ANSYS Workbench Verification Manualdocshare01.docshare.tips/files/22829/228290690.pdf · 2016. 7....

ANSYS Workbench Verification Manual

Release 15.0ANSYS, Inc.

November 2013Southpointe

275 Technology Drive

Canonsburg, PA 15317 ANSYS, Inc. is

certified to ISO

9001:[email protected]

http://www.ansys.com

(T) 724-746-3304

(F) 724-514-9494

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Published in the U.S.A.

Table of Contents

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Overview .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Index of Test Cases .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

I. DesignModeler Descriptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1. VMDM001: Extrude, Chamfer, and Blend Features .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

2. VMDM002: Cylinder using Revolve, Sweep, Extrude, and Skin-Loft ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3. VMDM003: Extrude, Revolve, Skin-Loft, and Sweep .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

II. Mechanical Application Descriptions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1. VMMECH001: Statically Indeterminate Reaction Force Analysis ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

2. VMMECH002: Rectangular Plate with Circular Hole Subjected to Tensile Loading .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3. VMMECH003: Modal Analysis of Annular Plate .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4. VMMECH004: Viscoplastic Analysis of a Body (Shear Deformation) ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

5. VMMECH005: Heat Transfer in a Composite Wall ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

6. VMMECH006: Heater with Nonlinear Conductivity ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

7.VMMECH007:Thermal Stress in a Bar with Temperature Dependent Conductivity ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

8. VMMECH008: Heat Transfer from a Cooling Spine .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

9. VMMECH009: Stress Tool for Long Bar with Compressive Load .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

10. VMMECH010: Modal Analysis of a Rectangular Plate .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

11. VMMECH011: Large Deflection of a Circular Plate with Uniform Pressure .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

12. VMMECH012: Buckling of a Stepped Rod .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

13. VMMECH013: Buckling of a Circular Arch .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

14. VMMECH014: Harmonic Response of a Single Degree of Freedom System ..... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

15.VMMECH015: Harmonic Response of Two Storied Building under Transverse Loading .... . . . . . . . . . . . . . . . . . . . . 45

16. VMMECH016: Fatigue Tool with Non-Proportional Loading for Normal Stress .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

17. VMMECH017: Thermal Stress Analysis with Remote Force and Thermal Loading .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

18. VMMECH018: A Bar Subjected to Tensile Load with Inertia Relief ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

19.VMMECH019: Mixed Model Subjected to Bending Loads with Solution Combination .... . . . . . . . . . . . . . . . . . . . . . 53

20. VMMECH020: Modal Analysis for Beams .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

21. VMMECH021: Buckling Analysis of Beams .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

22.VMMECH022: Structural Analysis with Advanced Contact Options .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

23. VMMECH023: Curved Beam Assembly with Multiple Loads .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

24. VMMECH024: Harmonic Response of a Single Degree of Freedom System for Beams .... . . . . . . . . . . . . . . . . . . . . . 63

25. VMMECH025: Stresses Due to Shrink Fit Between Two Cylinders .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

26. VMMECH026: Fatigue Analysis of a Rectangular Plate Subjected to Edge Moment .... . . . . . . . . . . . . . . . . . . . . . . . . . 67

27. VMMECH027: Thermal Analysis for Shells with Heat Flow and Given Temperature .... . . . . . . . . . . . . . . . . . . . . . . . . . . 69

28. VMMECH028: Bolt Pretension Load Applied on a Semi-Cylindrical Face .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

29. VMMECH029: Elasto-Plastic Analysis of a Rectangular Beam ..... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

30. VMMECH030: Bending of Long Plate Subjected to Moment - Plane Strain Model ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

31. VMMECH031: Long Bar with Uniform Force and Stress Tool - Plane Stress Model ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

32. VMMECH032: Radial Flow due to Internal Heat Generation in a Copper Disk - Axisymmetric Model ... . 79

33. VMMECH033: Electromagnetic Analysis of a C-Shaped Magnet .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

34. VMMECH034: Rubber cylinder pressed between two plates .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

35. VMMECH035: Thermal Stress in a Bar with Radiation ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

36. VMMECH036: Thermal Stress Analysis of a Rotating Bar using Temperature Dependant Density .... . . . . . 89

37. VMMECH037: Cooling of a Spherical Body .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

38. VMMECH038: Crashing Blocks Simulation with Transient Structural Analysis ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

39. VMMECH039: Transient Response of a Spring-mass System ..... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

40. VMMECH040: Deflection of Beam using Symmetry and Anti-Symmetry .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

41.VMMECH041: Brooks Coil with Winding for Periodic Symmetry ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

iiiRelease 15.0 - © SAS IP, Inc. All rights reserved. - Contains proprietary and confidential information

of ANSYS, Inc. and its subsidiaries and affiliates.

42.VMMECH042: Hydrostatic Pressure Applied on a Square Bar with Fully, Partially Submerged in a Flu-

id .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 Debond-

ing .... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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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Workbench Verification Manual

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 re-

peated. As a result, the tests offer a quick introduction to new features with which you may be unfamil-

iar.

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.

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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

Solution OptionsAnalysis TypeElement TypeTest Case Number

LinearStatic StructuralSolidVMMECH001


Free VibrationModalSolidVMMECH003

Nonlinear, Visco-

plastic Materials

StructuralSolidVMMECH004

LinearStatic ThermalSolidVMMECH005

NonlinearStatic ThermalSolidVMMECH006

Nonlinear Thermal

Stress

Static StructuralSolidVMMECH007

LinearStatic ThermalSolidVMMECH008


Free VibrationModalShellVMMECH010

Nonlinear, Large

Deformation

Static StructuralShellVMMECH011

BucklingSolidVMMECH012

BucklingShellVMMECH013

HarmonicSolidVMMECH014


FatigueStatic StructuralSolidVMMECH016

Linear Thermal

Stress


Linear, Inertia reliefStatic StructuralSolidVMMECH018

LinearStatic StructuralBeamVMMECH019

Shell

ModalBeamVMMECH020

BucklingBeamVMMECH021

Nonlinear, ContactStatic StructuralSolidVMMECH022

LinearStatic StructuralBeamVMMECH023

HarmonicBeamVMMECH024


FatigueStatic StructuralShellVMMECH026

Linear Thermal

Stress



Nonlinear, Plastic

Materials


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Introduction


Static Structural2-D Solid, Plane

Strain

VMMECH030


Stress

VMMECH031

Linear Thermal

Stress

Static Structural2-D Solid, Axisym-

metric

VMMECH032

ElectromagneticStatic StructuralSolidVMMECH033

Nonlinear, Large

Deformation


Coupled (Static

Thermal and Static

Stress)

SolidVMMECH035

Sequence LoadingStatic StructuralSolidVMMECH036

Transient Thermal2-D Solid, Axisym-

metric

VMMECH037

Flexible DynamicTransient StructuralSolidVMMECH038


Spring

Static StructuralBeamVMMECH040

ElectromagneticStatic StructuralSolidVMMECH041

Hydrostatic FluidStatic StructuralSolidVMMECH042

ModalBeamVMMECH043

Linear Thermal

Stress




Nonlinear, Plastic

Materials

Static Structural2-D Solid, Axisym-

metric

VMMECH047



Static StructuralAxisymmetric ShellVMMECH050


Rigid DynamicMultipoint Con-

straint

VMMECH052


straint

VMMECH042


straint

VMMECH054


straint

VMMECH055



Index of Test Cases



straint

VMMECH056


straint

VMMECH057


straint

VMMECH058

Static Structural2-D Plane Stress

Shell

VMMECH059


Multipoint Con-

straint



Nonlinear, Large

Deformation



Linear Thermal

Stress

Static StructuralSolid

Shell

VMMECH065



Nonlinear, Plastic

Materials


Strain

VMMECH068


Nonlinear, Large

Deformation

Static Structural2-D SolidVMMECH070

Static Thermal2-D Thermal SolidVMMECH071

Linear Thermal

Stress

Static Structural2-D Thermal SolidVMMECH072

ModalSolidVMMECH073

Rigid Body Dynam-

ics

Solid

Spring

VMMECH074



Static ThermalThermal ShellVMMECH077

Static Structural3-D SolidVMMECH078

3-D Gasket

ModalPipeVMMECH079

Mode Superposi-

tion

Transient DynamicSpring

Mass

VMMECH080


Introduction


ModalPipeVMMECH081

SpectralMass

Fracture MechanicsStatic StructuralSolidVMMECH082

Mode Superposi-

tion

Transient DynamicSpring, MassVMMECH083

Nonlinear, Hypere-

leastic


Composite MaterialStatic StructuralSolidVMMECH085


Submodeling (2D-

2D)

ModalLine BodyVMMECH087

Point Mass

Bearing Connection

Linear PerturbationStatic StructuralBeamVMMECH088

Modal

Harmonic

Contact-Based De-

bonding


Interface Delamina-

tion




Index of Test Cases

Part I: DesignModeler Descriptions

VMDM001: Extrude, Chamfer, and Blend Features

Overview

Extrude, Chamfer, and BlendFeature:

MillimeterDrawing Units:

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 rect-

angular 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.



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

Error (%)Design-

Modeler

TargetResults

0189561.95189561.95Volume (mm3)

-0.013544433.344439.29Surface Area (mm2)

05252Number of Faces

011Number of Bodies


VMDM001

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

Overview

Revolve, Sweep, Extrude, and Skin-LoftFeature:


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.



Results Comparison

Error (%)Design-

Modeler

TargetResults

01130973.31130973.3Volume (mm3)

081053.181053.1Surface Area (mm2)

033Number of Faces

011Number of Bodies


VMDM002

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

Overview

Extrude, Revolve, Skin-Loft, and SweepFeature:


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:



Vc = 47123.9 mm3 and Vd = 28352.8 mm

3.•

• 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

Error (%)Design-

Modeler

TargetResults

0825394.5825394.4Volume (mm3)

0101719.95101719.47Surface Area (mm2)

02222Number of Faces

044Number of Bodies


VMDM003

Part II: Mechanical Application Descriptions

VMMECH001: Statically Indeterminate Reaction Force Analysis

Overview

S. Timoshenko, Strength of Materials, Part 1, Elementary Theory

and Problems, 3rd Edition, CBS Publishers and Distributors, pg.

22 and 26

Reference:

Linear Static Structural AnalysisAnalysis

Type(s):

SolidElement

Type(s):

Test Case

An assembly of three prismatic bars is supported at both end faces and is axially loaded with forces F1

and F2. Force F1 is applied on the face between Parts 2 and 3 and F2 is applied on the face between

Parts 1 and 2. Apply advanced mesh control with element size of 0.5”.

Find reaction forces in the Y direction at the fixed supports.

Figure 4: Schematic

LoadingGeometric PropertiesMaterial Properties

Force F1 =

-1000 (Y direc-

tion)

Cross section of

all parts = 1” x

1”

E = 2.9008e7 psi

ν = 0.3

ρ = 0.28383 lbm/in3

Length of Part

1 = 4"Force F2 = -500

(Y direction)Length of Part

2 = 3"

Length of Part

3 = 3”



Results Comparison

Error (%)MechanicalTargetResults

0.127901.14900Y Reaction Force at Top

Fixed Support (lbf )

-0.190598.86600Y Reaction Force at Bottom

Fixed Support (lbf )


VMMECH001

VMMECH002: Rectangular Plate with Circular Hole Subjected to Tensile Loading

Overview

J. E. Shigley, Mechanical Engineering Design, McGraw-Hill, 1st

Edition, 1986, Table A-23, Figure A-23-1, pg. 673

Reference:


Type(s):

SolidElement

Type(s):

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


Pressure = -100

Pa

Length = 15 mE = 1000 Pa

Width = 5 mν = 0

Thickness = 1

m

Hole radius =

0.5 m

Results Comparison


0.864315.2312.5Maximum Normal X Stress

(Pa)



VMMECH003: Modal Analysis of Annular Plate

Overview

R. J. Blevins, Formula for Natural Frequency and Mode Shape,

Van Nostrand Reinhold Company Inc., 1979, Table 11-2, Case

4, pg. 247

Reference:

Free Vibration AnalysisAnalysis

Type(s):

SolidElement

Type(s):

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


Inner diameter

of inner plate =

20"

E = 2.9008e7 psi

ν = 0.3

ρ = 0.28383 lbm/in3

Inner diameter

of middle plate

= 28"

Inner diameter

of outer plate =

34"

Outer diameter

of outer plate =

40"




Thickness of all

plates = 1"

Results Comparison


-0.23310.21310.9111st Frequency Mode (Hz)

-0.78315.6318.0862nd Frequency Mode (Hz)

-0.77315.64318.0863rd Frequency Mode (Hz)

-1.38346.73351.5694th Frequency Mode (Hz)




VMMECH003

VMMECH004: Viscoplastic Analysis of a Body (Shear Deformation)

Overview

B. Lwo and G. M. Eggert, "An Implicit Stress Update Al-

gorithm Using a Plastic Predictor". Submitted to Computer

Reference:

Methods in Applied Mechanics and Engineering, January

1991.

