ANSYS Workbench Verification ManualANSYS, Inc. Southpointe 275 Technology Drive Canonsburg, PA 15317
[email protected] http://www.ansys.com (T) 724-746-3304 (F) 724-514-9494 Release 14.5 October 2012 ANSYS, Inc. is certified to ISO 9001:2008. Copyright and Trademark Information © 2012 SAS IP, Inc. All rights reserved. Unauthorized use, distribution or duplication is prohibited. ANSYS, ANSYS Workbench, Ansoft, AUTODYN, EKM, Engineering Knowledge Manager, CFX, FLUENT, HFSS 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 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. Disclaimer Notice THIS ANSYS SOFTWARE PRODUCT AND PROGRAM DOCUMENTATION INCLUDE TRADE SECRETS AND ARE CONFIDENTIAL AND PROPRIETARY PRODUCTS OF ANSYS, INC., ITS SUBSIDIARIES, OR LICENSORS. 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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, please contact ANSYS, Inc. Published in the U.S.A. Table of Contents Introduction ............................................................................................................................................... 1 Overview ............................................................................................................................................... 1 Index of Test Cases .................................................................................................................................. 2 I. DesignModeler Descriptions ................................................................................................................... 7 1. VMDM001: Extrude, Chamfer, and Blend Features ................................................................................. 9 2. VMDM002: Cylinder using Revolve, Sweep, Extrude, and Skin-Loft ....................................................... 11 3. VMDM003: Extrude, Revolve, Skin-Loft, and Sweep .............................................................................. 13 II. Mechanical Application Descriptions ................................................................................................... 15 1. VMMECH001: Statically Indeterminate Reaction Force Analysis ........................................................... 17 2. VMMECH002: Rectangular Plate with Circular Hole Subjected to Tensile Loading ................................. 19 3. VMMECH003: Modal Analysis of Annular Plate .................................................................................... 21 4. VMMECH004: Viscoplastic Analysis of a Body (Shear Deformation) ...................................................... 23 5. VMMECH005: Heat Transfer in a Composite Wall ................................................................................. 25 6. VMMECH006: Heater with Nonlinear Conductivity .............................................................................. 27 7. VMMECH007: Thermal Stress in a Bar with Temperature Dependent Conductivity ................................ 29 8. VMMECH008: Heat Transfer from a Cooling Spine ............................................................................... 31 9. VMMECH009: Stress Tool for Long Bar with Compressive Load ............................................................ 33 10. VMMECH010: Modal Analysis of a Rectangular Plate ......................................................................... 35 11. VMMECH011: Large Deflection of a Circular Plate with Uniform Pressure ........................................... 37 12. VMMECH012: Buckling of a Stepped Rod .......................................................................................... 39 13. VMMECH013: Buckling of a Circular Arch .......................................................................................... 41 14. VMMECH014: Harmonic Response of a Single Degree of Freedom System ......................................... 43 15. VMMECH015: Harmonic Response of Two Storied Building under Transverse Loading ........................ 45 16. VMMECH016: Fatigue Tool with Non-Proportional Loading for Normal Stress .................................... 47 17. VMMECH017: Thermal Stress Analysis with Remote Force and Thermal Loading ................................ 49 18. VMMECH018: A Bar Subjected to Tensile Load with Inertia Relief ....................................................... 51 19. VMMECH019: Mixed Model Subjected to Bending Loads with Solution Combination ......................... 53 20. VMMECH020: Modal Analysis for Beams ........................................................................................... 55 21. VMMECH021: Buckling Analysis of Beams ......................................................................................... 57 22. VMMECH022: Structural Analysis with Advanced Contact Options ..................................................... 59 23. VMMECH023: Curved Beam Assembly with Multiple Loads ............................................................... 61 24. VMMECH024: Harmonic Response of a Single Degree of Freedom System for Beams ......................... 63 25. VMMECH025: Stresses Due to Shrink Fit Between Two Cylinders ........................................................ 65 26. VMMECH026: Fatigue Analysis of a Rectangular Plate Subjected to Edge Moment ............................. 67 27. VMMECH027: Thermal Analysis for Shells with Heat Flow and Given Temperature .............................. 69 28. VMMECH028: Bolt Pretension Load Applied on a Semi-Cylindrical Face ............................................. 71 29. VMMECH029: Elasto-Plastic Analysis of a Rectangular Beam .............................................................. 73 30. VMMECH030: Bending of Long Plate Subjected to Moment - Plane Strain Model ............................... 75 31. VMMECH031: Long Bar with Uniform Force and Stress Tool - Plane Stress Model ................................ 77 32. VMMECH032: Radial Flow due to Internal Heat Generation in a Copper Disk - Axisymmetric Model .... 79 33. VMMECH033: Electromagnetic Analysis of a C-Shaped Magnet ......................................................... 81 34. VMMECH034: Rubber cylinder pressed between two plates .............................................................. 85 35. VMMECH035: Thermal Stress in a Bar with Radiation ........................................................................ 87 36. VMMECH036: Thermal Stress Analysis of a Rotating Bar using Temperature Dependant Density ......... 89 37. VMMECH037: Cooling of a Spherical Body ........................................................................................ 91 38. VMMECH038: Crashing Blocks Simulation with Transient Structural Analysis ...................................... 93 39. VMMECH039: Transient Response of a Spring-mass System ............................................................... 95 40. VMMECH040: Deflection of Beam using Symmetry and Anti-Symmetry ............................................. 97 41. VMMECH041: Brooks Coil with Winding for Periodic Symmetry ......................................................... 99 Release 14.5 - © SAS IP, Inc. All rights reserved. - Contains proprietary and confidential information of ANSYS, Inc. and its subsidiaries and affiliates. iii Verification Manual 42. VMMECH042: Hydrostatic Pressure Applied on a Square Bar with Fully, Partially Submerged in a Fluid ....................................................................................................................................................... 103 43. VMMECH043: Fundamental Frequency of a Simply-Supported Beam .............................................. 105 44. VMMECH044: Thermally Loaded Support Structure ......................................................................... 107 45. VMMECH045: Laterally Loaded Tapered Support Structure .............................................................. 109 46. VMMECH046: Pinched Cylinder ...................................................................................................... 111 47. VMMECH047: Plastic Compression of a Pipe Assembly .................................................................... 113 48. VMMECH048: Bending of a Tee-Shaped Beam ................................................................................. 115 49. VMMECH049: Combined Bending and Torsion of Beam ................................................................... 117 50. VMMECH050: Cylindrical Shell under Pressure ................................................................................. 119 51. VMMECH051: Bending of a Circular Plate Using Axisymmetric Elements .......................................... 121 52. VMMECH052: Velocity of Pistons for Trunnion Mechanism ............................................................... 123 53. VMMECH053: Simple Pendulum with SHM motion .......................................................................... 125 54. VMMECH054: Spinning Single Pendulum ........................................................................................ 127 55. VMMECH055: Projector mechanism- finding the acceleration of a point .......................................... 131 56. VMMECH056: Coriolis component of acceleration-Rotary engine problem ....................................... 133 57. VMMECH057: Calculation of velocity of slider and force by collar ..................................................... 135 58. VMMECH058: Reverse four bar linkage mechanism ......................................................................... 137 59. VMMECH059: Bending of a solid beam (Plane elements) ................................................................. 139 60. VMMECH060: Crank Slot joint simulation with flexible dynamic analysis .......................................... 141 61. VMMECH061: Out-of-plane bending of a curved bar ....................................................................... 145 62. VMMECH062: Stresses in a long cylinder ......................................................................................... 147 63. VMMECH063: Large deflection of a cantilever ................................................................................. 149 64. VMMECH064: Small deflection of a Belleville Spring ........................................................................ 151 65. VMMECH065: Thermal Expansion to Close a Gap at a Rigid Surface .................................................. 153 66. VMMECH066: Bending of a Tapered Plate ........................................................................................ 155 67. VMMECH067: Elongation of a Solid Tapered Bar .............................................................................. 157 68. VMMECH068: Plastic Loading of a Thick Walled Cylinder .................................................................. 159 69. VMMECH069: Barrel Vault Roof Under Self Weight ........................................................................... 161 70. VMMECH070: Hyperelastic Thick Cylinder Under Internal Pressure ................................................... 163 71. VMMECH071: Centerline Temperature of a Heat Generating Wire .................................................... 165 72. VMMECH072: Thermal Stresses in a Long Cylinder ........................................................................... 167 73. VMMECH073: Modal Analysis of a Cyclic Symmetric Annular Plate ................................................... 169 74. VMMECH074: Tension/Compression Only Springs ........................................................................... 171 75. VMMECH075: Harmonic Response of Two-Story Building under Transverse Loading ........................ 173 76. VMMECH076: Elongation of a Tapered Shell with Variable Thickness ............................................... 175 77. VMMECH077: Heat Transfer in a Bar with Variable Sheet Thickness .................................................. 177 78. VMMECH078: Gasket Material Under Uniaxial Compression Loading-3-D Analysis ........................... 179 79. VMMECH079: Natural Frequency of a Motor-Generator ................................................................... 183 80. VMMECH080: Transient Response of a Spring-mass System ............................................................. 185 81. VMMECH081: Statically Indeterminate Reaction Force Analysis ........................................................ 187 82. VMMECH082: Fracture Mechanics Stress for a Crack in a Plate .......................................................... 191 83. VMMECH083: Transient Response to a Step Excitation ..................................................................... 193 84. VMMECH084: Mullins Effect on a Rubber Tube Model Subjected to Tension Loading ........................ 197 85. VMMECH085: Bending of a Composite Beam .................................................................................. 199 III. Design Exploration Descriptions ....................................................................................................... 201 1. VMDX001: Optimization of L-Shaped Cantilever Beam under Axial Load ............................................ 203 2. VMDX002: Optimization of Bar with Temperature-Dependent Conductivity ....................................... 205 3. VMDX003: Optimization of Water Tank Column for Mass and Natural Frequency ................................ 207 4. VMDX004: Optimize frequency of plate with simply supported at all it's vertices ............................... 211 5. VMDX005: Optimization of buckling load multiplier with CAD parameters and Young's modulus ....... 213 iv Release 14.5 - © SAS IP, Inc. All rights reserved. - Contains proprietary and confidential information of ANSYS, Inc. and its subsidiaries and affiliates. The solutions for the test cases have been verified. however. or Design Exploration. The Educational version of Workbench should be able to solve most of these tests. 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. and output results can easily be changed and the solution repeated. Inc. Slightly different results may be obtained when different processor types or operating systems are used. Release 14. Many of them are classical engineering problems. The test results should be checked against the verified results in the documentation for each test. License limitations are not applicable to Workbench Education version but problem size may restrict the solution of some of the tests.© SAS IP. . The results in the databases can be cleared and the tests solved multiple times. an error rate of 3% or less has been the goal. material files. and its subsidiaries and affiliates. such as DesignModeler. Inc. The workbench analyses employ a balance between accuracy and solution time. Inc. The available tests are engineering problems that provide independent verification. All rights reserved.Contains proprietary and confidential information of ANSYS. For the tests. the tests offer a quick introduction to new features with which you may be unfamiliar. material properties. These results are reported in the test documentation. As a result. These zipped archives provide all of the necessary elements for running a test. You can use these tests to verify that your hardware is executing the ANSYS Workbench tests correctly. you may not be able to reproduce the results. including geometry parts. You are encouraged to use these tests as starting points when exploring new Workbench features. and workbench databases. 1 . These tests were run on an Intel Xeon processor using Microsoft Windows XP Professional. Download the ANSYS Workbench Verification Manual Archive Files. If you do not have the available licenses. The archive files for each of the Verification Manual tests are available at the Customer Portal. locate the archive and import it into Workbench. Inc. loads. 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. These differences have been examined and are considered acceptable.Introduction The following topics are discussed in this chapter: Overview 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. Geometries. certain differences may exist with regard to the references. Emag. Some test cases will require different licenses. usually a closed form equation.5 . offers the Workbench Verification and Validation package for users that must perform system validation. To open a test case in Workbench. Improved results can be obtained in some cases by employing a more refined finite element mesh but requires longer solution times. ANSYS. © SAS IP. Inc. Quality Assurance Group. . Large Deformation 2 Release 14. All rights reserved. If you are interested in contracting for such services contact the ANSYS. Inertia relief Linear Solution Options Linear Linear Free Vibration Nonlinear. Plastic Materials Linear Fatigue Linear Thermal Stress Nonlinear.Introduction This package automates the process of test execution and report generation. Contact Linear Analysis Type Static Structural Static Structural Modal Structural Static Thermal Static Thermal Static Structural Static Thermal Static Structural Modal Static Structural Buckling Buckling Harmonic Harmonic Static Structural Static Structural Static Structural Static Structural Fatigue Linear Thermal Stress Linear. Inc.Contains proprietary and confidential information of ANSYS. Index of Test Cases Test Case Number VMMECH001 VMMECH002 VMMECH003 VMMECH004 VMMECH005 VMMECH006 VMMECH007 VMMECH008 VMMECH009 VMMECH010 VMMECH011 VMMECH012 VMMECH013 VMMECH014 VMMECH015 VMMECH016 VMMECH017 VMMECH018 VMMECH019 Element Type Solid Solid Solid Solid Solid Solid Solid Solid Solid Shell Shell Solid Shell Solid Solid Solid Solid Solid Beam Shell VMMECH020 VMMECH021 VMMECH022 VMMECH023 VMMECH024 VMMECH025 VMMECH026 VMMECH027 VMMECH028 VMMECH029 Beam Beam Solid Beam Beam Solid Shell Shell Solid Solid Modal Buckling Static Structural Static Structural Harmonic Static Structural Static Structural Static Structural Static Structural Static Structural Nonlinear.5 . Viscoplastic Materials Linear Nonlinear Nonlinear Thermal Stress Linear Linear Free Vibration Nonlinear. Inc. . and its subsidiaries and affiliates. . and its subsidiaries and affiliates.Contains proprietary and confidential information of ANSYS. Axisymmetric Solid Solid Spring VMMECH040 VMMECH041 VMMECH042 VMMECH043 VMMECH044 VMMECH045 VMMECH046 VMMECH047 VMMECH048 VMMECH049 VMMECH050 VMMECH051 VMMECH052 VMMECH042 VMMECH054 VMMECH055 Beam Solid Solid Beam Beam Shell Shell 2-D Solid.© SAS IP.Index of Test Cases Test Case Number VMMECH030 VMMECH031 VMMECH032 VMMECH033 VMMECH034 VMMECH035 Element Type 2-D Solid. Plastic Materials Linear Thermal Stress Electromagnetic Hydrostatic Fluid Release 14. 3 . Axisymmetric Solid Solid Solid Analysis Type Static Structural Static Structural Static Structural Static Structural Static Structural Coupled (Static Thermal and Static Stress) Static Structural Transient Thermal Transient Structural Transient Structural Flexible Dynamic Flexible Dynamic Sequence Loading Linear Thermal Stress Electromagnetic Nonlinear. Inc. Inc. Axisymmetric Beam Beam Axisymmetric Shell Axisymmetric Shell Multipoint Constraint Multipoint Constraint Multipoint Constraint Multipoint Constraint Static Structural Static Structural Static Structural Modal Static Structural Static Structural Static Structural Static Structural Static Structural Static Structural Static Structural Static Structural Rigid Dynamic Rigid Dynamic Rigid Dynamic Rigid Dynamic Nonlinear. All rights reserved. Plane Stress 2-D Solid. Large Deformation Solution Options VMMECH036 VMMECH037 VMMECH038 VMMECH039 Solid 2-D Solid. Plane Strain 2-D Solid.5 . Contains proprietary and confidential information of ANSYS. Plastic Materials Modal Transient Dynamic Mode Superposition 4 .Introduction Test Case Number VMMECH056 VMMECH057 VMMECH058 VMMECH059 VMMECH060 Element Type Multipoint Constraint Multipoint Constraint Multipoint Constraint 2-D Plane Stress Shell Solid Multipoint Constraint VMMECH061 VMMECH062 VMMECH063 VMMECH064 VMMECH065 Beam Axisymmetric Shell Shell Shell Solid Shell VMMECH066 VMMECH067 VMMECH068 VMMECH069 VMMECH070 VMMECH071 VMMECH072 VMMECH073 VMMECH074 Shell Solid 2-D Solid. Large Deformation Static Structural Static Structural Static Structural Static Structural Static Structural Static Thermal Static Structural Modal Rigid Body Dynamics Harmonic Static Structural Static Thermal Static Structural Linear Thermal Stress Nonlinear. Plane Strain Shell 2-D Solid 2-D Thermal Solid 2-D Thermal Solid Solid Solid Spring VMMECH075 VMMECH076 VMMECH077 VMMECH078 Solid Shell Thermal Shell 3-D Solid 3-D Gasket VMMECH079 VMMECH080 Pipe Spring Mass Release 14. .