ANSYS Workbench Verification Manual

Transcription

1 ANSYS Workbench Verification Manual ANSYS, Inc. Southpointe 2600 ANSYS Drive Canonsburg, PA (T) (F) Release 18.2 August 2017 ANSYS, Inc. and ANSYS Europe, Ltd. are UL registered ISO 9001: 2008 companies.

2 Copyright and Trademark Information 2017 ANSYS, Inc. Unauthorized use, distribution or duplication is prohibited. ANSYS, ANSYS Workbench, AUTODYN, CFX, FLUENT and any and all ANSYS, Inc. brand, product, service and feature names, logos and slogans are registered trademarks or trademarks of ANSYS, Inc. or its subsidiaries located in the United States or other countries. ICEM CFD is a trademark used by ANSYS, Inc. under license. CFX is a trademark of Sony Corporation in Japan. All other brand, product, service and feature names or trademarks are the property of their respective owners. FLEXlm and FLEXnet are trademarks of Flexera Software LLC. Disclaimer Notice THIS ANSYS SOFTWARE PRODUCT AND PROGRAM DOCUMENTATION INCLUDE TRADE SECRETS AND ARE CONFID- ENTIAL AND PROPRIETARY PRODUCTS OF ANSYS, INC., ITS SUBSIDIARIES, OR LICENSORS. The software products and documentation are furnished by ANSYS, Inc., its subsidiaries, or affiliates under a software license agreement that contains provisions concerning non-disclosure, copying, length and nature of use, compliance with exporting laws, warranties, disclaimers, limitations of liability, and remedies, and other provisions. The software products and documentation may be used, disclosed, transferred, or copied only in accordance with the terms and conditions of that software license agreement. ANSYS, Inc. and ANSYS Europe, Ltd. are UL registered ISO 9001: 2008 companies. U.S. Government Rights For U.S. Government users, except as specifically granted by the ANSYS, Inc. software license agreement, the use, duplication, or disclosure by the United States Government is subject to restrictions stated in the ANSYS, Inc. software license agreement and FAR (for non-dod licenses). Third-Party Software See the legal information in the product help files for the complete Legal Notice for ANSYS proprietary software and third-party software. If you are unable to access the Legal Notice, contact ANSYS, Inc. Published in the U.S.A.

3 Table of Contents Introduction... 1 Overview... 1 AIM Test Cases... 2 Index of Test Cases... 2 I. DesignModeler Descriptions VMDM001: Extrude, Chamfer, and Blend Features VMDM002: Cylinder using Revolve, Sweep, Extrude, and Skin-Loft VMDM003: Extrude, Revolve, Skin-Loft, and Sweep VMDM004: Extrude, Revolve, Skin-Loft, and Sweep II. SpaceClaim Descriptions SCGEO1. VMSC_Geometry001: Alter Model Using Sketch, Pull, and Fill SCGEO2. VMSC_Geometry002: Alter Model Using Split Face, Split Body and Mirror SCGEO3. VMSC_Geometry003: Alter Model using Linear, Circular, and Two-Dimensional Patterns SCGEO4. VMSC_Geometry004: Alter Model Using Revolve, Chamfer, and Offset Faces SCGEO5. VMSC_Geometry005: Alter Model Using Blend and Project SCGEO6. VMSC_Geometry006: Alter Model Using Sheet Metal Tools and Operations SCPREP1. VMSC_Prepare001: Midsurface, Extend, Imprint, and Spot Weld SCPREP2. VMSC_Prepare002: Volume Extract and Enclosure SCPREP3. VMSC_Prepare003: Beam Create, Extract, and Orient SCPREP4. VMSC_Prepare004: Remove Rounds, Interference, Faces, and Short Edges SCREP1. VMSC_Repair001: Stitch, Gap, and Missing Faces SCREP2. VMSC_Repair002: Merge Faces, Small Faces, Simplify, Straighten SCREP3. VMSC_Repair003: Split Edges, Extra Edges, and Duplicates III. Application Descriptions VMMECH001: Statically Indeterminate Reaction Force Analysis VMMECH002: Rectangular Plate with Circular Hole Subjected to Tensile Loading VMMECH003: Modal Analysis of Annular Plate VMMECH004: Viscoplastic Analysis of a Body (Shear Deformation) VMMECH005: Heat Transfer in a Composite Wall VMMECH006: Heater with Nonlinear Conductivity VMMECH007:Thermal Stress in a Bar with Temperature Dependent Conductivity VMMECH008: Heat Transfer from a Cooling Spine VMMECH009: Stress Tool for Long Bar With Compressive Load VMMECH010: Modal Analysis of a Rectangular Plate VMMECH011: Large Deflection of a Circular Plate With Uniform Pressure VMMECH012: Buckling of a Stepped Rod VMMECH013: Buckling of a Circular Arch VMMECH014: Harmonic Response of a Single Degree of Freedom System VMMECH015: Harmonic Response of Two Storied Building Under Transverse Loading VMMECH016: Fatigue Tool with Non-Proportional Loading for Normal Stress VMMECH017: Thermal Stress Analysis with Remote Force and Thermal Loading VMMECH018: A Bar Subjected to Tensile Load with Inertia Relief VMMECH019: Mixed Model Subjected to Bending Loads with Solution Combination VMMECH020: Modal Analysis for Beams VMMECH021: Buckling Analysis of Beams VMMECH022: Structural Analysis with Advanced Contact Options VMMECH023: Curved Beam Assembly with Multiple Loads VMMECH024: Harmonic Response of a Single Degree of Freedom System for Beams VMMECH025: Stresses Due to Shrink Fit Between Two Cylinders VMMECH026: Fatigue Analysis of a Rectangular Plate Subjected to Edge Moment iii

4 Workbench Verification Manual 27. VMMECH027: Thermal Analysis for Shells with Heat Flow and Given Temperature VMMECH028: Bolt Pretension Load Applied on a Semi-Cylindrical Face VMMECH029: Elasto-Plastic Analysis of a Rectangular Beam VMMECH030: Bending of Long Plate Subjected to Moment - Plane Strain Model VMMECH031: Long Bar With Uniform Force and Stress Tool - Plane Stress Model VMMECH032: Radial Flow due to Internal Heat Generation in a Copper Disk - Axisymmetric Model VMMECH033: Spring Mass System Subjected to Enforced Motion with Displacement Base Excitations VMMECH034: Rubber Cylinder Pressed Between Two Plates VMMECH035: Thermal Stress in a Bar With Radiation VMMECH036: Thermal Stress Analysis of a Rotating Bar Using Temperature Dependant Density VMMECH037: Cooling of a Spherical Body VMMECH038: Crashing Blocks Simulation with Transient Structural Analysis VMMECH039: Transient Response of a Spring-Mass System VMMECH040: Deflection of Beam using Symmetry and Anti-Symmetry VMMECH041:T-Stress Calculation for a Crack in a Plate Using Pre-Meshed Crack VMMECH042: Hydrostatic Pressure Applied on a Square Bar Fully and Partially Submerged in a Fluid VMMECH043: Fundamental Frequency of a Simply-Supported Beam VMMECH044: Thermally Loaded Support Structure VMMECH045: Laterally Loaded Tapered Support Structure VMMECH046: Pinched Cylinder VMMECH047: Plastic Compression of a Pipe Assembly VMMECH048: Bending of a Tee-Shaped Beam VMMECH049: Combined Bending and Torsion of Beam VMMECH050: Cylindrical Shell Under Pressure VMMECH051: Bending of a Circular Plate Using Axisymmetric Elements VMMECH052: Velocity of Pistons for Trunnion Mechanism VMMECH053: Simple Pendulum With SHM Motion VMMECH054: Spinning Single Pendulum VMMECH055: Projector Mechanism - Finding the Acceleration of a Point VMMECH056: Coriolis Component of Acceleration (Rotary Engine Problem) VMMECH057: Calculation of Velocity of Slider and Force by Collar VMMECH058: Reverse Four Bar Linkage Mechanism VMMECH059: Bending of a Solid Beam (Plane Elements) VMMECH060: Crank Slot Joint Simulation with Flexible Dynamic Analysis VMMECH061: Out-of-Plane Bending of a Curved Bar VMMECH062: Stresses in a Long Cylinder VMMECH063: Large Deflection of a Cantilever VMMECH064: Small Deflection of a Belleville Spring VMMECH065:Thermal Expansion to Close a Gap at a Rigid Surface VMMECH066: Bending of a Tapered Plate VMMECH067: Elongation of a Solid Tapered Bar VMMECH068: Plastic Loading of a Thick Walled Cylinder VMMECH069: Barrel Vault Roof Under Self Weight VMMECH070: Hyperelastic Thick Cylinder Under Internal Pressure VMMECH071: Centerline Temperature of a Heat Generating Wire VMMECH072: Thermal Stresses in a Long Cylinder VMMECH073: Modal Analysis of a Cyclic Symmetric Annular Plate VMMECH074: Tension/Compression Only Springs VMMECH075: Harmonic Response of Two-Story Building under Transverse Loading VMMECH076: Elongation of a Tapered Shell With Variable Thickness iv

5 Workbench Verification Manual 77. VMMECH077: Heat Transfer in a Bar with Variable Sheet Thickness VMMECH078: Gasket Material Under Uniaxial Compression Loading-3-D Analysis VMMECH079: Natural Frequency of a Motor-Generator VMMECH080: Transient Response of a Spring-Mass System VMMECH081: Statically Indeterminate Reaction Force Analysis VMMECH082: Fracture Mechanics Stress for a Crack in a Plate VMMECH083: Transient Response to a Step Excitation VMMECH084: Mullins Effect on a Rubber Tube Model Subjected to Tension Loading VMMECH085: Bending of a Composite Beam VMMECH086: Stress Concentration at a Hole in a Plate VMMECH087: Campbell Diagrams and Critical Speeds Using Symmetric Orthotropic Bearings VMMECH088: Harmonic Response of a Guitar String VMMECH089: Delamination Analysis of a Double Cantilever Beam Using Contact-Based Debonding VMMECH090: Delamination Analysis of a Double Cantilever Beam Using Interface Delamination VMMECH091: Unbalanced Harmonic Response of a Shaft Carrying Single Rotor with Damping VMMECH092: Convection Treatment Problem for a Hollow Cylinder with Fluid Flow VMMECH093: C*-Integral Calculation for a Single-Edge Cracked Plate Using Pre-Meshed Crack VMMECH094: Residual Vector in Stand-Alone and Linked Mode-Superposition Harmonic Analysis VMMECH095: 2-D Double Cantilever Beam Problem VMMECH096: 2-D Fracture Problem Under Thermal Loading VMMECH097: Inclined Crack in 2-D Plate Under Uniform Tension Loading VMMECH098: 2-D End Notched Flexure Problem VMMECH099: Mode I Crack Growth Analysis of DCB Using Interface Delamination VMMECH100: 3-D Acoustic Modal Analysis with Temperature Change VMMECH101: Natural Frequency of a Submerged Ring IV. Design Exploration Descriptions VMDX001: Optimization of L-Shaped Cantilever Beam Under Axial Load VMDX002: Optimization of Bar with Temperature-Dependent Conductivity VMDX003: Optimization of Water Tank Column for Mass and Natural Frequency VMDX004: Optimization of Frequency for a Plate With Simple Support at all Vertices VMDX005: Optimization of Buckling Load Multiplier With CAD Parameters and Young's Modulus V. Explicit Dynamics Descriptions EXDVM2: Propagation of Shock and Rarefaction Waves in a Shock Tube EXDVM3: Flow of Gas Past an Infinite Two-Dimensional Wedge EXDVM4: Regular and Mach Deflections Off a Two-Dimensional Wedge EXDVM6: 3-D Taylor Cylinder Impact EXDVM7: 2-D Taylor Cylinder Impact VI. Aqwa Descriptions AQVM1: Hydrostatic Evaluation of a Floating Inverted Pyramid v

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7 Introduction The following topics are discussed in this chapter: Overview AIM Test Cases Index of Test Cases Overview This manual presents a collection of test cases that demonstrate a number of the capabilities of the Workbench analysis environment. The available tests are engineering problems that provide independent verification, usually a closed form equation. Many of them are classical engineering problems. The solutions for the test cases have been verified, however, certain differences may exist with regard to the references. These differences have been examined and are considered acceptable. The workbench analyses employ a balance between accuracy and solution time. Improved results can be obtained in some cases by employing a more refined finite element mesh but requires longer solution times. For the tests, an error rate of 3% or less has been the goal. These tests using Microsoft Windows 7 Enterprise 64-bit. These results are reported in the test documentation. Slightly different results may be obtained when different processor types or operating systems are used. The tests contained in this manual are a partial subset of the full set of tests that are run by ANSYS developers to ensure a high degree of quality for the Workbench product. The verification of the Workbench product is conducted in accordance with the written procedures that form a part of an overall Quality Assurance program at ANSYS, Inc. You are encouraged to use these tests as starting points when exploring new Workbench features. Geometries, material properties, loads, and output results can easily be changed and the solution repeated. As a result, the tests offer a quick introduction to new features with which you may be unfamiliar. Some test cases will require different licenses, such as DesignModeler, Emag, or Design Exploration. If you do not have the available licenses, you may not be able to reproduce the results. The Educational version of Workbench should be able to solve most of these tests. License limitations are not applicable to Workbench Education version but problem size may restrict the solution of some of the tests. The archive files for each of the Verification Manual tests are available at the Customer Portal. Download the ANSYS Workbench Verification Manual Archive Files. These zipped archives provide all of the necessary elements for running a test, including geometry parts, material files, and workbench databases. To open a test case in Workbench, locate the archive and import it into Workbench. You can use these tests to verify that your hardware is executing the ANSYS Workbench tests correctly. The results in the databases can be cleared and the tests solved multiple times. The test results should be checked against the verified results in the documentation for each test. 1

8 Introduction ANSYS, Inc. offers the Workbench Verification and Validation package for users that must perform system validation. This package automates the process of test execution and report generation. If you are interested in contracting for such services contact the ANSYS, Inc. Quality Assurance Group. AIM Test Cases A subset of test cases within this manual are supported by ANSYS AIM. These cases are divided into ANSYS and ANSYS AIM specific parts. The following test cases are supported by ANSYS AIM: VMMECH001 (p. 83) - Statically Indeterminate Reaction Force Analysis VMMECH002 (p. 87) - Rectangular Plate with Circular Hole Subjected to Tensile Loading VMMECH003 (p. 89) - Modal Analysis of Annular Plate VMMECH005 (p. 95) - Heat Transfer in a Composite Wall VMMECH007 (p. 99) - Thermal Stress in a Bar with Temperature Dependent Conductivity VMMECH006 (p. 97) - Heater with Nonlinear Conductivity VMMECH008 (p. 103) - Heat Transfer from a Cooling Spine VMMECH018 (p. 125) - A Bar Subjected to Tensile Load with Inertia Relief VMMECH025 (p. 139) - Stresses Due to Shrink Fit Between Two Cylinders VMMECH026 (p. 143) - Fatigue Analysis of a Rectangular Plate Subjected to Edge Moment VMMECH027 (p. 123) - Thermal Stress Analysis with Remote Force and Thermal Loading VMMECH032 (p. 157) - Radial Flow due to Internal Heat Generation in a Copper Disk VMMECH035 (p. 165) - Thermal Stress in a Bar with Radiation VMMECH037 (p. 171) - Cooling of a Spherical Body VMMECH050 (p. 197) - Cylindrical Shell Under Pressure VMMECH062 (p. 229) - Stresses in a Long Cylinder VMMECH067 (p. 241) - Elongation of a Solid Tapered Bar VMMECH071 (p. 251) - Centerline Temperature of a Heat Generating Wire VMMECH072 (p. 255) - Thermal Stresses in a Long Cylinder VMMECH073 (p. 259) - Modal Analysis of a Cyclic Symmetric Annular Plate Index of Test Cases Test Case Number Solver Element Type Analysis Type Solution Options VMMECH001 (p. 83) ANSYS AIM ANSYS Solid Static Structural Linear VMMECH002 (p. 87) ANSYS AIM ANSYS Solid Static Structural Linear VMMECH003 (p. 89) ANSYS Solid Modal Free Vibration VMMECH004 (p. 93) ANSYS Solid Structural Nonlinear, Viscoplastic Materials VMMECH005 (p. 95) ANSYS AIM Solid Static Thermal Linear 2

9 Index of Test Cases Test Case Number Solver Element Type Analysis Type Solution Options ANSYS VMMECH006 (p. 97) ANSYS AIM ANSYS Solid Static Thermal Nonlinear VMMECH007 (p. 99) ANSYS Solid Static Structural Nonlinear Thermal Stress VMMECH008 (p. 103) ANSYS AIM ANSYS Solid Static Thermal Linear VMMECH009 (p. 107) ANSYS Solid Static Structural Linear VMMECH010 (p. 109) ANSYS Shell Modal Free Vibration VMMECH011 (p. 111) ANSYS Shell Static Structural Nonlinear, Large Deformation VMMECH012 (p. 113) ANSYS Solid Buckling VMMECH013 (p. 115) ANSYS Shell Buckling VMMECH014 (p. 117) ANSYS Solid Harmonic VMMECH015 (p. 119) ANSYS Solid Harmonic VMMECH016 (p. 121) ANSYS Solid Static Structural Fatigue VMMECH017 (p. 123) ANSYS Solid Static Structural Linear Thermal Stress VMMECH018 (p. 125) ANSYS Solid Static Structural Linear, Inertia relief VMMECH019 (p. 127) ANSYS Beam Shell Static Structural Linear VMMECH020 (p. 129) ANSYS Beam Modal VMMECH021 (p. 131) ANSYS Beam Buckling VMMECH022 (p. 133) ANSYS Solid Static Structural Nonlinear, Contact VMMECH023 (p. 135) ANSYS Beam Static Structural Linear 3

10 Introduction Test Case Number Solver Element Type Analysis Type Solution Options VMMECH024 (p. 137) ANSYS Beam Harmonic VMMECH025 (p. 139) ANSYS Solid Static Structural Linear VMMECH026 (p. 143) ANSYS Shell Static Structural Fatigue VMMECH027 (p. 147) ANSYS Shell Static Structural Linear Thermal Stress VMMECH028 (p. 149) ANSYS Solid Static Structural VMMECH029 (p. 151) ANSYS Solid Static Structural Nonlinear, Plastic Materials VMMECH030 (p. 153) ANSYS 2-D Solid, Plane Strain Static Structural VMMECH031 (p. 155) ANSYS 2-D Solid, Plane Stress Static Structural VMMECH032 (p. 157) ANSYS 2-D Solid, Axisymmetric Static Structural Linear Thermal Stress VMMECH033 (p. 161) ANSYS Solid Static Structural Electromagnetic VMMECH034 (p. 163) ANSYS Solid Static Structural Nonlinear, Large Deformation VMMECH035 (p. 165) ANSYS AIM ANSYS Solid Coupled (Static Thermal and Static Stress) VMMECH036 (p. 169) ANSYS Solid Static Structural Sequence Loading VMMECH037 (p. 171) ANSYS 2-D Solid, Axisymmetric Transient Thermal VMMECH038 (p. 173) ANSYS Solid Transient Structural Flexible Dynamic VMMECH039 (p. 175) ANSYS Solid Spring Transient Structural Flexible Dynamic VMMECH040 (p. 177) ANSYS Beam Static Structural VMMECH041 (p. 179) ANSYS Solid Static Structural Electromagnetic VMMECH042 (p. 181) ANSYS Solid Static Structural Hydrostatic Fluid VMMECH043 (p. 183) ANSYS Beam Modal 4

11 Index of Test Cases Test Case Number Solver Element Type Analysis Type Solution Options VMMECH044 (p. 185) ANSYS Beam Static Structural Linear Thermal Stress VMMECH045 (p. 187) ANSYS Shell Static Structural VMMECH046 (p. 189) ANSYS Shell Static Structural VMMECH047 (p. 191) ANSYS 2-D Solid, Axisymmetric Static Structural Nonlinear, Plastic Materials VMMECH048 (p. 193) ANSYS Beam Static Structural VMMECH049 (p. 195) ANSYS Beam Static Structural VMMECH050 (p. 197) ANSYS Axisymmetric Shell Static Structural VMMECH051 (p. 201) ANSYS Axisymmetric Shell Static Structural VMMECH052 (p. 205) ANSYS Multipoint Constraint Rigid Dynamic VMMECH042 (p. 181) ANSYS Multipoint Constraint Rigid Dynamic VMMECH054 (p. 209) ANSYS Multipoint Constraint Rigid Dynamic VMMECH055 (p. 213) ANSYS Multipoint Constraint Rigid Dynamic VMMECH056 (p. 215) ANSYS Multipoint Constraint Rigid Dynamic VMMECH057 (p. 217) ANSYS Multipoint Constraint Rigid Dynamic VMMECH058 (p. 219) ANSYS Multipoint Constraint Rigid Dynamic VMMECH059 (p. 221) ANSYS 2-D Plane Stress Shell Static Structural VMMECH060 (p. 223) ANSYS Solid Multipoint Constraint Transient Structural Flexible Dynamic VMMECH061 (p. 227) ANSYS Beam Static Structural VMMECH062 (p. 229) ANSYS Axisymmetric Shell Static Structural VMMECH063 (p. 233) ANSYS Shell Static Structural Nonlinear, Large Deformation 5

12 Introduction Test Case Number Solver Element Type Analysis Type Solution Options VMMECH064 (p. 235) ANSYS Shell Static Structural VMMECH065 (p. 237) ANSYS Solid Shell Static Structural Linear Thermal Stress VMMECH066 (p. 239) ANSYS Shell Static Structural VMMECH067 (p. 241) ANSYS AIM ANSYS Solid Static Structural VMMECH068 (p. 245) ANSYS 2-D Solid, Plane Strain Static Structural Nonlinear, Plastic Materials VMMECH069 (p. 247) ANSYS Shell Static Structural VMMECH070 (p. 249) ANSYS 2-D Solid Static Structural Nonlinear, Large Deformation VMMECH071 (p. 251) ANSYS 2-D Thermal Solid Static Thermal VMMECH072 (p. 255) ANSYS 2-D Thermal Solid Static Structural Linear Thermal Stress VMMECH073 (p. 259) ANSYS AIM ANSYS Solid Modal VMMECH074 (p. 263) ANSYS Solid Spring Rigid Body Dynamics VMMECH075 (p. 265) ANSYS Solid Harmonic VMMECH076 (p. 267) ANSYS Shell Static Structural VMMECH077 (p. 269) ANSYS Thermal Shell Static Thermal VMMECH078 (p. 271) ANSYS 3-D Solid 3-D Gasket Static Structural VMMECH079 (p. 275) ANSYS Pipe Modal VMMECH080 (p. 277) ANSYS Spring Mass Transient Dynamic Mode-Superposition VMMECH081 (p. 279) ANSYS Pipe Modal 6

13 Index of Test Cases Test Case Number Solver Element Type Analysis Type Solution Options Mass Spectral VMMECH082 (p. 283) ANSYS Solid Static Structural Fracture Mechanics VMMECH083 (p. 285) ANSYS Spring, Mass Transient Dynamic Mode-Superposition VMMECH084 (p. 289) ANSYS Solid Static Structural Nonlinear, Hyperelastic VMMECH085 (p. 291) ANSYS Solid Static Structural Composite Material VMMECH086 (p. 293) ANSYS Solid Static Structural Submodeling (2D-2D) VMMECH087 (p. 297) ANSYS Line Body Point Mass Modal Bearing Connection VMMECH088 (p. 301) ANSYS Beam Static Structural Linear Perturbation Modal Harmonic VMMECH089 (p. 303) ANSYS Solid Static Structural Contact-Based Debonding VMMECH090 (p. 305) ANSYS Solid Static Structural Interface Delamination VMMECH091 (p. 307) ANSYS Beam Harmonic VMMECH092 (p. 311) ANSYS Coupled Thermal Pipe Static Structural Thermal Surface Thermal Solid VMMECH093 (p. 313) ANSYS Solid Static Structural Pre-Meshed Crack VMMECH094 (p. 315) ANSYS Spring-Damper Structural Mass Modal Harmonic VMMECH095 (p. 317) ANSYS Solid Static Structural Pre-Meshed Crack 7

14 Introduction Test Case Number Solver Element Type Analysis Type Solution Options VMMECH096 (p. 319) ANSYS Solid Static Structural VMMECH097 (p. 321) ANSYS Solid Static Structural Pre-Meshed Crack VMMECH098 (p. 323) ANSYS Solid Static Structural Pre-Meshed Crack VMMECH099 (p. 325) ANSYS Solid Static Structural 8

15 Part I: DesignModeler Descriptions

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17 VMDM001: Extrude, Chamfer, and Blend Features Overview Feature: Drawing Units: Extrude, Chamfer, and Blend Millimeter Test Case Create a Model using Extrude, Chamfer, and Blend features. A polygonal area is extruded 60 mm. A rectangular area of 30 mm x 40 mm [having a circular area of radius 6 mm subtracted] is extruded to 20 mm. Both resultant solids form one solid geometry. A rectangular area (24 mm x 5 mm) is subtracted from the solid. Two rectangular areas (40 mm x 10 mm) are extruded 10 mm and subtracted from solid. Two rectangular areas (25 mm x 40 mm) are extruded 40 mm and subtracted from solid. A Chamfer (10 mm x 10 mm) is given to 4 edges on the resultant solid. Four Oval areas are extruded and subtracted from Solid. Fillet (Radius 5 mm) is given to 4 edges using Blend Feature. Verify Volume of the resultant geometry. Figure 1: Final Model after creating Extrude, Chamfer, and Blend Calculations 1. Volume of Solid after extruding Polygonal Area: v1 = mm Volume of rectangular area having circular hole: v2 = mm 3. Net Volume = V = v1 + v2 = mm Volume of rectangular (24mm x 5mm) solid extruded 30mm using Cut Material = = mm 3. Net volume V = = mm Volume of two rectangular areas each 40mm x 10mm extruded 10mm = 8000 mm 3. 11

