ANSYS Workbench Verification Manual

ANSYS Workbench Verification Manual

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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

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Workbench 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 86. VMMECH086: Stress Concentration at a Hole in a Plate .................................................................... 201 87. VMMECH087: Campbell Diagrams and Critical Speeds Using Symmetric Orthotropic Bearings ......... 205 88. VMMECH088: Harmonic Response of a Guitar String ....................................................................... 209 89. VMMECH089: Delamination Analysis of a Double Cantilever Beam Using Contact-Based Debonding ..................................................................................................................................................... 211 90. VMMECH090: Delamination Analysis of a Double Cantilever Beam Using Interface Delamination ..... 213 III. Design Exploration Descriptions ....................................................................................................... 215

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Workbench Verification Manual 1. VMDX001: Optimization of L-Shaped Cantilever Beam under Axial Load ............................................ 217 2. VMDX002: Optimization of Bar with Temperature-Dependent Conductivity ....................................... 219 3. VMDX003: Optimization of Water Tank Column for Mass and Natural Frequency ................................ 221 4. VMDX004: Optimization of Frequency for a Plate with Simple Support at all Vertices ......................... 225 5. VMDX005: Optimization of Buckling Load Multiplier with CAD Parameters and Young's Modulus ....... 227

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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. The available tests are engineering problems that provide independent verification, usually a closed form equation. Many of them are classical engineering problems. The solutions for the test cases have been verified, however, certain differences may exist with regard to the references. These differences have been examined and are considered acceptable. The workbench analyses employ a balance between accuracy and solution time. Improved results can be obtained in some cases by employing a more refined finite element mesh but requires longer solution times. For the tests, an error rate of 3% or less has been the goal. These tests were run on an Intel Xeon processor using Microsoft Windows 7 Enterprise 64-bit . These results are reported in the test documentation. Slightly different results may be obtained when different processor types or operating systems are used. The tests contained in this manual are a partial subset of the full set of tests that are run by ANSYS developers to ensure a high degree of quality for the Workbench product. The verification of the Workbench product is conducted in accordance with the written procedures that form a part of an overall Quality Assurance program at ANSYS, Inc. You are encouraged to use these tests as starting points when exploring new Workbench features. Geometries, material properties, loads, and output results can easily be changed and the solution repeated. As a result, the tests offer a quick introduction to new features with which you may be unfamiliar. Some test cases will require different licenses, such as DesignModeler, Emag, or Design Exploration. If you do not have the available licenses, you may not be able to reproduce the results. The Educational version of Workbench should be able to solve most of these tests. License limitations are not applicable to Workbench Education version but problem size may restrict the solution of some of the tests. The archive files for each of the Verification Manual tests are available at the Customer Portal. Download the ANSYS Workbench Verification Manual Archive Files. These zipped archives provide all of the necessary elements for running a test, including geometry parts, material files, and workbench databases. To open a test case in Workbench, locate the archive and import it into Workbench. You can use these tests to verify that your hardware is executing the ANSYS Workbench tests correctly. The results in the databases can be cleared and the tests solved multiple times. The test results should be checked against the verified results in the documentation for each test. ANSYS, Inc. offers the Workbench Verification and Validation package for users that must perform system validation. Release 15.0 - © SAS IP, Inc. All rights reserved. - Contains proprietary and confidential information of ANSYS, Inc. and its subsidiaries and affiliates.

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Introduction This package automates the process of test execution and report generation. If you are interested in contracting for such services contact the ANSYS, Inc. Quality Assurance Group.

Index of Test Cases Test Case Number

Element Type

Analysis Type

Solution Options

VMMECH001

Solid

Static Structural

Linear

VMMECH002

Solid

Static Structural

Linear

VMMECH003

Solid

Modal

Free Vibration

VMMECH004

Solid

Structural

Nonlinear, Viscoplastic Materials

VMMECH005

Solid

Static Thermal

Linear

VMMECH006

Solid

Static Thermal

Nonlinear

VMMECH007

Solid

Static Structural

Nonlinear Thermal Stress

VMMECH008

Solid

Static Thermal

Linear

VMMECH009

Solid

Static Structural

Linear

VMMECH010

Shell

Modal

Free Vibration

VMMECH011

Shell

Static Structural

Nonlinear, Large Deformation

VMMECH012

Solid

Buckling

VMMECH013

Shell

Buckling

VMMECH014

Solid

Harmonic

VMMECH015

Solid

Harmonic

VMMECH016

Solid

Static Structural

Fatigue

VMMECH017

Solid

Static Structural

Linear Thermal Stress

VMMECH018

Solid

Static Structural

Linear, Inertia relief

VMMECH019

Beam

Static Structural

Linear

Shell VMMECH020

Beam

Modal

VMMECH021

Beam

Buckling

VMMECH022

Solid

Static Structural

Nonlinear, Contact

VMMECH023

Beam

Static Structural

Linear

VMMECH024

Beam

Harmonic

VMMECH025

Solid

Static Structural

Linear

VMMECH026

Shell

Static Structural

Fatigue

VMMECH027

Shell

Static Structural

Linear Thermal Stress

VMMECH028

Solid

Static Structural

VMMECH029

Solid

Static Structural

2

Nonlinear, Plastic Materials

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Index of Test Cases Test Case Number

Element Type

Analysis Type

Solution Options

VMMECH030

2-D Solid, Plane Strain

Static Structural

VMMECH031

2-D Solid, Plane Stress

Static Structural

VMMECH032

2-D Solid, Axisymmetric

Static Structural

Linear Thermal Stress

VMMECH033

Solid

Static Structural

Electromagnetic

VMMECH034

Solid

Static Structural

Nonlinear, Large Deformation

VMMECH035

Solid

Coupled (Static Thermal and Static Stress)

VMMECH036

Solid

Static Structural

VMMECH037

2-D Solid, Axisymmetric

Transient Thermal

VMMECH038

Solid

Transient Structural

Flexible Dynamic

VMMECH039

Solid

Transient Structural

Flexible Dynamic

Sequence Loading

Spring VMMECH040

Beam

Static Structural

VMMECH041

Solid

Static Structural

Electromagnetic

VMMECH042

Solid

Static Structural

Hydrostatic Fluid

VMMECH043

Beam

Modal

VMMECH044

Beam

Static Structural

VMMECH045

Shell

Static Structural

VMMECH046

Shell

Static Structural

VMMECH047

2-D Solid, Axisymmetric

Static Structural

VMMECH048

Beam

Static Structural

VMMECH049

Beam

Static Structural

VMMECH050

Axisymmetric Shell

Static Structural

VMMECH051

Axisymmetric Shell

Static Structural

VMMECH052

Multipoint Constraint

Rigid Dynamic

VMMECH042

Multipoint Constraint

Rigid Dynamic

VMMECH054

Multipoint Constraint

Rigid Dynamic

VMMECH055

Multipoint Constraint

Rigid Dynamic

Linear Thermal Stress

Nonlinear, Plastic Materials

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Introduction Test Case Number

Element Type

Analysis Type

VMMECH056

Multipoint Constraint

Rigid Dynamic

VMMECH057

Multipoint Constraint

Rigid Dynamic

VMMECH058

Multipoint Constraint

Rigid Dynamic

VMMECH059

2-D Plane Stress Shell

Static Structural

VMMECH060

Solid

Transient Structural

Solution Options

Flexible Dynamic

Multipoint Constraint VMMECH061

Beam

Static Structural

VMMECH062

Axisymmetric Shell

Static Structural

VMMECH063

Shell

Static Structural

VMMECH064

Shell

Static Structural

VMMECH065

Solid

Static Structural

Nonlinear, Large Deformation Linear Thermal Stress

Shell VMMECH066

Shell

Static Structural

VMMECH067

Solid

Static Structural

VMMECH068

2-D Solid, Plane Strain

Static Structural

VMMECH069

Shell

Static Structural

VMMECH070

2-D Solid

Static Structural

VMMECH071

2-D Thermal Solid

Static Thermal

VMMECH072

2-D Thermal Solid

Static Structural

VMMECH073

Solid

Modal

VMMECH074

Solid

Rigid Body Dynamics

Nonlinear, Plastic Materials Nonlinear, Large Deformation Linear Thermal Stress

Spring VMMECH075

Solid

Harmonic

VMMECH076

Shell

Static Structural

VMMECH077

Thermal Shell

Static Thermal

VMMECH078

3-D Solid

Static Structural

3-D Gasket VMMECH079

Pipe

Modal

VMMECH080

Spring

Transient Dynamic

Mode Superposition

Mass

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Index of Test Cases Test Case Number

Element Type

Analysis Type

Solution Options

Pipe

Modal

Mass

Spectral

VMMECH082

Solid

Static Structural

Fracture Mechanics

VMMECH083

Spring, Mass

Transient Dynamic

Mode Superposition

VMMECH084

Solid

Static Structural

Nonlinear, Hypereleastic

VMMECH085

Solid

Static Structural

Composite Material

VMMECH086

Solid

Static Structural

VMMECH081

Submodeling (2D2D) VMMECH087

Line Body

Modal

Point Mass Bearing Connection VMMECH088

Beam

Static Structural

Linear Perturbation

Modal Harmonic VMMECH089

Solid

Static Structural

Contact-Based Debonding

VMMECH090

Solid

Static Structural

Interface Delamination

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Part I: DesignModeler Descriptions

VMDM001: Extrude, Chamfer, and Blend Features

Overview Feature:

Extrude, Chamfer, and Blend

Drawing Units:

Millimeter

Test Case Create a Model using Extrude, Chamfer, and Blend features. A polygonal area is extruded 60 mm. A rectangular area of 30 mm x 40 mm [having a circular area of radius 6 mm subtracted] is extruded to 20 mm. Both resultant solids form one solid geometry. A rectangular area (24 mm x 5 mm) is subtracted from the solid. Two rectangular areas (40 mm x 10 mm) are extruded 10 mm and subtracted from solid. Two rectangular areas (25 mm x 40 mm) are extruded 40 mm and subtracted from solid. A Chamfer (10 mm x 10 mm) is given to 4 edges on the resultant solid. Four Oval areas are extruded and subtracted from Solid. Fillet (Radius 5 mm) is given to 4 edges using Blend Feature. Verify Volume of the resultant geometry. Figure 1: Final Model after creating Extrude, Chamfer, and Blend

Calculations 1. Volume of Solid after extruding Polygonal Area: v1 = 264000 mm3. 2. Volume of rectangular area having circular hole: v2 = 21738.05 mm3. Net Volume = V = v1 + v2 = 285738.05 mm3. 3. Volume of rectangular (24mm x 5mm) solid extruded 30mm using Cut Material = 3600 – 565.5 = 3034.5 mm3. Net volume V = 285738.05 – 3034.5 = 282703.5 mm3. 4. Volume of two rectangular areas each 40mm x 10mm extruded 10mm = 8000 mm3.

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VMDM001 Net volume V = 282703.5 – 8000 = 274703.5 mm3. 5. Volume of two rectangular areas 25mm x 40mm extruded 40mm = 80000 mm3. Net volume V = 274703.5 – 80000 = 194703.5 mm3. 6. Volume of four solids added due to Chamfer = 4 x 500 = 2000 mm3 Net volume V = 194703.5 + 2000 = 196703.5 mm3. 7. Volume of four oval areas extruded 10 mm = 7141.6 mm3. Net volume V = 196703.5 - 7141.6 = 189561.9 mm3. 8. Volume of 4 solids subtracted due to Blend of radius 5 mm = 429.2 mm3. Hence Net volume of final Solid body = V = 189561.9 – 429.2 = 189132.7 mm3.

Results Comparison Results

Target

DesignModeler

Error (%)

189561.95

189561.95

0

44439.29

44433.3

-0.0135

Number of Faces

52

52

0

Number of Bodies

1

1

0

Volume (mm3) 2

Surface Area (mm )

10

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VMDM002: Cylinder using Revolve, Sweep, Extrude, and Skin-Loft

Overview Feature:

Revolve, Sweep, Extrude, and Skin-Loft

Drawing Units:

Millimeter

Test Case Create a Model using Revolve, Sweep, Extrude, and Skin-Loft features. A Rectangular area (100 mm x 30 mm) is revolved about Z-Axis in 3600 to form a Cylinder. A circular area of radius 30 mm is swept 100 mm using Sweep feature. A circular area of radius 30 mm is extruded 100 mm. A solid cylinder is created using Skin-Loft feature between two coaxial circular areas each of radius 30 mm and 100 mm apart. Verify Volume of the resultant geometry. Figure 2: Final Model after creating Revolve, Sweep, Extrude, and Skin-Loft

Calculations 1. Volume of Cylinder created after Revolving Rectangular area (100 mm x 30 mm) = v1 = 282743.3 mm3. 2. Volume of Cylinder created when a circular area (Radius 30mm) is swept 100 mm = v2 = 282743.3 mm3. Net Volume = V = v1 + v2 = 282743.3 + 282743.3 = 565486.6 mm3. 3. Volume of Cylinder after extruding a circular area (Radius 30 mm) 100 mm = 282743.3 mm3. Net Volume = V = 565486.6 + 282743.3 = 848229.9 mm3. 4. Volume of Cylinder created after using Skin-Loft feature between two circular areas of Radius 30 mm and 100 mm apart. = 282743.3 mm3. Net Volume of the final Cylinder = 848229.9 + 282743.3 = 1130973.2 mm3.

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VMDM002

Results Comparison Results

Target

DesignModeler

Error (%)

1130973.3

1130973.3

0

81053.1

81053.1

0

Number of Faces

3

3

0

Number of Bodies

1

1

0

Volume (mm3) 2

Surface Area (mm )

12

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VMDM003: Extrude, Revolve, Skin-Loft, and Sweep

Overview Feature:

Extrude, Revolve, Skin-Loft, and Sweep

Drawing Units:

Millimeter

Test Case Create a Model using Extrude, Revolve, Skin-Loft, and Sweep. A rectangular area (103 mm x 88 mm) is extruded 100 mm to form a solid box. A circular area of radius 25 mm is revolved 900 using Revolve feature and keeping Thin/Surface option to Yes and 3 mm Inward and Outward Thickness. A solid is subtracted using Skin-Loft feature between two square areas (each of side 25 mm) and 100 mm apart. The two solid bodies are frozen using Freeze feature. A circular area of radius 25 mm is swept using Sweep feature and keeping Thin/Surface option to Yes and 3 mm Inward and Outward Thickness. Thus a total of 4 geometries are created. Verify the volume of the resulting geometry. Figure 3: Final Model after creating Extrude, Revolve, Skin-Loft and Sweep

Calculations 1. Volume of rectangular (103 mm x 88 mm) solid extruded 100mm = 906400 mm3. 2. Volume of solid after revolving circular area of Radius 25 mm through 900 = 29639.6 mm3. Net Volume of solid box, Va = 906400 - 29639.6 = 876760.3 mm3. 3. Volume of additional body created due to Revolve feature = Vb= 11134.15 mm3. 4. Volume of the rectangular box cut after Skin-Loft between two square areas each of side 25 mm = 62500 mm3. Net Volume of solid box becomes Va = 876760.3 – 62500 = 814260.3 mm3. 5. Volume of additional two bodies created due to Sweep feature: Release 15.0 - © SAS IP, Inc. All rights reserved. - Contains proprietary and confidential information of ANSYS, Inc. and its subsidiaries and affiliates.

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VMDM003 • Vc = 47123.9 mm3 and Vd = 28352.8 mm3. • And total volume that gets subtracted from box due to Sweep Feature = 75476.7 mm3. • Hence Net volume of box, Va = 814260.3 - 75476.7 = 738783.6 mm3. • Sum of volumes of all four bodies = Va+Vb+Vc+Vd = 738783.6 + 11134.15 + 47123.9 +28352.8 = 825394.4 mm3.

Results Comparison Results

Target

DesignModeler

Error (%)

825394.4

825394.5

0

101719.47

101719.95

0

Number of Faces

22

22

0

Number of Bodies

4

4

0

Volume (mm3) 2

Surface Area (mm )

14

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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

Analysis Type(s):

Linear Static Structural Analysis

Element Type(s):

Solid

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

Geometric Properties

E = 2.9008e7 psi ν = 0.3 ρ = 0.28383 lbm/in3

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)

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VMMECH001

Results Comparison Results

Target

Mechanical

Error (%)

Y Reaction Force at Top Fixed Support (lbf )

900

901.14

0.127

Y Reaction Force at Bottom Fixed Support (lbf )

600

598.86

-0.190

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VMMECH002: Rectangular Plate with Circular Hole Subjected to Tensile Loading

Overview Reference:

J. E. Shigley, Mechanical Engineering Design, McGraw-Hill, 1st Edition, 1986, Table A-23, Figure A-23-1, pg. 673

Analysis Type(s):

Linear Static Structural Analysis

Element Type(s):

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

Geometric Properties

E = 1000 Pa ν=0

Length = 15 m Width = 5 m Thickness = 1 m Hole radius = 0.5 m

Loading Pressure = -100 Pa

Results Comparison Results

Target

Mechanical

Error (%)

Maximum Normal X Stress (Pa)

312.5

315.2

0.864

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VMMECH003: Modal Analysis of Annular Plate

Overview Reference:

R. J. Blevins, Formula for Natural Frequency and Mode Shape, Van Nostrand Reinhold Company Inc., 1979, Table 11-2, Case 4, pg. 247

Analysis Type(s):

Free Vibration Analysis

Element Type(s):

Solid

Test Case An assembly of three annular plates has cylindrical support (fixed in the radial, tangential, and axial directions) applied on the cylindrical surface of the hole. Sizing control with element size of 0.5” is applied to the cylindrical surface of the hole. Find the first six modes of natural frequencies. Figure 6: Schematic

Material Properties

Geometric Properties

E = 2.9008e7 psi ν = 0.3 ρ = 0.28383 lbm/in3

Loading

Inner diameter of inner plate = 20" Inner diameter of middle plate = 28" Inner diameter of outer plate = 34" Outer diameter of outer plate = 40"

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VMMECH003 Material Properties

Geometric Properties

Loading

Thickness of all plates = 1"

Results Comparison Results

Target

Mechanical

Error (%)

1st Frequency Mode (Hz)

310.911

310.21

-0.23

2nd Frequency Mode (Hz)

318.086

315.6

-0.78

3rd Frequency Mode (Hz)

318.086

315.64

-0.77

4th Frequency Mode (Hz)

351.569

346.73

-1.38

5th Frequency Mode (Hz)

351.569

347.11

-1.27

6th Frequency Mode (Hz)

442.451

437.06

-1.22

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VMMECH004: Viscoplastic Analysis of a Body (Shear Deformation)

Overview Reference:

B. Lwo and G. M. Eggert, "An Implicit Stress Update Algorithm Using a Plastic Predictor". Submitted to Computer Methods in Applied Mechanics and Engineering, January 1991.

