 1 to Lesson 7 can be considered kinematic systems; i.e. given the mates and the applied motors, the motion of the systems was always determined and unique. However, as you will soon learn, all of these were also redundant leading to unique kinematic results (displacements, velocities and accelerations), but possibly non-unique dynamics results (for example, joint forces were not computed correctly because no unique solution existed). Redundant systems, i.e. systems with redundant constraints (alternatively we may call them over-constrained systems) are subject of this lesson.

What are redundancies?

Redundancies relate to modeling a real life system as a mathematical model and are an inherent problem in rigid body motion simulation. It is very important that you be aware of redundancies and how they can effect the simulation and results of a mechanism. At a base level, redundant constraints occur when more than one mate constrains a specific degree-of-freedom on a part.

Constraints in SolidWorks Motion remove degrees-of-freedom (DOF) from the system by adding algebraic equations to the governing system of DAE’s (Differential and Algebraic equations). Six algebraic equations used by SolidWorks Motion to represent DOF constrained by mates are as follows:

Equations 1-3 constrain translational DOF while equations 4-6 constrain rotational DOF, where “i” and “j” represent the first and second parts respectively. The above equations can be understood as follows: 1. X i – X j = 0 means that the global X-coordinate of the “i” part must always remain identical to the X-coordinate of the “j” part.

2. Y i – Y j = 0 means that the global Y-coordinate of the “i” part must always remain identical to the Y-coordinate of the “j” part.

3. Z i – Z j = 0 means that the global Z-coordinate of the “i” part must always remain identical to the Z-coordinate of the “j” part.

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Redundancies


4. Z i ⋅ X j = 0 means that the Z-axis of the “i” part always remains perpendicular to the X-axis of the “j” part (which means no rotation about the common Y-axis). 5. Z i ⋅ Y j = 0 means that the Z-axis of the “i” part always remains perpendicular to the Y-axis of the “j” part (which means no rotation about the common X-axis).

6. X i ⋅ Y j = 0 means that the X-axis of the “i” part always remains perpendicular to the Y-axis of the “j” part (which means no rotation about the common Z-axis). The "⋅ " notation in equations 4-6 signifies a dot product operation. Recall that when the dot product of two vectors is zero, the vectors are perpendicular.

Each Fixed mate in your model uses six equations (eq. 1-6), while a Concentric mate (of two spheres) uses three equations (eq 1-3), a Hinge mate uses five equations (eq. 1-5), etc.

Notice how each of these mates uses equations 1 and 2. Any such duplication of constrained DOF can lead to over constraining your system, or introduce what are known as redundant constraint equations. SolidWorks Motion outputs warning messages to try to help you understand which equations are redundant and therefore which DOF are unnecessarily removed. When you have a redundant constraint, you have two or more mates effectively fighting to control one specific degree-of-freedom. In simple cases, the solver will automatically remove a redundant constraint equation to stop the redundancy. In complex situations it may not remove the correct one for the mechanism, affecting the original design.

Important!

This leads to the simulation still running, but giving the wrong motion or answer.

Effects of Redundancies

There are two main failures due to redundancies: I

Simulation failure part way through a solution

As the solver progresses through a solution, it continually re-evaluates redundancies and removes them from the mechanism. Occasionally during the re-evaluation, different redundant constraints are removed based on the current positions and orientations. This can potentially lead to an inconsistent model. Because the solver is unable to understand the design intent of a mechanism, it can arbitrarily remove constraints which are mathematically valid, but not valid from a functional point of view. I

Incorrect force calculation

An example to illustrate this is covered in the next section.

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Before a simulation is actually run, the solver goes through the process of detecting if the mechanism contains redundancies. If it detects redundancies, it will try to remove them, and only if successful, will it continue to run the simulation. At each time step, it continues to reevaluate redundancies and removes them as needed.


How are redundancies removed in the solver?

There is a certain hierarchy by which redundancies are removed. The solver will remove redundancies based on the following order:

I I I

Rotational Constraint Translational Constraint Motion Inputs (Motors)

According to this hierarchy, the solver first looks for rotational constraints that can be removed to eliminate redundancies. If it cannot remove any rotational constraints, it will then try to remove translational constraints. If it cannot remove any translational constraints, it will then try to remove an input motion (as a last resort).

If all these attempts fail, the solver will abort with a message instructing the user to check for redundant or inconsistent constraints in the mechanism (or to see if it is in a locked position).

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Redundancies

Let's investigate this removal procedure with the help of a door mechanism. The most intuitive way to create mechanical connections consists in recreating the physical reality. For example, when you see a hinge, you want to model it with a hinge mate. If there are two hinges on the same part, like this door, and if you place two hinge mates, you create some redundancies.

Problem Description

We have a simple door consisting of a door and frame. The door is connected to the frame with two hinges. Determine the forces on the two hinges as a result of the weight of the door.


Case Study: Door Hinges

Frame

Door

Hinge

1


Open door from the Lesson08\Case Studies folder.

2

Float the door.

When the assembly is opened, both components are fixed and have zero degrees of freedom. Right-click the door and click Float.

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3

Add Hinge mates.


To make it easier to select faces on the hinge, move the door a small distance. Add a Hinge mate between the two halves of the upper hinge. It is not important if the mate is added as local or global.

Note

4

Add another Hinge mate.

Add a second hinge mate to the lower hinge.

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Redundancies

5

Check the door weight.


Check the mass properties of the door part. The door weighs 28,020.63 grams, so the vertical force of the door should be a 274.8 N.

6


7

Add gravity.


8

Change study properties.

Edit the motion study properties and set Frames

per second to 50.

Make sure that Replace redundant mates with bushings is cleared. We will discuss this option later in the lesson.

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Degrees of Freedom Calculation


Let us review how many degrees of freedom (DOF) are currently restricted by our mates. Because the frame is a fixed body it features zero DOF. The only floating body in the assembly is the door. Our mechanism may therefore feature up to 6 degrees of freedom.

The two hinge mates defined in the model are then each constraining 5 DOF.

Total Actual and Estimated DOF

The current DOF count for our system is therefore 6 – 2 × 5 = – 4 , i.e. our mechanism is over-constrained based on the simple DOF count. This simple count is referred to as approximate (or Gruebler) and is rather easy to obtain. It could indicate that our mechanism cannot move. It is obvious, however, that the door is allowed to rotate about the hinges and, in engineering sense, should not be over-constrained; using this engineering approach, our mechanism features +1 DOF (rotation about the hinges). This count, referred to as “actual” is more complex to obtain than the simple count introduced above. The number of redundant constraints in the system is therefore 6 – 2 × 5 – 1 = – 5 ; our system features 5 redundant constraints – constraints which are not needed from the mathematical point of view for the door to close and open. Indeed, by removing one of the hinges the kinematics of the system is unchanged.

9

Run the simulation.

Run the simulation for 1 second. There is no movement in this assembly.

We will now review the number of degrees of freedom as well as the number of redundancies with the help of functions within SolidWorks Motion.

Introducing: Degrees of Freedom Calculation

Rather than manually calculating the Degrees of Freedom, SolidWorks Motion can quickly calculate them for us.

Where to Find It

I

Right-click the local mate group and select Degrees of Freedom.

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Redundancies


10 Use the simulation panel to calculate degrees-of-freedom (DOF).

When the study completes, notice that in the Motion Study FeatureManager, the mate folder reads Mates (5 Redundancies), just as we calculated a while ago. Right click the local Mates folder and select Degrees of Freedom to open the dialog shown below. We can review the number of moving (floating) parts, number of mates (presented as joints), number of the estimated and actual DOF and the Total number of redundant constraints. SolidWorks Motion calculates five redundant constraints. The mechanism is overconstrained.

As mentioned above, the reason for this is that a second hinge mate is trying to constrain the same degree-of-freedom as the other hinge mate. Numerically, one hinge mate is sufficient to simulate the hinge condition. But this may not be enough, especially when reaction loads at both the hinges are to be calculated. In order to obtain a unique solution, the program is forced to remove the 5 redundant constraints. The selection is made internally without user intervention; the removed redundant DOF can also be found in the above list. We will now review the force solution in the joints to reveal the consequence of the redundancies.

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11 Plot reaction forces for the Hinge mates.


The weight of the door is approximately 274.8 Newtons. Gravity acts along the negative global Y direction. The two hinge mates should share this load equally. Let us verify this.

Create two plots to show the Y Component reaction force for the two hinges. When we define the plot, we will be warned:

The motion study has redundant constraints which can lead to invalid force results. Would you like to replace redundant constraints with bushings to ensure valid force results? Note that this will make the motion study slower to calculate.

As redundant mates are the subject of this lesson, we will first see what happens with the redundant constraints. Click No.

The reaction force on one of the hinge mates is zero which should not be the case.

On the other hinge mate the reaction force is 274.8 N.

It can be seen that one hinge mate carries the full force of 274.8 N, while the other one carries no load. The distribution of the forces between the two hinges is incorrect.

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Redundancies


Let us see why the simulation gave us such results. In Step 10, using the simulation panel, we calculated the DOF in the mechanism. Notice that one of the redundant constraints was mentioned as “Hinge2: Translation along Y”. This tells us that the mechanism is already constrained in the Y direction by the Hinge1 mate. The same degreeof-freedom is being constrained by the Hinge2 mate and will be ignored. Therefore, no results are calculated for the Y-direction reaction force on the Hinge2 mate.The entire weight of the door will then have to be reacted upon at the Hinge1 mate at simulation time. Likewise results for other redundant constraints will be ignored and hence turn out to be zero.

We will now see how this issue can be avoided by using the Flexible joints option.

Using Flexible Joints Option to Remove Redundancies

In the discussion on page 214 it was mentioned that the redundancies may lead to:

1. Simulation failure part way through a solution, and 2. Incorrect force calculation (distribution). The effect of point 1 can be minimized (though not avoided) by using mates closely representing the mechanical connections in the real product. For example, two hinges in the door-frame assembly could be mated with two hinge mates since they represent the real connection type the most closely. Alternatively, point 1 can be tackled by reducing the number of redundant constraints manually by using simpler mates such as point on axis and similar. In complex assemblies this can be, however, daunting task and may require iterative approach of mate design and the DOF calculation. For example, imagine that in our current example of the door, one hinge mate is deleted and the number of redundant constraints is then zero; the solution in the Y direction is then identical.

The effect of point 2 can be tackled by manually modifying the mates to remove the redundancies in the requested (or all locations) and readjusting the distribution of the reaction forces in the mates manually, or using the technique introduced in Lesson 7, flexible mates. To demonstrate the former, imagine that one redundant hinge mate is deleted from the simulation – all load is then carried by the remaining hinge mate. Knowing the geometry, we manually readjust the distribution equally into both mates. This approach may work in simple design and loads such as the current example of the door or many symmetrical mechanisms such as fork lift (analyzed in some of the exercises following this lesson). In the later approach, when flexible mates are used in place of mathematically rigid mates, stiffness of the mates in the respective directions decides on the distribution of the reaction forces.

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While this approach is still an approximation, it can provide a more realistic distribution of forces than the infinitely stiff case. When you make a mate flexible, the mechanism will be updated to have a bushing representation of the basic mate type instead of a rigid constraint. Mate motion and friction are not affected by using flexible mates.

Limitations of Flexible Mates

The following limitations may exist when using flexible mates: I I

I I

I

In some models, using bushings will slow down the solve time because of induced dynamic effects. We are not accounting for the stiffness of the part in the solution. Therefore, the distribution of loads due to part stiffness may differ from the bushing constraint solution. This bushing approach will ensure that force results are obtained at all mate locations. This limitation, however, exists in the case of rigid mates solutions as well. Advanced mates do not support mate flexibility. See the Help for a list of joints that can be made flexible. If the mechanism starts in a dynamic condition, there may be a spike in initial forces as the model reaches initial equilibrium (that you would not see with rigid joints). The spike is generated by initial conditions of the parts not balancing and the bushings resisting rapid changes in force/acceleration. If the model started with enforced motions (e.g., constant velocity), try ramping up motions from zero to the desired value over a time range to eliminate or minimize this (e.g., use a step function to ramp velocity from zero to a certain value over a time range). An optimum mate stiffness and damping characteristics may need to be entered. This may require an iterative approach.

The following joints can be made flexible: Fixed, Revolute, Translational, Cylindrical, Universal, Spherical, Planar, Orientation, In Line, Parallel Axis, In Plane, Perpendicular. In SolidWorks Motion Simulation, the flexibility in the mates can be introduced in two distinct ways.

1. Replace redundant mates with bushings option in the Motion Study Properties. This way, one set of global stiffness and damping characteristics is assigned to some algorithmically selected mates only. The decision on which mates are made flexible and which are kept rigid is done by an advanced algorithm and is fully automatic. This approach may work in most situations. 2. Assigning individual stiffness values to the selected (or all) mates manually. This technique will work in all situations, but can be time consuming. Local mates may be used with great advantage without altering the design intent of the assembly designer.

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Redundancies I

In the Motion Study Properties, select Replace redundant mates with bushings.


Where to Find It

When mates become flexible, the icon in the MotionManager tree.

.icon will appear next to the mate

Bushing Properties

When bushings are defined, their Translational and Rotational Stiffness and Damping may be defined.

Where to Find It

I

In the Motion Study Properties, click Bushing Parameters.

In the remainder of this lesson, we will use the Replace redundant mates with bushings option to correctly solve the door example.

Assigning individual stiffness to the selected mates manually as well as manual removal of the redundant constraints by building redundanciesfree assembly models are practised in the exercises following this lesson. Students are encouraged to complete all of the following exercises to fully understand this important subject.