18 VMDM001 Net volume V = = mm Volume of two rectangular areas 25mm x 40mm extruded 40mm = mm 3. Net volume V = = mm Volume of four solids added due to Chamfer = 4 x 500 = 2000 mm 3 Net volume V = = mm Volume of four oval areas extruded 10 mm = mm 3. Net volume V = = mm Volume of 4 solids subtracted due to Blend of radius 5 mm = mm 3. Hence Net volume of final Solid body = V = = mm 3. Results Comparison Results Volume (mm 3 ) Surface Area (mm 2 ) Number of Faces Number of Bodies Target DesignModeler Error (%) x

19 VMDM002: Cylinder using Revolve, Sweep, Extrude, and Skin-Loft Overview Feature: Drawing Units: Revolve, Sweep, Extrude, and Skin-Loft Millimeter Test Case Create a Model using Revolve, Sweep, Extrude, and Skin-Loft features. A Rectangular area (100 mm x 30 mm) is revolved about Z-Axis in 3600 to form a Cylinder. A circular area of radius 30 mm is swept 100 mm using Sweep feature. A circular area of radius 30 mm is extruded 100 mm. A solid cylinder is created using Skin-Loft feature between two coaxial circular areas each of radius 30 mm and 100 mm apart. Verify Volume of the resultant geometry. Figure 2: Final Model after creating Revolve, Sweep, Extrude, and Skin-Loft Calculations 1. Volume of Cylinder created after Revolving Rectangular area (100 mm x 30 mm) = v1 = mm Volume of Cylinder created when a circular area (Radius 30mm) is swept 100 mm = v2 = mm 3. Net Volume = V = v1 + v2 = = mm Volume of Cylinder after extruding a circular area (Radius 30 mm) 100 mm = mm 3. Net Volume = V = = mm Volume of Cylinder created after using Skin-Loft feature between two circular areas of Radius 30 mm and 100 mm apart. = mm 3. Net Volume of the final Cylinder = = mm 3. 13

20 VMDM002 Results Comparison Results Volume (mm 3 ) Surface Area (mm 2 ) Number of Faces Number of Bodies Target DesignModeler Error (%) x x

21 VMDM003: Extrude, Revolve, Skin-Loft, and Sweep Overview Feature: Drawing Units: Extrude, Revolve, Skin-Loft, and Sweep Millimeter Test Case Create a Model using Extrude, Revolve, Skin-Loft, and Sweep. A rectangular area (103 mm x 88 mm) is extruded 100 mm to form a solid box. A circular area of radius 25 mm is revolved 900 using Revolve feature and keeping Thin/Surface option to Yes and 3 mm Inward and Outward Thickness. A solid is subtracted using Skin-Loft feature between two square areas (each of side 25 mm) and 100 mm apart. The two solid bodies are frozen using Freeze feature. A circular area of radius 25 mm is swept using Sweep feature and keeping Thin/Surface option to Yes and 3 mm Inward and Outward Thickness. Thus a total of 4 geometries are created. Verify the volume of the resulting geometry. Figure 3: Final Model after creating Extrude, Revolve, Skin-Loft and Sweep Calculations 1. Volume of rectangular (103 mm x 88 mm) solid extruded 100mm = mm Volume of solid after revolving circular area of Radius 25 mm through 900 = mm 3. Net Volume of solid box, Va = = mm Volume of additional body created due to Revolve feature = Vb= mm Volume of the rectangular box cut after Skin-Loft between two square areas each of side 25 mm = mm 3. Net Volume of solid box becomes Va = = mm Volume of additional two bodies created due to Sweep feature: 15

22 VMDM003 Vc = mm 3 and Vd = mm 3. And total volume that gets subtracted from box due to Sweep Feature = mm 3. Hence Net volume of box, Va = = mm 3. Sum of volumes of all four bodies = Va+Vb+Vc+Vd = = mm 3. Results Comparison Results Volume (mm 3 ) Surface Area (mm 2 ) Number of Faces Number of Bodies Target DesignModeler Error (%) x x

23 VMDM004: Extrude, Revolve, Skin-Loft, and Sweep Overview Feature: Drawing Units: Extrude, Revolve, Skin-Loft, and Sweep Millimeter Test Case Create a model using Sketch, Plane, Extrude, Revolve, and Loft features in walled conditions. The total volume is created using the following procedures: Lofted volume (V L ): a hexagon denoted as Sketch 1 with an area of mm 2 is lofted through a circular profile with an area of mm 2 (Sketch 2) and finally through another circular profile with an area of mm 2 (Sketch 3). This results in a lofted body with a volume of x 10 6 mm 3. Shell volume (V S ): a ring with an outer radius of 60 mm and an inner radius or 50 mm is extruded 60 mm and connected to the lofted body. This results in a shell with a volume of x 10 5 mm 3. Dome volume (V D ): a circular arc that is oriented perpendicular to the extrusion is revolved 360 degrees to create a dome with a hollow cylindrical cap in the middle, and this body is placed on top of the shell. This results in a dome with a volume of x 10 5 mm 3. Removed volume (V R ): a cylindrical volume of x 10 4 mm 3 is removed from the total body by placing a circle centered at the X,Y origin and extruding it along the Z-axis. Verify the volume of the resulting geometry. Figure 4: Final Model After Creating Extrude, Revolve, and Loft Features in Walled Conditions 17

24 VMDM004 Calculations 1. Volume of dome (V D ) = x 10 6 mm 3 2. Volume of the shell (V S ) = x 10 5 mm 3 3. Volume of the loft (V L ) = x 10 6 mm 3 4. Volume of the removed cylinder (V R ) = x 10 4 mm 3 5. Total volume of solid body (V T ) = x x x x 10 4 = x 10 6 mm 3 Results Comparison Results Target DesignModeler Error (%) Volume of Loft (mm 3 ) Surface Area of Loft (mm 2 ) Volume of Extrusion & Loft (mm 3 ) Surface Area of Extrusion & Loft (mm 2 ) Volume of Revolution, Extrusion, and Loft (mm 3 ) Surface Area of Revolution, Extrusion, and Loft (mm 2 ) Volume of Total Body (mm 3 ) Surface Area of Total Body (mm 2 )

25 Part II: SpaceClaim Descriptions

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27 VMSC_Geometry001: Alter Model Using Sketch, Pull, and Fill Overview Feature: Drawing Units: Sketch, Pull, and Fill Millimeter Test Case A rectangular sketch of 50 mm x 35 mm and a circle with radius of 5 mm is drawn on the top face. The sketches are pulled for a distance of 20 mm. A cylindrical hole of radius 5 mm and depth of 20 mm is filled. The result is verified after each step using surface area and volume validations. Figure 5: Final Modal after Sketch, Pull, and Fill 21

28 VMSC_Geometry001 Figure 6: Original Model Results Comparison Figure 7: Model After Using Sketch 22

29 VMSC_Geometry001 Area of rectangle = 50 mm x 35 mm = 1750 mm 2 Area of circle = x 5 mm x 5 mm = mm 2 Result Target SCDM Error (%) Area of rectangular sketch (m 2 ) Area of circular sketch (m 2 ) x x Figure 8: Model After Using Pull Volume added by pulling rectangular sketch = 50 mm x 35 mm x 20 mm = mm 3 Volume added by pulling circular sketch = x 5 mm x 5 mm x 20 mm = mm 3 Original volme = mm 3 Final volume = = mm 3 Result Target SCDM Error (%) Total volume (m 3 )

30 VMSC_Geometry001 Figure 9: Model After Using Fill Volume added by filling cylindrical hole = x 5 mm x 5 mm x 20 mm = mm 3 Volume after pull = mm 3 Final volume = = mm 3 Result Target SCDM Error (%) Total volume (m 3 )

31 VMSC_Geometry002: Alter Model Using Split Face, Split Body and Mirror Overview Feature: Drawing Units: Split Face, Split Body, Mirror Millimeter Test Case The top face of a rectangular model is split using opposite corner edges as reference. The body is then split into two using the central vertical face. The smaller body is removed using the Split Body tool. The same face is used as a mirror plane to mirror the geometry. The result is verified after each step using surface area or volume validations. Figure 10: Final Model After Split Face, Split Body, and Mirror 25

32 VMSC_Geometry002 Figure 11: Original Model 26

33 VMSC_Geometry002 Results Comparison Figure 12: Model After Using Split Face Area of triangle after Split Face operation = 0.5 mm x 50 mm x 35 mm = 875 mm 2. Result Target SCDM Error (%) Area of triangular sketch (m 2 )

34 VMSC_Geometry002 Figure 13: Model After Using Split Body Volume removed by Split Body operation = 50 mm x 35 mm x 20 mm = mm 3 Original volume = mm 3 Final volume = = mm 3 Result Target SCDM Error (%) Total volume (m 3 ) 7 x x

35 VMSC_Geometry002 Figure 14: Model After Using Mirror Volume after Split Body operation = mm 3 Volume added due to Mirror operation = mm 3 Final volume = = mm 3 Result Target SCDM Error (%) Total volume (m 3 )

36 30

37 VMSC_Geometry003: Alter Model using Linear, Circular, and Two-Dimensional Patterns Overview Feature: Drawing Units: Linear, Circular, and Two-Dimensional Patterns Millimeters Test Case A linear pattern of 7 protrusions is created. Next, a circular pattern with equidistant angles and a total of 8 protrusions is made. Finally, a two-dimensional pattern with an x- and y-offset of 10 mm is made. The result is verified after each step using volume validations. Figure 15: Final Model After Using Linear, Circular, and Two-Dimensional Features 31

38 VMSC_Geometry003 Figure 16: Original Model 32

39 VMSC_Geometry003 Results Comparison Figure 17: Model After Using Linear Pattern Volume added by linear pattern operation = 6 x 5 mm x 5 mm x 20 mm = 3000 mm 3 Original volume of block (left) = mm 3 Final volume = = mm 3 Result Target SCDM Error (%) Total volume (m 3 ) 7.35 x x

40 VMSC_Geometry003 Figure 18: Model After Using Circular Pattern Volume added by Circular Pattern operation = x 2.5 mm x 2.5 mm x 20 mm x 7 = mm 3 Original volume of circular block (middle) = mm 3 Final volume = = mm 3 Result Target SCDM Error (%) Total volume (m 3 ) x x x

41 VMSC_Geometry003 Figure 19: Model After Using Two Dimensional Pattern Volume added by Two Dimensional Pattern operation = 13 x 5 mm x 5 mm x 20 mm = 6500 mm 3 Original volume of block (right) = mm 3 Final volume = = mm 3 Result Target SCDM Error (%) Total volume (m 3 ) 7.7 x x

42 36

43 VMSC_Geometry004: Alter Model Using Revolve, Chamfer, and Offset Faces Overview Feature: Drawing Units: Revolve, Chamfer, Offset Faces Millimeter Test Case The bottom face of a rectangular recess at the top-right of the model is revolved. A chamfer with sides of 3 mm and 4mm each is made on the top-left portion of the solid. An offset condition is specified for the protruding rectangular extrusions. They are pull for a distance of 20 mm to show the offset relationship. The result is verified after each stop using volume validations. Figure 20: Final Model After Revolve, Chamfer, and Offset Faces 37

44 VMSC_Geometry004 Figure 21: Original Model 38

45 VMSC_Geometry004 Results Comparison Figure 22: Model After Using Revolve Volume added by revolving rectangular face = = mm 3 Original volume = mm 3 Final volume = = mm 3 Result Target SCDM Error (%) Total volume (m 3 )

46 VMSC_Geometry004 Figure 23: Model After Using Chamfer Volume removed by Chamfer operation = (0.5 x 3 x 4 x 50) = 300 mm 3 Volume at the end of Revolve operation = mm 3 Final volume = = mm 3 Result Target SCDM Error (%) Total volume (m 3 )

47 VMSC_Geometry004 Figure 24: Model After Using Offset Faces Volume added by extruding protrusions = 2 x 15 x 15 x 20 = 9000 mm 3 Volume at the end of the Chamfer operation = mm 3 Final volume = = mm 3 Result Target SCDM Error (%) Total volume (m 3 )

48 42

49 VMSC_Geometry005: Alter Model Using Blend and Project Overview Feature: Drawing Units: Blend, Project Millimeter Test Case A rectangular surface of 50 x 35 mm is blended into a face of the same dimension, making a solid block. A circular surface is projected onto the face of the solid. The result is verified after each step using either surface area or volume validations. Figure 25: Final Model After Blend and Project 43

50 VMSC_Geometry005 Figure 26: Original Model 44

51 VMSC_Geometry005 Results Comparison Figure 27: Model After Using Blend Volume added by pulling rectangular sketch = 50 x 35 x 20 = mm 3 Original volume = mm 3 Final volume = = mm 3 Result Target SCDM Error (%) Total volume (m 3 )

52 VMSC_Geometry005 Figure 28: Model After Using Project Surface area of circular surface = x 5 x 5 = mm 2 Result Target SCDM Error (%) Surface area (m 2 ) x x

53 VMSC_Geometry006: Alter Model Using Sheet Metal Tools and Operations Overview Feature: Drawing Units: Junction-Specific Pull, Form, Bend, Split, Double Wall Millimeter Test Case Import a model, pull the left edge by 30 mm after specifying the resulting junction to be sharp. Create two forms, including one circular punch and one rectangular knock-out. Bend the sheet metal at 20 mm from the right-most edge. Split the created bend such that the edge length on the right is 20 mm. Create a double wall on the left-most pulled face. The result is verified after each step using volume or edge length calculations. Figure 29: Final Model After Using Sheet Metal Tools and Features 47

54 VMSC_Geometry006 Figure 30: Original Model 48

55 VMSC_Geometry006 Results Comparison Figure 31: Model After Using Pull Volume added by pulling edge = 31 x 50 x 1 = 1550 mm 3 Original volume = 3500 mm 3 Final volume = = 5050 mm 3 Result Target SCDM Error (%) Total volume (m 3 ) 5.05 x x

56 VMSC_Geometry006 Figure 32: Model After Using Form Volume removed by circular punch = x 2.5 x 2.5 = mm 3 Volume after Pull operation = 5050 mm 3 Final volume = = mm 3 Result Target SCDM Error (%) Total volume (m 3 ) x x

57 VMSC_Geometry006 Figure 33: Model After Using Bend Volume added due to bend = (50 x 2) = mm 3 Volume after Form operation = mm 3 Final volume = = mm 3 Result Target SCDM Error (%) Total volume (m 3 ) x x

58 VMSC_Geometry006 Figure 34: Model After Using Split Edge length of the shorter split region = 20 mm Result Target SCDM Error (%) Edge length (m)

59 VMSC_Geometry006 Figure 35: Model After Using Double Wall Volume added by double wall feature = = mm 3 Volume after Split operation = (0.1 x 19) - (0.2375) = mm 3 Final volume = = mm 3 Result Target SCDM Error (%) Total volume (m 3 ) x x

60 54

61 VMSC_Prepare001: Midsurface, Extend, Imprint, and Spot Weld Overview Feature: Drawing Units: Midsurface, Extend, Imprint, and Spot Weld Millimeter Test Case A solid model is midsurfaced and then the solid is hidden (midsurface thickness should be equal to solid thickness). 8 midsurface bodies are created. 5 untouched small midsurfaces are extended (surface area made larger). 3 of the extended midsurfaces are imprinted and 2 are spot welded onto the largest midsurface. Verify the existence of the midsurfaces and spot welds, and verify the geometry change due to the extending and imprinting operations. Note The target and SCDM values of surface area are measured from within SCDM. The target values of surface area are measured before extending the midsurfaces, and the SCDM values of surface area are measured after extending them. This part of the verification is to make sure the surface area does increase after the Extend operation. 55

62 VMSC_Prepare001 Figure 36: Final Model After Midsurface, Extend, Imprint and Spot Weld Operations Results Comparison Results Number of Midsurface Bodies Thickness of Midsurfaces Surface Area of Midsuface4 after extending Surface Area of Midsuface5 after extending Surface Area of Midsuface6 after extending Surface Area of Midsuface7 after extending Surface Area of Midsuface8 after extending Edge Count of Misdurface2 after imprinting Number of SpotWelds Number of Points for Each SpotWeld Target > > > > > SCDM 8 Error (%)

63 VMSC_Prepare002: Volume Extract and Enclosure Overview Feature: Drawing Units: Volume Extract and Enclosure Millimeter Test Case This test case is divided into several scenarios, with results for each. Scenario 1 Extract volume from an oil pan model, and verify the existence of the extracted volume. Figure 37: Extract Volume from Oil Pan Scenario 2 Modify the oil pan model by enlarging an internal cylindrical surface, then update the extracted volume. Verify the updated surface area of the counterpart of the cylindrical surface on the extracted volume. The target value is measured from the cylindrical surface of the oil pan model and the SCDM value is measured from the updated cylindrical surface (mating surface) of the extracted volume. 57

64 VMSC_Prepare002 Figure 38: Updated Extracted Volume Scenario 3 Create an enclosure volume from the cooling block object, then verify the existence of the enclosure volume. The target surface count is counted using the cooling block object and the SCDM surface count is counted using the enclosure volume. 58

65 VMSC_Prepare002 Figure 39: Enclosure Volume from the Cooling Block Object Scenario 4 Modify the cooling block object by cutting a cylindrical voume from the bottom base. Update the enclosure volume, verifying extra surfaces of the update volume. The target surface area is measured from the cooling block object and the SCDM surface area is measured from the enclosure volume (mating surfaces). 59

66 VMSC_Prepare002 Figure 40: Enclosure Volume from the Cooling Block Object Results Comparison Scenario 1 Results Name of the extracted volume Is it a closed body Target SCDM Error (%) Volume Volume N/A True True N/A Scenario 2 Results Surface area of the changed cylindrical surface Target SCDM Error (%) E-6 Scenario 3 Results Name of the Enclosure volume Target SCDM Error (%) Enclosure Enclosure N/A 60

67 VMSC_Prepare002 Results Target SCDM Error (%) Number of Face Count Scenario 4 Results Type of the 1st new surface on Enclosure volume Surface area of the planar surface Target Plane SCDM Plane Type of the 2nd new surface on Enclosure volume Cylinder Cylinder Surface area of the cylindrical surface 9.05 Error (%) N/A E N/A

68 62

69 VMSC_Prepare003: Beam Create, Extract, and Orient Overview Feature: Drawing Units: Beam Create, Extract, and Orient Millimeter Test Case This test case is divided into several scenarios, with results for each. Scenario 1 Create beams with circular profile and extract beams from existing solid. Verify the existence of the extracted volume. Figure 41: Created Beams and Extracted Beams Scenario 2 Orient one beam by changing its Section Anchor to Location, then move the section toward Y direction for mm. Verify the section location of the oriented beam. 63

70 VMSC_Prepare003 Figure 42: Orient Beam Results Comparison Scenario 1 Results Number of SketchCurve with Beam properties Number of Beams with section name of Circle Number of Beams with section name of Extracted Profile1 1 Number of Beams with section name of Extracted Profile2 2 Number of Beams with section name of Extracted Profile3 1 Scenario 2 Target SCDM Error (%) Results Anchor Location of Oriented Beam Target SCDM Error (%)

71 VMSC_Prepare004: Remove Rounds, Interference, Faces, and Short Edges Overview Feature: Drawing Units: Remove Rounds, Interference, Faces, and Short Edges Millimeter Test Case This test case is divided into several scenarios, with results for each. Scenario 1 Remove the rounds as shown in the figure, then verify the absence of these round faces. Figure 43: Remove Rounds Scenario 2 Remove the interference volume from the smaller object, then verify the cylindrical surface area and the circular edge length. The target values are measured from the larger object before removing interference, the SCDM values are measured from the mating feature of the smaller object after removing interference. 65

72 VMSC_Prepare004 Figure 44: Remove Interference Volume from Smaller Objects Scenario 3 Remove the round faces by using the Remove Faces tool, then verify the existence of the edge loop. 66

73 VMSC_Prepare004 Figure 45: Enclosure Volume from the Cooling Block Scenario 4 Remove the tiny edge using the Short Edge tool, then verify the absence of the tiny edge. Figure 46: Remove the Tiny Edge 67

74 VMSC_Prepare004 Results Comparison Scenario 1 Results Number of box selected round faces Target SCDM Error (%) Scenario 2 Results Surface area of the cylindrical surface on the changed object Edge length of the circular edge on the changed object Target SCDM Error (%) E Scenario 3 Results Number of newly created edges Target SCDM Error (%) Scenario 4 Results Number of box selected edges Target SCDM Error (%)

75 VMSC_Repair001: Stitch, Gap, and Missing Faces Overview Feature: Drawing Units: Stitch, Gap, and Missing Faces Millimeter Test Case Use the Stitch tool to stitch all the surface bodies into one single surface body, then use the Missing Faces tool to generate missing faces, then use the Gap tool to fill the gaps. Finally, the surface body will be solidified into a solid body. Verify the existence of the solid body. Figure 47: Solidify Surface Bodies into a Single Solid Body Results Comparison Results Number of solid body Is it a closed body Target SCDM Error (%) 1 True 1 True 0 N/A 69

76 70

77 VMSC_Repair002: Merge Faces, Small Faces, Simplify, Straighten Overview Feature: Drawing Units: Merge Faces, Small Faces, Simplify, Straighten Millimeter Test Case This test case is divided into several scenarios, with results for each. Scenario 1 Merge adjacent faces together, then verify the absence of the merged (removed) faces. Figure 48: Merge Faces into Adjacent Faces Scenario 2 Use the Small Faces tool to detect any small faces to remove, then verify the absence of the previous small faces. 71

78 VMSC_Repair002 Figure 49: Detect and Fix Small Faces Scenario 3 Use the Simplify tool to simplify the red face only (see the figure below). The Spline face is simplifed into a Sphere face. Verify the Surface Type of the simplified face. 72

79 VMSC_Repair002 Figure 50: Simplify the Red Face Scenario 4 Straighten the faces that are not exactly perpendicular to the adjacent faces, then verify the length of the face edges. The target value (baseline) was interactively measured in SCDM after confirming the straighten operation created perpendicular faces. Figure 51: Straighten the Highlighted Faces 73

80 VMSC_Repair002 Figure 52: Baseline Measurement After Confirming Straighten Operation (Right Edge) Figure 53: Baseline Measurement After Confirming Straighten Operation (Left Edge) 74

81 VMSC_Repair002 Results Comparison Scenario 1 Results Number of box selected faces Target SCDM Error (%) Scenario 2 Results Number of box selected faces Target SCDM Error (%) Scenario 3 Results Surface Type of the simplified face Target SCDM Error (%) Sphere Sphere N/A Scenario 4 Results Vertical edge length on both sides Target SCDM Error (%) Note In SC journal files, you will find edge length in meters. The values are all very close to the interactively measured baseline. 75

82 76

83 VMSC_Repair003: Split Edges, Extra Edges, and Duplicates Overview Feature: Drawing Units: Split Edges, Extra Edges, and Duplicates Millimeter Test Case This test case is divided into several scenarios, with results for each. Scenario 1 Use the Split Edges tool to find and fix the edges that are supposed to be combined together, thus reducing the edge numbers. Verify the edge count after using the tool. The target value was measured before the fix, so the SCDM value after the fix should be smaller. Figure 54: Find and Fix the Split Edges Scenario 2 Use the Extra Edges tool to detect and remove the extra eduge, then verify the removal. The target value was measured before the fix, so the SCDM value after the fix should be smaller. 77

84 VMSC_Repair003 Figure 55: Detect and Fix the Extra Edge Scenario 3 Use the Duplicate tool to find and remove duplicate surfaces. Verify the number of surface bodies after the fix to be zero. 78

85 VMSC_Repair003 Figure 56: Find and Remove Duplicate Surfaces Results Comparison Scenario 1 Results Number of edges after fix Target SCDM Error (%) < N/A Scenario 2 Results Number of edges after fix Target SCDM Error (%) < N/A 79

86 VMSC_Repair003 Scenario 3 Results Number of solid bodies after fix Number of surface bodies after fix Target SCDM Error (%)

87 Part III: Application Descriptions

88

89 VMMECH001: Statically Indeterminate Reaction Force Analysis Overview Reference: Solver(s): Analysis Type(s): Element Type(s): S. Timoshenko, Strength of Materials, Part 1, Elementary Theory and Problems, 3rd Edition, CBS Publishers and Distributors, pg. 22 and 26 ANSYS AIM ANSYS Linear Static Structural Analysis Solid Test Case An assembly of three prismatic bars is supported at both end faces and is axially loaded with forces F 1 and F 2. Force F 1 is applied on the face between Parts 2 and 3 and F 2 is applied on the face between Parts 1 and 2. Apply advanced mesh control with element size of 0.5. Find reaction forces in the Y direction at the fixed supports. Material Properties E = e7 psi ν = 0.3 ρ = lbm/in 3 Geometric Properties Cross section of all parts = 1 x 1 Length of Part 1 = 4" Length of Part 2 = 3" Length of Part 3 = 3 Loading Force F 1 = (Y direction) Force F 2 = -500 (Y direction) Analysis Assumptions and Modeling Notes for ANSYS The following figure shows the finite element model in ANSYS : 83

90 VMMECH001 Figure 57: ANSYS Schematic Results Comparison for ANSYS Results Target Error (%) Y Reaction Force at Top Fixed Support (lbf ) Y Reaction Force at Bottom Fixed Support (lbf ) Analysis Assumptions and Modeling Notes for ANSYS AIM The following figure shows the finite element model in ANSYS AIM: 84

91 VMMECH001 Figure 58: ANSYS AIM Schematic Results Comparison for ANSYS AIM Results Target AIM Error (%) Y Reaction Force at Top Fixed Support (lbf ) Y Reaction Force at Bottom Fixed Support (lbf )