Analysis Type(s):

Nonlinear Structural Analysis

Element Type(s):

Solid

Test Case A cubic shaped body made up of a viscoplastic material obeying Anand's law undergoes uniaxial shear deformation at a constant rate of 0.01 cm/s. The temperature of the body is maintained at 400°C. Find the shear load (Fx) required to maintain the deformation rate of 0.01 cm/sec at time equal to 20 seconds. Figure 7: Schematic

y Velocity = 0.01 cm/s

h

x h Problem Model Material Properties

Geometric Properties

Ex (Young's Modulus) = 60.6 GPa (Poisson's Ratio) = 0.4999 So = 29.7 MPa Q/R = 21.08999E3 K A = 1.91E7 s-1
= 7.0

h = 1 cm thickness = 1 cm

Loading Temp = 400°C = 673°K Velocity (x-direction) = 0.01 cm/sec @ y = 1 cm Time = 20 sec

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VMMECH004 Material Properties

Geometric Properties

Loading

m = 0.23348 ho = 1115.6 MPa

µ = 18.92 MPa
= 0.07049 a = 1.3

Results Comparison Results Fx, N

24

Target

Mechanical

Error (%)

845.00

-791.76

-6.3

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VMMECH005: Heat Transfer in a Composite Wall

Overview Reference:

F. Kreith, Principles of Heat Transfer, Harper and Row Publisher, 3rd Edition, 1976, Example 2-5, pg. 39

Analysis Type(s):

Linear Static Thermal Analysis

Element Type(s):

Solid

Test Case A furnace wall consists of two layers: fire brick and insulating brick. The temperature inside the furnace is 3000°F (Tf) and the inner surface convection coefficient is 3.333e-3 BTU/s ft2°F (hf). The ambient temperature is 80°F (Ta) and the outer surface convection coefficient is 5.556e-4 BTU/s ft2°F (ha). Find the Temperature Distribution. Figure 8: Schematic

Material Properties

Geometric Properties

Fire brick wall: k = 2.222e-4 BTU/s ft °F Insulating wall: k = 2.778e-5 BTU/s ft °F

Loading

Cross-section = 1" x 1" Fire brick wall thickness = 9" Insulating wall thickness = 5"

Results Comparison Results

Target

Mechanical

Error (%)

Minimum Temperature (°F)

336

336.68

0.202

Maximum Temperature (°F)

2957

2957.2

0.007

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VMMECH006: Heater with Nonlinear Conductivity

Overview Reference:

Vedat S. Arpaci, Conduction Heat Transfer, Addison-Wesley Book Series, 1966, pg. 130

Analysis Type(s):

Nonlinear Static Thermal Analysis

Element Type(s):

Solid

Test Case A liquid is boiled using the front face of a flat electric heater plate. The boiling temperature of the liquid is 212°F. The rear face of the heater is insulated. The internal energy generated electrically may be assumed to be uniform and is applied as internal heat generation. Find the maximum temperature and maximum total heat flux. Figure 9: Schematic

Material Properties

Geometric Properties

k = [0.01375 * (1 + 0.001 T)] BTU/s in°F Temperature (°F)

Conductivity (BTU/s in°F)

32

1.419e-002

1000

2.75e-002

Radius = 3.937” Thickness = 1”

Loading Front face temperature = 212°F Internal heat generation = 10 BTU/s in3

Results Comparison Results

Target

Mechanical

Error (%)

Maximum Temperature (°F)

476

480.58

0.96

Maximum Total Heat Flux (BTU/s in2)

10

9.9997

-0.003

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VMMECH007:Thermal Stress in a Bar with Temperature Dependent Conductivity

Overview Reference:

Any basic Heat Transfer book

Analysis Type(s):

Nonlinear Thermal Stress Analysis

Element Type(s):

Solid

Test Case A long bar has thermal conductivity that varies with temperature. The bar is constrained at both ends by frictionless surfaces. A temperature of T°C is applied at one end of the bar (End A). The reference temperature is 5°C. At the other end, a constant convection of h W/m2°C is applied. The ambient temperature is 5°C. Advanced mesh control with element size of 2 m is applied. Find the following: • Minimum temperature • Maximum thermal strain in z direction (on the two end faces) • Maximum deformation in z direction • Maximum heat flux in z direction at z = 20 m Figure 10: Schematic

Material Properties

Geometric Properties

E = 2e11 Pa ν=0 α = 1.5e-05 / °C k = 0.038*(1 + 0.00582*T) W/m °C Temperature (°C)

Conductivity (W/m °C)

5

3.91e-002

800

0.215

Length = 20 m Width = 2 m Breadth = 2 m

Loading Rear face temperature T = 100°C Film Coefficient h = 0.005 W/m2°C Ambient temperature = 5°C Reference temperature = 5°C

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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

Target

Mechanical

Error (%)

Minimum Temperature (°C)

38.02

38.014

-0.016

Maximum Thermal strain (z = 20) (m/m)

0.000495

0.00049521

0.042

Maximum Thermal strain (z = 0) (m/m)

0.001425

0.001425

0.000

Maximum Z Deformation (m)

0.00232

0.002341

0.905

Maximum Z Heat Flux (z = 20) (W/m2)

0.165

0.16507

0.042

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VMMECH008: Heat Transfer from a Cooling Spine

Overview Reference:

Kreith, F., Principles of Heat Transfer, Harper and Row, 3rd Edition, 1976, Equation 2-44a, pg. 59, Equation 2–45, pg. 60

Analysis Type(s):

Linear Static Thermal Analysis

Element Type(s):

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. The surface convection coefficient between the spine and the surrounding air is h, the air temper is Ta, and the tip of the spine is insulated. Apply advanced mesh control with element size of 0.025'. Find the heat conducted by the spine and the temperature of the tip. Figure 11: Schematic

Material Properties

Geometric Properties

Material Properties E = 4.177e9 psf ν = 0.3 Thermal conductivity k = 9.71e-3 BTU/s ft °F

Loading

Geometric Properties Cross section = 1.2” x 1.2” L = 8”

Loading Tw = 100°F Ta = 0°F h = 2.778e-4 BTU/s ft2 °F

Results Comparison Results

Target

Mechanical

Error (%)

Temperature of the Tip (°F)

79.0344

79.078

0.055

Heat Conducted by the Spine (Heat Reaction) (BTU/s)

6.364e-3

6.3614e-3

-0.041

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VMMECH009: Stress Tool for Long Bar with Compressive Load

Overview Reference:

Any basic Strength of Materials book

Analysis Type(s):

Linear Static Structural Analysis

Element Type(s):

Solid

Test Case A multibody of four bars connected end to end has one of the end faces fixed and a pressure is applied to the opposite face as given below. The multibody is used to nullify the numerical noise near the contact regions. Find the maximum equivalent stress for the whole multibody and the safety factor for each part using the maximum equivalent stress theory with tensile yield limit. Figure 12: Schematic

Material Properties Mater- E (Pa) ial

ν

Tensile Yield (Pa)

Part 1

1.93e11 0

2.07e8

Part 2

7.1e10 0

2.8e8

Part 3

2e11

0

2.5e8

Part 4

1.1e11 0

2.8e8

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

Loading Pressure = 2.5e8 Pa

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VMMECH009 Part 4: 2 m x 2 mx2m

Results Comparison Results

Target

Mechanical

Error (%)

Maximum Equivalent Stress (Pa)

2.5e8

2.5e8

0.000

Safety Factor for Part 1

0.828

0.828

0.000

Safety Factor for Part 2

1.12

1.12

0.000

Safety Factor for Part 3

1

1

0.000

Safety Factor for Part 4

1.12

1.12

0.000

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VMMECH010: Modal Analysis of a Rectangular Plate

Overview Reference:

Blevins, Formula for Natural Frequency and Mode Shape, Van Nostrand Reinhold Company Inc., 1979, Table 11-4, Case 11, pg. 256

Analysis Type(s):

Free Vibration Analysis

Element Type(s):

Shell

Test Case A rectangular plate is simply supported on both the smaller edges and fixed on one of the longer edges as shown below. Sizing mesh control with element size of 6.5 mm is applied on all the edges to get accurate results. Find the first five modes of natural frequency. Figure 13: Schematic

Material Properties

Geometric Properties

E = 2e11 Pa ν = 0.3

Length = 0.25 m Width = 0.1 m Thickness = 0.005 m

ρ = 7850 kg/m3

Loading

Results Comparison Results

Target

Mechanical

Error (%)

1st Frequency Mode (Hz)

595.7

590.03

-0.952

2nd Frequency Mode (Hz)

1129.55

1118.4

-0.987

3rd Frequency Mode (Hz)

2051.79

2038.1

-0.667

4th Frequency Mode (Hz)

2906.73

2879.3

-0.994

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VMMECH010 Results 5th Frequency Mode (Hz)

36

Target

Mechanical

Error (%)

3366.48

3350

-0.489

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VMMECH011: Large Deflection of a Circular Plate with Uniform Pressure

Overview Reference:

Timoshenko S.P., Woinowsky-Krieger S., Theory of Plates and Shells, McGraw-Hill, 2nd Edition, Article 97, equation 232, pg. 401

Analysis Type(s):

Nonlinear Structural Analysis (Large Deformation On)

Element Type(s):

Shell

Test Case A circular plate is subjected to a uniform pressure on its flat surface. The circular edge of the plate is fixed. To get accurate results, apply sizing control with element size of 5 mm on the circular edge. Find the total deformation at the center of the plate. Figure 14: Schematic

Material Properties

Geometric Properties

E = 2e11 Pa ν = 0.3

Loading

Radius = 0.25 m Thickness = 0.0025 m

Pressure = 6585.18 Pa

Results Comparison Results Total deformation (m)

Target

Mechanical

Error (%)

0.00125

0.0012374

-1.008

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VMMECH012: Buckling of a Stepped Rod

Overview Reference:

Warren C. Young, Roark's Formulas for Stress & Strains, McGraw Hill, 6th Edition, Table 34, Case 2a, pg. 672

Analysis Type(s):

Buckling Analysis

Element Type(s):

Solid

Test Case A stepped rod is fixed at one end face. It is axially loaded by two forces: a tensile load at the free end and a compressive load on the flat step face at the junction of the two cross sections. To get accurate results, apply sizing control with element size of 6.5 mm. Find the Load Multiplier for the First Buckling Mode. Figure 15: Schematic

Material Properties

Geometric Properties

E = 2e11 Pa ν = 0.3

Larger diameter = 0.011982 m Smaller diameter = 0.010 m Length of larger diameter = 0.2 m Length of smaller diameter = 0.1 m

Loading Force at free end = 1000 N Force at the flat step face = 2000 N Both forces are in the z direction

Results Comparison Results Load Multiplier

Target

Mechanical

Error (%)

22.5

22.958

2.0356

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VMMECH013: Buckling of a Circular Arch

Overview Reference:

Warren C. Young, Roark's Formulas for Stress Strains, McGraw Hill, 6th Edition, Table 34, Case 10, pg. 679

Analysis Type(s):

Buckling Analysis

Element Type(s):

Shell

Test Case A circular arch of a rectangular cross section (details given below) is subjected to a pressure load as shown below. Both the straight edges of the arch are fixed. Find the Load Multiplier for the first buckling mode. Figure 16: Schematic

Material Properties

Geometric Properties

E = 2e5 MPa ν=0

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

Mechanical

Error (%)

544

546.07

0.4

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VMMECH014: Harmonic Response of a Single Degree of Freedom System

Overview Reference:

Any basic Vibration Analysis book

Analysis Type(s):

Harmonic Analysis

Element Type(s):

Solid

Test Case An assembly where four cylinders represent massless springs in series and a point mass simulates a spring mass system. The flat end face of the cylinder (Shaft 1) is fixed. Harmonic force is applied on the end face of another cylinder (Shaft 4) as shown below. Find the z directional Deformation Frequency Response of the system on the face to which force is applied for the frequency range of 0 to 500 Hz for the following scenarios using Mode Superposition. Solution intervals = 20. • Scenario 1: Damping ratio = 0 • Scenario 2: Damping ratio = 0.05 Figure 17: Schematic

Material Properties Material

E (Pa)

ν

ρ (kg/m3)

Shaft 1

1.1e11

0.34

1e-8

Shaft 2

1.1e11

0.34

1e-8

Shaft 3

4.5e10

0.35

1e-8

Shaft 4

4.5e10

0.35

1e-8

Geometric Properties Each cylinder: Diameter = 20 mm Length = 50 mm

Loading Force = 1e7 N (Zdirection) Point Mass = 3.1044 Kg

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VMMECH014

Results Comparison Results

Target

Mechanical

Error (%)

0.1404

0.14123

0.591

Phase angle without damping (degrees)

180

180

0.000

Maximum Amplitude with damping (m)

0.14

0.1408

0.577

Phase angle with damping (degrees)

175.6

175.58

0.000

Maximum Amplitude without damping (m)

44

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VMMECH015: Harmonic Response of Two Storied Building under Transverse Loading

Overview Reference:

W. T. Thomson, Theory of Vibration with Applications, 3rd Edition, 1999, Example 6.4-1, pg. 166

Analysis Type(s):

Harmonic Analysis

Element Type(s):

Solid

Test Case A two-story building has two columns (2K and K) constituting stiffness elements and two slabs (2M and M) constituting mass elements. The material of the columns is assigned negligible density so as to make them as massless springs. The slabs are allowed to move only in the y direction by applying frictionless supports on all the faces of the slabs in the y direction. The end face of the column (2K) is fixed and a harmonic force is applied on the face of the slab (M) as shown in the figure below. Find the y directional Deformation Frequency Response of the system at 70 Hz on each of the vertices as shown below for the frequency range of 0 to 500 Hz using Mode Superposition. Use Solution intervals = 50. Figure 18: Schematic

Material Properties Material

E (Pa)

ν

ρ (kg/m3)

Block 2

2e18

0.3

7850

Shaft 2

4.5e10

0.35

1e-8

Block 1

2e18

0.3

15700

Shaft 1

9e10

0.35

1e-8

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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

Target

Mechanical

Error (%)

Maximum Amplitude for vertex A (m)

0.20853

0.21174

1.5

Maximum Amplitude for vertex B (m)

0.074902

0.075838

1.2

46

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VMMECH016: Fatigue Tool with Non-Proportional Loading for Normal Stress

Overview Reference:

Any basic Machine Design book

Analysis Type(s):

Fatigue Analysis

Element Type(s):

Solid

Test Case A bar of rectangular cross section has the following loading scenarios. • Scenario 1: One of the end faces is fixed and a force is applied on the opposite face as shown below in Figure 19: Scenario 1 (p. 47). • Scenario 2: Frictionless support is applied to all the faces of the three standard planes (faces not seen in Figure 20: Scenario 2 (p. 47)) and a pressure load is applied on the opposite faces in positive yand z-directions. Find the life, damage, and safety factor for the normal stresses in the x, y, and z directions for nonproportional fatigue using the Soderberg theory. Use a design life of 1e6 cycles, a fatigue strength factor or 1, a scale factor of 1, and 1 for coefficients of both the environments under Solution Combination. Figure 19: Scenario 1

Figure 20: Scenario 2

Material Properties E = 2e11 Pa ν = 0.3 Ultimate Tensile Strength = 4.6e8 Pa Release 15.0 - © SAS IP, Inc. All rights reserved. - Contains proprietary and confidential information of ANSYS, Inc. and its subsidiaries and affiliates.

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VMMECH016 Material Properties Yield Tensile Strength = 3.5e8 Pa Endurance Strength = 2.2998e6 Pa Number of Cycles

Alternating Stress (Pa)

1000

4.6e8

1e6

2.2998e6 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. The fatigue calculations use standard formulae for the Soderberg theory.

Results Comparison Results Stress Component - Component X

Stress Component - Component Y

Stress Component - Component Z

48

Target

Mechanical

Error (%)

Life

3335.1049 3329.9

-0.156

Damage

299.8406

300.31

0.157

Safety Factor

0.019

0.019025

0.132

Life

14765.7874 14653

-0.764

Damage

67.724

68.247

0.772

Safety Factor

0.04569

0.045378

-0.683

Life

14765.7874 14766

0.001

Damage

67.724

67.725

0.001

Safety Factor

0.04569

0.045696

0.013

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VMMECH017: Thermal Stress Analysis with Remote Force and Thermal Loading

Overview Reference:

Any basic Strength of Materials book

Analysis Type(s):

Linear Thermal Stress Analysis

Element Type(s):

Solid

Test Case A cylindrical rod assembly of four cylinders connected end to end has frictionless support applied on all the cylindrical surfaces and both the flat end faces are fixed. Other thermal and structural loads are as shown below. Find the Deformation in the x direction of the contact surface on which the remote force is applied. To get accurate results apply a global element size of 1.5 m. Figure 21: Schematic

Material Properties

Geometric Properties

E = 2e11 Pa ν=0 α = 1.2e-5/°C

Loading

Diameter = 2 m Lengths of cylinders in order from End A: 2 m, 5 m, 10 m, and 3 m.

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. Location of remote force = (7,0,0) m

Results Comparison Results Maximum X Deformation (m)

Target

Mechanical

Error (%)

0.101815

0.10025

-1.5

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50

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VMMECH018: A Bar Subjected to Tensile Load with Inertia Relief

Overview Reference:

Any basic Strength of Materials book

Analysis Type(s):

Linear Static Structural Analysis (Inertia Relief On)

Element Type(s):

Solid

Test Case A long bar assembly is fixed at one end and subjected to a tensile force at the other end as shown below. Turn on Inertia Relief. Find the deformation in the z direction Figure 22: Schematic

Material Properties

Geometric Properties

E = 2e11 Pa ν = 0.3

Cross-Section = 2mx2m Lengths of bars in order from End A: 2 m, 5 m, 10 m, and 3 m.