12 Make joints flexible.

Make all joints in the mechanism flexible, in the Motion Study Properties, select Replace redundant mates with bushings.

Click Bushing Parameters. If we were to change the stiffness and damping values of the hinges, we would do it here. To see the effect of the mates stiffness on the solution, complete the following exercises.

Click OK twice.


Notice how the mate icon in the Mates folder of the MotionManager changes. The yellow lightening indicates that the flexibility of this mate was forced by the software rather than manually specified by the user (as was the case in Lesson 7).

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14 Review results. The Y Component of the Reaction Force for both

mates now shows a value of 137.5 N.

The weight of the door has now been correctly shared by both of the Hinge mates.

Note

The approach to make selected (or all) mates flexible was introduced and practised in the previous lesson and will not be shown here.


Important!

Students are encouraged to review the following discussion and the exercises following this lesson. The subject of redundancies is not trivial and must be understood in order to correctly conduct dynamic analyses (analyses where correct force distribution is required).

How to Check For Redundancies

As mentioned previously, it is important to only supply sufficient constraints to obtain the required motion on any mechanism. Kinematic/Dynamic analysis needs only the necessary degrees-offreedom restrained in a system. A quick indication of whether a system is over-constrained is the Gruebler count. I

If the number is greater than zero, then the model is underconstrained (dynamic).

I

If the number is equal to zero, then the model is fully defined (kinematic).

I

If the number is less than zero, the model is over-constrained (redundant).

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Redundancies


An important aspect behind modeling mechanisms is in recognizing the restrained freedoms of connecting parts and making sure they are not repeated. This can be difficult in very complex assemblies, but will ensure you achieve the desired motion and force results. If this is not taken into consideration, redundant constraints will have been applied which may result in the simulation not working.

Typical Redundant Mechanisms

Several mechanisms are redundant by their nature. In the real world, assembly tolerance, slop, and stiffness make the mechanisms work, but in the mathematical world, they can be invalid. Below are a couple of examples of these mechanisms.

Dual Actuators Driving a Part

From a kinematic point of view, you only require one actuator to move a part. In the real world, pairs of actuators are used to provide balanced loads from side to side. The main problem in the motion simulation world is that motions are enforcing displacement in a specific degree-of-freedom. A specific degree-of-freedom is constrained by two actuators. Thus, by having two motions you are causing a redundancy. This can lead to two situations. One, that only one actuator carries the load and the other has no result, or two, that there is artificial load induced into the system (equal and opposite) that produces incorrect driving force results on the motions. Ways to work around this problem are to use non-rigid connections to link each actuator into the mechanism or to use a force based movement instead of a motion based movement.

Parallel Linkages

The scissors lift is a classic example of where one complete side of the mechanism is redundant but is done to provide balanced loads on both sides of the structure and to make the design easier. It is simpler to work with a lighter strut that only carries in-plane load than to design a heavier strut that must not only carry in-plane, but out-of-plane torsional loads. However, for mechanisms, it is easier to model only one side and let the other side “come along for the ride”. When analyzing these types of mechanisms, you can lock the duplicate parts together by attaching them together or using fixed joints. You then need to delete any duplicate constraints. When you

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obtain the joint loads, just remember to divide by two. Also, remember that out-of-plane moments should only be due to the non-symmetry of modeling one side, and the moment should equate to half of the reaction force times the distance between the two sides that raise the platform (see Exercise 14: Kinematic Mechanism on page 233).

Summary

In this lesson, we defined and familiarized ourselves with the concept of redundancies. Redundancies occur when identical degrees-offreedom in the assembly are constrained by multiple joints. Models with such redundant constraints are improperly defined and their solutions are likely to be incorrect (or impossible to obtain). The implication of redundancies was demonstrated in the first part of this lesson.

In reality, the rigidness of joints is only an idealized concept. As such, SolidWorks Motion enables users to disable such rigidity and specify some finite stiffness and damping along the constrained degrees-offreedom in the joint. This approach eliminates the problem of redundancies, but introduces additional parameters that must be specified (joint stiffness and damping). Flexible joints were the subject of the second part of this lesson. Students are encouraged to complete the following exercises.

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Exercise 12 Dynamic Systems

Exercise 12: Dynamic Systems


This exercise will demonstrate a simple dynamic system with four spheres dropping in a closed container.

This exercise reinforces the following skills: I

Project Description

Dynamic System on page 212.

Four aluminum balls are contained in a closed container and will fall under gravity. None of the components have mates and are free to interact with each other. Examine the motion of this dynamic system.

1

Open an assembly file. Open Vase with Spheres from the Lesson08\Exercises folder.

2


3

Add gravity.


4

Add contact.

Add contact between all components.

Specify Aluminum (Greasy) for both materials. Add Friction using the default values.

5

Run the simulation.

Run the study for 1 second.

6

Examine the results.

All but one ball falls through the container. This can be cause by either a very coarse time step or too coarse of a contact description.

7

Change study parameters.

Change the motion study properties to record 600 frames per second to save more data on the disk and specify Precise contact.

8

Run.

Re-run the simulation.

All four spheres are now contained within the Vase. As the spheres fall, they interact with each other and the vase.

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Dynamic Systems

9

Animate.


Play the study at 10% speed and examine the motion of the spheres. 10 Save and close the file.

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Exercise 13 Dynamic Systems 2

Exercise 13: Dynamic Systems 2


This is another exercise with a dynamic system. In this study we will compare our hand calculation of degrees of freedom with those calculated by SolidWorks Motion. We will also investigate the effects of changing the impact from elastic to plastic.


Project Description

Poisson Model (Restitution Coefficient) on page 110. Dynamic System on page 212.

Five spheres are attached to individual frames. One end sphere is pulled away from the others and released. Examine the motion of the five spheres with both elastic and plastic impact.

1


Open Momentum from the Lesson08\Exercises folder.

2

Calculate the DOF.

Calculate the DOF by hand. The DOF should be a positive number to confirm it is a dynamic system.

3

Create a new study.

Create a new study and name it Rest Coef = 1.0.

4

Add gravity.


5

Add contact sets.

Add four contact sets between the pairs of balls that will make impact. Clear Material and Friction.

For Elastic Properties, select Restitution coefficient and set it to 1.0 which is an elastic collision.

6


Record 25 frames per second. Select Use Precise contact.

7

Run the study.

Run study for 5 seconds.

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Dynamic Systems 2

8



We see nearly elastic contact. If it was exactly elastic, we would not see the interior balls move, however with the slight errors in the numerical methods used, we see some movement as the study progress.

9

Degrees of freedom.

Now that we have run the study, we can let SolidWorks Motion calculate the DOF so we can compare the results with that which we calculated by hand. We have five moving parts with six degrees of freedom for a total of thirty. The five hinge mates remove 25 degrees of freedom, leaving us with five degrees of freedom total.

10 Duplicate the study. Name the study Rest Coef=0.1. 11 Edit contacts.

Edit the five contacts to and change the Restitution coefficient to 0.1.

This is nearly plastic impact.

12 Run the study.


With plastic impact, once the first sphere makes contact, all the spheres move together as we would expect.


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Exercise 14 Kinematic Mechanism

The following exercise will demonstrate a kinematic mechanism. The basic characteristic of the kinematic mechanisms is a possibility of a single motion, irrespective of the applied forces and motors, contrary to the dynamics mechanisms (demonstrated in the previous exercises) where multiple motions may exist. The Scissor Lift demonstrated in the following exercise features no redundancies and one “Actual” DOF. We will “consume” this last DOF on the motor which will drive the mechanism.


Exercise 14: Kinematic Mechanism

This exercise reinforces the following skills: I I I I I

Project Description

Kinematic System on page 212. How are redundancies removed in the solver? on page 215. Effects of Redundancies on page 214. We will now review the force solution in the joints to reveal the consequence of the redundancies. on page 220. How to Check For Redundancies on page 225.

Analyze the mates utilized to built this assembly. Notice, that as suggested in the discussion on page 222, only half of the symmetrical mechanism is used for the mate definitions. The symmetrically located components move in phase with the mated components. Also note that the assembly features many geometric constraints (non-mechanical mates such as coincidence of a point and axis, or coincidence of two planes).

Mates on this side

No mates on this side

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SolidWorks 2011

Kinematic Mechanism


Examine the individual mates, such as Coincident14. Notice that this, like many of the mates in this assembly, is a geometrical constraint (point and face) rather than a mechanical mate (hinge).

Use of such mates requires existence of the reference entities and the building procedure can be time consuming; the DOF count must be checked after each rigid component is added until the whole assembly is completed. Due to the time constraints we will not build this assembly in its entirety; only its part showing the procedure is demonstrated in the following exercise.

1


Open Scissor_Lift from the Lesson08\Exercises\ Kinematic Mechanism folder.

2


Examine the existing mates and move the assembly. With the existing mates, the only motion allowed is that which moves the platform vertically.

3

New motion study.


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4

Add motor.


We will add one Linear Motor to the piston to drive the motion of the assembly. Add a linear motor to the piston.

Set the Motion to Oscillating, 100 mm at 0.5 Hz. with 0deg for Phase Shift. Set the motion to move relative the cylinder.

5


Set the study properties to record 50 Frames per second.

6

Run the study.


7

Degrees of freedom.

In the local mate group, right-click MateGroup1 and select Degrees of Freedom.

We now have zero Total DOF because of the addition of the motor.

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Kinematic Mechanism

8

Plot forces.


To see the consequences of modeling the assembly with geometric mates on just one side, we will plot the forces in two of the mates, Concentric14 and Coincident9.

Concentric14 Coincident9

Create plots of the Z Component of the Reaction Force for each mate in the global coordinate system.

9

Examine the plots.

The plot for mate Coincident9 shows a maximum force of 15,166 N. Because of the redundancies, this is actually the combined force on both sides of the assembly. At this joint, the real force will be half or about 7,583 N.

For Concentric14, the maximum combined force is 9,536 N which means that each side carries 4,768 N.

Note

The assumption that each side of the assembly carries half of the total force at each joint is based on symmetrical loading of the assembly.

10 Plot moments.

Create a plot of the X Component of the Reaction Moment for the Tangent3 mate.

This moment about the global X axis should be zero if both sides of the assembly carry the load symmetrically. This moment is only a product of the way this assembly was build and will be compensated by the reaction force on the opposite side.


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This exercise demonstrated a complicated model build without any redundancies. Such procedure may be time consuming; use of the geometric constraints such as coincidence of points and axes or planes is necessary and the step by step procedure with frequent DOF calculation is necessary. This procedure, applied to a part of the large assembly (sub-assembly) or the entire assembly may only be required if all other means of attempting solution failed.


Summary

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Exercise 15

SolidWorks 2011

Zero Redundancy Model-Part 1

Exercise 15: Zero Redundancy Model-Part 1


The following exercise will demonstrate a short section of the model building procedure for the mechanism with zero redundancies featured in the previous exercise. We will reuse the model from the previous exercise, Kinematic Mechanism, at an early stage of the model building phase. The model will feature one redundancy. The goal of this exercise is to remove the redundant constraint with the help of multiple geometric constraint (simple mates such as coincident of point and axis and similar). This exercise reinforces the following skills: I I I I

Project Description

Redundancies on page 210. Effects of Redundancies on page 214. How are redundancies removed in the solver? on page 215. How to Check For Redundancies on page 225.

The same scissor lift used in the previous exercise will be used to practice the procedure of removing and controlling the number of degrees of freedom in the model. We will start with just the base and first layer of scissors, remaining components have been suppressed. The components of interest in this exercise will be the cylinder and piston.

1

Open an assembly file. Open Scissor_Lift.sldasm from the Lesson08\Exercises\ Zero Redundancy Model folder.

The platform and layers 3 through 6 are suppressed.

2

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Run the analysis Exercise Study. The motor is already set up as it was in the previous exercise, so you can just click Calculate.

SolidWorks 2011

Exercise 15 Zero Redundancy Model-Part 1

3

Examine the Degrees of Freedom.


The local mate group shows one redundancy. Right-click MateGroup1 and click Degrees of Freedom.

With zero Total DOF, the mechanism will move as we expect.

We can also see that there is one redundant constraint and that the redundant constraint Concentric16, Rotation about X is removed.

4

Determine orientation.

This redundant constraint is about the local X axis, so how do we determine which way the local axes are oriented?

Create a plot based on the mate. Concentric16 is a mate between the cylinder and the piston. Create a plot of the Reaction Force, Y Component of the mate Concentric16. We do not actually need to complete the plot, but once the plot is set up, we can see the orientation Triad.

The X direction (red) is along the common axis of the two parts. Once we observe the direction, cancel the plot.

This concentric mate is redundant because neither the piston nor cylinder can rotate about this axis. The cylinder has a hinge mate to connect it to the Base, and the piston has a concentric mate to the cross rod.

5

Remove mate. The piston and cylinder need to stay concentric,

so we will not delete that mate. Instead, we will replace the hinge mate, which removes five degrees of freedom, with two less complicated mates.

Delete the mate Hinge1.

The end of the cylinder is now free to move.

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6

Add mate.


For the following mates, we will be mating points and axes, so make both visible.

To see the points you need to set the assembly mode to Resolved.

Note

There are already two points created in the hole at the end of the cylinder. Point1 is on the axis of the hole, half way between the parallel faces. Point2 is also on the axis of the hole, but coplanar with the side face. Add a Coincident mate between Point1 and the axis of the hole in the frame.

Add a second Coincident mate between Point2 and the inside face of the bracket on the frame.

You can add these mates as either local or global.

Note

7

Run.

Run the simulation and observe the results. The study appears to run correctly.