92 86

93 VMMECH002: Rectangular Plate with Circular Hole Subjected to Tensile Loading Overview Reference: Solver(s) Analysis Type(s): Element Type(s): J. E. Shigley, Engineering Design, McGraw-Hill, 1st Edition, 1986, Table A-23, Figure A-23-1, pg. 673 ANSYS AIM ANSYS Linear Static Structural Analysis Solid Test Case A rectangular plate with a circular hole is fixed along one of the end faces and a tensile pressure load is applied on the opposite face. A convergence with an allowable change of 10% is applied to account for the stress concentration near the hole. The Maximum Refinement Loops is set to 2 and the Refinement mesh control is added on the cylindrical surfaces of the hole with Refinement = 1. Find the Maximum Normal Stress in the x direction on the cylindrical surfaces of the hole. Material Properties E = 1000 Pa ν = 0 Geometric Properties Length = 15 m Width = 5 m Thickness = 1 m Hole radius = 0.5 m Loading Pressure = -100 Pa Analysis Assumptions and Modeling Notes for ANSYS The following figure shows the finite element model in ANSYS : Figure 59: ANSYS Schematic 87

94 VMMECH002 Results Comparison for ANSYS Results Target Error (%) Maximum Normal X Stress (Pa) Analysis Assumptions and Modeling Notes for ANSYS AIM The following figure shows the finite element model in ANSYS AIM: Figure 60: ANSYS AIM Schematic Results Comparison for ANSYS AIM Results Target AIM Error (%) Maximum Normal X Stress (Pa)

95 VMMECH003: Modal Analysis of Annular Plate Overview Reference: Solver(s) Analysis Type(s): Element Type(s): R. J. Blevins, Formula for Natural Frequency and Mode Shape, Van Nostrand Reinhold Company Inc., 1979, Table 11-2, Case 4, pg. 247 ANSYS AIM ANSYS Free Vibration Analysis Solid Test Case An assembly of three annular plates has cylindrical support (fixed in the radial, tangential, and axial directions) applied on the cylindrical surface of the hole. Sizing control with element size of 0.5 is applied to the cylindrical surface of the hole. Find the first six modes of natural frequencies. Material Properties E = e7 psi ν = 0.3 ρ = lbm/in 3 Geometric Properties Inner diameter of inner plate = 20" Inner diameter of middle plate = 28" Inner diameter of outer plate = 34" Outer diameter of outer plate = 40" Thickness of all plates = 1" Loading Analysis Assumptions and Modeling Notes for ANSYS The following figure shows the finite element model in ANSYS : 89

96 VMMECH003 Figure 61: ANSYS Schematic Results Comparison for ANSYS Results Target Error (%) 1st Frequency Mode (Hz) nd Frequency Mode (Hz) rd Frequency Mode (Hz) th Frequency Mode (Hz) th Frequency Mode (Hz) th Frequency Mode (Hz) Analysis Assumptions and Modeling Notes for ANSYS AIM The following figure shows the finite element model in ANSYS AIM: 90

97 VMMECH003 Figure 62: ANSYS AIM Schematic Results Comparison for ANSYS AIM Results Target AIM Error (%) 1st Frequency Mode (Hz) nd Frequency Mode (Hz) rd Frequency Mode (Hz) th Frequency Mode (Hz) th Frequency Mode (Hz) th Frequency Mode (Hz)

98 92

99 VMMECH004: Viscoplastic Analysis of a Body (Shear Deformation) Overview Reference: Solver(s) Analysis Type(s): Element Type(s): B. Lwo and G. M. Eggert, "An Implicit Stress Update Algorithm Using a Plastic Predictor". Submitted to Computer Methods in Applied Mechanics and Engineering, January ANSYS Nonlinear Structural Analysis Solid Test Case A cubic shaped body made up of a viscoplastic material obeying Anand's law undergoes uniaxial shear deformation at a constant rate of 0.01 cm/s. The temperature of the body is maintained at 400 C. Find the shear load (Fx) required to maintain the deformation rate of 0.01 cm/sec at time equal to 20 seconds. Figure 63: Schematic Material Properties Geometric Properties E x (Young's Modulus) = h = 1 cm 60.6 GPa thickness = 1 (Poisson's Ratio) = cm S o = 29.7 MPa Q/R = E3 K A = 1.91E7 s -1 = 7.0 m = h o = MPa = MPa = a = 1.3 Loading Temp = 400 C = 673 K Velocity (x-direction) = 0.01 y = 1 cm Time = 20 sec 93

100 VMMECH004 Results Comparison Results Target Error (%) F x, N

101 VMMECH005: Heat Transfer in a Composite Wall Overview Reference: Solver(s): Analysis Type(s): Element Type(s): F. Kreith, Principles of Heat Transfer, Harper and Row Publisher, 3rd Edition, 1976, Example 2-5, pg. 39 ANSYS AIM ANSYS Linear Static Thermal Analysis Solid Test Case A furnace wall consists of two layers: fire brick and insulating brick. The temperature inside the furnace is 3000 F (T f ) and the inner surface convection coefficient is x 10-3 BTU/s ft 2 F (h f ). The ambient temperature is 80 F (T a ) and the outer surface convection coefficient is x 10-4 BTU/s ft 2 F (h a ). Find the Temperature Distribution. Material Properties Fire brick wall: k = x 10-4 BTU/s ft F Insulating wall: k = x 10-5 BTU/s ft F Geometric Properties Cross-section = 1" x 1" Fire brick wall thickness = 9" Insulating wall thickness = 5" Loading Analysis Assumptions and Modeling Notes for ANSYS The following figure shows the finite element model in ANSYS : Figure 64: ANSYS Schematic Results Comparison for ANSYS Results Target Error (%) Minimum Temperature ( F) Maximum Temperature ( F)

102 VMMECH005 Analysis Assumptions and Modeling Notes for ANSYS AIM The following figure shows the finite element model in ANSYS AIM: Figure 65: ANSYS AIM Schematic Results Comparison for ANSYS AIM Results Target AIM Error (%) Minimum Temperature ( F) Maximum Temperature ( F)

103 VMMECH006: Heater with Nonlinear Conductivity Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Vedat S. Arpaci, Conduction Heat Transfer, Addison-Wesley Book Series, 1966, pg. 130 ANSYS AIM ANSYS Nonlinear Static Thermal Analysis Solid Test Case A liquid is boiled using the front face of a flat electric heater plate. The boiling temperature of the liquid is 212 F. The rear face of the heater is insulated. The internal energy generated electrically may be assumed to be uniform and is applied as internal heat generation. Find the maximum temperature and maximum total heat flux. Material Properties k = [ * ( T)] BTU/s in F Temperature ( F) Conductivity (BTU/s in F) 1.419e e-002 Geometric Properties Radius = Thickness = 1 Loading Front face temperature = 212 F Internal heat generation = 10 BTU/s in 3 Analysis Assumptions and Modeling Notes for ANSYS The following figure shows the finite element model in ANSYS : Figure 66: ANSYS Schematic 97

104 VMMECH006 Results Comparison for ANSYS Results Target Error (%) Maximum Temperature ( F) Maximum Total Heat Flux (BTU/s in 2 ) Analysis Assumptions and Modeling Notes for ANSYS AIM The following figure shows the finite element model in ANSYS AIM: Figure 67: ANSYS AIM Schematic Results Comparison for ANSYS AIM Results Target AIM Error (%) Maximum Temperature ( F) Maximum Total Heat Flux (BTU/s in 2 )

105 VMMECH007:Thermal Stress in a Bar with Temperature Dependent Conductivity Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any basic Heat Transfer book ANSYS AIM ANSYS Nonlinear Thermal Stress Analysis Solid Test Case A long bar has thermal conductivity that varies with temperature. The bar is constrained at both ends by frictionless surfaces. A temperature of T C is applied at one end of the bar (End A). The reference temperature is 5 C. At the other end, a constant convection of h W/m 2 C is applied. The ambient temperature is 5 C. Advanced mesh control with element size of 2 m is applied. Find the following: Minimum temperature Maximum thermal strain in z direction (on the two end faces) Maximum deformation in z direction Maximum heat flux in z direction at z = 20 m Temperature at a distance z from rear face is given by: Thermal strain in the z-direction in the bar is given by: Deformation in the z-direction is given by: Heat flux in the z-direction is given by: 99

106 VMMECH007 Material Properties E = 2 x Pa ν = 0 α = 1.5 x 10-5 / C k = * ( * T) W/m C Temperature ( C) Conductivity (W/m C) 3.91 x 10-2 Geometric Properties Length = 20 m Width = 2 m Breadth = 2 m Loading Rear face temperature T = 100 C Film Coefficient h = W/m 2 C Ambient temperature = 5 C 5 Reference temperature = 5 C Analysis Assumptions and Modeling Notes for ANSYS The following figure shows the finite element model in ANSYS : Figure 68: ANSYS Schematic Analysis Assumptions and Modeling Notes for ANSYS AIM The following figure shows the finite element model in ANSYS AIM: 100

107 VMMECH007 Figure 69: ANSYS AIM Structural Schematic Figure 70: ANSYS AIM Thermal Schematic 101

108 VMMECH007 Results Comparison for ANSYS Results Target Error (%) Minimum Temperature ( C) Maximum Thermal strain (z = 20) (m/m) Maximum Thermal strain (z = 0) (m/m) Maximum Z Deformation (m) Maximum Z Heat Flux (z = 20) (W/m 2 ) Results Comparison for ANSYS AIM Results Minimum Temperature ( C) Maximum Thermal strain (z = 20) (m/m) Maximum Thermal strain (z = 0) (m/m) Maximum Z Deformation (m) Maximum Z Heat Flux (z = 20) (W/m2) Target AIM Error (%)

109 VMMECH008: Heat Transfer from a Cooling Spine Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Kreith, F., Principles of Heat Transfer, Harper and Row, 3rd Edition, 1976, Equation 2-44a, pg. 59, Equation 2 45, pg. 60 ANSYS AIM ANSYS Linear Static Thermal Analysis Solid Test Case A steel cooling spine of cross-sectional area A and length L extend from a wall that is maintained at temperature T w. The surface convection coefficient between the spine and the surrounding air is h, the air temper is T a, and the tip of the spine is insulated. Apply advanced mesh control with element size of 0.025'. Find the heat conducted by the spine and the temperature of the tip. Figure 71: Schematic 103

110 VMMECH008 Figure 72: Model Setup for ANSYS AIM Material Properties E = x 10 9 psf ν = 0.3 Thermal conductivity k = 9.71 x 10-3 BTU/s ft F Geometric Properties Cross section = 1.2 x 1.2 L = 8 Loading T w = 100 F T a = 0 F h = x 10-4 BTU/s ft 2 F Results Comparison for ANSYS Results Target Error (%) Temperature of the Tip ( F) Heat Conducted by the Spine (Heat Reaction) (BTU/s) x x

111 VMMECH008 Results Comparison for ANSYS AIM Results Target AIM Error (%) Temperature of the Tip ( F) Heat Conducted by the Spine (Heat Reaction) (BTU/s) x x

112 106

113 VMMECH009: Stress Tool for Long Bar With Compressive Load Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any basic Strength of Materials book ANSYS Linear Static Structural Analysis Solid Test Case A multibody of four bars connected end to end has one of the end faces fixed and a pressure is applied to the opposite face as given below. The multibody is used to nullify the numerical noise near the contact regions. Find the maximum equivalent stress for the whole multibody and the safety factor for each part using the maximum equivalent stress theory with tensile yield limit. Figure 73: Schematic Material Properties Material E (Pa) ν Part e11 0 Part 2 7.1e10 0 Part 3 2e11 0 Part 4 1.1e11 0 Tensile Yield (Pa) 2.07e8 2.8e8 2.5e8 2.8e8 Geometric Properties Part 1: 2 m x 2 m x 3 m Part 2: 2 m x 2 m x 10 m Loading Pressure = 2.5e8 Pa 107

114 VMMECH009 Part 3: 2 m x 2 m x 5 m Part 4: 2 m x 2 m x 2 m Results Comparison Results Target Error (%) Maximum Equivalent Stress (Pa) 2.5e8 2.5e Safety Factor for Part Safety Factor for Part Safety Factor for Part Safety Factor for Part

115 VMMECH010: Modal Analysis of a Rectangular Plate Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Blevins, Formula for Natural Frequency and Mode Shape, Van Nostrand Reinhold Company Inc., 1979, Table 11-4, Case 11, pg. 256 ANSYS Free Vibration Analysis Shell Test Case A rectangular plate is simply supported on both the smaller edges and fixed on one of the longer edges as shown below. Sizing mesh control with element size of 6.5 mm is applied on all the edges to get accurate results. Find the first five modes of natural frequency. Figure 74: Schematic Material Properties Geometric Properties E = 2e11 Pa Length = 0.25 ν = 0.3 m ρ = 7850 kg/m 3 Width = 0.1 m Thickness = m Loading Results Comparison Results Target Error (%) 1st Frequency Mode (Hz) nd Frequency Mode (Hz) rd Frequency Mode (Hz)

116 VMMECH010 Results Target Error (%) 4th Frequency Mode (Hz) th Frequency Mode (Hz)

117 VMMECH011: Large Deflection of a Circular Plate With Uniform Pressure Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Timoshenko S.P., Woinowsky-Krieger S., Theory of Plates and Shells, McGraw-Hill, 2nd Edition, Article 97, equation 232, pg. 401 ANSYS Nonlinear Structural Analysis (Large Deformation On) Shell Test Case A circular plate is subjected to a uniform pressure on its flat surface. The circular edge of the plate is fixed. To get accurate results, apply sizing control with element size of 5 mm on the circular edge. Find the total deformation at the center of the plate. Figure 75: Schematic Material Properties E = 2e11 Pa ν = 0.3 Results Comparison Geometric Properties Radius = 0.25 m Thickness = m Loading Pressure = Pa Results Target Error (%) Total deformation (m)

118 112

119 VMMECH012: Buckling of a Stepped Rod Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Warren C. Young, Roark's Formulas for Stress & Strains, McGraw Hill, 6th Edition, Table 34, Case 2a, pg. 672 ANSYS Buckling Analysis Solid Test Case A stepped rod is fixed at one end face. It is axially loaded by two forces: a tensile load at the free end and a compressive load on the flat step face at the junction of the two cross sections. To get accurate results, apply sizing control with element size of 6.5 mm. Find the Load Multiplier for the First Buckling Mode. Figure 76: Schematic Material Properties E = 2e11 Pa ν = 0.3 Geometric Properties Larger diameter = m Smaller diameter = m Length of larger diameter = 0.2 m Length of smaller diameter = 0.1 m Loading Force at free end = 1000 N Force at the flat step face = N Both forces are in the z direction 113

120 VMMECH012 Results Comparison Results Target Error (%) Load Multiplier

121 VMMECH013: Buckling of a Circular Arch Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Warren C. Young, Roark's Formulas for Stress Strains, McGraw Hill, 6th Edition, Table 34, Case 10, pg. 679 ANSYS Buckling Analysis Shell Test Case A circular arch of a rectangular cross section (details given below) is subjected to a pressure load as shown below. Both the straight edges of the arch are fixed. Find the Load Multiplier for the first buckling mode. Figure 77: Schematic Material Properties E = 2e5 MPa ν = 0 Results Comparison Geometric Properties Arch cross-section = 5 mm x 50 mm Mean radius of arch = 50 mm Included angle = 90 Loading Pressure = 1 MPa Results Target Error (%) Load Multiplier

122 116

123 VMMECH014: Harmonic Response of a Single Degree of Freedom System Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any basic Vibration Analysis book ANSYS Harmonic Analysis Solid Test Case An assembly where four cylinders represent massless springs in series and a point mass simulates a spring mass system. The flat end face of the cylinder (Shaft 1) is fixed. Harmonic force is applied on the end face of another cylinder (Shaft 4) as shown below. Find the z directional Deformation Frequency Response of the system on the face to which force is applied for the frequency range of 0 to 500 Hz for the following scenarios using Mode-Superposition. Solution intervals = 20. Scenario 1: Damping ratio = 0 Scenario 2: Damping ratio = 0.05 Figure 78: Schematic Material Shaft 1 Shaft 2 Shaft 3 Shaft 4 Material Properties E (Pa) ν 1.1 x x x x ρ (kg/m 3 ) 1 x x x x 10-8 Geometric Properties Each cylinder: Diameter = 20 mm Length = 50 mm Loading Force = 1 x 10 7 N (Z-direction) 117

124 VMMECH014 Point Mass = Kg Results Comparison Results Target Error (%) Maximum Amplitude without damping (m) Phase angle without damping (degrees) Maximum Amplitude with damping (m) Phase angle with damping (degrees)

125 VMMECH015: Harmonic Response of Two Storied Building Under Transverse Loading Overview Reference: Solver(s): Analysis Type(s): Element Type(s): W. T. Thomson, Theory of Vibration with Applications, 3rd Edition, 1999, Example 6.4-1, pg. 166 ANSYS Harmonic Analysis Solid Test Case A two-story building has two columns (2K and K) constituting stiffness elements and two slabs (2M and M) constituting mass elements. The material of the columns is assigned negligible density so as to make them as massless springs. The slabs are allowed to move only in the y direction by applying frictionless supports on all the faces of the slabs in the y direction. The end face of the column (2K) is fixed and a harmonic force is applied on the face of the slab (M) as shown in the figure below. Find the y directional Deformation Frequency Response of the system at 70 Hz on each of the vertices as shown below for the frequency range of 0 to 500 Hz using Mode-Superposition. Use Solution intervals = 50. Figure 79: Schematic Material Block 2 Shaft 2 Block 1 Shaft 1 Material Properties E (Pa) ν 2e e e e ρ (kg/m 3 ) e e-8 119

126 VMMECH015 Geometric Loading Properties Force = -1e5 N (y Block 1 and 2: direction) 40 mm x 40 mm x 40 mm Shaft 1 and 2: 20 mm x 20 mm x 200 mm Results Comparison Results Target Error (%) Maximum Amplitude for vertex A (m) Maximum Amplitude for vertex B (m)

127 VMMECH016: Fatigue Tool with Non-Proportional Loading for Normal Stress Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any basic Machine Design book ANSYS Fatigue Analysis Solid Test Case A bar of rectangular cross section has the following loading scenarios. Scenario 1: One of the end faces is fixed and a force is applied on the opposite face as shown below in Figure 80: Scenario 1 (p. 121). Scenario 2: Frictionless support is applied to all the faces of the three standard planes (faces not seen in Figure 81: Scenario 2 (p. 121)) and a pressure load is applied on the opposite faces in positive y- and z-directions. Find the life, damage, and safety factor for the normal stresses in the x, y, and z directions for nonproportional fatigue using the Soderberg theory. Use a design life of 1e6 cycles, a fatigue strength factor or 1, a scale factor of 1, and 1 for coefficients of both the environments under Solution Combination. Figure 80: Scenario 1 Figure 81: Scenario 2 Material Properties E = 2e11 Pa ν =

128 VMMECH016 Material Properties Ultimate Tensile Strength = 4.6e8 Pa Yield Tensile Strength = 3.5e8 Pa Endurance Strength = e6 Pa Number of Cycles e6 Alternating Stress (Pa) 4.6e e6 Geometric Properties Bar: 20 m x 1 m x 1m Loading Scenario 1: Force = 2e6 N (y-direction) Scenario 2: Pressure = -1e8 Pa Analysis Non-proportional fatigue uses the corresponding results from the two scenarios as the maximum and minimum stresses for fatigue calculations. The fatigue calculations use standard formulae for the Soderberg theory. Results Comparison Results Target Error (%) Stress Component - Component Life X Damage Safety Factor Stress Component - Component Life Y Damage Safety Factor Stress Component - Component Life Z Damage Safety Factor

129 VMMECH017: Thermal Stress Analysis with Remote Force and Thermal Loading Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any basic Strength of Materials book ANSYS AIM ANSYS Linear Thermal Stress Analysis Solid Test Case A cylindrical rod assembly of four cylinders connected end to end has frictionless support applied on all the cylindrical surfaces and both the flat end faces are fixed. Other thermal and structural loads are as shown below. Find the Deformation in the x direction of the contact surface on which the remote force is applied. To get accurate results apply a global element size of 1.5 m. Material Properties E = 2 x Pa ν = 0 α = 1.2 x 10-5 / C Geometric Properties Diameter = 2 m Lengths of cylinders in order from End A: 2 m, 5 m, 10 m, and 3 m. Loading Given temperature (End A) = 1000 C Given temperature (End B) = 0 C Remote force = 1 x N applied on the contact surface at a distance 7 m from end A. Location of remote force = (7,0,0) m Analysis Assumptions and Modeling Notes for ANSYS The following figure shows the finite element model in ANSYS : Figure 82: ANSYS Schematic 123

130 VMMECH017 Results Comparison for ANSYS Results Target Error (%) Maximum X Deformation (m) Analysis Assumptions and Modeling Notes for ANSYS AIM The following figure shows the finite element model in ANSYS AIM: Figure 83: ANSYS AIM Schematic Results Comparison for ANSYS AIM Results Target AIM Error (%) Maximum X Deformation (m)

131 VMMECH018: A Bar Subjected to Tensile Load with Inertia Relief Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any basic Strength of Materials book ANSYS AIM ANSYS Linear Static Structural Analysis (Inertia Relief On) Solid Test Case A long bar assembly is fixed at one end and subjected to a tensile force at the other end as shown below. Turn on Inertia Relief. Find the deformation in the z direction Analysis where: L = total length of bar A = cross-section m = mass Material Properties Geometric Properties E = 2 x Pa Cross-Section = ν = m x 2 m Lengths of bars ρ = 7850 kg/m 3 in order from End A: 2 m, 5 m, 10 m, and 3 m. Loading Force P = 2 x 10 5 N (positive z direction) Analysis Assumptions and Modeling Notes for ANSYS The following figure shows the finite element model in ANSYS : 125

132 VMMECH018 Figure 84: ANSYS Schematic Results Comparison for ANSYS Results Target Error (%) Maximum Z Deformation (m) 2.5 x x Analysis Assumptions and Modeling Notes for ANSYS AIM The following figure shows the finite element model in ANSYS AIM: Figure 85: ANSYS AIM Schematic Results Comparison for ANSYS AIM Results Target AIM Error (%) Maximum Z Deformation (m) 2.5 x x

133 VMMECH019: Mixed Model Subjected to Bending Loads with Solution Combination Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any basic Strength of Materials book ANSYS Linear Static Structural Analysis Beam and Shell Test Case A mixed model (shell and beam) has one shell edge fixed as shown below. Bending loads are applied on the free vertex of the beam as given below. Apply a global element size of 80 mm to get accurate results. Scenario 1: Only a force load. Scenario 2: Only a moment load. Find the deformation in the y direction under Solution Combination with the coefficients for both the environments set to 1. Figure 86: Scenario 1 Figure 87: Scenario 2 Material Properties E = 2e5 Pa ν = 0 Geometric Properties Shell = 160 mm x 500 mm x 10 mm Loading Force F = -10 N (y direction) 127

134 VMMECH019 Material Properties Analysis Geometric Properties Beam rectangular cross section = 10 mm x 10 mm Beam length = 500 mm Loading Moment M = z-axis where: I = total bending length of the mixed model I = moment of inertia of the beam cross-section Results Comparison Results Target Error (%) Maximum Y-Deformation (mm)

135 VMMECH020: Modal Analysis for Beams Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any basic Vibration Analysis book ANSYS Modal Analysis Beam Test Case Two collinear beams form a spring mass system. The density of the longer beam is kept very low so that it acts as a massless spring and the smaller beam acts as a mass. The end vertex of the longer beam (acting as a spring) is fixed. The cross section details are as shown below. Find the natural frequency of the axial mode. Figure 88: Cross Section Details for Both Beams Figure 89: Schematic Material Properties Material E (Pa) ν ρ (kg/m 3 ) Spring 1.1e e-8 Mass 2e e5 129

136 VMMECH020 Geometric Properties Spring beam length = 500 mm Mass beam length = 5 mm Loading Results Comparison Results Target Error (%) Natural Frequency of Axial Mode (Hz)

137 VMMECH021: Buckling Analysis of Beams Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Warren C. Young, Roark's Formulas for Stress and Strains, McGraw Hill, 6th Edition, Table 34, Case 3a, pg. 675 ANSYS Buckling Analysis Beam Test Case A beam fixed at one end and is subjected to two compressive forces. One of the forces is applied on a portion of the beam of length 50 mm (L 1 ) from the fixed end and the other is applied on the free vertex, as shown below. Find the load multiplier for the first buckling mode. Figure 90: Schematic Material Properties Geometric Properties E = 2e11 Pa L 1 = 50 mm ν = 0.3 Total length = 200 mm Rectangular cross section = 10 mm x 10 mm Loading Force on L 1 = N (x direction) Force on free vertex = N (x direction) Results Comparison Results Target Error (%) Load Multiplier

138 132

139 VMMECH022: Structural Analysis with Advanced Contact Options Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any basic Strength of Material book ANSYS Nonlinear Static Structural Analysis Solid Test Case An assembly of two parts with a gap has a Frictionless Contact defined between the two parts. The end faces of both the parts are fixed and a given displacement is applied on the contact surface of Part 1 as shown below. Find the Normal stress and Directional deformation - both in the z direction for each part for the following scenarios: Scenario 1: Interface treatment - adjust to touch. Scenario 2: Interface treatment - add offset. Offset = 0 m. Scenario 3: Interface treatment - add offset. Offset = m. Scenario 4: Interface treatment - add offset. Offset = m. Validate all of the above scenarios for Augmented Lagrange and Pure Penalty formulations. Figure 91: Schematic Material Properties E = 2e11 Pa ν = 0 Geometric Properties Gap = m Dimensions for each part: 0.1 m x 0.1 m x 0.5m Loading Given displacement = (0, 0, ) m 133