Nonlinear Structural AnalysisAnalysis

Type(s):

SolidElement

Type(s):

Test Case

A cubic shaped body made up of a viscoplastic material obeying Anand's law undergoes uniaxial shear

deformation at a constant rate of 0.01 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

h

Velocity = 0.01 cm/s

Problem Model

h

x

y


Temp = 400°C

= 673°K

h = 1 cmEx (Young's Modulus) =

60.6 GPa thickness = 1

cm Velocity (x-direc-

tion) = 0.01 (Poisson's Ratio) =

0.4999cm/sec @ y = 1

cmSo = 29.7 MPa

Q/R = 21.08999E3 KTime = 20 sec

A = 1.91E7 s-1

= 7.0




m = 0.23348

ho = 1115.6 MPa

µ

= 18.92 MPa

= 0.07049

a = 1.3

Results Comparison


-6.3-791.76845.00Fx, N


VMMECH004

VMMECH005: Heat Transfer in a Composite Wall

Overview

F. Kreith, Principles of Heat Transfer, Harper and Row Publisher,

3rd Edition, 1976, Example 2-5, pg. 39

Reference:

Linear Static Thermal AnalysisAnalysis

Type(s):

SolidElement

Type(s):

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


Cross-section =

1" x 1"

Fire brick wall: k =

2.222e-4 BTU/s ft °F

Fire brick wall

thickness = 9"

Insulating wall: k =

2.778e-5 BTU/s ft °F

Insulating wall

thickness = 5"

Results Comparison


0.202336.68336Minimum Temperature (°F)

0.0072957.22957Maximum Temperature (°F)



VMMECH006: Heater with Nonlinear Conductivity

Overview

Vedat S. Arpaci, Conduction Heat Transfer, Addison-Wesley Book

Series, 1966, pg. 130

Reference:

Nonlinear Static Thermal AnalysisAnalysis Type(s):

SolidElement Type(s):

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 as-

sumed to be uniform and is applied as internal heat generation.

Find the maximum temperature and maximum total heat flux.

Figure 9: Schematic


Front face temperat-

ure = 212°F

k = [0.01375 * (1 + 0.001 T)]

BTU/s in°F

Radius = 3.937”

Thickness = 1”

Internal heat gener-

ation = 10 BTU/s

in3

Conductiv-

ity (BTU/s

in°F)

Temperat-

ure (°F)

1.419e-00232

2.75e-0021000

Results Comparison


0.96480.58476Maximum Temperature (°F)

-0.0039.999710Maximum Total Heat Flux

(BTU/s in2)



VMMECH007:Thermal Stress in a Bar with Temperature Dependent Conductivity

Overview

Any basic Heat Transfer bookReference:

Nonlinear Thermal Stress AnalysisAnalysis

Type(s):

SolidElement

Type(s):

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 tem-

perature 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


Rear face tem-

perature T =

100°C

Length = 20 mE = 2e11 Pa

Width = 2 mν = 0

Breadth = 2 mα = 1.5e-05 / °C

Film Coefficient

h = 0.005

W/m2°C

k = 0.038*(1 +

0.00582*T) W/m °C

Conductiv-

ity (W/m °C)

Temperat-

ure (°C) Ambient tem-

perature = 5°C3.91e-0025Reference tem-

perature = 5°C0.215800



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:

εT = × × −−5

Deformation in the z direction is given by:

= × × −−∫

0

Heat flux in the z direction is given by:

= × −

Results Comparison


-0.01638.01438.02Minimum Temperature (°C)

0.0420.000495210.000495Maximum Thermal strain (z

= 20) (m/m)

0.0000.0014250.001425Maximum Thermal strain (z

= 0) (m/m)

0.9050.0023410.00232Maximum Z Deformation

(m)

0.0420.165070.165Maximum Z Heat Flux (z =

20) (W/m2)


VMMECH007

VMMECH008: Heat Transfer from a Cooling Spine

Overview

Kreith, F., Principles of Heat Transfer, Harper and Row, 3rd Edition,

1976, Equation 2-44a, pg. 59, Equation 2–45, pg. 60

Reference:

Linear Static Thermal AnalysisAnalysis

Type(s):

SolidElement

Type(s):

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.



LoadingGeometric

Properties

Material Properties

Tw = 100°FE = 4.177e9 psf

Cross section =

1.2” x 1.2”

ν = 0.3 Ta = 0°FThermal conductiv-

ity k = 9.71e-3

BTU/s ft °F

h = 2.778e-4

BTU/s ft2 °F

L = 8”

Results Comparison


0.05579.07879.0344Temperature of the Tip (°F)

-0.0416.3614e-36.364e-3Heat Conducted by the

Spine (Heat Reaction)

(BTU/s)



VMMECH009: Stress Tool for Long Bar with Compressive Load

Overview

Any basic Strength of Materials bookReference:


Type(s):

SolidElement

Type(s):

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.


Material Properties

Tensile Yield

(Pa)

νE (Pa)Mater-

ial

2.07e801.93e11Part 1

2.8e807.1e10Part 2

2.5e802e11Part 3

2.8e801.1e11Part 4

LoadingGeometric Properties

Pressure = 2.5e8

Pa

Part 1: 2 m x 2

m x 3 m

Part 2: 2 m x 2

m x 10 m

Part 3: 2 m x 2

m x 5 m



Part 4: 2 m x 2

m x 2 m

Results Comparison


0.0002.5e82.5e8Maximum Equivalent Stress

(Pa)

0.0000.8280.828Safety Factor for Part 1


0.00011Safety Factor for Part 3



VMMECH009

VMMECH010: Modal Analysis of a Rectangular Plate

Overview

Blevins, Formula for Natural Frequency and Mode Shape, Van

Nostrand Reinhold Company Inc., 1979, Table 11-4, Case 11,

pg. 256

Reference:

Free Vibration AnalysisAnalysis

Type(s):

ShellElement

Type(s):

Test Case

A rectangular plate is simply supported on both the smaller edges and fixed on one of the longer edges

as shown below. 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.



Length = 0.25

m

E = 2e11 Pa

ν = 0.3

Width = 0.1 mρ = 7850 kg/m3

Thickness =

0.005 m

Results Comparison


-0.952590.03595.71st Frequency Mode (Hz)

-0.9871118.41129.552nd Frequency Mode (Hz)

-0.6672038.12051.793rd Frequency Mode (Hz)





-0.48933503366.485th Frequency Mode (Hz)


VMMECH010

VMMECH011: Large Deflection of a Circular Plate with Uniform Pressure

Overview

Timoshenko S.P., Woinowsky-Krieger S., Theory of Plates and

Shells, McGraw-Hill, 2nd Edition, Article 97, equation 232, pg.

401

Reference:

Nonlinear Structural Analysis (Large Deformation On)Analysis

Type(s):

ShellElement

Type(s):

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.



Pressure =

6585.18 Pa

Radius = 0.25

m

E = 2e11 Pa

ν = 0.3

Thickness =

0.0025 m

Results Comparison


-1.0080.00123740.00125Total deformation (m)



VMMECH012: Buckling of a Stepped Rod

Overview

Warren C. Young, Roark's Formulas for Stress & Strains, McGraw

Hill, 6th Edition, Table 34, Case 2a, pg. 672

Reference:

Buckling AnalysisAnalysis

Type(s):

SolidElement

Type(s):

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.



Force at free

end = 1000 N

Larger diameter

= 0.011982 m

E = 2e11 Pa

ν = 0.3

Force at the flat

step face = -

2000 N

Smaller diamet-

er = 0.010 m

Length of lar-

ger diameter =

0.2 m

Both forces are

in the z direc-

tionLength of smal-

ler diameter =

0.1 m

Results Comparison


2.035622.95822.5Load Multiplier



VMMECH013: Buckling of a Circular Arch

Overview

Warren C. Young, Roark's Formulas for Stress Strains, McGraw

Hill, 6th Edition, Table 34, Case 10, pg. 679

Reference:


Type(s):

ShellElement

Type(s):

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.



Pressure = 1

MPa

Arch cross-sec-

tion = 5 mm x

50 mm

E = 2e5 MPa

ν = 0

Mean radius of

arch = 50 mm

Included angle

= 90°

Results Comparison


0.4546.07544Load Multiplier



VMMECH014: Harmonic Response of a Single Degree of Freedom System

Overview

Any basic Vibration Analysis bookReference:

Harmonic AnalysisAnalysis

Type(s):

SolidElement

Type(s):

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


Material Properties

ρ (kg/m3

)νE (Pa)Material

1e-80.341.1e11Shaft 1

1e-80.341.1e11Shaft 2

1e-80.354.5e10Shaft 3

1e-80.354.5e10Shaft 4


Force = 1e7 N (Z-

direction)

Each cylinder:

Diameter = 20 mm

Point Mass =

3.1044 Kg

Length = 50 mm



Results Comparison


0.5910.141230.1404Maximum Amplitude

without damping (m)

0.000180180Phase angle without damp-

ing (degrees)

0.5770.14080.14Maximum Amplitude with

damping (m)

0.000175.58175.6Phase angle with damping

(degrees)


VMMECH014

VMMECH015: Harmonic Response of Two Storied Building under Transverse

Loading

Overview

W. T. Thomson, Theory of Vibration with Applications, 3rd Edition,

1999, Example 6.4-1, pg. 166

Reference:


Type(s):

SolidElement

Type(s):

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.


Material Properties

ρ (kg/m3

)νE (Pa)Material

78500.32e18Block 2

1e-80.354.5e10Shaft 2

157000.32e18Block 1

1e-80.359e10Shaft 1



LoadingGeometric Proper-

ties Force = -1e5 N (y

direction)Block 1 and 2:

40 mm x 40 mm x

40 mm

Shaft 1 and 2:

20 mm x 20 mm x

200 mm

Results Comparison


1.50.211740.20853Maximum Amplitude for

vertex A (m)


vertex B (m)


VMMECH015

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

Overview

Any basic Machine Design bookReference:

Fatigue AnalysisAnalysis

Type(s):

SolidElement

Type(s):

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 y-

and z-directions.

Find the life, damage, and safety factor for the normal stresses in the x, y, and z directions for non-

proportional 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


Material Properties

E = 2e11 Pa

ν = 0.3

Ultimate Tensile Strength = 4.6e8 Pa



Material Properties

Yield Tensile Strength = 3.5e8 Pa

Endurance Strength = 2.2998e6 Pa

Alternating Stress

(Pa)

Number of

Cycles

4.6e81000

2.2998e61e6

LoadingGeometric

Properties Scenario 1: Force

= 2e6 N (y-direc-

tion)

Bar: 20 m x 1 m

x 1m

Scenario 2: Pres-

sure = -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

Error

(%)

Mechanic-

al

TargetResults

-0.1563329.93335.1049LifeStress Component - Component

X 0.157300.31299.8406Damage

0.1320.0190250.019Safety

Factor

-0.7641465314765.7874LifeStress Component - Component

Y 0.77268.24767.724Damage

-0.6830.0453780.04569Safety

Factor

0.0011476614765.7874LifeStress Component - Component

Z 0.00167.72567.724Damage

0.0130.0456960.04569Safety

Factor


VMMECH016

VMMECH017: Thermal Stress Analysis with Remote Force and Thermal Loading

Overview


Linear Thermal Stress AnalysisAnalysis

Type(s):

SolidElement

Type(s):

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.