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Exercise 15 Zero Redundancy Model-Part 1

Check the DOF. Right-click MateGroup1 and click Degrees of Freedom.


8

There is still one actual degree of freedom, but there should be none for correct results.

9

Determine the problem.

Try to drag the piston along the cross_rod. When you do, you can see the cylinder rotate.

10 Edit mate.

Remove point2

Add edge

The mate between Point2 and the face is not enough to stop the rotation, so we will have to raise the level of the mate to remove an addition degree of freedom.

Edit the Coincident mate and replace Point2 with the edge shown.

11 Run.

Run the study and observe the results.

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12 Check the DOF. Right-click MateGroup1 and click Degrees of Freedom.

We now have zero DOF as we should for a kinematic system.


If you are going on to the next exercise, leave the assembly open, otherwise save and close the files.

Summary

242

This exercise demonstrated how a mate with redundant constraint is detected, removed and replaced with a combination of simpler geometrical constraints such as coincidence of a point and an axis. As was mentioned in the previous exercise, this technique requires additional reference entities (points, axes) and the building procedure can be lengthy. It should be used if all other techniques failed to give the desired solution. In general, it is easier for the solver to obtain the solution for models without redundancies then for models with multiple redundancies.

SolidWorks 2011

Exercise 16 Zero Redundancy Model-Part 2 (Optional)

Exercise 16: Zero Redundancy Model-Part 2 (Optional)


This exercise continues the process started in the previous exercise. The remaining parts and subassemblies with the mates in the Scissor_Lift assembly need to be add to achieve zero degrees of freedom for correct results.

This exercise reinforces the following skills: I I I I

Project Description


Add mates to the assembly to achieve zero degrees of freedom.

1


Continue working on the model prepared in the previous exercise. If the assembly is not open, open Scissor_Lift from the Lesson08\ Exercises\Zero Redundancy Model folder. Complete the previous exercise first and then continue with this exercise.

2

Unsuppress layers.

Unsuppress the sub-assemblies layer_3 and layer_4.

3

Repair mates.

Mate them to the rest of the assembly so that the mechanism operates as required with redundant constraints and actual DOF count both equal to zero.

Left side

Right side

Continue building mates on the left side as indicated in the figure.

4

Continue.

Unsuppress the sub-assemblies layer_5 and layer_6. Continue adding mates to achieve zero degrees of freedom.

5

Continue.

Unsuppress platform.

Continue adding mates to achieve zero degrees of freedom.

6


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Exercise 17

SolidWorks 2011

Removing Redundancies with Bushings

Exercise 17: Removing Redundancies with Bushings


In this exercise, we will use a model that has mates applied symmetrically in preparation for exporting results to SolidWorks Simulation. To remove the redundancies, we will add bushings and explore the effects of different damping values. This exercise reinforces the following skills: I I I I

Project Description

How are redundancies removed in the solver? on page 215. How are redundancies removed in the solver? on page 215. Using Flexible Joints Option to Remove Redundancies on page 222. Bushing Properties on page 224.

This is the same Scissor Lift assembly used in the previous exercises except that the components are mated differently.

1


Open Scissor_Lift from the Lesson08\Exercises\Redundancies Removal with Bushings\completed-low stiffness folder.

2


The approach to the assembly mates is different in this model. Notice that most mates, namely the Concentric mates, represent the real mechanical connections closely. Coincident mates only ensure that the assembly does not move sideways.

Notice also that, unlike previous exercises, mates are applied on both sides of the symmetry.

This method is appropriate when we Concentric36 Concentric37 want to import the results in SolidWorks Simulation to get stress results in the different components, or if we want to see the correct force distribution at all mate locations on the model.

The problem with this mating scheme however, is that we are going to generate a large number of redundancies that will have to be removed.

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Exercise 17 Removing Redundancies with Bushings


We saw in a previous lesson that we could have SolidWorks Motion treat all redundant mates as bushings. We can also manually configure mates to act as bushings.

3

Add bushings manually.

To save time, each concentric mate has already been configured as a bushing. Edit one of the Concentric mates. Select the Analysis tab.

Notice that Bushing has been selected and the values set as follows: Translational I I I I

Select Isotropic Stiffness = 5,000 N/mm Damping = 20.0 N-s/mm Force = 0 N-mm

Torsional I I I I

Select Isotropic Stiffness = 100 N-mm/deg Damping = 20.0 N-mm-s/deg Toque = 0.0 N-mm

These values for stiffness and damping are very low for a practical system, however we will start here to see the effect on the mechanism.

Notice that each mate that has been defined as a bushing now features the bushing icon shown in the MateGroups.

4

Run.

Notice that the motion is not smooth.

Play the animation back at a slower speed and watch the action of the individual joints.

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5

Examine the plots.

Concentric20

Concentric21


Plots of the reaction force in the Z direction have already be created for mates Concentric20 and Concentric21. These mates are on opposite sides of the assembly. Notice that the plots are exactly the same.

6

Create additional plots.

Concentric36

Concentric37

Create additional plots of the force in the Z direction for mates Concentric36 and Concentric37. These mates are between the lower cross arms and the brackets on the frame.

Note

When we create these plots, we still get a warning message about redundancies. This will be explained in upcoming steps. Click Yes to dismiss the message.

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As with the previous set of mates these plots are also identical. While they are identical, they are not the sinusoidal shape that matches the driving motor due to the low stiffness of the joints.

Why do we still have redundancies?

When we created the plots, we were warned that redundancies still exist in the model. If we check the degrees of freedom we see that there are 11 redundant constraints. If you examine the mates, not all concentric mates were made flexible. For instance, the mates between the piston, cylinder and frame (that were changed in the previous exercise) are still hinge and concentric mates in this model. We do not have to change the mates in this problem because we are not interested in the forces for those components.

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Examine the other mates that have not been made flexible. These mate concern forces or motion in the global Y direction (across the plane of symmetry). As we expect these forces are going to be zero, we are not concerned with these forces and do not have to take the time to remove these redundancies.

7 8

Save and close the file. Open an assembly file.

Open Scissor_Lift from the Lesson08\Exercises\Redundancies Removal with Bushings\completed-optimum stiffness folder.

9


This is exactly the same assembly as used in the previous steps except that the stiffness of the flexible mates has been changed.

10 Examine the bushings. Edit one of the Concentric mates that is flexible.

Select the Analysis tab.

Notice that Bushing has been selected and the values set as follows: Translational I I I I

Select Isotropic Stiffness = 100,000 N/mm Damping = 2000.0 N-s/mm Force = 0 N-mm

Torsional I I I I

Select Isotropic Stiffness = 100 N-mm/deg Damping = 20.0 N-mm-s/deg Toque = 0.0 N-mm

These values for stiffness and damping are more realistic for a practical system then those used in the previous example.

11 Run the study.

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Exercise 17 Removing Redundancies with Bushings



The plots for the same four mates examined in the previous example have already be generated. As before, the plots for the symmetric pairs of mates are identical.

With the higher stiffness, we can see that after the initial acceleration, the motion is sinusoidal.

Add the two values for the maximum force (ignoring the initial spike in the magnitude) in mates Concentric20 and Concentric21, they should be approximately 9,500 N which compares favorably with the result obtain in Exercise 14: Kinematic Mechanism.

Also compare the values for the maximum force in mates Concentric36 and Concentric37, they should be approximately 15,000 N which also compares favorably with the result obtain in Exercise 14: Kinematic Mechanism.

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From these results, we can see that the forces were equal when we had all the mates on one side of the model to the total force when we removed the redundancies and split the force to the two sides.


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Exercise 18 Catapult

Exercise 18: Catapult


This exercise will further examine the use of local flexible mates to properly calculate the forces where multiple supports are used.

We will use the same catapult model that we saw in Lesson 3. With a lot of redundancies, SolidWorks Motion will solve the kinematics correctly, however the force distribution may be incorrect. This exercise reinforces the following skills: I I I I

Project Description


Calculate the forces on the pivots between the arm and counterweight.

1


Open Catapult-assembly from Lesson08\Exercises\Catapult folder.

This assembly has been set up and run in the study named original

study with results.

2


The counterweight is connected to the arm with a single mate, ConcentricB. There is also Coincident4 which holds the counterweight centered on the arm.

3

Play the simulation.

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Exercise 18

SolidWorks 2011

Catapult

4

Degrees of Freedom.


This assembly has 54 redundant constraints, but it runs without problems. While the kinematic problem is solved, the problem with the solution is that the resulting forces may not be distributed properly.

5

Create plot.

Create a plot of the global Y direction resultant force on the mate ConcentricB. We can observe a force of about -1.22 N while the arm is being rotated into position, lifting the counterweight.

Depending on our engineering judgement, this result may be good enough if we are confident that the load is shared equally as we can just divide the result in half to get the correct force at each pivot. If this is not good enough, then we must make the mates flexible to distribute the force correctly.

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6

Add another mate.


We are interested in the force in each of the pivots between the arm and counterweight, so we need to have a mate on the other pivot.

Select the Model tab, then add a Concentric mate to the other pivot. Rename this mate Concentric C.

7

Run.

Make sure that Replace redundant mate with bushings is cleared, then rerun the study.

8

Create plot.

Create an additional plot showing the reaction force in the Y direction (global coordinates) for the mate Concentric C.

We can see that the force is distributed evenly between the two mates. This, however, may be just a coincidence as the distribution will depends on how the software removes the redundancies. We will not use flexible mates to ensure the correct force redistribution.

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Exercise 18

SolidWorks 2011

Catapult

Create local flexible mates. Edit mates ConcentricB and ConcentricC.


9

Select the Analysis tab and select Bushing. Keep the default values.

10 Run.

Make sure that Replace redundant mates with bushings is cleared, then, rerun the study.

We will still have a lot of redundancies in the model, however these do not affect the results we are interested in at the two pivots.


The plots now show that the force is divided over the two mates.

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12 Change scale.


To make them easier to read, modify the two plots to show the Y Axis: I I I

Start Point = -2 End Point = 1.0 Major Units = 0.5.

We can see that the forces are exactly the same.


255

PR Do E no RE t c LE op AS y E or D di RA st F rib T ut e Exercise 18 Catapult

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PR Do E no RE t c LE op AS y E or D di RA st F rib T ut e Lesson 9 Export to FEA

Objectives


Create an Action Only Moment.

I

Export loads from SolidWorks Motion to FEA Simulation.

I

Run the structural analysis in SolidWorks Simulation.

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Export to FEA

Determining the forces on a part is generally not the end of the analysis of a part. Usually, the forces obtained are to be used in finite element analysis to determine the strength, displacement and Factor of Safety of the individual parts. SolidWorks Motion and SolidWorks Simulation work together to make the exporting of data from SolidWorks Motion to SolidWorks Simulation seamless.

Case Study: Drive Shaft

The Drive Shaft assembly is composed of 5 sub-assemblies, and 2 single parts. SolidWorks Motion will be used to determine the forces acting on one component, the Journal-cross and then using SolidWorks Simulation, we will determine the stress and displacements of the part.

Project Description

The universal joint is required to transmit a torque of 15,000,000 Nmm at a speed of 2800 RPM. Determine the stress and deflection of the part Journal_Cross_output.


Exporting Results

Output _housing Output_shaft

Journal-cross_output

Driveshaft

Output_housing

Driveshaft

Input_housing

Journal_cross_input

Input_shaft

Driveshaft

Input_housing

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Lesson 9 Export to FEA


I

Create the motion study.


Using the known data as input, create a motions study. I

Run the motion study.

The motion study is calculated to determine the forces on the part or parts in question.

I

Export loads to Analysis.

From SolidWorks Motion, export the loads directly to SolidWorks Simulation.

I

Open the part for analysis.

Open the specific part in its own window.

I

Run the FEA simulation.

Complete the boundary conditions in SolidWorks Simulation and run the analysis.

I


Use the results to determine if design changes are needed.

1

Open an assembly file. Open Drive_shaft_assembly from the Lesson09\Case Studies

folder.

2

Insert a new motion study.

Make sure the units are in MMGS.

3

Add a motor. Add a Rotary Motor to the Input_shaft. Make it turn at 16,800 deg/ sec (2,800 RPM).

Note the direction of rotation. It doesn’t matter which direction it turns only that we know the direction so that we can add the Action Only moment in the opposite direction in the next step.

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Export to FEA

Add a force. Apply an Action only Torque on the Output_shaft. This is a torque


4

that opposes the rotation, so set the direction opposite the motor added in the previous step. Input a value of 15,000,000 N-mm for the torque.

5

Study properties.

We are going to run the study for only 0.05 seconds, so we will need a high frame rate to capture enough information. Set the frame rate at 2,000. This will give us 101 frames.

6

Run.

Run the study for 0.05 seconds.

The following message will indicate that the current setting for the Number of Frames parameter seem to be excessive and may negatively impact the performance:

The playback speed or frames per second for this motion study will result in poor performance given the current study duration. Would you like these settings adjusted for better playback?

Click No to complete the simulation with the current settings.

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7

Calculate DOF.


We can see that there are zero degrees of freedom, so we have a kinematic system. Close the Degrees of Freedom window.

8

Examine the mates.

This assembly has zero DOF because of the way it was built. If you examine the individual mates, many of them are point to point or point to line to avoid removing too many degrees of freedom.

9

Plot results.

Create plots of the Angular Velocity Magnitude for both the input and output shaft. We can see that both shafts are turning at 16,800 deg/sec which was the input speed.

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Export to FEA


10 Plot the angular velocity of the driveshaft. Create a plot of the Angular Velocity Magnitude of the driveshaft. We

can see the expected variation of the velocity caused by the offset angle between the input and output.