© SAS IP. All rights reserved. Inc. Inc. Analysis Type Rigid Dynamic Rigid Dynamic Rigid Dynamic Static Structural Transient Structural Solution Options Flexible Dynamic Static Structural Static Structural Static Structural Static Structural Static Structural Linear Thermal Stress Nonlinear. Large Deformation Nonlinear. and its subsidiaries and affiliates.5 . and its subsidiaries and affiliates. . Inc. All rights reserved. Inc.Index of Test Cases Test Case Number VMMECH081 Pipe Mass VMMECH082 VMMECH083 VMMECH084 VMMECH085 Solid Spring. 5 .© SAS IP. Mass Solid Solid Element Type Analysis Type Modal Spectral Static Structural Transient Dynamic Static Structural Static Structural Fracture Mechanics Mode Superposition Nonlinear.Contains proprietary and confidential information of ANSYS. Hypereleastic Composite Material Solution Options Release 14.5 . Inc.6 Release 14.© SAS IP. and its subsidiaries and affiliates. .Contains proprietary and confidential information of ANSYS. .5 . Inc. All rights reserved. Part I: DesignModeler Descriptions . . 5 mm3. and Blend Millimeter Test Case Create a Model using Extrude.05 – 3034. Two rectangular areas (25 mm x 40 mm) are extruded 40 mm and subtracted from solid.VMDM001: Extrude. Inc.5 = 3034. Four Oval areas are extruded and subtracted from Solid. Net volume V = 285738. Chamfer. Net Volume = V = v1 + v2 = 285738. A rectangular area (24 mm x 5 mm) is subtracted from the solid. Chamfer. A polygonal area is extruded 60 mm. 3. Chamfer. . and Blend Features Overview Feature: Drawing Units: Extrude.5 . and Blend features. Volume of Solid after extruding Polygonal Area: v1 = 264000 mm3. Release 14.05 mm3. Volume of rectangular (24mm x 5mm) solid extruded 30mm using Cut Material = 3600 – 565. A Chamfer (10 mm x 10 mm) is given to 4 edges on the resultant solid.05 mm3. Figure 1: Final Model after creating Extrude. All rights reserved. Verify Volume of the resultant geometry. Both resultant solids form one solid geometry. 2.Contains proprietary and confidential information of ANSYS. 4. Inc. A rectangular area of 30 mm x 40 mm [having a circular area of radius 6 mm subtracted] is extruded to 20 mm. and Blend Calculations 1. Chamfer. Two rectangular areas (40 mm x 10 mm) are extruded 10 mm and subtracted from solid. Volume of two rectangular areas each 40mm x 10mm extruded 10mm = 8000 mm3. and its subsidiaries and affiliates. Volume of rectangular area having circular hole: v2 = 21738.5 = 282703.© SAS IP. Fillet (Radius 5 mm) is given to 4 edges using Blend Feature.5 mm3. 9 . 5 mm3. Net volume V = 196703. Volume of two rectangular areas 25mm x 40mm extruded 40mm = 80000 mm3.2 mm3.5 mm3.VMDM001 Net volume V = 282703.7141. 5.7 44261. Volume of 4 solids subtracted due to Blend of radius 5 mm = 429.2 = 189132. Volume of four oval areas extruded 10 mm = 7141.5 + 2000 = 196703. 7.7 44261. and its subsidiaries and affiliates.6 52 1 Error (%) 0 0. Hence Net volume of final Solid body = V = 189561. All rights reserved.5 – 8000 = 274703.7 mm3.5 – 80000 = 194703.001 0 0 10 Release 14.6 mm3.5 . Inc. Volume of four solids added due to Chamfer = 4 x 500 = 2000 mm3 Net volume V = 194703.Contains proprietary and confidential information of ANSYS. 8. .29 52 1 DesignModeler 189132.5 mm3.© SAS IP.6 = 189561.9 mm3. . 6. Net volume V = 274703.9 – 429.5 . Inc. Results Comparison Results Volume (mm3) Surface Area (mm ) Number of Faces Number of Bodies 2 Target 189132. All rights reserved. Volume of Cylinder created when a circular area (Radius 30mm) is swept 100 mm = v2 = 282743. 4. and Skin-Loft Overview Feature: Drawing Units: Revolve. Inc. A circular area of radius 30 mm is extruded 100 mm. Release 14. Extrude. = 282743.3 mm3.6 + 282743.6 mm3. Net Volume of the final Cylinder = 848229.3 mm3.3 mm3. A solid cylinder is created using Skin-Loft feature between two coaxial circular areas each of radius 30 mm and 100 mm apart. 11 . Verify Volume of the resultant geometry.3 = 565486.3 + 282743.5 .2 mm3. and Skin-Loft Millimeter Test Case Create a Model using Revolve. and Skin-Loft Calculations 1. Volume of Cylinder created after using Skin-Loft feature between two circular areas of Radius 30 mm and 100 mm apart. Sweep. Volume of Cylinder after extruding a circular area (Radius 30 mm) 100 mm = 282743. Net Volume = V = 565486. 2. Sweep. .9 mm3. A Rectangular area (100 mm x 30 mm) is revolved about Z-Axis in 3600 to form a Cylinder. and Skin-Loft features.VMDM002: Cylinder using Revolve. Extrude. Figure 2: Final Model after creating Revolve. Volume of Cylinder created after Revolving Rectangular area (100 mm x 30 mm) = v1 = 282743.3 mm3. Sweep.Contains proprietary and confidential information of ANSYS. Extrude.9 + 282743. Inc.© SAS IP. Net Volume = V = v1 + v2 = 282743.3 = 1130973.3 = 848229. Extrude. Sweep. 3. and its subsidiaries and affiliates. A circular area of radius 30 mm is swept 100 mm using Sweep feature. 1 3 1 DesignModeler 1130973.© SAS IP.VMDM002 Results Comparison Results Volume (mm3) Surface Area (mm ) Number of Faces Number of Bodies 2 Target 1130973. All rights reserved. .3 81053. Inc.5 .Contains proprietary and confidential information of ANSYS. Inc.3 81053. and its subsidiaries and affiliates.1 3 1 Error (%) 0 0 0 0 12 Release 14. . Contains proprietary and confidential information of ANSYS. All rights reserved. Va = 906400 . Verify the volume of the resulting geometry.3 mm3. Figure 3: Final Model after creating Extrude. Thus a total of 4 geometries are created. Volume of additional body created due to Revolve feature = Vb= 11134. A rectangular area (103 mm x 88 mm) is extruded 100 mm to form a solid box. Skin-Loft.3 mm3. and Sweep Millimeter Test Case Create a Model using Extrude. Net Volume of solid box becomes Va = 876760. 5. Volume of rectangular (103 mm x 88 mm) solid extruded 100mm = 906400 mm3. The two solid bodies are frozen using Freeze feature.© SAS IP.29639.VMDM003: Extrude. Revolve. Volume of additional two bodies created due to Sweep feature: Release 14. 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.6 mm3. Skin-Loft and Sweep Calculations 1.5 . 2. Net Volume of solid box. 3. 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. Revolve. Skin-Loft.6 = 876760.15 mm3. Inc. . and its subsidiaries and affiliates. Volume of solid after revolving circular area of Radius 25 mm through 900 = 29639.3 – 62500 = 814260. 4. Revolve. 13 . and Sweep Overview Feature: Drawing Units: Extrude. Revolve. Volume of the rectangular box cut after Skin-Loft between two square areas each of side 25 mm = 62500 mm3. Skin-Loft. A solid is subtracted using Skin-Loft feature between two square areas (each of side 25 mm) and 100 mm apart. Inc. and Sweep. All rights reserved.VMDM003 • Vc = 47123.6 mm3.9 +28352.8 = 825394. and its subsidiaries and affiliates. Inc.47 22 4 DesignModeler 825394.8 mm3.7 mm3. • Hence Net volume of box.3 . Results Comparison Results Volume (mm3) Surface Area (mm ) Number of Faces Number of Bodies 2 Target 825394. .5 101719.© SAS IP.Contains proprietary and confidential information of ANSYS.15 + 47123.4 101719.75476. Va = 814260. .6 + 11134.4 mm3.5 . • And total volume that gets subtracted from box due to Sweep Feature = 75476.7 = 738783.95 22 4 Error (%) 0 0 0 0 14 Release 14. Inc.9 mm3 and Vd = 28352. • Sum of volumes of all four bodies = Va+Vb+Vc+Vd = 738783. Part II: Mechanical Application Descriptions . VMMECH001: Statically Indeterminate Reaction Force Analysis Overview Reference: S. Timoshenko, Strength of Materials, Part 1, Elementary Theory and Problems, 3rd Edition, CBS Publishers and Distributors, pg. 22 and 26 Linear Static Structural Analysis Solid Analysis Type(s): Element Type(s): Test Case An assembly of three prismatic bars is supported at both end faces and is axially loaded with forces F1 and F2. Force F1 is applied on the face between Parts 2 and 3 and F2 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. Figure 4: Schematic Material Properties E = 2.9008e7 psi ν = 0.3 ρ = 0.28383 lbm/in3 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 F1 = -1000 (Y direction) Force F2 = -500 (Y direction) Release 14.5 - © SAS IP, Inc. All rights reserved. - Contains proprietary and confidential information of ANSYS, Inc. and its subsidiaries and affiliates. 17 VMMECH001 Results Comparison Results Y Reaction Force at Top Fixed Support (lbf ) Y Reaction Force at Bottom Fixed Support (lbf ) Target 900 600 Mechanical 901.14 598.86 Error (%) 0.127 -0.190 18 Release 14.5 - © SAS IP, Inc. All rights reserved. - Contains proprietary and confidential information of ANSYS, Inc. and its subsidiaries and affiliates. VMMECH002: Rectangular Plate with Circular Hole Subjected to Tensile Loading Overview Reference: Analysis Type(s): Element Type(s): J. E. Shigley, Mechanical Engineering Design, McGraw-Hill, 1st Edition, 1986, Table A-23, Figure A-23-1, pg. 673 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. Figure 5: Schematic 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 Results Comparison Results Maximum Normal X Stress (Pa) Target 312.5 Mechanical 313.24 Error (%) 0.24 Release 14.5 - © SAS IP, Inc. All rights reserved. - Contains proprietary and confidential information of ANSYS, Inc. and its subsidiaries and affiliates. 19 Contains proprietary and confidential information of ANSYS. Inc.5 . . and its subsidiaries and affiliates. Inc. All rights reserved.© SAS IP.20 Release 14. . 28383 lbm/in3 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" Loading Release 14. Inc. Sizing control with element size of 0. and its subsidiaries and affiliates.© SAS IP.9008e7 psi ν = 0.5” is applied to the cylindrical surface of the hole. Van Nostrand Reinhold Company Inc. . Blevins. 21 .5 .VMMECH003: Modal Analysis of Annular Plate Overview Reference: R. pg. Case 4. 247 Free Vibration Analysis Solid Analysis Type(s): Element Type(s): Test Case An assembly of three annular plates has cylindrical support (fixed in the radial. Inc. 1979. J. Formula for Natural Frequency and Mode Shape. tangential. Find the first six modes of natural frequencies. Table 11-2. Figure 6: Schematic Material Properties E = 2. and axial directions) applied on the cylindrical surface of the hole.3 ρ = 0. All rights reserved.Contains proprietary and confidential information of ANSYS.. All rights reserved. .3 -1.63 346.4 -1.09 437.2 22 Release 14.086 318.2 -0.19 315.086 351.569 442. Inc.569 351. and its subsidiaries and affiliates.911 318.451 Mechanical 310. Inc.72 347.VMMECH003 Material Properties Geometric Properties Thickness of all plates = 1" Loading Results Comparison Results 1st Frequency Mode (Hz) 2nd Frequency Mode (Hz) 3rd Frequency Mode (Hz) 4th Frequency Mode (Hz) 5th Frequency Mode (Hz) 6th Frequency Mode (Hz) Target 310.5 .Contains proprietary and confidential information of ANSYS.05 Error (%) -0.8 -1.58 315. .8 -0.© SAS IP. 23 . The temperature of the body is maintained at 400°C. "An Implicit Stress Update Algorithm Using a Plastic Predictor". Nonlinear Structural Analysis Solid Analysis Type(s): Element Type(s): 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 0.6 GPa (Poisson's Ratio) = 0. Find the shear load (Fx) required to maintain the deformation rate of 0.01 cm/sec at time equal to 20 seconds. All rights reserved.7 MPa Q/R = 21.5 . Eggert. M. January 1991.01 cm/s.08999E3 K A = 1.01 cm/s h x h Problem Model Material Properties Ex (Young's Modulus) = 60. Inc.VMMECH004: Viscoplastic Analysis of a Body (Shear Deformation) Overview Reference: B. Figure 7: Schematic y Velocity = 0. and its subsidiaries and affiliates. Submitted to Computer Methods in Applied Mechanics and Engineering.4999 So = 29. .01 cm/sec @ y = 1 cm Time = 20 sec Release 14. Lwo and G.Contains proprietary and confidential information of ANSYS.© SAS IP.91E7 s-1 ¡ = 7.0 Geometric Properties h = 1 cm thickness = 1 cm Loading Temp = 400°C = 673°K Velocity (x-direction) = 0. Inc. 5 . and its subsidiaries and affiliates. Inc. Inc.00 Mechanical -791.6 MPa Geometric Properties Loading = 18.© SAS IP. N Target 845.23348 ho = 1115.92 MPa ¡ = 0. . . All rights reserved.3 Results Comparison Results Fx.3 24 Release 14.07049 a = 1.VMMECH004 Material Properties m = 0.Contains proprietary and confidential information of ANSYS.76 Error (%) -6. 556e-4 BTU/s ft2°F (ha). All rights reserved. The temperature inside the furnace is 3000°F (Tf) and the inner surface convection coefficient is 3. 39 Linear Static Thermal Analysis Solid Test Case A furnace wall consists of two layers: fire brick and insulating brick. Inc.68 2957.5 . Find the Temperature Distribution.Contains proprietary and confidential information of ANSYS.205 0. The ambient temperature is 80°F (Ta) and the outer surface convection coefficient is 5.778e-5 BTU/s ft °F Geometric Properties Cross-section = 1" x 1" Fire brick wall thickness = 9" Insulating wall thickness = 5" Loading Results Comparison Results Minimum Temperature (°F) Maximum Temperature (°F) Target 336 2957 Mechanical 336. Harper and Row Publisher. . 1976.2 Error (%) 0. Principles of Heat Transfer. Figure 8: Schematic Material Properties Fire brick wall: k = 2.333e-3 BTU/s ft2°F (hf).007 Release 14. Inc.222e-4 BTU/s ft °F Insulating wall: k = 2.© SAS IP. Kreith. and its subsidiaries and affiliates. 25 . 3rd Edition.VMMECH005: Heat Transfer in a Composite Wall Overview Reference: Analysis Type(s): Element Type(s): F. Example 2-5. pg. All rights reserved.© SAS IP. Inc.26 Release 14.5 . . and its subsidiaries and affiliates. Inc.Contains proprietary and confidential information of ANSYS. . The internal energy generated electrically may be assumed to be uniform and is applied as internal heat generation. Figure 9: Schematic Material Properties k = [0. 27 . 1966. Arpaci. The boiling temperature of the liquid is 212°F.003 Release 14. 130 Nonlinear Static Thermal Analysis Solid Test Case A liquid is boiled using the front face of a flat electric heater plate. . All rights reserved. Find the maximum temperature and maximum total heat flux. Inc.75e-002 Geometric Properties Radius = 3. and its subsidiaries and affiliates. Inc.58 9.© SAS IP. Addison-Wesley Book Series.96 -0. Conduction Heat Transfer.001 T)] BTU/s in°F Temperature (°F) 32 1000 Conductivity (BTU/s in°F) 1.419e-002 2. pg.937” Thickness = 1” Loading Front face temperature = 212°F Internal heat generation = 10 BTU/s in3 Results Comparison Results Maximum Temperature (°F) Maximum Total Heat Flux (BTU/s in2) Target 476 10 Mechanical 480. The rear face of the heater is insulated.01375 * (1 + 0.Contains proprietary and confidential information of ANSYS.5 .VMMECH006: Heater with Nonlinear Conductivity Overview Reference: Analysis Type(s): Element Type(s): Vedat S.9997 Error (%) 0. Inc. Inc. and its subsidiaries and affiliates.© SAS IP.Contains proprietary and confidential information of ANSYS. . All rights reserved.28 Release 14.5 . . 91e-002 0. 29 . All rights reserved.5 . The reference temperature is 5°C. At the other end. A temperature of T°C is applied at one end of the bar (End A). and its subsidiaries and affiliates.038*(1 + 0. Advanced mesh control with element size of 2 m is applied.Contains proprietary and confidential information of ANSYS. Inc.215 Geometric Properties Length = 20 m Width = 2 m Breadth = 2 m Loading Rear face temperature T = 100°C Film Coefficient h = 0. a constant convection of h W/m2°C is applied.00582*T) W/m °C Temperature (°C) 5 800 Conductivity (W/m °C) 3. The bar is constrained at both ends by frictionless surfaces. Inc.VMMECH007:Thermal Stress in a Bar with Temperature Dependent Conductivity Overview Reference: Analysis Type(s): Element Type(s): Any basic Heat Transfer book Nonlinear Thermal Stress Analysis Solid Test Case A long bar has thermal conductivity that varies with temperature.© SAS IP. The ambient temperature is 5°C.005 W/m2°C Ambient temperature = 5°C Reference temperature = 5°C Release 14. 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 Figure 10: Schematic Material Properties E = 2e11 Pa ν=0 α = 1. .5e-05 / °C k = 0. 5 .000 0.000495 0. All rights reserved. Inc.© SAS IP.02 0. . Inc.016 0.165 Mechanical 38.VMMECH007 Analysis Temperature at a distance "z" from rear face is given by: z =− + − Thermal strain in the z direction in the bar is given by: ε T = × −5 × − Deformation in the z direction is given by: ¡ =∫ ¡ × −¢ × 0 ¡ − Heat flux in the z direction is given by: = × £ − Results Comparison 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 38.Contains proprietary and confidential information of ANSYS.905 0.00232 0.00049521 0.001425 0. and its subsidiaries and affiliates.002341 0.014 0.042 30 Release 14.16507 Error (%) -0.042 0. .001425 0. Equation 2-44a.055 -0.3614e-3 Error (%) 0. 31 .025'. Harper and Row.364e-3 Mechanical 79. F. 59.VMMECH008: Heat Transfer from a Cooling Spine Overview Reference: Analysis Type(s): Element Type(s): Kreith.2” L = 8” Loading Loading Tw = 100°F Ta = 0°F h = 2. Equation 2–45.© SAS IP. . Find the heat conducted by the spine and the temperature of the tip. and the tip of the spine is insulated. Inc.Contains proprietary and confidential information of ANSYS.078 6. 3rd Edition.177e9 psf ν = 0. All rights reserved. Apply advanced mesh control with element size of 0. 60 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 Tw. pg..778e-4 BTU/s ft2 °F Results Comparison Results Temperature of the Tip (°F) Heat Conducted by the Spine (Heat Reaction) (BTU/s) Target 79. Figure 11: Schematic Material Properties Material Properties E = 4.2” x 1.0344 6. and its subsidiaries and affiliates. 1976. Inc.041 Release 14. The surface convection coefficient between the spine and the surrounding air is h.5 .71e-3 BTU/s ft °F Geometric Properties Geometric Properties Cross section = 1. the air temper is Ta. pg.3 Thermal conductivity k = 9. Principles of Heat Transfer. 5 . .© SAS IP.Contains proprietary and confidential information of ANSYS. Inc. . All rights reserved. and its subsidiaries and affiliates.32 Release 14. Inc. 1e10 0 2e11 0 1.Contains proprietary and confidential information of ANSYS. 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.© SAS IP.5 .8e8 2.5e8 Pa 1. All rights reserved. Inc. 33 . and its subsidiaries and affiliates.93e11 0 7. .E (Pa) ial Part 1 Part 2 Part 3 Part 4 ν Tensile Yield (Pa) 2. Figure 12: Schematic Material Properties Mater.8e8 Loading Pressure = 2. Inc.07e8 2.VMMECH009: Stress Tool for Long Bar with Compressive Load Overview Reference: Analysis Type(s): Element Type(s): Any basic Strength of Materials book 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.5e8 2.1e11 0 Geometric Properties Part 1: 2 m x 2 mx3m Part 2: 2 m x 2 m x 10 m Part 3: 2 m x 2 mx5m Release 14. The multibody is used to nullify the numerical noise near the contact regions. 000 0. .Contains proprietary and confidential information of ANSYS.000 0.5e8 0.828 1.5e8 0.12 Mechanical 2. Inc.12 Error (%) 0.12 1 1. All rights reserved.12 1 1. .VMMECH009 Part 4: 2 m x 2 mx2m Results Comparison Results Maximum Equivalent Stress (Pa) Safety Factor for Part 1 Safety Factor for Part 2 Safety Factor for Part 3 Safety Factor for Part 4 Target 2.828 1. Inc.000 0.© SAS IP.5 .000 0. and its subsidiaries and affiliates.000 34 Release 14. Sizing mesh control with element size of 6. Inc. .4 2038.3 ρ = 7850 kg/m3 Geometric Properties Length = 0.5 .3 Error (%) -0.55 2051.1 m Thickness = 0. pg.Contains proprietary and confidential information of ANSYS.994 35 Release 14.03 1118.987 -0.952 -0.79 2906. Van Nostrand Reinhold Company Inc. 1979. Table 11-4.1 2879..7 1129. Case 11.© SAS IP.73 Mechanical 590.5 mm is applied on all the edges to get accurate results.667 -0. and its subsidiaries and affiliates. 256 Free Vibration Analysis Shell Analysis Type(s): Element Type(s): Test Case A rectangular plate is simply supported on both the smaller edges and fixed on one of the longer edges as shown below. Inc.005 m Loading Results Comparison Results 1st Frequency Mode (Hz) 2nd Frequency Mode (Hz) 3rd Frequency Mode (Hz) 4th Frequency Mode (Hz) Target 595. Formula for Natural Frequency and Mode Shape. . Figure 13: Schematic Material Properties E = 2e11 Pa ν = 0. All rights reserved.VMMECH010: Modal Analysis of a Rectangular Plate Overview Reference: Blevins. Find the first five modes of natural frequency.25 m Width = 0. Inc. . and its subsidiaries and affiliates. All rights reserved. Inc.VMMECH010 Results 5th Frequency Mode (Hz) Target 3366.489 36 Release 14. .