Given temperature

(End A) = 1000°C

Diameter = 2 mE = 2e11 Pa

Lengths of cylin-

ders in order

ν = 0

Given temperature

(End B) = 0°C

α = 1.2e-5/°C

from End A: 2

m, 5 m, 10 m,

and 3 m.

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


-1.50.100250.101815Maximum X Deformation

(m)



VMMECH018: A Bar Subjected to Tensile Load with Inertia Relief

Overview


Linear Static Structural Analysis (Inertia Relief On)Analysis

Type(s):

SolidElement

Type(s):

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



Force P = 2e5 N

(positive z direc-

tion)

Cross-Section =

2 m x 2 m

E = 2e11 Pa

ν = 0.3

Lengths of bars

in order fromρ = 7850 kg/m

3

End A: 2 m, 5

m, 10 m, and 3

m.

Analysis

δρ

z = −2

where:

L = total length of bar

A = cross-section

m = mass



Results Comparison


0.1722.5043E-062.5e-6Maximum Z Deformation

(m)


VMMECH018

VMMECH019: Mixed Model Subjected to Bending Loads with Solution Combination

Overview



Type(s):

Beam and ShellElement

Type(s):

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.




Force F = -10 N (y

direction)

Shell = 160 mm

x 500 mm x 10

mm

E = 2e5 Pa

ν = 0




Moment M = -

4035 Nmm @ z-ax-

is

Beam rectangu-

lar cross section

= 10 mm x 10

mm

Beam length =

500 mm

Analysis

δy = +3 2l l

where:

I = total bending length of the mixed model

I = moment of inertia of the beam cross-section

Results Comparison


0.929-7.2542-7.18742Maximum Y-Deformation

(mm)


VMMECH019

VMMECH020: Modal Analysis for Beams

Overview


Modal AnalysisAnalysis

Type(s):

BeamElement

Type(s):

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


Material Properties

ρ (kg/m3)νE (Pa)Material

1e-80.341.1e11Spring

7.85e502e11Mass




Spring beam length =

500 mm

Mass beam length = 5

mm

Results Comparison


0.1601190.51188.6Natural Frequency of Axial

Mode (Hz)


VMMECH020

VMMECH021: Buckling Analysis of Beams

Overview

Warren C. Young, Roark's Formulas for Stress and Strains, McGraw

Hill, 6th Edition, Table 34, Case 3a, pg. 675

Reference:


Type(s):

BeamElement

Type(s):

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.



Force on L1 =

-1000 N (x direc-

tion)

L1 = 50 mmE = 2e11 Pa

ν = 0.3 Total length =

200 mm

Force on free ver-

tex = -1000 N (x

direction)

Rectangular

cross section =

10 mm x 10

mm

Results Comparison


-0.40710.19810.2397Load Multiplier



VMMECH022: Structural Analysis with Advanced Contact Options

Overview

Any basic Strength of Material bookReference:

Nonlinear Static Structural AnalysisAnalysis

Type(s):

SolidElement

Type(s):

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.



Given displace-

ment = (0, 0,

0.0006) m

Gap = 0.0005 mE = 2e11 Pa

Dimensions for

each part: 0.1

m x 0.1 m x

0.5m

ν = 0

Results Comparison

The same results are obtained for both Augmented Lagrange and Pure Penalty formulations.



Er-

ror

(%)

Mechan-

ical

Tar-

get

Results

0.0006e-46e-4Maximum directional z de-

formation Part 1 (m)

Adjust To Touch

-0.3575.9786e-

4

6e-4Maximum directional z de-


0.0002.4e82.4e8Maximum normal z stress

Part 1 (Pa)

-0.354-2.3915e8-2.4e8Maximum normal z stress

Part 2 (Pa)



Add Offset. Offset = 0 m

-0.3560.99644e-

4

1e-4Maximum directional z de-



Part 1 (Pa)

-0.355-3.9858e7-4e7Maximum normal z stress

Part 2 (Pa)



Add Offset. Offset =

0.001 m

-0.3551.0961e-

3

1.1e-

3

Maximum directional z de-



Part 1 (Pa)

-0.357-4.3843e8-4.4e8Maximum normal z stress

Part 2 (Pa)



Add Offset. Offset = -

0.001 m

0.00000Maximum directional z de-



Part 1 (Pa)

000Maximum normal z stress

Part 2 (Pa)


VMMECH022

VMMECH023: Curved Beam Assembly with Multiple Loads

Overview



Type(s):

BeamElement

Type(s):

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.


Equivalent Loading:


Force F = -1000 N

(y direction)

For each beam:Beam 1:

Cross-section =

10 mm x 10

mm

E1 = 1.1e5 MPa

Moment M = -

10000 Nmm

(about z-axis)

ν1 = 0

ρ1 = 8.3e-6

kg/mm3

Radius r = 105

mm




UDL w = -5 N/mm

(y direction) on

both beams

Included angle

= 45°

Beam 2:

E2 = 2e5 MPa

ν2 = 0This UDL is applied

as an edge forceρ2 = 7.85e-6

kg/mm3

on each beam

with magnitude =

-5 (2 x 3.14 x 105)

/ 8 = -412.334 N

Analysis

The deflection in the y direction is in the direction of the applied force F and is given by:

δ

ω

= −

+ +

+

1

3 2 4

2

3 2 4+ +

ω

where:

δ = deflection at free end in the y direction

I = moment of inertia of the cross-section of both beams

Results Comparison


0.619-8.4688-8.416664Minimum Y Deformation

(mm)


VMMECH023

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

Beams

Overview



Type(s):

BeamElement

Type(s):

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.


Material Properties

ρ

(kg/m3)

νE

(Pa)

Mater-

ial

1e-80.341.1e11Spring

7.85e502e11Mass


Harmonic force F

= 1 e6 N (z-direc-

tion)

Cross-section of

each beam:

Outer radius =

10 mm




Inner radius =

5 mm

Length of

longer beam =

100 mm

Length of

shorter beam =

5 mm

Results Comparison

Er-

ror

(%)

Mechan-

ical

TargetResults

-0.8594.078e-34.11332e-

3

Maximum z directional deforma-

tion without damping (m)

Mode Superposi-

tion

-0.8764.0765e-

3

4.11252e-

3


tion with damping (m)

-0.0034.1132e-

3

4.11332e-

3


tion without damping (m)

Full Method

-1.0464.0695e-

3

4.11252e-

3


tion with damping (m)


VMMECH024

VMMECH025: Stresses Due to Shrink Fit Between Two Cylinders

Overview

Stephen P. Timoshenko, Strength of Materials, Part 2 - Advanced

Theory and Problems, 3rd Edition, pg. 208-214

Reference:


Type(s):

SolidElement

Type(s):

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 coordin-

ate system.

To simulate interference, set Contact Type to Rough with interface treatment set to add offset

with Offset = 0.



P = 30000 psiInner Cylinder:Both cylinders

are made of ri = 4”

the same mater-

ialro = 6.005”

Ri = 6”E = 3e7psi




Ro = 8”ν = 0

ρ = 0.28383

lbm/in3

Length of both

cylinders = 5”

Results Comparison


1.03573835396.67Maximum normal y stress, inner

cylinder (psi)

0.04227942281.09Maximum normal y stress, outer

cylinder (psi)

Note

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


VMMECH025

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

Overview

Any standard Machine Design and Strength of Materials bookReference:

Fatigue AnalysisAnalysis

Type(s):

ShellElement

Type(s):

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.


Material Properties

E = 2e11 Pa

ν = 0.0

Ultimate tensile strength =

1.29e9 Pa

Endurance strength = 1.38e8 Pa

Yield Strenth = 2.5e8 Pa

Alternating Stresses

(Pa)

No. of Cycles

1.08e91000

1.38e81e6


Moment M = 0.15

Nm (counterclock-

wise @ z-axis)

Length L = 12e-

3 m

Width W = 1e-3

m




Thickness T = 1

e-3 m

Results Comparison

Er-

ror

(%)

Mechan-

ical

TargetResults

0.0009e89e8Maximum normal x-stress (Pa)

0.2796.4981e-

4

6.48e-4Maximum total deformation (m)

0.0200.153330.1533Safety factorSN-Goodman

0.0051844.41844.3Life

0.0200.153330.1533Safety factorSN-Soderberg

0.0051844.41844.3Life

0.0200.153330.1533Safety FactorSN-Gerber

0.0051844.41844.3Life


VMMECH026

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

Overview

Any standard Thermal Analysis bookReference:

Thermal Stress AnalysisAnalysis

Type(s):

ShellElement

Type(s):

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.



Heat flow Q = 5 WLength L = 0.2

m

E = 2e11 Pa

Given Temperature

= 20°C

ν = 0.0

Width W = 0.05

m

Coefficient of

thermal expan-

sion α = 1.2e-

5/°C

Thickness T =

0.005 m

Thermal con-

ductivity k =

60.5 W/m°C

Analysis

Heat Reaction = -(Total heat generated)

Heat flow due to conduction is given by:

h l=−



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 hh= −− ×

1

Thermal deformation in the z-direction is given by:

δ α

l

= −∫

0

Results Comparison

Er-

ror

(%)

Mechan-

ical

TargetResults

0.00086.11686.1157Maximum Temperature (°C)

0.0002e42e4Maximum Total Heat Flux (W/m2)

0.7817.9958e-

5

7.93386e-

5

Maximum Total Deformation (m)

0.000-5-5Heat Reaction (W)


VMMECH027

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

Overview

Any standard Strength of Materials bookReference:

Static Structural AnalysisAnalysis

Type(s):

SolidElement

Type(s):

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.



Pretension as pre-

load = 19.635 N

Length L = 1 mE = 2e11 Pa

Diameter D =

0.05 m

ν = 0.0

(equal to adjust-

ment 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



Results Comparison

Er-

ror

(%)

Mechan-

ical

TargetResults

0.004-5.0002E-

08

-5.00E-

08

Minimum z-directional deformation

(m)

-0.9964.9502E-

08

5.00E-08Maximum z-directional deformation

(m)

0.0001.00E-071.00E-07Adjustment Reaction (m)


VMMECH028

VMMECH029: Elasto-Plastic Analysis of a Rectangular Beam

Overview

Timoshenko S., Strength of Materials, Part II, Advanced Theory

and Problems, Third Edition, Article 64, pp. 349

Reference:

Static Plastic AnalysisAnalysis

Type(s):

SolidElement

Type(s):

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



M = 1.0 Myp to 1.5

Myp

Length L = 10”E = 3e7 psi

Width b = 1”ν = 0.0

Height h = 2”σyp = 36000 psi (Myp = 24000 lbf -

in)



Analysis

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

Results Comparison

Error

(%)

MechanicalTargetM/Myp

Equivalent

Stress (psi)

StateEquivalent

Stress (psi)

State

0.16436059fully

elastic

36000fully

elastic

1

0.80036288elastic-

plastic

36000elastic-

plastic

1.25

-solution not

converged

plasticsolution not

converged

plastic1.5


VMMECH029

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

Overview


Plane Strain AnalysisAnalysis

Type(s):

2D Structural SolidElement

Type(s):

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.



Moment M = 5000

lbf-in

Length L =

1000”

E = 2.9e7 psi

ν = 0.0

Width W = 40”

Thickness T =

1”

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.



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

Results Comparison


0.0003030Normal Stress

0.0003030Maximum Normal Stress in

the X-Direction (psi)

0.0180.16553e-20.1655e-2Maximum Total Deformation

(in)

0.000-5-5Reaction Moment (lbf-in)


VMMECH030

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

Overview


Plane Stress AnalysisAnalysis Type(s):

2D Structural SolidElement Type(s):

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.