11 Plot the required torque.

Create a plot of the torque of the input rotary motor. This is the torque required by the motor to move the shaft at this load.

FEA Export

Motion Simulation enables you to apply all of the necessary resulting quantities (forces, moments, accelerations etc.) onto the load bearing faces and solve for the for the stress and deformation analysis (SolidWorks Simulation module is required for the deformation solution). This way, Motion Simulation simplifies the transient problem using the rigid body dynamics approach and solves for the parts’ accelerations and joint reaction forces. Then, in SolidWorks Simulation, these loads are applied on the bearing faces and the stress analysis problem is solved. Motion simulation enables you to apply the loads and solve the deformation analysis using SolidWorks Simulation in two distinct ways:

I

Direct solution, where the setup, solution and the post-processing

is performed directly in the Motion Simulation interface.

I

Export of the loads into SolidWorks Simulation. Deformation

solution is performed using SolidWorks Simulation interface.

Both approaches are presented in this lesson.

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The applied (or exported) forces are transferred onto the faces only; edges and points are not allowed. Any face used in the mate definition in SolidWorks is also assumed to be the load bearing area for the applied (or exported) loads. If other entity types (points, edges) are used in the mate definition, load bearing faces have to be specified under the Analysis tab.


Load Bearing Faces

Mate location

Default initial location of the mate in motion analysis is determined using the first entity in the definition of the mate. For example, in the mate definition shown in the figure above, the initial mate location is at the center of Face<1>@Input_shaft-1/universal_bearing-1. Optionally, this can be changed by selecting a new entity in the Mate location field. Changing the location of the mate may change the motion analysis results and the resulting reaction forces somewhat; the impact of this change varies from case to case. It is recommended that you change the mate location if the initial configuration is not suitable. This can be especially important when using the motion loads for the finite element analysis using the SolidWorks Simulation modulus.

Motion Simulation also exports the body loads due to the accelerations of the parts. Similarly to the joint reaction forces, body loads are exported at each (or all) requested time step. The load bearing faces as well as the new mate locations have to be input before running the motion analysis.

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Lesson 9

SolidWorks 2011

Export to FEA

This section of the lesson demonstrates how to properly prepare a part subjected to the motion loads for the finite element analysis in SolidWorks Simulation. First, the correct load bearing faces and mate locations will be defined. Then, the motion loads are imported in the SolidWorks Simulation, where the finite element analysis and the postprocessing are performed.


Export of Loads

12 Isolate on journal_cross<1>. This is the journal_cross on the input side of the driveshaft. Isolating this

component is done just to make it easier to see the part. We are interested in computing the stresses and displacements of this part.

Examine the four mates of this part. None of the mating entities are faces, but rather points or axes. This will require us to specify the faces where the forces will be transferred for each of these four mates. Click Exit Isolate.

13 Specify the load bearing faces. Edit the first mate, Coincident24.


Select Load Bearing Faces.

Click Isolate components. This will hide all components except those associated with this mate.

Use the Select Other command to select the exterior surface of the journal_cross and the internal cylindrical face of the universal_bearing. The two parts are shown in the exploded view below for clarity.

Because the faces are touching, Treat as Bonded if touching is automatically selected. Clear Treat as Bonded if touching.

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As mentioned in the previous discussion block, the default initial mate location is determined from the first entity in the mate definition, center of the Face<1>@Input_shaft-1/universal_bearing-1. Because these two parts are in permanent contact and do not translate significantly relative to each other, the default location of the mate at the center of the above face is acceptable and would not have to be changed. It is, nevertheless, a good habit to place the initial location to the most ideal location, especially if we intend to follow with the finite element stress analysis of a part. To practice this we will change the location for all four journal_cross-1 mates. Selecting the mate location is optional. You can select either of the points that define the mate, but it is not necessary.

Click OK and Exit Isolate.

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14 Second load bearing face. Edit mate Coincident25.


Select Load Bearing Faces. Click Isolate components.

Select the two faces shown, one on the journal_cross and the other on the attachment flange.

As the faces are not touching, the option to Treat as Bonded if

touching is not available.

15 Define remaining load bearing faces.

Repeat the above procedure on the remaining two mates, Coincident26 and Coincident28.

16 Re-run the analysis and Save the assembly.

After the mate locations have been changed, the motion analysis must be re-calculated.

We will now proceed with a stress analysis of the journal_cross-1 component.

SolidWorks Simulation Users Only

266

SolidWorks Simulation may read the Motion loads for a single time step or a multiple time steps at once. In the latter case a design scenario feature of the Simulation software is used to run multiple analyses at all requested time steps. Design Study enables us to locate the critical time instance where the part exhibits the largest stresses and deformations.

SolidWorks 2011



17 Import motion loads.

Ensure that SolidWorks Simulation is added inside SolidWorks. Click Import Motion Loads on the Simulation menu.

Select the Motion Study from the list that you used to create the forces.

Select journal_cross-1 in Available assembly components, then click > to move it to the Selected components box. Click Multiple frame study.

In the Start (Frame No.) box, type 80. The End box should already be 101.

Click OK. This will import and save the load data to the CWR file for the journal_cross-1 part, and define the design study.

The above specifications define design study with 22 sets. Each set has its loads defined from the motion loads developed at the time instant of the frame associated with that set.

18 Open the part. Open the part journal_cross-1 in its own window.

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19 Select the SolidWorks Simulation study. A new static study named CM-ALT-Frames-80-101-1 has been

added. The numbers 80, 101 and 1 in the study name refer to the starting and ending frame numbers and the frame increment, respectively.

20 Select the Design Study.

A new design study named CM-ALT-Frames-80-101-1 has also been added. You can review the list of the parameters along with their values that have been imported from SolidWorks Motion. 22 scenarios corresponding to the frames 80 to 101 have been created.

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21 Apply material.


The definition of the static study needs to be completed.

Back in the static study, apply Alloy Steel to the part. In the Simulation Study tree, right-click the journal_cross part and click Apply/ Edit material.

Select Alloy Steel from the SolidWorks Materials library files. Click OK.

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22 Mesh the part. Right-click Mesh in the Simulation Study tree and click Create Mesh.

Under Advanced, verify that Draft Quality Mesh is cleared.

Move the Mesh Density slider to set the Maximum Element Size close to the value of 30 mm. Click OK and the model will mesh.

23 Study properties.

Right-click the study icon and click Properties.

Because this part is self equilibrated, Use inertial relief is on by default. Click OK.

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The inertial relief is one of the options used to stabilize selfequilibrated problems in the finite element analysis. The detailed discussion of the option is a subject of the SolidWorks Simulation course.


Note

24 Run Design Study.

Select the design study tab and click the Run button.

The 22 different sets of data will be solved sequentially.

25 Global maximum for von Mises stress.

Global maximums indicate the maximum values over all 22 scenarios.

In the design study tree, right-click Results and Graphs and select Define Design History Graph.

Click Constrains for the Y-Axis and select VON: von Mises Stress. Click OK.

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The graph shows the variation of the maximum von Mises stress in the journal_cross-1 part across all 22 scenarios. We can observe that the largest value of 5.07 e8 N/m2 (507 MPa), reached in scenarios 1 and 22, is smaller than the yield strength of the material (620.4 MPa).

27 Global maximum for resultant displacement.

Create a similar graph showing the global maximum of the resultant displacements.

The maximum displacement is 0.12 mm.

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28 Von Mises Stress plot for design scenario #15.


Design study stores full results for all computed scenarios.

In the design study, click the column corresponding to Scenario 15 to access the results.

Under Results and Graphs, double-click the VON: von Mises Stress plot.

The maximum von Mises stress magnitude in scenario #15 is 491 MPa.

29 Resultant Displacement plot for design scenario #15.

The maximum resultant displacement in scenario 15 is 0.12 mm.

30 Save and close the journal_cross part file.

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Export to FEA

This section of the lesson demonstrates how to perform stress analysis on a part directly in the SolidWorks Motion interface.

Important!

The correct load bearing faces and the mate locations specified in steps 13 to 15 must be specified for the direct stress solution in the Motion Simulation as well.

Note

SolidWorks Simulation modulus must be activated in order to obtain the stress solution.


Direct Solution in SolidWorks Motion

r

31 Simulation setup.

In the Drive_Shaft_Assembly motion study, click the Simulation Setup icon . In the Part for Simulation field select the journal_cross<1> on the input side of the driveshaft.

Specify 0.0395s and 0.05s for the Simulation Start Time and Simulation End Time, respectively. Click Add Time to add the time range to the Simulation Time Steps and Time Ranges field. Under Advanced, move the mesh slider to set the Mesh Density Scale Factor to 0.95 to generate finer mesh. Click OK.

The following message will display:

Do you want to assign material to the part?

Click Yes to open the Material window.

32 Material.

Similarly to step 21, specify Alloy Steel. Click Apply and Close.

33 Solve finite element simulation.

Click the Calculate Simulation Results button

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34 Stress results at 0.045s.


To show the result plot, move the time line to 0.045s.

Note

The specified time must fall within the time range requested in step 31.

Set the Results Plot button to show the von Mises Stress Plot.

The legend indicates the maximum stress of approximately 542 MPa. However, because journal_cross<1> is shown in the context of the whole assembly the stress contours are not easily visible. In this case clearer plot will be provided when we isolate the part.

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35 Isolate on journal_cross<1>.

The stress contours are now visible. The indicated maximum of 542 MPa is below the yield strength of the material, 620.4 MPa.

36 Factor of Safety at 0.045s.

Follow steps 34 to 35 and show the plot of Factor of safety.

The minimum factor of safety indicated in the plot is 1.14 (620.4/ 542=1.62).

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37 Deformation results at 0.045s.

The maximum resultant displacement at time 0.045s is 0.37 mm.

38 Results at different times.

Move the time line to any other time step. The contour plots will update automatically.

Note

The specified time must again fall within the time range requested in step 31.

39 Animate and show the overall maximums.

To set the legend to show the overall maximum for the requested analysis time range and to see the animation, click the Play button.

The maximum resultant displacement over the entire requested analysis interval (0.035s - 0.05s) is 0.38 mm.


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This lesson showed the procedure for the application of the joint and body loads computed in the Motion Simulation in the finite element stress analysis. In the first part we solved the rigid body dynamics problem and obtained the necessary joint and body loads. Then, load bearing faces and mate initial locations were specified in the assembly mate definitions.


Summary

In the second part the loads from multiple time steps were applied on the selected part and the stress analysis was carried out. Two procedures are currently available: direct stress solution in the Motion Simulation interface, or the export of the motion loads in the SolidWorks Simulation. In the later case, the stress solution is carried out in the SolidWorks Simulation interface with the help of the design study feature.

The above procedures allowed us to locate the extreme stress in the part of the rotating drive shaft assembly. Displacements, factor of safety and other results available in the SolidWorks Simulation are available and were shown in this lesson.

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Exercise 19 Export to FEA

Exercise 19: Export to FEA


In this exercise, we will export the loads for a latch mechanism to SolidWorks Simulation and conduct an analysis of the part. This exercise reinforces the following skills: I

Project Description

Exporting Results on page 258.

Determine the maximum stress and deflection on the part J_Spring.

1


Open Full latch mechanism from the Lesson09\Exercises folder. This is the same assembly used in Lesson 4. The motion study is already set up and has been run.

2

Play the study. Click Play (do not run) just to refresh your memory about how the

mechanism works.

3

Specify load bearing faces. Locate mate Concentric6. This is the mate that is used as the pivot for

the spring.

Edit the mate and specify the four faces shown as the load bearing faces. The two parts are shown in exploded view for clarity.

For the Mate location, select the edge of the split surface on either the clip or pin.

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4



Because the contact faces as well as the mate locations were changed, the motion simulation needs to be recalculated.

Contact Forces

While the forces in the mates can be imported to SolidWorks Simulation automatically, the contact forces cannot and must be defined manually.

We will first determine where the contact forces are maximum through observing the plots created in SolidWorks Motion. We will then determine the frame at which this maximum force occurs so that we only have to output the data for a single frame. We must also determine the directions along which these forces must be applied.

5

Examine the plot of contact force.

The plot of the magnitude contact force between the J-spring and the keeper is already created. We can see that the maximum force occurs at about 2.4 seconds.

Fmagnitude, max

6

Create additional plots.

Plot the contact forces for both the X (red) and Z (blue) components. We expect the Y (green) component to be zero, or nearly so, therefore it is not important to the analysis.

Note

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By default, the forces are output in the assembly global coordinate system.

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You do not have to select the actual contact faces, only the components.


Note

Fx, max

Fz, max

Ignoring the short duration peaks, notice that at the point where the X component is maximum, the overall magnitude is not at its maximum. We will run the analysis at the point where the overall magnitude is maximum. For additional practice, run the analysis at the point where the X component is maximum.

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7

Create an additional plot.


Create a plot of the joint reaction force for the mate linking the J_spring to the knurled_pin, Concentric6.

Modify the Y axis of the plot so that the End Point is 50.

Compare this plot to the first plot of the contact force. Both plots should be exactly the same, i.e. the magnitude of this force must be the same as the contact force.

Close the plot.

8

Modify the plot to show frames.

Modify the contact force magnitude plot to show the X axis in Frames.

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9

Change the axis range.


To better see the area of interest, modify X axis of the plot to show from frame 320 to frame 340.

To make the graph easier to read, change the X axis major and minor units to 10 and 5 respectively.

We can see that the most extreme loading occurs at about frame 325. When we export the motion loads to SolidWorks Simulation, we will export the data from just this one frame.