© SAS IP.Contains proprietary and confidential information of ANSYS.48 Mechanical 3350 Error (%) -0.5 . © SAS IP. Theory of Plates and Shells.1 Release 14. and its subsidiaries and affiliates.25 m Thickness = 0. To get accurate results.0025 m Loading Pressure = 6585.VMMECH011: Large Deflection of a Circular Plate with Uniform Pressure Overview Reference: Timoshenko S. pg. equation 232. 2nd Edition. apply sizing control with element size of 5 mm on the circular edge. Find the total deformation at the center of the plate..18 Pa Results Comparison Results Total deformation (m) Target 0.5 . 37 .3 Geometric Properties Radius = 0. McGraw-Hill.. Article 97. The circular edge of the plate is fixed. All rights reserved. Inc.0012362 Error (%) -1.Contains proprietary and confidential information of ANSYS. Woinowsky-Krieger S. Inc. 401 Nonlinear Structural Analysis (Large Deformation On) Shell Analysis Type(s): Element Type(s): Test Case A circular plate is subjected to a uniform pressure on its flat surface. .00125 Mechanical 0. Figure 14: Schematic Material Properties E = 2e11 Pa ν = 0.P. 5 .© SAS IP.Contains proprietary and confidential information of ANSYS. . Inc. All rights reserved.38 Release 14. and its subsidiaries and affiliates. Inc. . 5 mm. Figure 15: Schematic Material Properties E = 2e11 Pa ν = 0. Young. apply sizing control with element size of 6. Table 34.903 Error (%) 1. 39 . To get accurate results. 672 Buckling Analysis Solid Test Case A stepped rod is fixed at one end face.© SAS IP.010 m Length of larger diameter = 0. McGraw Hill.1 m Loading Force at free end = 1000 N Force at the flat step face = 2000 N Both forces are in the z direction Results Comparison Results Load Multiplier Target 22. .3 Geometric Properties Larger diameter = 0.011982 m Smaller diameter = 0. and its subsidiaries and affiliates. Inc. 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. Case 2a.Contains proprietary and confidential information of ANSYS. 6th Edition.VMMECH012: Buckling of a Stepped Rod Overview Reference: Analysis Type(s): Element Type(s): Warren C. Roark's Formulas for Stress & Strains.5 .5 Mechanical 22.2 m Length of smaller diameter = 0. Find the Load Multiplier for the First Buckling Mode. Inc. pg. All rights reserved.8 Release 14. © SAS IP.40 Release 14. Inc. All rights reserved. .5 . Inc. and its subsidiaries and affiliates.Contains proprietary and confidential information of ANSYS. . pg.4 Release 14. Table 34. Case 10.Contains proprietary and confidential information of ANSYS.5 .07 Error (%) 0. and its subsidiaries and affiliates. Both the straight edges of the arch are fixed. Figure 16: Schematic Material Properties E = 2e5 MPa ν=0 Geometric Properties Arch cross-section = 5 mm x 50 mm Mean radius of arch = 50 mm Included angle = 90° Loading Pressure = 1 MPa Results Comparison Results Load Multiplier Target 544 Mechanical 546. Inc. Inc. McGraw Hill.© SAS IP. . All rights reserved. 41 . 679 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. Find the Load Multiplier for the first buckling mode.VMMECH013: Buckling of a Circular Arch Overview Reference: Analysis Type(s): Element Type(s): Warren C. 6th Edition. Roark's Formulas for Stress Strains. Young. and its subsidiaries and affiliates.© SAS IP.42 Release 14. All rights reserved. . . Inc.Contains proprietary and confidential information of ANSYS.5 . Inc. 1044 Kg ρ (kg/m3) 1e-8 1e-8 1e-8 1e-8 Geometric Properties Each cylinder: Diameter = 20 mm Length = 50 mm Release 14. 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.35 0. All rights reserved.Contains proprietary and confidential information of ANSYS.5e10 ν 0.5e10 4.34 0.© SAS IP. Solution intervals = 20.1e11 4.05 Figure 17: Schematic Material Properties Material Shaft 1 Shaft 2 Shaft 3 Shaft 4 E (Pa) 1.35 Loading Force = 1e7 N (Zdirection) Point Mass = 3. 43 . The flat end face of the cylinder (Shaft 1) is fixed.5 . • Scenario 1: Damping ratio = 0 • Scenario 2: Damping ratio = 0.1e11 1. Inc. Harmonic force is applied on the end face of another cylinder (Shaft 4) as shown below.VMMECH014: Harmonic Response of a Single Degree of Freedom System Overview Reference: Analysis Type(s): Element Type(s): Any basic Vibration Analysis book Harmonic Analysis Solid Test Case An assembly where four cylinders represent massless springs in series and a point mass simulates a spring mass system. and its subsidiaries and affiliates. . Inc.34 0. Contains proprietary and confidential information of ANSYS. All rights reserved.000 44 Release 14.591 0. . .14 175.1404 180 0.5 .000 0.14123 180 0.6 Mechanical 0. and its subsidiaries and affiliates.577 0. Inc. Inc.© SAS IP.VMMECH014 Results Comparison Results Maximum Amplitude without damping (m) Phase angle without damping (degrees) Maximum Amplitude with damping (m) Phase angle with damping (degrees) Target 0.1408 175.58 Error (%) 0. Figure 18: Schematic Material Properties Material Block 2 Shaft 2 Block 1 Shaft 1 E (Pa) 2e18 4. T.3 0.© SAS IP.4-1. 45 .VMMECH015: Harmonic Response of Two Storied Building under Transverse Loading Overview Reference: Analysis Type(s): Element Type(s): W. The material of the columns is assigned negligible density so as to make them as massless springs. Inc. Example 6.5e10 2e18 9e10 ν 0. and its subsidiaries and affiliates. Inc. All rights reserved. 166 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. pg. Use Solution intervals = 50.35 0.35 ρ (kg/m3) 7850 1e-8 15700 1e-8 Release 14.3 0.5 . 3rd Edition. Thomson.Contains proprietary and confidential information of ANSYS. 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. 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. 1999. . Theory of Vibration with Applications. 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. © SAS IP.5 1.2 46 Release 14.20853 0. Inc. Inc. All rights reserved. and its subsidiaries and affiliates. . .5 .21172 0.Contains proprietary and confidential information of ANSYS.VMMECH015 Geometric Properties Block 1 and 2: 40 mm x 40 mm x 40 mm Shaft 1 and 2: 20 mm x 20 mm x 200 mm Loading Force = -1e5 N (y direction) Results Comparison Results Maximum Amplitude for vertex A (m) Maximum Amplitude for vertex B (m) Target 0.074902 Mechanical 0.075836 Error (%) 1. and 1 for coefficients of both the environments under Solution Combination.5 .3 Ultimate Tensile Strength = 4. Inc. a fatigue strength factor or 1. • Scenario 1: One of the end faces is fixed and a force is applied on the opposite face as shown below in Figure 19: Scenario 1 (p. All rights reserved. 47)) and a pressure load is applied on the opposite faces in positive yand z-directions. and its subsidiaries and affiliates. Figure 19: Scenario 1 Figure 20: Scenario 2 Material Properties E = 2e11 Pa ν = 0.Contains proprietary and confidential information of ANSYS. Inc. • Scenario 2: Frictionless support is applied to all the faces of the three standard planes (faces not seen in Figure 20: Scenario 2 (p. Find the life.VMMECH016: Fatigue Tool with Non-Proportional Loading for Normal Stress Overview Reference: Analysis Type(s): Element Type(s): Any basic Machine Design book Fatigue Analysis Solid Test Case A bar of rectangular cross section has the following loading scenarios. a scale factor of 1. Use a design life of 1e6 cycles.6e8 Pa Release 14. and z directions for nonproportional fatigue using the Soderberg theory. and safety factor for the normal stresses in the x. . damage. 47). 47 .© SAS IP. y. 04569 67.683 0.725 0.013 3335. and its subsidiaries and affiliates.6e8 2.© SAS IP. .045696 48 Release 14.Component Z Life Damage Safety Factor Target Mechanical 300.772 -0.9 299.2998e6 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.157 0.001 0.764 0.001 0.724 0.7874 14766 67.132 -0.019 14765.31 0.045378 14765.04569 68. All rights reserved.2998e6 Pa Number of Cycles 1000 1e6 Alternating Stress (Pa) 4.247 0.5 .Component X Life Damage Safety Factor Stress Component . Results Comparison Results Stress Component . .Component Y Life Damage Safety Factor Stress Component . The fatigue calculations use standard formulae for the Soderberg theory.Contains proprietary and confidential information of ANSYS. Inc.VMMECH016 Material Properties Yield Tensile Strength = 3.156 0.1049 3329.5e8 Pa Endurance Strength = 2. Inc.019025 Error (%) -0.724 0.7874 14653 67.8406 0. 0. 49 . Inc. . Location of remote force = (7. All rights reserved.VMMECH017: Thermal Stress Analysis with Remote Force and Thermal Loading Overview Reference: Analysis Type(s): Element Type(s): Any basic Strength of Materials book 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.5 m.101815 Mechanical 0.10025 Error (%) -1. To get accurate results apply a global element size of 1. and 3 m.© SAS IP. Inc. Figure 21: Schematic Material Properties E = 2e11 Pa ν=0 α = 1.0) m Results Comparison Results Maximum X Deformation (m) Target 0. Loading Given temperature (End A) = 1000°C Given temperature (End B) = 0°C Remote force = 1e10 N applied on the contact surface at a distance 7 m from end A. Find the Deformation in the x direction of the contact surface on which the remote force is applied. 10 m. 5 m. and its subsidiaries and affiliates. Other thermal and structural loads are as shown below.5 Release 14.Contains proprietary and confidential information of ANSYS.5 .2e-5/°C Geometric Properties Diameter = 2 m Lengths of cylinders in order from End A: 2 m. © SAS IP.5 . All rights reserved. Inc. .50 Release 14.Contains proprietary and confidential information of ANSYS. Inc. . and its subsidiaries and affiliates. 5 m. and 3 m.3 ρ = 7850 kg/m3 Geometric Properties Cross-Section = 2mx2m Lengths of bars in order from End A: 2 m. Find the deformation in the z direction Figure 22: Schematic Material Properties E = 2e11 Pa ν = 0.© SAS IP. and its subsidiaries and affiliates. Inc.5 . 51 . Turn on Inertia Relief. 10 m. Loading Force P = 2e5 N (positive z direction) Analysis δz = where: L = total length of bar A = cross-section m = mass − 2ρ Release 14. All rights reserved.VMMECH018: A Bar Subjected to Tensile Load with Inertia Relief Overview Reference: Analysis Type(s): Element Type(s): Any basic Strength of Materials book 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. . Inc.Contains proprietary and confidential information of ANSYS. and its subsidiaries and affiliates. . All rights reserved.172 52 Release 14.5e-6 Mechanical 2. Inc.5043E-06 Error (%) 0.5 .© SAS IP.Contains proprietary and confidential information of ANSYS. Inc. .VMMECH018 Results Comparison Results Maximum Z Deformation (m) Target 2. and its subsidiaries and affiliates. 53 . .Contains proprietary and confidential information of ANSYS. Find the deformation in the y direction under Solution Combination with the coefficients for both the environments set to 1. Inc. Inc. • Scenario 1: Only a force load. Figure 23: Scenario 1 Figure 24: 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) Release 14.5 . • Scenario 2: Only a moment load.© SAS IP.VMMECH019: Mixed Model Subjected to Bending Loads with Solution Combination Overview Reference: Analysis Type(s): Element Type(s): Any basic Strength of Materials book Linear Static Structural Analysis Beam and Shell Test Case A mixed model (shell and beam) has one shell edge fixed as shown below. All rights reserved. Apply a global element size of 80 mm to get accurate results. Bending loads are applied on the free vertex of the beam as given below. 2542 Error (%) 0.VMMECH019 Material Properties Geometric Properties Beam rectangular cross section = 10 mm x 10 mm Beam length = 500 mm Loading Moment M = 4035 Nmm @ z-axis Analysis δy = where: I = total bending length of the mixed model I = moment of inertia of the beam cross-section l3 + l2 Results Comparison Results Maximum Y-Deformation (mm) Target -7. .18742 Mechanical -7. All rights reserved. . Inc.5 .929 54 Release 14. Inc. and its subsidiaries and affiliates.© SAS IP.Contains proprietary and confidential information of ANSYS. The cross section details are as shown below.85e5 Release 14. .34 0 ρ (kg/m3) 1e-8 7. All rights reserved. and its subsidiaries and affiliates.1e11 2e11 ν 0.Contains proprietary and confidential information of ANSYS. Figure 25: Cross Section Details for Both Beams Figure 26: Schematic Material Properties Material Spring Mass E (Pa) 1. The end vertex of the longer beam (acting as a spring) is fixed.5 .VMMECH020: Modal Analysis for Beams Overview Reference: Analysis Type(s): Element Type(s): Any basic Vibration Analysis book Modal Analysis Beam Test Case Two collinear beams form a spring mass system. Find the natural frequency of the axial mode. Inc. Inc. 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. 55 .© SAS IP. 160 56 Release 14. . .VMMECH020 Geometric Properties Spring beam length = 500 mm Mass beam length = 5 mm Loading Results Comparison Results Natural Frequency of Axial Mode (Hz) Target 1188.Contains proprietary and confidential information of ANSYS.© SAS IP. All rights reserved.5 .5 Error (%) 0.6 Mechanical 1190. Inc. Inc. and its subsidiaries and affiliates. 3 Geometric Properties L1 = 50 mm Total length = 200 mm Rectangular cross section = 10 mm x 10 mm Loading Force on L1 = -1000 N (x direction) Force on free vertex = -1000 N (x direction) Results Comparison Results Load Multiplier Target 10. McGraw Hill.© SAS IP. 6th Edition. Inc. Young. and its subsidiaries and affiliates.VMMECH021: Buckling Analysis of Beams Overview Reference: Analysis Type(s): Element Type(s): Warren C. 675 Buckling Analysis Beam Test Case A beam fixed at one end and is subjected to two compressive forces.5 .Contains proprietary and confidential information of ANSYS. Case 3a. as shown below.407 Release 14. Figure 27: Schematic Material Properties E = 2e11 Pa ν = 0. Roark's Formulas for Stress and Strains. Find the load multiplier for the first buckling mode.2397 Mechanical 10.198 Error (%) -0. All rights reserved. One of the forces is applied on a portion of the beam of length 50 mm (L1) from the fixed end and the other is applied on the free vertex. Inc. . Table 34. pg. 57 . Inc. .Contains proprietary and confidential information of ANSYS. All rights reserved. Inc.© SAS IP.58 Release 14. and its subsidiaries and affiliates. .5 . 1 m x 0. 0.add offset. • Scenario 2: Interface treatment . Release 14.adjust to touch.Contains proprietary and confidential information of ANSYS. All rights reserved.both in the z direction for each part for the following scenarios: • Scenario 1: Interface treatment . . Offset = 0 m. and its subsidiaries and affiliates. Offset = 0. • Scenario 4: Interface treatment .add offset. 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.VMMECH022: Structural Analysis with Advanced Contact Options Overview Reference: Analysis Type(s): Element Type(s): Any basic Strength of Material book Nonlinear Static Structural Analysis Solid Test Case An assembly of two parts with a gap has a Frictionless Contact defined between the two parts.0005 m Dimensions for each part: 0.5 . Find the Normal stress and Directional deformation . Offset = -0.5m Loading Given displacement = (0.0006) m Results Comparison The same results are obtained for both Augmented Lagrange and Pure Penalty formulations. 0. 59 . Figure 28: Schematic Material Properties E = 2e11 Pa ν=0 Geometric Properties Gap = 0.1 m x 0. Validate all of the above scenarios for Augmented Lagrange and Pure Penalty formulations.© SAS IP.001 m.add offset. • Scenario 3: Interface treatment .001 m. Inc. Inc. Inc.9786e4 2.VMMECH022 Results Target 6e-4 6e-4 2. Offset = 0.000 Add Offset.4e8 -4e7 6e-4 1.4e8 0 6e-4 0 2.000 Adjust To Touch 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) -2.356 4 2.Contains proprietary and confidential information of ANSYS. . and its subsidiaries and affiliates.4e8 Mechanical 6e-4 5.357 6e-4 0 2.-0.9858e7 -0.000 -0.1e3 2.354 6e-4 1e-4 2.000 0 Add Offset.355 6e-4 1.4e8 -2.4e8 Error (%) 0.357 0.001 m 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) 60 Release 14.3915e8 -0. Inc. Offset = 0 m 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) 0.000 Add Offset.4e8 0 0.4e8 0. All rights reserved. .© SAS IP.000 -0.5 .001 m 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) -4.000 0.99644e. Offset = 0.4e8 -4.000 -3.000 0.0961e3 2.4e8 0.4e8 6e-4 0.3843e8 -0.355 0. . Inc. Inc. has a square cross-section.Contains proprietary and confidential information of ANSYS.5 . It is fixed at one end and at the free end a Force F and a Moment M are applied. 61 . Figure 29: Schematic Equivalent Loading: Material Properties Beam 1: E1 = 1.© SAS IP.3e-6 kg/mm3 Geometric Properties For each beam: Cross-section = 10 mm x 10 mm Radius r = 105 mm Loading Force F = -1000 N (y direction) Moment M = 10000 Nmm (about z-axis) Release 14.1e5 MPa ν1 = 0 ρ1 = 8. See the figure below for details. each having an included angle of 45°. a UDL of "w " N / mm is applied on both the beams. Find the deformation of the free end in the y direction. Also. Use a global element size of 30 mm to get accurate results. and its subsidiaries and affiliates.VMMECH023: Curved Beam Assembly with Multiple Loads Overview Reference: Analysis Type(s): Element Type(s): Any basic Strength of Materials book Linear Static Structural Analysis Beam Test Case An assembly of two curved beams. All rights reserved. Inc.85e-6 kg/mm3 Geometric Properties Included angle = 45° Loading UDL w = -5 N/mm (y direction) on both beams This UDL is applied as an edge force on each beam with magnitude = -5 (2 x 3.619 62 Release 14.4688 Error (%) 0.334 N Analysis The deflection in the y direction is in the direction of the applied force F and is given by: 3 1 δ = − 3 + 2 where: δ = deflection at free end in the y direction I = moment of inertia of the cross-section of both beams + + 2 2 + 4ω + 4ω Results Comparison Results Minimum Y Deformation (mm) Target -8.5 .Contains proprietary and confidential information of ANSYS. . . Inc.© SAS IP.14 x 105) / 8 = -412.VMMECH023 Material Properties Beam 2: E2 = 2e5 MPa ν2 = 0 ρ2 = 7. and its subsidiaries and affiliates.416664 Mechanical -8. All rights reserved. Figure 30: Schematic Material Properties Mater. Both beams have hollow circular cross-sections.VMMECH024: Harmonic Response of a Single Degree of Freedom System for Beams Overview Reference: Analysis Type(s): Element Type(s): Any basic Vibration Analysis book Harmonic Analysis Beam Test Case Two collinear beams form a spring-mass system.© SAS IP. Inc. Inc. All rights reserved. .1e11 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. and its subsidiaries and affiliates.Contains proprietary and confidential information of ANSYS. • Scenario 1: Damping ratio = 0 • Scenario 2: Damping ratio = 0. The end vertex of the longer beam (acting as a spring) is fixed.34 1e-8 0 Geometric Properties Cross-section of each beam: Outer radius = 10 mm Release 14. 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.85e5 Loading Harmonic force F = 1 e6 N (z-direction) Spring 1.E ial (Pa) Mass 2e11 ν ρ (kg/m3) 7. A Harmonic force F is applied on the free vertex of the shorter beam in z direction. 63 . as indicated below.5 . Use both Mode Superposition and Full Method. VMMECH024 Geometric Properties Inner radius = 5 mm Length of longer beam = 100 mm Length of shorter beam = 5 mm Loading Results Comparison Results Target Mechanical Error (%) -0. All rights reserved. .0765etion with damping (m) 3 3 Full Method Maximum z directional deforma.0695etion with damping (m) 3 3 64 Release 14.4.Contains proprietary and confidential information of ANSYS.4.4.11252e. Inc.003 -1.4. .11332e.11252e. Inc.4. and its subsidiaries and affiliates.078e-3 tion tion without damping (m) 3 Maximum z directional deforma.4.1132etion without damping (m) 3 3 Maximum z directional deforma.859 -0.4.11332e.Maximum z directional deforma.© SAS IP.046 Mode Superposi.5 .4.876 -0. To get accurate results. Part 2 . . Inc.Contains proprietary and confidential information of ANSYS. Strength of Materials. 65 . All rights reserved. An internal pressure of P is applied on the inner surface of the inner cylinder. Find the maximum tangential stresses in both cylinders. set Contact Type to Rough with interface treatment set to add offset with Offset = 0. To simulate interference. They are free to move in radial and tangential directions.VMMECH025: Stresses Due to Shrink Fit Between Two Cylinders Overview Reference: Analysis Type(s): Element Type(s): Stephen P.5 . Inc. apply a global element size of 0. 