Material Properties

Tensile Yield (Pa)νE (Pa)Material

2.07e801.93e11Part 1

2.8e807.1e10Part 2

2.5e802e11Part 3

2.8e801.1e11Part 4


Force = 1e9 N in

the negative x-direc-

tion

Part 1: 2 m x 2 m x

3 m

Part 2: 2 m x 2 m x

10 m

Part 3: 2 m x 2 m x

5 m

Part 4: 2 m x 2 m x

2 m



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


0.0002.5e82.5e8Maximum Equivalent Stress (Pa)

0.0000.8280.828Safety FactorPart 1

0.000-0.172-0.172Safety Margin

0.0581.20771.207Safety Ratio

0.0001.121.12Safety FactorPart 4

0.0000.120.12Safety Margin

0.0960.892860.892Safety Ratio


VMMECH031

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

Axisymmetric Model

Overview

Any basic Heat Transfer bookReference:

Axisymmetric AnalysisAnalysis

Type(s):

2D Structural SolidElement

Type(s):

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 ac-

curate results, set the advanced mesh control element size to 0.002 m.

Find the disk temperature and heat flux at inner and outer radii.



Q = 100 W (Internal

Heat Generation =

39788735.77 W/m3)

Ri = 10 mmE = 1.1e11 Pa

ν = 0.34 Ro = 60 mmThermal conductiv-

ity k = 401.0 W/m-

°C

t = 8 mmFilm coefficient h =

1105 W/m2-°C

Surrounding tem-

perature 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.



Figure 42: Axisymmetric Model

Results Comparison


-0.01038.89638.9Maximum Temperature (°C)

0.02330.00730Minimum Temperature (°C)

-0.5541978401.98943e5Maximum Heat Flux (W/m2)

-0.0183315133157Minimum Heat Flux (W/m2)


VMMECH032

VMMECH033: Electromagnetic Analysis of a C-Shaped Magnet

Overview

J. A. Edminster, Theory and Problems of Electromagnetics, Tata

McGraw Hill, 2nd Edition, Example 11.9, pg. 181

Reference:

Electromagnetic AnalysisAnalysis

Type(s):

SolidElement

Type(s):

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 44: Coil Details in cm

Figure 45: Current and Voltage

Figure 46: Flux Parallel Applied on 9 Outer Faces

Material Properties

Electric Res-

istivity (ohm-

m)

Relative Per-

meability

Density

(kg/m3)

Poisson's Ra-

tio

Young's Modu-

lus (Pa)

01001e7Air Body

2e-7183000.341.1e11Coil

050078500.32e11Core


Voltage = 0 VGiven in Figure 44: Coil

Details in cm (p. 82) Current = 0.1 A


VMMECH033


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


0.1280.0406624.061e-2Total Flux Density (T)

0.1143235732320.0585Total Field Intensity (A/m)



VMMECH033

VMMECH034: Rubber cylinder pressed between two plates

Overview

T. Tussman, K.J. Bathe, "A Finite Element Formulation for Non-

linear Incompressible Elastic and Inelastic Analysis", Computers

and Structures, Vol. 26 Nos 1/2, 1987, pp. 357-409

Reference:

Nonlinear Static Structural Analysis (Large Deformation ON)Analysis

Type(s):

SolidElement

Type(s):

Test Case

A rubber cylinder is pressed between two rigid plates using a maximum imposed displacement of δmax.

Determine the total deformation.



Displacement in Y direction

= -0.1m

Solid1:Solid1:

0.05 m x 0.01m x 0.4 mE = 2e11Pa

ν = 0.3

ρ = 7850 kg/m3

Solid2: Quarter Circular CylinderSolid2: Mooney-Rivlin Constants

Radius = 0.2 mC10 = 2.93e5 Pa

Length = 0.05mC01 = 1.77e5 Pa

Incompressibility Para-

meter D1 1/Pa =0

Analysis

Due to geometric and loading symmetry, the analysis can be performed using one quarter of the cross

section.

• Frictionless supports are applied on 3 faces (X = 0, Z = 0 and Z = 0.05 m).



• Given displacement of 0.1m 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


00.165260.165285Total Deformation (m)


VMMECH034

VMMECH035: Thermal Stress in a Bar with Radiation

Overview

Any Basic Heat transfer and Strength of Materials bookReference:

Coupled Analysis (Static Thermal and Static Stress)Analysis

Type(s):

SolidElement

Type(s):

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.



Heat Flow = 2500W on Part

4

Part 1: 2 m x 2 m x

2 m

E = 2.0e11 Pa

v = 0

Heat Flux = 625 W/m2 on

Part 4

Part 2: 2 m x 2 m x

5 mα = 1.2 x 10

-5 1/°C

k = 60.5 W/m°CPart 3: 2 m x 2 m x

10 mRadiation = 20°C, 0.3

Part 4: 2 m x 2 m x

3 m

Analysis

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



(Heat flowing through body) = (Heat Conducted through body) = (Heat Radiated out of the Surface)

i.e. Q = Qr =QC = 5000 W.

Heat Radiated out of the body Q A T T Wr = ∗ ∗ ∗ −ε σ α( ) ;2

4 4

gives T2 = 260.16°C.

Heat Conducted through the body

K

bc =

∗ ∗ −

1

gives T1 = 673.38°C.

Thermal strain is given by,

ε α m m_ax . . . /= = × × − = ×∆ − − 0 673 38 7 8 656 0

ε α in = = × × − = ×∆ − − 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,

B C CD = −−

= −

FGH HI

FGH HI JFL MF

JLFGH HI JL FFMN

N NN N

Only half-length is considered for calculating deformation, since deformation is symmetric

δ ε ε= +∫ O RS

U

VXYZ[\] ^VZu`VuZ\]

d

fj

k

o

δ = × × − − +− ×

×

−∫ p q w q q yq

q|

~ ~ ~~

~

∫

|

δ = × −

Results Comparison


0673.49673.38Temperature on Part 4(°C)

0260.15260.16Temperature on Part 1 (°C)

07.8179e-37.81656e-3Maximum Thermal Strain

(m/m)

00.00285782.85792e-3Minimum Thermal Strain

(m/m)

-4.6-1.0183e9-1.067448e9Normal Stress in Z direction

(Pa)

1.4-0.012572-0.0123966Directional Deformation in

Z direction (m)


VMMECH035

VMMECH036: Thermal Stress Analysis of a Rotating Bar using Temperature

Dependant Density

Overview

Any Basic Strength of Materials bookReference:

Static Stress Analysis (Sequence Loading)Analysis

Type(s):

SolidElement

Type(s):

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.


LoadingGeometric

Properties

Material Properties

Rotational Velocity (rad/s) in

steps:

Part 1:

2 m x 2

m x 20

m

E = 1 x 106 Pa

α = 1 x 10-5

1/°C

ν = 0 1. (1, 0, 0)

Density kg/m3

Temperature °C2. (0.5, 0, 0)

30503. (0.25, 0, 0)60100

90150Thermal Condition °C

1. 50°C

2. 100°C

3. 150°C



Analysis

Rotational Stress = =∫ ρω ρω2

0

2 2

Thm = × × ∆α

Df = 3ρω

= × × ∆α

!"#$ Equ%v#$&'" (")&** +!"#"%!'#$ (")&** ,&)-#$ (= = +σ./.14 "")&**

56789 :;<6=>87?6@ A6787?6@89 :;<6=>87?6@ 5B;=>89 :;= = +δCFCGH <<6=>87?6@

Results Comparison


0.0406502.66500Step 1Equivalent Stress (Pa)

0.0324001.34000Step 2

0.0192625.52625Step 3

00.090.09Step 1Total Deformation (m)

00.060.06Step 2

00.0450.045Step 3


VMMECH036

VMMECH037: Cooling of a Spherical Body

Overview

F. Kreith, "Principles of Heat Transfer", 2nd Printing, International

Textbook Co., Scranton, PA, 1959, pg. 143, ex. 4-5.

Reference:

Transient Thermal AnalysisAnalysis

Type(s):

PlaneElement

Type(s):

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



Convection applied

on Edge = 5.5556e-

4 BTU/s-ft2-°F

Quarter Circular lamina

Radius = 0.16667 ft

K = (1/3) BTU/hr-ft-°F

ρ = 62 lb/ft3

c = 1.075 Btu/lb-°F

Ambient Temperat-

ure for Convection

= 25°F



Analysis

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

Results Comparison


2.45728.68828Temperature at the Centre

of body after 21600s (°F)


VMMECH037

VMMECH038: Crashing Blocks Simulation with Transient Structural Analysis

Overview

Any basic Kinematics book.Reference:

Flexible Dynamic AnalysisAnalysis

Type(s):

SolidElement

Type(s):

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.



Left Block Initial

Velocity = 100

mm/s (X direction)

Left Block = 3mm x

2mm x 5mm

E = 2e5 MPa

ν = 0.3

Middle Block = 2.5mm

x 2mm x 3mmρ = 7.85e-6kg/mm

3

Right Block =3mm x

6mm x 4mm




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


1.533.80933.3Velocity of Left Block after

impact (mm/sec)

-0.8132.38133.4Velocity of Middle Block

after impact (mm/sec)


VMMECH038

VMMECH039: Transient Response of a Spring-mass System

Overview

R. K. Vierck, Vibration Analysis, 2nd Edition, Harper & Row Pub-

lishers, New York, NY, 1979, sec. 5-8.

Reference:


Type(s):

Solid and SpringElement

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.



F0 = 50 N2 Blocks = 2m x 2m x

2m

E = 2e11 Pa

γ = 0.3 td = 1.8 secLength of L1 spring =

6mρ = 0.25 kg/m

3

k1 = 6 N/mLength of L2 spring =

7mk2 = 16 N/m

m1 = 2 kg

m2 = 2kg



Results Comparison


-114.33514.48Y1, m (@ t = 1.3s)

-1.93.91513.99Y2, m (@ t = 1.3s)

118.51118.32Y1, m (@ t = 2.4s)

0.96.19716.14Y2, m (@ t = 2.4s)


VMMECH039

VMMECH040: Deflection of Beam using Symmetry and Anti-Symmetry

Overview

Any Basic Strength of Materials BookReference:


Type(s):

BeamElement

Type(s):

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



Force = -1000 N (Y-

direction) at 8m

Bar = 1m x 1m x 24mE = 2e11 Pa

γ = 0

from Simply Suppor-

ted endρ = 0.001 kg/m

3

Analysis

Scenario 1: Considering Symmetry

δ =× ××

3

Scenario 2: Considering Anti-Symmetry



δ =× ×

×−

× ××

3 3

Results Comparison


0.019-2.5695e-5-2.569e-5Scenario 1: Directional De-

formation in Y-direction (m)

1.856-1.7383e-6-1.70662e-6Scenario 2: Directional De-

formation in Y-direction (m)


VMMECH040

VMMECH041: Brooks Coil with Winding for Periodic Symmetry

Overview

W. Boast, Principles of Electric and Magnetic Fields, 1948 Harpers

Brothers, Page 242, Equation 12.05.

Reference:

Electromagnetic AnalysisAnalysis

Type(s):

SolidElement

Type(s):

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



Figure 55: Schematic Diagram

Material Properties

Electric Res-

istivity

(ohm-m)

Relative Per-

meability

Density

(kg/m3)

Poisson's Ra-

tio

Young's

Modulus

(Pa)

01001e7DSVM41_MAT1

(Emag Part)

2e-7183000.341.1e11DSVM41_MAT2

(Winding Body)

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)

=× ×

+ ×=

× × × ×

× + × × ×

−

− −

7

3 3

π

-= ×


VMMECH041

Results Comparison


-0.30.00198481.99e-3Total Flux Density (T)



VMMECH041

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

Submerged in a Fluid

Overview

Any Basic Strength of Materials BookReference:


Type(s):

SolidElement

Type(s):

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



Hydrostatic Pres-

sure Acceleration =

Long bar = 20m x 2m

x 2m

E = 2e11 Pa

γ = 0

-10 m/s2 (Z direc-

tion)

ρ = 7850 kg/m3

Surface Location:




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

Maximum bending moment = =

− 2

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


-1.08885293008.4375e6Normal Stress (Partially Sub-

merged) (Pa)

0.6893.5241e73.50e7Normal Stress (Fully Sub-

merged) (Pa)


VMMECH042

VMMECH043: Fundamental Frequency of a Simply-Supported Beam

Overview

W. T. Thompson, Vibration Theory and Applications, 2nd Printing,

Prentice-Hall, Inc., Englewood Cliffs, NJ, 1965, pg. 18, ex. 1.5-1

Reference:

Modal AnalysisAnalysis

Type(s):

BeamElement

Type(s):

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.