ρ = 7850 kg/m3

Loading Force P = 2e5 N (positive z direction)

Analysis δz =

−

2ρ

where: L = total length of bar A = cross-section m = mass

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VMMECH018

Results Comparison Results Maximum Z Deformation (m)

52

Target

Mechanical

Error (%)

2.5e-6

2.5043E-06

0.172

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VMMECH019: Mixed Model Subjected to Bending Loads with Solution Combination

Overview Reference:

Any basic Strength of Materials book

Analysis Type(s):

Linear Static Structural Analysis

Element Type(s):

Beam and Shell

Test Case A mixed model (shell and beam) has one shell edge fixed as shown below. Bending loads are applied on the free vertex of the beam as given below. Apply a global element size of 80 mm to get accurate results. • Scenario 1: Only a force load. • Scenario 2: Only a moment load. Find the deformation in the y direction under Solution Combination with the coefficients for both the environments set to 1. Figure 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)

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VMMECH019 Material Properties

Geometric Properties

Loading

Beam rectangular cross section = 10 mm x 10 mm Beam length = 500 mm

Moment M = 4035 Nmm @ z-axis

Analysis δy =

l3

+

l2

where: I = total bending length of the mixed model I = moment of inertia of the beam cross-section

Results Comparison Results Maximum Y-Deformation (mm)

54

Target

Mechanical

Error (%)

-7.18742

-7.2542

0.929

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VMMECH020: Modal Analysis for Beams

Overview Reference:

Any basic Vibration Analysis book

Analysis Type(s):

Modal Analysis

Element Type(s):

Beam

Test Case Two collinear beams form a spring mass system. The density of the longer beam is kept very low so that it acts as a massless spring and the smaller beam acts as a mass. The end vertex of the longer beam (acting as a spring) is fixed. The cross section details are as shown below. Find the natural frequency of the axial mode. Figure 25: Cross Section Details for Both Beams

Figure 26: Schematic

Material Properties Material

E (Pa)

ν

ρ (kg/m3)

Spring

1.1e11

0.34

1e-8

Mass

2e11

0

7.85e5

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55

VMMECH020 Geometric Properties

Loading

Spring beam length = 500 mm Mass beam length = 5 mm

Results Comparison Results

Target

Mechanical

Error (%)

Natural Frequency of Axial Mode (Hz)

1188.6

1190.5

0.160

56

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VMMECH021: Buckling Analysis of Beams

Overview Reference:

Warren C. Young, Roark's Formulas for Stress and Strains, McGraw Hill, 6th Edition, Table 34, Case 3a, pg. 675

Analysis Type(s):

Buckling Analysis

Element Type(s):

Beam

Test Case A beam fixed at one end and is subjected to two compressive forces. One of the forces is applied on a portion of the beam of length 50 mm (L1) from the fixed end and the other is applied on the free vertex, as shown below. Find the load multiplier for the first buckling mode. Figure 27: Schematic

Material Properties

Geometric Properties

E = 2e11 Pa ν = 0.3

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

Mechanical

Error (%)

10.2397

10.198

-0.407

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57

58

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VMMECH022: Structural Analysis with Advanced Contact Options

Overview Reference:

Any basic Strength of Material book

Analysis Type(s):

Nonlinear Static Structural Analysis

Element Type(s):

Solid

Test Case An assembly of two parts with a gap has a Frictionless Contact defined between the two parts. The end faces of both the parts are fixed and a given displacement is applied on the contact surface of Part 1 as shown below. Find the Normal stress and Directional deformation - both in the z direction for each part for the following scenarios: • Scenario 1: Interface treatment - adjust to touch. • Scenario 2: Interface treatment - add offset. Offset = 0 m. • Scenario 3: Interface treatment - add offset. Offset = 0.001 m. • Scenario 4: Interface treatment - add offset. Offset = -0.001 m. Validate all of the above scenarios for Augmented Lagrange and Pure Penalty formulations. Figure 28: Schematic

Material Properties

Geometric Properties

E = 2e11 Pa ν=0

Gap = 0.0005 m Dimensions for each part: 0.1 m x 0.1 m x 0.5m

Loading Given displacement = (0, 0, 0.0006) m

Results Comparison The same results are obtained for both Augmented Lagrange and Pure Penalty formulations. Release 15.0 - © SAS IP, Inc. All rights reserved. - Contains proprietary and confidential information of ANSYS, Inc. and its subsidiaries and affiliates.

59

VMMECH022 Results

Adjust To Touch

Add Offset. Offset = 0 m

Add Offset. Offset = 0.001 m

Add Offset. Offset = 0.001 m

60

Target

Mechanical

Error (%)

Maximum directional z deformation Part 1 (m)

6e-4

6e-4

0.000

Maximum directional z deformation Part 2 (m)

6e-4

5.9786e4

-0.357

Maximum normal z stress Part 1 (Pa)

2.4e8

2.4e8

0.000

Maximum normal z stress Part 2 (Pa)

-2.4e8 -2.3915e8 -0.354

Maximum directional z deformation Part 1 (m)

6e-4

Maximum directional z deformation Part 2 (m)

1e-4

Maximum normal z stress Part 1 (Pa)

2.4e8

Maximum normal z stress Part 2 (Pa)

-4e7

Maximum directional z deformation Part 1 (m)

6e-4

6e-4

0.000

Maximum directional z deformation Part 2 (m)

1.1e3

1.0961e3

-0.355

Maximum normal z stress Part 1 (Pa)

2.4e8

2.4e8

0.000

Maximum normal z stress Part 2 (Pa)

-4.4e8 -4.3843e8 -0.357

Maximum directional z deformation Part 1 (m)

6e-4

6e-4

0.000

Maximum directional z deformation Part 2 (m)

0

0

0.000

Maximum normal z stress Part 1 (Pa)

2.4e8

2.4e8

0.000

Maximum normal z stress Part 2 (Pa)

0

0

0

6e-4

0.000

0.99644e- -0.356 4 2.4e8

0.000

-3.9858e7 -0.355

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VMMECH023: Curved Beam Assembly with Multiple Loads

Overview Reference:

Any basic Strength of Materials book

Analysis Type(s):

Linear Static Structural Analysis

Element Type(s):

Beam

Test Case An assembly of two curved beams, each having an included angle of 45°, has a square cross-section. It is fixed at one end and at the free end a Force F and a Moment M are applied. Also, a UDL of "w " N / mm is applied on both the beams. Use a global element size of 30 mm to get accurate results. See the figure below for details. Find the deformation of the free end in the y direction. Figure 29: Schematic

Equivalent Loading:

Material Properties Beam 1: E1 = 1.1e5 MPa ν1 = 0 ρ1 = 8.3e-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)

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61

VMMECH023 Material Properties

Geometric Properties

Loading

Included angle = 45°

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 3.14 x 105) / 8 = -412.334 N

Beam 2: E2 = 2e5 MPa ν2 = 0 ρ2 = 7.85e-6 kg/mm3

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

2

+ +

     

+ 4ω 2

+ 4ω

where: δ = deflection at free end in the y direction I = moment of inertia of the cross-section of both beams

Results Comparison Results Minimum Y Deformation (mm)

62

Target

Mechanical

Error (%)

-8.416664

-8.4688

0.619

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VMMECH024: Harmonic Response of a Single Degree of Freedom System for Beams

Overview Reference:

Any basic Vibration Analysis book

Analysis Type(s):

Harmonic Analysis

Element Type(s):

Beam

Test Case Two collinear beams form a spring-mass system. The density of the longer beam is kept very low so that it acts as a massless spring and the smaller beam acts as a mass. The end vertex of the longer beam (acting as a spring) is fixed. A Harmonic force F is applied on the free vertex of the shorter beam in z direction. Both beams have hollow circular cross-sections, as indicated below. • Scenario 1: Damping ratio = 0 • Scenario 2: Damping ratio = 0.05 Find the z directional deformation of the vertex where force is applied at frequency F = 500 Hz for the above scenarios with solution intervals = 25 and a frequency range of 0 to 2000 Hz. Use both Mode Superposition and Full Method. Figure 30: Schematic

Material Properties Mater- E ial (Pa)

ν

ρ (kg/m3)

Spring 1.1e11 0.34 1e-8 Mass

2e11

0

7.85e5

Geometric Properties Cross-section of each beam: Outer radius = 10 mm

Loading Harmonic force F = 1 e6 N (z-direction)

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63

VMMECH024 Geometric Properties

Loading

Inner radius = 5 mm Length of longer beam = 100 mm Length of shorter beam = 5 mm

Results Comparison Results

Target

Mechanical

Error (%)

Mode Superposi- Maximum z directional deforma- 4.11332e- 4.078e-3 tion tion without damping (m) 3

-0.859

Maximum z directional deforma- 4.11252e- 4.0765etion with damping (m) 3 3

-0.876

Maximum z directional deforma- 4.11332e- 4.1132etion without damping (m) 3 3

-0.003

Maximum z directional deforma- 4.11252e- 4.0695etion with damping (m) 3 3

-1.046

Full Method

64

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VMMECH025: Stresses Due to Shrink Fit Between Two Cylinders

Overview Reference:

Stephen P. Timoshenko, Strength of Materials, Part 2 - Advanced Theory and Problems, 3rd Edition, pg. 208-214

Analysis Type(s):

Linear Static Structural Analysis

Element Type(s):

Solid

Test Case One hollow cylinder is shrink fitted inside another. Both cylinders have length L and both the flat faces of each cylinder are constrained in the axial direction. They are free to move in radial and tangential directions. An internal pressure of P is applied on the inner surface of the inner cylinder. To get accurate results, apply a global element size of 0.8 inches. Find the maximum tangential stresses in both cylinders.

Note Tangential stresses can be obtained in the Mechanical application using a cylindrical coordinate system. To simulate interference, set Contact Type to Rough with interface treatment set to add offset with Offset = 0. 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

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65

VMMECH025 Material Properties

Geometric Properties

ν=0 ρ = 0.28383

Ro = 8” Length of both cylinders = 5”

lbm/in3

Loading

Results Comparison Results

Target

Mechanical

Error (%)

Maximum normal y stress, inner cylinder (psi)

35396.67

35738

1.0

Maximum normal y stress, outer cylinder (psi)

42281.09

42279

0.0

Note Here y corresponds to θ direction of a cylindrical coordinate system.

66

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VMMECH026: Fatigue Analysis of a Rectangular Plate Subjected to Edge Moment

Overview Reference:

Any standard Machine Design and Strength of Materials book

Analysis Type(s):

Fatigue Analysis

Element Type(s):

Shell

Test Case A plate of length L, width W, and thickness T is fixed along the width on one edge and a moment M is applied on the opposite edge about the z-axis. Find the maximum Bending Stress (Normal X Stress) and maximum Total Deformation of the plate. Also find the part life and the factor of safety using Goodman, Soderberg, & Gerber criteria. Use the x-stress component. Consider load type as fully reversed and a Design Life of 1e6 cycles, Fatigue Strength factor of 1, and Scale factor of 1. Figure 32: Schematic

Material Properties E = 2e11 Pa ν = 0.0 Ultimate tensile strength = 1.29e9 Pa Endurance strength = 1.38e8 Pa Yield Strenth = 2.5e8 Pa No. of Cycles

Alternating Stresses (Pa)

1000

1.08e9

1e6

1.38e8

Geometric Properties Length L = 12e3m Width W = 1e-3 m

Loading Moment M = 0.15 Nm (counterclockwise @ z-axis)

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67

VMMECH026 Geometric Properties

Loading

Thickness T = 1 e-3 m

Results Comparison Results

Target

Mechanical

Error (%)

9e8

9e8

0.000

Maximum total deformation (m)

6.48e-4

6.4981e4

0.279

SN-Goodman

Safety factor

0.1533

0.15333

0.020

Life

1844.3

1844.4

0.005

Safety factor

0.1533

0.15333

0.020

Life

1844.3

1844.4

0.005

Safety Factor

0.1533

0.15333

0.020

Life

1844.3

1844.4

0.005

Maximum normal x-stress (Pa)

SN-Soderberg SN-Gerber

68

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VMMECH027:Thermal Analysis for Shells with Heat Flow and Given Temperature

Overview Reference:

Any standard Thermal Analysis book

Analysis Type(s):

Thermal Stress Analysis

Element Type(s):

Shell

Test Case A plate of length (L), width (W), and thickness (T) is fixed along the width on one edge and heat flow (Q) is applied on the same edge. The opposite edge is subjected to a temperature of 20 °C. Ambient temperature is 20 °C. To get accurate results, apply a sizing control with element size = 2.5e-2 m. Find the maximum temperature, maximum total heat flux, maximum total deformation, and heat reaction at the given temperature. Figure 33: Schematic

Material Properties

Geometric Properties

Loading

E = 2e11 Pa ν = 0.0 Coefficient of thermal expansion α = 1.2e5/°C Thermal conductivity k = 60.5 W/m°C

Length L = 0.2 m Width W = 0.05 m Thickness T = 0.005 m

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

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69

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: l

δ = ∫α 0

−


Results Comparison Results

Maximum Temperature (°C) 2

Maximum Total Heat Flux (W/m ) Maximum Total Deformation (m) Heat Reaction (W)

70

Target

Mechanical

Error (%)

86.1157

86.116

0.000

2e4

2e4

0.000

7.93386e- 7.9958e5 5 -5

-5

0.781 0.000

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VMMECH028: Bolt Pretension Load Applied on a Semi-Cylindrical Face

Overview Reference:

Any standard Strength of Materials book

Analysis Type(s):

Static Structural Analysis

Element Type(s):

Solid

Test Case A semi-cylinder is fixed at both the end faces. The longitudinal faces have frictionless support. A bolt pretension load is applied on the semi-cylindrical face. To get accurate results, apply sizing control with element size of 0.01 m. Find the Z directional deformation and the adjustment reaction due to the bolt pretension load. Figure 34: Schematic

Material Properties

Geometric Properties

E = 2e11 Pa ν = 0.0

Length L = 1 m Diameter D = 0.05 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. Therefore the z-directional deformation due to pretension is given by: ×

δPretension = = δ
 ×

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71

VMMECH028

Results Comparison Results

Target

Mechanical

Error (%)

Minimum z-directional deformation (m)

-5.00E08

-5.0002E- 0.004 08

Maximum z-directional deformation 5.00E-08 (m)

4.9502E08

-0.996

Adjustment Reaction (m)

1.00E-07

0.000

72

1.00E-07

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VMMECH029: Elasto-Plastic Analysis of a Rectangular Beam

Overview Reference:

Timoshenko S., Strength of Materials, Part II, Advanced Theory and Problems, Third Edition, Article 64, pp. 349

Analysis Type(s):

Static Plastic Analysis

Element Type(s):

Solid

Test Case A rectangular beam is loaded in pure bending. For an elastic-perfectly-plastic stress-strain behavior, show that the beam remains elastic at M = Myp = σypbh2 / 6 and becomes completely plastic at M = Mult = 1.5 Myp. To get accurate results, set the advanced mesh control element size to 0.5 inches. Figure 35: Stress-Strain Curve

Figure 36: Schematic

Material Properties E = 3e7 psi ν = 0.0 σyp = 36000 psi

Geometric Properties Length L = 10” Width b = 1” Height h = 2”

Loading M = 1.0 Myp to 1.5 Myp (Myp = 24000 lbf in)

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73

VMMECH029

Analysis The load is applied in three increments: M1 = 24000 lbf-in, M2 = 30000 lbf-in, and M3 = 36000 lbf-in.

Results Comparison M/Myp

Target

Mechanical

Error (%)

State

Equivalent Stress (psi)

State

Equivalent Stress (psi)

1

fully elastic

36000

fully elastic

36059

0.164

1.25

elasticplastic

36000

elasticplastic

36288

0.800

1.5

plastic

solution not converged

plastic

solution not converged

-

74

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VMMECH030: Bending of Long Plate Subjected to Moment - Plane Strain Model

Overview Reference:

Any standard Strength of Materials book

Analysis Type(s):

Plane Strain Analysis

Element Type(s):

2D Structural Solid

Test Case A long, rectangular plate is fixed along the longitudinal face and the opposite face is subjected to a moment of 5000 lbf-in about the z-axis. To get accurate results, set the advanced mesh control element size to 0.5 inches. Find X normal stress at a distance of 0.5 inches from the fixed support. Also find total deformation and reaction moment. Figure 37: Schematic

Material Properties

Geometric Properties

E = 2.9e7 psi ν = 0.0

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), the above problem can be analyzed as a plane strain problem. Therefore, the moment applied will be per unit length (5000/1000 = 5 lbf-in). Analysis takes into account the unit length in the z-direction.