208-214 Linear Static Structural Analysis Solid Test Case One hollow cylinder is shrink fitted inside another. Timoshenko. 3rd Edition. pg. Both cylinders have length L and both the flat faces of each cylinder are constrained in the axial direction.Advanced Theory and Problems. Figure 31: Schematic Material Properties Both cylinders are made of the same material E = 3e7psi Geometric Properties Inner Cylinder: ri = 4” ro = 6.005” Ri = 6” Loading P = 30000 psi Release 14. Note Tangential stresses can be obtained in the Mechanical application using a cylindrical coordinate system. and its subsidiaries and affiliates.© SAS IP.8 inches. © SAS IP.5 .VMMECH025 Material Properties ν=0 ρ = 0. Inc.0 0.67 42281.0 Note Here y corresponds to θ direction of a cylindrical coordinate system. Inc.Contains proprietary and confidential information of ANSYS. .09 Mechanical 35738 42279 Error (%) 1. and its subsidiaries and affiliates.28383 lbm/in3 Geometric Properties Ro = 8” Length of both cylinders = 5” Loading Results Comparison Results Maximum normal y stress. inner cylinder (psi) Maximum normal y stress. outer cylinder (psi) Target 35396. All rights reserved. . 66 Release 14. Contains proprietary and confidential information of ANSYS.5 . 67 .5e8 Pa No. and its subsidiaries and affiliates. Find the maximum Bending Stress (Normal X Stress) and maximum Total Deformation of the plate. Inc. Consider load type as fully reversed and a Design Life of 1e6 cycles. All rights reserved. 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.38e8 Loading Moment M = 0. and Scale factor of 1. & Gerber criteria.0 Ultimate tensile strength = 1. . Use the x-stress component. Figure 32: Schematic Material Properties E = 2e11 Pa ν = 0. Fatigue Strength factor of 1. Also find the part life and the factor of safety using Goodman.15 Nm (counterclockwise @ z-axis) Geometric Properties Length L = 12e3m Width W = 1e-3 m Release 14. of Cycles 1000 1e6 Alternating Stresses (Pa) 1.VMMECH026: Fatigue Analysis of a Rectangular Plate Subjected to Edge Moment Overview Reference: Analysis Type(s): Element Type(s): Any standard Machine Design and Strength of Materials book Fatigue Analysis Shell Test Case A plate of length L.© SAS IP.08e9 1. width W. Inc.29e9 Pa Endurance strength = 1. Soderberg.38e8 Pa Yield Strenth = 2. 4 0.15333 1844. Inc.3 0.1533 1844.5 .020 0.005 0.© SAS IP.3 0.15333 1844.5002e4 0.3 68 Release 14.48e-4 0.VMMECH026 Geometric Properties Thickness T = 1 e-3 m Loading Results Comparison Results Target Mechanical 9e8 6.4 0.005 Maximum normal x-stress (Pa) Maximum total deformation (m) SN-Goodman SN-Soderberg SN-Gerber Safety factor Life Safety factor Life Safety Factor Life 9e8 6.020 0.Contains proprietary and confidential information of ANSYS. All rights reserved.020 0.1533 1844. Inc.311 0.4 Error (%) 0.1533 1844. . and its subsidiaries and affiliates.005 0.000 0. .15333 1844. To get accurate results.5e-2 m.05 m Thickness T = 0.© SAS IP. maximum total heat flux. and its subsidiaries and affiliates. All rights reserved. and thickness (T) is fixed along the width on one edge and heat flow (Q) is applied on the same edge.2 m Width W = 0. width (W). Figure 33: Schematic Material Properties E = 2e11 Pa ν = 0. Find the maximum temperature. maximum total deformation. Inc.005 m Loading Heat flow Q = 5 W Given Temperature = 20°C Analysis Heat Reaction = -(Total heat generated) Heat flow due to conduction is given by: = h− l Release 14.Contains proprietary and confidential information of ANSYS. and heat reaction at the given temperature.2e5/°C Thermal conductivity k = 60. Inc.5 W/m°C Geometric Properties Length L = 0. . 69 .5 . apply a sizing control with element size = 2. The opposite edge is subjected to a temperature of 20 °C.0 Coefficient of thermal expansion α = 1. Ambient temperature is 20 °C.VMMECH027:Thermal Analysis for Shells with Heat Flow and Given Temperature Overview Reference: Analysis Type(s): Element Type(s): Any standard Thermal Analysis book Thermal Stress Analysis Shell Test Case A plate of length (L). .1157 2e4 7.000 0.7.9958e5 5 -5 -5 70 Release 14.© SAS IP. . Inc.000 Maximum Temperature (°C) Maximum Total Heat Flux (W/m ) Maximum Total Deformation (m) Heat Reaction (W) 2 86. All rights reserved.VMMECH027 where: Th = maximum temperature T1 = given temperature Total heat flux is: = Temperature at a variable distance z from the fixed support is given by: z = h − h − 1 × Thermal deformation in the z-direction is given by: δ = ∫ α − ¡ 0 l Results Comparison Results Target Mechanical 86.116 2e4 Error (%) 0. and its subsidiaries and affiliates.781 0.Contains proprietary and confidential information of ANSYS.93386e.5 .000 0. Inc. Find the Z directional deformation and the adjustment reaction due to the bolt pretension load.05 m Loading Pretension as preload = 19.635 N (equal to adjustment of 1e-7 m) Analysis The bolt pretension load applied as a preload is distributed equally to both halves of the bar.© SAS IP. and its subsidiaries and affiliates. 71 . The longitudinal faces have frictionless support.01 m. apply sizing control with element size of 0. . To get accurate results. Figure 34: Schematic Material Properties E = 2e11 Pa ν = 0. Inc.5 . Therefore the z-directional deformation due to pretension is given by: δPretension = = δ ¡¢£¢¤¥¦§¤ × × Release 14.VMMECH028: Bolt Pretension Load Applied on a Semi-Cylindrical Face Overview Reference: Analysis Type(s): Element Type(s): Any standard Strength of Materials book Static Structural Analysis Solid Test Case A semi-cylinder is fixed at both the end faces. All rights reserved.0 Geometric Properties Length L = 1 m Diameter D = 0. Inc.Contains proprietary and confidential information of ANSYS. A bolt pretension load is applied on the semi-cylindrical face. © SAS IP.996 0.0.9502E08 1.VMMECH028 Results Comparison Results Target Mechanical Error (%) Minimum z-directional deformation (m) -5.5 . All rights reserved.Contains proprietary and confidential information of ANSYS.004 08 4.00E08 -5.00E-08 (m) Adjustment Reaction (m) 1.000 Maximum z-directional deformation 5. .00E-07 -0. Inc. Inc.0002E. and its subsidiaries and affiliates.00E-07 72 Release 14. . Inc.. Third Edition.VMMECH029: Elasto-Plastic Analysis of a Rectangular Beam Overview Reference: Analysis Type(s): Element Type(s): Timoshenko S. and its subsidiaries and affiliates. All rights reserved.Contains proprietary and confidential information of ANSYS. set the advanced mesh control element size to 0.© SAS IP. pp. Article 64.0 Myp to 1. For an elastic-perfectly-plastic stress-strain behavior. show that the beam remains elastic at M = Myp = σypbh2 / 6 and becomes completely plastic at M = Mult = 1. . 349 Static Plastic Analysis Solid Test Case A rectangular beam is loaded in pure bending. Figure 35: Stress-Strain Curve Figure 36: Schematic Material Properties E = 3e7 psi ν = 0. Strength of Materials.5 inches. Inc. Advanced Theory and Problems. Part II.0 σyp = 36000 psi Geometric Properties Length L = 10” Width b = 1” Height h = 2” Loading M = 1.5 Myp.5 Myp (Myp = 24000 lbf in) Release 14. To get accurate results. 73 .5 . 25 1.© SAS IP.800 Error (%) 74 Release 14. Inc. All rights reserved. Results Comparison M/Myp State 1 1.VMMECH029 Analysis The load is applied in three increments: M1 = 24000 lbf-in. Inc. . M2 = 30000 lbf-in.Contains proprietary and confidential information of ANSYS.5 .164 0. and its subsidiaries and affiliates. and M3 = 36000 lbf-in. .5 fully elastic elasticplastic plastic Target Equivalent Stress (psi) 36000 36000 solution not converged State fully elastic elasticplastic plastic Mechanical Equivalent Stress (psi) 36059 36288 solution not converged 0. 9e7 psi ν = 0. 75 . set the advanced mesh control element size to 0. Therefore. the moment applied will be per unit length (5000/1000 = 5 lbf-in). Inc. Figure 37: Schematic Material Properties E = 2. . Analysis takes into account the unit length in the z-direction. 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.© SAS IP. All rights reserved.5 . the above problem can be analyzed as a plane strain problem.Contains proprietary and confidential information of ANSYS. Find X normal stress at a distance of 0.5 inches.VMMECH030: Bending of Long Plate Subjected to Moment . and its subsidiaries and affiliates.0 Geometric Properties Length L = 1000” Width W = 40” Thickness T = 1” Loading Moment M = 5000 lbf-in Analysis Since the loading is uniform and in one plane (the x-y plane). To get accurate results. Release 14.Plane Strain Model Overview Reference: Analysis Type(s): Element Type(s): Any standard Strength of Materials book Plane Strain Analysis 2D Structural Solid Test Case A long. Inc.5 inches from the fixed support. Also find total deformation and reaction moment. 16553e-2 -5 Error (%) 0.000 76 Release 14.018 0. Inc.000 0. . Inc. All rights reserved.1655e-2 -5 Mechanical 30 30 0.5 .Contains proprietary and confidential information of ANSYS.© SAS IP.VMMECH030 Figure 38: Plane Strain Model (analyzing any cross section (40” x 1”) along the length) Results Comparison Results Normal Stress Maximum Normal Stress in the X-Direction (psi) Maximum Total Deformation (in) Reaction Moment (lbf-in) Target 30 30 0.000 0. . and its subsidiaries and affiliates. 5 . Inc.93e11 7. Figure 39: Schematic Material Properties Material Part 1 Part 2 Part 3 Part 4 E (Pa) 1.8e8 2. and its subsidiaries and affiliates. To get accurate results.© SAS IP.5e8 2.VMMECH031: Long Bar with Uniform Force and Stress Tool .1e10 2e11 1. . Find the maximum equivalent stress for the whole assembly and safety factor.1e11 ν 0 0 0 0 Loading Force = 1e9 N in the negative x-direction Tensile Yield (Pa) 2.8e8 Geometric Properties Part 1: 2 3m Part 2: 2 10 m Part 3: 2 5m Part 4: 2 2m mx2mx mx2mx mx2mx mx2mx Release 14. safety margin. All rights reserved.Plane Stress Model Overview Reference: Analysis Type(s): Element Type(s): Any standard Strength of Materials book Plane Stress Analysis 2D Structural Solid Test Case A long. 77 .07e8 2. Inc.Contains proprietary and confidential information of ANSYS. and safety ratio for the first and last part using the maximum equivalent stress theory with Tensile Yield Limit. rectangular bar assembly is fixed at one of the faces and the opposite face is subjected to a compressive force. set the advanced mesh control element size to 1 m. 172 1.000 0. .89286 Error (%) 0.12 0. All rights reserved.5 .096 78 Release 14.12 0. and its subsidiaries and affiliates.000 0. Analysis is done considering thickness of 2 m along z-direction Figure 40: Plane Stress Model (Analyzing any cross section along Z) Results Comparison Results Maximum Equivalent Stress (Pa) Part 1 Safety Factor Safety Margin Safety Ratio Part 4 Safety Factor Safety Margin Safety Ratio Target 2.828 -0.2077 1.5e8 0.12 0.000 0.Contains proprietary and confidential information of ANSYS. Inc.828 -0.000 0.207 1.892 Mechanical 2.172 1.VMMECH031 Analysis Since the loading is uniform and in one plane. .5e8 0. the above problem can be analyzed as a plane stress problem. Inc.058 0.000 0.© SAS IP.12 0. Inc. the above problem can be analyzed as an axisymmetric problem. Figure 41: Schematic Material Properties E = 1.Contains proprietary and confidential information of ANSYS.© SAS IP. All rights reserved. To get accurate results. Release 14. set the advanced mesh control element size to 0. The cable delivers a total heat flow of Q to the disk. .34 Thermal conductivity k = 401. Find the disk temperature and heat flux at inner and outer radii.5 . Inc.0 W/m°C Geometric Properties Ri = 10 mm Ro = 60 mm t = 8 mm Loading Q = 100 W (Internal Heat Generation = 39788735.002 m. It has a heat-generating copper coaxial cable (of radius Ri) passing through its center.1e11 Pa ν = 0. and its subsidiaries and affiliates. 79 .VMMECH032: Radial Flow due to Internal Heat Generation in a Copper Disk Axisymmetric Model Overview Reference: Analysis Type(s): Element Type(s): Any basic Heat Transfer book Axisymmetric Analysis 2D Structural Solid Test Case A copper disk with thickness t and radii Ri and Ro is insulated on the flat faces.77 W/m3) Film coefficient h = 1105 W/m2-°C Surrounding temperature To = 0°C Analysis Because the geometry and loading are symmetric about the y-axis. The surrounding air is at a temperature of To with convective film coefficient h. 98943e5 33157 Mechanical 38.554 -0.© SAS IP.9 30 1.896 30. and its subsidiaries and affiliates.VMMECH032 Figure 42: Axisymmetric Model Results Comparison Results Maximum Temperature (°C) Minimum Temperature (°C) Maximum Heat Flux (W/m ) Minimum Heat Flux (W/m ) 2 2 Target 38. . Inc.023 -0.010 0.018 80 Release 14.Contains proprietary and confidential information of ANSYS. Inc. . All rights reserved.5 .007 197840 33151 Error (%) -0. 181 Electromagnetic Analysis Solid Test Case A C-shaped magnet has a coil with 400 turns and a cross section of the core with area 4 cm2. Edminster.003 m. Figure 43: Schematic Release 14. Inc. .2 cm and the coil details are given in Figure 44: Coil Details in cm (p.1 A flows through the coil. The air gap is 0. 2nd Edition.© SAS IP. 82). 82). A current of 0. Theory and Problems of Electromagnetics. Example 11. Inc. To get accurate results.VMMECH033: Electromagnetic Analysis of a C-Shaped Magnet Overview Reference: Analysis Type(s): Element Type(s): J.Contains proprietary and confidential information of ANSYS.5 . 81 . Find the total flux density and total field intensity. Tata McGraw Hill. A.9. All rights reserved. pg. and its subsidiaries and affiliates. set the advanced mesh control element size to 0. Flux parallel is applied on the nine outer faces as shown in Figure 46: Flux Parallel Applied on 9 Outer Faces (p. 34 0. Inc.© SAS IP.3 Density (kg/m3) 0 8300 7850 Loading Voltage = 0 V Current = 0. .1e11 2e11 0 0. and its subsidiaries and affiliates.Poisson's Ralus (Pa) tio Air Body Coil Core 1e7 1.VMMECH033 Figure 44: Coil Details in cm Figure 45: Current and Voltage Figure 46: Flux Parallel Applied on 9 Outer Faces Material Properties Young's Modu.Contains proprietary and confidential information of ANSYS. 82) 82 Release 14. All rights reserved. .5 . Inc.1 A Relative Permeability 1 1 500 Electric Resistivity (ohmm) 0 2e-7 0 Geometric Properties Given in Figure 44: Coil Details in cm (p. 05 0.0585 Mechanical 0.04 Release 14. Inc.061e-2 32320. we have the following equation: Magnetic flux is: φ= + ¡ µ µ¡ ¡ where: N = number of turns I = current Lc = mean core length La = air gap Ac = cross-sectional area of core Aa = apparent area of air gap µc = permeability of core µa = permeability of air The air-gap average flux density is given by: ¢ = φ ¢ The air-gap average filed intensity is given by: £ = £ µ£ Results Comparison Results Total Flux Density (T) Total Field Intensity (A/m) Target 4.© SAS IP. 83 .VMMECH033 Geometric Properties Depth = 2cm Loading Analysis Using the analogy of Ohm's law of Magnetism. and its subsidiaries and affiliates.Contains proprietary and confidential information of ANSYS. .040629 32332 Error (%) 0. All rights reserved. Inc.5 . 5 . Inc. and its subsidiaries and affiliates.Contains proprietary and confidential information of ANSYS. .84 Release 14. . All rights reserved. Inc.© SAS IP. Contains proprietary and confidential information of ANSYS. All rights reserved.© SAS IP.4 m Loading Displacement in Y direction = -0. K. pp.3 ρ = 7850 kg/m3 Geometric Properties Solid1: 0. 26 Nos 1/2.77e5 Pa Incompressibility Parameter D1 1/Pa =0 Radius = 0. Figure 47: Schematic Material Properties Solid1: E = 2e11Pa ν = 0. Release 14. Tussman.05m Analysis Due to geometric and loading symmetry.1m Solid2: Mooney-Rivlin Constants Solid2: Quarter Circular Cylinder C10 = 2. .J.2 m Length = 0.VMMECH034: Rubber cylinder pressed between two plates Overview Reference: T. Vol. 357-409 Nonlinear Static Structural Analysis (Large Deformation ON) Solid Analysis Type(s): Element Type(s): Test Case A rubber cylinder is pressed between two rigid plates using a maximum imposed displacement of δmax. and its subsidiaries and affiliates. Inc.01m x 0. the analysis can be performed using one quarter of the cross section. 85 .05 m x 0. Z = 0 and Z = 0.93e5 Pa C01 = 1. Determine the total deformation. Bathe.05 m).5 . • Frictionless supports are applied on 3 faces (X = 0. "A Finite Element Formulation for Nonlinear Incompressible Elastic and Inelastic Analysis". 1987. Inc. Computers and Structures. All rights reserved. Inc.16526 Error (%) 0 86 Release 14.© SAS IP. Inc. and its subsidiaries and affiliates.1m is applied on the top surface. .165285 Mechanical 0. .5 . • Augmented Lagrange is used for Contact formulation.Contains proprietary and confidential information of ANSYS. • The bottom surface of Solid1 is completely fixed.VMMECH034 • Given displacement of 0. Results Comparison Results Total Deformation (m) Target 0. • Frictionless Contact with Contact stiffness factor of 100 is used to simulate the rigid target. 2 x 10-5 1/°C k = 60. • 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. . Ambient Temperature is maintained at 20°C.3.Contains proprietary and confidential information of ANSYS. All rights reserved. Figure 48: Schematic Material Properties E = 2. Find the following: • Temperatures on End Faces.5 . and its subsidiaries and affiliates.VMMECH035: Thermal Stress in a Bar with Radiation Overview Reference: Analysis Type(s): Element Type(s): Any Basic Heat transfer and Strength of Materials book Coupled Analysis (Static Thermal and Static Stress) Solid Test Case Heat of magnitude 2500 W and Heat Flux of magnitude 625 W/m2 is flowing through a long bar (2 x 2 x 20) m in an axial direction.© SAS IP.5 W/m°C Geometric Properties Part 1: 2 2m Part 2: 2 5m Part 3: 2 10 m Part 4: 2 3m mx2mx mx2mx mx2mx mx2mx Loading Heat Flow = 2500W on Part 4 Heat Flux = 625 W/m2 on Part 4 Radiation = 20°C. 87 . Inc.0e11 Pa v=0 α = 1. 0.3 Analysis (Heat flowing through body) Q = (Heat Flow) + (Heat Flux * Area) = 5000 W Release 14. and radiating out from the other face having emissivity 0. Inc. 38 − .38°C.e. × 0 − × 673. Heat Conducted through the body Thermal strain is given by. ε¨ _ ax = α∆© = . ε in = α∆ = × −! × − = The compressive stress introduced is given by. Qr = ε ∗ σ ∗ A ∗ ( T 2 gives T2 = 260.VMMECH035 (Heat flowing through body) = (Heat Conducted through body) = (Heat Radiated out of the Surface) i.16°C. Heat Radiated out of the body 4 − Tα 4 ) W. Q = Qr =QC = 5000 W. c = K ∗ ¡ ∗ ¤¢1 − ¢§ ¥ £¦ b gives T1 = 673. 6 1.012572 Error (%) 0 0 0 0 -4. since deformation is symmetric dUo k | | −}q × } δ = ∫ p}q~ × } −