10 Modify plots.

Change the X axis to Frames for both the X and Z contact force plots.

11 Export forces. In step 6 we determined that the

two directions of interest were X and Z as shown in the image. We will just export these as the Y direction should essentially be zero. Export both the X and Z contact forces for the J_spring to CSV files. Right-click each plot and click Export CSV.

Each file will get a default name and be saved in the same directory as the assembly.

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12 Examine the output data.


Open each of the two CSV files and note the values at frame 325.

13 Export Motion Loads.

When the calculation completes, save the result and export the loads for the J_spring for the frame 325 only.

14 Open the part. Open the J_spring part in

its own window.

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15 Simulation study.


Select the simulation tab for the new study CM1-ALT-Frame325.

The mate loads have been imported into the part, but we will have to apply the contact forces manually.

Notice the directions of the global coordinate system for this part are different from the assembly. The X direction in the assembly is the Y direction in the part and the Z direction in the assembly is the X direction in the part. When we apply the contact forces to this part, we will have to insure we are using the correct force for the direction on the part.

16 Apply X contact force. Apply a force of -34.80615053 (from the exported values in the CSV file) to the indicated face. Select the Right plane to define the

direction. As the negative value of the force was the reaction force, we have to reverse its direction to be correct in the part.

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17 Apply Z contact force. Apply a force of 9.569044324 to the indicated edge. Select the Top

plane to define the direction. Reverse its direction to be correct in the part.

18 Apply material.

Apply the material Alloy Steel to the part in the Simulation Study tree.

19 Mesh the model. Right-click Mesh in the

Simulation Study tree and click Create Mesh. Use the default settings. Click OK.

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20 Run the study.


Right-click the study and click Run. We will get a warning that says:

Warning: There is a significant external imbalance force in the X-direction which will be balanced by the application of opposing inertia forces. Unless you model is under such a force or under marginally imbalance forces, application of Inertia Relief may alter the characteristics of your model.

This message is the result of exporting the loads from the motion simulation and entering values by hand. This part can therefore be considered as nearly self equilibrated. Click Yes. Click Yes.

21 Stress Plot.

Examine the stress plot. We can see that the maximum stress 150 MPa and is on the underside of the J_spring.

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22 Factor of Safety. Create a Factor of Safety plot to determine if the part is yielding.

Right-click the Results folder and click Define Factor of Safety Plot.

Use the default values to create a plot that shows the Factor of safety distribution.

Click OK.


We can see that the minimum Factor of Safety is 4.12, so the part is not yielding.


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Objectives


Understand and run event based simulation.

I

Apply servo motors.

I

Create events with specific timing and logic.

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Event Based Simulation

This lesson introduces the event based motion simulation of the mechanism, which incorporates the event-triggered control.

Case Study: Sorting Device

The sorting device shown in the figure is used to sort two types of boxes: yellow with the hole and the solid brown. Each type should be moved to the corresponding bay. Event based simulation will be used to simulate this mechanism.

Problem Description

The mechanism used to sort the boxes into the respective bays consists of six parts. The vertical motion of the boxes is caused by the gravity. The horizontal motions are then driven by a set of three pistons with servo motors. Motors actuate the motion based on a set of sensors controlling the box type and their position in the mechanism.



Simulate a mechanism placing each box type into its respective bay.

1


Open Sorting device from the Lesson10\Case Studies folder.

2

Verify the units.


3

Create a new Motion study. Name the study Sorting device.

Servo motors

Servo motors are both rotational and linear motor features driving mechanisms in event based simulation. Their motion is, however, not prescribed directly in the Motor FeatureManager. It is controlled via an event based simulation interface, and it can be triggered based on various criteria such as proximity of a certain part in the system.

Introducing: Servo Motors

Servo motors are used as motion drivers in the event based simulations.

Where to Find It

I

290

On the MotionManager toolbar, click Motor select Servo Motor.

. Under Motion

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Lesson 10 Event Based Simulation

Servo motor #1. Define linear Servo Motor for Actuator<1>.


4

Click the Motor icon and select Linear Motor (Actuator).

Select the indicated face for both Motor Location and Motor

Direction.

Under Motion select Servo Motor and Displacement.

Rename this simulation component to Actuator 1. Click OK.

5

Servo motor #2 and #3.

Similarly, define two more linear, displacement based servo motors for Actuator<2> and Actuator<3>.

Rename the two motors to Actuator 2 and Actuator 3, correspondingly. Click OK.

Sensors

Sensors can be used to trigger events or stop them. Three different sensor types can be used in event based simulations: I

Interference detection sensor for detecting collisions.

I

Proximity sensor, which detects motion of a body crossing a line.

I

Dimension sensor used to detect the relative position of component

from dimensions.

Introducing: Sensor s

Sensors can be used to trigger or stop the motion in event based simulation.

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Where to Find It

In the SolidWorks FeatureManager, right-click Sensors and select Add Sensor. On the Evaluate tab click the Sensor button .


I I

6

Proximity sensor #1.

Two proximity sensors are used to control the system. Sensor 1 is used to detect the solid box on the bottom platform of the holder. Sensor 2 is then tracking the hollow box.

Sensor 1

Sensor 2

Define Proximity sensor detecting the presence of the solid box on the platform.

Select the indicated face of sensor 1 for Proximity sensor location. The

Proximity sensor direction

field may remain empty to keep the default vertical direction.

Select the two solid boxes for the Components to track field. Enter 12 mm for Proximity sensor range. Click OK.

Rename this sensor to Sensor 1.

A 12 mm range was used to trigger the necessary event when the box reaches the horizontal platform of the holder. Since the thickness of the platform is 10 mm, any sensor ray longer than 10 mm will trigger an event as the box approaches the platform.

Note

7

Proximity sensor #2.

Similarly, define proximity sensor #2 to track the boxes with the hole. Rename this sensor to Sensor 2.

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8

Contacts.


Define the following four solid bodies contacts. Contact group 1 (Acrylic)

All in one group (Acrylic)

Contact group 2 (Steel (Dry))

Contact group 1 (Acrylic)

Contact group 1 (Acrylic)



Whenever possible, use contact groups to simplify the contact solution.

9

Gravity. Define Gravity in the negative Y direction.

Task

Event based simulation requires a set of tasks, triggered by sensors, and ordered sequentially or overlapping in time. Each task is defined by a triggering event and its associated task action, which controls or defines motion during the task.

Triggers

Each task is triggered with a triggering condition. The triggering condition can depend on the status of a sensor, or start or end of some other task in the sequence.

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The following is a list of actions which can be specified in a task definition.


Task Action

I I I I

Timeline View vs. Event-based Motion View

Stop . Stop the motion of a component. Motors . Turn on or off any motor, or change a constant speed of

motor according to the selected profile. Forces . Apply or stop applying any force, or change a constant force according to a selected profile. Mates . Toggle the suppression of a selected mate.

To define the task, Motion Simulation offers an Event-based Motion View which can be accessed through the corresponding button on the MotionManager toolbar . This view is used to define tasks and design the logic of the system. Tasks design table

Tasks time sequence and logical relationship

Timeline based view provides classical Motion simulation view with the keys, indicating the beginning, end and change in the behavior of the simulation components. The sequence of the keys in time is generated when the event based simulation computations complete and is also an important result of the simulation.

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Task controls and defines motion of the components during the simulation. It is defined by a triggering event and its associated action.


Introducing: Task

Where to Find It

I

On the MotionManager toolbar, click the Event-based Motion button. To add a task, click the Click here to add line on View the bottom of the task list table.

10 Event-based motion view. Switch to the Event-based Motion View. 11 Task #1 - Name and Triggers.

The first task for the system is to move the lowest solid box along the holder platform to the position, where Actuator 2 may push it into Bay 1. This task will be triggered when the bottom solid box activates the proximity Sensor 1. Because this sensor triggers an event when the solid box is 2 mm above the platform, and to provide enough time for Actuator 2 to fully retract, a 0.1s time delay for this task will be specified. Click Click here to add line to add a new task line. Enter Push solid box for Name.

Click here to add a new task line

Click the selection button in the Trigger field to open the Trigger dialog window. Select Sensor 1.

Click OK to close the Trigger dialog window.

Back in the Event-based Motion View, complete the Triggers section by setting Condition to Alert On and Time/Delay to 0.1s (see the figure in the next step).

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12 Task #1 - Action.


The task definition will complete with the specification of action. In this case the action comprises of Actuator 1 motor pushing the solid box 75 mm along the platform. (this is an ideal position for subsequent action of Actuator 2). Select Actuator 1 from the Motors in the Feature field, Change for Action, 75mm for Value, 1s for Duration and choose Harmonic for Profile.

13 Task #2 - retracting Actuator 1. Define the second task to retract Actuator 1. This task should be triggered after the task #1, Push solid box, completes and its duration is 0.2s.

Name this task Retract Actuator 1.

14 Task #3 - pushing solid box into Bay 1. Define the third task, this time for Actuator 2, pushing the solid box into Bay 1. This task comprises of a 50mm extension of Actuator 2 in 0.6s.

Similarly to the task #2, Retract Actuator 1, the task #3 is triggered by the completion of the task #1, Push solid box.

Rename this task to Push solid box to Bay 1.

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15 Task #4 - retracting Actuator 2. Similarly to step 13, define the task #4 retracting Actuator 2 in 0.1s. This task is triggered by the completion of task Push solid box to Bay 1.

Rename this task Retract Actuator 2.

16 Tasks for boxes with hole. Follow steps 11 to 15 and specify similar tasks to move the box with the hole into Bay 2.

To move the box with hole next to Actuator 3, extend Actuator 1 by 130 mm in 1.2s with a 0.1s delay. Then retract Actuator 1 in 0.3s.

Use the same time and distance values for Actuator 3 as those which were used for Actuator 2 in steps 14 and 15. Give the new tasks names similar to those used in steps 11 to 15.

17 Simulation Properties. Set the Frames per second to 200 and check Use Precise Contact.

Under Advanced Options, set Maximum Integrator Step Size to 0.05s.

Note

To speed up the simulation and because we are not interested in the force results, the maximum integrator step size value can be relaxed.

18 Calculate simulation for 7 seconds.

It takes approximately 15 minutes for the simulation to complete.

19 Animate. Animate the final motion of the system.

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20 Timeline View. Switch to the Timeline View. Here you can see the result of the event

based simulation. Each key indicates the beginning, end or a change in the motion of the system components. It also indicates the duration of the whole cycle.

The Timeline View provides insight into the duration of the whole operation. Each task start and end is identified with a time key. Possible action following this simulation may be a change in the velocities of the actuators in order to optimize the system, change of material in order to change the effect of friction, change of the design to better stack the boxes in the bays, or similar.


Summary

In this lesson, we introduced the event based motion simulation to model and optimize the behavior of mechanical systems. We modeled an operation of the sorting device mechanism used to move two different box types into their respective bays.

The event based simulation is a sequence of tasks triggered by sensors, ordered sequentially or overlapping in time. Each task is defined by a triggering event and its associated task action which controls or defines motion during the task. The triggering condition can depend on the status of a sensor, or the start or end of some other task in the sequence. The triggered action can be applied to motors, forces, mates or can completely stop the motion of a moving part. A special type of motor, servo motor, was introduces and utilized.

The tasks with their respective actions, sequence and the logic is designed in the Event-based Motion View interface introduced in this lesson.

The result of the simulation, the animation and the duration as well as time sequencing of the entire operation was shown. This result can then be used to modify the system parameters (kinematic parameters of the actuators, for example) to optimize it.

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Objectives


Create a function based force.

I

Export loads to SolidWorks Simulation.

I

Complete an analysis project from motion to FEA.

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Design Project (Optional)

This lesson is in two parts, in the first section it is up to the individual students to solve the problem of the Surgical Shear. In the second part of the lesson, the complete solution will be shown.


Design Project

The Case Study is also in two parts:

The overall problem is to determine the suitability of the design of the handle used in the surgical shear. Part 1: Develop loads on the parts based on a motion study. This will require the development of a force function to simulate the resistance of the catheter being cut by the surgical shear. Part 2: Conduct an FEA simulation of the handle, using the loads developed in part 1.

Case Study: Surgical Shear Part 1

The Surgical Shear is used to cut arteries and catheters. It consists of a fixed blade and a moving blade.

Because the surgical shear has to be operated by many people in the medical industry, it is important to estimate the handle force that will be sufficient to generate the required cutting force.

In this part of the lesson, we will mate the components, create a motion study and develop a force function to simulate the blades cutting through a catheter.

Problem Description

The mechanism is composed of seven parts. The fixed_cutter is stationary and the rotation of the handle provides the motion of the moving_cutter. The latch rotates with respect to the fixed_cutter and is inserted into the moving_cutter. A spring maintains the moving_cutter in an open configuration when no force is applied to the handle. The removable blades are attached to the fixed_cutter and moving_cutter parts.

When the surgeon squeezes the handle, it will rotate 12 degrees and move the blade. The spring is used to help to return the shear to the open position. It is assumed that it takes a surgeon approximately one second to cut the catheter.

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Determine the design suitability of the handle part. fixed_cutter

latch

handle_link

blade1

blade2

moving_cutter

handle

Force to Cut the Catheter

From experimentation, the force to cut a 3 mm catheter was determined and plotted in the graph below. The X axis shows the travel of the blade starting at Point 1(X=0mm) where the blade contacts the catheter. The cutting force increases slowly at first as the catheter is compressed and then climbs more rapidly as we approach the point when the cutting begins. At Point 2 (X=1.5mm) the blade starts cutting the catheter and the force reduces quickly as the cut portion of the catheter returns to its round shape.