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VMMECH030 Figure 38: Plane Strain Model (analyzing any cross section (40” x 1”) along the length)

Results Comparison Results

Target

Mechanical

Error (%)

Normal Stress

30

30

0.000

Maximum Normal Stress in the X-Direction (psi)

30

30

0.000

Maximum Total Deformation (in)

0.1655e-2

0.16553e-2

0.018

Reaction Moment (lbf-in)

-5

-5

0.000

76

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VMMECH031: Long Bar with Uniform Force and Stress Tool - Plane Stress Model

Overview Reference:

Any standard Strength of Materials book

Analysis Type(s):

Plane Stress Analysis

Element Type(s):

2D Structural Solid

Test Case A long, rectangular bar assembly is fixed at one of the faces and the opposite face is subjected to a compressive force. To get accurate results, set the advanced mesh control element size to 1 m. Find the maximum equivalent stress for the whole assembly and safety factor, safety margin, and safety ratio for the first and last part using the maximum equivalent stress theory with Tensile Yield Limit. Figure 39: Schematic

Material Properties Material

E (Pa)

ν

Tensile Yield (Pa)

Part 1

1.93e11

0

2.07e8

Part 2

7.1e10

0

2.8e8

Part 3

2e11

0

2.5e8

Part 4

1.1e11

0

2.8e8

Geometric Properties Part 1: 2 3m Part 2: 2 10 m Part 3: 2 5m Part 4: 2 2m

mx2mx mx2mx

Loading Force = 1e9 N in the negative x-direction

mx2mx mx2mx

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VMMECH031

Analysis Since the loading is uniform and in one plane, the above problem can be analyzed as a plane stress problem. Analysis is done considering thickness of 2 m along z-direction Figure 40: Plane Stress Model (Analyzing any cross section along Z)

Results Comparison Results

Target

Mechanical

Error (%)

Maximum Equivalent Stress (Pa)

2.5e8

2.5e8

0.000

Safety Factor

0.828

0.828

0.000

Safety Margin

-0.172

-0.172

0.000

Safety Ratio

1.207

1.2077

0.058

Safety Factor

1.12

1.12

0.000

Safety Margin

0.12

0.12

0.000

Safety Ratio

0.892

0.89286

0.096

Part 1

Part 4

78

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VMMECH032: Radial Flow due to Internal Heat Generation in a Copper Disk Axisymmetric Model

Overview Reference:

Any basic Heat Transfer book

Analysis Type(s):

Axisymmetric Analysis

Element Type(s):

2D Structural Solid

Test Case A copper disk with thickness t and radii Ri and Ro is insulated on the flat faces. It has a heat-generating copper coaxial cable (of radius Ri) passing through its center. The cable delivers a total heat flow of Q to the disk. The surrounding air is at a temperature of To with convective film coefficient h. To get accurate results, set the advanced mesh control element size to 0.002 m. Find the disk temperature and heat flux at inner and outer radii. Figure 41: Schematic

Material Properties

Geometric Properties

E = 1.1e11 Pa ν = 0.34 Thermal conductivity k = 401.0 W/m°C

Ri = 10 mm Ro = 60 mm t = 8 mm

Loading Q = 100 W (Internal Heat Generation = 39788735.77 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 above problem can be analyzed as an axisymmetric problem.

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VMMECH032 Figure 42: Axisymmetric Model

Results Comparison Results

Target

Mechanical

Error (%)

Maximum Temperature (°C)

38.9

38.896

-0.010

Minimum Temperature (°C)

30

30.007

0.023

2

1.98943e5

197840

-0.554

2

33157

33151

-0.018

Maximum Heat Flux (W/m ) Minimum Heat Flux (W/m )

80

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VMMECH033: Electromagnetic Analysis of a C-Shaped Magnet

Overview Reference:

J. A. Edminster, Theory and Problems of Electromagnetics, Tata McGraw Hill, 2nd Edition, Example 11.9, pg. 181

Analysis Type(s):

Electromagnetic Analysis

Element Type(s):

Solid

Test Case A C-shaped magnet has a coil with 400 turns and a cross section of the core with area 4 cm2. A current of 0.1 A flows through the coil. The air gap is 0.2 cm and the coil details are given in Figure 44: Coil Details in cm (p. 82). Flux parallel is applied on the nine outer faces as shown in Figure 46: Flux Parallel Applied on 9 Outer Faces (p. 82). To get accurate results, set the advanced mesh control element size to 0.003 m. Find the total flux density and total field intensity. Figure 43: Schematic

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81

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- Poisson's Ralus (Pa) tio

Density (kg/m3)

Relative Permeability

Electric Resistivity (ohmm)

Air Body

1e7

0

0

1

0

Coil

1.1e11

0.34

8300

1

2e-7

Core

2e11

0.3

7850

500

0

Geometric Properties Given in Figure 44: Coil Details in cm (p. 82) 82

Loading Voltage = 0 V Current = 0.1 A

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VMMECH033 Geometric Properties

Loading

Depth = 2cm

Analysis Using the analogy of Ohm's law of Magnetism, 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

Mechanical

Error (%)

4.061e-2

0.040662

0.128

32320.0585

32357

0.114

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83

84

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VMMECH034: Rubber cylinder pressed between two plates

Overview Reference:

T. Tussman, K.J. Bathe, "A Finite Element Formulation for Nonlinear Incompressible Elastic and Inelastic Analysis", Computers and Structures, Vol. 26 Nos 1/2, 1987, pp. 357-409

Analysis Type(s):

Nonlinear Static Structural Analysis (Large Deformation ON)

Element Type(s):

Solid

Test Case A rubber cylinder is pressed between two rigid plates using a maximum imposed displacement of δmax. Determine the total deformation. Figure 47: Schematic

Material Properties

Geometric Properties

Loading

Solid1:

Solid1:

Displacement in Y direction = -0.1m

E = 2e11Pa ν = 0.3

0.05 m x 0.01m x 0.4 m

ρ = 7850 kg/m3 Solid2: Mooney-Rivlin Constants Solid2: Quarter Circular Cylinder C10 = 2.93e5 Pa C01 = 1.77e5 Pa Incompressibility Parameter D1 1/Pa =0

Radius = 0.2 m Length = 0.05m

Analysis Due to geometric and loading symmetry, the analysis can be performed using one quarter of the cross section. • Frictionless supports are applied on 3 faces (X = 0, Z = 0 and Z = 0.05 m). Release 15.0 - © SAS IP, Inc. All rights reserved. - Contains proprietary and confidential information of ANSYS, Inc. and its subsidiaries and affiliates.

85

VMMECH034 • Given displacement of 0.1m is applied on the top surface. • The bottom surface of Solid1 is completely fixed. • Frictionless Contact with Contact stiffness factor of 100 is used to simulate the rigid target. • Augmented Lagrange is used for Contact formulation.

Results Comparison Results Total Deformation (m)

86

Target

Mechanical

Error (%)

0.165285

0.16526

0

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VMMECH035: Thermal Stress in a Bar with Radiation

Overview Reference:

Any Basic Heat transfer and Strength of Materials book

Analysis Type(s):

Coupled Analysis (Static Thermal and Static Stress)

Element Type(s):

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, and radiating out from the other face having emissivity 0.3; Ambient Temperature is maintained at 20°C. Find the following: • Temperatures on End Faces. • Thermal strain and Directional deformation and Normal Stress in Z direction if both the end faces have frictionless supports and Reference temperature of 22°C. Figure 48: Schematic

Material Properties

Geometric Properties

E = 2.0e11 Pa v=0 α = 1.2 x 10-5 1/°C k = 60.5 W/m°C

Part 1: 2 2m Part 2: 2 5m Part 3: 2 10 m Part 4: 2 3m

Loading

mx2mx

Heat Flow = 2500W on Part 4

mx2mx

Heat Flux = 625 W/m2 on Part 4 Radiation = 20°C, 0.3

mx2mx mx2mx

Analysis (Heat flowing through body) Q = (Heat Flow) + (Heat Flux * Area) = 5000 W

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87

VMMECH035 (Heat flowing through body) = (Heat Conducted through body) = (Heat Radiated out of the Surface) i.e. Q = Qr =QC = 5000 W. Heat Radiated out of the body

Qr = ε ∗ σ ∗ A ∗ (T24 − Tα4 ) W;

Heat Conducted through the body

c=

gives T2 = 260.16°C.

K ∗
∗ 1 −    b

gives T1 = 673.38°C.

Thermal strain is given by, ε _ ax = α∆ = . × 0 − ×  673.38 −  = 7.8 656 × 0 − m / m

ε in = α∆ =  ×  −! ×   −  = 

9 × −"   

The compressive stress introduced is given by, σz = -#vg$the%&#l$st%#'* × E = -+,:>? × +:@ P# Temperature at a distance z from the face with higher temperature is given by,

BD = FGHNHI −  FGHNHI − JFLNMF  C = FGHNHI − JLNFFMC JL   Only half-length is considered for calculating deformation, since deformation is symmetric

dUo k ‹|‡ ‹ | ‡  −}q€„„ƒ × }ˆ  δ = ∫ p}q~ × } −… × w €‚q‚ƒ − ~q€€}‰ − ~~yq{ Š‰ + ∫   Š‰ † †  ~ × }ŒŒ  δ = ”Ž‘‘ × ’ −“ • δ = ∫ Oε VXYZ[\] + ε^VZu`VuZ\] RSfj

Results Comparison Results

Target

Mechanical

Error (%)

Temperature on Part 4(°C)

673.38

673.49

0

Temperature on Part 1 (°C)

260.16

260.15

0

Maximum Thermal Strain (m/m)

7.81656e-3

7.8179e-3

0

Minimum Thermal Strain (m/m)

2.85792e-3

0.0028578

0

Normal Stress in Z direction (Pa)

-1.067448e9

-1.0183e9

-4.6

Directional Deformation in Z direction (m)

-0.0123966

-0.012572

1.4

88

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VMMECH036: Thermal Stress Analysis of a Rotating Bar using Temperature Dependant Density

Overview Reference:

Any Basic Strength of Materials book

Analysis Type(s):

Static Stress Analysis (Sequence Loading)

Element Type(s):

Solid

Test Case A Bar (2 m x 2m x 20m) with one end fixed and with a rotational velocity about X axis at location (1, 1, 0) is subjected to a Uniform Temperature (Thermal Condition Load) in three steps. For all the steps, Reference Temperature is 0°C. Frictionless Support is applied on all the longitudinal faces. Figure 49: Schematic

Material Properties

Geometric Properties

E = 1 x 106 Pa α = 1 x 10-5 1/°C ν=0 Temperature °C

Density kg/m3

Part 1: 2mx2 m x 20 m

Loading Rotational Velocity (rad/s) in steps: 1. (1, 0, 0) 2. (0.5, 0, 0)

50 100 150

30 60 90

3. (0.25, 0, 0) Thermal Condition °C 1. 50°C 2. 100°C 3. 150°C

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89

VMMECH036

Analysis Rotational Stress

= ∫ ρω

=

2

ρω

2 2

0

Th
m 


=

×α×∆ ρω

 3

   Df 

 

= ×α×∆

!"#$ Equ%v#$&'" (")&**

56789 :;87?6@

=δ

=

=σ

CFCGH

./.14

=

=

+!"#"%!'#$ (")&**

A6787?6@89 :;87?6@

+ +

,&)-#$ ( " )&**

5B;=>89 :; 87?6@

Results Comparison Results Equivalent Stress (Pa)

Total Deformation (m)

90

Target

Mechanical

Error (%)

Step 1

6500

6502.6

0.040

Step 2

4000

4001.3

0.032

Step 3

2625

2625.5

0.019

Step 1

0.09

0.09

0

Step 2

0.06

0.06

0

Step 3

0.045

0.045

0

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VMMECH037: Cooling of a Spherical Body

Overview Reference:

F. Kreith, "Principles of Heat Transfer", 2nd Printing, International Textbook Co., Scranton, PA, 1959, pg. 143, ex. 4-5.

Analysis Type(s):

Transient Thermal Analysis

Element Type(s):

Plane

Test Case Determine the temperature at the center of a spherical body, initially at a temperature T0, when exposed to an environment having a temperature Te for a period of 6 hours (21600 s). The surface convection coefficient is h. • Initial temperature, T0 = 65 °F • Surface temperature, Te = 25°F • Convection coefficient h = 5.5556e-4 BTU/s-ft2-°F • Time, t = 21600 seconds • Radius of the sphere ro = 2 in = 1/6 ft Figure 50: Schematic

Material Properties

Geometric Properties

K = (1/3) BTU/hr-ft-°F 3

ρ = 62 lb/ft c = 1.075 Btu/lb-°F

Quarter Circular lamina Radius = 0.16667 ft

Loading Convection applied on Edge = 5.5556e4 BTU/s-ft2-°F Ambient Temperature for Convection = 25°F

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91

VMMECH037

Analysis Since the problem is axisymmetric, only a 2-D quarter model is used.

Results Comparison Results

Target

Mechanical

Error (%)

Temperature at the Centre of body after 21600s (°F)

28

28.688

2.457

92

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VMMECH038: Crashing Blocks Simulation with Transient Structural Analysis

Overview Reference:

Any basic Kinematics book.

Analysis Type(s):

Flexible Dynamic Analysis

Element Type(s):

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 1.1775e-4 kg. All three blocks are resting on Base. Frictionless supports are applied as shown in the figure and also on the bottom faces of left and middle blocks. Right block is fixed using Fixed Support and the base is fixed by applying Fixed Joint. Find the velocity of both the moving blocks after impact. Figure 51: Schematic

Material Properties

Geometric Properties

E = 2e5 MPa ν = 0.3 ρ = 7.85e-6kg/mm3

Left Block = 3mm x 2mm x 5mm Middle Block = 2.5mm x 2mm x 3mm Right Block =3mm x 6mm x 4mm

Loading Left Block Initial Velocity = 100 mm/s (X direction)

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93

VMMECH038 Material Properties

Geometric Properties

Loading

Base = 3mm x 8.607mm x 75.15mm

Analysis For Perfectly Elastic Collision between the blocks, mL (γLi - γLf) = mM (γMf - γMi) . . . . . . . . . . . . . . . . . . . .I γLi + γLf = γMf + γMi. . . . . . . . . . . . . . . . . . . . . . . . . . . . . II mL, mM = Mass of Left and Middle Block in kg γLi, 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, γLf = 33.3 mm/sec γMf = 133.34 mm/sec

Results Comparison Results

Target

Mechanical

Error (%)

Velocity of Left Block after impact (mm/sec)

33.3

33.809

1.5

Velocity of Middle Block after impact (mm/sec)

133.4

132.38

-0.8

94

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VMMECH039: Transient Response of a Spring-mass System

Overview Reference:

R. K. Vierck, Vibration Analysis, 2nd Edition, Harper & Row Publishers, New York, NY, 1979, sec. 5-8.

Analysis Type(s):

Flexible Dynamic Analysis

Element Type(s):

Solid and Spring

Test Case A system containing two masses, m1 and m2, and two springs of stiffness k1 and k2 is subjected to a pulse load F(t) on mass 1. Determine the displacement response of the system for the load history shown. Figure 52: Schematic

Material Properties E = 2e11 Pa γ = 0.3 ρ = 0.25 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 = 1.8 sec

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95

VMMECH039

Results Comparison Results

Target

Mechanical

Error (%)

Y1, m (@ t = 1.3s)

14.48

14.335

-1

Y2, m (@ t = 1.3s)

3.99

3.9151

-1.9

Y1, m (@ t = 2.4s)

18.32

18.511

1

Y2, m (@ t = 2.4s)

6.14

6.1971

0.9

96

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VMMECH040: Deflection of Beam using Symmetry and Anti-Symmetry

Overview Reference:

Any Basic Strength of Materials Book

Analysis Type(s):

Static Structural Analysis

Element Type(s):

Beam

Test Case A long bar 1m X 1m X 24m with simply supported ends is subjected to lateral load of 1000 N at a distance of 8m from one end. Find Deformation at the 8m from simply Supported end. Scenario 1: Considering Symmetry Scenario 2: Considering Anti-Symmetry Figure 53: Schematic

Material Properties

Geometric Properties

E = 2e11 Pa γ=0

Bar = 1m x 1m x 24m

ρ = 0.001 kg/m3

Loading Force = -1000 N (Ydirection) at 8m from Simply Supported end

Analysis Scenario 1: Considering Symmetry δ=

× × 3 ×

Scenario 2: Considering Anti-Symmetry

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97

VMMECH040

δ=

×

× 3 − ×

× × 3 ×

Results Comparison Results

Target

Mechanical

Error (%)

Scenario 1: Directional Deformation in Y-direction (m)

-2.569e-5

-2.5695e-5

0.019

Scenario 2: Directional Deformation in Y-direction (m)

-1.70662e-6

-1.7383e-6

1.856

98

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VMMECH041: Brooks Coil with Winding for Periodic Symmetry

Overview Reference:

W. Boast, Principles of Electric and Magnetic Fields, 1948 Harpers Brothers, Page 242, Equation 12.05.

Analysis Type(s):

Electromagnetic Analysis

Element Type(s):

Solid

Test Case The winding body is enclosed in an Air Body. The radius of Coil is 30 mm and cross section is 20 mm X 20 mm. The number of turns is 200 and current is 0.5 A. "Flux Parallel" is applied on all the 7 outer surfaces. Periodic Symmetry is applied on two faces. The dimensions of the air body are such that it encloses the coil. Find the Total Flux Density. Figure 54: Dimensions of Body

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99

VMMECH041 Figure 55: Schematic Diagram

Material Properties Young's Modulus (Pa)

Poisson's Ratio

Density (kg/m3)

Relative Permeability

Electric Resistivity (ohm-m)

DSVM41_MAT1 (Emag Part)

1e7

0

0

1

0

DSVM41_MAT2 (Winding Body)

1.1e11

0.34

8300

1

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) =

× × + ×

=

100

×

× π×

= ×

−3

−7

+ ×

× × ×

×

−3

-


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VMMECH041

Results Comparison Results Total Flux Density (T)

Target

Mechanical

Error (%)

1.99e-3

0.0019848

-0.3

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101

102

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VMMECH042: Hydrostatic Pressure Applied on a Square Bar with Fully, Partially Submerged in a Fluid

Overview Reference:

Any Basic Strength of Materials Book

Analysis Type(s):

Static Structural Analysis

Element Type(s):

Solid

Test Case Long bar 20m x 2m x 2m is immersed in a fluid and is fixed at one end. Fluid density is 1000 kg/m3 and Hydrostatic acceleration is 10 m/s2 in negative Z direction. Hydrostatic pressure is applied on a longitudinal face normal to X-axis at different locations as given in the scenarios below. Find normal stress in Z direction of square bar. Scenario 1: Square bar is partially immersed in the fluid up to 15 m in Z direction from the fixed support. Scenario 2: Square bar is fully immersed in the fluid up to 25 m in Z direction from the fixed support Figure 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:

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103

VMMECH042 Material Properties

Geometric Properties

Loading Scenario 1: (2,1,5) m Scenario 2: (2,1,-5) m

Analysis Scenario 1: Partialy Submerged (Pressure distribution in triangular form) Pressure distribution on square bar in triangular form, one end is maximum and other end is zero Pressure on square bar = P = ρ x g x h Load per meter is w = P x L −

=

2

Maximum bending moment = Normal stress = Bending stress = Maximum bending moment / Sectional Modulus Scenario 2: Fully Submerged (Pressure distribution in trapezoidal form)    Maximum bending moment = 

1

+

   

where: W1 = Maximum Load per meter (@ 25m) W2 = Minimum Load per meter (@ 5m) Normal stress = Bending stress = Maximum bending moment / Sectional Modulus

Results Comparison Results

Target

Mechanical

Error (%)

Normal Stress (Partially Submerged) (Pa)

8.4375e6

8529300

-1.088

3.50e7

3.5241e7

0.689

Normal Stress (Fully Submerged) (Pa)

104

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VMMECH043: Fundamental Frequency of a Simply-Supported Beam

Overview Reference:

W. T. Thompson, Vibration Theory and Applications, 2nd Printing, Prentice-Hall, Inc., Englewood Cliffs, NJ, 1965, pg. 18, ex. 1.5-1

Analysis Type(s):

Modal Analysis

Element Type(s):

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. Figure 57: Schematic

Material Properties

Geometric Properties

E = 3e7 psi ρ=0.2836 lb/in3

Loading

ℓ = 80 in A = 4 in2 h = 2 in I = 1.3333 in4

Results Comparison Results Frequency (Hz)

Target

Mechanical

Error (%)

28.766

28.613

0.532

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105

106

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VMMECH044: Thermally Loaded Support Structure

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. 30, problem 9.