× w q − ~q} − ~~yq{ + ∫ ~ × } δ = × − δ = ∫ Oε VXYZ[\] + ε^VZu`VuZ\] RSfj Results Comparison Results Temperature on Part 4(°C) Temperature on Part 1 (°C) Maximum Thermal Strain (m/m) Minimum Thermal Strain (m/m) Normal Stress in Z direction (Pa) Directional Deformation in Z direction (m) Target 673.© SAS IP.81656e-3 2.067448e9 -0.0123966 Mechanical 673.:<=>>? × +:@ P# Temperature at a distance z from the face with higher temperature is given by.38 260. and its subsidiaries and affiliates.4 88 Release 14. Inc.Contains proprietary and confidential information of ANSYS.8179e-3 0.8 656 × 0 − m / m 9 × −" σz = -#vg$the%&#l$st%#'* × E = -+.49 260.16 7.0028578 -1. Inc.5 . = 7.0183e9 -0.85792e-3 -1. FGHNHI − JFLNMF C = FGHNHI − JLNFFMC BD = FGHNHI − JL Only half-length is considered for calculating deformation. All rights reserved.15 7. . . 0) Thermal Condition °C 1. 150°C Release 14. 100°C 3. (1.25. Inc. For all the steps. 89 . Reference Temperature is 0°C.5 . 1. 0) is subjected to a Uniform Temperature (Thermal Condition Load) in three steps.5. . (0.VMMECH036: Thermal Stress Analysis of a Rotating Bar using Temperature Dependant Density Overview Reference: Analysis Type(s): Element Type(s): Any Basic Strength of Materials book 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.Contains proprietary and confidential information of ANSYS. 0. Inc. 0) 2.© SAS IP. Figure 49: Schematic Material Properties E = 1 x 106 Pa α = 1 x 10-5 1/°C ν=0 Temperature °C 50 100 150 Density kg/m3 30 60 90 Geometric Properties Part 1: 2mx2 m x 20 m Loading Rotational Velocity (rad/s) in steps: 1. 0. (0. 0) 3. 50°C 2. All rights reserved. 0. and its subsidiaries and affiliates. Frictionless Support is applied on all the longitudinal faces. VMMECH036 Analysis Rotational Stress = ∫ ρω 0 2 = ρω 2 2 Th¡¢m£¤ ¥¦¢¡§§ = ×α×∆ = ρω 3 ¨© . © Df© . &)-#$ ( " )&** 56789 :.09 0.<6=>87?6@ 5B.09 0.032 0.06 0. Inc./. All rights reserved.5 .045 Error (%) 0. and its subsidiaries and affiliates.14 !"#$ Equ%v#$&'" (")&** = +!"#"%!'#$ (")&** + + .06 0.045 Mechanical 6502. .5 0.© SAS IP. .019 0 0 0 90 Release 14.<6=>87?6@ =δ CFCGH = A6787?6@89 :.040 0.3 2625.6 4001.Contains proprietary and confidential information of ANSYS.=>89 :.© = ×α×∆ =σ . Inc. <6=>87?6@ Results Comparison Results Equivalent Stress (Pa) Step 1 Step 2 Step 3 Total Deformation (m) Step 1 Step 2 Step 3 Target 6500 4000 2625 0. . Inc. International Textbook Co.5556e4 BTU/s-ft2-°F Ambient Temperature for Convection = 25°F Release 14. • Initial temperature. initially at a temperature T0. All rights reserved.5 . T0 = 65 °F • Surface temperature. Te = 25°F • Convection coefficient h = 5.5556e-4 BTU/s-ft2-°F • Time. 2nd Printing. The surface convection coefficient is h. 143. .© SAS IP.VMMECH037: Cooling of a Spherical Body Overview Reference: Analysis Type(s): Element Type(s): F.Contains proprietary and confidential information of ANSYS. Inc. "Principles of Heat Transfer".16667 ft Loading Convection applied on Edge = 5. ex. PA. Scranton. when exposed to an environment having a temperature Te for a period of 6 hours (21600 s). and its subsidiaries and affiliates. 91 . pg. t = 21600 seconds • Radius of the sphere ro = 2 in = 1/6 ft Figure 50: Schematic Material Properties K = (1/3) BTU/hr-ft-°F ρ = 62 lb/ft c = 1. 1959.075 Btu/lb-°F 3 Geometric Properties Quarter Circular lamina Radius = 0. Kreith. Transient Thermal Analysis Plane Test Case Determine the temperature at the center of a spherical body. 4-5. Inc.© SAS IP. . All rights reserved.457 92 Release 14. Inc. and its subsidiaries and affiliates.5 . only a 2-D quarter model is used.VMMECH037 Analysis Since the problem is axisymmetric.Contains proprietary and confidential information of ANSYS. Results Comparison Results Temperature at the Centre of body after 21600s (°F) Target 28 Mechanical 28. .688 Error (%) 2. All three blocks are resting on Base. Find the velocity of both the moving blocks after impact. and its subsidiaries and affiliates.Contains proprietary and confidential information of ANSYS.VMMECH038: Crashing Blocks Simulation with Transient Structural Analysis Overview Reference: Analysis Type(s): Element Type(s): Any basic Kinematics book.© SAS IP.1775e-4 kg. .85e-6kg/mm3 Geometric Properties Left Block = 3mm x 2mm x 5mm Middle Block = 2. 93 . Right block is fixed using Fixed Support and the base is fixed by applying Fixed Joint. All rights reserved.5mm x 2mm x 3mm Right Block =3mm x 6mm x 4mm Loading Left Block Initial Velocity = 100 mm/s (X direction) Release 14. Inc. Frictionless supports are applied as shown in the figure and also on the bottom faces of left and middle blocks. Inc. Figure 51: Schematic Material Properties E = 2e5 MPa ν = 0. Flexible Dynamic Analysis Solid Test Case Left Block of mass 2.355e-4 kg is given a constant initial velocity of 100 mm/sec to collide with the middle block1of mass 1.5 .3 ρ = 7. . . . .5 -0.813 132. . .3 mm/sec γMf = 133. . . . . . . . . . II mL. . . . .607mm x 75.γLf) = mM (γMf .γMi) . . .5 . . . . mL (γLi . .4 Mechanical 33. mM = Mass of Left and Middle Block in kg γLi. . . . .3 133. Inc. . Inc. γLf = 33. . . .VMMECH038 Material Properties Geometric Properties Base = 3mm x 8. .34 mm/sec Results Comparison Results Velocity of Left Block after impact (mm/sec) Velocity of Middle Block after impact (mm/sec) Target 33. . . . .8 94 Release 14. .15mm Loading Analysis For Perfectly Elastic Collision between the blocks. . . .37 Error (%) 1. .I γLi + γLf = γMf + γMi. . . . . and its subsidiaries and affiliates. All rights reserved.Contains proprietary and confidential information of ANSYS. γLf = Initial and Final Velocity of the Left Block in mm/sec γMi = Initial velocity of Middle Block in mm/sec = 0 as it is at rest γMf = Velocity of Middle Block after impact in mm/sec Solving I and II. .© SAS IP. . sec. 1979.© SAS IP.Contains proprietary and confidential information of ANSYS. 5-8.8 sec Release 14. Flexible Dynamic Analysis Solid and Spring Test Case A system containing two masses. m1 and m2. Figure 52: Schematic Material Properties E = 2e11 Pa γ = 0. Determine the displacement response of the system for the load history shown.VMMECH039: Transient Response of a Spring-mass System Overview Reference: Analysis Type(s): Element Type(s): R. 2nd Edition. K. 95 . and its subsidiaries and affiliates. and two springs of stiffness k1 and k2 is subjected to a pulse load F(t) on mass 1. Inc. All rights reserved. Harper & Row Publishers. Vierck.3 ρ = 0. Inc.5 . Vibration Analysis. New York. NY.25 kg/m3 k1 = 6 N/m k2 = 16 N/m m1 = 2 kg m2 = 2kg Geometric Properties 2 Blocks = 2m x 2m x 2m Length of L1 spring = 6m Length of L2 spring = 7m Loading F0 = 50 N td = 1. . 51 6. m (@ t = 2. Inc. m (@ t = 2.99 18. m (@ t = 1.9152 18. .4s) Target 14.1976 Error (%) -1 -1. Inc.9 1 0. m (@ t = 1.© SAS IP.4s) Y2.335 3. and its subsidiaries and affiliates.3s) Y2.32 6. All rights reserved.14 Mechanical 14.9 96 Release 14.48 3.Contains proprietary and confidential information of ANSYS.3s) Y1. .5 .VMMECH039 Results Comparison Results Y1. © SAS IP. . All rights reserved.001 kg/m3 Geometric Properties Bar = 1m x 1m x 24m Loading Force = -1000 N (Ydirection) at 8m from Simply Supported end Analysis Scenario 1: Considering Symmetry δ= × × 3 × Scenario 2: Considering Anti-Symmetry Release 14. and its subsidiaries and affiliates.5 .Contains proprietary and confidential information of ANSYS. 97 . Inc. Find Deformation at the 8m from simply Supported end. Scenario 1: Considering Symmetry Scenario 2: Considering Anti-Symmetry Figure 53: Schematic Material Properties E = 2e11 Pa γ=0 ρ = 0.VMMECH040: Deflection of Beam using Symmetry and Anti-Symmetry Overview Reference: Analysis Type(s): Element Type(s): Any Basic Strength of Materials Book 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. Inc. and its subsidiaries and affiliates.7383e-6 Error (%) 0.VMMECH040 × × 3 − × × × 3 × δ= Results Comparison Results Scenario 1: Directional Deformation in Y-direction (m) Scenario 2: Directional Deformation in Y-direction (m) Target -2. Inc.856 98 Release 14. .70662e-6 Mechanical -2. Inc. All rights reserved.019 1.Contains proprietary and confidential information of ANSYS.569e-5 -1. .5695e-5 -1.© SAS IP.5 . and its subsidiaries and affiliates.05. The radius of Coil is 30 mm and cross section is 20 mm X 20 mm. Electromagnetic Analysis Solid Test Case The winding body is enclosed in an Air Body.5 A. Equation 12. Periodic Symmetry is applied on two faces. The number of turns is 200 and current is 0. The dimensions of the air body are such that it encloses the coil. "Flux Parallel" is applied on all the 7 outer surfaces. 1948 Harpers Brothers. Page 242. . 99 . Inc. Find the Total Flux Density. Inc. Figure 54: Dimensions of Body Release 14. Boast.5 .Contains proprietary and confidential information of ANSYS.VMMECH041: Brooks Coil with Winding for Periodic Symmetry Overview Reference: Analysis Type(s): Element Type(s): W.© SAS IP. All rights reserved. Principles of Electric and Magnetic Fields. .34 Density (kg/m3) 0 8300 Relative Permeability 1 1 Electric Resistivity (ohm-m) 0 2e-7 Analysis Flux Density = = × × 2+ × 2 where: N = number of turns (1) I = current per turn (100) mu = (4 x π x 10-7) S = width of coil (20e-3m) R = radius to midspan of coil (3*S/2) = = × × + × -¡ = × × π× −3 −7 × × × × −3 + × × 100 Release 14. Inc. Inc.1e11 Poisson's Ratio 0 0.5 . All rights reserved. and its subsidiaries and affiliates.Contains proprietary and confidential information of ANSYS. .© SAS IP.VMMECH041 Figure 55: Schematic Diagram Material Properties Young's Modulus (Pa) DSVM41_MAT1 (Emag Part) DSVM41_MAT2 (Winding Body) 1e7 1. 0019848 Error (%) -0.99e-3 Mechanical 0. Inc.Contains proprietary and confidential information of ANSYS. 101 .5 . and its subsidiaries and affiliates.© SAS IP.3 Release 14. Inc.VMMECH041 Results Comparison Results Total Flux Density (T) Target 1. All rights reserved. . Inc.© SAS IP.102 Release 14. . Inc.Contains proprietary and confidential information of ANSYS. All rights reserved.5 . . and its subsidiaries and affiliates. and its subsidiaries and affiliates. 103 .Contains proprietary and confidential information of ANSYS. Fluid density is 1000 kg/m3 and Hydrostatic acceleration is 10 m/s2 in negative Z direction. Scenario 1: Square bar is partially immersed in the fluid up to 15 m in Z direction from the fixed support. Inc. Hydrostatic pressure is applied on a longitudinal face normal to X-axis at different locations as given in the scenarios below.5 . Inc.© SAS IP. Partially Submerged in a Fluid Overview Reference: Analysis Type(s): Element Type(s): Any Basic Strength of Materials Book Static Structural Analysis Solid Test Case Long bar 20m x 2m x 2m is immersed in a fluid and is fixed at one end.VMMECH042: Hydrostatic Pressure Applied on a Square Bar with Fully. Scenario 2: Square bar is fully immersed in the fluid up to 25 m in Z direction from the fixed support Figure 56: Schematic Material Properties E = 2e11 Pa γ=0 ρ = 7850 kg/m3 Geometric Properties Long bar = 20m x 2m x 2m Loading Hydrostatic Pressure Acceleration = -10 m/s2 (Z direction) Surface Location: Release 14. All rights reserved. Find normal stress in Z direction of square bar. . 5241e7 Error (%) -1.50e7 Mechanical 8529300 3.© SAS IP.VMMECH042 Material Properties Geometric Properties Loading Scenario 1: (2. 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: W1 = Maximum Load per meter (@ 25m) W2 = Minimum Load per meter (@ 5m) Normal stress = Bending stress = Maximum bending moment / Sectional Modulus 1 − 2 + Results Comparison Results Normal Stress (Partially Submerged) (Pa) Normal Stress (Fully Submerged) (Pa) Target 8.4375e6 3.088 0.1.-5) m Analysis Scenario 1: Partialy Submerged (Pressure distribution in triangular form) Pressure distribution on square bar in triangular form.Contains proprietary and confidential information of ANSYS.5 .1. All rights reserved.5) m Scenario 2: (2. and its subsidiaries and affiliates. . Inc. .689 104 Release 14. Inc. VMMECH043: Fundamental Frequency of a Simply-Supported Beam Overview Reference: Analysis Type(s): Element Type(s): W. ex. Vibration Theory and Applications. Inc. NJ.. Inc. pg. 2nd Printing. 1965.2836 lb/in3 Geometric Properties ℓ = 80 in A = 4 in2 h = 2 in I = 1.766 Mechanical 28. 18. . Figure 57: Schematic Material Properties E = 3e7 psi ρ=0. 1.532 Release 14.5-1 Modal Analysis Beam Test Case Determine the fundamental frequency f of a simply-supported beam of length ℓ = 80 in and uniform cross-section A = 4 in2 as shown below. Inc. and its subsidiaries and affiliates.613 Error (%) 0. All rights reserved.Contains proprietary and confidential information of ANSYS. Thompson. 105 .3333 in4 Loading Results Comparison Results Frequency (Hz) Target 28.© SAS IP. T. Englewood Cliffs. Prentice-Hall.5 . 5 .© SAS IP. Inc. . . Inc. All rights reserved.Contains proprietary and confidential information of ANSYS.106 Release 14. and its subsidiaries and affiliates. . and its subsidiaries and affiliates. . Strength of Material. Elementary Theory and Problems. NY.Contains proprietary and confidential information of ANSYS. 30. 1955. pg. The entire assembly is subjected to a temperature rise of ∆T.2e-6 / °F VMSIM044_material_steel: Geometric Properties A = 0. 107 .6e7 psi νc = 0 αc = 9. The wires have a cross-sectional area of A. the spacing between the wires is 10 in and the reference temperature is 70°F. Inc. Find the stresses in the copper and steel wire of the structure shown below.VMMECH044: Thermally Loaded Support Structure Overview Reference: S. Linear Thermal Stress Analysis Beam Analysis Type(s): Element Type(s): Test Case An assembly of three vertical wires has a rigid horizontal beam on which a vertically downward force Q is acting. Figure 58: Schematic Material Properties VMSIM044_material_rigid: Er = 3e16 psi νr = 0 VMSIM044_material_copper: Ec = 1. Length of the wires is 20 in. All rights reserved. 3rd Edition. D. Part I. Inc. Timoshenko. problem 9. Van Nostrand Co.5 .. Inc.© SAS IP. The rigid beam is connected to the wires by spot welds. New York.1 in2 Loading Q = 4000 lb (Y direction) ∆T = 10°F Release 14. 236 108 Release 14. and its subsidiaries and affiliates.© SAS IP. Inc.Contains proprietary and confidential information of ANSYS. Inc. All rights reserved. .VMMECH044 Material Properties Es = 3e7 psi νs = 0 αs = 7e-6 / °F Geometric Properties Loading Results Comparison Results Stress in steel (psi) Stress in copper (psi) Target 19695 10152 Mechanical 19637 10176 Error (%) -0.294 0. .5 . problem 7. H. Static Structural Analysis Shell Analysis Type(s): Element Type(s): Test Case A cantilever beam of thickness t and length ℓ has a depth which tapers uniformly from d at the tip to 3d at the wall. N. 109 . McGraw-Hill Book Co.18. All rights reserved. New York. An Introduction to the Mechanics of Solids. Dahl.. Figure 59: Schematic Material Properties Es = 3e7 psi νs = 0 Geometric Properties ℓ = 50 in d = 3 in t = 2 in Loading F = 4000 lb (Y direction) Results Comparison Results Bending stress at mid length (psi) Target 8333 Mechanical 8373. . pg. NY.© SAS IP. as shown..5 . Find the maximum bending stress at the mid-length (X = ℓ ).6 Error (%) 0.Contains proprietary and confidential information of ANSYS. It is loaded by a force F at the tip. Crandall. Inc. C. and its subsidiaries and affiliates. 1959. Inc. 342.VMMECH045: Laterally Loaded Tapered Support Structure Overview Reference: S. Inc.5 Release 14. Inc.© SAS IP. Inc.110 Release 14. All rights reserved.5 . . . and its subsidiaries and affiliates.Contains proprietary and confidential information of ANSYS. 7 No. 1981. 111 . A one-eighth symmetry model is used. 1973.094 in Loading F = 100 lbf (Y direction) Release 14. Concepts and Applications of Finite Element Analysis. pp.© SAS IP. Vol.. NY. Takemoto. Figure 60: Schematic Material Properties Es = 10. All rights reserved. "Some Modifications of an Isoparametric Shell Element". Cook. Analysis Type(s): Element Type(s): Static Structural Analysis Shell Test Case A thin-walled cylinder is pinched by a force F at the middle of the cylinder length. 284-287 H. . Determine the radial displacement δ at the point where F is applied. Inc.5 .953 in t = 0. R. New York. and its subsidiaries and affiliates. One-fourth of the load is applied due to symmetry. 3.5e6 psi νs = 0.VMMECH046: Pinched Cylinder Overview Reference: R. International Journal for Numerical Methods in Engineering. D. Inc. Cook. John Wiley and Sons. D.Contains proprietary and confidential information of ANSYS. 2nd Edition. Inc.3125 Geometric Properties ℓ = 10. The ends of the cylinder are free edges.35 in r = 4. All rights reserved. Inc.11376 Error (%) -0. .© SAS IP.VMMECH046 Analysis Due to symmetrical boundary and loading conditions.5 . . one-eighth model is used and one-fourth of the load is applied.Contains proprietary and confidential information of ANSYS. Inc.1 112 Release 14. and its subsidiaries and affiliates.1139 Mechanical –0. Results Comparison Results Deflection (in) Target -0. VMMECH047: Plastic Compression of a Pipe Assembly Overview Reference: 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. 5.1. Plastic Structural Analysis Axisymmetric Analysis Type(s): Element Type(s): Test Case Two coaxial tubes, the inner one of 1020 CR steel and cross-sectional area As, and the outer one of 2024-T4 aluminum alloy and of area Aa, 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 61: Schematic Material Properties VMSIM047_CR_steel: Es = 26,875,000 psi σ(yp)s = 86,000 psi VMSIM047_T4_aluminum alloy: Geometric Properties ℓ = 10 in Steel: Inside radius = 1.9781692 in Loading 1st Load step: δ = 0.032 in 2nd Load step: δ = -0.05 in 3rd Load step: δ = -0.10 in Release 14.5 - © SAS IP, Inc. All rights reserved. - Contains proprietary and confidential information of ANSYS, Inc. and its subsidiaries and affiliates. 113 VMMECH047 Material Properties Ea = 11,000,000 psi σ(yp)a = 55,000 psi ν = 0.3 Geometric Properties Wall thickness = 0.5 in Aluminum: Inside radius = 3.5697185 in Wall thickness = 0.5 in Loading Analysis 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 Deflection @ 0.032 in Load, lb for Deflection @ 0.05 in Load, lb for Deflection @ 0.1 in Target 1.0244e6 1.262e6 1.262e6 Mechanical 1033900 1262800 1267200 Error (%) 0.9 0.1 0.412 114 Release 14.5 - © SAS IP, Inc. All rights reserved. - Contains proprietary and confidential information of ANSYS, Inc. and its subsidiaries and affiliates. VMMECH048: Bending of a Tee-Shaped Beam Overview Reference: 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. 7.2. Static Structural Analysis Beam Analysis Type(s): Element Type(s): Test Case Find the maximum tensile and compressive bending stresses in an unsymmetrical T beam subjected to uniform bending Mz, with dimensions and geometric properties as shown below. Figure 62: Schematic Material Properties E = 3e7 psi Geometric Properties b = 1.5 in h = 8 in y = 6 in Area = 60 in2 Iz = 2000 in4 Loading Mz = 100,000 lbf-in (Z direction) Results Comparison Results StressBEND, Bottom (psi) StressBEND, Top (psi) Target 300 -700 Mechanical 300 -700 Error (%) 0 0 115 Release 14.5 - © SAS IP, Inc. All rights reserved. - Contains proprietary and confidential information of ANSYS, Inc. and its subsidiaries and affiliates. 116 Release 14.5 - © SAS IP, Inc. All rights reserved. - Contains proprietary and confidential information of ANSYS, Inc. and its subsidiaries and affiliates. problem 2. Strength of Material.3 Geometric Properties ℓ = 25 ft r = 2.33508 in d = 3 ft Loading F = 250 lb (Y direction) M = 9000 lbf-in (Z direction) Release 14. Static Structural Analysis Beam Analysis Type(s): Element Type(s): Test Case A vertical bar of length ℓ and radius r is subjected to the action of a horizontal force F acting at a distance d from the axis of the bar. Elementary Theory and Problems. D.© SAS IP. Inc.. New York. pg. NY.VMMECH049: Combined Bending and Torsion of Beam Overview Reference: S.5 . and its subsidiaries and affiliates. Inc. 1955. 3rd Edition. Part I.Contains proprietary and confidential information of ANSYS. Figure 63: Problem Sketch Figure 64: Schematic Material Properties E = 3e7 psi ν = 0. Inc. 117 .. Timoshenko. All rights reserved. . 299. Van Nostrand Co. Determine the maximum principal stress σmax. VMMECH049 Results Comparison Results Principal stressmax (psi) Target 7527 Mechanical 7515.5 Error (%) -0.153 118 Release 14.5 - © SAS IP, Inc. All rights reserved. - Contains proprietary and confidential information of ANSYS, Inc. and its subsidiaries and affiliates. VMMECH050: Cylindrical Shell under Pressure Overview Reference: 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, 1981. Analysis Type(s): Element Type(s): 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 σy and the hoop stress σz in the vessel at the mid-thickness of the wall. Figure 65: Schematic Release 14.5 - © SAS IP, Inc. All rights reserved. - Contains proprietary and confidential information of ANSYS, Inc. and its subsidiaries and affiliates. 119 VMMECH050 Material Properties E = 3e7 psi ν = 0.3 Geometric Properties t = 1 in d = 120 in Loading P = 500 psi (radial direction) Analysis An axial force of 5654866.8 lb ((Pπd2)/4) is applied to simulate the closed-end effect. Results Comparison Results Stressy (psi) Stressz (psi) Target 15000 30000 Mechanical 15000 30002 Error (%) 0 0.007 120 Release 14.5 - © SAS IP, Inc. All rights reserved. - Contains proprietary and confidential information of ANSYS, Inc. and its subsidiaries and affiliates. VMMECH051: Bending of a Circular Plate Using Axisymmetric Elements Overview Reference: 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. Static Structural Analysis Axisymmetric Shell element Analysis Type(s): Element Type(s): 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 σmax for each case. Case 1: Uniform loading P, clamped edge Case 2: Concentrated center loading F, clamped edge Figure 66: Schematic Case 1: Case 2: Release 14.5 - © SAS IP, Inc. All rights reserved. - Contains proprietary and confidential information of ANSYS, Inc. and its subsidiaries and affiliates. 121 and its subsidiaries and affiliates.VMMECH051 Material Properties E = 3e7 psi ν = 0.