From Point 3 to Point 4, the remaining thickness is cut and at Point 4 the cut is complete.

Point 2

Point 1

Point 3

Point 4

X

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The preceding graph contains the variation of the cutting force from the experiment. In order to input this curve in SolidWorks Motion Simulation each segment has to be expressed as function of the catheter location (location of the cutting blade); in the above graph this location is expressed through the variable “x”. Notice that it assumes values from 0 mm (cutting blade touches the catheter) to 3mm (cutting blade completes the cut). Each segment is therefore expressed as a linear function shown in the graph.

Note

The above cutting force curve as measured from the experiment is expressed as function of the blade location (not the function of time). Time dependent data is not available since, in general, it depends on how fast the cutting operation is completed and how the input force from the surgeon’s hand varies in time. While the input of the time dependent force would be trivial (this procedure was practiced multiple times throughout this course), input of the location dependent force is more challenging. Note also, that with certain assumption the above location dependent function can be converted in time dependent input. However, to demonstrate the more complex case (which may be required in some analyses) we will use the location driven input.

As we will see, inputting the number of data segments into the force expression can be tedious. We will simplify this curve as shown in red, with only three segments; this should be enough to reasonably simulate the cutting force.

First Segment

Second Segment

Third Segment

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Each of the three segments can be defined by a linear equation:


First Segment: y = 7.333333 x

Second Segment: y = -80 x + 131

Third Segment: y = -2.14286 x + 6.42857

Self Guided Problem - Part 1

In this section, it is up to each student to solve this motion part of the problem. A basic outline of the procedure is provided as a guide, but the details of the steps are left up to the student.


The basic steps are: I

Add mates.

Add the appropriate mates to insure that the mechanism operates as specified.

I

Determine the cutting force.

The action/reaction force on the blades while cutting a catheter has been experimentally determined and is not linear. From the experimental data, develop an expression to simulate this force on the blades.

I

Run the motion analysis.

Run the study and create the appropriate plots.

I

Analyze the mechanics. Interference detection is run to make sure that the mechanism will move through its full range of motion. Load paths are examined to ensure that the correct forces have been calculated in the simulation.

You can view the Surgical_shear.avi movie to help you understand the mechanism motion.

Note

The steps below outline the procedure to form a road map of the necessary steps:

1


Open Surgical_shear from the Lesson11\Case Studies folder. When opened, the components are not mated.

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2

Mate the components.


It is up to the student to determine the best method to mate the components to reflect the mechanical operation of this mechanism and reduce redundancies.

Keep in mind that the part of interest in this project is the handle.

3

Add motion drivers.

Add appropriate motors and springs to capture the design motion (see the problem description).

4

Develop a distance based force.

The action/reaction force developed by cutting the catheter is not linear. An expression for this force, based on the position of the blades, must be developed to simulate the experimentally determined forces.

5

Analyze the results.

Create plots and check for interference. Modify parts as necessary.

Self Guided Problem - Part 2

In this section, it is up to each student to solve this FEA part of the problem. A basic outline of the procedure is provided as a guide, but the details of the steps are left up to the student.


I

Export loads to SolidWorks Simulation.

Once we have the loads calculated, they are exported to SolidWorks Simulation to evaluate the suitability of the parts.

I

Replace motion drivers.

Some motion drivers such as motors need to be replaced with forces or moments in order to run a static analysis.

I

Analyze part.

Analyze the part using SolidWorks Simulation to determine the suitability of the part based on strength and deflection.

I

Refine part.

If the analysis determines that the part is not suitable as designed, modify the part as necessary and re-run the analysis.

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Problem Solution - Part 1


In this section, the complete solution to this problem is provided. 1


Open Surgical_shear from the Lesson11\Case Studies folder.

Mate names are normally not important, however to ensure that mates described in the text are consistent with the model, specific mate names are given in the following steps. If you apply mates in a different order, just rename the mates to be consistent with the images.

Note

2

Lock mates.

The two blades are rigidly connected to the fixed and moving cutters, so the appropriate mate would be the lock mate.

Lock1

Lock2

3

Coincident mates.

Coincident1 Coincident2

The moving_cutter slides along the outside faces of the fixed_cutter. Two Coincident mates, between the faces shown and the corresponding faces on the moving_cutter, can be used to hold this relationship.

fixed_cutter

4

Mating the linkage.

Concentric1

We need three mates to connect the linkage.

A Hinge mate can be used to connect the handle to the fixed_cutter, and another Hinge mate to connect the handle_link to the moving_cutter.

The remaining mate between the handle and handle_link should be a Concentric mate, preferably using faces, to avoid over defining the mates.

moving_cutter

Hinge1

Hinge2

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5

Mating the latch mechanism.


The latch mechanism needs two different mates. A Hinge mate is used to control the rotation and position with respect to the fixed_cutter. The selected surfaces are shown in the image (the moving_cutter has been hidden).

A Cam mate could be used to mate the boss on the latch to the slot in the moving_cutter. In our initial solution to this problem, we will not use a Cam mate, but will use Contact in the motion study.

6

Cam

Set the initial position.

Before creating the motion study, we need the blades to be 7.25 mm apart as the initial position. Use a Position Only mate to set the distance.

Now that everything else is positioned, we need to also make sure that the boss on the latch is touching the appropriate surface. Later, when we add contact and a spring, those conditions will force the boss onto the surface. We want to make sure however, that when the motion study starts, the boss doesn’t have to move onto the surface and create a transient condition that would not occur in the physical part.

Temporarily Fix the moving_cutter and use a Tangent Position Only mate to set the boss on the latch against the surface of the slot. When finished, Float the moving_cutter.

7

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8

Add a spring.


A linear spring is used to connect the latch to fixed_cutter.

The spring has a stiffness of

0.175 N/mm and free length of 40 mm.

9

Add Contact. Add Solid Bodies contact between the latch and the moving_cutter.

Specify Steel (Dry) for the material.

Select Friction, both kinematic and static. Click OK.

10 Add a rotary motor.

When the surgeon uses the shear, the squeezing of the handle will rotate it 12 degrees. It takes approximately 1 second to squeeze and open the shear. To simulate this action, we will add a Rotary Motor. The motor parameters should be Oscillating, 12deg at 1 Hz with 0deg Phase Shift.

11 Motion Study Properties.

Set the properties to record 100 Frames per second. Select Use precise contact.

Verify that Replace redundant mates with bushings is cleared.

12 Run the simulation for 1 second.

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Now, we need to create an action/ reaction force that will simulate the resistance effect of cutting the catheter. We will consider a catheter with 3 mm diameter.


Creating the Force Function

The force will have to be defined in several steps based on the physical conditions. Before creating an expression, we should be able to describe the motion in words:

Step 1: During the initial movement of the blade, there is no force as the blade is moving through open air.

Step 2: Once the blade contacts the catheter, there is a resistance as the catheter is compressed before it is actually cut. Step 3: The catheter is cut and the force is rapidly reduced.

Step 4: The catheter is cut, but the blade continues forward without resistance. Step 5: The blade moves back to the starting position without resistance.

Steps 1, 4 and 5 above are easy as they are just zero, while our real problem is defining the force in Steps 2 and 3.

Force to Cut the Catheter

The experimental data was shown on page 301. The two graphs are repeated below. Point 2

Point 1

Point 3

Point 4

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The simplified Cutting Force plots is shown again with the equations for the three segments.

First Segment

Second Segment

Third Segment

Each of the three segments can be defined by a linear equation: First Segment: y = 7.333333 x

Second Segment: y = -80 x + 131

Third Segment: y = -2.14286 x + 6.42857


We will build the full expression piece by piece to see how it is constructed. I I I I I I

Create a variable for the location of the cutting blade as it cuts through the catheter (variable x in the above graph). Plot the force function for the first segment Plot the force function for the first and second segments Plot the force function for the first, second and third segments Terminate the force function when the cutting blade cuts through the catheter completely (x=3mm) Set the force function to zero in the second part of the cutting process when the cutting blade moves in the opposite direction.

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13 Determine the blade clearance.

Measure distance between blades. The distance is 7.25 mm, so if the catheter is 3mm, then at the start there is a clearance of 4.25 mm.

14 Create plot of the displacement between blades.

Select the two vertices in the order shown. If you select them in the reverse order, the plot will be reversed. As the force is a function of blade position, we need to know the position of the blades. By creating the plot, we have a variable to use in the expression.

Important!

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1 2

The remainder of this section of the lesson assumes that this is the first linear displacement plot and therefore its name is LinearDisplacement1. Likewise the force we are about to add will be Force1 and the linear velocity plot created later will be LinearVelocity1. If you have created other plots or forces and the plot names you obtain in these steps are different, you must either rename your plots or substitute your plot names as appropriate.

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The goal in the next section is to develop an action/re-action force between the two blades that represents the force necessary to cut the catheter.


Creating the Force Expression

Rather than apply the force directly to the blades while developing the force expression, we will use a dummy force that will not affect the outcome of the motion analysis. We are going to apply this force to the fixed_cutter. As SolidWorks Motion is a rigid body analysis tool, any force applied to a fixed part can have no effect on the motion analysis.

15 Add a force.

This force does not affect the results as it is applied to a non-moving part. We will use it to develop the full expression for action/reaction force.

Define this force as a function equal to the Linear Displacement1 defined in the previous plot.

16 Run.

Run the simulation for 1 second.

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17 Create a plot. Plot the Y Component for the Reaction Force. The force is now

directly related to the position of the moving blade.

Force = 0

w


The above force starts at -7.25 because the blades are 7.25 mm apart at the beginning of the simulation. The force is zero when the distance between the blades is zero. In our simulation however, we want the force to be zero when the blades are 3 mm apart (when the blade first contacts the catheter). Change the force expression to be {LinearDisplacement1} + 3.

19 Re-run the simulation.

The force now starts at -4.25 because that is the distance from the blade to the catheter. The force is now zero at contact between the blade and catheter.

Force = 0

Important!

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This last expression, {LinearDisplacement1} + 3, therefore defines the X variable used in the expressions on page 309.

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The expression we are going to develop is: IF({Linear Velocity1}:IF({Linear Displacement1}:IF({Linear Displacement1}+3:0,0,7.333333*({Linear Displacement1}+3))+IF({Linear Displacement1}+1.5:0,0,7.333333*({Linear Displacement1}+3)-80*({Linear Displacement1}+3)+131)+IF({Linear Displacement1}+1.4:0,0,80*({Linear Displacement1}+3)-1312.142868*({Linear Displacement1}+3)+6.42857),0,0),0,0)


Force Expression

While this expression, at first, looks complicated, it is just a nested set of IF statements.

IF Statement

The IF statement is used to define an output based on the sign of an input variable. It is in the form of: IF (Input variable: A, B, C)

When the value of the Input variable is negative, output the value A. When the value of the Input variable is zero, output the value B.

When the value of the Input variable is positive, output the value C. The Input variable, A, B and C can all be either fixed values or expressions.

In the expression above, we can see that in all the IF statements, there are only two different input variables, LinearVelocity1 and LinearDisplacement1.

Developing the Expression

The first thing is to define the point where the blade first touches the catheter. At this point and before, the force must be zero. From our measurements, when the blades are open, they are 7.25 mm apart, and the catheter is 3 mm in diameter. Therefore, the contact occurs when {LinearDisplacement1}+3=0. We determined this in Step 18. Therefore, the Force in the first segment will be:

IF({Linear Displacement1}+3:0,0,7.333333*({Linear Displacement1}+3))

This says, I I I

When the value of {Linear Displacement1}+3 is negative, the expression equals zero. When the value {Linear Displacement1}+3 is zero, then the expression equals zero. When the value of {Linear Displacement1}+3 is positive, then the value of the expression will be 7.333333 * ({Linear Displacement1}+3). The value of 7.333333 comes from the experimental data and is the slope of the curve in Segment 1.

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20 Input the expression.

Edit the force.

Input the above expression. Remember to enter the variable Linear Displacement1, you can double-click it in the list below the expression entry box.


Examine the plot of the force.

The plot is correct from distance zero to 5.75 (Point 2 in the graph on page 308). At that point, the force continues to climb, so we need add to the IF statement to define Segment 2 (see the graph on page 309).

When you first look at the plot, it may not look correct as we used the equation of a straight line (7.333333 * ({Linear Displacement1}+3). Remember however that the linear equation is based on displacement while the plot is versus time. As the blade motion is not linear, the plot is correct. To get through Segment 2 we need to add more to the expression so it will be: IF({Linear Displacement1}+3:0,0,7.333333*({Linear Displacement1}+3))+IF({Linear Displacement1}+1.5:0,0,7.333333*({Linear Displacement1}+3)-80*({Linear Displacement1}+3)+131)

If you examine the expression, it is the sum of two IF statements. The first IF statement is: IF({Linear Displacement1}+3:0,0,7.333333*({Linear Displacement1}+3)

This is the part we already had. We then add to it:

IF({Linear Displacement1}+1.5:0,0,-7.333333*({Linear Displacement1}+3)-80*({Linear Displacement1}+3)+131)

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This says, when {Linear Displacement1}+1.5 is either negative or zero, its value will be zero. In other words, until the blade displacement is 5.75 (Point 2), this part of the expression has no effect. Once positive, the value will be:

-7.333333*({Linear Displacement1}+3)-80*({Linear Displacement1}+3)+131

The first part of this is the negative of the first expression, so it is used to zero the effect of the first expression. The second part of the expression is the equation for the force in segment 2: -80*({Linear Displacement1}+3)+131


Edit the force.

Input the above expression.


Examine the plot of the force. For now, we are just interested in the area circled.