Analysis Type(s):

Linear Thermal Stress Analysis

Element Type(s):

Beam

Test Case An assembly of three vertical wires has a rigid horizontal beam on which a vertically downward force Q is acting. Length of the wires is 20 in, the spacing between the wires is 10 in and the reference temperature is 70 °F. The entire assembly is subjected to a temperature rise of ∆T. Find the stresses in the copper and steel wire of the structure shown below. The wires have a cross-sectional area of A. Figure 58: Schematic

Material Properties

Geometric Properties

VMSIM044_material_rigid: Er = 3e16 psi νr = 0 VMSIM044_material_copper:

A = 0.1 in2

Loading Q = 4000 lb (Y direction) ∆T = 10 °F

Ec = 1.6e7 psi νc = 0 αc = 9.2e-6 / °F VMSIM044_material_steel:

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107

VMMECH044 Material Properties

Geometric Properties

Loading

Es = 3e7 psi νs = 0 αs = 7e-6 / °F

Results Comparison Results

Target

Mechanical

Error (%)

Stress in steel (psi)

19695

19695

0.00

Stress in copper (psi)

10152

10152

0.00

108

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VMMECH045: Laterally Loaded Tapered Support Structure

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. 342, problem 7.18.

Analysis Type(s):

Static Structural Analysis

Element Type(s):

Shell

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. It is loaded by a force F at the tip, as shown. Find the maximum bending stress at the mid-length (X = ℓ ). Figure 59: Schematic

Material Properties

Geometric Properties

Es = 3e7 psi νs = 0

Loading F = 4000 lb (Y direction)

ℓ = 50 in d = 3 in t = 2 in

Results Comparison Results Bending stress at mid length (psi)

Target

Mechanical

Error (%)

8333

8373.7

0.5

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109

110

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VMMECH046: Pinched Cylinder

Overview Reference:

R. D. Cook, Concepts and Applications of Finite Element Analysis, 2nd Edition, John Wiley and Sons, Inc., New York, NY, 1981, pp. 284-287 H. Takemoto, R. D. Cook, "Some Modifications of an Isoparametric Shell Element", International Journal for Numerical Methods in Engineering, Vol. 7 No. 3, 1973.

Analysis Type(s):

Static Structural Analysis

Element Type(s):

Shell

Test Case A thin-walled cylinder is pinched by a force F at the middle of the cylinder length. Determine the radial displacement δ at the point where F is applied. The ends of the cylinder are free edges. A one-eighth symmetry model is used. One-fourth of the load is applied due to symmetry. Figure 60: Schematic

Material Properties Es = 10.5e6 psi νs = 0.3125

Geometric Properties ℓ = 10.35 in r = 4.953 in t = 0.094 in

Loading F = 100 lbf (Y direction)

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111

VMMECH046

Analysis Due to symmetrical boundary and loading conditions, one-eighth model is used and one-fourth of the load is applied.

Results Comparison Results Deflection (in)

112

Target

Mechanical

Error (%)

-0.1139

–0.11376

-0.1

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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.

Analysis Type(s):

Plastic Structural Analysis

Element Type(s):

Axisymmetric

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

Geometric Properties

VMSIM047_CR_steel:

ℓ = 10 in Steel:

Es = 26,875,000 psi σ(yp)s = 86,000 psi VMSIM047_T4_aluminum alloy:

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

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113

VMMECH047 Material Properties

Geometric Properties

Ea = 11,000,000 psi σ(yp)a = 55,000 psi ν = 0.3

Loading

Wall thickness = 0.5 in Aluminum: Inside radius = 3.5697185 in Wall thickness = 0.5 in

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

Target

Mechanical

Error (%)

Load, lb for Deflection @ 0.032 in

1.0244e6

1033900

0.9

Load, lb for Deflection @ 0.05 in

1.262e6

1262800

0.1

Load, lb for Deflection @ 0.1 in

1.262e6

1267200

0.412

114

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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.

Analysis Type(s):

Static Structural Analysis

Element Type(s):

Beam

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

Geometric Properties

E = 3e7 psi

Loading

b = 1.5 in h = 8 in y = 6 in

Mz = 100,000 lbf-in (Z direction)

Area = 60 in2 Iz = 2000 in4

Results Comparison Results

Target

Mechanical

Error (%)

StressBEND, Bottom (psi)

300

300

0

StressBEND, Top (psi)

-700

-700

0

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115

116

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VMMECH049: Combined Bending and Torsion of Beam

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. 299, problem 2.

Analysis Type(s):

Static Structural Analysis

Element Type(s):

Beam

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. Determine the maximum principal stress σmax. Figure 63: Problem Sketch

Figure 64: Schematic

Material Properties E = 3e7 psi ν = 0.3

Geometric Properties ℓ = 25 ft r = 2.33508 in d = 3 ft

Loading F = 250 lb (Y direction) M = 9000 lbf-in (Z direction)

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117

VMMECH049

Results Comparison Results Principal stressmax (psi)

118

Target

Mechanical

Error (%)

7527

7515.5

-0.153

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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):

Static Structural Analysis

Element Type(s):

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

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119

VMMECH050

Material Properties

Geometric Properties

E = 3e7 psi ν = 0.3

Loading

t = 1 in d = 120 in

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

Target

Mechanical

Error (%)

Stressy (psi)

15000

15000

0

Stressz (psi)

30000

30002

0.007

120

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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.

Analysis Type(s):

Static Structural Analysis

Element Type(s):

Axisymmetric Shell element

Test Case A flat circular plate of radius r and thickness t is subject to various edge constraints and surface loadings. Determine the deflection δ at the middle and the maximum stress 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:

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121

VMMECH051

Material Properties

Geometric Properties

E = 3e7 psi ν = 0.3

r = 40 in t = 1 in

Loading Case 1: P = 6 psi Case 2: F = -7539.82 lb (y direction)

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

Target

Mechanical

Error (%)

Case 1:

Deflection (in) Stressmax (psi)

-0.08736 7200

-0.087114 7212.8

-0.282 0.178

Case 2:

Deflection (in) Stressmax (psi)

-0.08736 3600

-0.088025 3607.9

0.761 0.219

122

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VMMECH052: Velocity of Pistons for Trunnion Mechanism

Overview Reference:

Any Basic Kinematics book

Analysis Type(s):

Rigid Dynamic Analysis

Element Type(s):

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 = 12.57 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

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123

VMMECH052 Material Properties

Geometric Properties

E = 2e11 Pa ν = 0.3

AB = 400 mm AC = 150 mm CE = 350 mm EF = 300 mm

Loading Constant angular velocity at crank = 12.57 rad/s

Analysis Analysis done using graphical solution. Consider the Space Diagram, Velocity Diagram at the 180° from Initial Position. Figure 68: Schematic

Results Comparison Results

Target

Mechanical

Error (%)

Velocity of Piston (F) m/s

501.8

497.04

-0.949

Velocity of Piston (B) m/s

955

959.72

0.494

124

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VMMECH053: Simple Pendulum with SHM motion

Overview Reference:

Any Basic Kinematics book

Analysis Type(s):

Rigid Dynamic Analysis

Element Type(s):

Multipoint Constraint Element

Test Case A simple pendulum as shown in Figure 69: Schematic (p. 125) has a SHM motion about its hinged point given by the following equation: θ = 1.571*sin (0.5235*t) rad The hinge point coordinates are: 1. Hinge point = (0, 0, -35.56) mm Find the relative angular acceleration of pendulum after t = 3s. Figure 69: Schematic

Material Properties

Geometric Properties

E = 2000000 MPa ν = 0.3

Hinge point = (0, 0, 35.56) mm

Loading Rotation θ = 1.571*sin (0.5235*t) rad

Analysis The pendulum is having SHM motion in X-Z plane about the hinge. Angular acceleration of pendulum:

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125

VMMECH053 α=

ω 2

α=−

Results Comparison Results

Target

Mechanical

Error (%)

Relative angular acceleration of pendulum after t = 3s (rad/s2)

-0.433

-0.43054

-0.568

126

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VMMECH054: Spinning Single Pendulum

Overview Reference:

Any Basic Kinematics book

Analysis Type(s):

Rigid Dynamic Analysis

Element Type(s):

Multipoint Constraint Element

Test Case A uniform bar A is connected to a vertical shaft by a revolute joint. The vertical shaft is rotating around its vertical axis at a constant velocity Ω. A point mass M is attached at the tip of the bar in the figure below. The length of bar A is L. Its mass is m, its rotational inertia to its principal axis are Jx, Jy, Jz. The angle of the bar A to the vertical axis is denoted as . The motion equation has been established as follows.

( J z + 14 ml 2 + Ml 2 )θɺɺ − ( J x − J y + 14 ml 2 + Ml 2 ) Ω2

θ

θ + mgl

θ + Mgl

θ=

The problem is solved for { { during the first second of motion. The WB/Mechanical results are compared to a fourth order Runge-Kutta solution.

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127

VMMECH054 Figure 70: Schematic

Material Properties

Geometric Properties

Loading

L= 2.2361 m

= tan-1(1,2)


=0

Ω = 17.1522

m = 551.45 kg M = 100.0 kg Jx = 229.97 kg-m2 Jy = 2.7293 kg-m2 Jz = 229.97 kg-m2

Results Comparison Results

 at 0.5 sec  at 0.5 sec

128

Target

Mechanical

Error (%)

-1.3233

-1.3233

0.0

116.1368

116.1368

0.0

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VMMECH054 Results

Target

Mechanical

Error (%)

-2.6755

-2.6755

0.0

119.8471

119.8471

0.0

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

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129

130

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VMMECH055: Projector mechanism- finding the acceleration of a point

Overview Reference:

Any Basic Kinematics book

Analysis Type(s):

Rigid Dynamic Analysis

Element Type(s):

Multipoint Constraint Element

Test Case The mechanism shown in figure is used to pull a movie through a projector. The mechanism is driven by the drive wheel rotating at a constant -58.643 rad/s. The link lengths of all the links are constant as given below. • 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. Determine the acceleration of point C with a change of angle of link AB (θ1) from 0 to 60° in counter clockwise direction. Figure 73: Schematic

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131

VMMECH055 Material Properties

Geometric Properties

E = 2e11 Pa ν = 0.3

Loading

r1 = 18 mm r2 = 48 mm x = 45 mm y = 28 mm

Constant rotational velocity = -58.643 rad/s

Analysis

Linear acceleration of point C is given by αc = 2

θ1 − θ2 + θ2

Results Comparison Results

Target

Mechanical

Error (%)

Relative acceleration (θ1 = 10) mm/s2

-12.06

-12.043

-0.141

Relative acceleration (θ1 = 30) mm/s2

1.317

1.3168

-0.015

Relative acceleration (θ1 = 60) mm/s2

6.739

6.7386

-0.006

132

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VMMECH056: Coriolis component of acceleration-Rotary engine problem

Overview Reference:

Any Basic Kinematics book

Analysis Type(s):

Rigid Dynamic Analysis

Element Type(s):

Multipoint Constraint Element

Test Case Kinematics diagram of one of the cylinders of a rotary engine is shown below. OA is 50mm long and fixed at point o. The length of the connecting rod AB is 125mm. The line of stroke OB is inclined at 50° to the vertical. The cylinders are rotating at a uniform speed of 300 rpm in a clockwise direction, about the fixed center O. Find Angular acceleration of the connecting rod. Figure 74: Schematic

Material Properties E = 2e11 Pa ν = 0.3

Geometric Properties Connecting rod AB is 125mm Crank OA is 50mm long OB is inclined at 50° to the vertical.

Loading Constant rotational velocity = 300 rpm

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133

VMMECH056

Analysis

Angular acceleration of the connecting rod is given by: α AB =

αt 3

Results Comparison Results

Target

Mechanical

Error (%)

Angular acceleration (radian/s2)

294.52

294.53

0

134

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VMMECH057: Calculation of velocity of slider and force by collar

Overview Reference:

Beer-Johnston ‘Vector Mechanics for Engineers’ Statics & Dynamics (In SI Units), 7th Edition, TATA McGRAW HILL Edition 2004, Problem 13.73, Page No: 793

Analysis Type(s):

Rigid Dynamic Analysis

Element Type(s):

Multipoint Constraint Element

Test Case A 1.2 Kg collar is attached to a spring and slides without friction along a circular rod in a vertical plane. The spring has an undeformed length of 105 mm and a constant K = 300 N/m. Knowing that the collar is at rest at "C" and is given a slight push to get it moving. Length OP = 75 mm. Length OB = 180 mm. Determine the force exerted by the rod on the collar as it passes through point "A" and "B". Figure 75: Schematic

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135

VMMECH057 Material Properties

Geometric Properties

E = 2e11 Pa ν = 0.3

Loading

Spring:

Gravitational acceleration = -9.8066

Undeformed length = 105 mm Stiffness K = 300 N/m

m/s2 (Y Direction)

Results Comparison Results

Target

Mechanical

Error (%)

At point A (N)

14.88

14.992

0.753

At point B (N)

-23.6

-23.667

0.3

136

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VMMECH058: Reverse four bar linkage mechanism

Overview Reference:

Results are simulated using MATLAB

Analysis Type(s):

Rigid Dynamic Analysis

Element Type(s):

Multipoint Constraint Element

Test Case The figure (below) shows a reverse four bar linkage consisting of uniform rigid links PQ, QR, and RS and ground PS. Link PQ is connected with revolute joints to links QR and PS at points Q and P, respectively. Link RS is connected with revolute joints to links QR and PS at points R and S, respectively. The link lengths of all the links are constant as given below. • Fixed Link PS length r1 = 0.5m • Crank Link PQ length r2 = 0.15m • Link QR length r3 = 0.4m • Link RS length r4 = 0.45m • Gravity g = 9.81m/sec2 Determine the angular accelerations, angular velocity and rotation of link RS at joint R.

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137

VMMECH058 Figure 76: Schematic

Material Properties

Geometric Properties

E = 2e11 Pa ν = 0.3

Loading

Link PS length r1 = 0.5m Link PQ length r2 = 0.15m Link QR length r3 = 0.4m Link RS length r4 = 0.45m

Gravitational acceleration = -9.8066 m/s2 (Y Direction)

Analysis Results are obtained using MATLAB.

Results Comparison Results

Target

Mechanical

Error (%)

Angular Acceleration (rad/s )

39.6

39.336

-0.671

Angular Velocity (rad/sec)

-5.16

-5.1247

-0.7

Rotation (rad)

-0.36

-0.36255

0.7

2

138

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VMMECH059: Bending of a solid beam (Plane elements)

Overview Reference:

R. J. Roark, Formulas for Stress and Strain, 4th Edition, McGrawHill Book Co., Inc., New York, NY, 1965, pp. 104, 106.

Analysis Type(s):

Static Structural Analysis

Element Type(s):

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, determine the deflection δ at the free end and the bending stress σBend at a distance d from the wall at the outside fiber. Figure 77: Schematic

Case 1:

Case 2:

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139

VMMECH059

Material Properties

Geometric Properties

E = 30 x 106 psi ν = 0.3

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, the above problem can be analyzed as a plane stress problem.

Results Comparison Results

Target

Mechanical

Error (%)

Case 1:

Deflection (in) StressBend (psi)

0.00500 -3000

0.00500 -3000

0 0

Case 2:

Deflection (in) StressBend (psi)

0.00500 -4050

0.0051232 -4051.5

2 0

140

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VMMECH060: Crank Slot joint simulation with flexible dynamic analysis

Overview Reference:

Mechanical APDL Multibody Analysis

Analysis Type(s):

Flexible Dynamic Analysis

Element Type(s):

Solid and Multipoint Constraint Element

Test Case The figure shows crank slot model consists of a base and two rods. The two rods are attached to each other and the base with three bolts. The base of the model is fixed to the ground via a fixed joint and Bolt3 connected with slot joint to base. Define Rod1 and Rod2 as a flexible body and run the crank slot analysis using a Flexible Dynamic Analysis. Determine the Equivalent (von Mises) Stress for both flexible rods. Figure 78: Schematic

Material Properties E = 2e5 MPa ν = 0.3

Geometric Properties Rod1 length = 75mm Rod2 length = 115mm

Loading Constant angular acceleration at base to Bolt1 = 25 rad/s2

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141

VMMECH060

Analysis Figure 79: Contour Plot

Figure 80: Equivalent (von Mises) Stress

Figure 81: Total Force at Base to Bolt1

Results Comparison Results

Target

Mechanical

Error (%)

Equivalent (von Mises) Stress (MPa)

0.398

0.40834

2.6

142

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VMMECH060 Results Force @ Bolt1 (N)

Target

Mechanical

Error (%)

7.67

7.6808

0.141

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143

144

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VMMECH061: Out-of-plane bending of a curved bar

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. 412, eq. 241.

Analysis Type(s):

Static Structural Analysis

Element Type(s):

Beam

Test Case A portion of a horizontal circular ring, built-in at A, is loaded by a vertical (Z) load F applied at the end B. The ring has a solid circular cross-section of diameter d. Determine the deflection δ at end B and the maximum bending stress σBend. Figure 82: Schematic

Material Properties

Geometric Properties

Loading

r = 100 in d = 2 in θ = 90°

E = 30 x 106 psi ν = 0.3

F = -50 lb (Z direction)

Results Comparison Results Deflection (in)

Target

Mechanical

Error (%)

-2.648

-2.655

0.264

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145

VMMECH061 Results StressBend (psi)

146

Target

Mechanical

Error (%)

6366.0

6399.2

0.522

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VMMECH062: Stresses in a long cylinder

Overview Reference:

S. Timoshenko, Strength of Material, Part II, Elementary Theory and Problems, 3rd Edition, D. Van Nostrand Co., Inc., New York, NY, 1956, pg. 213, problem 1 and pg. 213, article 42.