140 VMMECH022 Results Comparison The same results are obtained for both Augmented Lagrange and Pure Penalty formulations. Adjust To Touch Add Offset. Offset = 0 m Add Offset. Offset = m Add Offset. Offset = m Results Maximum directional z deformation Part 1 (m) Maximum directional z deformation Part 2 (m) Maximum normal z stress Part 1 (Pa) Maximum normal z stress Part 2 (Pa) Maximum directional z deformation Part 1 (m) Maximum directional z deformation Part 2 (m) Maximum normal z stress Part 1 (Pa) Maximum normal z stress Part 2 (Pa) Maximum directional z deformation Part 1 (m) Maximum directional z deformation Part 2 (m) Maximum normal z stress Part 1 (Pa) Maximum normal z stress Part 2 (Pa) Maximum directional z deformation Part 1 (m) Maximum directional z deformation Part 2 (m) Maximum normal z stress Part 1 (Pa) Maximum normal z stress Part 2 (Pa) Target Error (%) 6e-4 6e-4 2.4e8-2.4e8 6e-4 1e-4 2.4e8-4e7 6e-4 1.1e-3 2.4e8-4.4e8 6e e8 0 6e e-4 2.4e e8 6e e e e7 6e e-3 2.4e e8 6e e

141 VMMECH023: Curved Beam Assembly with Multiple Loads Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any basic Strength of Materials book ANSYS Linear Static Structural Analysis Beam Test Case An assembly of two curved beams, each having an included angle of 45, has a square cross-section. It is fixed at one end and at the free end a Force F and a Moment M are applied. Also, a UDL of "w " N / mm is applied on both the beams. Use a global element size of 30 mm to get accurate results. See the figure below for details. Find the deformation of the free end in the y direction. Figure 92: Schematic Equivalent Loading: Material Properties Beam 1: E 1 = 1.1e5 MPa ν 1 = 0 Geometric Properties For each beam: Cross-section = 10 mm x 10 mm Loading Force F = N (y direction) Moment M = Nmm (about z-axis) 135

142 VMMECH023 Material Properties ρ 1 = 8.3e-6 Geometric Properties Radius r = 105 mm Included angle = 45 Loading UDL w = -5 N/mm (y direction) on both beams This UDL is applied as an edge force kg/mm 3 Beam 2: E 2 = 2e5 MPa ν 2 = 0 on each beam ρ 2 = 7.85e-6 with magnitude = kg/mm 3-5 (2 x 3.14 x 105) / 8 = N Analysis The deflection in the y direction is in the direction of the applied force F and is given by: where: δ = deflection at free end in the y direction I = moment of inertia of the cross-section of both beams Results Comparison Results Target Error (%) Minimum Y Deformation (mm)

143 VMMECH024: Harmonic Response of a Single Degree of Freedom System for Beams Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any basic Vibration Analysis book ANSYS Harmonic Analysis Beam Test Case Two collinear beams form a spring-mass system. The density of the longer beam is kept very low so that it acts as a massless spring and the smaller beam acts as a mass. The end vertex of the longer beam (acting as a spring) is fixed. A Harmonic force F is applied on the free vertex of the shorter beam in z direction. Both beams have hollow circular cross-sections, as indicated below. Scenario 1: Damping ratio = 0 Scenario 2: Damping ratio = 0.05 Find the z directional deformation of the vertex where force is applied at frequency F = 500 Hz for the above scenarios with solution intervals = 25 and a frequency range of 0 to 2000 Hz. Use both Mode Superposition and Full Method. Figure 93: Schematic Material Properties Material E (Pa) Spring Mass ν 1.1e e11 0 ρ (kg/m 3 ) 1e e5 Geometric Properties Cross-section of each beam: Outer radius = 10 mm Loading Harmonic force F = 1 e6 N (z-direction) 137

144 VMMECH024 Geometric Properties Inner radius = 5 mm Length of longer beam = 100 mm Length of shorter beam = 5 mm Loading Results Comparison Results Mode-SuperpositionMaximum z directional deformation without damping (m) Full Method Maximum z directional deformation with damping (m) Maximum z directional deformation without damping (m) Maximum z directional deformation with damping (m) Target x x x x 10-3 Error (%) x x x x

145 VMMECH025: Stresses Due to Shrink Fit Between Two Cylinders Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Stephen P. Timoshenko, Strength of Materials, Part 2 - Advanced Theory and Problems, 3rd Edition, pg ANSYS AIM ANSYS Linear Static Structural Analysis Solid Test Case One hollow cylinder is shrink fitted inside another. Both cylinders have length L and both the flat faces of each cylinder are constrained in the axial direction. They are free to move in radial and tangential directions. An internal pressure of P is applied on the inner surface of the inner cylinder. To get accurate results, apply a global element size of 0.8 inches. Find the maximum tangential stresses in both cylinders. Material Properties Geometric Properties Both cylinders Inner Cylinder: are made of r i = 4 the same r o = material R i = 6 E = 3 x 10 7 psi R o = 8 ν = 0 ρ = lbm/in 3 Length of both cylinders = 5 Loading P = psi Analysis Assumptions and Modeling Notes for ANSYS The following figure shows the finite element model in ANSYS : 139

146 VMMECH025 Figure 94: Schematic Note Tangential stresses can be obtained in the application using a cylindrical coordinate system. To simulate interference, set Contact Type to Rough with interface treatment set to add offset with Offset = 0. Analysis Assumptions and Modeling Notes for ANSYS AIM The following figure shows the finite element model in ANSYS AIM: 140

147 VMMECH025 Figure 95: ANSYS AIM Schematic Results Comparison for ANSYS Results Target Error (%) Maximum normal y stress, inner cylinder (psi) Maximum normal y stress, outer cylinder (psi) Note Here y corresponds to θ direction of a cylindrical coordinate system. Results Comparison for ANSYS AIM Results Maximum Normal Y Stress, Inner Cylinder (psi) Maximum Normal Y Stress, Outer Cylinder (psi) Target AIM Error (%)

148 142

149 VMMECH026: Fatigue Analysis of a Rectangular Plate Subjected to Edge Moment Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any standard Machine Design and Strength of Materials book ANSYS AIM ANSYS Fatigue Analysis Shell Test Case A plate of length L, width W, and thickness T is fixed along the width on one edge and a moment M is applied on the opposite edge about the z-axis. Find the maximum Bending Stress (Normal X Stress) and maximum Total Deformation of the plate. Also find the part life and the factor of safety using Goodman, Soderberg, & Gerber criteria. Use the x-stress component. Consider load type as fully reversed and a Design Life of 1e6 cycles, Fatigue Strength factor of 1, and Scale factor of 1. Figure 96: Schematic for ANSYS 143

150 VMMECH026 Figure 97: Schematic for ANSYS AIM Material Properties E = 2 x Pa ν = 0.0 Ultimate tensile strength = 1.29 x 10 9 Pa Endurance strength = No. of Cycles x 10 8 Pa Yield Strenth = 2.5 x 10 8 Pa 1 x 10 6 Alternating Stresses (Pa) 1.08 x x 10 8 Geometric Properties Length L = 12 x 10-3 m Width W = 1 x 10-3 m Thickness T = 1 x 10-3 m Loading Moment M = 0.15 Nm z-axis) Results Comparison for ANSYS Results Maximum normal x-stress (Pa) Maximum total deformation (m) SN-Goodman Safety factor Life Target 9 x x Error (%) 9 x x

151 VMMECH026 Results Target Error (%) SN-Soderberg Safety factor Life SN-Gerber Safety Factor Life Results Comparison for ANSYS AIM Results Maximum normal x-stress (Pa) Target 9 x10 8 AIM 9 x 10 8 Error (%) Maximum total deformation (m) 6.48 x x

152 146

153 VMMECH027:Thermal Analysis for Shells with Heat Flow and Given Temperature Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any standard Thermal Analysis book ANSYS Thermal Stress Analysis Shell Test Case A plate of length (L), width (W), and thickness (T) is fixed along the width on one edge and heat flow (Q) is applied on the same edge. The opposite edge is subjected to a temperature of 20 C. Ambient temperature is 20 C. To get accurate results, apply a sizing control with element size = 2.5e-2 m. Find the maximum temperature, maximum total heat flux, maximum total deformation, and heat reaction at the given temperature. Figure 98: Schematic Material Properties E = 2e11 Pa ν = 0.0 Coefficient of thermal expansion α = 1.2e-5/ C Thermal conductivity k = 60.5 W/m C Analysis Geometric Properties Length L = 0.2 m Width W = 0.05 m Thickness T = m Loading Heat flow Q = 5 W Given Temperature = 20 C Heat Reaction = -(Total heat generated) Heat flow due to conduction is given by: 147

154 VMMECH027 where: T h = maximum temperature T 1 = given temperature Total heat flux is: Temperature at a variable distance z from the fixed support is given by: Thermal deformation in the z-direction is given by: Results Comparison Results Maximum Temperature ( C) Maximum Total Heat Flux (W/m 2 ) Maximum Total Deformation (m) Heat Reaction (W) Target e4 Error (%) e e e

155 VMMECH028: Bolt Pretension Load Applied on a Semi-Cylindrical Face Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any standard Strength of Materials book ANSYS Static Structural Analysis Solid Test Case A semi-cylinder is fixed at both the end faces. The longitudinal faces have frictionless support. A bolt pretension load is applied on the semi-cylindrical face. To get accurate results, apply sizing control with element size of 0.01 m. Find the Z directional deformation and the adjustment reaction due to the bolt pretension load. Figure 99: Schematic Material Properties E = 2e11 Pa ν = 0.0 Analysis Geometric Properties Length L = 1 m Diameter D = 0.05 m Loading Pretension as preload = N (equal to adjustment of 1e-7 m) The bolt pretension load applied as a preload is distributed equally to both halves of the bar. Therefore the z-directional deformation due to pretension is given by: 149

156 VMMECH028 Results Comparison Results Minimum z-directional deformation (m) Maximum z-directional deformation (m) Adjustment Reaction (m) Target -5.00E E E-07 Error (%) E E E

157 VMMECH029: Elasto-Plastic Analysis of a Rectangular Beam Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Timoshenko S., Strength of Materials, Part II, Advanced Theory and Problems, Third Edition, Article 64, pp. 349 ANSYS Static Plastic Analysis Solid Test Case A rectangular beam is loaded in pure bending. For an elastic-perfectly-plastic stress-strain behavior, show that the beam remains elastic at M = Myp = σ yp bh 2 / 6 and becomes completely plastic at M = Mult = 1.5 Myp. To get accurate results, set the advanced mesh control element size to 0.5 inches. Figure 100: Stress-Strain Curve Figure 101: Schematic Material Properties Geometric Properties Loading E = 3e7 psi Length L = 10 M = 1.0 M yp to 1.5 ν = 0.0 Width b = 1 M yp σ yp = psi Height h = 2 (M yp = lbf - in) 151

158 VMMECH029 Analysis The load is applied in three increments: M 1 = lbf-in, M 2 = lbf-in, and M 3 = lbf-in. Results Comparison M/M yp Target Error (%) State Equivalent Stress (psi) State Equivalent Stress (psi) 1 fully elastic fully elastic elastic-plastic elastic-plastic plastic solution not converged plastic solution not converged - 152

159 VMMECH030: Bending of Long Plate Subjected to Moment - Plane Strain Model Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any standard Strength of Materials book ANSYS Plane Strain Analysis 2D Structural Solid Test Case A long, rectangular plate is fixed along the longitudinal face and the opposite face is subjected to a moment of 5000 lbf-in about the z-axis. To get accurate results, set the advanced mesh control element size to 0.5 inches. Find X normal stress at a distance of 0.5 inches from the fixed support. Also find total deformation and reaction moment. Figure 102: Schematic Material Properties E = 2.9e7 psi ν = 0.0 Analysis Geometric Properties Length L = 1000 Width W = 40 Thickness T = 1 Loading Moment M = 5000 lbf-in Since the loading is uniform and in one plane (the x-y plane), the above problem can be analyzed as a plane strain problem. Therefore, the moment applied will be per unit length (5000/1000 = 5 lbf-in). Analysis takes into account the unit length in the z-direction. 153

160 VMMECH030 Figure 103: Plane Strain Model (analyzing any cross section (40 x 1 ) along the length) Results Comparison Results Target Error (%) Normal Stress Maximum Normal Stress in the X-Direction (psi) Maximum Total Deformation (in) e e Reaction Moment (lbf-in)

161 VMMECH031: Long Bar With Uniform Force and Stress Tool - Plane Stress Model Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any standard Strength of Materials book ANSYS Plane Stress Analysis 2D Structural Solid Test Case A long, rectangular bar assembly is fixed at one of the faces and the opposite face is subjected to a compressive force. To get accurate results, set the advanced mesh control element size to 1 m. Find the maximum equivalent stress for the whole assembly and safety factor, safety margin, and safety ratio for the first and last part using the maximum equivalent stress theory with Tensile Yield Limit. Figure 104: Schematic Material Properties Material E (Pa) ν Tensile Yield (Pa) Part e e8 Part 2 7.1e e8 Part 3 2e e8 Part 4 1.1e e8 Geometric Properties Part 1: 2 m x 2 m x 3 m Part 2: 2 m x 2 m x 10 m Part 3: 2 m x 2 m x 5 m Part 4: 2 m x 2 m x 2 m Loading Force = 1e9 N in the negative x-direction 155

162 VMMECH031 Analysis Since the loading is uniform and in one plane, the above problem can be analyzed as a plane stress problem. Analysis is done considering thickness of 2 m along z-direction Figure 105: Plane Stress Model (Analyzing any cross section along Z) Results Comparison Results Target Error (%) Maximum Equivalent Stress (Pa) 2.5e8 2.5e Part 1 Safety Factor Safety Margin Safety Ratio Part 4 Safety Factor Safety Margin Safety Ratio

163 VMMECH032: Radial Flow due to Internal Heat Generation in a Copper Disk - Axisymmetric Model Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any basic Heat Transfer book ANSYS AIM ANSYS Axisymmetric Analysis 2D Structural Solid Test Case A copper disk with thickness t and radii R i and R o is insulated on the flat faces. It has a heat-generating copper coaxial cable (of radius R i ) passing through its center. The cable delivers a total heat flow of Q to the disk. The surrounding air is at a temperature of T o with convective film coefficient h. To get accurate results, set the advanced mesh control element size to m. Find the disk temperature and heat flux at inner and outer radii. Figure 106: Schematic Material Properties E = 1.1 x Pa ν = 0.34 Thermal conductivity k = W/m- C Geometric Properties R i = 10 mm R o = 60 mm t = 8 mm Loading Q = 100 W (Internal Heat Generation = W/m 3 ) Film coefficient h = 1105 W/m 2 - C Surrounding temperature T o = 0 C 157

164 VMMECH032 Analysis Assumptions and Modeling Notes for ANSYS Because the geometry and loading are symmetric about the y-axis, the above problem can be analyzed as an axisymmetric problem in ANSYS. Figure 107: Axisymmetric Model Analysis Assumptions and Modeling Notes for ANSYS AIM The following figure shows the finite element model in ANSYS AIM: Figure 108: Model in ANSYS AIM Results Comparison for ANSYS Results Target Error (%) Maximum Temperature ( C) Minimum Temperature ( C)

165 VMMECH032 Results Target Error (%) Maximum Heat Flux (W/m 2 ) Minimum Heat Flux (W/m 2 ) Results Comparison for ANSYS AIM Results Maximum Disk Temperature (C) Minimum Disk Temperature (C) Maximum Heat Flux (W/m2) Minimum Heat Flux (W/m2) Target AIM Error (%)

166 160

167 VMMECH033: Spring Mass System Subjected to Enforced Motion with Displacement Base Excitations Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Thompson, W.T., Theory of Vibration with Applications, 3rd Edition, Chapter 3, pp , 1999 ANSYS Harmonic Analysis Spring-Damper Surface Test Case A vehicle has a mass of 500 kg (applied as distributed mass) and the total spring constant of its suspension system is N/m. The profile of the road is approximated as a sine wave of amplitude 10 mm and a wavelength of 1.5 m. Determine the amplitude of oscillations of the mass: When driven at critical speed and having damping factor of 0.5 When driven at 50 km/h and having damping factor of 0.4 Figure 109: Schematic Material Properties E = 1 x Pa = 1 x kg/m 3 Geometric Properties Mass of vehicle, m = 500 kg Loading Sinusoidal base excitation of amplitude Y = 10 mm and wavelength = 1.5 m 161

168 VMMECH033 Material Properties Analysis Geometric Properties Stiffness of spring, K = N/m Loading Natural circular frequency of the system, Critical damping coefficient, Damping ratio, Circular frequency of forced vibration, Absolute amplitude of vibration, Relative amplitude of vibration, Absolute phase angle, where Relative phase angle, m = mass of vehicle V = speed of vehicle Y = amplitude of sine wave = wavelength C = damping coefficient Results Comparison Results Target Error (%) Critical speed ( = Hz), damping factor ( ) of 0.5 = mm = -45 or 135 = mm = mm = -45 = mm 0% 0% 0% = -90 or 90 = -90 0% 50 km/h ( = 9.26 Hz), = mm = mm 0% damping factor ( ) of 0.4 = or = mm = = mm 0% 0% or = % 162

169 VMMECH034: Rubber Cylinder Pressed Between Two Plates Overview Reference: Solver(s): Analysis Type(s): Element Type(s): T. Tussman, K.J. Bathe, "A Finite Element Formulation for Nonlinear Incompressible Elastic and Inelastic Analysis", Computers and Structures, Vol. 26 Nos 1/2, 1987, pp ANSYS Nonlinear Static Structural Analysis (Large Deformation ON) Solid Test Case A rubber cylinder is pressed between two rigid plates using a maximum imposed displacement of δ max. Determine the total deformation. Figure 110: Schematic Material Properties Solid1: E = 2 x Pa ν = 0.3 ρ = 7850 kg/m 3 Solid2: Mooney-Rivlin Constants C10 = 2.93 x 10 5 Pa C01 = 1.77 x 10 5 Pa Incompressibility Parameter D1 1/Pa = 0 Geometric Properties Solid1: 0.05 m x 0.01 m x 0.4 m Solid2: Quarter Circular Cylinder Radius = 0.2 m Length = 0.05 m Loading Displacement in Y direction = -0.1 m 163

170 VMMECH034 Analysis Due to geometric and loading symmetry, the analysis can be performed using one quarter of the cross section. Frictionless supports are applied on 3 faces (X = 0, Z = 0 and Z = 0.05 m). Given displacement of 0.1 m is applied on the top surface. The bottom surface of Solid1 is completely fixed. Frictionless Contact with Contact stiffness factor of 100 is used to simulate the rigid target. Augmented Lagrange is used for Contact formulation. Results Comparison Results Target Error (%) Total Deformation (m)

171 VMMECH035: Thermal Stress in a Bar With Radiation Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any Basic Heat transfer and Strength of Materials book ANSYS AIM ANSYS Coupled Analysis (Static Thermal and Static Stress) Solid Test Case Heat of magnitude 2500 W and Heat Flux of magnitude 625 W/m 2 is flowing through a long bar (2 x 2 x 20) m in an axial direction, and radiating out from the other face having emissivity 0.3; Ambient temperature is maintained at 20 C. Find the following: Temperatures on End Faces. Thermal strain and Directional deformation and Normal Stress in Z direction if both the end faces have frictionless supports and Reference temperature of 22 C. Analysis (Heat flowing through body) Q = (Heat Flow) + (Heat Flux * Area) = 5000 W (Heat flowing through body) = (Heat Conducted through body) = (Heat Radiated out of the Surface) i.e. Q = Q r =Q C = 5000 W. Heat Radiated out of the body gives T 2 = C. Heat Conducted through the body gives T 1 = C. Thermal strain is given by: The compressive stress introduced is given by: Temperature at a distance z from the face with higher temperature is given by: Only half-length is considered for calculating deformation, since deformation is symmetric: 165

172 VMMECH035 Material Properties E = 2.0 x Pa v = 0 α = 1.2 x / C k = 60.5 W/m C Geometric Properties Part 1: 2 m x 2 m x 2 m Part 2: 2 m x 2 m x 5 m Part 3: 2 m x 2 m x 10 m Part 4: 2 m x 2 m x 3 m Loading Heat Flow = 2500W on Part 4 Heat Flux = 625 W/m 2 on Part 4 Radiation = 20 C, 0.3 Analysis Assumptions and Modeling Notes for ANSYS The following figure shows the finite element model in ANSYS : Figure 111: Schematic in ANSYS Results Comparison for ANSYS Results Target Error (%) Temperature on Part 4 ( C) Temperature on Part 1 ( C) Maximum Thermal Strain (m/m) x x Minimum Thermal Strain (m/m) x Normal Stress in Z direction (Pa) x x Directional Deformation in Z direction (m)

173 VMMECH035 Analysis Assumptions and Modeling Notes for ANSYS AIM The following figure shows the finite element model in ANSYS AIM: Figure 112: Schematic in ANSYS AIM Results Comparison for ANSYS AIM Results Target AIM Error (%) Temperature on Part 4 ( C) x 10-4 Temperature on Part 1 ( C) Maximum Thermal Strain (m/m) x x Minimum Thermal Strain (m/m) x Directional Deformation in Z direction (m)

174 168

175 VMMECH036: Thermal Stress Analysis of a Rotating Bar Using Temperature Dependant Density Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any Basic Strength of Materials book ANSYS Static Stress Analysis (Sequence Loading) Solid Test Case A Bar (2 m x 2m x 20m) with one end fixed and with a rotational velocity about X axis at location (1, 1, 0) is subjected to a Uniform Temperature (Thermal Condition Load) in three steps. For all the steps, Reference Temperature is 0 C. Frictionless Support is applied on all the longitudinal faces. Figure 113: Schematic Material Properties E = 1 x 10 6 Pa Geometric Properties Part 1: 2 m x 2 m x 20 m Loading Rotational Velocity (rad/s) in steps: α = 1 x / C ν = 0 1. (1, 0, 0) Temperature C Density kg/m 3 2. (0.5, 0, 0) (0.25, 0, 0) Thermal Condition C C C C 169

176 VMMECH036 Analysis Results Comparison Results Target Error (%) Equivalent Stress (Pa) Step Step Step Total Deformation (m) Step Step Step

177 VMMECH037: Cooling of a Spherical Body Overview Reference: Solver(s): Analysis Type(s): Element Type(s): F. Kreith, "Principles of Heat Transfer", 2nd Printing, International Textbook Co., Scranton, PA, 1959, pg. 143, ex ANSYS AIM ANSYS Transient Thermal Analysis Plane Test Case Determine the temperature at the center of a spherical body, initially at a temperature T 0, when exposed to an environment having a temperature T e for a period of 6 hours (21600 s). The surface convection coefficient is h. Initial temperature, T 0 = 65 F Surface temperature, T e = 25 F Convection coefficient h = e-4 BTU/s-ft 2 - F Time, t = seconds Radius of the sphere r o = 2 in = 1/6 ft Material Properties K = (1/3) BTU/hr-ft- F ρ = 62 lb/ft 3 c = Btu/lb- F Geometric Properties Quarter Circular lamina Radius = ft Loading Convection applied on Edge = e-4 BTU/s-ft 2 - F Ambient Temperature for Convection = 25 F Analysis Assumptions and Modeling Notes for ANSYS Since the problem is axisymmetric, only a 2-D quarter model is used in ANSYS. 171

178 VMMECH037 Figure 114: Schematic for ANSYS Analysis Assumptions and Modeling Notes for ANSYS AIM The following figure shows the finite element model in ANSYS AIM: Figure 115: Schematic for ANSYS AIM Results Comparison for ANSYS Results Target Error (%) Temperature at the center of body after 21600s ( F) Results Comparison for ANSYS AIM Results Temperature at the center of body after 21600s ( F) Target AIM Error (%)

179 VMMECH038: Crashing Blocks Simulation with Transient Structural Analysis Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any basic kinematics book ANSYS Flexible Dynamic Analysis Solid Test Case Three blocks are resting on a base. A left block of mass x 10-4 kg is given a constant initial velocity of 100 mm/sec to collide with a middle block of mass x 10-4 kg. Frictionless supports are applied as shown in the figure and also on the bottom faces of the left and middle blocks. The right block is fixed using a fixed support and the base is fixed by applying a fixed joint. Find the velocity of both the moving blocks after impact. Figure 116: Schematic Material Properties E = 2 x 10 5 MPa ν = 0.3 ρ = 7.85 x 10-6 kg/mm 3 Geometric Properties Left block = 3 mm x 2 mm x 5 mm Middle block = 2.5 mm x 2 mm x 3 mm Loading Left block initial velocity = 100 mm/s (X direction) 173

180 VMMECH038 Material Properties Analysis Geometric Properties Right block =3 mm x 6 mm x 4 mm Base = 3 mm x mm x mm Loading For a perfectly elastic collision between the blocks, the following equations are true: (1) where: and = mass of the left and middle blocks in kg and = initial and final velocity of the left block in mm/sec = initial velocity of the middle block in mm/sec = final velocity of the middle block after impact in mm/sec Results Comparison Results Target Error (%) Velocity of the left block after impact (mm/sec) Velocity of the middle block after impact (mm/sec)

181 VMMECH039: Transient Response of a Spring-Mass System Overview Reference: Solver(s): Analysis Type(s): Element Type(s): R. K. Vierck, Vibration Analysis, 2nd Edition, Harper & Row Publishers, New York, NY, 1979, sec ANSYS Flexible Dynamic Analysis Solid and Spring Test Case A system containing two masses, m 1 and m 2, and two springs of stiffness k 1 and k 2 is subjected to a pulse load F(t) on mass 1. Determine the displacement response of the system for the load history shown. Figure 117: Schematic Material Properties E = 2 x Pa γ = 0.3 ρ = 0.25 kg/m 3 k 1 = 6 N/m k 2 = 16 N/m m 1 = 2 kg Geometric Properties 2 Blocks = 2 m x 2 m x 2 m Length of L1 spring = 6 m Length of L2 spring = 7 m Loading F 0 = 50 N t d = 1.8 sec 175