Edit the Y axis so that it shows form -11 to +11. This will make it easier to see the area of interest.

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The plot is correct from distance zero to 5.85 (Point 3 in the graph on page 308). At that point, the force needs to reduce at a different rate based on Segment 3 of our experimental data. So, we will again add to the IF statement to define Segment 3. To get through Segment 3 we need to add more to the expression so it will be: IF({Linear Displacement1}+3:0,0,7.333333*({Linear Displacement1}+3))+IF({Linear Displacement1}+1.5:0,0,7.333333*({Linear Displacement1}+3)-80*({Linear Displacement1}+3)+131)+IF({Linear Displacement1}+1.4:0,0,80*({Linear Displacement1}+3)-1312.142868*({Linear Displacement1}+3)+6.42857)

Again, the first part of this expression is what we had before. The new statement is:

IF({Linear Displacement1}+1.4:0,0,80*({Linear Displacement1}+3)131-2.142868*({Linear Displacement1}+3)+6.42857)

The first part, IF({Linear Displacement1}+1.4:0,0,80*({Linear Displacement1}+3)-131is again just the negative of the previous part of expression to zero it out. The remaining part: 2.142868*({Linear Displacement1}+3)+6.42857)

defines the curve of Segment 3.


Edit the force.




The plot is now correct from zero until the blade finishes the cut (Point 4). We now have to add another IF statement that will make the force zero from this point until the end of the blade travel.

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Let’s call the entire expression we have developed thus far Force1. The IF statement we need is then: IF ({LinearDisplacement1}: Force1, 0, 0)

When the above expression is negative (blades have not touched yet), use the entire Force function. If it is zero (the cut is complete) or negative (blades overlapping), then the force will be zero. The full expression will now be:

IF({Linear Displacement1}:IF({Linear Displacement1}+3:0,0,7.333333*({Linear Displacement1}+3))+IF({Linear Displacement1}+1.5:0,0,7.333333*({Linear Displacement1}+3)-80*({Linear Displacement1}+3)+131)+IF({Linear Displacement1}+1.4:0,0,80*({Linear Displacement1}+3)-1312.142868*({Linear Displacement1}+3)+6.42857),0,0)

Let’s call this expression Force2.


Edit the force.


27 Run the simulation. 28 Edit the plot.

Change the Y axis back to automatic scaling.

The plot is now correct for the forward travel of the blade, but forces are mirrored on the blade retraction where they should instead be zero.

To solve this problem, we will add another IF statement, based on the velocity of the blade that will only use the previously define force function (Force2) when the blade velocity is negative. That is the portion of the motion when the surgeon is squeezing the handle and the blades are closing. When he releases the handle and the spring is returning the blades to the open position, the blade velocity will be positive.

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29 Create a new plot.


Create a plot of the Linear Velocity, X Component of the vertex of the blade shown.

We only want the force to equal the force function when the velocity is negative. Once the velocity is zero or positive, it should be zero.

The new IF statement is:

IF({Linear Velocity1}:Force2,0,0)

In the above expression Force2 is used to represent the entire force function we have already defined. We can see in the expression that it will only be used when the velocity is negative. When the blades stop moving and returns, the force will be zero. If we insert the previous expression for Force2, we get:

IF({Linear Velocity1}:IF({Linear Displacement1}:IF({Linear Displacement1}+3:0,0,7.333333*({Linear Displacement1}+3))+IF({Linear Displacement1}+1.5:0,0,7.333333*({Linear Displacement1}+3)-80*({Linear Displacement1}+3)+131)+IF({Linear Displacement1}+1.4:0,0,80*({Linear Displacement1}+3)-1312.142868*({Linear Displacement1}+3)+6.42857),0,0),0,0)


Edit the force.


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The plot is now correct for the entire motion of the blade. Its shape is now the same as the input data from the experiment.

32 Edit the force.

Now that the force expression is properly defined, we need to have it applied to the blades as an Action/Reaction force. Change the force from Action only to Action & reaction. Select the two blade vertices as shown. Click OK.

33 Modify plot.

Edit the force plot and change it to show the X Component. The original force was in the Y direction while the direction between the to blades is the X direction.

Make sure that the force is positive. If it is negative, switch the order of the vertices in the action/reaction force definition (Step 32).


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In the second part of this Case Study, we will examine the design of the handle of the surgical shear.


Case Study: Surgical Shear Part 2

We have already run the motion analysis to determine the loads.

Problem Description

Determine the stresses on the handle part based on the maximum loading found in the motion analysis. Evaluate the handle part for suitability based on the results.


The basic steps are: I

Evaluate the redundancies.

There were several redundancies in the motion study. Each must be evaluated to determine its effect on the loads needed for the FEA problem.

I

Interference detection.

The assembly must be checked to make sure that parts only contact each other where designed and that there are no contacts that will stop the assembly from working properly.

I

Export loads.

Exports the loads from SolidWorks Motion to SolidWorks Simulation.

I

Evaluate the imported loads.

The motion loads may not be correct for simulation. Each is evaluated to make sure it is correct for the FEA process.

I

Replace imported loads with local loads.

Loads that were not suitable for FEA must be replaced with the appropriate load or fixture.

I

Run the simulation.

I

Evaluate the results.

All results need to be evaluated to make sure the part works as intended without failure.

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Redundancies.


1

When we were solving the motion simulation, we got several warnings about the redundancies. Right-click the local mategroup and click Degrees of Freedom.

There are three redundancies.

Coincident2, has some rotations removed, but that is OK as we are not

concerned with the forces in this mate; it is connecting components other than the handle.

Concentric1 has both rotations removed. As this is a mate connecting the handle to an adjacent part it must be checked carefully.

Note that your list of removed degrees of freedom may be slightly different. This will, however, have no effect on the conclusions regarding the effect of redundancies made in this step.

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2

Examine the mechanical connection.


The line of action will be through the center of each part. Because this connection is not symmetrical, it does not allow these two forces to act directly on each other (on the same line of action), there is a small offset distance which creates a moment.

handle_link

Concentric1

This effect is minimized, but not eliminated, by the offset cut in each of the two parts.

handle

In the physical model, the two hinge mates and the concentric mate would all have some stiffness which would lead to the redistribution of the torsional moments between the three connections.

TOP View

As these torsional moments will be very small, because they were minimized by the cutouts, we can ignore this. We will assume that the pins in the two hinges (between the handle, handle_link and the fixed_cutter) are very stiff, taking on the torsional loads.

3

Create a plot.

Create a plot of each of the X, Y and Z reaction moment of the hinge mate (Hinge1) between the handle and the fixed_cutter.

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The last component of the moment is zero because it is the axial direction.

The X and Y moments are not zero and have some significant values. Also note that the moments do not occur at the time when the maximum cutting force is generated (0.25 seconds), but rather some time later at about 0.50 seconds.

4

Check interferences.

Before exporting the loads, we need to determine why the large moment is generated.

Check interference between the latch and moving_cutter. Right-click the assembly icon in the Motion Study tree and click Check Interference. Check from frame 1 to 115 in increments of 1.

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5



Most of the interferences are very small (volume <0.01 mm3) and are due to tiny penetrations in the contact. To locate the individual volumes, select an interference in the table and click Zoom to Selection .

If we continue to examine the list of interferences, we will find some that are several cubic millimeters. Zoom in on one of these larger interferences and we will see that at some point, the latch penetrates the moving_cutter. To fix this problem, the cutout in the moving cutter has to be increased in size.

6

Modify the part.

Open the moving_cutter in its own window.

Edit Sketch3 for Cut-Extrude2.

Change the dimensions to increase the size of the slot by 3 mm. Before

After

7

Re-check interferences.

Return to the assembly and re-check the interferences.

There should now only be the small contact interferences.

8


Plot force and moment of hinge connecting the handle to the fixed_cutter.

9

Locate the maximum force and moment.

Create plots for both the X and Y Force and Moment for the hinge.

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We have a high force and moment generated at about 0.14 seconds. This is not the point where the maximum cutting force is located which is at 0.24 seconds and shown by the red arrows. You can show the cutting force plot to verify this location.

10 Examine the latch.

If we examine the latch as the simulation progresses, we can see that maximum forces and moments are generated when the pin in the latch goes around the slot path. When the handle is first squeezed, there is a jump in force as static friction is overcome. As the spring expands rapidly, the spring works against the forward movement of the moving_cutter. As the pin gets to the turning point in the slot path, the forces continue to rise until the pin gets to the horizontal section of the slot path when the force becomes perpendicular to the path and held by contact.

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We can see that the maximum force is not caused by cutting the catheter, but rather by the spring used to retract the mechanism.

11 Determine the frame where maximum force occurs.

For the plot of the X Component force, change the X axis scale to frames.

Change the Start and End values to 10 and 20 so that we can see the exact frame.

From this we can see that the maximum force occurs at frame 15.

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12 Moment of the motor.


In the motion simulation, we used a rotary motor to move the mechanism. When we do the stress analysis, we are going to have to replace this motor with a force that represents the force applied by the surgeon on the handle. In order to calculate that force, we need to know the maximum moment generated by the motor. Create a plot of the Z Component, Motor Torque of the Rotary Motor. Note the peak torque of 5233 N-mm.

13 Export to FEA. Click Simulation, Import Motion Loads from the menu.

Note

The motion simulation results must be saved before importing the motion loads. Export the loads for the handle at frame 15.

14 Open the handle part. Open the handle in its own window.

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15 Examine loads.


Imported from SolidWorks Motion should be gravity, a centrifugal load and two remote loads.

The remote load from the handle_link is OK, however the loads from the rotary motor are not. When in use, the handle is squeezed by the surgeon which applies a force directly on the surface of the handle. Therefore, we will have to remove the loads from the motor and replace it with a force.

16 Add a loading force.

We will apply the force at the edge shown in the image. Measure the distance from the pivot hole to the edge. It is approximately 50 mm. The force we need to apply will be computed from the torque we measured in Step 11, 5233/50=104.66 N.

Apply a force of 104.66 N, normal to Plane5, on edge of handle. Reverse the direction if necessary to insure the correct rotation of the handle.

Suppress the remote load from the motor.

Note

Your torque value may be slightly different. In such case it is necessary to update the value of the loading force.

17 Restrain the model.

Suppress the remote load at the pivot point and replace it with a Fixed

Hinge fixture.

18 Add material.

Apply the material Alloy Steel.

19 Mesh.

Mesh the model with a high quality mesh at the default settings.

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We will get a warning: There is a significant external imbalance force in the Ydirection which will be balanced by the application of opposing inertia forces. Unless your model is under such a force or under marginally imbalance force, application of Inertia Relief may alter the characteristics of your model.

The problem is that we have manually added the force, which is approximate, so this warning is expected. Click Yes.

We will get another warning:

Excessive displacements were calculated in this model. If your system is properly restrained, consider using the Large Displacement option to improve the accuracy of the calculations. Otherwise continue with the current settings and review the causes of these displacements.

Since the external forces are in slight imbalance, the handle wants to rotate about the hinge support. This is an unavoidable consequence having no implication on the resulting stresses and deformations. The handle will rotate as a rigid body. Click No to continue with the linear solution.


The maximum stress about 252 MPa and is located at the sharp edge under the pivot.

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A closer examination shows that the highest stress is at the connection of the cylindrical sections to the pivot. A more detail analysis of this area might be appropriate since this is a stress singularity location. Given the yield strength of 620 MPa, the stresses in the handle are acceptable.

22 Create a Factor of Safety plot.

The plot shows that the Factor of Safety is close to 2.5, so the design is acceptable.

Note

The above plot has the upper limit set to 100.


Summary

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In this lesson, we analyzed a surgical shear assembly utilized to cut a catheter. Since a catheter is not a rigid object, its resistance against the cut is expressed by a reaction force acting on the blades. In rigid body dynamics software, we cannot simulate the deformation and separation of the flexible catheter and, therefore, modeling of an equivalent catheter reaction force is required.

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Real experimental data for a catheter cutting force was used to construct a complex, position dependent, expression for the equivalent action/reaction force. Due to the complexity of the problem, the expression involved multiple uses of the IF statement. Various graphs were generated at the end of each analysis. Motion mate forces were then imported into SolidWorks Simulation for the finite element stress analysis. The analysis indicated that the handle component is designed with a satisfactory safety factor of 2.5.

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PR Do E no RE t c LE op AS y E or D di RA st F rib T ut e Appendix A Motion Study Convergence Solutions and Advanced Options

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Motion Study Convergence Solutions and Advanced Options

Complex assemblies with many redundancies or problems featuring scenarios numerically difficult to overcome (for example instability points featured in Lesson 4, fast changing motions or high velocity impacts to name a few) may cause the solver to fail to converge and the solution may terminate before reaching the end. Convergence issues are unavoidable consequences of the numerical simulation and certain expertise may be required to overcome them. On occasions we may foresee that a complex assembly will pose difficulty and will need more attention. In general, however, it is difficult to predict when the convergence issues may occur. The following text should help you understand the basics of solving the above mentioned difficulties. It introduces some of the advanced software features not used during the regular part of the course.


Convergence

When the SolidWorks Motion solver faces convergence problems, the motion study terminates and the following window opens:

There are a few possible reasons for the convergence issues; we will review them in the next paragraphs.

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Appendix A Motion Study Convergence Solutions and Advanced Options

A set of coupled differential and algebraic equations (DAE) define the equations of motion of a SolidWorks Motion model. A solution to the equations of motion is obtained by solving these equations using an integrator. The integrator obtains the solution in two stages: first it predicts the solution at the next time step based on the past history and then it corrects that solution based on the state data at that time until the solution is within the desired accuracy level.