Analysis Type(s):

Static Structural Analysis

Element Type(s):

Axisymmetric Shell

Test Case A long thick-walled cylinder is initially subjected to an internal pressure p. Determine the radial displacement δr at the inner surface, the radial stress σr, and tangential stress σt, at the inner and outer surfaces and at the middle wall thickness. Internal pressure is then removed and the cylinder is subjected to a rotation ω about its center line. Determine the radial σr and tangential σt stresses at the inner wall and at an interior point located at r = Xi. Figure 83: Schematic

Case 1:

Case 2:

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147

VMMECH062

Material Properties

Geometric Properties

E = 30 x 106 psi ν = 0.3 ρ = 0.281826 lbm/in3

Loading

a = 4 in b = 8 in Xi = 5.43 in

Case 1: Pressure = 30,000 psi (radial direction) Case 2: Rotational velocity = 1000 rad/s (Y direction)

Analysis Because the geometry and loading are symmetric about the y-axis, the above problem can be analyzed as an axisymmetric problem.

Results Comparison Results

Target

Mechanical

Error (%)

Case 1:

Displacementr, in (r = 4 in) Stressr, psi (r = 4 in) Stressr, psi (r = 6 in) Stresst, psi (r = 8 in) Stresst, psi (r = 4 in) Stresst, psi (r = 6 in) Stresst, psi (r = 8 in)

0.0078666 -30000. -7778. 0 50000. 27778. 20000.

0.0076267 -29988 -7775.3 0.79611 49988 27775 19999

-3.05 -0.04 -0.035 --0.024 -0.011 -0.005

Case 2:

Stressr, psi (r = 4 in) Stresst, psi (r = 4 in) Stressr, psi (r = 5.43 in) Stresst, psi (r = 5.43 in)

0 40588. 4753. 29436.

6.5483 41672 4933.7 29719

-2.671 3.802 0.961

148

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VMMECH063: Large deflection of a cantilever

Overview Reference:

K. J. Bathe, E. N. Dvorkin, "A Formulation of General Shell Elements - The Use of Mixed Interpolation of Tensorial Components”, International Journal for Numerical Methods in Engineering, Vol. 22 No. 3, 1986, pg. 720.

Analysis Type(s):

Static Structural Analysis

Element Type(s):

Shell

Test Case A cantilever plate of length ℓ , width b and thickness t is fixed at one end and subjected to a pure bending moment M at the free end. Determine the true (large deflection) free-end displacements and the top surface stress at the fixed end using shell elements. Figure 84: Schematic

Material Properties E = 1800 N/mm2 ν = 0.0

Geometric Properties ℓ = 12 mm b = 1 mm t = 1 mm

Loading M = 15.708 N-mm (Y direction)

Analysis Large deformation is used to simulate the problem.

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149

VMMECH063

Results Comparison Results

Target

Mechanical

Error (%)

Directional Deformation Xdirection (mm)

-2.9

-2.9354

1.221

Directional Deformation Zdirection (mm)

-6.5

-6.608

1.662

94.25

94.266

0.017

Normal Stress X-direction (N/mm2)

150

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VMMECH064: Small deflection of a Belleville Spring

Overview Reference:

S. Timoshenko, Strength of Material, Part II, Elementary Theory and Problems, 3rd Edition, D. Van Nostrand Co., Inc., New York, NY, 1956, pg. 143, problem 2.

Analysis Type(s):

Static Structural Analysis

Element Type(s):

Shell

Test Case The conical ring shown below represents an element of a Belleville spring. Determine the deflection y produced by a load F per unit length on the inner edge of the ring. Figure 85: Schematic

Material Properties

Geometric Properties

Loading

a = 1 in b = 1.5 in t = 0.1 in β = 7° = 0.12217 rad

E = 30 x 106 psi ν = 0.0

Line pressure = 100 lb/in (Y direction)

Results Comparison Results

Target

Mechanical

Error (%)

Directional Deformation Ydirection (in)

-0.0028205

-0.0029273

3.8

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151

152

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VMMECH065: Thermal Expansion to Close a Gap at a Rigid Surface

Overview Reference:

C. O. Harris, Introduction to Stress Analysis, The Macmillan Co., New York, NY, 1959, pg. 58, problem 8.

Analysis Type(s):

Static Thermal Stress Analysis

Element Type(s):

Solid and Shell

Test Case An aluminum-alloy bar is initially at a temperature of 70°F. Calculate the stresses and the thermal strain in the bar after it has been heated to 170°F. The supports are assumed to be rigid. Use a global mesh size of 0.25 in. Figure 86: Schematic

Material Properties E = 10.5 x 106 psi -5

α = 1.25 x 10 /°F

Geometric Properties ℓ = 3 in. δ = 0.002 in.

Loading ∆t = 170°F - 70°F

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153

VMMECH065 Material Properties

Geometric Properties

Loading

ν = 0.0

Results Comparison Results Normal Stress Y (psi) Thermal Strain Y (in/in)

154

Target

Mechanical

Error (%)

-6125

-6122.4

0

1.25e-003

1.25e-003

0

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VMMECH066: Bending of a Tapered Plate

Overview Reference:

C. O. Harris, Introduction to Stress Analysis, The Macmillan Co., New York, NY, 1959, pg. 114, problem 61.

Analysis Type(s):

Static Structural Analysis

Element Type(s):

Shell

Test Case A tapered cantilever plate of rectangular cross-section is subjected to a load F at its tip. Find the maximum deflection δ and the maximum principal stress σ1 in the plate. Use a global mesh size of 0.75 in. Figure 87: Schematic

Material Properties E = 30 x 106 psi ν = 0.0

Geometric Properties L = 20 in d = 3 in t = 0.5 in

Loading F = 10 lbf

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155

VMMECH066

Results Comparison Results

Target

Mechanical

Error (%)

Maximum Principal Stress (psi)

1600

1614.7

0.9

Directional Deformation Z (in)

-0.042667

-0.042746

-0.2

156

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VMMECH067: Elongation of a Solid Tapered Bar

Overview Reference:

C. O. Harris, Introduction to Stress Analysis, The Macmillan Co., New York, NY, 1959, pg. 237, problem 4.

Analysis Type(s):

Static Structural Analysis

Element Type(s):

Solid

Test Case A tapered aluminum alloy bar of square cross-section and length L is suspended from a ceiling. An axial load F is applied to the free end of the bar. Determine the maximum axial deflection δ in the bar and the axial stress σy at mid-length (Y = L/2). Use a global mesh size of 0.5 in. Figure 88: Schematic

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157

VMMECH067

Material Properties

Geometric Properties

Loading

L = 10 in d = 2 in

E = 10.4 x 106 psi ν = 0.3

F = 10000 lbf

Results Comparison Results

Target

Mechanical

Error (%)

Directional Deformation Y (in)

0.0048077

0.0048215

- 0.287

Normal Stress Y at L/2 (psi)

4444

4463

- 0.428

158

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VMMECH068: Plastic Loading of a Thick Walled Cylinder

Overview Reference:

S. Timoshenko, Strength of Material, Part II, Elementary Theory and Problems, 3rd Edition, D. Van Nostrand Co., Inc., New York, NY, 1956, pg. 388, article 70.

Analysis Type(s):

Static, Plastic Analysis (Plane Strain)

Element Type(s):

2-D Structural Solid

Test Case A long thick-walled cylinder is subjected to an internal pressure p (with no end cap load). Determine the radial stress, σr, and the tangential (hoop) stress, σt, at locations near the inner and outer surfaces of the cylinder for a pressure, pel, just below the yield strength of the material, a fully elastic material condition. Determine the effective (von Mises) stress, σeff, at the same locations for a pressure, pult, which brings the entire cylinder wall into a state of plastic flow. Use a global mesh size of 0.4 in along with a Mapped Face Meshing. Figure 89: Schematic

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159

VMMECH068 Material Properties

Geometric Properties

Loading

a = 4 in b = 8 in

E = 30 x 106 psi σyp = 30,000 psi ν = 0.3

pel = 12,990 psi pult = 24,011 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. Symmetry conditions are used on the edges perpendicular to X and Y axes. Load is applied in two steps as shown in the above table. The stresses are calculated at a distance of r = 4.4 in and 7.6 in, w.r.t a cylindrical coordinate system whose origin is same as that of the global coordinate system.

Results Comparison Results Fully Elastic

Fully Plastic

160

Target

Mechanical

Error (%)

Stressr, psi (X = 4.4 in)

-9984

-9948.8

-0.4

Stresst, psi (X = 4.4 in)

18645

18609

-0.2

Stressr, psi (X = 7.6 in)

-468

-469.1

0.2

Stresst, psi (X = 7.6 in)

9128

9129.1

0

Stresseff, psi (X = 4.4 in)

30000

30000

0

Stresseff, psi (X = 7.6 in)

30000

30000

0.00

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VMMECH069: Barrel Vault Roof Under Self Weight

Overview Reference:

R. D. Cook, Concepts and Applications of Finite Element Analysis, 2nd Edition, John Wiley and Sons, Inc., New York, NY, 1981, pp. 284-287.

Analysis Type(s):

Static Analysis

Element Type(s):

Shell

Test Case A cylindrical shell roof of density ρ is subjected to a loading of its own weight. The roof is supported by walls at each end and is free along the sides. Find the x and y displacements at point A and the top and bottom stresses at points A and B. Express stresses in the cylindrical coordinate system. Use a global mesh size of 4 m. Figure 90: Schematic

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161

VMMECH069

Material Properties

Geometric Properties

E = 4.32 x 108 N/m2 ν = 0.3 ρ = 36.7347 kg/m3

Loading

t = 0.25 m r = 25 m ℓ = 50 m Θ = 40°

g = 9.8 m/s2

Analysis A one-fourth symmetry model is used. Displacements, UX and UY, and the longitudinal rotation, ROTZ, are constrained at the roof end to model the support wall.

Results Comparison Results

Target

Mechanical

Error (%)

Directional Deformation Y @ A, m

-0.3019

-0.30903

2.362

Directional Deformation X @ A, m

-0.1593

-0.16267

2.116

Stressz, Top @ A, Pa

215570

223680

3.762

Stressz, Bottom @ A, Pa

340700

350030

2.738

Stressangle, Top @ B, Pa

191230

184270

-3.639

Stressangle, Bottom @ B, Pa

-218740

-210980

-3.5

162

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VMMECH070: Hyperelastic Thick Cylinder Under Internal Pressure

Overview Reference:

J. T. Oden, Finite Elements of Nonlinear Continua, McGraw-Hill Book Co., Inc., New York, NY, 1972, pp. 325-331.

Analysis Type(s):

Static, Large Deflection Analysis

Element Type(s):

2-D Structural Solid Elements

Test Case An infinitely long cylinder is made of Mooney-Rivlin type material. An internal pressure of Pi is applied. Find the radial displacement at the inner radius and the radial stress at radius R = 8.16 in. Use a global mesh size of 1 in along with a Mapped Face Meshing. Figure 91: Schematic

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163

VMMECH070

Material Properties

Geometric Properties

Mooney-Rivlin material coefficients:

Loading

ri = 7.0 in ro = 18.625 in

Pi = 150 psi

C10 = 80 psi C01 = 20 psi D1 = 0 /psi

Analysis Because of the loading conditions and the infinite length, this problem is solved as a plane strain problem. A one-fourth symmetry model is used. The total pressure is applied in two load increments 90 and 150 psi. Stress and Deformation are expressed in cylindrical coordinate system.

Results Comparison Results

Target

Mechanical

Error (%)

Deformation at inner radius in radial direction, in

7.18

7.1819

0.026

Radial Stress at r = 8.16 in, psi

-122

-122

0

164

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VMMECH071: Centerline Temperature of a Heat Generating Wire

Overview Reference:

W. M. Rohsenow, H. Y. Choi, Heat, Mass and Momentum Transfer, 2nd Printing, Prentice-Hall, Inc., Englewood Cliffs, NJ, 1963, pg. 106, ex. 6.5.

Analysis Type(s):

Thermal Analysis

Element Type(s):

2-D Thermal Solid Elements

Test Case Determine the centerline temperature TcL and the surface temperature Ts of a bare steel wire generating heat at the rate Q. The surface convection coefficient between the wire and the air (at temperature Ta) is h. Also, determine the heat dissipation rate q. Use a global mesh size of 0.02 ft along with a Mapped Face Meshing. Figure 92: Schematic

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165

VMMECH071

Material Properties

Geometric Properties

-3

k = 3.6111 x 10 Btu/sft-°F

Loading

ro = 0.03125 ft

h = 1.3889 x 10-3 Btu/s-ft2-°F Ta = 70°F Q = 30.92 Btu/s-ft3

Analysis Because of the symmetry in loading conditions and in the geometry, this problem is solved as an axisymmetric problem. The solution is based on a wire 1 foot long.

Results Comparison Results

Target

Mechanical

Error (%)

Centerline Temperature, °F

419.9

419.94

0.01

Surface Temperature, °F

417.9

417.85

0.012

-0.094861

-0.094861

0.00

Heat dissipation rate, BTU/s

166

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VMMECH072: Thermal Stresses in a Long Cylinder

Overview Reference:

S. Timoshenko, Strength of Material, Part II, Elementary Theory and Problems, 3rd Edition, D. Van Nostrand Co, Inc., New York, NY, 1956, pg. 234, problem 1.

Analysis Type(s):

Thermal Stress Analysis

Element Type(s):

2-D Thermal Solid Elements

Test Case A long thick-walled cylinder is maintained at a temperature Ti on the inner surface and To on the outer surface. Determine the temperature distribution through the wall thickness. 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. Figure 93: Schematic

Material Properties E = 30 x 106 psi

Geometric Properties a = 0.1875 in b = 0.625 in

Loading Ti = -1°F

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167

VMMECH072 Material Properties

Geometric Properties

Loading To = 0°F

α = 1.435 x 10-5/°F ν = 0.3 k = 8.333e-4 Btu/s-in-°F

Analysis Because of the symmetry in loading conditions and in the geometry, this problem is solved as an axisymmetric problem. The axial length is arbitrary and it is taken has 0.1 in. Nodal coupling is used in the static stress analysis. Model is used for the thermal and stress solutions.

Results Comparison Thermal Analysis

Target

Mechanical

Error (%)

T,°F (at X = 0.1875 in)

-1.0000

-1.0000

0

T,°F (at X = 0.2788 in)

-0.67037

-0.67052

0.022

T,°F (at X = 0.625 in)

0

0

0

Static Analysis

Target

Mechanical

Error (%)

Stressa, psi (at X = 0.1875 in)

420.42

416.06

-1.037

Stresst, psi (at X = 0.1875 in)

420.42

405.31

-3.594

Stressa, psi (at X = 0.625 in)

-194.58

-195.06

0.247

Stresst, psi (at X = 0.625 in)

-194.58

-195.01

0.221

168

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VMMECH073: Modal Analysis of a Cyclic Symmetric Annular Plate

Overview Reference:

R. D. Blevins, Formulas for Natural Frequency and Mode Shape, New York, NY, VanNostrand Reinhold Publishing Inc., 1979, PP. 246-247, 286-287.

Analysis Type(s):

Mode-Frequency Analysis

Element Type(s):

Solid

Test Case The fundamental natural frequency of an annular plate is determined using a mode-frequency analysis. The lower bound is calculated from the natural frequency of the annular plates that are free on the inner radius and fixed on the outer. The bounds for the plate frequency are compared to the theoretical results. Figure 94: Schematic

37 cm

0.5cm

100 cm

Material Properties

Geometric Properties

E = 7.1 x 105 kg/cm2 ν = 0.3

100 cm

Loading

Outside Radius (a) = 50 cm

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169

VMMECH073 Material Properties ρ = 2.79 x 10

-9

Geometric Properties

kg/cm2

-6

γ = 1.415 x 10 kgsec2/cm3

Loading

Inside Radius (b) = 18.5 cm Thickness (h) = 0.5 cm Sector Angle = 30°

Analysis Assumptions and Modeling Notes According to Blevins, the lower bound for the fundamental natural frequency of the annular plate is found using the formula presented in Table 11-2 of the reference:         = 
   −  





Where, λ2 = 4.80

Results Comparison Results Frequency (Hz)

170

Target 23.38

Mechanical

Error (%)

23.0747648726393 -1.305539

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(1)

VMMECH074: Tension/Compression Only Springs

Overview Reference:

Rao, Singiresu S. Mechanical Vibrations. 4th ed. Singapore: Prentice Hall, 2004. 20.

Analysis Type(s):

Rigid Body Dynamic Spring Analysis

Element Type(s):

Solid

Test Case This test calculates the elastic forces of both tension and compression only springs. A compression only spring uses a negative (compressive) displacement. A tension only spring uses a positive (tensile) displacement. Both spring types are analyzed in tension and compression loading. The detection of the spring state being in tension or compression is determined by the non-linear solver. Figure 95: Schematic

0.5 m

0.5 m 1m natural length

Tensile (x 1 ) Material Properties

Compressive (x 2 ) Geometric Properties

k = 1.0e7 N/m x1 = 0.5 m x2 = -0.5 m m = 7850 kg

Loading

Lo = 1 m

Analysis Assumptions and Modeling Notes Hooke’s Law: Elastic Force = Spring Constant * Displacement F = k*x Spring 1: Compression Only spring Release 15.0 - © SAS IP, Inc. All rights reserved. - Contains proprietary and confidential information of ANSYS, Inc. and its subsidiaries and affiliates.