© SAS IP.08736 7200 -0.3 Geometric Properties r = 40 in t = 1 in Loading Case 1: P = 6 psi Case 2: F = -7539.087114 7212. the above problem can be analyzed as an axisymmetric problem.282 0.82 lb (y direction) Analysis Because the geometry and loading are symmetric about the y-axis. .Contains proprietary and confidential information of ANSYS.219 122 Release 14. All rights reserved.8 -0. Inc.9 Error (%) -0.08736 3600 Mechanical -0.761 0. Inc.5 .088025 3607. . Results Comparison Results Case 1: Case 2: Deflection (in) Stressmax (psi) Deflection (in) Stressmax (psi) Target -0.178 0. All rights reserved. Inc.Contains proprietary and confidential information of ANSYS.5 . and its subsidiaries and affiliates. .57 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 67: Schematic Release 14.VMMECH052: Velocity of Pistons for Trunnion Mechanism Overview Reference: Analysis Type(s): Element Type(s): Any Basic Kinematics book 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 = 12. Inc. 123 .© SAS IP. VMMECH052 Material Properties E = 2e11 Pa ν = 0.72 Error (%) -0. Inc.8 955 Mechanical 497. Consider the Space Diagram. All rights reserved.Contains proprietary and confidential information of ANSYS.3 Geometric Properties AB = 400 mm AC = 150 mm CE = 350 mm EF = 300 mm Loading Constant angular velocity at crank = 12. . Inc.57 rad/s Analysis Analysis done using graphical solution. Figure 68: Schematic Results Comparison Results Velocity of Piston (F) m/s Velocity of Piston (B) m/s Target 501. .949 0. Velocity Diagram at the 180° from Initial Position. and its subsidiaries and affiliates.04 959.494 124 Release 14.© SAS IP.5 . 571*sin (0. 125 .5235*t) rad The hinge point coordinates are: 1.3 Geometric Properties Hinge point = (0. and its subsidiaries and affiliates.VMMECH053: Simple Pendulum with SHM motion Overview Reference: Analysis Type(s): Element Type(s): Any Basic Kinematics book Rigid Dynamic Analysis Multipoint Constraint Element Test Case A simple pendulum as shown in Figure 69: Schematic (p. 0.5235*t) rad Analysis The pendulum is having SHM motion in X-Z plane about the hinge. Hinge point = (0. .© SAS IP. All rights reserved.56) mm Find the relative angular acceleration of pendulum after t = 3s. 0. Inc.Contains proprietary and confidential information of ANSYS. Angular acceleration of pendulum: Release 14.5 .56) mm Loading Rotation θ = 1. 35.571*sin (0. -35. Figure 69: Schematic Material Properties E = 2000000 MPa ν = 0. 125) has a SHM motion about its hinged point given by the following equation: θ = 1. Inc. 5 . Inc. All rights reserved.43054 Error (%) -0.433 Mechanical -0.© SAS IP. . Inc.568 126 Release 14. . and its subsidiaries and affiliates.VMMECH053 α= α=− ω 2 Results Comparison Results Relative angular acceleration of pendulum after t = 3s (rad/s2) Target -0.Contains proprietary and confidential information of ANSYS. The WB/Mechanical results are compared to a fourth order Runge-Kutta solution. Jy. Its mass is m. 127 .Contains proprietary and confidential information of ANSYS. ɺɺ − ( J − J + 1 ml 2 + Ml 2 ) Ω 2 ( J z + 14 ml 2 + Ml 2 )θ x y 4 θ θ + mgl θ + Mgl θ= The problem is solved for { { during the first second of motion.© SAS IP. The angle of the bar A to the vertical axis is denoted as . A point mass M is attached at the tip of the bar in the figure below. All rights reserved. The vertical shaft is rotating around its vertical axis at a constant velocity Ω. its rotational inertia to its principal axis are Jx. . and its subsidiaries and affiliates. The length of bar A is L. Jz. The motion equation has been established as follows. Inc. Release 14.VMMECH054: Spinning Single Pendulum Overview Reference: Analysis Type(s): Element Type(s): Any Basic Kinematics book Rigid Dynamic Analysis Multipoint Constraint Element Test Case A uniform bar A is connected to a vertical shaft by a revolute joint. Inc.5 . 2361 m m = 551.1368 Mechanical -1. All rights reserved.0 ¢ at 0.3233 116.0 0.VMMECH054 Figure 70: Schematic Material Properties Geometric Properties L= 2.97 kg-m2 Jy = 2.Contains proprietary and confidential information of ANSYS. . .3233 116. Inc.1368 Error (%) 0.5 .97 kg-m2 Loading Ω = 17. Inc.5 sec £ at 0.© SAS IP.1522 = tan-1(1.45 kg M = 100.5 sec 128 Release 14.7293 kg-m2 Jz = 229.0 kg Jx = 229. and its subsidiaries and affiliates.2) ¡ =0 Results Comparison Results Target -1. All rights reserved.Contains proprietary and confidential information of ANSYS. . and its subsidiaries and affiliates.© SAS IP.6755 119.0 at 1.0 sec ¡ at 1.0 sec Figure 71: Plot of¢ from 0 to 1 sec Figure 72: Plot of£ from 0 to 1 sec Release 14.VMMECH054 Results Target -2. Inc.5 .6755 119. 129 .0 0. Inc.8471 Mechanical -2.8471 Error (%) 0. All rights reserved.130 Release 14.Contains proprietary and confidential information of ANSYS.5 . . and its subsidiaries and affiliates. Inc. Inc.© SAS IP. . Inc. .5 . and its subsidiaries and affiliates. The mechanism is driven by the drive wheel rotating at a constant -58. Figure 73: Schematic Release 14.© SAS IP. • Link AB length r1 = 18mm • Link BC length r2 = 48mm • Length BX = x = 45 mm and CX = y = 28 mm The horizontal distance between A and C is length=34 mm. All rights reserved.Contains proprietary and confidential information of ANSYS. 131 . The link lengths of all the links are constant as given below.643 rad/s. Determine the acceleration of point C with a change of angle of link AB (θ1) from 0 to 60° in counter clockwise direction.VMMECH055: Projector mechanism.finding the acceleration of a point Overview Reference: Analysis Type(s): Element Type(s): Any Basic Kinematics book Rigid Dynamic Analysis Multipoint Constraint Element Test Case The mechanism shown in figure is used to pull a movie through a projector. Inc. and its subsidiaries and affiliates.7386 Error (%) -0.© SAS IP.317 6.015 -0.06 1.3 Geometric Properties r1 = 18 mm r2 = 48 mm x = 45 mm y = 28 mm Loading Constant rotational velocity = -58. Inc.VMMECH055 Material Properties E = 2e11 Pa ν = 0.5 . Inc.Contains proprietary and confidential information of ANSYS. All rights reserved. .739 Mechanical -12.643 rad/s Analysis Linear acceleration of point C is given by αc = 2 θ1 − θ2 + θ2 Results Comparison Results Relative acceleration (θ1 = 10) mm/s2 Relative acceleration (θ1 = 30) mm/s2 Relative acceleration (θ1 = 60) mm/s2 Target -12.006 132 Release 14.141 -0.3168 6. .043 1. The cylinders are rotating at a uniform speed of 300 rpm in a clockwise direction. Inc. Find Angular acceleration of the connecting rod. . Loading Constant rotational velocity = 300 rpm Release 14.© SAS IP. OA is 50mm long and fixed at point o. The length of the connecting rod AB is 125mm. about the fixed center O.Contains proprietary and confidential information of ANSYS.VMMECH056: Coriolis component of acceleration-Rotary engine problem Overview Reference: Analysis Type(s): Element Type(s): Any Basic Kinematics book Rigid Dynamic Analysis Multipoint Constraint Element Test Case Kinematics diagram of one of the cylinders of a rotary engine is shown below.5 . The line of stroke OB is inclined at 50° to the vertical.3 Geometric Properties Connecting rod AB is 125mm Crank OA is 50mm long OB is inclined at 50° to the vertical. All rights reserved. and its subsidiaries and affiliates. Figure 74: Schematic Material Properties E = 2e11 Pa ν = 0. Inc. 133 . © SAS IP.Contains proprietary and confidential information of ANSYS.5 . All rights reserved.VMMECH056 Analysis Angular acceleration of the connecting rod is given by: α AB = αt 3 Results Comparison Results Angular acceleration (radian/s2) Target 294.53 Error (%) 0 134 Release 14. . Inc. and its subsidiaries and affiliates.52 Mechanical 294. . Inc. Knowing that the collar is at rest at "C" and is given a slight push to get it moving.Contains proprietary and confidential information of ANSYS.© SAS IP. Inc. Inc. Figure 75: Schematic Release 14. All rights reserved. Length OP = 75 mm. TATA McGRAW HILL Edition 2004. and its subsidiaries and affiliates.2 Kg collar is attached to a spring and slides without friction along a circular rod in a vertical plane. 7th Edition.VMMECH057: Calculation of velocity of slider and force by collar Overview Reference: Beer-Johnston ‘Vector Mechanics for Engineers’ Statics & Dynamics (In SI Units). Determine the force exerted by the rod on the collar as it passes through point "A" and "B". Page No: 793 Rigid Dynamic Analysis Multipoint Constraint Element Analysis Type(s): Element Type(s): Test Case A 1. .5 .73. Length OB = 180 mm. The spring has an undeformed length of 105 mm and a constant K = 300 N/m. 135 . Problem 13. © SAS IP.Contains proprietary and confidential information of ANSYS. All rights reserved.667 Error (%) 0. .3 Geometric Properties Spring: Undeformed length = 105 mm Stiffness K = 300 N/m Loading Gravitational acceleration = -9.88 -23.6 Mechanical 14. Inc.3 136 Release 14.992 -23. .8066 m/s2 (Y Direction) Results Comparison Results At point A (N) At point B (N) Target 14. Inc. and its subsidiaries and affiliates.753 0.VMMECH057 Material Properties E = 2e11 Pa ν = 0.5 . VMMECH058: Reverse four bar linkage mechanism Overview Reference: Analysis Type(s): Element Type(s): Results are simulated using MATLAB Rigid Dynamic Analysis Multipoint Constraint Element Test Case The figure (below) shows a reverse four bar linkage consisting of uniform rigid links PQ.81m/sec2 Determine the angular accelerations. The link lengths of all the links are constant as given below. QR. Inc.© SAS IP.5 . Inc.5m • Crank Link PQ length r2 = 0.15m • Link QR length r3 = 0. . • Fixed Link PS length r1 = 0. Release 14. respectively. 137 . and RS and ground PS.45m • Gravity g = 9.4m • Link RS length r4 = 0. All rights reserved. angular velocity and rotation of link RS at joint R. Link PQ is connected with revolute joints to links QR and PS at points Q and P.Contains proprietary and confidential information of ANSYS. respectively. Link RS is connected with revolute joints to links QR and PS at points R and S. and its subsidiaries and affiliates. and its subsidiaries and affiliates.7 138 Release 14.VMMECH058 Figure 76: Schematic Material Properties E = 2e11 Pa ν = 0.8066 m/s2 (Y Direction) Analysis Results are obtained using MATLAB.671 -0.45m Loading Gravitational acceleration = -9.© SAS IP.3 Geometric Properties Link PS length r1 = 0.36 Mechanical 39.15m Link QR length r3 = 0.1247 -0. Inc.7 0.4m Link RS length r4 = 0. .Contains proprietary and confidential information of ANSYS.16 -0. All rights reserved.6 -5. Results Comparison Results Angular Acceleration (rad/s ) Angular Velocity (rad/sec) Rotation (rad) 2 Target 39.5m Link PQ length r2 = 0.36255 Error (%) -0.5 . . Inc.336 -5. . 1965. Figure 77: Schematic Case 1: Case 2: Release 14. Inc.VMMECH059: Bending of a solid beam (Plane elements) Overview Reference: Analysis Type(s): Element Type(s): R. and its subsidiaries and affiliates.. Inc. Static Structural Analysis 2-D Plane Stress Shell element Test Case A beam of length ℓ 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. Formulas for Stress and Strain.. 106. 139 . 104. pp. New York. Inc. J. determine the deflection δ at the free end and the bending stress σBend at a distance d from the wall at the outside fiber. McGrawHill Book Co.© SAS IP. NY. 4th Edition.Contains proprietary and confidential information of ANSYS. Roark.5 . All rights reserved. 00500 -4050 Mechanical 0.0051232 -4051. .3 Geometric Properties ℓ = 10 in h = 2 in d = 1 in Loading Case 1: M = 2000 ibf-in (Z direction) Case 2: F = 300 lb (Y direction) Analysis Since the loading is uniform and in one plane. and its subsidiaries and affiliates.5 Error (%) 0 0 2 0 140 Release 14.© SAS IP. All rights reserved. . Results Comparison Results Case 1: Case 2: Deflection (in) StressBend (psi) Deflection (in) StressBend (psi) Target 0.VMMECH059 Material Properties E = 30 x 106 psi ν = 0.00500 -3000 0.5 . Inc.00500 -3000 0. the above problem can be analyzed as a plane stress problem.Contains proprietary and confidential information of ANSYS. Inc. and its subsidiaries and affiliates. Figure 78: Schematic Material Properties E = 2e5 MPa ν = 0. Inc.5 .VMMECH060: Crank Slot joint simulation with flexible dynamic analysis Overview Reference: Analysis Type(s): Element Type(s): Mechanical APDL Multibody Analysis Flexible Dynamic Analysis Solid and Multipoint Constraint Element Test Case The figure shows crank slot model consists of a base and two rods. Determine the Equivalent (von Mises) Stress for both flexible rods. The base of the model is fixed to the ground via a fixed joint and Bolt3 connected with slot joint to base.Contains proprietary and confidential information of ANSYS. Define Rod1 and Rod2 as a flexible body and run the crank slot analysis using a Flexible Dynamic Analysis. Inc.3 Geometric Properties Rod1 length = 75mm Rod2 length = 115mm Loading Constant angular acceleration at base to Bolt1 = 25 rad/s2 Release 14. . All rights reserved.© SAS IP. The two rods are attached to each other and the base with three bolts. 141 . 40834 Error (%) 2. Inc.VMMECH060 Analysis Figure 79: Contour Plot Figure 80: Equivalent (von Mises) Stress Figure 81: Total Force at Base to Bolt1 Results Comparison Results Equivalent (von Mises) Stress (MPa) Target 0.Contains proprietary and confidential information of ANSYS. . .5 .6 142 Release 14.398 Mechanical 0. and its subsidiaries and affiliates. Inc. All rights reserved.© SAS IP. and its subsidiaries and affiliates.Contains proprietary and confidential information of ANSYS.© SAS IP.141 Release 14. .67 Mechanical 7. Inc.5 .VMMECH060 Results Force @ Bolt1 (N) Target 7.6808 Error (%) 0. All rights reserved. 143 . Inc. Inc.Contains proprietary and confidential information of ANSYS. .© SAS IP. All rights reserved. .144 Release 14. and its subsidiaries and affiliates. Inc.5 . . New York.5 . 241. All rights reserved. pg. Strength of Material.264 Release 14. Inc. D.Contains proprietary and confidential information of ANSYS. eq. is loaded by a vertical (Z) load F applied at the end B. 145 .3 Geometric Properties r = 100 in d = 2 in θ = 90° Loading F = -50 lb (Z direction) Results Comparison Results Deflection (in) Target -2. 3rd Edition. Part I. Inc. 412. Timoshenko. Inc. 1955.VMMECH061: Out-of-plane bending of a curved bar Overview Reference: S.655 Error (%) 0.648 Mechanical -2. built-in at A. Van Nostrand Co. Elementary Theory and Problems. Determine the deflection δ at end B and the maximum bending stress σBend. Static Structural Analysis Beam Analysis Type(s): Element Type(s): Test Case A portion of a horizontal circular ring. The ring has a solid circular cross-section of diameter d. .. Figure 82: Schematic Material Properties E = 30 x 106 psi ν = 0. NY. and its subsidiaries and affiliates.© SAS IP. 0 Mechanical 6399. Inc.© SAS IP. .5 . Inc. . All rights reserved.VMMECH061 Results StressBend (psi) Target 6366.Contains proprietary and confidential information of ANSYS.522 146 Release 14.2 Error (%) 0. and its subsidiaries and affiliates. the radial stress σr. problem 1 and pg.. article 42.Contains proprietary and confidential information of ANSYS. 1956. 147 . Internal pressure is then removed and the cylinder is subjected to a rotation ω about its center line. Timoshenko. Elementary Theory and Problems. Strength of Material. 213. . Determine the radial σr and tangential σt stresses at the inner wall and at an interior point located at r = Xi.5 . D. and its subsidiaries and affiliates. Static Structural Analysis Axisymmetric Shell Analysis Type(s): Element Type(s): Test Case A long thick-walled cylinder is initially subjected to an internal pressure p. Van Nostrand Co. Determine the radial displacement δr at the inner surface. and tangential stress σt.. Inc. pg. 213. at the inner and outer surfaces and at the middle wall thickness. NY. Inc. 3rd Edition.© SAS IP.VMMECH062: Stresses in a long cylinder Overview Reference: S. Inc. Figure 83: Schematic Case 1: Case 2: Release 14. New York. All rights reserved. Part II. . psi (r = 4 in) Stresst.035 --0.005 Case 2: 0 40588. 4753. psi (r = 4 in) Stresst. Results Comparison Results Case 1: Displacementr. in (r = 4 in) Stressr. 6. and its subsidiaries and affiliates.43 in) Stresst. 29436.024 -0. psi (r = 5.000 psi (radial direction) Case 2: Rotational velocity = 1000 rad/s (Y direction) Analysis Because the geometry and loading are symmetric about the y-axis. .79611 49988 27775 19999 Error (%) -3.961 148 Release 14.0078666 -30000. psi (r = 5. -7778.671 3. psi (r = 6 in) Stresst.43 in) Target 0. psi (r = 4 in) Stressr. 20000. psi (r = 8 in) Stresst.Contains proprietary and confidential information of ANSYS. psi (r = 4 in) Stressr.© SAS IP.011 -0.802 0.VMMECH062 Material Properties E = 30 x 106 psi ν = 0. the above problem can be analyzed as an axisymmetric problem.0076267 -29988 -7775.05 -0.3 ρ = 0. All rights reserved. 0 50000. Mechanical 0. psi (r = 8 in) Stressr.5 . 27778.43 in Loading Case 1: Pressure = 30. Inc.04 -0. psi (r = 6 in) Stresst.281826 lbm/in3 Geometric Properties a = 4 in b = 8 in Xi = 5.3 0.5483 41672 4933.7 29719 -2. Inc. width b and thickness t is fixed at one end and subjected to a pure bending moment M at the free end. Release 14. E.5 . Static Structural Analysis Shell Analysis Type(s): Element Type(s): Test Case A cantilever plate of length ℓ .© SAS IP. and its subsidiaries and affiliates.708 N-mm (Y direction) Analysis Large deformation is used to simulate the problem. Figure 84: Schematic Material Properties E = 1800 N/mm2 ν = 0.0 Geometric Properties ℓ = 12 mm b = 1 mm t = 1 mm Loading M = 15. 720. Bathe. 22 No. Dvorkin.Contains proprietary and confidential information of ANSYS.The Use of Mixed Interpolation of Tensorial Components” . . 149 . 1986. "A Formulation of General Shell Elements . Vol. Determine the true (large deflection) free-end displacements and the top surface stress at the fixed end using shell elements. Inc. pg. International Journal for Numerical Methods in Engineering.VMMECH063: Large deflection of a cantilever Overview Reference: K. Inc. J. All rights reserved. N. 3. Inc.608 94.Contains proprietary and confidential information of ANSYS.221 1. .662 0.9354 -6. All rights reserved. .266 Error (%) 1.5 .25 Mechanical -2.© SAS IP.9 -6. Inc.017 150 Release 14. and its subsidiaries and affiliates.5 94.VMMECH063 Results Comparison Results Directional Deformation Xdirection (mm) Directional Deformation Zdirection (mm) Normal Stress X-direction (N/mm2) Target -2. 0 Geometric Properties a = 1 in b = 1.5 .1 in β = 7° = 0. New York.© SAS IP. Timoshenko. Figure 85: Schematic Material Properties E = 30 x 106 psi ν = 0. problem 2.VMMECH064: Small deflection of a Belleville Spring Overview Reference: S. All rights reserved. .8 Release 14..5 in t = 0..12217 rad Loading Line pressure = 100 lb/in (Y direction) Results Comparison Results Directional Deformation Ydirection (in) Target -0.0029273 Error (%) 3. Elementary Theory and Problems.0028205 Mechanical -0. 1956. Van Nostrand Co. 151 . NY.Contains proprietary and confidential information of ANSYS. Determine the deflection y produced by a load F per unit length on the inner edge of the ring. Strength of Material. and its subsidiaries and affiliates. Part II. pg. D. 3rd Edition. Inc. Static Structural Analysis Shell Analysis Type(s): Element Type(s): Test Case The conical ring shown below represents an element of a Belleville spring. Inc. Inc. 143. Inc. All rights reserved.© SAS IP.Contains proprietary and confidential information of ANSYS.5 . and its subsidiaries and affiliates. Inc. . .152 Release 14. pg. Calculate the stresses and the thermal strain in the bar after it has been heated to 170°F.Contains proprietary and confidential information of ANSYS.5 x 106 psi α = 1. Static Thermal Stress Analysis Solid and Shell Test Case An aluminum-alloy bar is initially at a temperature of 70°F.5 .. Inc. .002 in. New York. Use a global mesh size of 0.25 in. Inc. Loading ∆t = 170°F .VMMECH065: Thermal Expansion to Close a Gap at a Rigid Surface Overview Reference: Analysis Type(s): Element Type(s): C. and its subsidiaries and affiliates. Introduction to Stress Analysis.25 x 10 /°F -5 Geometric Properties ℓ = 3 in. O. NY. δ = 0. The Macmillan Co. 153 . problem 8. Harris.70°F Release 14. Figure 86: Schematic Material Properties E = 10. All rights reserved. 58. 1959.© SAS IP. The supports are assumed to be rigid. Inc. . .© SAS IP. All rights reserved.VMMECH065 Material Properties ν = 0.25e-003 Mechanical -6122.Contains proprietary and confidential information of ANSYS.4 1.0 Geometric Properties Loading Results Comparison Results Normal Stress Y (psi) Thermal Strain Y (in/in) Target -6125 1. Inc. and its subsidiaries and affiliates.5 .25e-003 Error (%) 0 0 154 Release 14. Harris.. problem 61. NY. Figure 87: Schematic Material Properties E = 30 x 106 psi ν = 0.Contains proprietary and confidential information of ANSYS. and its subsidiaries and affiliates. 