ℓ = 80 inE = 3e7 psi

ρ=0.2836 lb/in3

A = 4 in2

h = 2 in

I = 1.3333 in4

Results Comparison


0.53228.61328.766Frequency (Hz)



VMMECH044: Thermally Loaded Support Structure

Overview

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.

Reference:

Linear Thermal Stress AnalysisAnalysis

Type(s):

BeamElement

Type(s):

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 tem-

perature 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.



VMSIM044_material_rigid: Q = 4000 lb (Y dir-

ection)A = 0.1 in

2

Er = 3e16 psi ∆T = 10 °F

νr = 0

VMSIM044_material_copper:

Ec = 1.6e7 psi

νc = 0

αc = 9.2e-6 / °F

VMSIM044_material_steel:




Es = 3e7 psi

νs = 0

αs = 7e-6 / °F

Results Comparison


0.001969519695Stress in steel (psi)

0.001015210152Stress in copper (psi)


VMMECH044

VMMECH045: Laterally Loaded Tapered Support Structure

Overview

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.

Reference:


Type(s):

ShellElement

Type(s):

Test Case

A cantilever beam of thickness t and length ℓ has a depth which tapers uniformly from d at the tip to

3d at the wall. It is loaded by a force F at the tip, as shown. Find the maximum bending stress at the

mid-length (X = ℓ ).



F = 4000 lb (Y direc-

tion)ℓ = 50 inEs = 3e7 psi

νs = 0 d = 3 in

t = 2 in

Results Comparison


0.58373.78333Bending stress at mid length (psi)



VMMECH046: Pinched Cylinder

Overview

R. D. Cook, Concepts and Applications of Finite Element

Analysis, 2nd Edition, John Wiley and Sons, Inc., New York,

NY, 1981, pp. 284-287

Reference:

H. Takemoto, R. D. Cook, "Some Modifications of an Isopa-

rametric Shell Element", International Journal for Numerical

Methods in Engineering, Vol. 7 No. 3, 1973.


Type(s):

ShellElement

Type(s):

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.



F = 100 lbf (Y direc-

tion)ℓ = 10.35 inEs = 10.5e6 psi

νs = 0.3125 r = 4.953 in

t = 0.094 in



Analysis

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

load is applied.

Results Comparison


-0.1–0.11376-0.1139Deflection (in)


VMMECH046

VMMECH047: Plastic Compression of a Pipe Assembly

Overview



ex. 5.1.

Reference:

Plastic Structural AnalysisAnalysis

Type(s):

AxisymmetricElement

Type(s):

Test Case

Two coaxial tubes, the inner one of 1020 CR steel and cross-sectional area As, and the outer one of

2024-T4 aluminum alloy and of area Aa, are compressed between heavy, flat end plates, as shown below.

Determine the load-deflection curve of the assembly as it is compressed into the plastic region by an

axial displacement. Assume that the end plates are so stiff that both tubes are shortened by exactly

the same amount.



VMSIM047_CR_steel: 1st Load step: δ = -

0.032 inℓ = 10 in

Steel:Es = 26,875,000 psi 2nd Load step: δ =

-0.05 inσ(yp)s = 86,000 psi Inside radius =

1.9781692 in 3rd Load step: δ =

-0.10 inVMSIM047_T4_aluminum alloy:




Ea = 11,000,000 psi Wall thickness = 0.5

inσ(yp)a = 55,000 psiAluminum:ν = 0.3

Inside radius =

3.5697185 in

Wall thickness = 0.5

in

Analysis



Results Comparison


0.910339001.0244e6Load, lb for Deflection @ 0.032 in




VMMECH047

VMMECH048: Bending of a Tee-Shaped Beam

Overview



ex. 7.2.

Reference:


Type(s):

BeamElement

Type(s):

Test Case

Find the maximum tensile and compressive bending stresses in an unsymmetrical T beam subjected

to uniform bending Mz, with dimensions and geometric properties as shown below.



Mz = 100,000 lbf-in

(Z direction)

b = 1.5 inE = 3e7 psi

h = 8 in

y = 6 in

Area = 60 in2

Iz = 2000 in4

Results Comparison


0300300StressBEND, Bottom (psi)

0-700-700StressBEND, Top (psi)



VMMECH049: Combined Bending and Torsion of Beam

Overview



NY, 1955, pg. 299, problem 2.

Reference:


Type(s):

BeamElement

Type(s):

Test Case

A vertical bar of length ℓ and radius r is subjected to the action of a horizontal force F acting at a dis-

tance d from the axis of the bar. Determine the maximum principal stress σmax.

Figure 63: Problem Sketch




tion)ℓ = 25 ftE = 3e7 psi

ν = 0.3 r = 2.33508 inM = 9000 lbf-in (Z

direction)d = 3 ft



Results Comparison


-0.1537515.57527Principal stressmax (psi)


VMMECH049

VMMECH050: Cylindrical Shell under Pressure

Overview


and Problems, 3rd Edition, D. Van Nostrand Co., Inc., New

York, NY, 1955, pg. 45, article 11.

Reference:

A. C. Ugural, S. K. Fenster, Advanced Strength and Applied

Elasticity, Elsevier, 1981.


Type(s):

Axisymmetric Shell elementElement

Type(s):

Test Case

A long cylindrical pressure vessel of mean diameter d and wall thickness t has closed ends and is sub-

jected 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.





P = 500 psi (radial

direction)

t = 1 inE = 3e7 psi

d = 120 inν = 0.3

Analysis

An axial force of 5654866.8 lb ((Pπd2)/4) is applied to simulate the closed-end effect.

Results Comparison


01500015000Stressy (psi)

0.0073000230000Stressz (psi)


VMMECH050

VMMECH051: Bending of a Circular Plate Using Axisymmetric Elements

Overview

S. Timoshenko, Strength of Material, Part II, Elementary Theory


NY, 1956, pp. 96, 97, and 103.

Reference:


Type(s):

Axisymmetric Shell elementElement

Type(s):

Test Case

A flat circular plate of radius r and thickness t is subject to various edge constraints and surface loadings.

Determine the deflection δ at the middle and the maximum stress σmax for each case.

Case 1: Uniform loading P, clamped edge

Case 2: Concentrated center loading F, clamped edge


Case 1:

Case 2:




Case 1:r = 40 inE = 3e7 psi

t = 1 inν = 0.3P = 6 psi

Case 2:

F = -7539.82 lb (y

direction)

Analysis



Results Comparison


Case 1: -0.282-0.087114-0.08736Deflection (in)

0.1787212.87200Stressmax (psi)

Case 2: 0.761-0.088025-0.08736Deflection (in)

0.2193607.93600Stressmax (psi)


VMMECH051

VMMECH052: Velocity of Pistons for Trunnion Mechanism

Overview

Any Basic Kinematics bookReference:

Rigid Dynamic AnalysisAnalysis

Type(s):

Multipoint Constraint ElementElement

Type(s):

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





Constant angular

velocity at crank =

12.57 rad/s

AB = 400 mmE = 2e11 Pa

AC = 150 mmν = 0.3

CE = 350 mm

EF = 300 mm

Analysis

Analysis done using graphical solution.

Consider the Space Diagram, Velocity Diagram at the 180° from Initial Position.


Results Comparison


-0.949497.04501.8Velocity of Piston (F) m/s

0.494959.72955Velocity of Piston (B) m/s


VMMECH052

VMMECH053: Simple Pendulum with SHM motion

Overview



Type(s):


Type(s):

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.



Rotation θ =

1.571*sin (0.5235*t)

rad

Hinge point = (0, 0, -

35.56) mm

E = 2000000 MPa

ν = 0.3

Analysis

The pendulum is having SHM motion in X-Z plane about the hinge.

Angular acceleration of pendulum:



αω

α

=

= − 2

Results Comparison


-0.568-0.43054-0.433Relative angular acceleration

of pendulum after t = 3s

(rad/s2)


VMMECH053

VMMECH054: Spinning Single Pendulum

Overview



Type(s):


Type(s):

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 ml Ml J J ml Ml mgl Mgz x y+ +( ) − − + +( ) + +14

2 2 14

2 2 2ɺɺθ θ θ θΩ ll 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.





L= 2.2361 m Ω = 17.1522 = tan-1

(1,2)

m = 551.45 kg = 0

M = 100.0 kg

Jx = 229.97 kg-m2

Jy = 2.7293 kg-m2

Jz = 229.97 kg-m2

Results Comparison


0.0-1.3233-1.3233 at 0.5 sec

0.0116.1368116.1368 at 0.5 sec


VMMECH054


0.0-2.6755-2.6755 at 1.0 sec

0.0119.8471119.8471 at 1.0 sec

Figure 71: Plot of from 0 to 1 sec

Figure 72: Plot of from 0 to 1 sec



VMMECH054

VMMECH055: Projector mechanism- finding the acceleration of a point

Overview



Type(s):


Type(s):

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.





Constant rotational

velocity = -58.643

rad/s

r1 = 18 mmE = 2e11 Pa

ν = 0.3 r2 = 48 mm

x = 45 mm

y = 28 mm

Analysis

Linear acceleration of point C is given by

αθ θ

θc =−+

2 1 2

2

Results Comparison


-0.141-12.043-12.06Relative acceleration (θ1 =

10) mm/s2

-0.0151.31681.317Relative acceleration (θ1 =

30) mm/s2

-0.0066.73866.739Relative acceleration (θ1 =

60) mm/s2


VMMECH055

VMMECH056: Coriolis component of acceleration-Rotary engine problem

Overview



Type(s):


Type(s):

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.



Constant rotational

velocity = 300 rpm

Connecting rod AB is

125mm Crank OA is

50mm long

E = 2e11 Pa

ν = 0.3

OB is inclined at 50° to

the vertical.



Analysis

Angular acceleration of the connecting rod is given by:

αα

AB

t

=3

Results Comparison


0294.53294.52Angular acceleration (radi-

an/s2)


VMMECH056

VMMECH057: Calculation of velocity of slider and force by collar

Overview

Beer-Johnston ‘Vector Mechanics for Engineers’ Statics & Dy-

namics (In SI Units), 7th Edition, TATA McGRAW HILL Edition

2004, Problem 13.73, Page No: 793

Reference:


Type(s):


Type(s):

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".





Spring: Gravitational accel-

eration = -9.8066

m/s2 (Y Direction)

E = 2e11 Pa

ν = 0.3Undeformed length =

105 mm

Stiffness K = 300 N/m

Results Comparison


0.75314.99214.88At point A (N)

0.3-23.667-23.6At point B (N)


VMMECH057

VMMECH058: Reverse four bar linkage mechanism

Overview

Results are simulated using MATLABReference:


Type(s):


Type(s):

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.





Gravitational accel-

eration = -9.8066

m/s2 (Y Direction)

Link PS length r1 =

0.5m

E = 2e11 Pa

ν = 0.3

Link PQ length r2 =

0.15m

Link QR length r3 =

0.4m

Link RS length r4 =

0.45m

Analysis

Results are obtained using MATLAB.

Results Comparison


-0.67139.33639.6Angular Acceleration (rad/s2)

-0.7-5.1247-5.16Angular Velocity (rad/sec)

0.7-0.36255-0.36Rotation (rad)


VMMECH058

VMMECH059: Bending of a solid beam (Plane elements)

Overview

R. J. Roark, Formulas for Stress and Strain, 4th Edition, McGraw-

Hill Book Co., Inc., New York, NY, 1965, pp. 104, 106.