171

VMMECH074 Spring 2: Tension Only spring

Results Comparison Tensile Displacement (x1) Results

Target

Mechanical

Error (%)

Elastic Force (N) Spring 1

0

0

0

Elastic Force (N) Spring 2

5.0e6

5.0e6

0

Target

Mechanical

Error (%)

Elastic Force (N) Spring 1

-5.0e6

-5.0e6

0

Elastic Force (N) Spring 2

0

0

0

Compressive Displacement (x2) Results

172

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VMMECH075: Harmonic Response of Two-Story Building under Transverse Loading

Overview Reference:

W. T. Thomson, Theory of Vibration with Applications, 3rd Edition, 1999, Example 6.4-1, pg. 166

Analysis Type(s):

Harmonic Analysis

Element Type(s):

Solid

Test Case A two-story building has two columns (2K and K) constituting stiffness elements and two slabs (2M and M) constituting mass elements. Find the y directional deformation frequency response of the system at 70 Hz on each of the vertices for the frequency range of 0 to 500 Hz using mode superposition as the solution method. Figure 96: Schematic

Material Properties Material

E (Pa)

ν

ρ (kg/m3)

Block 2

2e18

0.3

7850

Shaft 2

4.5e10

0.35

1e-8

Block 1

2e18

0.3

15700

Shaft 1

9e10

0.35

1e-8

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)

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173

VMMECH075

Analysis Assumptions and Modeling Notes The material of the columns is assigned negligible density to make them as massless springs. The slabs are allowed to move only in the y direction by applying frictionless supports on all the faces of the slabs in the y direction. The end face of the column (2K) is fixed and a harmonic force is applied on the face of the slab (M) as shown in Figure 96: Schematic (p. 173). Set the solution intervals to 50. Add the frictionless support and fixed support in a modal system, and then link the modal system to a harmonic response system.

Note There are frictionless supports on 8 faces of the geometry shown.

Results Comparison Results

Target

Mechanical

Error (%)

Maximum Amplitude for Vertex A (m)

0.20853

0.2119

1.6

Maximum Amplitude for Vertex B (m)

0.074902

0.075859

1.3

174

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VMMECH076: Elongation of a Tapered Shell with Variable Thickness

Overview Reference:

C. O. Harris, Introduction to Stress Analysis, The Macmillan Co., New York, NY, 1959, pg. 237, problem 4.

Analysis Type(s):

Static Structural Analysis

Element Type(s):

Shell

Test Case A tapered aluminum alloy plate of length L with varying thickness across length is suspended from a ceiling. An axial load F is applied to the free end of the plate. Determine the maximum axial deflection δ in the plate and the axial stress σy at mid-length (Y = L/2). Use a global mesh size of 0.5 in with mapped-face meshing. Figure 97: Schematic

Material Properties E = 10.4 x 106 psi ν = 0.3

Geometric Properties Tapered plate:

Loading F = 10000 lbf

L = 10 in Base width = 2 in Top width = 1 in Thickness varying from 2 in to 1 in from base to top.

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175

VMMECH076

Results Comparison Results

Target

Mechanical

Error (%)

Directional Deformation Y (in)

0.0048077

0.0048137

-0.1246

Normal Stress Y at L/2 (psi)

4444

4454.6

-0.2379

176

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VMMECH077: Heat Transfer in a Bar with Variable Sheet Thickness

Overview Reference:

For basic equation: Frank P. Incropera and David P. DeWitt, Heat and Mass Transfer, John Wiley & Sons, Inc, 2002, 5th Edition pg. 5.

Analysis Type(s):

Static Thermal Analysis

Element Type(s):

Shell

Test Case A 10 x 50 mm plate with a thickness varying from 1 mm to 4 mm is maintained at temperatures of 100 °C and 200 °C as shown below. Find the following: • Temperatures at mid of the surface. • Heat flow reactions on end edges. Figure 98: Schematic

Material Properties

Geometric Properties

E = 2.0e11 Pa v=0 α = 1.2 x 10-5 1/°C k = 60.5 W/m°C

Plate Dimensions : 10 X 50 mm. Thickness Variation : 1 mm to 4 mm

Loading Temperature (T1) on edge (@ 1mm thickness) = 100 °C Temperature (T2)on edge (@ 4mm thickness) = 200 °C

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177

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: +  −  

 =



Inserting Equation 3 (p. 178) in equation Equation 2 (p. 178) and integrating the equation from 0 to L,   −    −   

=

   

(3)

(4)

Temperature at any point y is given as:

!=

" & # & '( ,,. + $ & ) %# − %*+

(5)

Results Comparison Results

Target

Mechanical

Error (%)

Heat reaction at T1 (W)

2.618

2.6188

0.00

Heat reaction at T2 (W)

-2.618

-2.6188

0.00

166.083

166.09

0.00

Temperature at mid of surface (°C)

178

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VMMECH078: Gasket Material Under Uniaxial Compression Loading-3-D Analysis

Overview Reference:

Any Nonlinear Material Verification Text

Analysis Type(s):

Static Analysis (ANTYPE=0)

Element Type(s):

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.

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179

VMMECH078

180

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VMMECH078

Material Properties

Geometric Properties

E = 104728E6 Pa ν = 0.21

Loading

L=1m T = 0.02 m

P1 = 44006400 Pa P2 = 157147000 Pa

Analysis A 3-D analysis is performed first using a mesh of 4 x 4 gasket elements. In order to simulate the loadingunloading behavior of gasket material, the model is first loaded with a pressure P1 and unloaded and then loaded with a pressure P2 and unloaded. The pressure-closure responses simulated are compared to the material definition. Because of convergence issues, the model could not be unloaded to 0 Pa and was instead unloaded to 100 Pa.

Results Comparison Target

Mechanical

Error (%)

Gasket Pressure and Closure at End of 1st Loading: GK-PRES

4.4006E+07

4.4006E+07

0

GK-CLOS

4.064E-04

4.064E-04

0

Gasket Pressure and Closure at End of 2nd Loading: GK-PRES

1.5715E+08

1.5715E+08

0

GK-CLOS

6.8327E-04

6.8327E-04

0

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181

182

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VMMECH079: Natural Frequency of a Motor-Generator

Overview Reference:

W. T. Thomson. “Vibration Theory and Applications”. 2nd Printing, Prentice-Hall, Inc., Englewood Cliffs, NJ. pg. 10, ex. 1.3-3. 1965.

Analysis Type(s):

Mode-Frequency Analysis

Element Type(s):

Pipe Element

Test Case A small generator of mass m is driven by a main engine through a solid steel shaft of diameter d. If the polar moment of inertia of the generator rotor is J, determine the natural frequency f in torsion. Assume that the engine is large compared to the rotor so that the engine end of the shaft may be assumed to be fixed. Neglect the mass of the shaft also. Figure 99: Schematic

Material Properties

Geometric Properties

Loading

d = .375 in

E = 31.2 x 106 psi

ℓ = 8.00 in

2

m = 1 lb-sec /in

J = .031 lb-in-sec2

Results Comparison Results

Target

Mechanical

Error (%)

Lower Order F, Hz

48.781

48.781

0

Higher Order F, Hz

48.781

48.781

0

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183

184

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VMMECH080: Transient Response of a Spring-mass System

Overview Reference:

R. K. Vierck. “Vibration Analysis”. 2nd Edition. Harper & Row Publishers, New York, NY, 1979. sec. 5-8.

Analysis Type(s):

Transient Dynamic Mode Superposition Analysis

Element Type(s):

Test Case A system containing two masses, m1 and m2, and two springs of stiffness k1 and k2 is subjected to a pulse load F(t) on mass 1. Determine the displacement response of the system for the load history shown. 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 = 1.8 sec

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185

VMMECH080

Results Comparison Results

Target

Mechanical

Error (%)

Y1 , m (@ t = 1.3s)

14.48

14.349

-0.9

Y2 , m (@ t = 1.3s)

3.99

3.9478

-1.1

Y1 , m (@ t = 2.4s)

18.32

18.097

-1.2

Y2 , m (@ t = 2.4s)

6.14

6.094

-0.7

186

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VMMECH081: Statically Indeterminate Reaction Force Analysis

Overview Reference:

P.Bezler, M. Hartzman, and M. Reich. Dynamic Analysis of Uniform Support Motion Response Spectrum Method, (NUREG/CR-1677), Brookhaven National Laboratory, August 1980. Problem 2. Pages 48-80.

Analysis Type(s):

Modal analysis Spectral analysis

Element Type(s):

Elastic straight pipe elements Structural Mass element

Test Case This benchmark problem contains three-dimensional multi-branched piping systems. The total mass of the system is represented by structural mass elements specified at individual nodes. Modal and response spectrum analyses are performed on the piping model. Frequencies obtained from modal solve and the nodal/element solution obtained from spectrum solve are compared against reference results. The NUREG intermodal/interspatial results are used for comparison. Figure 101: Schematic

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187

VMMECH081 Material Properties

Geometric Properties

Loading

Pipe Elements:

Straight Pipe:

Acceleration response spectrum curve defined by SV and FREQ commands.

Outer Diameter = 2.375 in Wall Thickness = 0.154 in

E = 27.8999 x 106 psi. Nu = 0.3 Density = 2.587991718e10 lb-sec2/in4 Mass Elements (lb-sec2/in): (Mass is isotropic) Mass @ node 1: M = 0.447000518e-01 Mass @ node 2: M = 0.447000518e-01 Mass @ node 3: M = 0.447000518e-01 Mass @ node 4: M = 0.447000518e-01 Mass @ node 5: M = 0.432699275e-01 Mass @ node 6: M = 0.893995859e-02 Mass @ node 7: M = 0.432699275e-01 Mass @ node 8: M = 0.893995859e-02 Mass @ node 9: M = 0.893995859e-02 Mass @ node 10: M = 0.432699275e-01 Mass @ node 11: M = 0.893995859e-02 Mass @ node 12: M = 0.432699275e-01 Mass @ node 13: M = 0.893995859e-02 Mass @ node 14: M = 0.893995859e-02

Results Comparison Results

Target

Mechanical

Error (%)

1

8.712

8.7121

0.00

2

8.806

8.8091

0.04

3

17.510

17.509

0.01

4

40.370

40.368

0.00

5

41.630

41.642

0.03

188

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VMMECH081 Results Node

Target

Mechanical

Error (%)

UX at node6

0.46186

0.46186

0.00

UY at node8

0.0025747

0.0025747

0.00

UZ at node8

0.446591

0.44949

0.65

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189

190

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VMMECH082: Fracture Mechanics Stress for a Crack in a Plate

Overview Reference:

W.F.Brown, Jr., J.E.Srawley, Plane strain crack toughness testing of high strength metallic materials, ASTM STP-410, (1966).

Analysis Type(s):

Static Structural Analysis

Element Type(s):

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

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191

VMMECH082

Material Properties

Geometric Properties

Loading

E = 30 x 106 psi

a = 1 in

σ0 = 0.5641895 psi

ν = 0.3

b = 5 in h = 5 in t = 0.25 in

Results Comparison Results

Target

Mechanical

Error (%)

Stress Intensity KI

1.0249

1.0504

2.5

192

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VMMECH083: Transient Response to a Step Excitation

Overview Reference:

W. T. Thomson, Vibration Theory and Applications, 2nd Printing, Prentice-Hall, Inc., Englewood Cliffs, NJ, 1965, pg. 102, article 4.3.

Analysis Type(s):

Mode-Superposition Transient Dynamic Analysis

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. Determine the displacement u at time t for damping ratio, ξ = 0.5. Figure 103: Schematic

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193

VMMECH083

Material Properties

Loading

m = 0.5 lb-

F= 200 lb

2

sec /in k = 200 lb/in

Analysis Assumptions and Modeling Notes The damping coefficient c is calculated as 2ξ sqrt(km) = 10 lb-sec/in for ξ = 0.5.

194

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VMMECH083

Results Comparison Results

Target

Mechanical

Error (%)

Total Def Max (ξ = 0.5) Time = 0.20 sec

1.1531

1.1544

0.1

Figure 104: Maximum Deformation vs. Time (damped)

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195

196

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VMMECH084: Mullins Effect on a Rubber Tube Model Subjected to Tension Loading

Overview Reference:

.W.Ogden, et al., “A Pseudo-elastic Model for the Mullins Effect in Filled Rubber", Royal Society of London Proceedings Series A., (1989), pg: 2861-2877.

Analysis Type(s):

Static Analysis

Element Type(s):

Solid

Test Case An axisymmetric rubber plate made of Neo-Hookean material is modeled with radius R and height H. The model is subjected to cyclic displacement loading on the top surface. The axial stress obtained at different load steps is compared against the reference solution. Figure 105: Schematic

Material Properties

Geometric Properties

Loading

Neo-Hookean Constants:

R = 0.5m

One cycle of loading

µ = 8 MPa

H = 1m Step 1: λ = 1.5

Ogden-Roxburgh Mullins Constants:

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197

VMMECH084 Material Properties

Geometric Properties

r = 2.104 m = 30.45 β =0.2

Loading Step 2: λ = 2.0 Step 3: λ = 3.0 Step 4: λ = 2.0 Step 5: λ = 1.5 Step 6: λ = 1.0

Results Comparison Results

Axial Stress (Pa)

Stretch λ

Target

Mechanical

Error (%)

1.5

12.666

12.667

0.008

2.0

28.000

28.000

0.0

3.0

69.333

69.333

0.0

2.0

20.819

20.823

0.019

1.5

8.660

8.6704

0.12

1.0

0.000

0.0

0.0

Figure 106: Variation of Axial Stress

198

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VMMECH085: Bending of a Composite Beam

Overview Reference:

R. J. Roark, W. C. Young, Formulas for Stress and Strain, McGraw-Hill Book Co., Inc., New York, NY, 1975, pg. 112114, article 7.2.

Analysis Type(s):

Static Analysis

Element Type(s):

Solid

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, and a bending moment My at the free-end. Ei and αi correspond to the Young's modulus and thermal coefficient of expansion for layer i, respectively. Determine the free-end displacement δ (in the Z-direction) and the X-direction stresses at the top and bottom surfaces of the layered beam. Figure 107: Schematic

Material Properties

Geometric Properties

Loading

MAT1:

ℓ = 8 in

To = 100°F

E1 = 1.2 x 6

10 psi α1 = 1.8 x 10-4 in/in/°F

w = 0.5 in t1 = 0.2 in t2 = 0.1 in

Tref = 0°F My = 10.0 inlb

MAT2:

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199

VMMECH085 Material Properties

Geometric Properties

Loading

E2 = 0.4 x 106 psi α2 = 0.6 x 10-4 in/in/°F

Results Comparison Results

Target

Mechanical

Error (%)

Displacement, in

-0.832

-0.832

0.0

StressxTOP , psi

1731

1731

0.0

StressxBOT , psi

2258

2258

0.0

200

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VMMECH086: Stress Concentration at a Hole in a Plate

Overview Reference:

R. J. Roark, Formulas for Stress and Strain, 4th Edition, McGraw-Hill Book Co., Inc., New York, NY, 1965, pg. 384

Analysis Type(s):

Static Structural, Submodeling (2D-2D)

Element Type(s):

Solid

Test Case Determine the maximum stress at a circular hole cut into a rectangular plate loaded with uniform tension P. Figure 108: Plate Problem Sketch

Material Properties

Geometric Properties L = 12 in d = 1 in t = 1 in

E = 30 x 106 psi υ = 0.3

Loading P = 1000 psi

Analysis Assumptions and Modeling Notes Due to symmetry, only a quarter sector of the plate is modeled. The reference result is from an infinitely long plate. Using a transferred load from the coarse model, the submodel result closely approximates the fine model.

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201

VMMECH086

Results Comparison 2D-2D Results Results

Target

Mechanical

Error (%)

Fine Model

Equivalent Stress - Max

3018

3025.7

0.255

Coarse Model

Equivalent Stress - Max

3018

2272.1

-24.715

Submodel

Equivalent Stress - Max

3018

3032.8

0.490

Figure 109: 2D-2D Fine Model Equivalent Stress

Figure 110: 2D-2D Coarse Model Equivalent Stress

202

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VMMECH086 Figure 111: 2D-2D Submodel Equivalent Stress

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203

204

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VMMECH087: Campbell Diagrams and Critical Speeds Using Symmetric Orthotropic Bearings

Overview Reference:

Nelson, H.D., McVaugh, J.M., “The Dynamics of RotorBearing Systems Using Finite Elements”, Journal of Engineering for Industry, Vol 98, pp. 593-600, 1976

Analysis Type(s):

Modal Analysis

Element Type(s):

Line Body Point Mass Bearing Connection

Test Case A rotor-bearing system is analyzed to determine the forward and backward whirl speeds. The distributed rotor is modeled as a configuration of six elements, with each element composed of subelements. See Table 1: Geometric Data of Rotor-Bearing Elements (p. 205) for a list of the geometric data of the individual elements. Two symmetric orthotropic bearings are located at positions four and six. A modal analysis is performed on the rotor-bearing system with multiple load steps to determine the whirl speeds and Campbell values for the system. Figure 112: Rotor-Bearing Configuration

Table 1: Geometric Data of Rotor-Bearing Elements Element Number 1 2 3

Subelement number

Axial Distance to Subelement

Inner Diameter (cm)

Outer Diameter (cm)

1

0.00

0.51

2

1.27

1.02

1

5.08

0.76

2

7.62

2.03

1

8.89

2.03

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205

VMMECH087 Element Number

4 5 6

Subelement number

Axial Distance to Subelement

Inner Diameter (cm)

Outer Diameter (cm)

2

10.16

3.30

3

10.67

1.52

3.30

4

11.43

1.78

2.54

5

12.70

2.54

6

13.46

1.27

1

16.51

1.27

2

19.05

1.52

1

22.86

1.52

2

26.67

1.27

1

28.70

1.27

2

30.48

3.81

3

31.50

2.03

4

34.54

1.52

2.03

Material Properties

Geometric Properties

Loading

Shaft

Refer to Table 1: Geometric Data of Rotor-Bearing Elements (p. 205)

Rotational Velocity

E11 = 2.078 x 1011 Pa G12 = 1.0 x 1014 Pa Density = 7806 kg/m3 Mass Element Mass = 1.401 kg Polar inertia = 0.002 kg⋅m2 Diametral inertia = 0.00136 kg⋅m2 Bearing Element

Spin (1) = 1000 RPM Spin (2) = 20000 RPM Spin (3) = 40000 RPM Spin (4) = 60000 RPM Spin (5) = 80000 RPM Spin (6) = 100000 RPM

Spring coefficients K11 = K22 = 3.503 x 107 N/m K12 = K21 = -8.756 x 106 N/m

Analysis Assumptions and Modeling Notes A modal analysis is performed on the rotor-bearing system with QR Damp methods using pipe elements (PIPE288) to determine the whirl speeds and Campbell values. A point mass is used to model the rigid disk (concentrated mass). Two symmetric orthotropic bearings are used to assemble the rotor system. No shear effect is included in the rotor-bearing system. The displacement and rotation along and around the X-axis is constrained so that the rotor-bearing system does not have any torsion or traction related displacements.