182 VMMECH039 Material Properties m 2 = 2kg Geometric Properties Loading Analysis Assumptions and Modeling Notes The step drop is modeled as F 0 = 50 t = 1.8 sec and F 0 = 0 t = sec. Results Comparison Results Target Error (%) Y 1, m (@ t = 1.3s) Y 2, m (@ t = 1.3s) Y 1, m (@ t = 2.4s) Y 2, m (@ t = 2.4s)

183 VMMECH040: Deflection of Beam using Symmetry and Anti-Symmetry Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any Basic Strength of Materials Book ANSYS Static Structural Analysis Beam Test Case A long bar 1m X 1m X 24m with simply supported ends is subjected to lateral load of 1000 N at a distance of 8m from one end. Find Deformation at the 8m from simply Supported end. Scenario 1: Considering Symmetry Scenario 2: Considering Anti-Symmetry Figure 118: Schematic Material Properties E = 2e11 Pa γ = 0 ρ = kg/m 3 Analysis Geometric Properties Bar = 1m x 1m x 24m Loading Force = N (Y-direction) at 8m from Simply Supported end Scenario 1: Considering Symmetry Scenario 2: Considering Anti-Symmetry 177

184 VMMECH040 Results Comparison Results Target Error (%) Scenario 1: Directional Deformation in Y-direction (m) e e Scenario 2: Directional Deformation in Y-direction (m) e e

185 VMMECH041: T-Stress Calculation for a Crack in a Plate Using Pre-Meshed Crack Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Fett, T., Stress Intensity Factors, T-Stresses, Weight Functions, Institute of Ceramics in ANSYS Static Analysis Solid Test Case. Symmetry boundary con- A rectangular plate with a center crack is subjected to an end tensile load ditions are considered and T-Stress is determined using Pre-Meshed Crack. Figure 119: Two-Dimensional Fracture Problem Sketch Material Properties Young's Modulus (E) = MPa Coefficient of friction ( ) = 0.3 Geometric Properties a = 10 mm a / W = 0.2 H / W = 0.75 Loading Tensile stress ( ) = 100 MPa 179

186 VMMECH041 Analysis Assumptions and Modeling Notes The problem is solved using 2-D solid elements with plane-strain behavior. A one-quarter plate is modeled and symmetric boundary conditions are considered. The crack tip nodes and the number of paths surrounding the crack tip nodes are defined using Pre-Meshed Crack. The plate is subjected to a tensile stress and the T-Stress is computed for the crack tip nodes. The crack front and the path surrounding the crack front are defined using Pre-Meshed Crack. Results Comparison Result Target Error (%) T-Stress

187 VMMECH042: Hydrostatic Pressure Applied on a Square Bar Fully and Partially Submerged in a Fluid Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any Basic Strength of Materials Book ANSYS Static Structural Analysis Solid Test Case Long bar 20m x 2m x 2m is immersed in a fluid and is fixed at one end. Fluid density is 1000 kg/m 3 and Hydrostatic acceleration is 10 m/s 2 in negative Z direction. Hydrostatic pressure is applied on a longitudinal face normal to X-axis at different locations as given in the scenarios below. Find normal stress in Z direction of square bar. Scenario 1: Square bar is partially immersed in the fluid up to 15 m in Z direction from the fixed support. Scenario 2: Square bar is fully immersed in the fluid up to 25 m in Z direction from the fixed support Figure 120: Schematic Material Properties E = 2e11 Pa γ = 0 ρ = 7850 kg/m 3 Geometric Properties Long bar = 20m x 2m x 2m Loading Hydrostatic Pressure Acceleration = -10 m/s 2 (Z direction) Surface Location: 181

188 VMMECH042 Material Properties Analysis Geometric Properties Loading Scenario 1: (2,1,5) m Scenario 2: (2,1,-5) m Scenario 1: Partially Submerged (Pressure distribution in triangular form) Pressure distribution on square bar in triangular form, one end is maximum and other end is zero Pressure on square bar = P = ρ x g x h Load per meter is w = P x L Maximum bending moment = Normal stress = Bending stress = Maximum bending moment / Sectional Modulus Scenario 2: Fully Submerged (Pressure distribution in trapezoidal form) Maximum bending moment = where: W 1 = Maximum Load per meter (@ 25m) W 2 = Minimum Load per meter (@ 5m) Normal stress = Bending stress = Maximum bending moment / Sectional Modulus Results Comparison Results Target Error (%) Normal Stress (Partially Submerged) (Pa) e Normal Stress (Fully Submerged) (Pa) 3.50e e

189 VMMECH043: Fundamental Frequency of a Simply-Supported Beam Overview Reference: Solver(s): Analysis Type(s): Element Type(s): W. T. Thompson, Vibration Theory and Applications, 2nd Printing, Prentice-Hall, Inc., Englewood Cliffs, NJ, 1965, pg. 18, ex ANSYS Modal Analysis Beam Test Case Determine the fundamental frequency f of a simply-supported beam of length cross-section A = 4 in 2 as shown below. = 80 in and uniform Figure 121: Schematic Material Properties E = 3 x 10 7 psi ρ= lb/in 3 Results Comparison Geometric Properties = 80 in A = 4 in 2 h = 2 in I = in 4 Loading Results Target Error (%) Frequency (Hz)

190 184

191 VMMECH044: Thermally Loaded Support Structure Overview Reference: Solver(s): Analysis Type(s): Element Type(s): S. Timoshenko, Strength of Material, Part I, Elementary Theory and Problems, 3rd Edition, D. Van Nostrand Co., Inc., New York, NY, 1955, pg. 30, problem 9. ANSYS Linear Thermal Stress Analysis Beam Test Case An assembly of three vertical wires has a rigid horizontal beam on which a vertically downward force Q is acting. Length of the wires is 20 in, the spacing between the wires is 10 in and the reference temperature is 70 F. The entire assembly is subjected to a temperature rise of ΔT. Find the stresses in the copper and steel wire of the structure shown below. The wires have a cross-sectional area of A. Figure 122: Schematic Material Properties Geometric Properties Loading VMSIM044_material_rigid: A = 0.1 in 2 Q = 4000 lb (Y direction) E r = 3e16 psi ΔT = 10 F ν r = 0 VMSIM044_material_copper: E c = 1.6e7 psi ν c = 0 α c = 9.2e-6 / F VMSIM044_material_steel: 185

192 VMMECH044 Material Properties E s = 3e7 psi ν s = 0 α s = 7e-6 / F Results Comparison Geometric Properties Loading Results Target Error (%) Stress in steel (psi) Stress in copper (psi)

193 VMMECH045: Laterally Loaded Tapered Support Structure Overview Reference: Solver(s): Analysis Type(s): Element Type(s): S. H. Crandall, N. C. Dahl, An Introduction to the Mechanics of Solids, McGraw-Hill Book Co., Inc., New York, NY, 1959, pg. 342, problem ANSYS Static Structural Analysis Shell Test Case A cantilever beam of thickness t and length has a depth which tapers uniformly from d at the tip to 3d at the wall. It is loaded by a force F at the tip, as shown. Find the maximum bending stress at the mid-length (X = ). Figure 123: Schematic Material Properties E s = 3 x 10 7 psi ν s = 0 Results Comparison Geometric Properties = 50 in d = 3 in t = 2 in Loading F = 4000 lb (Y direction) Results Target Error (%) Bending stress at mid length (psi)

194 188

195 VMMECH046: Pinched Cylinder Overview Reference: Solver(s): Analysis Type(s): Element Type(s): R. D. Cook, Concepts and Applications of Finite Element Analysis, 2nd Edition, John Wiley and Sons, Inc., New York, NY, 1981, pp H. Takemoto, R. D. Cook, "Some Modifications of an Isoparametric Shell Element", International Journal for Numerical Methods in Engineering, Vol. 7 No. 3, ANSYS Static Structural Analysis Shell Test Case A thin-walled cylinder is pinched by a force F at the middle of the cylinder length. Determine the radial displacement δ at the point where F is applied. The ends of the cylinder are free edges. A one-eighth symmetry model is used. One-fourth of the load is applied due to symmetry. Figure 124: Schematic Material Properties E s = 10.5e6 psi ν s = Geometric Properties = in r = in t = in Loading F = 100 lbf (Y direction) 189

196 VMMECH046 Analysis Due to symmetrical boundary and loading conditions, one-eighth model is used and one-fourth of the load is applied. Results Comparison Results Target Error (%) Deflection (in)

197 VMMECH047: Plastic Compression of a Pipe Assembly Overview Reference: Solver(s): Analysis Type(s): Element Type(s): S. H. Crandall, N. C. Dahl, An Introduction to the Mechanics of Solids, McGraw-Hill Book Co., Inc., New York, NY, 1959, pg. 180, ex ANSYS Plastic Structural Analysis Axisymmetric Test Case Two coaxial tubes, the inner one of 1020 CR steel and cross-sectional area A s, and the outer one of 2024-T4 aluminum alloy and of area A a, are compressed between heavy, flat end plates, as shown below. Determine the load-deflection curve of the assembly as it is compressed into the plastic region by an axial displacement. Assume that the end plates are so stiff that both tubes are shortened by exactly the same amount. Figure 125: Schematic Material Properties Geometric Properties Loading VMSIM047_CR_steel: = 10 in Steel: 1st Load step: δ = in E s = 26,875,000 psi 2nd Load step: δ = σ (yp)s = 86,000 psi Inside radius = in in 191

198 VMMECH047 Material Properties VMSIM047_T4_aluminum alloy: E a = 11,000,000 psi σ (yp)a = 55,000 psi ν = 0.3 Analysis Geometric Properties Wall thickness = 0.5 in Aluminum: Inside radius = in Wall thickness = 0.5 in Loading 3rd Load step: δ = in Because the geometry and loading are symmetric about the y-axis, the above problem can be analyzed as an axisymmetric problem. Results Comparison Results Load, lb for in Load, lb for 0.05 in Load, lb for 0.1 in Target x x x Error (%)

199 VMMECH048: Bending of a Tee-Shaped Beam Overview Reference: Solver(s): Analysis Type(s): Element Type(s): S. H. Crandall, N. C. Dahl, An Introduction to the Mechanics of Solids, McGraw-Hill Book Co., Inc., New York, NY, 1959, pg. 294, ex ANSYS Static Structural Analysis Beam Test Case Find the maximum tensile and compressive bending stresses in an unsymmetrical T beam subjected to uniform bending M z, with dimensions and geometric properties as shown below. Figure 126: Schematic Material Properties E = 3e7 psi Results Comparison Geometric Properties b = 1.5 in h = 8 in y = 6 in Area = 60 in 2 I z = 2000 in 4 Loading M z = 100,000 lbf-in (Z direction) Results Target Error (%) Stress BEND, Bottom (psi)

200 VMMECH048 Results Target Error (%) Stress BEND, Top (psi)

201 VMMECH049: Combined Bending and Torsion of Beam Overview Reference: Solver(s): Analysis Type(s): Element Type(s): S. Timoshenko, Strength of Material, Part I, Elementary Theory and Problems, 3rd Edition, D. Van Nostrand Co., Inc., New York, NY, 1955, pg. 299, problem 2. ANSYS Static Structural Analysis Beam Test Case A vertical bar of length and radius r is subjected to the action of a horizontal force F acting at a distance d from the axis of the bar. Determine the maximum principal stress σ max. Figure 127: Problem Sketch Figure 128: Schematic Material Properties E = 3e7 psi ν = 0.3 Geometric Properties = 25 ft r = in d = 3 ft Loading F = 250 lb (Y direction) M = 9000 lbf-in (Z direction) 195

202 VMMECH049 Results Comparison Results Target Error (%) Principal stress max (psi)

203 VMMECH050: Cylindrical Shell Under Pressure Overview Reference: Solver(s): Analysis Type(s): Element Type(s): S. Timoshenko, Strength of Material, Part I, Elementary Theory and Problems, 3rd Edition, D. Van Nostrand Co., Inc., New York, NY, 1955, pg. 45, article 11. A. C. Ugural, S. K. Fenster, Advanced Strength and Applied Elasticity, Elsevier, ANSYS AIM ANSYS Static Structural Analysis Axisymmetric Shell element Test Case A long cylindrical pressure vessel of mean diameter d and wall thickness t has closed ends and is subjected to an internal pressure P. Determine the axial stress σ y and the hoop stress σ z in the vessel at the mid-thickness of the wall. An axial force of lb ((Pπd2)/4) is applied to simulate the closed-end effect. Figure 129: Schematic Material Properties E = 3e7 psi ν = 0.3 Geometric Properties t = 1 in d = 120 in Loading P = 500 psi (radial direction) 197

204 VMMECH050 Analysis Assumptions and Modeling Notes for ANSYS The following figure shows the finite element model for ANSYS : Figure 130: Schematic for ANSYS Analysis Assumptions and Modeling Notes for ANSYS AIM The following figure shows the finite element model for ANSYS AIM: 198

205 VMMECH050 Figure 131: Schematic for ANSYS AIM Results Comparison for ANSYS Results Target Error (%) Stress y (psi) Stress z (psi) Results Comparison for ANSYS AIM Results Stress y (psi) Stress z (psi) Target AIM Error (%)

206 200

207 VMMECH051: Bending of a Circular Plate Using Axisymmetric Elements Overview Reference: Solver(s): Analysis Type(s): Element Type(s): S. Timoshenko, Strength of Material, Part II, Elementary Theory and Problems, 3rd Edition, D. Van Nostrand Co., Inc., New York, NY, 1956, pp. 96, 97, and 103. ANSYS AIM ANSYS Static Structural Analysis Axisymmetric Shell element Test Case A flat circular plate of radius r and thickness t is subject to various edge constraints and surface loadings. Determine the deflection δ at the middle and the maximum stress σ max for each case. Case 1: Uniform loading P, clamped edge Case 2: Concentrated center loading F, clamped edge Figure 132: Schematic Material Properties E = 3e7 psi ν = 0.3 Geometric Properties r = 40 in t = 1 in Loading Case 1: P = 6 psi Case 2: F = lb (y direction) Analysis Assumptions and Modeling Notes for ANSYS Because the geometry and loading are symmetric about the y-axis, the above problem can be analyzed as an axisymmetric problem. 201

208 VMMECH051 Figure 133: Schematic for ANSYS Case 1: Case 2: Analysis Assumptions and Modeling Notes for ANSYS AIM The following figure shows the finite element model for ANSYS AIM: Figure 134: Schematic for ANSYS AIM 202

209 VMMECH051 Results Comparison for ANSYS Results Target Error (%) Case 1: Deflection (in) Stress max (psi) Case 2: Deflection (in) Stress max (psi) Results Comparison for ANSYS AIM Results Case 1: Case 2: Target AIM Error (%) Deflection (in) Stressmax (psi) Deflection (in) Stressmax (psi) 203

210 204

211 VMMECH052: Velocity of Pistons for Trunnion Mechanism Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any Basic Kinematics book ANSYS Rigid Dynamic Analysis Multipoint Constraint Element Test Case The Trunnion mechanism has the following data (all distances are center-to-center distances): Crank radius OA = 100 mm and is oriented at 30 deg to Global Y Axis AB = 400 mm AC = 150 mm CE = 350 mm EF = 300 mm Constant Angular Velocity at Crank = rad/s Center of Trunnion is at distance of 200 mm from line of stroke of Piston B horizontally and 300 mm vertical from Center of Crank Find the Velocity of Piston (F) at the 180 deg from Initial Position Find the Velocity of Piston (B) at the 180 deg from Initial Position Figure 135: Schematic 205

212 VMMECH052 Material Properties E = 2e11 Pa ν = 0.3 Analysis Geometric Properties AB = 400 mm AC = 150 mm CE = 350 mm EF = 300 mm Loading Constant angular velocity at crank = rad/s Analysis done using graphical solution. Consider the Space Diagram, Velocity Diagram at the 180 from Initial Position. Figure 136: Schematic Results Comparison Results Target Error (%) Velocity of Piston (F) mm/s Velocity of Piston (B) mm/s

213 VMMECH053: Simple Pendulum With SHM Motion Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any Basic Kinematics book ANSYS Rigid Dynamic Analysis Multipoint Constraint Element Test Case A simple pendulum as shown in Figure 137: Schematic (p. 207) has a SHM motion about its hinged point given by the following equation: θ = 1.571*sin (0.5235*t) rad The hinge point coordinates are: 1. Hinge point = (0, 0, ) mm Find the relative angular acceleration of pendulum after t = 3s. Figure 137: Schematic Material Properties E = MPa ν = 0.3 Analysis Geometric Properties Hinge point = (0, 0, ) mm Loading Rotation θ = 1.571*sin (0.5235*t) rad The pendulum is having SHM motion in X-Z plane about the hinge. Angular acceleration of pendulum: 207

214 VMMECH053 Results Comparison Results Target Error (%) Relative angular acceleration of pendulum after t = 3s (rad/s 2 )

215 VMMECH054: Spinning Single Pendulum Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any Basic Kinematics book ANSYS Rigid Dynamic Analysis Multipoint Constraint Element Test Case A uniform bar A is connected to a vertical shaft by a revolute joint. The vertical shaft is rotating around its vertical axis at a constant velocity Ω. A point mass M is attached at the tip of the bar in the figure below. The length of bar A is L. Its mass is m, its rotational inertia to its principal axis are J x, J y, J z. The angle of the bar A to the vertical axis is denoted as as follows.. The motion equation has been established The problem is solved for during the first second of motion. The WB/ results are compared to a fourth order Runge-Kutta solution. 209

216 VMMECH054 Figure 138: Schematic Material Properties = tan -1 (1,2) = 0 Geometric Properties Loading L= m Ω = m = kg M = kg J x = kg-m 2 J y = kg-m 2 J z = kg-m 2 Results Comparison Results Target Error (%) at 0.5 sec at 0.5 sec at 1.0 sec

217 VMMECH054 Results Target Error (%) at 1.0 sec Figure 139: Plot of from 0 to 1 sec Figure 140: Plot of from 0 to 1 sec 211

218 212

219 VMMECH055: Projector Mechanism - Finding the Acceleration of a Point Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any Basic Kinematics book ANSYS Rigid Dynamic Analysis Multipoint Constraint Element Test Case The mechanism shown in figure is used to pull a movie through a projector. The mechanism is driven by the drive wheel rotating at a constant rad/s. The link lengths of all the links are constant as given below. Link AB length r 1 = 18mm Link BC length r 2 = 48mm Length BX = x = 45 mm and CX = y = 28 mm The horizontal distance between A and C is length=34 mm. Determine the acceleration of point C with a change of angle of link AB (θ 1 ) from 0 to 60 in counter clockwise direction. Figure 141: Schematic 213

220 VMMECH055 Material Properties E = 2e11 Pa ν = 0.3 Analysis Geometric Properties r 1 = 18 mm r 2 = 48 mm x = 45 mm y = 28 mm Loading Constant rotational velocity = rad/s Linear acceleration of point C is given by Results Comparison Results Target Error (%) Relative acceleration (θ 1 = 10) mm/s Relative acceleration (θ 1 = Relative acceleration (θ 1 = ) mm/s 2 60) mm/s 2 214

221 VMMECH056: Coriolis Component of Acceleration (Rotary Engine Problem) Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any Basic Kinematics book ANSYS Rigid Dynamic Analysis Multipoint Constraint Element Test Case Kinematics diagram of one of the cylinders of a rotary engine is shown below. OA is 50mm long and fixed at point o. The length of the connecting rod AB is 125mm. The line of stroke OB is inclined at 50 to the vertical. The cylinders are rotating at a uniform speed of 300 rpm in a clockwise direction, about the fixed center O. Find Angular acceleration of the connecting rod. Figure 142: Schematic Material Properties E = 2e11 Pa ν = 0.3 Geometric Properties Connecting rod AB is 125mm Crank OA is 50mm long OB is inclined at 50 to the vertical. Loading Constant rotational velocity = 300 rpm 215

222 VMMECH056 Analysis Angular acceleration of the connecting rod is given by: Results Comparison Results Target Error (%) Angular acceleration (radian/s 2 )

223 VMMECH057: Calculation of Velocity of Slider and Force by Collar Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Beer-Johnston Vector Mechanics for Engineers Statics & Dynamics (In SI Units), 7th Edition, TATA McGRAW HILL Edition 2004, Problem 13.73, Page No: 793 ANSYS Rigid Dynamic Analysis Multipoint Constraint Element Test Case A 1.2 Kg collar is attached to a spring and slides without friction along a circular rod in a vertical plane. The spring has an undeformed length of 105 mm and a constant K = 300 N/m. Knowing that the collar is at rest at "C" and is given a slight push to get it moving. Length OP = 75 mm. Length OB = 180 mm. Determine the force exerted by the rod on the collar as it passes through point "A" and "B". Figure 143: Schematic 217

224 VMMECH057 Material Properties E = 2e11 Pa ν = 0.3 Results Comparison Geometric Properties Spring: Undeformed length = 105 mm Stiffness K = 300 N/m Loading Gravitational acceleration = m/s 2 (Y Direction) Results Target Error (%) At point A (N) At point B (N)

225 VMMECH058: Reverse Four Bar Linkage Mechanism Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Results are simulated using MATLAB ANSYS Rigid Dynamic Analysis Multipoint Constraint Element Test Case The figure (below) shows a reverse four bar linkage consisting of uniform rigid links PQ, QR, and RS and ground PS. Link PQ is connected with revolute joints to links QR and PS at points Q and P, respectively. Link RS is connected with revolute joints to links QR and PS at points R and S, respectively. The link lengths of all the links are constant as given below. Fixed Link PS length r 1 = 0.5 m Crank Link PQ length r 2 = 0.15 m Link QR length r 3 = 0.4 m Link RS length r 4 = 0.45 m Gravity g = 9.81 m/sec 2 Determine the angular accelerations, angular velocity and rotation of link RS at joint R. 219

226 VMMECH058 Figure 144: Schematic Material Properties E = 2 x Pa ν = 0.3 Analysis Geometric Properties Link PS length r 1 = 0.5 m Link PQ length r 2 = 0.15 m Link QR length r 3 = 0.4 m Link RS length r 4 = 0.45 m Loading Gravitational acceleration = m/s 2 (Y Direction) Results are obtained using MATLAB. Results Comparison Results Target Error (%) Angular Acceleration (rad/s 2 ) Angular Velocity (rad/sec) Rotation (rad)

227 VMMECH059: Bending of a Solid Beam (Plane Elements) Overview Reference: Solver(s): Analysis Type(s): Element Type(s): R. J. Roark, Formulas for Stress and Strain, 4th Edition, McGraw-Hill Book Co., Inc., New York, NY, 1965, pp. 104, 106. ANSYS Static Structural Analysis 2-D Plane Stress Shell element Test Case A beam of length and height h is built-in at one end and loaded at the free end with: a moment M a shear force F For each case, determine the deflection δ at the free end and the bending stress σ Bend at a distance d from the wall at the outside fiber. Figure 145: Schematic Case 1: Case 2: 221

228 VMMECH059 Material Properties E = 30 x 10 6 psi ν = 0.3 Analysis Geometric Properties = 10 in h = 2 in d = 1 in Loading Case 1: M = 2000 ibf-in (Z direction) Case 2: F = 300 lb (Y direction) Since the loading is uniform and in one plane, the above problem can be analyzed as a plane stress problem. Results Comparison Results Target Error (%) Case 1: Deflection (in) Stress Bend (psi) Case 2: Deflection (in) Stress Bend (psi)

229 VMMECH060: Crank Slot Joint Simulation with Flexible Dynamic Analysis Overview Reference: Solver(s): Analysis Type(s): Element Type(s): APDL Multibody Analysis ANSYS Flexible Dynamic Analysis Solid and Multipoint Constraint Element Test Case The figure shows crank slot model consists of a base and two rods. The two rods are attached to each other and the base with three bolts. The base of the model is fixed to the ground via a fixed joint and Bolt3 connected with slot joint to base. Define Rod1 and Rod2 as a flexible body and run the crank slot analysis using a Flexible Dynamic Analysis. Determine the Equivalent (von Mises) Stress for both flexible rods. Figure 146: Schematic Material Properties E = 2 x 10 5 MPa ν = 0.3 Geometric Properties Rod1 length = 75 mm Rod2 length = 115 mm Loading Constant angular acceleration at base to Bolt1 = 25 rad/s 2 223

230 VMMECH060 Analysis Figure 147: Contour Plot Figure 148: Equivalent (von Mises) Stress Figure 149: Total Force at Base to Bolt1 Results Comparison Results Target Error (%) Equivalent (von Mises) Stress (MPa)

231 VMMECH060 Results Target Error (%) Bolt1 (N)

232 226

233 VMMECH061: Out-of-Plane Bending of a Curved Bar Overview Reference: Solver(s): Analysis Type(s): Element Type(s): S. Timoshenko, Strength of Material, Part I, Elementary Theory and Problems, 3rd Edition, D. Van Nostrand Co., Inc., New York, NY, 1955, pg. 412, eq ANSYS Static Structural Analysis Beam Test Case A portion of a horizontal circular ring, built-in at A, is loaded by a vertical (Z) load F applied at the end B. The ring has a solid circular cross-section of diameter d. Determine the deflection δ at end B and the maximum bending stress σ Bend. Figure 150: Schematic Material Properties E = 30 x 10 6 psi ν = 0.3 Geometric Properties r = 100 in d = 2 in θ = 90 Loading F = -50 lb (Z direction) 227