Reference:


Type(s):

2-D Plane Stress Shell elementElement

Type(s):

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.


Case 1:

Case 2:




Case 1:ℓ = 10 inE = 30 x 106 psi

ν = 0.3 h = 2 inM = 2000 ibf-in (Z

direction)d = 1 in

Case 2:


tion)

Analysis

Since the loading is uniform and in one plane, the above problem can be analyzed as a plane stress

problem.

Results Comparison


Case 1: 00.005000.00500Deflection (in)

0-3000-3000StressBend (psi)

Case 2: 20.00512320.00500Deflection (in)

0-4051.5-4050StressBend (psi)


VMMECH059

VMMECH060: Crank Slot joint simulation with flexible dynamic analysis

Overview

Mechanical APDL Multibody AnalysisReference:


Type(s):

Solid and Multipoint Constraint ElementElement

Type(s):

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.



Constant angular

acceleration at base

to Bolt1 = 25 rad/s2

Rod1 length = 75mmE = 2e5 MPa

Rod2 length = 115mmν = 0.3



Analysis

Figure 79: Contour Plot

Figure 80: Equivalent (von Mises) Stress

Figure 81: Total Force at Base to Bolt1

Results Comparison


2.60.408340.398Equivalent (von Mises) Stress

(MPa)


VMMECH060


0.1417.68087.67Force @ Bolt1 (N)



VMMECH060

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

Overview



NY, 1955, pg. 412, eq. 241.

Reference:


Type(s):

BeamElement

Type(s):

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.



F = -50 lb (Z direc-

tion)

r = 100 inE = 30 x 106 psi

d = 2 inν = 0.3θ = 90°

Results Comparison


0.264-2.655-2.648Deflection (in)




0.5226399.26366.0StressBend (psi)


VMMECH061

VMMECH062: Stresses in a long cylinder

Overview



NY, 1956, pg. 213, problem 1 and pg. 213, article 42.

Reference:


Type(s):

Axisymmetric ShellElement

Type(s):

Test Case

A long thick-walled cylinder is initially subjected to an internal pressure p. Determine the radial displace-

ment 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.


Case 1:

Case 2:




Case 1:a = 4 inE = 30 x 106 psi

b = 8 inν = 0.3Pressure = 30,000

psi (radial direction)Xi = 5.43 in

ρ = 0.281826 lbm/in3

Case 2:

Rotational velocity

= 1000 rad/s (Y dir-

ection)

Analysis



Results Comparison


Case 1: -3.050.00762670.0078666Displacementr, in (r

= 4 in) -0.04-29988-30000.

-0.035-7775.3-7778.Stressr, psi (r = 4 in)--0.796110Stressr, psi (r = 6 in)

-0.0244998850000.Stresst, psi (r = 8 in)

-0.0112777527778.Stresst, psi (r = 4 in) -0.0051999920000.Stresst, psi (r = 6 in)

Stresst, psi (r = 8 in)

Case 2: --6.54830Stressr, psi (r = 4 in)

2.6714167240588.Stresst, psi (r = 4 in)3.8024933.74753.Stressr, psi (r = 5.43

in)0.9612971929436.

Stresst, psi (r = 5.43

in)


VMMECH062

VMMECH063: Large deflection of a cantilever

Overview

K. J. Bathe, E. N. Dvorkin, "A Formulation of General Shell Ele-

ments - The Use of Mixed Interpolation of Tensorial Compon-

Reference:

ents”, International Journal for Numerical Methods in Engineering,

Vol. 22 No. 3, 1986, pg. 720.


Type(s):

ShellElement

Type(s):

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.



M = 15.708 N-mm

(Y direction)ℓ = 12 mmE = 1800 N/mm

2

ν = 0.0 b = 1 mm

t = 1 mm

Analysis

Large deformation is used to simulate the problem.



Results Comparison


1.221-2.9354-2.9Directional Deformation X-

direction (mm)

1.662-6.608-6.5Directional Deformation Z-

direction (mm)

0.01794.26694.25Normal Stress X-direction

(N/mm2)


VMMECH063

VMMECH064: Small deflection of a Belleville Spring

Overview



NY, 1956, pg. 143, problem 2.

Reference:


Type(s):

ShellElement

Type(s):

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.



Line pressure = -

100 lb/in (Y direc-

tion)

a = 1 inE = 30 x 106 psi

b = 1.5 inν = 0.0t = 0.1 in

β = 7° = 0.12217 rad

Results Comparison


3.8-0.0029273-0.0028205Directional Deformation Y-

direction (in)



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

Overview

C. O. Harris, Introduction to Stress Analysis, The Macmillan Co.,

New York, NY, 1959, pg. 58, problem 8.

Reference:

Static Thermal Stress AnalysisAnalysis

Type(s):

Solid and ShellElement

Type(s):

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.



∆t = 170°F - 70°Fℓ = 3 in.E = 10.5 x 106 psi

α = 1.25 x 10-5

/°F δ = 0.002 in.




ν = 0.0

Results Comparison


0-6122.4-6125Normal Stress Y (psi)

01.25e-0031.25e-003Thermal Strain Y (in/in)


VMMECH065

VMMECH066: Bending of a Tapered Plate

Overview



Reference:


Type(s):

ShellElement

Type(s):

Test Case

A tapered cantilever plate of rectangular cross-section is subjected to a load F at its tip. Find the max-

imum deflection δ and the maximum principal stress σ1 in the plate. Use a global mesh size of 0.75 in.



F = 10 lbfL = 20 inE = 30 x 106 psi

d = 3 inν = 0.0t = 0.5 in



Results Comparison


0.91614.71600Maximum Principal Stress

(psi)

-0.2-0.042746-0.042667Directional Deformation Z

(in)


VMMECH066

VMMECH067: Elongation of a Solid Tapered Bar

Overview



Reference:


Type(s):

SolidElement

Type(s):

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.





F = 10000 lbfL = 10 inE = 10.4 x 106 psi

d = 2 inν = 0.3

Results Comparison


- 0.2870.00482150.0048077Directional Deformation Y

(in)

- 0.42844634444Normal Stress Y at L/2 (psi)


VMMECH067

VMMECH068: Plastic Loading of a Thick Walled Cylinder

Overview



NY, 1956, pg. 388, article 70.

Reference:

Static, Plastic Analysis (Plane Strain)Analysis

Type(s):

2-D Structural SolidElement

Type(s):

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.





pel = 12,990 psia = 4 inE = 30 x 106 psi

b = 8 in pult = 24,011 psiσyp = 30,000 psi

ν = 0.3

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

Error

(%)

Mechan-

ical

TargetResults

-0.4-9948.8-9984Stressr, psi (X = 4.4 in)Fully Elastic

-0.21860918645Stresst, psi (X = 4.4 in)

0.2-469.1-468Stressr, psi (X = 7.6 in)

09129.19128Stresst, psi (X = 7.6 in)

03000030000Stresseff, psi (X = 4.4 in)Fully Plastic

0.003000030000Stresseff, psi (X = 7.6 in)


VMMECH068

VMMECH069: Barrel Vault Roof Under Self Weight

Overview

R. D. Cook, Concepts and Applications of Finite Element Analysis,

2nd Edition, John Wiley and Sons, Inc., New York, NY, 1981, pp.

284-287.

Reference:

Static AnalysisAnalysis

Type(s):

ShellElement

Type(s):

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.





g = 9.8 m/s2t = 0.25 mE = 4.32 x 10

8 N/m

2

r = 25 mν = 0.3ℓ = 50 mρ = 36.7347 kg/m

3

Θ = 40°

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


2.362-0.30903-0.3019Directional Deformation Y

@ A, m

2.116-0.16267-0.1593Directional Deformation X

@ A, m

3.762223680215570Stressz, Top @ A, Pa

2.738350030340700Stressz, Bottom @ A, Pa

-3.639184270191230Stressangle, Top @ B, Pa

-3.5-210980-218740Stressangle, Bottom @ B, Pa


VMMECH069

VMMECH070: Hyperelastic Thick Cylinder Under Internal Pressure

Overview

J. T. Oden, Finite Elements of Nonlinear Continua, McGraw-Hill

Book Co., Inc., New York, NY, 1972, pp. 325-331.

Reference:

Static, Large Deflection AnalysisAnalysis

Type(s):

2-D Structural Solid ElementsElement

Type(s):

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.





Mooney-Rivlin material coeffi-

cients:

Pi = 150 psiri = 7.0 in

ro = 18.625 in

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


0.0267.18197.18Deformation at inner radius

in radial direction, in

0-122-122Radial Stress at r = 8.16 in,

psi


VMMECH070

VMMECH071: Centerline Temperature of a Heat Generating Wire

Overview

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.

Reference:

Thermal AnalysisAnalysis

Type(s):

2-D Thermal Solid ElementsElement

Type(s):

Test Case

Determine the centerline temperature TcL and the surface temperature Ts of a bare steel wire generating

heat at the rate Q. 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.





h = 1.3889 x 10-3

Btu/s-ft2-°F

ro = 0.03125 ftk = 3.6111 x 10-3

Btu/s-

ft-°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


0.01419.94419.9Centerline Temperature, °F

0.012417.85417.9Surface Temperature, °F

0.00-0.094861-0.094861Heat dissipation rate, BTU/s


VMMECH071

VMMECH072: Thermal Stresses in a Long Cylinder

Overview


and Problems, 3rd Edition, D. Van Nostrand Co, Inc., New York,

NY, 1956, pg. 234, problem 1.

Reference:

Thermal Stress AnalysisAnalysis

Type(s):

2-D Thermal Solid ElementsElement

Type(s):

Test Case

A long thick-walled cylinder is maintained at a temperature Ti on the inner surface and To on the outer

surface. 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.



Ti = -1°Fa = 0.1875 inE = 30 x 106 psi

b = 0.625 in




To = 0°Fα = 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

Error (%)MechanicalTargetThermal Analysis

0-1.0000-1.0000T,°F (at X = 0.1875 in)

0.022-0.67052-0.67037T,°F (at X = 0.2788 in)

000T,°F (at X = 0.625 in)

Error (%)MechanicalTargetStatic Analysis

-1.037416.06420.42Stressa, psi (at X = 0.1875 in)

-3.594405.31420.42Stresst, psi (at X = 0.1875 in)

0.247-195.06-194.58Stressa, psi (at X = 0.625 in)

0.221-195.01-194.58Stresst, psi (at X = 0.625 in)


VMMECH072

VMMECH073: Modal Analysis of a Cyclic Symmetric Annular Plate

Overview

R. D. Blevins, Formulas for Natural Frequency and Mode Shape,

New York, NY, VanNostrand Reinhold Publishing Inc., 1979, PP.

246-247, 286-287.

Reference:

Mode-Frequency AnalysisAnalysis

Type(s):

SolidElement

Type(s):

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.


0.5cm

37 cm

100 cm100 cm


Outside Radius (a) = 50

cmE = 7.1 x 10

5 kg/cm

2

ν = 0.3




Inside Radius (b) = 18.5

cmρ = 2.79 x 10

-9 kg/cm

2

γ = 1.415 x 10-6

kg-

sec2/cm

3Thickness (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:

(1)=

−

Where,

λ2 = 4.80

Results Comparison


-1.30553923.074764872639323.38Frequency (Hz)


VMMECH073

VMMECH074: Tension/Compression Only Springs

Overview

Rao, Singiresu S. Mechanical Vibrations. 4th ed. Singapore:

Prentice Hall, 2004. 20.

Reference:

Rigid Body Dynamic Spring AnalysisAnalysis

Type(s):

SolidElement

Type(s):

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) dis-

placement. 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.