155 . .© SAS IP. The Macmillan Co. Use a global mesh size of 0. Static Structural Analysis Shell Test Case A tapered cantilever plate of rectangular cross-section is subjected to a load F at its tip. Introduction to Stress Analysis. Find the maximum deflection δ and the maximum principal stress σ1 in the plate. 1959.5 . O.5 in Loading F = 10 lbf Release 14. 114. All rights reserved. New York.0 Geometric Properties L = 20 in d = 3 in t = 0.VMMECH066: Bending of a Tapered Plate Overview Reference: Analysis Type(s): Element Type(s): C. pg.75 in. Inc. Inc. 9 -0.5 . and its subsidiaries and affiliates. Inc.Contains proprietary and confidential information of ANSYS. Inc.7 -0. . All rights reserved.2 156 Release 14.042667 Mechanical 1614.© SAS IP.042746 Error (%) 0. .VMMECH066 Results Comparison Results Maximum Principal Stress (psi) Directional Deformation Z (in) Target 1600 -0. Static Structural Analysis Solid Test Case A tapered aluminum alloy bar of square cross-section and length L is suspended from a ceiling. 237. 1959. Use a global mesh size of 0. The Macmillan Co.© SAS IP. 157 . . Inc. Introduction to Stress Analysis. New York. All rights reserved. Determine the maximum axial deflection δ in the bar and the axial stress σy at mid-length (Y = L/2). O.5 in.5 . An axial load F is applied to the free end of the bar.VMMECH067: Elongation of a Solid Tapered Bar Overview Reference: Analysis Type(s): Element Type(s): C. problem 4. Harris. Inc. NY.Contains proprietary and confidential information of ANSYS. and its subsidiaries and affiliates. pg.. Figure 88: Schematic Release 14. 0048077 4444 Mechanical 0.428 158 Release 14.4 x 106 psi ν = 0. Inc. . Inc.0.0. .0048215 4463 Error (%) .5 .3 Geometric Properties L = 10 in d = 2 in Loading F = 10000 lbf Results Comparison Results Directional Deformation Y (in) Normal Stress Y at L/2 (psi) Target 0.© SAS IP.VMMECH067 Material Properties E = 10.287 . All rights reserved.Contains proprietary and confidential information of ANSYS. and its subsidiaries and affiliates. Inc. Use a global mesh size of 0. Timoshenko. just below the yield strength of the material. and the tangential (hoop) stress.5 . . at the same locations for a pressure. pg.. Static. Elementary Theory and Problems. which brings the entire cylinder wall into a state of plastic flow. and its subsidiaries and affiliates. New York. a fully elastic material condition. pel. NY. article 70. 1956. Van Nostrand Co. All rights reserved. D. Figure 89: Schematic Release 14.Contains proprietary and confidential information of ANSYS. Plastic Analysis (Plane Strain) 2-D Structural Solid Analysis Type(s): Element Type(s): Test Case A long thick-walled cylinder is subjected to an internal pressure p (with no end cap load). Part II. at locations near the inner and outer surfaces of the cylinder for a pressure.. σt. 3rd Edition.VMMECH068: Plastic Loading of a Thick Walled Cylinder Overview Reference: S. 159 . σr. Inc.4 in along with a Mapped Face Meshing. σeff. Determine the effective (von Mises) stress. 388.© SAS IP. pult. Inc. Determine the radial stress. Strength of Material. 990 psi pult = 24. and its subsidiaries and affiliates. psi (X = 4.r.4 18608 -469.4 in) Stresseff.4 -0.4 in) Stressr.6 in) Stresst. All rights reserved. Inc.6 in.3 Geometric Properties a = 4 in b = 8 in Loading pel = 12.6 in) Fully Plastic Stresseff.6 in) Target -9984 18608 -468 9128 30000 30000 Mechanical -9948.4 in and 7.4 in) Stresst. Symmetry conditions are used on the edges perpendicular to X and Y axes.011 psi Analysis This problem is modeled as a plane strain problem with only a quarter of the cross-section as shown in the above figures. psi (X = 4.Contains proprietary and confidential information of ANSYS. Inc. psi (X = 7. .2 30000 30000 Error (%) -0. Load is applied in two steps as shown in the above table. w. psi (X = 4.2 0. psi (X = 7.3 0 0 0. psi (X = 7. Results Comparison Results Fully Elastic Stressr. .5 .VMMECH068 Material Properties E = 30 x 106 psi σyp = 30.00 160 Release 14.24 9129.t a cylindrical coordinate system whose origin is same as that of the global coordinate system.© SAS IP. The stresses are calculated at a distance of r = 4.000 psi ν = 0. and its subsidiaries and affiliates. Static Analysis Shell Analysis Type(s): Element Type(s): Test Case A cylindrical shell roof of density ρ is subjected to a loading of its own weight. John Wiley and Sons.Contains proprietary and confidential information of ANSYS. D. The roof is supported by walls at each end and is free along the sides.VMMECH069: Barrel Vault Roof Under Self Weight Overview Reference: R. 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. Inc. All rights reserved. Concepts and Applications of Finite Element Analysis. pp. New York. Figure 90: Schematic Release 14. 161 . 1981..© SAS IP. . 284-287. Inc. Inc. Cook. NY. 2nd Edition.5 . 7347 kg/m3 Geometric Properties t = 0.116 3.362 2. Top @ A. Displacements. .639 -3.© SAS IP. Pa Stressz.738 -3.25 m r = 25 m ℓ = 50 m Θ = 40° Loading g = 9.762 2.3 ρ = 36.5 . Bottom @ B. and the longitudinal rotation. Pa Stressangle. ROTZ.1593 215570 340700 191230 -218740 Mechanical -0.32 x 108 N/m2 ν = 0. . Pa Target -0.30903 -0.8 m/s2 Analysis A one-fourth symmetry model is used. Pa Stressangle.Contains proprietary and confidential information of ANSYS. and its subsidiaries and affiliates. Inc. Bottom @ A. are constrained at the roof end to model the support wall. All rights reserved. UX and UY. Inc.3019 -0. Results Comparison Results Directional Deformation Y @ A.5 162 Release 14.16267 223680 350030 184270 -210980 Error (%) 2. m Directional Deformation X @ A. m Stressz. Top @ B.VMMECH069 Material Properties E = 4. Finite Elements of Nonlinear Continua.Contains proprietary and confidential information of ANSYS. 163 . Figure 91: Schematic Release 14. Inc.© SAS IP. 325-331.. and its subsidiaries and affiliates. T. . Use a global mesh size of 1 in along with a Mapped Face Meshing.5 . McGraw-Hill Book Co. An internal pressure of Pi is applied.VMMECH070: Hyperelastic Thick Cylinder Under Internal Pressure Overview Reference: Analysis Type(s): Element Type(s): J.. pp. Inc. 1972. New York. Find the radial displacement at the inner radius and the radial stress at radius R = 8. Oden. All rights reserved. NY. Static. Inc. Large Deflection Analysis 2-D Structural Solid Elements Test Case An infinitely long cylinder is made of Mooney-Rivlin type material.16 in. All rights reserved.VMMECH070 Material Properties Mooney-Rivlin material coefficients: C10 = 80 psi C01 = 20 psi D1 = 0 /psi Geometric Properties ri = 7.625 in Loading Pi = 150 psi Analysis Because of the loading conditions and the infinite length. Inc.016 164 Release 14.Contains proprietary and confidential information of ANSYS.5 .16 in.0 in ro = 18. . A one-fourth symmetry model is used.1819 -121.98 Error (%) 0. in Radial Stress at r = 8. Inc. . The total pressure is applied in two load increments 90 and 150 psi.18 -122 Mechanical 7. this problem is solved as a plane strain problem. Results Comparison Results Deformation at inner radius in radial direction. Stress and Deformation are expressed in cylindrical coordinate system. psi Target 7.026 0. and its subsidiaries and affiliates.© SAS IP. © SAS IP.Contains proprietary and confidential information of ANSYS. Choi. Y. The surface convection coefficient between the wire and the air (at temperature Ta) is h. .5 . pg. 165 . Rohsenow. 6. Prentice-Hall. 2nd Printing. Inc. Also.. Inc. Englewood Cliffs.5. 1963. H. 106. NJ.VMMECH071: Centerline Temperature of a Heat Generating Wire Overview Reference: W. Figure 92: Schematic Release 14. M. Heat. Thermal Analysis 2-D Thermal Solid Elements Analysis Type(s): Element Type(s): Test Case Determine the centerline temperature TcL and the surface temperature Ts of a bare steel wire generating heat at the rate Q. determine the heat dissipation rate q.02 ft along with a Mapped Face Meshing. and its subsidiaries and affiliates. ex. Inc. Mass and Momentum Transfer. All rights reserved. Use a global mesh size of 0. 9 417.94 417.Contains proprietary and confidential information of ANSYS.92 Btu/s-ft3 Analysis Because of the symmetry in loading conditions and in the geometry.03125 ft Loading h = 1. Inc.094861 Error (%) 0.9 -0.012 0.01 0.5 . Results Comparison Results Centerline Temperature. Inc. and its subsidiaries and affiliates. BTU/s Target 419. All rights reserved.094861 Mechanical 419.VMMECH071 Material Properties k = 3.6111 x 10 Btu/sft-°F -3 Geometric Properties ro = 0.3889 x 10-3 Btu/s-ft2-°F Ta = 70°F Q = 30.85 -0. °F Heat dissipation rate. . this problem is solved as an axisymmetric problem.00 166 Release 14. °F Surface Temperature. The solution is based on a wire 1 foot long. .© SAS IP. 1875 in b = 0. Also determine the axial stress σa and the tangential (hoop) stress σt at the inner and outer surfaces Edge sizing is used for all edges and edge behavior is defined as hard. 3rd Edition.625 in Loading Ti = -1°F Release 14. Elementary Theory and Problems. NY.5 . 234. . Determine the temperature distribution through the wall thickness. New York. Strength of Material. Inc. D.VMMECH072: Thermal Stresses in a Long Cylinder Overview Reference: S. 167 . and its subsidiaries and affiliates.. problem 1. 1956. Inc. Van Nostrand Co. Timoshenko. Inc. Thermal Stress Analysis 2-D Thermal Solid Elements Analysis Type(s): Element Type(s): Test Case A long thick-walled cylinder is maintained at a temperature Ti on the inner surface and To on the outer surface. All rights reserved. pg.© SAS IP. Figure 93: Schematic Material Properties E = 30 x 106 psi Geometric Properties a = 0. Part II.Contains proprietary and confidential information of ANSYS. .298 168 Release 14.5 . Nodal coupling is used in the static stress analysis. psi (at X = 0.Contains proprietary and confidential information of ANSYS.25 -195.42 420.625 in) Static Analysis Stressa.034 0 Error (%) -0. Model is used for the thermal and stress solutions.6706 0 Mechanical 416.°F (at X = 0.333e-4 Btu/s-in-°F Geometric Properties Loading To = 0°F Analysis Because of the symmetry in loading conditions and in the geometry.16 Error (%) 0 0.1875 in) T. All rights reserved. psi (at X = 0.344 0.58 -194.0000 -0.3 k = 8.58 Mechanical -1.68 406.1875 in) Stresst. Results Comparison Thermal Analysis T. The axial length is arbitrary and it is taken has 0. psi (at X = 0. .1 in.67037 0 Target 420.240 0.8 -195. psi (at X = 0.1875 in) Stressa.625 in) Target -1.435 x 10-5/°F ν = 0.42 -194. Inc.625 in) Stresst. and its subsidiaries and affiliates.°F (at X = 0.VMMECH072 Material Properties α = 1. Inc.°F (at X = 0.890 -3.0000 -0.2788 in) T.© SAS IP. this problem is solved as an axisymmetric problem. 169 .VMMECH073: Modal Analysis of a Cyclic Symmetric Annular Plate Overview Reference: R.3 Geometric Properties Outside Radius (a) = 50 cm Loading Release 14. New York.5cm 100 cm 100 cm Material Properties E = 7.. PP. Blevins. and its subsidiaries and affiliates. Formulas for Natural Frequency and Mode Shape. 246-247. D. Mode-Frequency Analysis Solid Analysis Type(s): Element Type(s): Test Case The fundamental natural frequency of an annular plate is determined using a mode-frequency analysis. Figure 94: Schematic 37 cm 0. 1979.5 .1 x 105 kg/cm2 ν = 0. 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. . NY.Contains proprietary and confidential information of ANSYS.© SAS IP. The bounds for the plate frequency are compared to the theoretical results. VanNostrand Reinhold Publishing Inc. Inc. All rights reserved. 286-287. Inc. 415 x 10 kg- Analysis Assumptions and Modeling Notes According to Blevins.© SAS IP.VMMECH073 Material Properties ρ = 2. . Inc. λ2 = 4. Inc. 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: £ ¤ © = §¡¢ © ¨§¥ ¨ − ¦ © © (1) Where.306031 170 Release 14.38 Mechanical Error (%) 23.Contains proprietary and confidential information of ANSYS.5 cm Sector Angle = 30° Loading kg/cm2 γ = 1. . All rights reserved.5 . and its subsidiaries and affiliates.0746499526039 -1.80 Results Comparison Results Frequency (Hz) Target 23.5 cm Thickness (h) = 0.79 x 10 sec2/cm3 -9 -6 Geometric Properties Inside Radius (b) = 18. All rights reserved. Rigid Body Dynamic Spring Analysis Solid Test Case This test calculates the elastic forces of both tension and compression only springs. Inc. Mechanical Vibrations. Figure 95: Schematic 0. The detection of the spring state being in tension or compression is determined by the non-linear solver.0e7 N/m x1 = 0. . and its subsidiaries and affiliates. 2004. Singapore: Prentice Hall.© SAS IP. Singiresu S.VMMECH074: Tension/Compression Only Springs Overview Reference: Analysis Type(s): Element Type(s): Rao.5 m 1m natural length Tensile (x 1 ) Material Properties k = 1. Both spring types are analyzed in tension and compression loading. 171 .Contains proprietary and confidential information of ANSYS. Inc. A compression only spring uses a negative (compressive) displacement.5 m x2 = -0.5 m 0.5 .5 m m = 7850 kg Compressive (x 2 ) Geometric Properties Lo = 1 m Loading Analysis Assumptions and Modeling Notes Hooke’s Law: Elastic Force = Spring Constant * Displacement F = k*x Spring 1: Compression Only spring Release 14. A tension only spring uses a positive (tensile) displacement. 20. 4th ed. Contains proprietary and confidential information of ANSYS. All rights reserved. Inc. Inc. . and its subsidiaries and affiliates.5 .0e6 Error (%) 0 0 172 Release 14. .© SAS IP.0e6 0 Error (%) 0 0 Target 0 5.0e6 0 Mechanical -5.0e6 Mechanical 0 5.VMMECH074 Spring 2: Tension Only spring Results Comparison Tensile Displacement (x1) Results Elastic Force (N) Spring 1 Elastic Force (N) Spring 2 Compressive Displacement (x2) Results Elastic Force (N) Spring 1 Elastic Force (N) Spring 2 Target -5. 3 0. 1999. 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.3 0.VMMECH075: Harmonic Response of Two-Story Building under Transverse Loading Overview Reference: Analysis Type(s): Element Type(s): W. pg.© SAS IP. T. All rights reserved. Inc. 3rd Edition.35 Loading Force = -1e5 N (y direction) ρ (kg/m3) 7850 1e-8 15700 1e-8 Release 14.5 . and its subsidiaries and affiliates. Figure 96: Schematic Material Properties Material Block 2 Shaft 2 Block 1 Shaft 1 Geometric Properties Block 1 and 2: 40 mm x 40 mm x 40 mm Shaft 1 and 2: 20 mm x 20 mm x 200 mm E (Pa) 2e18 4. Thomson. 166 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.4-1. .35 0.5e10 2e18 9e10 ν 0. Inc.Contains proprietary and confidential information of ANSYS. 173 . Theory of Vibration with Applications. Example 6. . Inc.074902 Mechanical 0. All rights reserved.075859 Error (%) 1.3 174 Release 14. and then link the modal system to a harmonic response system.VMMECH075 Analysis Assumptions and Modeling Notes The material of the columns is assigned negligible density to make them as massless springs.5 . 173). Inc. Add the frictionless support and fixed support in a modal system. Set the solution intervals to 50. Results Comparison Results Maximum Amplitude for Vertex A (m) Maximum Amplitude for Vertex B (m) Target 0. 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 96: Schematic (p. .2119 0.6 1. 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.20853 0. Note There are frictionless supports on 8 faces of the geometry shown. and its subsidiaries and affiliates.© SAS IP.Contains proprietary and confidential information of ANSYS. Inc. O. . Use a global mesh size of 0.. pg. Static Structural Analysis Shell Test Case A tapered aluminum alloy plate of length L with varying thickness across length is suspended from a ceiling.VMMECH076: Elongation of a Tapered Shell with Variable Thickness Overview Reference: Analysis Type(s): Element Type(s): C. Inc. 1959.3 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. and its subsidiaries and affiliates. 175 . An axial load F is applied to the free end of the plate.Contains proprietary and confidential information of ANSYS. problem 4. Harris. Figure 97: Schematic Material Properties E = 10.4 x 106 psi ν = 0. All rights reserved.5 in with mapped-face meshing. 237. New York.© SAS IP.5 . Loading F = 10000 lbf Release 14. Introduction to Stress Analysis. Determine the maximum axial deflection δ in the plate and the axial stress σy at mid-length (Y = L/2). The Macmillan Co. NY. 0048077 4444 Mechanical 0.© SAS IP.0048137 4454.5 .1246 -0. Inc. . Inc.VMMECH076 Results Comparison Results Directional Deformation Y (in) Normal Stress Y at L/2 (psi) Target 0.2379 176 Release 14.Contains proprietary and confidential information of ANSYS.6 Error (%) -0. . and its subsidiaries and affiliates. All rights reserved. Static Thermal Analysis Shell Analysis Type(s): Element Type(s): 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. 2002. Incropera and David P. All rights reserved. Find the following: • Temperatures at mid of the surface. Inc.VMMECH077: Heat Transfer in a Bar with Variable Sheet Thickness Overview Reference: For basic equation: Frank P. Thickness Variation : 1 mm to 4 mm Loading Temperature (T1) on edge (@ 1mm thickness) = 100 °C Temperature (T2)on edge (@ 4mm thickness) = 200 °C Release 14. John Wiley & Sons. Heat and Mass Transfer. Figure 98: Schematic Material Properties E = 2. and its subsidiaries and affiliates. 5th Edition pg. DeWitt.2 x 10-5 1/°C k = 60. . Inc.5 W/m°C Geometric Properties Plate Dimensions : 10 X 50 mm. Inc. • Heat flow reactions on end edges.© SAS IP.0e11 Pa v=0 α = 1. 5. 177 .Contains proprietary and confidential information of ANSYS.5 . VMMECH077 Analysis Heat flow due to conduction is given by: = ¢£ ¤¥ ¡ ¦ §¨ = (2) The area for conduction varies from A1 to A2. The area Ay at any distance y is given as: ©+ © − © (3) = Inserting Equation 3 (p. 178) in equation Equation 2 (p. 178) and integrating the equation from 0 to L, − − (4) Temperature at any point y is given as: != " & # & '( ,,. + $ & ) %# − %*+ (5) Results Comparison Results Heat reaction at T1 (W) Heat reaction at T2 (W) Temperature at mid of surface (°C) Target 2.618 -2.618 166.083 Mechanical 2.6188 -2.6188 166.09 Error (%) 0.00 0.00 0.00 178 Release 14.5 - © SAS IP, Inc. All rights reserved. - Contains proprietary and confidential information of ANSYS, Inc. and its subsidiaries and affiliates. VMMECH078: Gasket Material Under Uniaxial Compression Loading-3-D Analysis Overview Reference: Analysis Type(s): Element Type(s): Any Nonlinear Material Verification Text 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. Release 14.5 - © SAS IP, Inc. All rights reserved. - Contains proprietary and confidential information of ANSYS, Inc. and its subsidiaries and affiliates. 179 .Contains proprietary and confidential information of ANSYS. Inc.VMMECH078 180 Release 14. Inc.© SAS IP.5 . and its subsidiaries and affiliates. All rights reserved. . the model is first loaded with a pressure P1 and unloaded and then loaded with a pressure P2 and unloaded. . All rights reserved.Contains proprietary and confidential information of ANSYS. 181 . In order to simulate the loadingunloading behavior of gasket material.5715E+08 6. Inc.VMMECH078 Material Properties E = 104728E6 Pa ν = 0. the model could not be unloaded to 0 Pa and was instead unloaded to 100 Pa.02 m Loading P1 = 44006400 Pa P2 = 157147000 Pa Analysis A 3-D analysis is performed first using a mesh of 4 x 4 gasket elements.8327E-04 Mechanical 4.5715E+08 6.8327E-04 Error (%) 0 0 0 0 Gasket Pressure and Closure at End of 1st Loading: Gasket Pressure and Closure at End of 2nd Loading: Release 14. and its subsidiaries and affiliates.064E-04 1. The pressure-closure responses simulated are compared to the material definition.21 Geometric Properties L=1m T = 0.4006E+07 4. Because of convergence issues.4006E+07 4.© SAS IP. Results Comparison Target GK-PRES GK-CLOS GK-PRES GK-CLOS 4.064E-04 1.5 . Inc. .Contains proprietary and confidential information of ANSYS. All rights reserved. . Inc.© SAS IP. Inc. and its subsidiaries and affiliates.182 Release 14.5 . Assume that the engine is large compared to the rotor so that the engine end of the shaft may be assumed to be fixed.1 Release 14. Hz Target 48.3-3.Contains proprietary and confidential information of ANSYS.VMMECH079: Natural Frequency of a Motor-Generator Overview Reference: Analysis Type(s): Element Type(s): W. All rights reserved. 10. .781 48. Englewood Cliffs. NJ.