234 VMMECH061 Results Comparison Results Target Error (%) Deflection (in) Stress Bend (psi)

235 VMMECH062: Stresses in a Long Cylinder Overview Reference: Solver(s): Analysis Type(s): Element Type(s): S. Timoshenko, Strength of Material, Part II, Elementary Theory and Problems, 3rd Edition, D. Van Nostrand Co., Inc., New York, NY, 1956, pg. 213, problem 1 and pg. 213, article 42. ANSYS AIM ANSYS Static Structural Analysis Axisymmetric Shell Test Case A long thick-walled cylinder is initially subjected to an internal pressure p. Determine the radial displacement δ r at the inner surface, the radial stress σ r, and tangential stress σ t, at the inner and outer surfaces and at the middle wall thickness. Internal pressure is then removed and the cylinder is subjected to a rotation ω about its center line. Determine the radial σ r and tangential σ t stresses at the inner wall and at an interior point located at r = X i. Figure 151: Schematic Material Properties Geometric Properties Loading E = 30 x 10 6 psi a = 4 in Case 1: ν = 0.3 b = 8 in X i = 5.43 in Pressure = 30,000 ρ = lbm/in 3 psi (radial direction) Case 2: Rotational velocity = 1000 rad/s (Y direction) 229

236 VMMECH062 Analysis Assumptions and Modeling Notes for ANSYS Because the geometry and loading are symmetric about the y-axis, the above problem can be analyzed as an axisymmetric problem in ANSYS. Figure 152: Schematic for ANSYS Case 1: Case 2: Analysis Assumptions and Modeling Notes for ANSYS AIM The following figure shows the finite element model in ANSYS AIM: 230

237 VMMECH062 Figure 153: Schematic for ANSYS AIM Results Comparison for ANSYS Case 1: Case 2: Results Target Error (%) Displacement r, in (r = 4 in) Stress r, psi (r = 4 in) Stress r, psi (r = 6 in) Stress t, psi (r = 8 in) Stress t, psi (r = 4 in) Stress t, psi (r = 6 in) Stress t, psi (r = 8 in) Stress r, psi (r = 4 in) Stress t, psi (r = 4 in) Stress r, psi (r = 5.43 in) Stress t, psi (r = 5.43 in) Results Comparison for ANSYS AIM Results Target AIM Error (%) Case 1: Displacement r, in (r = in)

238 VMMECH062 Results Case 2: Target AIM Error (%) Stress r, psi (r = 4 in) Stress t, psi (r = 8 in) Stress r, psi (r = 4 in) Stress t, psi (r = 8 in) Stress r, psi (r = 4 in) Stress t, psi (r = 4 in) 232

239 VMMECH063: Large Deflection of a Cantilever Overview Reference: Solver(s): Analysis Type(s): Element Type(s): K. J. Bathe, E. N. Dvorkin, "A Formulation of General Shell Elements - The Use of Mixed Interpolation of Tensorial Components, International Journal for Numerical Methods in Engineering, Vol. 22 No. 3, 1986, pg ANSYS Static Structural Analysis Shell Test Case A cantilever plate of length, width b, and thickness t is fixed at one end and subjected to a pure bending moment M at the free end. Determine the true (large deflection) free-end displacements and the top surface stress at the fixed end using shell elements. Figure 154: Schematic Material Properties E = 1800 N/mm 2 ν = 0.0 Analysis Geometric Properties = 12 mm b = 1 mm t = 1 mm Loading M = N-mm (Y direction) Large deformation is used to simulate the problem. 233

240 VMMECH063 Results Comparison Results Target Error (%) Directional Deformation X-direction (mm) Directional Deformation Z-direction (mm) Normal Stress X-direction (N/mm 2 )

241 VMMECH064: Small Deflection of a Belleville Spring Overview Reference: Solver(s): Analysis Type(s): Element Type(s): S. Timoshenko, Strength of Material, Part II, Elementary Theory and Problems, 3rd Edition, D. Van Nostrand Co., Inc., New York, NY, 1956, pg. 143, problem 2. ANSYS Static Structural Analysis Shell Test Case The conical ring shown below represents an element of a Belleville spring. Determine the deflection y produced by a load F per unit length on the inner edge of the ring. Figure 155: Schematic Material Properties E = 30 x 10 6 psi ν = 0.0 Results Comparison Geometric Properties a = 1 in b = 1.5 in t = 0.2 in β = 7 = rad Loading Line pressure = -100 lb/in (Y direction) Results Target Error (%) Directional Deformation Y-direction (in)

242 236

243 VMMECH065: Thermal Expansion to Close a Gap at a Rigid Surface Overview Reference: Solver(s): Analysis Type(s): Element Type(s): C. O. Harris, Introduction to Stress Analysis, The Macmillan Co., New York, NY, 1959, pg. 58, problem 8. ANSYS Static Thermal Stress Analysis Solid and Shell Test Case An aluminum-alloy bar is initially at a temperature of 70 F. Calculate the stresses and the thermal strain in the bar after it has been heated to 170 F. The supports are assumed to be rigid. Use a global mesh size of 0.25 in. Figure 156: Schematic Material Properties E = 10.5 x 10 6 psi Geometric Properties = 3 in. Loading Δt = 170 F - 70 F 237

244 VMMECH065 Material Properties α = 1.25 x 10-5 / F ν = 0.0 Results Comparison Geometric Properties δ = in. Loading Results Target Error (%) Normal Stress Y (psi) Thermal Strain Y (in/in) 1.25 x x

245 VMMECH066: Bending of a Tapered Plate Overview Reference: Solver(s): Analysis Type(s): Element Type(s): C. O. Harris, Introduction to Stress Analysis, The Macmillan Co., New York, NY, 1959, pg. 114, problem 61. ANSYS Static Structural Analysis Shell Test Case A tapered cantilever plate of rectangular cross-section is subjected to a load F at its tip. Find the maximum deflection δ and the maximum principal stress σ 1 in the plate. Use a global mesh size of 0.75 in. Figure 157: Schematic Material Properties E = 30 x 10 6 psi ν = 0.0 Geometric Properties L = 20 in d = 3 in t = 0.5 in Loading F = 10 lbf 239

246 VMMECH066 Results Comparison Results Target Error (%) Maximum Principal Stress (psi) Directional Deformation Z (in)

247 VMMECH067: Elongation of a Solid Tapered Bar Overview Reference: Solver(s): Analysis Type(s): Element Type(s): C. O. Harris, Introduction to Stress Analysis, The Macmillan Co., New York, NY, 1959, pg. 237, problem 4. ANSYS AIM ANSYS Static Structural Analysis Solid Test Case A tapered aluminum alloy bar of square cross-section and length L is suspended from a ceiling. An axial load F is applied to the free end of the bar. Determine the maximum axial deflection δ in the bar and the axial stress σ y at mid-length (Y = L/2). Use a global mesh size of 0.5 in. Figure 158: Problem Sketch Material Properties E = 10.4 x 10 6 psi ν = 0.3 Geometric Properties L = 10 in d = 2 in Loading F = lbf 241

248 VMMECH067 Analysis Assumptions and Modeling Notes for ANSYS The following figure shows the finite element model in ANSYS : Figure 159: ANSYS Schematic Results Comparison for ANSYS Results Target Error (%) Directional Deformation Y (in) Normal Stress Y at L/2 (psi) Analysis Assumptions and Modeling Notes for ANSYS AIM The following figure shows the finite element model in ANSYS AIM: 242

249 VMMECH067 Figure 160: ANSYS AIM Schematic Results Comparison for ANSYS AIM Results Target AIM Error (%) Directional Deformation Y (in) Normal Stress Y at L/2 (psi)

250 244

251 VMMECH068: Plastic Loading of a Thick Walled Cylinder Overview Reference: Solver(s): Analysis Type(s): Element Type(s): S. Timoshenko, Strength of Material, Part II, Elementary Theory and Problems, 3rd Edition, D. Van Nostrand Co., Inc., New York, NY, 1956, pg. 388, article 70. ANSYS Static, Plastic Analysis (Plane Strain) 2-D Structural Solid Test Case A long thick-walled cylinder is subjected to an internal pressure p (with no end cap load). Determine the radial stress, σ r, and the tangential (hoop) stress, σ t, at locations near the inner and outer surfaces of the cylinder for a pressure, p el, just below the yield strength of the material, a fully elastic material condition. Determine the effective (von Mises) stress, σ eff, at the same locations for a pressure, p ult, which brings the entire cylinder wall into a state of plastic flow. Use a global mesh size of 0.4 in along with a mapped face meshing. Figure 161: Schematic 245

252 VMMECH068 Material Properties E = 30 x 10 6 psi σ yp = 30,000 psi ν = 0.3 Analysis Geometric Properties a = 4 in b = 8 in Loading p el = 12,990 psi p ult = 24,011 psi This problem is modeled as a plane strain problem with only a quarter of the cross-section as shown in the above figures. Symmetry conditions are used on the edges perpendicular to X and Y axes. Load is applied in two steps as shown in the above table. The stresses are calculated at a distance of r = 4.4 in and 7.6 in, w.r.t a cylindrical coordinate system whose origin is same as that of the global coordinate system. Results Comparison Results Target Error (%) Fully Elastic Stress r, psi (X = 4.4 in) Stress t, psi (X = 4.4 in) Stress r, psi (X = 7.6 in) Stress t, psi (X = 7.6 in) Fully Plastic Stress eff, psi (X = 4.4 in) Stress eff, psi (X = 7.6 in)

253 VMMECH069: Barrel Vault Roof Under Self Weight Overview Reference: Solver(s): Analysis Type(s): Element Type(s): R. D. Cook, Concepts and Applications of Finite Element Analysis, 2nd Edition, John Wiley and Sons, Inc., New York, NY, 1981, pp ANSYS Static Analysis Shell Test Case A cylindrical shell roof of density ρ is subjected to a loading of its own weight. The roof is supported by walls at each end and is free along the sides. Find the x and y displacements at point A and the top and bottom stresses at points A and B. Express stresses in the cylindrical coordinate system. Use a global mesh size of 4 m. Figure 162: Schematic 247

254 VMMECH069 Material Properties Geometric Properties E = 4.32 x 10 8 N/m 2 t = 0.25 m ν = 0.3 r = 25 m = 50 m ρ = kg/m 3 Θ = 40 Loading g = 9.8 m/s 2 Analysis A one-fourth symmetry model is used. Displacements, UX and UY, and the longitudinal rotation, ROTZ, are constrained at the roof end to model the support wall. Results Comparison Results Target Error (%) Directional Deformation A, m Directional Deformation A, m Stress z, A, Pa Stress z, A, Pa Stress angle, B, Pa Stress angle, B, Pa

255 VMMECH070: Hyperelastic Thick Cylinder Under Internal Pressure Overview Reference: Solver(s): Analysis Type(s): Element Type(s): J. T. Oden, Finite Elements of Nonlinear Continua, McGraw-Hill Book Co., Inc., New York, NY, 1972, pp ANSYS Static, Large Deflection Analysis 2-D Structural Solid Elements Test Case An infinitely long cylinder is made of Mooney-Rivlin type material. An internal pressure of P i is applied. Find the radial displacement at the inner radius and the radial stress at radius R = 8.16 in. Use a global mesh size of 1 in along with a mapped face meshing. Figure 163: Schematic 249

256 VMMECH070 Material Properties Mooney-Rivlin material coefficients: C10 = 80 psi C01 = 20 psi D1 = 0 /psi Analysis Geometric Properties r i = 7.0 in r o = in Loading P i = 150 psi Because of the loading conditions and the infinite length, this problem is solved as a plane strain problem. A one-fourth symmetry model is used. The total pressure is applied in two load increments 90 and 150 psi. Stress and Deformation are expressed in cylindrical coordinate system. Results Comparison Results Target Error (%) Deformation at inner radius in radial direction, in Radial Stress at r = 8.16 in, psi

257 VMMECH071: Centerline Temperature of a Heat Generating Wire Overview Reference: Solver(s): Analysis Type(s): Element Type(s): W. M. Rohsenow, H. Y. Choi, Heat, Mass and Momentum Transfer, 2nd Printing, Prentice-Hall, Inc., Englewood Cliffs, NJ, 1963, pg. 106, ex ANSYS AIM ANSYS Thermal Analysis 2-D Thermal Solid Elements Test Case Determine the centerline temperature T cl and the surface temperature T s of a bare steel wire generating heat at the rate Q. The surface convection coefficient between the wire and the air (at temperature T a ) is h. Also, determine the heat dissipation rate q. Use a global mesh size of 0.02 ft along with a mapped face meshing. Figure 164: Schematic Material Properties k = x 10-3 Btu/s-ft- F Geometric Properties r o = ft Loading h = x 10-3 Btu/s-ft 2 - F T a = 70 F Q = Btu/s-ft 3 Analysis Assumptions and Modeling Notes for ANSYS Because of the symmetry in loading conditions and in the geometry, this problem is solved as an axisymmetric problem in ANSYS. The solution is based on a wire 1 foot long. 251

258 VMMECH071 Figure 165: Schematic for ANSYS Analysis Assumptions and Modeling Notes for ANSYS AIM The following figure shows the finite element model in ANSYS AIM: 252

259 VMMECH071 Figure 166: Schematic for ANSYS AIM Results Comparison for ANSYS Results Target Error (%) Centerline Temperature, F Surface Temperature, F Heat dissipation rate, BTU/s Results Comparison for ANSYS AIM Results Maximum Centerline Temperature, F Maximum Surface Temperature, F Target AIM Error (%)

260 254

261 VMMECH072: Thermal Stresses in a Long Cylinder Overview Reference: Solver(s): Analysis Type(s): Element Type(s): S. Timoshenko, Strength of Material, Part II, Elementary Theory and Problems, 3rd Edition, D. Van Nostrand Co, Inc., New York, NY, 1956, pg. 234, problem 1. ANSYS AIM ANSYS Thermal Stress Analysis 2-D Thermal Solid Elements Test Case A long thick-walled cylinder is maintained at a temperature T i on the inner surface and T o on the outer surface. Determine the temperature distribution through the wall thickness. Also determine the axial stress σ a and the tangential (hoop) stress σ t at the inner and outer surfaces Edge sizing is used for all edges and edge behavior is defined as hard. Figure 167: Schematic Material Properties E = 30 x 10 6 psi α = x 10-5 / F ν = 0.3 k = x 10-4 Btu/s-in- F Geometric Properties Loading a = in T i = -1 F b = in T o = 0 F Analysis Assumptions and Modeling Notes for ANSYS Because of the symmetry in loading conditions and in the geometry, this problem is solved as an axisymmetric problem in ANSYS. The axial length is arbitrary and it is taken has 0.1 in. Nodal coupling is used in the static stress analysis. Model is used for the thermal and stress solutions. 255

262 VMMECH072 Figure 168: Schematic in ANSYS Analysis Assumptions and Modeling Notes for ANSYS AIM The following figure shows the finite element model in ANSYS AIM: 256

263 VMMECH072 Figure 169: Schematic in ANSYS AIM Results Comparison for ANSYS Thermal Analysis Target Error (%) T, F (at X = in) T, F (at X = in) T, F (at X = in) Static Analysis Target Error (%) Stress a, psi (at X = in) Stress t, psi (at X = in) Stress a, psi (at X = in) Stress t, psi (at X = in) Results Comparison for ANSYS AIM Thermal Analysis T, F (at X = in) T, F (at X = in) -1 0 Target -1 0 AIM 0 0 Error (%) 257

264 VMMECH072 Static Analysis Stress a, psi (at X = in) Stress t, psi (at X = in) Stress a, psi (at X = in) Stress t, psi (at X = in) Target AIM Error (%)

265 VMMECH073: Modal Analysis of a Cyclic Symmetric Annular Plate Overview Reference: Solver(s): Analysis Type(s): Element Type(s): R. D. Blevins, Formulas for Natural Frequency and Mode Shape, New York, NY, VanNostrand Reinhold Publishing Inc., 1979, PP , ANSYS AIM ANSYS Mode-Frequency Analysis Solid Test Case The fundamental natural frequency of an annular plate is determined using a mode-frequency analysis. The lower bound is calculated from the natural frequency of the annular plates that are free on the inner radius and fixed on the outer. The bounds for the plate frequency are compared to the theoretical results. Figure 170: Problem Sketch Analysis According to Blevins, the lower bound for the fundamental natural frequency of the annular plate is found using the formula presented in Table 11-2 of the reference: (2) where: λ 2 = 4.80 Material Properties E = 7.1 x 10 5 kg/cm 2 ν = 0.3 ρ = 2.79 x 10-9 kg/cm 2 γ = x 10-6 kg-sec 2 /cm 3 Geometric Properties Outside Radius (a) = 50 cm Inside Radius (b) = 18.5 cm Thickness (h) = 0.5 cm Sector Angle = 30 Loading 259

266 VMMECH073 Analysis Assumptions and Modeling Notes for ANSYS The following figure shows the finite element model in ANSYS : Figure 171: ANSYS Schematic Results Comparison for ANSYS Results Target Error (%) Frequency (Hz) Analysis Assumptions and Modeling Notes for ANSYS AIM The following figure shows the finite element model in ANSYS AIM: 260

267 VMMECH073 Figure 172: ANSYS AIM Schematic Results Comparison for ANSYS AIM Results Target AIM Error (%) Frequency (Hz)

268 262

269 VMMECH074: Tension/Compression Only Springs Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Rao, Singiresu S. Vibrations. 4th ed. Singapore: Prentice Hall, ANSYS Rigid Body Dynamic Spring Analysis Solid Test Case This test calculates the elastic forces of both tension and compression only springs. A compression only spring uses a negative (compressive) displacement. A tension only spring uses a positive (tensile) displacement. Both spring types are analyzed in tension and compression loading. The detection of the spring state being in tension or compression is determined by the non-linear solver. Figure 173: Schematic Material Properties k = 1.0e7 N/m x 1 = 0.5 m x 2 = -0.5 m m = 7850 kg Geometric Properties L o = 1 m Loading Analysis Assumptions and Modeling Notes Hooke s Law: Elastic Force = Spring Constant * Displacement F = k*x Spring 1: Compression Only spring Spring 2: Tension Only spring 263

270 VMMECH074 Results Comparison Tensile Displacement (x 1 ) Results Target Error (%) Elastic Force (N) Spring Elastic Force (N) Spring 2 5.0e6 5.0e6 0 Compressive Displacement (x 2 ) Results Target Error (%) Elastic Force (N) Spring 1-5.0e6-5.0e6 0 Elastic Force (N) Spring

271 VMMECH075: Harmonic Response of Two-Story Building under Transverse Loading Overview Reference: Solver(s): Analysis Type(s): Element Type(s): W. T. Thomson, Theory of Vibration with Applications, 3rd Edition, 1999, Example 6.4-1, pg. 166 ANSYS Harmonic Analysis Solid Test Case A two-story building has two columns (2K and K) constituting stiffness elements and two slabs (2M and M) constituting mass elements. Find the y directional deformation frequency response of the system at 70 Hz on each of the vertices for the frequency range of 0 to 500 Hz using mode-superposition as the solution method. Figure 174: Schematic Material Properties Material E (Pa) ν ρ (kg/m3) Block 2 2 x Shaft x x 10-8 Block 1 2 x Shaft 1 9 x x 10-8 Geometric Properties Loading Block 1 and 2: 40 mm x 40 mm x 40 mm Shaft 1 and 2: 20 mm x 20 mm x 200 mm Force = -1 x 10 5 N (y direction) 265

272 VMMECH075 Analysis Assumptions and Modeling Notes The material of the columns is assigned negligible density to make them as massless springs. The slabs are allowed to move only in the y-direction by applying frictionless supports on all the faces of the slabs in the y direction. The end face of the column (2K) is fixed and a harmonic force is applied on the face of the slab (M) as shown in Figure 174: Schematic (p. 265). Set the solution intervals to 50. Add the frictionless support and fixed support in a modal system, and then link the modal system to a harmonic response system. Note There are frictionless supports on 8 faces of the geometry shown. Results Comparison Results Target Error (%) Maximum Amplitude for Vertex A (m) Maximum Amplitude for Vertex B (m)

273 VMMECH076: Elongation of a Tapered Shell With Variable Thickness Overview Reference: Solver(s): Analysis Type(s): Element Type(s): C. O. Harris, Introduction to Stress Analysis, The Macmillan Co., New York, NY, 1959, pg. 237, problem 4. ANSYS Static Structural Analysis Shell Test Case A tapered aluminum alloy plate of length L with varying thickness across length is suspended from a ceiling. An axial load F is applied to the free end of the plate. Determine the maximum axial deflection δ in the plate and the axial stress σy at mid-length (Y = L/2). Use a global mesh size of 0.5 in with mapped face meshing. Figure 175: Schematic Material Properties E = 10.4 x 10 6 psi ν = 0.3 Geometric Properties Tapered plate: L = 10 in Base width = 2 in Top width = 1 in Thickness varying from 2 in to 1 in from base to top. Loading F = lbf 267

274 VMMECH076 Results Comparison Results Target Error (%) Directional Deformation Y (in) Normal Stress Y at L/2 (psi)

275 VMMECH077: Heat Transfer in a Bar with Variable Sheet Thickness Overview Reference: Solver(s): Analysis Type(s): Element Type(s): For basic equation: Frank P. Incropera and David P. DeWitt, Heat and Mass Transfer, John Wiley & Sons, Inc, 2002, 5 th Edition pg. 5. ANSYS Static Thermal Analysis Shell Test Case A 10 x 50 mm plate with a thickness varying from 1 mm to 4 mm is maintained at temperatures of 100 C and 200 C as shown below. Find the following: Temperatures at mid of the surface. Heat flow reactions on end edges. Figure 176: Schematic Material Properties E = 2.0 x Pa v = 0 α = 1.2 x / C Geometric Properties Plate Dimensions : 10 X 50 mm. Thickness Variation : 1 mm to 4 mm Loading Temperature (T1) on edge (@ 1mm thickness) = 100 C 269

276 VMMECH077 Material Properties k = 60.5 W/m C Analysis Geometric Properties Loading Temperature (T2)on edge (@ 4mm thickness) = 200 C Heat flow due to conduction is given by: (3) The area for conduction varies from A1 to A2. The area Ay at any distance y is given as: (4) Inserting Equation 4 (p. 270) in equation Equation 3 (p. 270) and integrating the equation from 0 to L, (5) Temperature at any point y is given as: (6) Results Comparison Results Target Error (%) Heat reaction at T1 (W) Heat reaction at T2 (W) Temperature at mid of surface ( C)

277 VMMECH078: Gasket Material Under Uniaxial Compression Loading-3-D Analysis Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any Nonlinear Material Verification Text ANSYS Static Analysis (ANTYPE=0) 3-D Structural Solid Elements 3-D Gasket Elements Test Case A thin interface layer of thickness t is defined between two blocks of length and width l placed on top of each other. The blocks are constrained on the left and bottom and back faces. The blocks are loaded with pressure P on the top face. Determine the pressure-closure response for gasket elements. 271

278 VMMECH

279 VMMECH078 Material Properties E = x 10 6 Pa ν = 0.21 Analysis Geometric Properties L = 1 m T = 0.02 m Loading P1 = Pa P2 = Pa A 3-D analysis is performed first using a mesh of 4 x 4 gasket elements. In order to simulate the loadingunloading behavior of gasket material, the model is first loaded with a pressure P1 and unloaded and then loaded with a pressure P2 and unloaded. The pressure-closure responses simulated are compared to the material definition. Because of convergence issues, the model could not be unloaded to 0 Pa and was instead unloaded to 100 Pa. Results Comparison Target Gasket Pressure and Closure at End of 1st Loading: GK-PRES GK-CLOS x x x x 10 4 Gasket Pressure and Closure at End of 2nd Loading: GK-PRES GK-CLOS x x x x 10 4 Error (%)

280 274

281 VMMECH079: Natural Frequency of a Motor-Generator Overview Reference: Solver(s): Analysis Type(s): Element Type(s): W. T. Thomson. Vibration Theory and Applications. 2nd Printing, Prentice-Hall, Inc., Englewood Cliffs, NJ. pg. 10, ex ANSYS Mode-Frequency Analysis Pipe Element Test Case A small generator of mass m is driven by a main engine through a solid steel shaft of diameter d. If the polar moment of inertia of the generator rotor is J, determine the natural frequency f in torsion. Assume that the engine is large compared to the rotor so that the engine end of the shaft may be assumed to be fixed. Neglect the mass of the shaft also. Figure 177: Schematic Material Properties E = 31.2 x 10 6 psi m = 1 lb-sec 2 /in Results Comparison Geometric Properties d =.375 in = 8.00 in J =.031 lb-in-sec 2 Loading Results Target Error (%) Lower Order F, Hz Higher Order F, Hz

282 276

283 VMMECH080: Transient Response of a Spring-Mass System Overview Reference: Solver(s): Analysis Type(s): Element Type(s): R. K. Vierck. Vibration Analysis. 2nd Edition. Harper & Row Publishers, New York, NY, sec ANSYS Transient Dynamic Mode-Superposition Analysis Test Case A system containing two masses, m 1 and m 2, and two springs of stiffness k 1 and k 2 is subjected to a pulse load F(t) on mass 1. Determine the displacement response of the system for the load history shown. Figure 178: Schematic Material Properties k 1 = 6 N/m k 2 = 16 N/m m 1 = 2 Kg Geometric Properties Loading F 0 = 50 N 277

284 VMMECH080 Material Properties m 2 = 2 Kg Geometric Properties Loading t d = 1.8 sec Results Comparison Results Target Error (%) Y 1, m (@ t = 1.3s) Y 2, m (@ t = 1.3s) Y 1, m (@ t = 2.4s) Y 2, m (@ t = 2.4s)