1m

natural

length

0.5 m

0.5 m

Tensile (x ) Compressive (x )1 2


Lo = 1 mk = 1.0e7 N/m

x1 = 0.5 m

x2 = -0.5 m

m = 7850 kg


Hooke’s Law:

Elastic Force = Spring Constant * Displacement

F = k*x

Spring 1: Compression Only spring



Spring 2: Tension Only spring

Results Comparison

Tensile Displacement (x1)


000Elastic Force (N) Spring 1

05.0e65.0e6Elastic Force (N) Spring 2

Compressive Displacement (x2)


0-5.0e6-5.0e6Elastic Force (N) Spring 1

000Elastic Force (N) Spring 2


VMMECH074

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

Overview

W. T. Thomson, Theory of Vibration with Applications, 3rd Edition,

1999, Example 6.4-1, pg. 166

Reference:


Type(s):

SolidElement

Type(s):

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.


Material Properties

ρ (kg/m3)νE (Pa)Material

78500.32e18Block 2

1e-80.354.5e10Shaft 2

157000.32e18Block 1

1e-80.359e10Shaft 1


Force = -1e5 N (y direction)Block 1 and 2:

40 mm x 40 mm x 40 mm

Shaft 1 and 2:

20 mm x 20 mm x 200 mm




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



Vertex A (m)


Vertex B (m)


VMMECH075

VMMECH076: Elongation of a Tapered Shell with Variable Thickness

Overview



Reference:


Type(s):

ShellElement

Type(s):

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.



Tapered plate: F = 10000 lbfE = 10.4 x 106 psi

ν = 0.3L = 10 in

Base width = 2 in

Top width = 1 in

Thickness varying from

2 in to 1 in from base

to top.



Results Comparison


-0.12460.00481370.0048077Directional Deformation Y

(in)

-0.23794454.64444Normal Stress Y at L/2 (psi)


VMMECH076

VMMECH077: Heat Transfer in a Bar with Variable Sheet Thickness

Overview

For basic equation: Frank P. Incropera and David P. DeWitt,

Heat and Mass Transfer, John Wiley & Sons, Inc, 2002, 5th

Edition

pg. 5.

Reference:

Static Thermal AnalysisAnalysis

Type(s):

ShellElement

Type(s):

Test Case

A 10 x 50 mm plate with a thickness varying from 1 mm to 4 mm is maintained at temperatures of 100

°C and 200 °C as shown below. Find the following:

• Temperatures at mid of the surface.

• Heat flow reactions on end edges.



Temperature (T1)

on edge (@ 1mm

thickness) = 100 °C

Plate Dimensions : 10

X 50 mm.

E = 2.0e11 Pa

v = 0

Thickness Variation : 1

mm to 4 mmα = 1.2 x 10

-5 1/°C

Temperature (T2)on

edge (@ 4mm

thickness) = 200 °C

k = 60.5 W/m°C



Analysis

Heat flow due to conduction is given by:

(2)=

The area for conduction varies from A1 to A2. The area Ay at any distance y is given as:

(3)= + −

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

(4)= − −

Temperature at any point y is given as:

(5)= +−

!

" #

$ %# %

& & '(

& ) *+

,-

,.

Results Comparison


0.002.61882.618Heat reaction at T1 (W)

0.00-2.6188-2.618Heat reaction at T2 (W)

0.00166.09166.083Temperature at mid of sur-

face (°C)


VMMECH077

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

Overview

Any Nonlinear Material Verification TextReference:

Static Analysis (ANTYPE=0)Analysis

Type(s):

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.




VMMECH078


P1 = 44006400 PaL = 1 mE = 104728E6 Pa

P2 = 157147000 PaT = 0.02 mν = 0.21

Analysis

A 3-D analysis is performed first using a mesh of 4 x 4 gasket elements. In order to simulate the loading-

unloading 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

Error (%)MechanicalTarget

Gasket Pressure and Closure at End of 1st Loading:

04.4006E+074.4006E+07GK-PRES

04.064E-044.064E-04GK-CLOS

Gasket Pressure and Closure at End of 2nd Loading:

01.5715E+081.5715E+08GK-PRES

06.8327E-046.8327E-04GK-CLOS



VMMECH078

VMMECH079: Natural Frequency of a Motor-Generator

Overview

W. T. Thomson. “Vibration Theory and Applications”. 2nd Printing,

Prentice-Hall, Inc., Englewood Cliffs, NJ. pg. 10, ex. 1.3-3. 1965.

Reference:

Mode-Frequency AnalysisAnalysis

Type(s):

Pipe ElementElement

Type(s):

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.



d = .375 inE = 31.2 x 106 psi

ℓ = 8.00 inm = 1 lb-sec

2/in

J = .031 lb-in-sec2

Results Comparison


048.78148.781Lower Order F, Hz

048.78148.781Higher Order F, Hz



VMMECH080: Transient Response of a Spring-mass System

Overview

R. K. Vierck. “Vibration Analysis”. 2nd Edition. Harper & Row

Publishers, New York, NY, 1979. sec. 5-8.

Reference:

Transient Dynamic Mode Superposition AnalysisAnalysis

Type(s):

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.



ties

Material Proper-

ties

F0 =

50

N

k1 = 6 N/m

k2 = 16 N/m

m1 = 2 Kg td =

1.8

secm2 = 2 Kg



Results Comparison


-0.914.34914.48Y1 , m (@ t = 1.3s)

-1.13.94783.99Y2 , m (@ t = 1.3s)

-1.218.09718.32Y1 , m (@ t = 2.4s)

-0.76.0946.14Y2 , m (@ t = 2.4s)


VMMECH080

VMMECH081: Statically Indeterminate Reaction Force Analysis

Overview

P.Bezler, M. Hartzman, and M. Reich. Dynamic Analysis of Uniform

Support Motion Response Spectrum Method, (NUREG/CR-1677),

Reference:

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.





Acceleration response spectrum

curve defined

by SV and FREQ commands.

Straight Pipe:Pipe Elements:

Outer Dia-

meter =

2.375 in

E = 27.8999 x 106 psi.

Nu = 0.3

Density = 2.587991718e-

10 lb-sec2/in

4 Wall Thick-

ness =

0.154 inMass Elements (lb-sec

2/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


0.008.71218.7121

0.048.80918.8062

0.0117.50917.5103

0.0040.36840.3704

0.0341.64241.6305


VMMECH081

Error (%)MechanicalTargetResults Node

0.000.461860.46186UX at node6

0.000.00257470.0025747UY at node8

0.650.449490.446591UZ at node8



VMMECH081

VMMECH082: Fracture Mechanics Stress for a Crack in a Plate

Overview

W.F.Brown, Jr., J.E.Srawley, Plane strain crack toughness testing

of high strength metallic materials, ASTM STP-410, (1966).

Reference:


Type(s):

SolidElement

Type(s):

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.





ties

Material Proper-

ties

σ0 =

0.5641895

psi

a = 1 in

b = 5 in

h = 5 in

E = 30 x 106

psi

ν = 0.3

t = 0.25 in

Results Comparison


2.51.05041.0249Stress Intensity KI


VMMECH082

VMMECH083: Transient Response to a Step Excitation

Overview

W. T. Thomson, Vibration Theory and Applications, 2nd

Printing, Prentice-Hall, Inc., Englewood Cliffs, NJ, 1965, pg.

102, article 4.3.

Reference:

Mode-Superposition Transient Dynamic AnalysisAnalysis

Type(s):

Element

Type(s):

Test Case

A spring-mass-damping system that is initially at rest is subjected to a step force change F acting on

the mass. Determine the displacement u at time t for damping ratio, ξ = 0.5.




Load-

ing

Material Proper-

ties

F =

200

lb

m = 0.5 lb-

sec2/in

k = 200 lb/in


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


VMMECH083

Results Comparison


0.11.15441.1531Total Def Max (ξ = 0.5) Time

= 0.20 sec

Figure 104: Maximum Deformation vs. Time (damped)



VMMECH083

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

Overview

.W.Ogden, et al., “A Pseudo-elastic Model for the Mullins

Effect in Filled Rubber", Royal Society of London Proceed-

ings Series A., (1989), pg: 2861-2877.

Reference:


Type(s):

SolidElement

Type(s):

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.



ties

Material Properties

One cycle of

loading

R = 0.5m

H = 1m

Neo-Hookean Constants:

µ = 8 MPa

Step 1: λ = 1.5Ogden-Roxburgh Mullins

Constants:




ties

Material Properties

Step 2: λ = 2.0r = 2.104

Step 3: λ = 3.0m = 30.45

β =0.2

Step 4: λ = 2.0

Step 5: λ = 1.5

Step 6: λ = 1.0

Results Comparison

Axial Stress

(Pa)

Results

Error (%)MechanicalTargetStretch λ

0.00812.66712.6661.5

0.028.00028.0002.0

0.069.33369.3333.0

0.01920.82320.8192.0

0.128.67048.6601.5

0.00.00.0001.0

Figure 106: Variation of Axial Stress


VMMECH084

VMMECH085: Bending of a Composite Beam

Overview

R. J. Roark, W. C. Young, Formulas for Stress and Strain,

McGraw-Hill Book Co., Inc., New York, NY, 1975, pg. 112-

114, article 7.2.

Reference:


Type(s):

SolidElement

Type(s):

Test Case

A beam of length ℓ and width w made up of two layers of different materials is subjected to a uniform

rise in temperature from Tref to To, 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.



ties

Material Properties

To =

100°F

ℓ = 8 in

w = 0.5 in

MAT1:

E1 =

1.2 x

106 psi

Tref =

0°F

My =

10.0 in-

lb

t1 = 0.2 in

t2 = 0.1 inα1 =

1.8 x

10-4

in/in/°F

MAT2:




ties

Material Properties

E2 =

0.4 x

106 psi

α2 =

0.6 x

10-4

in/in/°F

Results Comparison


0.0-0.832-0.832Displacement, in

0.017311731StressxTOP , psi

0.022582258StressxBOT , psi


VMMECH085

VMMECH086: Stress Concentration at a Hole in a Plate

Overview

R. J. Roark, Formulas for Stress and Strain, 4th Edition, Mc-

Graw-Hill Book Co., Inc., New York, NY, 1965, pg. 384

Reference:

Static Structural, Submodeling (2D-2D)Analysis Type(s):


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


P = 1000 psiL = 12 inE = 30 x 106 psi

d = 1 inυ = 0.3t = 1 in


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.



Results Comparison

2D-2D Results


0.2553025.73018Equivalent Stress - MaxFine Model

-24.7152272.13018Equivalent Stress - MaxCoarse Model

0.4903032.83018Equivalent Stress - MaxSubmodel

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

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


VMMECH086

Figure 111: 2D-2D Submodel Equivalent Stress



VMMECH086

VMMECH087: Campbell Diagrams and Critical Speeds Using Symmetric Orthotropic

Bearings

Overview

Nelson, H.D., McVaugh, J.M., “The Dynamics of Rotor-

Bearing Systems Using Finite Elements”, Journal of Engin-

eering for Industry, Vol 98, pp. 593-600, 1976

Reference:

Modal AnalysisAnalysis Type(s):

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 indi-

vidual 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

Outer Diamet-

er (cm)

Inner Diameter

(cm)

Axial Distance to

Subelement

Subelement

number

Element Number

0.510.0011

1.021.272

0.765.0812

2.037.622

2.038.8913



Outer Diamet-

er (cm)

Inner Diameter

(cm)

Axial Distance to

Subelement

Subelement

number

Element Number

3.3010.162

3.301.5210.673

2.541.7811.434

2.5412.705

1.2713.466

1.2716.5114

1.5219.052

1.5222.8615

1.2726.672

1.2728.7016

3.8130.482

2.0331.503

2.031.5234.544


Rotational VelocityRefer to Table 1: Geomet-

ric Data of Rotor-Bearing

Elements (p. 205)

Shaft

Spin (1) =

1000 RPME11 = 2.078 x 10

11 Pa

G12 = 1.0 x 1014

Pa Spin (2) =

20000 RPMDensity = 7806 kg/m3

Spin (3) =

40000 RPMMass Element

Spin (4) =

60000 RPMMass = 1.401 kg

Polar inertia = 0.002 kg⋅m2

Spin (5) =

80000 RPMDiametral inertia = 0.00136 kg⋅m2

Spin (6) =

100000 RPMBearing Element

Spring coefficients

K11 = K22 = 3.503 x 107 N/m

K12 = K21 = -8.756 x 106 N/m


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.