206

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VMMECH087 Backward and forward whirl speeds for slope = 1 @ 100000 RPM are determined from the modal analysis.

Results Comparison Target

Mechanical

Ratio

Backward and forward whirl speeds for slope = 1 @ 100000 RPM PIPE288 Mode 1 (BW)

10747.0000

10793.4

1.004

Mode 2 (FW)

19665.0000

19560.0

0.995

Mode 3 (BW)

39077.0000

39668.4

1.015

Mode 4 (FW)

47549.0000

48207.0

1.014

Figure 113: Campbell Diagram for Rotor-Bearing System

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207

208

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VMMECH088: Harmonic Response of a Guitar String

Overview Reference:

Blevins, R.D., Formulas for Natural Frequency and Mode Shape, Nostrand Reinhold Co., New York, NY, 1979, pg. 90, tab. 7-1

Analysis Type(s):

Static Structural Linear Perturbed Modal Linear Perturbed Harmonic

Element Type(s):

Beam

Test Case A uniform stainless steel guitar string of length l and diameter d is stretched between two rigid supports by a tensioning force F1, which is required to tune the string to the E note of a C scale. The string is then struck near the quarter point with a force F2. Determine the fundamental frequency, f1. Also, show that only the odd-numbered frequencies produce a response at the midpoint of the string for this excitation. Material Properties

Geometric Properties l = 710 mm c = 165 mm d = 0.254 mm

E = 190 x 109 Pa ρ = 7920 kg/m3

Loading F1 = 84 N F2 = 1 N

Analysis Assumptions and Modeling Notes Enough elements are selected so that the model can be used to adequately characterize the string dynamics. The stress stiffening capability of the elements is used. Linear perturbed harmonic analysis determines the displacement response to the lateral force F2. Figure 114: Guitar String Problem

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209

VMMECH088

Results Comparison Target

Mechanical

Ratio

Modal

f, Hz

322.2

322.621

1.001

Frequency Response

f1, (322.2 Hz)

Response

Response, 320 < f < 328

-

f2, (644.4 Hz)

No Response

No Response

-

f3, (966.6 Hz)

Response

Response, 966 < f < 974

-

f4, (1288.8 Hz)

No Response

No Response

-

f5, (1611.0 Hz)

Response

Response, 1611 < f < 1619

-

f6, (1933.2 Hz)

No Response

No Response

-

Figure 115: String Midpoint Displacement Amplitude

210

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VMMECH089: Delamination Analysis of a Double Cantilever Beam Using Contact-Based Debonding

Overview Reference:

Alfano, G., Crisfield, M.A., “Finite Element Interface Models for the Delamination Analysis o Mechanical and Computation Issues”, International Journal for Numerical Methods in Engine 1736, 2001

Analysis Type(s):

Static Structural

Element Type(s):

Solid

Test Case A double cantilever beam of length l, width w, and height h with an initial crack of length a at the free end is subjected to a maximum vertical displacement Umax at the top and bottom free end nodes. Determine the vertical reaction at point P, based on the vertical displacement using the contact-based debonding capability. Figure 116: Double Cantilever Beam Sketch

Material Properties

Geometric Properties

Composite E11 = 135.3 GPa E22 = 9.0 GPa E33 = 9.0 GPa G12 = 5.2 GPa ν12 = 0.24 ν13 = 0.24 ν23 = 0.46

l = 100 mm a = 30 mm h = 3 mm w = 20 mm

Loading Umax = 10 mm

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211

VMMECH089 Material Properties

Geometric Properties

Loading

C5 = 1.0 x 10-5

Analysis Assumptions and Modeling Notes A double cantilever beam is analyzed under displacement control using 2-D plane strain formulation elements. An imposed displacement of Uy = 10 mm acts at the top and bottom free vertex. Contact debonding is inserted at the interface. Defined fracture-energy based debonding material is used to define the material for contact debonding. Equivalent separation-distance based debonding material is also used for the contact debonding object. Based on the interface material parameters used, results obtained using Mechanical are compared to results shown in Figure 15(a) of the reference material.

Results Comparison Target

Mechanical

Ratio

Max RFORCE and corresponding displacement using debonding RFORCE FY (N)

50.677

50.677

1.000

DISP UY (mm)

1.50

1.50

1.000

RFORCE and corresponding displacement U = 10.0 using debonding RFORCE FY (N)

24.553

24.553

1.000

DISP UY (mm)

10.00

10.00

1.000

212

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VMMECH090: Delamination Analysis of a Double Cantilever Beam Using Interface Delamination

Overview Reference:

Alfano, G., Crisfield, M.A., “Finite Element Interface Models for the Delamination Analysis o Engineering, Vol 50, pp. 1701-1736, 2001

Analysis Type(s):

Static Structural

Element Type(s):

Solid

Test Case A double cantilever beam of length l, width w, and height h with an initial crack of length a at the free end is subjected to a maximum vertical displacement Umax at the top and bottom free end nodes. Determine the vertical reaction at point P based on the vertical displacement for the interface model. Figure 117: Double Cantilever Beam Sketch

Material Properties

Geometric Properties

Composite E11 = 135.3 GPa E22 = 9.0 GPa E33 = 9.0 GPa G12 = 5.2 GPa ν12 = 0.24 ν13 = 0.24 ν23 = 0.46

l = 100 mm a = 30 mm h = 3 mm w = 20 mm

Loading Umax = 10 mm

Interface C1 (maximum stress) = 25 MPa C2 (normal separation) = 0.004 mm C3 (shear separation) = 1000 mm

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213

VMMECH090

Analysis Assumptions and Modeling Notes A double cantilever beam is analyzed under displacement load using interface elements for delamination and 2-D plane strain formulation elements. An imposed displacement of Uy = 10 mm acts at the top and bottom free vertex. An Interface Delamination object is inserted to model delamination. Equivalent material constants are used for the interface material, as Mechanical uses the exponential form of the cohesive zone model and the reference uses a bilinear constitutive model.

Results Comparison Lower Order Results Target

Mechanical

Ratio

Max RFORCE and corresponding DISP: RFORCE FY (N)

60.00

60.069

1.001

DISP UY (mm)

1.00

1.000

1.000

End RFORCE and corresponding DISP RFORCE FY (N)

24.00

24.288

1.012

DISP UY (mm)

10.00

10.00

1.00

Higher Order Results Target

Mechanical

Ratio

Max RFORCE and corresponding DISP RFORCE FY (N)

60.00

60.063

1.001

DISP UY (mm)

1.00

1.000

1.000

End RFORCE and corresponding DISP RFORCE FY (N)

24.00

24.289

1.012

DISP UY (mm)

10.00

10.00

1.00

214

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Part III: Design Exploration Descriptions

VMDX001: Optimization of L-Shaped Cantilever Beam under Axial Load

Overview Reference:

From the Basic Principle

Analysis Type(s):

Goal Driven Optimization

Element Type(s):

3-D Solid

Test Case An L-shaped beam with dimensions 30 x 25 mm with 4 mm as the rib thickness and 300 mm in length has the surface fixed at one end. A force of 10,000 N is then applied to the opposite end of the beam. Input Parameters:

Width, Height, and Length (CAD Geometry)

Response Parameters:

Volume, Stress, and Deflection

Figure 118: 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

Loading Fixed Support Force F = 10000 N (Z direction)

Parameter

Type

Limits

Desired Value

Importance

Width

Input

20 mm ≤ W ≤ 30 mm

No Preference

High

Height

Input

25 mm ≤ H ≤ 35 mm

No Preference

High

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217

VMDX001 Parameter

Type

Limits

Desired Value

Importance

Length

Input

250 mm ≤ L ≤ 350 mm

No Preference

High

Volume

Output

n/a

Minimum Possible

Low

Stress

Output

n/a

Minimum Possible

High

Deflection

Output

n/a

Minimum Possible

High

Analysis Beam volume: =

+

+

Maximum axial deformation under load F: =

× +

=

−2

× +

Normal stress along Z-direction: σ=

=

+

+

Combined objective function becomes: Φ=

×

−5

+

+

+

+

+

+

+

+

−

Minimizing ϕ we get dimensions as: L = Length = 0.250 m W = Width = 0.030 m H = Height = 0.035 m

Results Comparison Results Volume (V) Deformation (D) Stress (σ)

218

Target

DesignXplorer Error (%) 6.9E-05 m3

0.0

4.5290e-5 m

4.5339E-05 m

0.10862

3.62319e7 Pa

3.623065E07 Pa

0.00046

6.9e-5 m

3

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VMDX002: Optimization of Bar with Temperature-Dependent Conductivity

Overview Reference:

From the Basic Principle

Analysis Type(s):

Goal Driven Optimization

Element Type(s):

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 = 0.038, a = 0.00582. The bar is constrained on all faces by frictionless support. A temperature of 100°C is applied at one end of the bar. The reference temperature is 5°C. At the other end, a constant convection coefficient of 0.005 W/m2°C is applied. The ambient temperature is 5°C. Input Parameters:

Convection coefficient, coefficient of thermal expansion and length

Response Parameters:

Temperature (scoped on end face), thermal strain

Figure 119: Schematic

Material Properties E = 2e11 Pa ν=0 α = 1.5E-05/°C K = k0*(1 + a*T) W/m-°C k0 = 0.038 a = 0.00582

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

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219

VMDX002 Parameter

Type

Limits

Desired Value

Importance

Length (l)

Input

15 m ≤ l ≤ 25 m

No Preference

Low

Convection coefficient (h)

Input

0.004 W/m2°C ≤ h ≤ 0.006 W/m2°C

No Preference

Low

Coefficient of temperature expansion (α)

Input

1.4e-5/°C ≤ α ≤ 1.6e-5/°C

No Preference

Low

Temperature (T)

Output

n/a

Minimum Possible

High

Thermal strain (ε)

Output

n/a

Minimum Possible

High

Analysis Temperature: s

=−

a

−

+

7 22 a

×

+

×

6

a

+

Thermal strain: ε=α

−
=α

−

Combined objective function becomes,

Φ=

− α   

+



−

+ ×

  

+



×



+

 −  

Minimizing ϕ we get input parameters as: l = beam length = 25 m h = convection coefficient = 0.006 W/m2°C α = coefficient of thermal expansion = 1.4e-5/°C

Results Comparison Results

Target

Length (l)

DesignXplorer Error (%)

25 m

25 m

0

0.006 W/m2°C

0.006 W/m2°C

0

Coefficient of thermal expan- 1.4e-5/°C sion (α)

1.4e-5/°C

0

Temperature (T)

29.6528°C

29.553°C

-0.3278

Thermal strain (ε)

3.4514e-4 m/m

3.437e-4 m/m

-0.4115

Convection coefficient (h)

220

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α−

VMDX003: Optimization of Water Tank Column for Mass and Natural Frequency

Overview Reference:

S. S. Rao, Optimization Theory and Application Second edition, example 1.10, page 28-30

Analysis Type(s):

Goal Driven Optimization with APDL

Element Type(s):

3-D Solid

Test Case A uniform column of rectangular cross section b and d m is to be constructed for supporting a water tank of mass M. It is required to: 1. minimize the mass of the column for economy 2. maximize the natural frequency of transverse vibration of the system for avoiding possible resonance due to wind. Design the column to avoid failure due to direct compression (should be less than maximum permissible compressive stress) and buckling (should be greater than direct compressive stress). Assume the maximum permissible compressive stress as σmax. The design vector is defined as: T=

=

T

where: b = width of cross-section of column d = depth of cross-section of column Input Parameters:

Width and Height

Response Parameters:

Mass, Natural Frequency, Direct Stress, Buckling Stress

Material Properties

Geometric Perperties

Loading

E = 3e10 Pa

Width, b = 0.4 m

Mass of water tank M = 1000000 Kg

ρ = 2300

Depth, d =1.2 m

Kg/m

Acceleration due to gravity =

3

Length, I = 20 m

9.81 m/s2

σ max = 4.1e7 Pa Sample Size:

10000

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221

VMDX003 Results

Target

DesignXplorer Error (%)

Width b

0.36102 m

0.36102 m

0.000

Depth d

1.3181 m

1.318137 m

0.002

Mass of column M

21890 kg

21889.77 kg

–0.001

0.87834 rad/sec

0.87816 rad/sec

-0.020

Direct stress

2.0386e7 Pa

2.0383e7 Pa

-0.015

Buckling stress

6.1526e6 Pa

6.15174e6 Pa

-0.013

Natural frequency w

Analysis Minimize:

Mass of the column =

= ρ× × ×

Maximize:   N
r  rqy 
r  vr  vibr
i 
 w
r
k w =   3   × ×  

1/ 2   × × 3   + ×ρ× × ×  

Subject to constraints:  × D _S  = σx −   ×  π × d Bg_S  =   

 ≥  ×   − ×   

×  ≥ × 

Required objective is obtained by having: b = 0.36102 m d = 1.3181 m M = (minimum) = 21890 kg W = (maximum) = 0.87834 rad/sec Direct stress = 2.0386e7 Pa Buckling stress = 6.1526e6 Pa

Results Comparison Results

Target

DesignXplorer Error (%)

Width b

0.36102 m

0.36102 m

0.000

Depth d

1.3181 m

1.318137 m

0.002807

Mass of column M

21890 kg

21890.1957 kg

-0.00089

222

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VMDX003 Results Natural frequency w

Target

DesignXplorer Error (%)

0.87834 rad/sec

0.87816 rad/sec

-0.02074

Direct stress

2.0386e7 Pa

2.0383e7 Pa

-0.01277

Buckling stress

6.1526e6 Pa

6.15174e6 Pa

-0.0139

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223

224

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VMDX004: Optimization of Frequency for a Plate with Simple Support at all Vertices

Overview Reference:

Blevins, Formula for Natural Frequency and Mode Shape, Van Nostrand Reinhold Company Inc., 1979, pg. 269-271

Analysis Type(s):

Goal Driven Optimization

Element Type(s):

3-D Shell

Test Case A square plate of side 250 mm and thickness 5 mm is simply supported on all its vertices. Input Parameters:

Young's modulus, Poisson's ratio and density

Response Parameters:

First natural frequency

Figure 120: Schematic

Material Properties E = 2e5 MPa ν = 0.3 ρ = 7.850 e-6 kg/mm3

Geometric Properties Length a = 250 mm Width b = 250 mm Thickness h = 5 mm

Loading All vertices are simply supported

Parameter

Type

Constraints

Desired Value

Importance

Young's Modulus E

Input

1.8e11 Pa ≤ E ≤ 2.2e11Pa

No Preference

Low

Poisson's Ratio µ

Input

0.27 ≤ µ ≤ 0.30

No Preference

Low

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225

VMDX004 Parameter

Type

Constraints

Desired Value

Importance

Density ρ

Input

7065 kg/m3 ≤ ρ ≤ 8635 kg/m3

No Preference

Low

First Natural Frequency w

Output

N/a

Minimum Possible

High

Analysis First Natural Frequency:

=

2

π

2

  

1/ 2

3

ρ

− ν2

  

Objective function becomes: 


φ=

−

     ρ − ν


Minimizing ϕ we get dimensions as: Young's Modulus E = 1.8e11 Pa Poisson's Ratio µ = 0.27 Density ρ = 8635 kg/m3 First Natural Frequency w = 124.0913 rad/s

Results Comparison Results

Target

Young's Modulus E

1.8e11 Pa

Poisson's Ratio µ

1.8e11 Pa

0.27

Density ρ

8635 kg/m

First Natural Frequency w

226

DesignXplorer Error (%)

124.0913 rad/s

0.00

0.27 3

8635 kg/m

0.00 3

123.36 rad/s

0.00 -0.5894

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VMDX005: Optimization of Buckling Load Multiplier with CAD Parameters and Young's Modulus

Overview Reference:

Timoshenko, Strength of Materials, Part 2 (Advanced theory and problems), pg. 167–168

Analysis Type(s):

Goal Driven Optimization

Element Type(s):

3-D Solid

Test Case The cantilever bar of length 25 feet is loaded by uniformly distributed axial force p = 11 lbf on one of the vertical face of the bar in negative Z-direction. The bar has a cross-sectional area A is 0.0625 ft2. Input Parameters:

Side of Square C/S , Length of Cantilever Bar and Young's Modulus

Response Parameters:

Load Multiplier of the First Buckling Mode

Optimization Method:

Genetic Algroithm

Sample Size:

200

Figure 121: Schematic

Material Properties E = 4.1771e 9 psf ν = 0.3

Geometric Properties Cross-section of square = 0.25 ft. x 0.25 ft. Length of bar = 25 ft.

ρ = 490.45 lbm/ft3

Loading Fixed support on one face, Force = 11 lbf (Negative Z-direction) on top face

Parameter

Type

Constraints

Desired Value

Importance

Cross-section side

Input

0.225 ft. ≤ a ≤ 0.275 ft.

No Preference

N/A

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227

VMDX005 Parameter

Type

Constraints

Desired Value

Importance

Length

Input

22.5 ft. ≤ l ≤ 27.5 ft.

No Preference

N/A

Young's Modulus

Input

3.7594e9 psf ≤ E ≤ 4.5948e9 psf

No Preference

N/A

First buckling mode load multiplier

Output

N/A

Maximum Possible

N/A

Analysis Assuming that under the action of uniform axial load a slight lateral bucking occurs. The expression for deflection is: π

=δ −

The critical load is given by,

cr

=

cr

=

π2

=

2

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: ×

 

×

=

×

 

Combined objective function becomes: Φ=

−

×

−5 E.a  l



Minimizing ϕ we get dimensions as:

228

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VMDX005 Cross-section side a = 0.275 ft. Length l = 22.5 ft. Young's Modulus E = 4.5948e9 psf Buckling load multiplier = 3083.32

Results Comparison Results

Target

First buckling mode load multiplier

3083.32

DesignXplorer Error (%) 3036.07

-1.532

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229

230

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