285 VMMECH081: Statically Indeterminate Reaction Force Analysis Overview Reference: Solver(s): Analysis Type(s): Element Type(s): P.Bezler, M. Hartzman, and M. Reich. Dynamic Analysis of Uniform Support Motion Response Spectrum Method, (NUREG/CR-1677), Brookhaven National Laboratory, August Problem 2. Pages ANSYS Modal analysis Spectral analysis Elastic straight pipe elements Structural Mass element Test Case This benchmark problem contains three-dimensional multi-branched piping systems. The total mass of the system is represented by structural mass elements specified at individual nodes. Modal and response spectrum analyses are performed on the piping model. Frequencies obtained from modal solve and the nodal/element solution obtained from spectrum solve are compared against reference results. The NUREG intermodal/interspatial results are used for comparison. This benchmark problem is also presented in VM-NR a in the APDL Verification Manual. Figure 179: Schematic 279

286 VMMECH081 Material Properties Pipe Elements: Geometric Properties Straight Pipe: Outer Diameter = in E = x 10 6 psi. Nu = 0.3 Density = x lb-sec 2 /in 4 Wall Mass Elements (lb-sec 2 /in): Thickness = in Loading Acceleration response spectrum curve defined by SV and FREQ commands. (Mass is isotropic) node 1: M = x 10-1 node 2: M = x 10-1 node 3: M = x 10-1 node 4: M = x 10-1 node 5: M = x 10-1 node 6: M = x 10-2 node 7: M = x 10-1 node 8: M = x 10-2 node 9: M = x 10-2 node 10: M = x 10-1 node 11: M = x 10-2 node 12: M = x 10-1 node 13: M = x 10-2 node 14: M = x 10-2 Results Comparison Results Target Error (%)

287 VMMECH081 Results Target Error (%) Results Node Target Error (%) UX at node UY at node UZ at node

288 282

289 VMMECH082: Fracture Mechanics Stress for a Crack in a Plate Overview Reference: Solver(s): Analysis Type(s): Element Type(s): W.F.Brown, Jr., J.E.Srawley, Plane strain crack toughness testing of high strength metallic materials, ASTM STP-410, (1966). ANSYS Static Structural Analysis Solid Test Case A long plate with a center crack is subjected to an end tensile stress 0 as shown in problem sketch. Symmetry boundary conditions are considered and the fracture mechanics stress intensity factor KI is determined. This problem is also presented in VM256 in the APDL Verification Manual. Figure 180: Schematic 283

290 VMMECH082 Material Properties E = 30 x 10 6 psi ν = 0.3 Geometric Properties a = 1 in b = 5 in h = 5 in t = 0.25 in Loading σ 0 = psi Results Comparison Results Target Error (%) Stress Intensity KI

291 VMMECH083: Transient Response to a Step Excitation Overview Reference: Solver(s): Analysis Type(s): Element Type(s): W. T. Thomson, Vibration Theory and Applications, 2nd Printing, Prentice-Hall, Inc., Englewood Cliffs, NJ, 1965, pg. 102, article 4.3. ANSYS Mode-Superposition Transient Dynamic Analysis Test Case A spring-mass-damping system that is initially at rest is subjected to a step force change F acting on the mass. Determine the displacement u at time t for damping ratio, ξ = 0.5. This problem is also presented in VM75 in the APDL Verification Manual. Figure 181: Schematic 285

292 VMMECH083 Material Properties m = 0.5 lb-sec 2 /in k = 200 lb/in Loading F = 200 lb Analysis Assumptions and Modeling Notes The damping coefficient c is calculated as lb-sec/in for. 286

293 VMMECH083 Results Comparison Results Target Error (%) Total Def Max (ξ = 0.5) Time = 0.20 sec Figure 182: Maximum Deformation vs. Time (damped) 287

294 288

295 VMMECH084: Mullins Effect on a Rubber Tube Model Subjected to Tension Loading Overview Reference: Solver(s): Analysis Type(s): Element Type(s):.W.Ogden, et al., A Pseudo-elastic Model for the Mullins Effect in Filled Rubber", Royal Society of London Proceedings Series A., (1989), pg: ANSYS Static Analysis Solid Test Case An axisymmetric rubber plate made of Neo-Hookean material is modeled with radius R and height H. The model is subjected to cyclic displacement loading on the top surface. The axial stress obtained at different load steps is compared against the reference solution. This problem is also presented in VM268 in the APDL Verification Manual. Figure 183: Schematic Material Properties Neo-Hookean Constants: µ = 8 MPa Geometric Properties R = 0.5m H = 1m Loading One cycle of loading 289

296 VMMECH084 Material Properties Geometric Properties Loading Step 1: λ = 1.5 Ogden-Roxburgh Mullins Constants: Step 2: λ = 2.0 r = m = β =0.2 Step 3: λ = 3.0 Step 4: λ = 2.0 Step 5: λ = 1.5 Step 6: λ = 1.0 Results Comparison Results Axial Stress (Pa) Stretch λ Target Error (%) Figure 184: Variation of Axial Stress 290

297 VMMECH085: Bending of a Composite Beam Overview Reference: Solver(s): Analysis Type(s): Element Type(s): R. J. Roark, W. C. Young, Formulas for Stress and Strain, McGraw-Hill Book Co., Inc., New York, NY, 1975, pg , article 7.2. ANSYS Static Analysis Solid Test Case A beam of length and width w made up of two layers of different materials is subjected to a uniform rise in temperature from T ref to T o, and a bending moment M y at the free-end. E i and α i correspond to the Young's modulus and thermal coefficient of expansion for layer i, respectively. Determine the free-end displacement δ (in the Z-direction) and the X-direction stresses at the top and bottom surfaces of the layered beam. This problem is also presented in VM144 in the APDL Verification Manual. Figure 185: Schematic Material Properties MAT1: E 1 = 1.2 Geometric Properties = 8 in w = 0.5 in Loading T o = 100 F x 10 6 psi T ref = t 1 = 0.2 in α 1 = 0 F 1.8 x t 2 = 0.1 in 10-4 M y = 10.0 in/in/ F in-lb 291

298 VMMECH085 Material Properties MAT2: E 2 = 0.4 x 10 6 psi α 2 = 0.6 x 10-4 in/in/ F Geometric Properties Loading Results Comparison Results Target Error (%) Displacement, in Stress x TOP, psi Stress x BOT, psi

299 VMMECH086: Stress Concentration at a Hole in a Plate Overview Reference: Solver(s): Analysis Type(s): Element Type(s): R. J. Roark, Formulas for Stress and Strain, 4th Edition, McGraw-Hill Book Co., Inc., New York, NY, 1965, pg. 384 ANSYS Static Structural, Submodeling (2D-2D) Solid Test Case Determine the maximum stress at a circular hole cut into a rectangular plate loaded with uniform tension P. This problem is also presented in VM142 in the APDL Verification Manual. Figure 186: Plate Problem Sketch Material Properties E = 30 x 10 6 psi υ = 0.3 Geometric Properties L = 12 in d = 1 in t = 1 in Loading P = 1000 psi Analysis Assumptions and Modeling Notes Due to symmetry, only a quarter sector of the plate is modeled. The reference result is from an infinitely long plate. Using a transferred load from the coarse model, the submodel result closely approximates the fine model. 293

300 VMMECH086 Results Comparison 2D-2D Results Results Target Error (%) Fine Model Equivalent Stress - Max Coarse Model Equivalent Stress - Max Submodel Equivalent Stress - Max Figure 187: 2D-2D Fine Model Equivalent Stress Figure 188: 2D-2D Coarse Model Equivalent Stress 294

301 VMMECH086 Figure 189: 2D-2D Submodel Equivalent Stress 295

302 296

303 VMMECH087: Campbell Diagrams and Critical Speeds Using Symmetric Orthotropic Bearings Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Nelson, H.D., McVaugh, J.M., The Dynamics of Rotor-Bearing Systems Using Finite Elements, Journal of Engineering for Industry, Vol 98, pp , 1976 ANSYS Modal Analysis Line Body Point Mass Bearing Connection Test Case A rotor-bearing system is analyzed to determine the forward and backward whirl speeds. The distributed rotor is modeled as a configuration of six elements, with each element composed of subelements. See Table 1: Geometric Data of Rotor-Bearing Elements (p. 297) for a list of the geometric data of the individual elements. Two symmetric orthotropic bearings are located at positions four and six. A modal analysis is performed on the rotor-bearing system with multiple load steps to determine the whirl speeds and Campbell values for the system. This problem is also presented in VM254 in the APDL Verification Manual. Figure 190: Rotor-Bearing Configuration Table 1: Geometric Data of Rotor-Bearing Elements Element Number Subelement number Axial Distance to Subelement Inner Diameter (cm) Outer Diameter (cm)

304 VMMECH087 Element Number Subelement number Axial Distance to Subelement Inner Diameter (cm) Outer Diameter (cm) Material Properties Geometric Properties Loading Shaft Refer to Table 1: Geometric Data Rotational Velocity E Spin (1) = 11 = x Pa of Rotor-Bearing Elements (p. 297) 1000 RPM G 12 = 1.0 x Pa Spin (2) = Density = 7806 kg/m RPM Mass Element Spin (3) = RPM Mass = kg Spin (4) = RPM Polar inertia = kg m 2 Spin (5) = Diametral inertia = kg m RPM Spin (6) = Bearing Element RPM Spring coefficients K11 = K22 = x 10 7 N/m K12 = K21 = x 10 6 N/m Analysis Assumptions and Modeling Notes A modal analysis is performed on the rotor-bearing system with QR Damp methods using pipe elements (PIPE288) to determine the whirl speeds and Campbell values. 298

305 A point mass is used to model the rigid disk (concentrated mass). Two symmetric orthotropic bearings are used to assemble the rotor system. No shear effect is included in the rotor-bearing system. The displacement and rotation along and around the X-axis is constrained so that the rotor-bearing system does not have any torsion or traction related displacements. Backward and forward whirl speeds for slope = RPM are determined from the modal analysis. Results Comparison Target Error (%) Backward and forward whirl speeds for slope = RPM VMMECH087 RPM = Hz * 60 PIPE288 Mode 1 (BW) Mode 2 (FW) Mode 3 (BW) Mode 4 (FW) Figure 191: Campbell Diagram for Rotor-Bearing System 299

306 300

307 VMMECH088: Harmonic Response of a Guitar String Overview Reference: Solver(s): Analysis Type(s): Blevins, R.D., Formulas for Natural Frequency and Mode Shape, Nostrand Reinhold Co., New York, NY, 1979, pg. 90, tab. 7-1 ANSYS Static Structural Linear Perturbed Modal Element Type(s): Linear Perturbed Harmonic Beam Test Case A uniform stainless steel guitar string of length l and diameter d is stretched between two rigid supports by a tensioning force F 1, which is required to tune the string to the E note of a C scale. The string is then struck near the quarter point with a force F 2. Determine the fundamental frequency, f 1. Also, show that only the odd-numbered frequencies produce a response at the midpoint of the string for this excitation. This problem is also presented in VM76 in the APDL Verification Manual. Material Properties E = 190 x 10 9 Pa ρ = 7920 kg/m 3 Geometric Properties l = 710 mm c = 165 mm d = mm Loading F 1 = 84 N F 2 = 1 N Analysis Assumptions and Modeling Notes Enough elements are selected so that the model can be used to adequately characterize the string dynamics. The stress stiffening capability of the elements is used. Linear perturbed harmonic analysis determines the displacement response to the lateral force F

308 VMMECH088 Figure 192: Guitar String Problem Results Comparison Target Error (%) Modal f, Hz Frequency Response f 1, (322.2 Hz) Response Response, 320 < f < f 2, (644.4 Hz) No Response No Response - f 3, (966.6 Hz) Response Response, 966 < f < f 4, ( Hz) No Response No Response - f 5, ( Hz) Response Response, 1611 < f < f 6, ( Hz) No Response No Response - Figure 193: String Midpoint Displacement Amplitude 302

309 VMMECH089: Delamination Analysis of a Double Cantilever Beam Using Contact-Based Debonding Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Alfano, G., Crisfield, M.A., Finite Element Interface Models for the Delamination Analysis of La and Computation Issues, International Journal for Numerical Methods in Engineerin , 2001 ANSYS Static Structural Solid Test Case A double cantilever beam of length l, width w, and height h with an initial crack of length a at the free end is subjected to a maximum vertical displacement U max at the top and bottom free end nodes. Determine the vertical reaction at point P, based on the vertical displacement using the contact-based debonding capability. This problem is also presented in VM255 in the APDL Verification Manual. Figure 194: Double Cantilever Beam Sketch Material Properties Composite E 11 = GPa E 22 = 9.0 GPa E 33 = 9.0 GPa G 12 = 5.2 GPa ν 12 = 0.24 ν 13 = 0.24 ν 23 = 0.46 Geometric Properties l = 100 mm a = 30 mm h = 3 mm w = 20 mm Loading U max = 10 mm 303

310 VMMECH089 Material Properties Interface C1 = 1.7 MPa C2 = 0.28 N/mm C5 = 1.0 x 10-5 Geometric Properties Loading Analysis Assumptions and Modeling Notes A double cantilever beam is analyzed under displacement control using 2-D plane strain formulation elements. An imposed displacement of U y = 10 mm acts at the top and bottom free vertex. Contact debonding is inserted at the interface. Defined fracture-energy based debonding material is used to define the material for contact debonding. Equivalent separation-distance based debonding material is also used for the contact debonding object. Based on the interface material parameters used, results obtained using are compared to results shown in Figure 15(a) of the reference material. Results Comparison Target Error (%) Max RFORCE and corresponding displacement using debonding RFORCE FY (N) DISP UY (mm) RFORCE and corresponding displacement U = 10.0 using debonding RFORCE FY (N) DISP UY (mm)

311 VMMECH090: Delamination Analysis of a Double Cantilever Beam Using Interface Delamination Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Alfano, G., Crisfield, M.A., Finite Element Interface Models for the Delamination Analysis of La Engineering, Vol 50, pp , 2001 ANSYS Static Structural Solid Test Case A double cantilever beam of length l, width w, and height h with an initial crack of length a at the free end is subjected to a maximum vertical displacement U max at the top and bottom free end nodes. Determine the vertical reaction at point P based on the vertical displacement for the interface model. This problem is also presented in VM248 in the APDL Verification Manual. Figure 195: Double Cantilever Beam Sketch Material Properties Composite Interface E 11 = GPa E 22 = 9.0 GPa E 33 = 9.0 GPa G 12 = 5.2 GPa ν 12 = 0.24 ν 13 = 0.24 ν 23 = 0.46 C1 (maximum stress) = 25 MPa Geometric Properties l = 100 mm a = 30 mm h = 3 mm w = 20 mm Loading U max = 10 mm 305

312 VMMECH090 Material Properties C2 (normal separation) = mm C3 (shear separation) = 1000 mm Geometric Properties Loading Analysis Assumptions and Modeling Notes A double cantilever beam is analyzed under displacement load using interface elements for delamination and 2-D plane strain formulation elements. An imposed displacement of U y = 10 mm acts at the top and bottom free vertex. An Interface Delamination object is inserted to model delamination. Equivalent material constants are used for the interface material, as uses the exponential form of the cohesive zone model and the reference uses a bilinear constitutive model. Results Comparison Lower Order Results Target Max RFORCE and corresponding DISP: RFORCE FY (N) DISP UY (mm) End RFORCE and corresponding DISP RFORCE FY (N) DISP UY (mm) Higher Order Results Target Max RFORCE and corresponding DISP RFORCE FY (N) DISP UY (mm) End RFORCE and corresponding DISP RFORCE FY (N) DISP UY (mm) Error (%) Ratio

313 VMMECH091: Unbalanced Harmonic Response of a Shaft Carrying Single Rotor with Damping Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Any basic vibration analysis book. ANSYS Harmonic Beam Test Case A disc with mass of 5 kg is mounted midway on a simply supported shaft of diameter 0.01 m and length 0.5 m. All bodies are cylindrical. The center of gravity of the disc is 5 x 10-3 m away from the geometric center. The equivalent viscous damping at the center of the disc is 40 N sec/m. The shaft rotates axially at 740 RPM. Find the frequency response in the Y direction at the midpoint of the shaft. Figure 196: Schematic Material Properties E = 2 x Pa ρ = 1 x kg/m 3 Geometric Properties Length of shaft, L = 0.5 m Diameter of shaft, D = 0.01 m Loading Point Mass at midpoint of shaft, m = 5 kg Rotating radius (eccentricity), e = 5 x 10-3 m 307

314 VMMECH091 Material Properties Geometric Properties Loading Rotating force, (m x e) = 2.5 x 10-2 kg m Analysis Assumptions and Modeling Notes Static deflection of shaft, Weight of shaft, Stiffness of shaft, Critical speed, Critical damping coefficient, Damping ratio, The amplitude of vibration is therefore: where = eccentricity = speed of shaft = damping coefficient = mass of shaft Results Comparison Results Frequency response (Y) at the midpoint of the 740 RPM ( Hz) (m) Target Error (%)

315 VMMECH091 Figure 197: Bode Plot 309

316 310

317 VMMECH092: Convection Treatment Problem for a Hollow Cylinder with Fluid Flow Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Arpaci, V.S., Selamet, A., and Kao, S.H., Introduction to Heat Transfer, 2000, pp ANSYS Static Structural Coupled Thermal Pipe Thermal Surface Thermal Solid Test Case A hollow cylinder is modeled with an inner radius of m, an outer radius of 0.02 m, and a length of 0.1 m. Fluid is made to flow through the cylinder to simulate the convection problem. Surface effect elements with film coefficients are used in between the fluid and cylinder to include the convection loads. The inlet temperature of the fluid, mass flow rate of the fluid, and the bulk temperature at the outer cylinder surface are defined. A static analysis is performed on the model to determine the nodal temperature of the fluid elements. This problem is also presented in VM271 in the APDL Verification Manual. Figure 198: Schematic 311

318 VMMECH092 Material Properties Fluid: Specific heat = J/(kg C) Thermal conductivity = 1.0 x W/(m C) Cylinder: Thermal conductivity = 1000 W/(m C) Geometric Properties Loading Inner radius, r 1 = Inlet temperature m of fluid, T inlet = Outer radius, r 2 = 700 C 0.02 m Temperature at Length, l = 0.1 m the outer cylinder surface, T bulk = 2000 C Film coefficients for surface element = 300 W/m 2 C Mass flow rate for fluid = 7.2 kg/s Analysis Assumptions and Modeling Notes The line body model type is set to thermal fluid and the discretization type is set to upwind/exponential in a 3-D steady state thermal analysis. The existing convection is scoped to the inner face of the cylinder and is modified to consider fluid flow with fluid flow edge(s) of line bodies. The mass flow rate is applied using line body edge. The problem is solved using MKS as the unit system. Results are evaluated at different locations on a line body using nodal named selections. Results Comparison Result Temperature ( Y = 0.0 Temperature ( Y = 3.33 x Target Error (%) Temperature ( Y = 6.66 x Temperature ( Y =

319 VMMECH093: C*-Integral Calculation for a Single-Edge Cracked Plate Using Pre-Meshed Crack Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Kanninen, M.F., Popelar, C.H., Advanced Fracture Mechanics, Oxford University Press, p. 550, 198 ANSYS Static Analysis 2-D Structural Solid Element Test Case A rectangular plate with an edge crack is subjected to a tensile load. The load is applied instantly and held for 1000 hours. The C* integral is then determined using Pre-Meshed Crack and compared against the reference solution. Figure 199: Two-Dimensional Fraction Problem Sketch Material Properties E= MPa = 0.3 Parameters for strain hardening creep C1 = 5 x C2 = 3 C3 = 0 Geometric Properties L = 100 mm a = 2.5 mm b = 20 mm c = 17.5 mm Loading = 150 MPa 313

320 VMMECH093 Material Properties C4 = 0 Geometric Properties Loading Analysis Assumptions and Modeling Notes The problem is solved using 2-D structural solid element with plane strain element behavior. One half of the plate is modeled and symmetric boundary conditions are considered. The crack tip nodes and the number of paths surrounding the crack tip nodes are defined using Pre-Meshed Crack. The computation is completed in two steps. In the first step, the instal loading is applied on the top edge. In the second tep, the load is held for 1000 hours. The C* integral is computed for the crack tip nodes by averaging the C* integral values from contour 2 to contour 9. The reference value for the C* integral is determined using the following equation: where A = 5 x a = 2.5 c = 17.5 C 1 = 17.5 h 1 = 8.57 (according to the reference material) P = 3000 Results Comparison Result Target Error (%) C*-Integral

321 VMMECH094: Residual Vector in Stand-Alone and Linked Mode-Superposition Harmonic Analysis Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Dickens, J.M., Nakagawa, J.M., Wittbrodt, M.J., "A Critique of Mode Acceleration and Modal Tru 1997 ANSYS Modal Analysis Harmonic Analysis Spring-Damper Structural Mass Test Case A mode-superposition harmonic analysis (stand-alone and linked) with an excitation frequency range of 3-70 Hz and a force load along the X-direction is performed on a spring-mass model, extracting one mode and residual vector. The spring-mass model is represented using Springs and Point Masses. Refer to APDL VM149 for more details. Figure 200: Spring-Mass Model Material Properties String stiffness (K) = 1000 N/m Mass 1 = 1.0 kg Mass 2 = 0.5 kg Damping ratio = 2% Geometric Properties Total length = 5 m Loading Force along X-direction at node 4 = 1 N Excitation frequency: 3-70 Hz Analysis Assumptions and Modeling Notes The spring-mass model is represented using 2-D Spring-Damper and 2-D Structural Mass elements without rotary inertia. The X and Y-axes of the spring model are inverted from the reference model to model it along the global X-axis. The model is fixed at both ends, and the displacement along the Y- direction is constrained on all nodes. In order to obtain four distinct modes, the mass at node 5 is set to half the value of the other three masses defined at nodes 2, 3, and 4. A modal analysis is performed 315

322 VMMECH094 first using the Block Lanczos eigensolver. A mode-superposition harmonic analysis is then performed with an excitation frequency range of 3-70 Hz and a force load along the X-direction at node 4. A constant damping ratio of 0.02 (2%) is defined in the analysis. Results Comparison Result Target Error (%) Standalone Results UX_MAX (m) F_MAX (N) UX_MAX (Hz) F_MAX2 (N) Linked Results UX_MAX (m) F_MAX (N) UX_MAX (Hz) F_MAX2 (N)

323 VMMECH095: 2-D Double Cantilever Beam Problem Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Madell, J.F., et al., "Prediction of Delamination in Wind Turbine Blade Structural Details", Journa ANSYS Static Structural 2-D Structural Elements Test Case A double cantilever beam is modeled with composite material. One end of the beam is fixed, and the other end is loaded with two external forces of the same magnitude of the same magnitude but opposite directions as shown in the problem schematic. G computation for the cracked tip is conducted for VCCT using a pre-meshed crack object and compared with Equation 1 in the reference. Figure 201: Problem Schematic Material Properties E = 210 GPa = 0.3 Geometric Properties L = 100 a = 60 mm h = 5 mm b = 1 mm Loading P = 10 N Analysis Assumptions and Modeling Notes The problem is solved using 2-D element with plain strain element behavior. A pre-meshed crack object is defined. The plate is subjected to vertical loading the middle. G values are computed for the crack tip node. Results Comparison Result Target Error (%) G

324 318

325 VMMECH096: 2-D Fracture Problem Under Thermal Loading Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Wilson, W.K., et al., "The Use of the J-Integral in Thermal Stress Crack Problems", International J ANSYS Static Structural 2-D Structural Elements Test Case An edge-cracked strip is modeled with its ends constrained. The strip is subject to a linear temperature gradient through its thickness, starting at zero at the mid-thickness and reaching final value T0 at the right edge. Stress intensity factor for the cracked strip is calculated and compared against the reference value. Figure 202: Problem Schematic L W T0 -T0 0 Temperature Distribution Material Properties E = 1 x 10 5 Pa = 0.3 Geometric Properties Crack length = 1 mm L = 4 mm Thermal expansion, = 1 x 10-4 W = 2 mm Loading T0 = 10 Analysis Assumptions and Modeling Notes The problem is solved using 2-D structural elements with plain strain element behavior. Results Comparison Result Target Error (%) Stress intensity, KI

326 320

327 VMMECH097: Inclined Crack in 2-D Plate Under Uniform Tension Loading Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Anderson, T.L., "Fracture Mechanics: Fundamentals and Applications", CRC Press, Boca Raton, F ANSYS Static Structural 2-D Structural Elements Test Case A 2-D plate with length L is subjected to uniform tension loading. An inclined crack of length 2a is modeled with an angle of between the crack surface and loading direction. Stress intensity factor is calculated and compared against analytical value. Figure 203: Problem Sketch L 2a L Material Properties E = 210 GPa = 0.3 Geometric Properties L = 0.3 m Crack length, 2a = 0.09 m = 30 Loading = 10 MPa Analysis Assumptions and Modeling Notes The problem is solved using 2-D elements with plain strain element behavior. The plate is constrained along the X-direction at X = 0 and along the Y-direction at Y = 0. Stress intensity factors K1 and K2 are computed using a Pre-Meshed Crack object and compared against the analytical value. Results Comparison Results Target Error (%) KI_Right tip (Mode1) % KII_Right tip (Mode2) % 321

328 322

329 VMMECH098: 2-D End Notched Flexure Problem Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Mandell, J.F., et al., "Prediction of Delamination in Wind Turbine Blade Structural Details", Journ ANSYS Static Structural 2-D Structural Elements Test Case A beam is clamped at one end and contains a delamination of length a at the other end. A load P is applied in the middle to cause crack growth. VCCT G result is computed using a Pre-Meshed Crack object and compared against the analytical solution (equation 2 in reference). Figure 204: Problem Sketch a P 2L P/2 P/2 Material Properties E = MPa = 0.3 Geometric Properties L = 30 mm a = 10 mm Loading P = 10 N Analysis Assumptions and Modeling Notes The problem is solved using 2-D elements with plain strain element behavior. Results Comparison Result Target Error (%) G % 323