VMMECH087

Backward and forward whirl speeds for slope = 1 @ 100000 RPM are determined from the modal ana-

lysis.

Results Comparison

RatioMechanicalTarget

Backward and forward whirl speeds for slope = 1 @ 100000 RPM

PIPE288

1.00410793.410747.0000Mode 1 (BW)

0.99519560.019665.0000Mode 2 (FW)

1.01539668.439077.0000Mode 3 (BW)

1.01448207.047549.0000Mode 4 (FW)

Figure 113: Campbell Diagram for Rotor-Bearing System



VMMECH087

VMMECH088: Harmonic Response of a Guitar String

Overview

Blevins, R.D., Formulas for Natural Frequency and Mode

Shape, Nostrand Reinhold Co., New York, NY, 1979, pg. 90,

tab. 7-1

Reference:

Static StructuralAnalysis Type(s):

Linear Perturbed Modal

Linear Perturbed Harmonic

BeamElement Type(s):

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 ex-

citation.


F1 = 84 Nl = 710 mmE = 190 x 109 Pa

c = 165 mm F2 = 1 Nρ = 7920 kg/m3

d = 0.254 mm


Enough elements are selected so that the model can be used to adequately characterize the string dy-

namics. The stress stiffening capability of the elements is used. Linear perturbed harmonic analysis de-

termines the displacement response to the lateral force F2.

Figure 114: Guitar String Problem



Results Comparison


1.001322.621322.2f, HzModal

-Response, 320 < f <

328

Responsef1, (322.2 Hz)Frequency Re-

sponse

-No ResponseNo Responsef2, (644.4 Hz)


974

Responsef3, (966.6 Hz)



1619

Responsef5, (1611.0 Hz)


Figure 115: String Midpoint Displacement Amplitude


VMMECH088

VMMECH089: Delamination Analysis of a Double Cantilever Beam Using

Contact-Based Debonding

Overview

Alfano, G., Crisfield, M.A., “Finite Element Interface Models for the Delamination Analysis of Lamina

Mechanical and Computation Issues”, International Journal for Numerical Methods in Engineer

1736, 2001

Reference:



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. De-

termine the vertical reaction at point P, based on the vertical displacement using the contact-based

debonding capability.

Figure 116: Double Cantilever Beam Sketch


Umax = 10 mmComposite l = 100 mm

a = 30 mmE11 = 135.3 GPa h = 3 mm

E22 = 9.0 GPa w = 20 mm

E33 = 9.0 GPa

G12 = 5.2 GPa

ν12 = 0.24

ν13 = 0.24

ν23 = 0.46

Interface

C1 = 1.7 MPa

C2 = 0.28 N/mm




C5 = 1.0 x 10-5


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


Max RFORCE and corresponding displacement using debonding

1.00050.67750.677RFORCE FY (N)

1.0001.501.50DISP UY (mm)

RFORCE and corresponding displacement U = 10.0 using debonding

1.00024.55324.553RFORCE FY (N)

1.00010.0010.00DISP UY (mm)


VMMECH089

VMMECH090: Delamination Analysis of a Double Cantilever Beam Using Interface

Delamination

Overview

Alfano, G., Crisfield, M.A., “Finite Element Interface Models for the Delamination Analysis of Lamina

Engineering, Vol 50, pp. 1701-1736, 2001

Reference:



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. De-

termine the vertical reaction at point P based on the vertical displacement for the interface model.

Figure 117: Double Cantilever Beam Sketch


Umax = 10 mmComposite l = 100 mm

a = 30 mmE11 = 135.3 GPa h = 3 mm

E22 = 9.0 GPa w = 20 mm

E33 = 9.0 GPa

G12 = 5.2 GPa

ν12 = 0.24

ν13 = 0.24

ν23 = 0.46

Interface

C1 (maximum stress) = 25 MPa

C2 (normal separation) = 0.004 mm

C3 (shear separation) = 1000 mm




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

RatioMechanic-

al

Target

Max RFORCE and corresponding DISP:

1.00160.06960.00RFORCE FY

(N)

1.0001.0001.00DISP UY

(mm)

End RFORCE and corresponding DISP

1.01224.28824.00RFORCE FY

(N)

1.0010.0010.00DISP UY

(mm)

Higher Order Results

RatioMechanic-

al

Target

Max RFORCE and corresponding DISP

1.00160.06360.00RFORCE FY

(N)

1.0001.0001.00DISP UY

(mm)

End RFORCE and corresponding DISP

1.01224.28924.00RFORCE FY

(N)

1.0010.0010.00DISP UY

(mm)


VMMECH090

Part III: Design Exploration Descriptions

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

Overview

From the Basic PrincipleReference:

Goal Driven OptimizationAnalysis

Type(s):

3-D SolidElement

Type(s):

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


LoadingGeometric

Properties

Material Properties

Fixed SupportE = 2e11 Pa

Force F =

10000 N (Z dir-

ection)

Width = 25 mmν = 0

Height = 30

mmρ = 7850 kg/m

3

Rib Thickness =

4 mm

Length = 300

mm

ImportanceDesired ValueLimitsTypeParameter

HighNo Preference20 mm ≤ W ≤ 30

mm

InputWidth

HighNo Preference25 mm ≤ H ≤ 35

mm

InputHeight




HighNo Preference250 mm ≤ L ≤ 350

mm

InputLength

LowMinimum Possiblen/aOutputVolume

HighMinimum Possiblen/aOutputStress

HighMinimum Possiblen/aOutputDeflection

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

Error (%)DesignXplorerTargetResults

0.06.9E-05 m3

6.9e-5 m3

Volume (V)

0.108624.5339E-05

m

4.5290e-5 mDeformation (D)

0.000463.623065E-

07 Pa

3.62319e7

Pa

Stress (σ)


VMDX001

VMDX002: Optimization of Bar with Temperature-Dependent Conductivity

Overview

From the Basic PrincipleReference:


Type(s):

3-D SolidElement

Type(s):

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


LoadingGeometric

Properties

Material Properties

Frictionless Support (on

all faces)

E = 2e11 Pa

Breadth B = 2

m

ν = 0

Reference temperature

= 5°C

α = 1.5E-05/°C

Width W = 2 mK = k0*(1 + a*T)

W/m-°C Temperature on end

face T = 100°C

Length L = 20

mk0 = 0.038Convection on other

end facea = 0.00582

Convection coefficient

h = 5e-3 W/m2°C

Ambient temperature

Ta = 5°C




LowNo Preference15 m ≤ l ≤ 25 mInputLength (l)

LowNo Preference0.004 W/m2°C ≤ h

≤ 0.006 W/m2°C

InputConvection

coefficient

(h)

LowNo Preference1.4e-5/°C ≤ α ≤1.6e-5/°C

InputCoefficient

of temperat-

ure expan-

sion (α)

HighMinimum Possiblen/aOutputTemperature

(T)

HighMinimum Possiblen/aOutputThermal

strain (ε)

Analysis

Temperature:

s a a a= − − + × + × +7 2 2 6

Thermal strain:

ε α α= − = −

Combined objective function becomes,

Φ = +− − +

× α

++ × +

− −

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


025 m25 mLength (l)

00.006

W/m2°C

0.006

W/m2°C

Convection coefficient (h)

01.4e-5/°C1.4e-5/°CCoefficient of thermal expan-

sion (α)

-0.327829.553°C29.6528°CTemperature (T)

-0.41153.437e-4

m/m

3.4514e-4

m/m

Thermal strain (ε)


VMDX002

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

Overview

S. S. Rao, Optimization Theory and Application Second edition,

example 1.10, page 28-30

Reference:

Goal Driven Optimization with APDLAnalysis

Type(s):

3-D SolidElement

Type(s):

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 max-

imum 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

LoadingGeometric PerpertiesMaterial Proper-

ties

Mass of water tank M = 1000000

Kg

Width, b = 0.4 m

Depth, d =1.2 m

E = 3e10 Pa

ρ = 2300

Kg/m3 Acceleration due to gravity =

9.81 m/s2Length, I = 20 m

σ max =

4.1e7 Pa

Sample Size: 10000




0.0000.36102 m0.36102 mWidth b

0.0021.318137 m1.3181 mDepth d

–0.00121889.77 kg21890 kgMass of column M

-0.0200.87816

rad/sec

0.87834

rad/sec

Natural frequency w

-0.0152.0383e7 Pa2.0386e7 PaDirect stress

-0.0136.15174e6

Pa

6.1526e6 PaBuckling stress

Analysis

Minimize:

Mass of the column = = × × ×ρ

Maximize:

Nr rqy rvr vibri wr k w =× ×

× × + × × × ×

3

3

1 2

ρ

/

Subject to constraints:

D _S

d Bg_S

= −××

≥

=× ×

σ

π

x

×

−

××

≥

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


0.0000.36102 m0.36102 mWidth b

0.0028071.318137 m1.3181 mDepth d

-0.0008921890.1957

kg

21890 kgMass of column M


VMDX003


-0.020740.87816

rad/sec

0.87834

rad/sec

Natural frequency w

-0.012772.0383e7 Pa2.0386e7 PaDirect stress

-0.01396.15174e6

Pa

6.1526e6 PaBuckling stress



VMDX003

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

Vertices

Overview

Blevins, Formula for Natural Frequency and Mode Shape, Van

Nostrand Reinhold Company Inc., 1979, pg. 269-271

Reference:


Type(s):

3-D ShellElement

Type(s):

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


LoadingGeometric

Properties

Material Properties

All vertices are

simply suppor-

ted

E = 2e5 MPa

Length a = 250

mm

ν = 0.3

ρ = 7.850 e-6

kg/mm3 Width b = 250

mm

Thickness h = 5

mm

ImportanceDesired ValueConstraintsTypeParameter

LowNo Preference1.8e11 Pa ≤ E ≤2.2e11Pa

InputYoung's Modulus E

LowNo Preference0.27 ≤ µ ≤ 0.30InputPoisson's Ratio µ




LowNo Preference7065 kg/m3 ≤ ρ ≤

8635 kg/m3

InputDensity ρ

HighMinimum PossibleN/aOutputFirst Natural Fre-

quency w

Analysis

First Natural Frequency:

=−

2

2

3

2

1 2/

π ρ ν

Objective function becomes:

φρ ν

= −−


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


0.001.8e11 Pa1.8e11 PaYoung's Modulus E

0.000.270.27Poisson's Ratio µ

0.008635 kg/m3

8635 kg/m3

Density ρ

-0.5894123.36 rad/s124.0913

rad/s

First Natural Frequency w


VMDX004

VMDX005: Optimization of Buckling Load Multiplier with CAD Parameters and

Young's Modulus

Overview

Timoshenko, Strength of Materials, Part 2 (Advanced theory and

problems), pg. 167–168

Reference:


Type(s):

3-D SolidElement

Type(s):

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


LoadingGeometric

Properties

Material Properties

Fixed support

on one face,

E = 4.1771e 9 psf

Cross-section of

square = 0.25

ft. x 0.25 ft.

ν = 0.3

Force = 11 lbf

(Negative Z-dir-

ection) on top

face

ρ = 490.45 lbm/ft3

Length of bar =

25 ft.


N/ANo Preference0.225 ft. ≤ a ≤0.275 ft.

InputCross-section side




N/ANo Preference22.5 ft. ≤ l ≤ 27.5

ft.

InputLength

N/ANo Preference3.7594e9 psf ≤ E ≤4.5948e9 psf

InputYoung's Modulus

N/AMaximum PossibleN/AOutputFirst buckling mode

load multiplier

Analysis

Assuming that under the action of uniform axial load a slight lateral bucking occurs.

The expression for deflection is:

= −δπ

The critical load is given by,

cr cr= = =2

2

2

π

where:

q = force per unit length

The first critical buckling load is:

= =

=

ππ

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



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


-1.5323036.073083.32First buckling mode load

multiplier



VMDX005

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