Accuracy

The Accuracy setting controls how accurate you want your solution to be. There is a trade-off between accuracy and performance. If the Accuracy setting is towards the High end, then the motion solver may take a long time to compute the solution. On the other hand, if this setting is towards the Low end, then the results may not be very accurate.

While the default value of 0.0001 fits most situations; it may need to be changed if sudden changes in the system occur. In such situations, the predictor provides an incorrect initial guess to the corrector resulting in large error and failure. This value may need to be reduced when sudden and discontinuous changes occur during simulation; such as sudden changes in the force or motor magnitudes, use of non-differentiable intrinsic functions in the statements (IF, MIN, MAX, SIGN, MOD, and DIM), friction and similar.

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The advanced options listed below can be accessed by clicking the Advanced Options in the Motion Study Properties shown in the figure above.

Integrator Type

The SolidWorks Motion solver solves the DAE equations of motion by integrating the differential equations in such a way that the algebraic constraint equations are also satisfied at every time step. The speed of the solution depends upon the numerical stiffness of these equations; the stiffer the equations the slower the solution. A set of ordinary differential equations are characterized as numerically stiff when there is a wide spread between high and low frequency eigenvalues, with the high-frequency eigenvalues being overdamped. Special efficient integration methods are required to solve numerically stiff differential equations because usual methods for solving differential equations perform poorly and are too slow.

The SolidWorks Motion solver offers three stiff integration methods for computing motion:

GSTIFF

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GSTIFF WSTIFF SI2

The GSTIFF integration method developed by C. W. Gear is a variable order, variable step size integration method. It is the default method used by the SolidWorks Motion solver. The GSTIFF method is a fast and accurate method for computing displacements for a wide range of motion analysis problems.

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WSTIFF is another variable order, variable step size stiff integrator. Both methods are very similar in formulation and behavior. Both of them use a backwards difference formulation. The only difference is that the coefficients used internally by GSTIFF are calculated assuming a constant step size whereas in WSTIFF, these coefficients are a function of the step size. If the step size changes suddenly during integration, GSTIFF introduces a small error in the solution whereas WSTIFF can handle it without any loss of accuracy. So the problems run more smoothly in WSTIFF. Sudden step size changes occur whenever there are discontinuous forces, discontinuous motions or abrupt events such as 3D contacts in the model.


WSTIFF

Stabilized Index Two (SI2)

The Stabilized Index Two (SI2) method offered in SolidWorks Motion is a modification of the GSTIFF integration method. This method provides better error control over velocity and acceleration terms in the equations of motion. Provided the motion is sufficiently smooth, SI2 velocity and acceleration results are more accurate than those computed with GSTIFF or WSTIFF, even for motions with high frequency oscillations. SI2 is also more accurate with smaller step sizes, but is significantly slower.

Integrator Settings

With each integrator, there are several settings that control the step size and number of integration steps.

Maximum Iterations

Maximum Iterations parameter controls the maximum number of iterations the SolidWorks Motion solver may use to converge to a solution. The default value of 25 iterations should suffice for most problems. It is not recommended to increase this parameter substantially; this parameter is typically not the cause of the failure.

Initial Integrator Step Size

Initial Integrator Step Size controls the value of the step at the first solution instance. If your simulation faces some difficulties at the initial stages of the solution, consider reducing this value. Typically, this parameter does not need to be changed.

Minimum Integrator Step Size

During the integration process, if the simulation error is too large the integrator reduces the time step and attempts the solution again until the desired accuracy is satisfied. The integrator will not reduce the step size beyond the Minimum Integrator Step Size. The default magnitude is acceptable for most of the simulation and does not need to be changed.

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The Maximum Integrator Step Size controls the value of the largest time step the integrator may take during the solution. Increasing the Maximum Integrator Step Size speeds up the solution, and reduces the time required to solve the model. But if this value is too big, there is a chance the solver may take too large a step and enter a region from which it may not recover and hence fail to converge. Reducing this value has no effect on the accuracy of the solution. When using GSTIFF integrator, velocities and accelerations may have discontinuities for larger values of the integrator time step. You can reduce this error by reducing the maximum integrator step size. If you know that the motion is smooth and there are no such abrupt changes, you can increase this value to speed up the solution. When facing convergence problems, modifying this parameter may help.


Maximum Integrator Step Size

If there are abrupt changes in forces or motions happening over small time durations that you may need to reduce the maximum integrator step size to make sure that the integrator does not miss such events. You may want to reduce this value if you have contact between a solid body and a thin body, and the solver fails to recognize this contact. This can happen, for example, if you have a ball bouncing on a thin plate. Depending upon your model parameters, it is possible that the ball may pass through the plate without recognizing the contact between them. In such a case, reducing the Maximum Integrator Step Size forces the solver to take smaller steps so that it does not miss the contact incidence between the two bodies. Reducing this value will slow down the integrator but it has no effect on the accuracy of the results. On the other hand, if you know that the motion is smooth and there are no such abrupt changes, you can increase this value to speed up the solution.

Jacobian Reevaluation

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The Jacobian Matrix is a matrix of partial derivatives required to solve the linearized approximation of the original nonlinear equations of motion during the Newton-Raphson iteration procedure. Users may find it useful to view this matrix similar to what the stiffness matrix is in the finite element analysis. The default setting, the most accurate and also the most time consuming is the re-evaluation of the Jacobian Matrix at every iteration. While reducing the re-evaluation speeds up the solution, it should only be done when changes in the assembly motion are slow. Setting of this parameter has no effect on the accuracy, but too low a setting may cause the integrator to fail.

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The most common parameters that need to be adjusted when you face convergence difficulties are Accuracy, Maximum Integrator Step Size, and Contact Resolution. If chancing none of the above parameters help the convergence, make sure that your inputs are smooth and differentiable, especially the user input expressions with mathematical functions. It is advisable to use STEP function rather than IF statement.


Conclusion

On occasions redundant constraint may cause the integrator to fail because the solver is having difficulty satisfying the constraints. The most likely cause for such a failure is an inconsistently defined or illbehaved model. In these situations try to eliminate the redundancy or the mating relationships in the assembly.

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Mate Friction

Friction is a force that occurs in mates and parts in contact. When parts are in contact, friction is calculated based on the static and dynamic coefficients of friction and the normal force acting on the part. Mate friction is more complex, because the size of the mate can affect the magnitude of friction.


Mate Friction

In 1699, Amontons rediscovered Leonardo da Vinci's two laws of friction: the frictional force is directly proportional to the normal load, and the size of the bodies does not affect the friction [Bowden and Tabor, 1950, 1974]. Engineers have relied on Amontons' laws, extensively and routinely for three centuries. Contrary to popular belief, the size of the bodies does affect the friction forces in the case of mate friction. Mate friction is a resistive, sliding, surface force between parts that must be overcome for the parts to move with respect to one another. The force develops due to contact between the surfaces and the loads acting on the connection. For a pin in a hole, mate friction is experienced as an additional torque restricting the pin from rotating with respect to the hole. Mate friction is not anything more than standard friction between bodies; however it takes into account aspects of components’ geometries in determining the net frictional forces acting. For example, think of a pin in a hole, but with a little slope. In the first image shown below, the pin is resting in the hole under a centrally located force. This is the equivalent of a pure bearing load. The force needed to slide the pin back and forth is only dependent on the vertical load. The torque needed to rotate the pin is dependent on this force, but also on the radius of the pin (see second image below). In this example, the radius of the pin has no effect on the magnitude of the friction force, but does have an effect on the moment required to overcome friction to rotate the pin (mu.r.F)

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Appendix B Mate Friction


Now, consider the case where there is an additional moment on the pin. The moment forces the pin to rotate, becoming supported at the outer edges of the hole (w). The moment is reacted as a force couple (M/w). Dividing the bearing load (F) between the ends, results in local force of F/2 + M/w. Frictional forces are accumulative so you can sum these force couples to get the total force upon which friction is based (F+2M/w). It is a simple extension of this to derive the torque necessary to rotate the pin as mu*r(F+2M/w).

This influence of the bending moment of a mate is an important factor in mate friction. You can see that if the hole supporting the pin is not thick (in terms of w), the moment component tends to be very high. If the hole supporting the pin is very thick, the moment component tends towards zero.

Concentric, coincident and many other SolidWorks mates support friction. When friction effects are enabled for these mates, a force is induced that opposes the motion of the mates and is a function of the reaction forces acting on the mate.

Where to Find It

Concentric (Spherical) Mate Friction Model

For the purpose of calculating friction effects, a Concentric (Spherical) mate is modeled as a ball rotating in a socket. Some portion of the ball's surface area is in contact with the socket. Dimension d is the diameter of the ball.

In the SolidWorks mate property manager, Analysis tab, Friction dialog.

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Appendix B

SolidWorks 2011

Mate Friction

For the purpose of calculating friction effects, a Coincident (translational) mate is modeled as a rectangular bar sliding in a rectangular sleeve. Dimension h is the height of the rectangular bar. Dimension w is the width of the rectangular bar. Dimension l is the length of the bar that is in contact with the sleeve.

Concentric Mate Friction Model

For the purpose of calculating friction effects, a Concentric mate is modeled as a snug fit pin rotating and sliding in a hole. Dimension r is the radius of the pin, and Dimension l is the length of the pin that is in contact with the hole.


Coincident Translational Mate Friction Model

Concentric mate friction model can only be activated for faces. No edges are allowed.

Coincident Mate (Planar) Friction Model

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For the purpose of calculating friction effects, this is modeled as one block sliding and rotating across the surface of another block. Dimensions l and w are the length and width of the sliding block. Dimension r is the radius of a circle, centered at the center of the block face that circumscribes the face of the sliding block that is in contact with the other block.

SolidWorks 2011

Appendix B Mate Friction

For the purpose of calculating friction effects, a universal joint is modeled as a cylindrical cross piece rotating in a set of end caps. Dimension r is the radius of the bearing end cap. Dimension w is the height of the cross pieces.


Universal Joint Friction Model

Friction Results Reported

Joint Type

Friction Force

Friction Moment

Concentric (two faces)

Yes

Yes

Concentric (two spheres)

No

Yes

Universal

No

Yes

Coincident (translational)

Yes

No

Coincident (planar)

Yes

Yes

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PR Do E no RE t c LE op AS y E or D di RA st F rib T ut e Appendix B

Mate Friction

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SolidWorks 2011

PR Do E no RE t c LE op AS y E or D di RA st F rib T ut e Index

A accuracy 335 action 294 action and reaction forces 17 action forces 17 applied forces 18

E event based motion view 294 event based simulation 290 export results to FEA 262 trace path curves 170

B bushings defining 204 properties 224

F fixed parts 7 flexible joints 194, 204 bushings 204 flexible mates limitations 223 floating parts 8 force expression 313 force function 105, 308 forces 17 action and reaction 17 action only 17 applied 18 closing 123 contact 100, 154, 280 definition 18 impact 110 frames per second 26 friction 76, 342 contact 84 kinematic coefficient 84 static coefficient 84 Function Builder 45

C CAM 166 desmodromic 179 profile 167 rocker 185 chart properties 119 closing force 123 coefficient of restitution 110 constraint forces 8 constraint mapping 8 contact 76, 82 curve to curve 155, 159 contact forces 154 forces 100, 280 friction 84 precise contact 123 precise geometry 115 solid bodies 159 tessellated geometry 115 contact, solid bodies - contact 109 convergence 334 Cycle Based Motion 173 D damper translational 87 damping coefficient c 111 penetration d 111 degrees of freedom 219 calculation 219 estimated 219 total actual 219 dynamic systems 212, 229, 231

G gravity 8, 16 GSTIFF 124, 336

I IF statement 313 impact force 110 exponent e 111 stiffness k 111 instability 118 integrator settings 337 integrator types 123, 336 GSTIFF 124, 336 SI2 124, 337 WSTIFF 124, 337 interference detection 81 J jacobian 338

Jerk 46 joints flexible 194, 204 rigid 194

K kinematic coefficient of friction 84 kinematic systems 212, 233 L linear spring magnitude - spring force 86 load bearing faces 263 M mass properties 131 mate friction 342 concentric 344 planar 344 results 345 spherical 343 translational 344 universal 345 mates 8, 342 modifying plots 119 motion driving 14, 45, 173 motion study properties 26 frames per second 26 precise contact 123 motor servo motor 290 motors 8, 14 fixing motion 101 force function 105 force types 103 functional expressions 103

P Path Mate 144 PathMate 144 plots modifying 119 resizing 21 Plotting Kinematic Results 54 plotting results 21 Poisson model 110 postprocessing 89 precise contact 123 precise geometry 115

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Index


properties bushings 224 chart 119 mass 131 motion study 26 proximity sensor 291

SolidWorks 2011

R redundancies check 225 defined 213 effects 214 mechanisms 226 removal by solver 215 removal with bushings 244 removal with flexible joints 222 removed in the solver 215 zero 238, 243 redundant mechanisms 226 resizing plots 21 restitution coefficient 110 results export 258 export to FEA 262 plotting 21 rigid body 7 rigid joints 194 S sensors 291 proximity 291 servo motor 290 SI2 337 SI2 (Stabilized Index Two) 124 spring force magnitude 86 translational 85 spring force 86 static coefficient of friction 84 STEP function 105 T task 293, 295 action 294 triggers 293 tessellated geometry 115 timeline view 294 toe angle 201 trace path 169 export curves 170 translational damper 87 translational spring 85 triggers 293 V view event based 294 timeline 294 W-Z WSTIFF 124, 337

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SolidWorks Motion Tutorials

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