330 324

331 VMMECH099: Mode I Crack Growth Analysis of DCB Using Interface Delamination Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Krueger, R., "Application of Benchmark Examples to Assess the Single and Mixed-Mode Static ANSYS Static Structural 2-D Structural Elements Test Case A double cantilever beam is fixed at one end and contains a crack of length a 0 at the other end. Equal and opposite displacements are applied to top and bottom vertices to cause crack growth. Critical load and corresponding displacement values for Mode I failure are computed using the VCCT method of interface delamination and compared against results shown in Figure 14 of the reference material. Figure 205: Problem Schematic Material Properties T300/1076 Unidirectional Graphite/Epoxy Prepreg E 11 = 139 GPa E 22 = GPa E 33 = GPa = 0.3 = 0.3 = G Ic = kj/m 2 G IIc = kj/m 2 n (material constant) = 1.62 Geometric Properties B = 25 mm 2h = 3 mm 2L = 150 mm a 0 = 20 mm Loading d = 1.96 mm 325

332 VMMECH099 Analysis Assumptions and Modeling Notes The problem is solved using 2-D elements with plain strain element behavior. This problem is solved using two methods of VCCT-based interface delamination, node matching, and matched meshing. Results Comparison Result Target Error (%) VCCT (G1) Node Matching VCCT (G1) Matched Meshing

333 VMMECH100: 3-D Acoustic Modal Analysis with Temperature Change Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Oberg, C.L., Ryan, N.W., Baer, A.D., "A Study of T-Burner Behavior", AIAA Journal, Vol. 6, No. 6., pp ANSYS Modal Analysis Solid Test Case A cylindrical region (single-ended T-burner) is filled with propellant. The temperature distribution in the propellant is discontinuous, where 31% of the length is occupied by "cool" gas, with the remaining volume filled with "hot" gas. Determine the ratio of the amplitudes of pressure at the two ends of the T-burner. This problem is also presented in VM157 in the APDL Verification Manual. Figure 206: Project Schematic Material Properties Geometric Properties Loading Cool Gas: R = 1.5 in T c = F L = 9 in T h = F = 1.1 x 10-7 lbf s 2 /in 4 = in/s 2 327

334 VMMECH100 Material Properties Hot Gas: = 2.2 x 10-8 lbf s 2 /in 4 = in/s 2 Reference pressure = 14.7 psi Geometric Properties Loading Analysis Assumptions and Modeling Notes The T-burner is represented as a cylindrical volume. The cool gas occupies 31% of the length at F, while the hot gas of F occupies the remaining volume. The first non-zero mode is of interest, and the mode is essentially 1-D in nature. Using the ideal gas relationship, the ratio of the temperatures can be used to determine the ratio of the speed of sound and density: The temperature is applied in the cool and hot regions with a discontinuity at 31% of the length. Based on the reference, the relative amplitude at the hot end should be Results Comparison Result Ratio of hot/cold amplitude with higher order tetrahedral mesh Target Error (%)

335 VMMECH101: Natural Frequency of a Submerged Ring Overview Reference: Solver(s): Analysis Type(s): Element Type(s): Schroeder, E.A., Marcus, M.S., "Finite Element Solution of Fluid Structure Interaction Problems" ANSYS Modal Analysis Solid Test Case A steel ring is submerged in compressible fluid (water). Determine the lowest natural frequency for x- y plane bending modes of the fluid-structure system. Figure 207: Problem Schematic Material Properties Steel: E = 30 x 10 6 PSI = 0.3 = slugs per cubic inch (per reference) Geometric Properties a = 10 in b = 30 in t = 0.25 in Water: C = in/sec (speed of sound in water) = slugs per cubic inch (per reference) 329

336 VMMECH101 Analysis Assumptions and Modeling Notes The acoustic media (water) is assumed to extend to a radius of 30 inches. An acoustic pressure of zero on the outer surface of the acoustic media is modeled with a negligible value of 1 x 10-6 PSI. Results Comparison Result Frequency, Hz (higher order) Frequency, Hz (lower order) Target Error (%) Figure 208: Total Deformation of First Mode 330

337 Part IV: Design Exploration Descriptions

338

339 VMDX001: Optimization of L-Shaped Cantilever Beam Under Axial Load Overview Reference: Analysis Type(s): Element Type(s): From the Basic Principle Goal Driven Optimization 3-D Solid Test Case An L-shaped beam with dimensions 30 x 25 mm with 4 mm as the rib thickness and 300 mm in length has the surface fixed at one end. A force of 10,000 N is then applied to the opposite end of the beam. Input Parameters: Width, Height, and Length (CAD Geometry) Response Parameters: Volume, Stress, and Deflection Figure 209: Schematic Material Properties E = 2e11 Pa ν = 0 ρ = 7850 kg/m 3 Geometric Properties Width = 25 mm Height = 30 mm Rib Thickness = 4 mm Length = 300 mm Loading Fixed Support Force F = N (Z direction) Parameter Type Limits Desired Value Importance Width Input 20 mm W 30 mm No Preference High Height Input 25 mm H 35 mm No Preference High 333

340 VMDX001 Parameter Type Limits Desired Value Importance Length Input 250 mm L 350 mm No Preference High Volume Output n/a Minimum Possible Low Stress Output n/a Minimum Possible High Deflection Output n/a Minimum Possible High Analysis Beam volume: Maximum axial deformation under load F: Normal stress along Z-direction: Combined objective function becomes: Minimizing ϕ we get dimensions as: L = Length = m W = Width = m H = Height = m Results Comparison Results Target DesignXplorer Error (%) Volume (V) 6.9e-5 m 3 6.9E-05 m Deformation (D) e-5 m E-05 m Stress (σ) e7 Pa E-07 Pa

341 VMDX002: Optimization of Bar with Temperature-Dependent Conductivity Overview Reference: Analysis Type(s): Element Type(s): From the Basic Principle Goal Driven Optimization 3-D Solid Test Case A long bar 2 X 2 X 20 m is made up of material having thermal conductivity linearly varying with the temperature K = k 0 *(1 + a*t) W/m- C, k 0 = 0.038, a = The bar is constrained on all faces by frictionless support. A temperature of 100 C is applied at one end of the bar. The reference temperature is 5 C. At the other end, a constant convection coefficient of W/m 2 C is applied. The ambient temperature is 5 C. Input Parameters: Convection coefficient, coefficient of thermal expansion and length Response Parameters: Temperature (scoped on end face), thermal strain Figure 210: Schematic Material Properties E = 2e11 Pa ν = 0 α = 1.5E-05/ C K = k 0 *(1 + a*t) W/m- C k 0 = a = Geometric Properties Breadth B = 2 m Width W = 2 m Length L = 20 m Loading Frictionless Support (on all faces) Reference temperature = 5 C Temperature on end face T = 100 C Convection on other end face Convection coefficient h = 5e-3 W/m 2 C Ambient temperature Ta = 5 C 335

342 VMDX002 Parameter Type Limits Desired Value Importance Length (l) Input 15 m l 25 m No Preference Low Convection coefficient (h) Input W/m 2 C h W/m 2 C No Preference Low Coefficient of temperature expansion (α) Input 1.4e-5/ C 1.6e-5/ C α No Preference Low Temperature (T) Output n/a Minimum Possible High Thermal strain (ε) Output n/a Minimum Possible High Analysis Temperature: Thermal strain: Combined objective function becomes, Minimizing ϕ we get input parameters as: l = beam length = 25 m h = convection coefficient = W/m 2 C α = coefficient of thermal expansion = 1.4e-5/ C Results Comparison Results Target DesignXplorer Error (%) Length (l) 25 m 25 m 0 Convection coefficient (h) W/m 2 C W/m 2 C 0 Coefficient of thermal expansion (α) 1.4e-5/ C 1.4e-5/ C 0 Temperature (T) C C Thermal strain (ε) e-4 m/m 3.437e-4 m/m

343 VMDX003: Optimization of Water Tank Column for Mass and Natural Frequency Overview Reference: Analysis Type(s): Element Type(s): S. S. Rao, Optimization Theory and Application Second edition, example 1.10, page Goal Driven Optimization with APDL 3-D Solid Test Case A uniform column of rectangular cross section b and d m is to be constructed for supporting a water tank of mass M. It is required to: 1. minimize the mass of the column for economy 2. maximize the natural frequency of transverse vibration of the system for avoiding possible resonance due to wind. Design the column to avoid failure due to direct compression (should be less than maximum permissible compressive stress) and buckling (should be greater than direct compressive stress). Assume the maximum permissible compressive stress as σ max. The design vector is defined as: where: b = width of cross-section of column d = depth of cross-section of column Input Parameters: Width and Height Response Parameters: Mass, Natural Frequency, Direct Stress, Buckling Stress Material Properties Geometric Properties Loading E = 3e10 Pa Width, b = 0.4 m Mass of water tank M = Kg ρ = 2300 Depth, d =1.2 m Kg/m 3 Acceleration due to gravity = 9.81 Length, I = 20 m m/s 2 σ max = 4.1e7 Pa Sample Size:

344 VMDX003 Results Target DesignXplorer Error (%) Width b m m Depth d m m Mass of column M kg kg Natural frequency w rad/sec rad/sec Direct stress e7 Pa e7 Pa Buckling stress e6 Pa e6 Pa Analysis Minimize: Maximize: Subject to constraints: Required objective is obtained by having: b = m d = m M = (minimum) = kg W = (maximum) = rad/sec Direct stress = e7 Pa Buckling stress = e6 Pa Results Comparison Results Target DesignXplorer Error (%) Width b m m Depth d m m Mass of column M kg kg Natural frequency w rad/sec rad/sec Direct stress e7 Pa e7 Pa

345 VMDX003 Results Target DesignXplorer Error (%) Buckling stress e6 Pa e6 Pa

346 340

347 VMDX004: Optimization of Frequency for a Plate With Simple Support at all Vertices Overview Reference: Analysis Type(s): Element Type(s): Blevins, Formula for Natural Frequency and Mode Shape, Van Nostrand Reinhold Company Inc., 1979, pg Goal Driven Optimization 3-D Shell Test Case A square plate of side 250 mm and thickness 5 mm is simply supported on all its vertices. Input Parameters: Young's modulus, Poisson's ratio and density Response Parameters: First natural frequency Figure 211: Schematic Material Properties Geometric E = 2e5 MPa Properties ν = 0.3 Length a = 250 ρ = e-6 mm kg/mm 3 Width b = 250 mm Thickness h = 5 mm Loading All vertices are simply supported Parameter Type Constraints Desired Value Importance Young's Modulus E Input 1.8e11 Pa E 2.2e11Pa No Preference Low Poisson's Ratio μ Input 0.27 μ 0.30 No Preference Low 341

348 VMDX004 Parameter Type Constraints Desired Value Importance Density ρ Input 7065 kg/m 3 ρ 8635 kg/m 3 No Preference Low First Natural Frequency w Output N/a Minimum Possible High Analysis First Natural Frequency: Objective function becomes: Minimizing ϕ we get dimensions as: Young's Modulus E = 1.8e11 Pa Poisson's Ratio μ = 0.27 Density ρ = 8635 kg/m 3 First Natural Frequency w = rad/s Results Comparison Results Target DesignXplorer Error (%) Young's Modulus E 1.8e11 Pa 1.8e11 Pa 0.00 Poisson's Ratio μ Density ρ 8635 kg/m kg/m First Natural Frequency w rad/s rad/s

349 VMDX005: Optimization of Buckling Load Multiplier With CAD Parameters and Young's Modulus Overview Reference: Analysis Type(s): Element Type(s): Timoshenko, Strength of Materials, Part 2 (Advanced theory and problems), pg Goal Driven Optimization 3-D Solid Test Case The cantilever bar of length 25 feet is loaded by uniformly distributed axial force p = 11 lbf on one of the vertical face of the bar in negative Z-direction. The bar has a cross-sectional area A is ft 2. Input Parameters: Side of Square C/S, Length of Cantilever Bar and Young's Modulus Response Parameters: Optimization Method: Load Multiplier of the First Buckling Mode Genetic Algorithm Sample Size: 200 Figure 212: Schematic Material Properties E = e 9 psf ν = 0.3 ρ = lbm/ft 3 Geometric Properties Cross-section of square = 0.25 ft. x 0.25 ft. Length of bar = 25 ft. Loading Fixed support on one face, Force = 11 lbf (Negative Z-direction) on top face Parameter Type Constraints Desired Value Importance Cross-section side Input ft. a ft. No Preference N/A 343

350 VMDX005 Parameter Type Constraints Desired Value Importance Length Input 22.5 ft. l 27.5 ft. No Preference N/A Young's Modulus Input e9 psf E e9 psf No Preference N/A First buckling mode load multiplier Output N/A Maximum Possible N/A Analysis Assuming that under the action of uniform axial load a slight lateral bucking occurs. The expression for deflection is: The critical load is given by, where: q = force per unit length The first critical buckling load is: The load multiplier is given by the ratio of critical load to applied load. The first buckling multiplier is: Combined objective function becomes: Minimizing ϕ we get dimensions as: Cross-section side a = ft. Length l = 22.5 ft. Young's Modulus E = e9 psf Buckling load multiplier =

351 VMDX005 Results Comparison Results Target DesignXplorer Error (%) First buckling mode load multiplier

352 346

353 Part V: Explicit Dynamics Descriptions

354

355 EXDVM2: Propagation of Shock and Rarefaction Waves in a Shock Tube Overview Reference: Analysis Type(s): Elements: Boundary Conditions: Structural Interactions: Fluid-Structure Interactions: Bonds: Materials: Harlow, Francis H. et al., Fluid Dynamics A LASL Monograph, LA-4700, June Autodyn 3D Multi-material Euler Quad Euler Ideal Gas Quad Default Wall No No No Ideal Gas Test Case A 1000 mm (1000 element) shock tube is divided into two sections be a diaphragm at its midpoint. Initially, gas is at rest on both sides. To the left of the diaphragm, the gas is initially at a higher density and pressure than on the right. Both sides of the diaphragm have the same internal energy. At t = 0, the diaphragm is removed and the gas in the tube is observed. A shock is observed moving from left to right, as well as a a contact discontinuity moving from the left to the right, and a rarefaction wave moving from the right to the left. There is no significant length to the system. The appearance of the system at a later time is a magnification of an earlier appearance (similarity solution). Material Properties Heat capacity ratio, Density of gas on left of diaphragm, g/ml Density of gas on right of column, g/ml Pressure of gas on left of diaphragm, MPa Pressure of gas on right of column, MPa Internal energy, J/kg Geometric Properties Length of shock tube = 1000 mm Loading 349

356 EXDVM2 Analysis Assumptions and Modeling Notes Two identical 1000 mm shock tubes are modeled using 1000 elements for each tube. One tube uses the Multi-Material Euler solver and the other uses the Ideal Gas (Blast) solver. Because of a limitation on creating Euler inflow boundary conditions, two interacting parts are modeled, one with Euler inflow properties and the other with Ideal Gas properties. The calculation is run to t = 0.8 ms, the time at which the shock front nears the right boundary. Results Comparison Applying the theoretical solution to the initial conditions used for the simulation, the following values are obtained: Final pressure in tube, MPa Density of gas on left of diaphragm, g/ml Density of gas on right of diaphragm, g/ml Continuous pressure, MPa Comparing 28 profiles of pressure and density along the length of the shock tube at t = 0.8 ms show that these theoretical values are closely matched by the simulation. The results using the Ideal Gas (Blast) solver show steeper rise times for the shock and density discontinuity, reflecting the more accurate 2 nd order solution calculated by this solver. Figure 213: Pressure Along the Shock Tube at t = 0.8 ms 350

357 EXDVM2 Figure 214: Density of Gas Along the Shock Tube at t = 0.8 ms 351

358 352

359 EXDVM3: Flow of Gas Past an Infinite Two-Dimensional Wedge Overview Reference: Analysis Type(s): Elements: Boundary Conditions: Structural Interactions: Fluid-Structure Interactions: Bonds: Materials: Harlow, Francis H. et al., Fluid Dynamics A LASL Monograph, LA-4700, June Autodyn 2D, Planar Symmetry Multi-material Euler Quad Euler Ideal Gas Quad Euler Inflow, Euler Outflow No Yes, fully coupled No Ideal Gas Test Case Perform a dynamic flow analysis of a Mach 2 gas flowing past a 2-D wedge until a steady-state configuration is obtained. An attached shock is formed, the angle of which is determined by the flow conditions. Since the appearance of the configuration is independent of magnification, there is no significant length to the system. 353

360 EXDVM3 Figure 215: Model of Ideal Gas Flowing Past a Two-Dimensional Wedge Material Properties Heat capacity ratio, Density of gas, g/ml Pressure of gas, MPa Internal energy, J/kg Flow velocity, m/s Geometric Properties Angle of wedge = 25 Loading Analysis Assumptions and Modeling Notes Two 200 x 200 mm square regions are modeled using two 200 x 200 element Euler Parts. One part uses the Euler Multi-Material solver, the other part uses the Ideal Gas (Blast) solver. Single element fixed Lagrange parts are superimposed on top of each Euler part to impose a 25 degree wedge on the flow. 354

361 An inflow boundary condition is applied to the left boundary and an outflow boundary condition is applied to the right boundary of the Euler parts. The default wall boundary condition is maintained elsewhere. The calculation is run for 3000 cycles, by which time a steady-state condition is achieved. Results Comparison EXDVM3 The sound speed of the gas is given by m/s. The incoming flow Mach number is therefore. According to the reference document, the angle of the detached shock is approximately 49. Figure 216: Pressure of the Gas in Steady-State Condition 355

362 EXDVM3 Figure 217: Density of the Gas in Steady-State Condition 356

363 EXDVM4: Regular and Mach Deflections Off a Two-Dimensional Wedge Overview Reference: Analysis Type(s): Elements: Boundary Conditions: Structural Interactions: Fluid-Structure Interactions: Bonds: Materials: Harlow, Francis H. et al., Fluid Dynamics A LASL Monograph, LA-4700, June Autodyn 2D, Planar Symmetry Multi-material Euler Quad Euler Ideal Gas Quad Euler Inflow, Euler Outflow No Yes No Ideal Gas Test Case Perform a dynamic flow analysis of a Mach 1.1 shock reflection off a 45 wedge, and a Mach 2 shock reflection off a 25 wedge. The first analysis should produce a regular reflection, and the second a Mach reflection. Since the appearance of the configuration is independent of magnification, there is no significant length to the system. 357

364 EXDVM4 Figure 218: Mach 1.1 and Mach 2.0 Problem Sketches for the Ideal Gas Solver and Multi-Material Euler Solver Material Properties Initial Conditions Heat capacity ratio, Density of gas, g/ml Pressure of gas, MPa Internal energy, Mach 1.1 Flow J/kg Velocity of gas, m/s Pressure of gas, MPa Density of gas, g/ml Internal energy, J/kg Geometric Properties Mach 1.1 Flow Angle of wedge = 45 Mach 2.0 Flow Angle of wedge = 25 Loading 358

365 EXDVM4 Material Properties Mach 2.0 Flow Velocity of gas, m/s Pressure of gas, MPa Density of gas, g/ml Internal energy, J/kg Geometric Properties Loading Analysis Assumptions and Modeling Notes Four Euler parts are creating, two using the Euler Multi-Material solver and two using the Ideal Gas (Blast) solver. Single element fixed Lagrange parts are superimposed on top of each Euler part to impose a wedge on the flow. Two of the wedges have a 45 degree angle, and two have a 25 degree angle. An inflow boundary condition is applied to the left boundary of each Euler part. Mach 1.1 flow conditions are used for the parts containing the 45 wedge, and Mach 2.0 flow conditions are used for the parts containing the 25 wedge. The calculation is run for 0.85 ms, by which time the shock has progressed almost to the right edge of the Euler parts. Results Comparison Results shown below are from the end of the simulation (t = 0.85 ms). Both plots are from the same result, but have the pressure scales adjusted to show good resolution of the desired reflection. 359

366 EXDVM4 Figure 219: Regular Reflection Off the 45 Wedge 360

367 EXDVM4 Figure 220: Mach Reflection Off the 25 Wedge 361

368 362

369 EXDVM6: 3-D Taylor Cylinder Impact Overview Reference: Analysis Type(s): Elements: Boundary Conditions: Structural Interactions: Fluid-Structure Interactions: Bonds: Materials: No theoretical solution, experimental results and code-comparisons are available: Taylor, G.I., The Use of Flat Ended Projectiles for Determining Yield Stress, Part I: Theoretical Considerations, Proceedings of the Royal Society (London), Vol. 194, pp , 1948 Lacy, J.M., Novascone, S.R., Richins, W.D., and Larson, T.K., A Method for Selecting Software for Dynamic Event Analysis II: The Taylor Anvil and Dynamic Brazilian Tests, Proceedings of the 16 th International Conference on Nuclear Engineering, INL/CON , Idaho National Laboratory, 2008 Autodyn 3D 8-Node Linear Interpolated Reduced Integration Hex Fixed Constraint Yes, trajectory contact No No Copper Test Case The Taylor cylinder impact test uses a right circular cylinder of a test material which impacts a theoretically rigid target. In this test, an OHFC copper cylinder, cm in diameter and 2.54 cm in length, impacts a rigid plate at cm/s. 363

370 EXDVM6 Figure 221: Finite Element Model of Copper Taylor Cylinder with 1/4 Symmetry Material Properties Cylinder material = copper Geometric Properties Diameter of cylinder = cm Length of cylinder = 2.54 cm Loading Impact velocity = 190 m/s Analysis Assumptions and Modeling Notes Two parts are created to model the copper cylinder and the rigid plate. Quarter symmetry is used to reduce simulation time. Material data for copper is obtained from the Explicit Materials data source in Engineering Data. This data is the same as the material data used in the code comparison reference. A 0.2 mm element size is used to mesh the cylinder. The rigid plate is modeled with a single element. Trajectory contact is used to compute the impact of the cylinder on the plate. The initial velocity of the cylinder is 190 m/s and the simulation is run for 8 x 10-5 s. Results Comparison The final cylinder profile is similar to the profile shown for the Autodyn results shown in the code comparison reference. The cylinder radius in the impact plane agrees well with the experimentally obtained values, as well as with other simulation programs. The final cylindar length in this simulation is greater than the experimental value, but agrees well with the length computed by the other simulation programs using the same material model for copper. The plot below shows the final share of the deformed copper cylinder at the end of the simulation, compared to the original shape (semi-transparent). 364

371 EXDVM6 Figure 222: Final Deformed Shape of the Taylor Cylinder Figure 223: Profile of Final Deformed Shape of Taylor Cylinder With Scale 365

372 366

373 EXDVM7: 2-D Taylor Cylinder Impact Overview Reference: Analysis Type(s): Elements: Boundary Conditions: Structural Interactions: Fluid-Structure Interactions: Bonds: Materials: No theoretical solution, experimental results and code-comparisons are available: Taylor, G.I., The Use of Flat Ended Projectiles for Determining Yield Stress, Part I: Theoretical Considerations, Proceedings of the Royal Society (London), Vol. 194, pp , 1948 Lacy, J.M., Novascone, S.R., Richins, W.D., and Larson, T.K., A Method for Selecting Software for Dynamic Event Analysis II: The Taylor Anvil and Dynamic Brazilian Tests, Proceedings of the 16 th International Conference on Nuclear Engineering, INL/CON , Idaho National Laboratory, 2008 Autodyn 2D 4-Node Hex Fixed Constraint Yes, proximity-based contact No No Copper Test Case The Taylor cylinder impact test uses a right circular cylinder of a test material which impacts a theoretically rigid target. In this test, an OHFC copper cylinder, cm in diameter and 2.54 cm in length, impacts a rigid plate at cm/s. 367

374 EXDVM7 Figure 224: Finite Element Model of a Copper Taylor Cylinder with 2-D Axial Symmetry Material Properties Cylinder material = copper Geometric Properties Diameter of cylinder = cm Length of cylinder = 2.54 cm Loading Impact velocity = 190 m/s Analysis Assumptions and Modeling Notes Two parts are created to model the copper cylinder and the rigid plate. 2-D axial symmetry is used to reduce simulation time. Material data for copper is obtained from the Explicit Materials data source in Engineering Data. This data is the same as the material data used in the code comparison reference. A ~0.15 mm element size is used to mesh the cylinder. The rigid plate is modeled with a single element. Proximity-based contact is used to compute the impact of the cylinder on the plate. The initial velocity of the cylinder is 190 m/s and the simulation is run for 8 x 10-5 s. Results Comparison The final cylinder profile is similar to the profile shown for the Autodyn results shown in the code comparison reference. The cylinder radius in the impact plane agrees well with the experimentally obtained values, as well as with other simulation programs. The final cylindar length in this simulation is greater than the experimental value, but agrees well with the length computed by the other simulation programs using the same material model for copper. The plot below shows the final share of the deformed copper cylinder (rotated through 360 ) at the end of the simulation. 368

375 EXDVM7 Figure 225: Final Deformed Shape of the Taylor Cylinder (Rotated Through 360 ) Figure 226: Profile of Final Deformed Shape of Taylor Cylinder With Scale 369

376 370

377 Part VI: Aqwa Descriptions

378

379 AQVM1: Hydrostatic Evaluation of a Floating Inverted Pyramid Overview Reference: Analysis Type(s): Floating Structures: a guide for the design and analysis, Ed. N.D.P. Barltrop, Oilfield Publications Ltd., 1998 Hydrostatic Test Case An inverted pyramid is floating in static equilibrium in a large body of water with a density of 1025 kg/m 3. Calculate the hydrostatic properties for this model. Figure 227: Problem Sketch of a Floating Inverted Pyramid 373

380 AQVM1 Figure 228: Finite Element Mesh of a Floating Inverted Pyramid Material Properties Density of water = 1025 kg/m 3 Results Comparison Geometric Properties Center of gravity at Z = 0 Loading Acceleration due to gravity = 9.8 m/s 2 Waterplane area Displaced volume A w Target Aqwa x x 10 4 Center of B x buoyancy B y B z Center of LCF floatation TCF Waterplane x x inertia x x Ratio