© SAS IP.375 in Loading ℓ = 8.1 -0.5 . T. Prentice-Hall. Figure 99: Schematic Material Properties E = 31. Thomson.. determine the natural frequency f in torsion.73 48. Neglect the mass of the shaft also. 1. 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. 2nd Printing.031 lb-in-sec2 Results Comparison Results Lower Order F. Inc.2 x 106 psi m = 1 lb-sec /in 2 Geometric Properties d = . 1965. ex. and its subsidiaries and affiliates. Inc. pg. Hz Higher Order F.73 Error (%) -0.00 in J = . “Vibration Theory and Applications” . 183 . If the polar moment of inertia of the generator rotor is J.781 Mechanical 48. Inc. . All rights reserved. Inc.© SAS IP. and its subsidiaries and affiliates. .5 . Inc.Contains proprietary and confidential information of ANSYS.184 Release 14. All rights reserved. Harper & Row Publishers. sec. 2nd Edition.8 sec Release 14. Determine the displacement response of the system for the load history shown.5 . and its subsidiaries and affiliates. and two springs of stiffness k1 and k2 is subjected to a pulse load F(t) on mass 1. m1 and m2. Inc. 1979. “Vibration Analysis” .Contains proprietary and confidential information of ANSYS.VMMECH080: Transient Response of a Spring-mass System Overview Reference: Analysis Type(s): Element Type(s): R.© SAS IP. . Vierck. NY. New York. Transient Dynamic Mode Superposition Analysis Test Case A system containing two masses. 185 . K. 5-8. Inc. Figure 100: Schematic Material Properties k1 = 6 N/m k2 = 16 N/m m1 = 2 Kg m2 = 2 Kg Geometric Properties Loading F0 = 50 N td = 1. 0 -1. All rights reserved. m (@ t = 2.347 3. and its subsidiaries and affiliates.5 .Contains proprietary and confidential information of ANSYS. .32 6.4s) Target 14.48 3. Inc.© SAS IP. Inc. m (@ t = 2.093 6. m (@ t = 1.4s) Y2 . m (@ t = 1.14 Mechanical 14.VMMECH080 Results Comparison Results Y1 .99 18.3s) Y1 .9 -1. .2 -0.099 Error (%) -0.3s) Y2 .6 186 Release 14.9493 18. © SAS IP. Hartzman. and its subsidiaries and affiliates.Bezler. August 1980. Dynamic Analysis of Uniform Support Motion Response Spectrum Method. Modal analysis Spectral analysis Elastic straight pipe elements Structural Mass element Analysis Type(s): Element Type(s): Test Case This benchmark problem contains three-dimensional multi-branched piping systems. M. . Problem 2. All rights reserved. 187 . Figure 101: Schematic Release 14. Inc. The NUREG intermodal/interspatial results are used for comparison. Brookhaven National Laboratory. Frequencies obtained from modal solve and the nodal/element solution obtained from spectrum solve are compared against reference results.5 . The total mass of the system is represented by structural mass elements specified at individual nodes. Reich.Contains proprietary and confidential information of ANSYS. and M. Inc. Modal and response spectrum analyses are performed on the piping model.VMMECH081: Statically Indeterminate Reaction Force Analysis Overview Reference: P. Pages 48-80. (NUREG/CR-1677). 893995859e-02 Mass @ node 9: M = 0.806 17.3 Density = 2.447000518e-01 Mass @ node 3: M = 0.893995859e-02 Mass @ node 12: M = 0. Inc. Results Comparison Results 1 2 3 4 5 Target 8.712 8.8999 x 106 psi.Contains proprietary and confidential information of ANSYS.© SAS IP. .00 188 Release 14.00 1.375 in Wall Thickness = 0.00 1.893995859e-02 Mass @ node 7: M = 0.432699275e-01 Mass @ node 13: M = 0.VMMECH081 Material Properties Pipe Elements: E = 27. All rights reserved.587991718e10 lb-sec2/in4 Mass Elements (lb-sec2/in): (Mass is isotropic) Mass @ node 1: M = 0.370 41.711 8.806 17.154 in Loading Acceleration response spectrum curve defined by SV and FREQ commands. and its subsidiaries and affiliates.447000518e-01 Mass @ node 2: M = 0. .507 40.432699275e-01 Mass @ node 11: M = 0.366 41.00 1.432699275e-01 Mass @ node 8: M = 0.625 Error (%) 1. Inc.630 Mechanical 8.447000518e-01 Mass @ node 4: M = 0. Nu = 0.893995859e-02 Mass @ node 14: M = 0.893995859e-02 Mass @ node 10: M = 0.893995859e-02 Geometric Properties Straight Pipe: Outer Diameter = 2.510 40.00 1.432699275e-01 Mass @ node 6: M = 0.447000518e-01 Mass @ node 5: M = 0.5 . and its subsidiaries and affiliates.002 0.461 0.00 1. All rights reserved. Inc.99 1.Contains proprietary and confidential information of ANSYS.461 0. Inc.002 0.00 Release 14.VMMECH081 Results Node UX at node6 UY at node8 UZ at node8 Target 0.5 . 189 .© SAS IP.446 Mechanical 0. .450 Error (%) 0. 190 Release 14.Contains proprietary and confidential information of ANSYS. Inc. Inc. All rights reserved. and its subsidiaries and affiliates.© SAS IP. .5 . . ASTM STP-410. Plane strain crack toughness testing of high strength metallic materials. Inc. J. 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. Figure 102: Schematic Release 14. Jr. 191 . Inc.© SAS IP..Contains proprietary and confidential information of ANSYS. . All rights reserved.VMMECH082: Fracture Mechanics Stress for a Crack in a Plate Overview Reference: Analysis Type(s): Element Type(s): W.Srawley. (1966).Brown.E. and its subsidiaries and affiliates.5 .F. . and its subsidiaries and affiliates.VMMECH082 Material Properties E = 30 x 106 psi ν = 0.25 in Loading σ0 = 0.© SAS IP.0504 Error (%) 2. All rights reserved.Contains proprietary and confidential information of ANSYS.3 Geometric Properties a = 1 in b = 5 in h = 5 in t = 0.5641895 psi Results Comparison Results Stress Intensity KI Target 1. Inc. Inc.5 192 Release 14.5 .0249 Mechanical 1. . pg. Figure 103: Schematic Release 14.VMMECH083: Transient Response to a Step Excitation Overview Reference: W. All rights reserved. Mode-Superposition Transient Dynamic Analysis Analysis Type(s): Element Type(s): Test Case A spring-mass-damping system that is initially at rest is subjected to a step force change F acting on the mass. Inc. Inc. Prentice-Hall.3. 2nd Printing. Inc. . Englewood Cliffs. T.5. ξ = 0. NJ. 193 .. 102. 1965. Determine the displacement u at time t for damping ratio.5 . Thomson. article 4. Vibration Theory and Applications.Contains proprietary and confidential information of ANSYS.© SAS IP. and its subsidiaries and affiliates. 194 Release 14.Contains proprietary and confidential information of ANSYS. Inc.5 lbsec /in k = 200 lb/in 2 Loading F= 200 lb Analysis Assumptions and Modeling Notes The damping coefficient c is calculated as 2ξ sqrt(km) = 10 lb-sec/in for ξ = 0.5 . .VMMECH083 Material Properties m = 0. .5. Inc. All rights reserved. and its subsidiaries and affiliates.© SAS IP. Inc.1531 Mechanical 1.Contains proprietary and confidential information of ANSYS. All rights reserved.1544 Error (%) 0. and its subsidiaries and affiliates.© SAS IP. Inc. Time (damped) Release 14.1 Figure 104: Maximum Deformation vs.20 sec Target 1. 195 .VMMECH083 Results Comparison Results Total Def Max (ξ = 0. .5) Time = 0.5 . and its subsidiaries and affiliates.5 .Contains proprietary and confidential information of ANSYS. Inc.© SAS IP. . . Inc. All rights reserved.196 Release 14. Inc. The model is subjected to cyclic displacement loading on the top surface. and its subsidiaries and affiliates.VMMECH084: Mullins Effect on a Rubber Tube Model Subjected to Tension Loading Overview Reference: . 197 .5 .Ogden. Static Analysis Solid Analysis Type(s): Element Type(s): Test Case An axisymmetric rubber plate made of Neo-Hookean material is modeled with radius R and height H. pg: 2861-2877..W. All rights reserved. Figure 105: Schematic Material Properties Neo-Hookean Constants: µ = 8 MPa Ogden-Roxburgh Mullins Constants: Geometric Properties R = 0. et al.© SAS IP. Inc. (1989).5m H = 1m Loading One cycle of loading Step 1: λ = 1. .Contains proprietary and confidential information of ANSYS.. Royal Society of London Proceedings Series A.5 Release 14. The axial stress obtained at different load steps is compared against the reference solution. “A Pseudo-elastic Model for the Mullins Effect in Filled Rubber". 0 Figure 106: Variation of Axial Stress 198 Release 14.VMMECH084 Material Properties r = 2.0 3.333 20. Inc.000 69.0 2.0 Step 5: λ = 1.45 β =0.0 Step 3: λ = 3.5 2.5 Step 6: λ = 1.823 8.0 Step 4: λ = 2.667 28.© SAS IP.019 0.333 20.008 0.104 m = 30. .Contains proprietary and confidential information of ANSYS. .0 Axial Stress (Pa) Target 12. All rights reserved.2 Geometric Properties Loading Step 2: λ = 2.0 0.666 28.0 Error (%) 0.000 Mechanical 12.0 Results Comparison Results Stretch λ 1.819 8.5 1.5 .0 1. and its subsidiaries and affiliates.660 0.000 69.0 0.12 0. Inc.6704 0. 2 x 10 psi α1 = 1. Ei and αi correspond to the Young's modulus and thermal coefficient of expansion for layer i.1 in Release 14.2 in t2 = 0. pg.5 in t1 = 0. Figure 107: Schematic Material Properties MAT1: E1 = 1. and a bending moment My at the free-end. respectively. New York. . 112114. article 7. Inc. Roark. 1975. Static Analysis Solid Analysis Type(s): Element Type(s): Test Case A beam of length ℓ and width w made up of two layers of different materials is subjected to a uniform rise in temperature from Tref to To. W.8 x 10-4 in/in/°F MAT2: 6 Geometric Properties Loading To = 100°F Tref = 0°F My = 10.Contains proprietary and confidential information of ANSYS.2. All rights reserved.© SAS IP. and its subsidiaries and affiliates. McGraw-Hill Book Co.. Inc. NY. Inc. Young.5 . C. Formulas for Stress and Strain. 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.0 inlb ℓ = 8 in w = 0.. 199 .VMMECH085: Bending of a Composite Beam Overview Reference: R. J. VMMECH085 Material Properties E2 = 0.© SAS IP. in StressxTOP .Contains proprietary and confidential information of ANSYS. and its subsidiaries and affiliates. All rights reserved.6 x 10-4 in/in/°F Geometric Properties Loading Results Comparison Results Displacement. .0 0. .832 1731 2258 Error (%) 0.832 1731 2258 Mechanical -0. Inc.4 x 106 psi α2 = 0. psi StressxBOT .0 200 Release 14.0 0.5 . psi Target -0. Inc. Part III: Design Exploration Descriptions . . All rights reserved. A force of 10. Inc.000 N is then applied to the opposite end of the beam.Contains proprietary and confidential information of ANSYS. Height. Stress. and its subsidiaries and affiliates.© SAS IP. and Length (CAD Geometry) Volume. and Deflection Response Parameters: Figure 108: Schematic Material Properties E = 2e11 Pa ν=0 ρ = 7850 kg/m3 Geometric Properties Width = 25 mm Height = 30 mm Rib Thickness = 4 mm Length = 300 mm Limits 20 mm ≤ W ≤ 30 mm 25 mm ≤ H ≤ 35 mm Loading Fixed Support Force F = 10000 N (Z direction) Parameter Width Height Type Input Input Desired Value No Preference No Preference Importance High High Release 14. . Input Parameters: Width. 203 .5 . Inc.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. 030 m H = Height = 0. Inc.9E-05 m3 4.00046 4.10862 0. All rights reserved. . . Inc. and its subsidiaries and affiliates.VMDX001 Parameter Length Volume Stress Deflection Type Input Output Output Output Limits 250 mm ≤ L ≤ 350 mm n/a n/a n/a Desired Value No Preference Minimum Possible Minimum Possible Minimum Possible Importance High Low High High Analysis Beam volume: = + + Maximum axial deformation under load F: = = × + −2 × + Normal stress along Z-direction: σ= = + + Combined objective function becomes: Φ= × −5 + + + + + + + + − Minimizing ϕ we get dimensions as: L = Length = 0.Contains proprietary and confidential information of ANSYS.5290e-5 m 3.© SAS IP.035 m Results Comparison Results Volume (V) Deformation (D) Stress (σ) Target 6.5 .9e-5 m 3 DesignXplorer Error (%) 6.623065E07 Pa 0.5339E-05 m 3.250 m W = Width = 0.62319e7 Pa 204 Release 14.0 0. At the other end. The reference temperature is 5°C. coefficient of thermal expansion and length Temperature (scoped on end face). Inc.038 a = 0. Inc. and its subsidiaries and affiliates.5 . .00582.© SAS IP. a constant convection coefficient of 0. Input Parameters: Convection coefficient. 205 . The ambient temperature is 5°C.038.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 = k0*(1 + a*T) W/m-°C. k0 = 0.005 W/m2°C is applied.5E-05/°C K = k0*(1 + a*T) W/m-°C k0 = 0.Contains proprietary and confidential information of ANSYS. a = 0. thermal strain Response Parameters: Figure 109: Schematic Material Properties E = 2e11 Pa ν=0 α = 1. A temperature of 100°C is applied at one end of the bar. The bar is constrained on all faces by frictionless support. All rights reserved.00582 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/m2°C Ambient temperature Ta = 5°C Release 14. . − α ¤ Φ= + − × + ¢ ££ ¤ + × ¥ ¤ + − α− Minimizing ϕ we get input parameters as: l = beam length = 25 m h = convection coefficient = 0.437e-4 m/m 0 0 0 -0.4e-5/°C sion (α) Temperature (T) Thermal strain (ε) 29. Inc.6e-5/°C Desired Value No Preference No Preference Importance Low Low Input No Preference Low Output Output n/a n/a Minimum Possible Minimum Possible High High Analysis Temperature: s =− a − + × 7 22 a + × 6 a + Thermal strain: ε=α − ¡ =α − Combined objective function becomes.4e-5/°C 29.006 W/m2°C α = coefficient of thermal expansion = 1.553°C 3.Contains proprietary and confidential information of ANSYS.4514e-4 m/m 206 Release 14. All rights reserved.4e-5/°C Results Comparison Results Length (l) Convection coefficient (h) Target 25 m 0.004 W/m2°C ≤ h ≤ 0. and its subsidiaries and affiliates.© SAS IP.006 W/m2°C DesignXplorer Error (%) 25 m 0. Inc.6528°C 3.1.VMDX002 Parameter Length (l) Convection coefficient (h) Coefficient of temperature expansion (α) Temperature (T) Thermal strain (ε) Type Input Input Limits 15 m ≤ l ≤ 25 m 0.4115 Coefficient of thermal expan.006 W/m2°C 1.006 W/m2°C 1. .3278 -0.4e-5/°C ≤ α ≤ 1.5 . 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 Mass. . Natural Frequency. Optimization Theory and Application Second edition. Inc.© SAS IP. S.5 . d =1. 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). Rao. page 28-30 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. Assume the maximum permissible compressive stress as σmax. minimize the mass of the column for economy 2. I = 20 m 9.1e7 Pa Sample Size: 10000 Release 14. It is required to: 1. maximize the natural frequency of transverse vibration of the system for avoiding possible resonance due to wind.VMDX003: Optimization of Water Tank Column for Mass and Natural Frequency Overview Reference: Analysis Type(s): Element Type(s): S.Contains proprietary and confidential information of ANSYS. Direct Stress. Inc. 207 . Buckling Stress Loading Mass of water tank M = 1000000 Kg Acceleration due to gravity = Length. All rights reserved.4 m Depth. example 1. and its subsidiaries and affiliates.2 m σ max = 4.10. b = 0.81 m/s2 T= T Response Parameters: Material Properties E = 3e10 Pa ρ = 2300 Kg/m 3 Geometric Perperties Width. 318137 m 21889.020 -0.0383e7 Pa 6.000 0.002 –0.013 Analysis Minimize: Mass of the column = Maximize: = ρ× × × N ¡¢r £ ¤r¥q¢¥¦§y ¨¤ ¡r ¦©v¥r©¥ vibr ¡i¨¦© ¨¤ ¡ ¥ w ¡¥r ¡ ¦ ¦k w = 3 × × Subject to constraints: × D .36102 m 1.87834 rad/sec 2.1526e6 Pa DesignXplorer Error (%) 0.0386e7 Pa 6.15174e6 Pa 0.3181 m 21890 kg 0.001 -0.77 kg 0.VMDX003 Results Width b Depth d Mass of column M Natural frequency w Direct stress Buckling stress Target 0.87816 rad/sec 2.015 -0.36102 m 1. _S. = σx − × π × d B g_S. 36102 m d = 1.77 kg 0. = ≥ × − × × ≥ × 1/ 2 × × 3 + ×ρ× × × Required objective is obtained by having: b = 0.Contains proprietary and confidential information of ANSYS.318137 m 21889.87834 rad/sec Direct stress = 2.1526e6 Pa Results Comparison Results Width b Depth d Mass of column M Target 0. Inc. .36102 m 1.3181 m M = (minimum) = 21890 kg W = (maximum) = 0.3181 m 21890 kg DesignXplorer Error (%) 0.00104 208 Release 14.0386e7 Pa Buckling stress = 6. Inc.5 .36102 m 1.© SAS IP. and its subsidiaries and affiliates. All rights reserved.002807 -0. .000 0. All rights reserved.0386e7 Pa 6.0139 Release 14. Inc.Contains proprietary and confidential information of ANSYS.02102 -0.© SAS IP.01277 -0.5 .87834 rad/sec 2.0383e7 Pa 6.87816 rad/sec 2.VMDX003 Results Natural frequency w Direct stress Buckling stress Target 0. . 209 . and its subsidiaries and affiliates.1526e6 Pa DesignXplorer Error (%) 0.15174e6 Pa -0. Inc. Inc. All rights reserved. Inc. .Contains proprietary and confidential information of ANSYS.210 Release 14. and its subsidiaries and affiliates.5 .© SAS IP. . Poisson's ratio and density First natural frequency Response Parameters: Figure 110: Schematic Material Properties E = 2e5 MPa ν = 0.8e11 Pa ≤ E ≤ 2.2e11Pa 0. 269-271 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.Contains proprietary and confidential information of ANSYS. pg.3 ρ = 7.© SAS IP.27 ≤ µ ≤ 0.. Van Nostrand Reinhold Company Inc. 211 . and its subsidiaries and affiliates. Input Parameters: Young's modulus. Formula for Natural Frequency and Mode Shape.VMDX004: Optimize frequency of plate with simply supported at all it's vertices Overview Reference: Analysis Type(s): Element Type(s): Blevins.30 7065 kg/m3 ≤ ρ ≤ 8635 kg/m3 Loading All vertices are simply supported Parameter Young's Modulus E Poisson's Ratio µ Density ρ Desired Value No Preference No Preference No Preference Importance Low Low Low Release 14. .5 .850 e-6 kg/mm3 Geometric Properties Length a = 250 mm Width b = 250 mm Thickness h = 5 mm Type Input Input Input Constraints 1. Inc. Inc. All rights reserved. 1979. VMDX004 Parameter First Natural Frequency w Type Output Constraints N/a Desired Value Minimum Possible Importance High Analysis First Natural Frequency: 2 = π 2 3 ρ − ν2 1/ 2 Objective function becomes: ¡ ρ −ν ¢ ¡ φ= − Minimizing ϕ we get dimensions as: Young's Modulus E = 1.27235 8615. .27 8635 kg/m 124.Contains proprietary and confidential information of ANSYS. Inc.69 rad/s 0.7 kg/m3 123.869 -0.321 212 Release 14.179 0.8e11 Pa 0.223 -0.0913 rad/s 3 DesignXplorer Error (%) 1.© SAS IP.27 Density ρ = 8635 kg/m3 First Natural Frequency w = 124. and its subsidiaries and affiliates.0913 rad/s Results Comparison Results Young's Modulus E Poisson's Ratio µ Density ρ First Natural Frequency w Target 1.8e11 Pa Poisson's Ratio µ = 0.5 .8032e11 Pa 0. All rights reserved. Inc. . Input Parameters: Side of Square C/S . The bar has a cross-sectional area A is 0. Strength of Materials.VMDX005: Optimization of buckling load multiplier with CAD parameters and Young's modulus Overview Reference: Analysis Type(s): Element Type(s): Timoshenko. All rights reserved. Inc. .© SAS IP.5 . Length of Cantilever Bar and Young's Modulus Load Multiplier of the First Buckling Mode Genetic Algroithm Response Parameters: Optimization Method: Sample Size: 200 Figure 111: Schematic Material Properties E = 4.45 lbm/ft3 Geometric Properties Cross-section of square = 0. Length of bar = 25 ft.Contains proprietary and confidential information of ANSYS. and its subsidiaries and affiliates. pg. Part 2 (Advanced theory and problems).3 ρ = 490.25 ft. 213 .275 ft. 167–168 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. ≤ a ≤ 0. Type Input Constraints 0. Loading Fixed support on one face.1771e 9 psf ν = 0. Force = 11 lbf (Negative Z-direction) on top face Desired Value No Preference Importance N/A Parameter Cross-section side Release 14. Inc.0625 ft2.25 ft. x 0.225 ft. 3. and its subsidiaries and affiliates. .Contains proprietary and confidential information of ANSYS.5948e9 psf N/A Desired Value No Preference No Preference Maximum Possible Importance N/A N/A N/A Analysis Assuming that under the action of uniform axial load a slight lateral bucking occurs.7594e9 psf ≤ E ≤ 4.5 ft. ≤ l ≤ 27.© SAS IP. . Inc. Inc.5 ft. All rights reserved.a ¨ l § Minimizing ϕ we get dimensions as: 214 Release 14. × = × ¥ ¦ Combined objective function becomes: Φ= − × −5 E. = = = π2 2 cr cr 2 where: q = force per unit length The first critical buckling load is: π¢ ¢ ¡ = π¢ = 4 ¢ ¢ = × 4 ¢ £¤ The load multiplier is given by the ratio of critical load to applied load The first buckling multiplier is: × ¥ ¦ .5 . The expression for deflection is: =δ − π The critical load is given by.VMDX005 Parameter Length Young's Modulus First buckling mode load multiplier Type Input Input Output Constraints 22. Contains proprietary and confidential information of ANSYS.55 -1.32 DesignXplorer Error (%) 3035. Young's Modulus E = 4.© SAS IP.5948e9 psf Buckling load multiplier = 3083.32 Results Comparison Results First buckling mode load multiplier Target 3083. All rights reserved. and its subsidiaries and affiliates. Inc.275 ft.5 ft. Inc. 215 .5 .549 Release 14. . Length l = 22.VMDX005 Cross-section side a = 0. Inc. All rights reserved.Contains proprietary and confidential information of ANSYS.© SAS IP. and its subsidiaries and affiliates.5 .216 Release 14. . Inc. .