ANSYS Workbench Verification Manual

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ANSYS Workbench Verification Manual

Release 15.0 ANSYS, Inc. November 2013 Southpointe 275 Technology Drive

Canonsburg, PA 15317 ANSYS, Inc. is

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Introduction

The following topics are discussed in this chapter:

Overview

Index of Test Cases

Overview

This manual presents a collection of test cases that demonstrate a number of the capabilities of the Workbench analysis environment. The available tests are engineering problems that provide independent verification, usually a closed form equation. Many of them are classical engineering problems.

The solutions for the test cases have been verified, however, certain differences may exist with regard to the references. These differences have been examined and are considered acceptable. The workbench analyses employ a balance between accuracy and solution time. Improved results can be obtained in some cases by employing a more refined finite element mesh but requires longer solution times. For the tests, an error rate of 3% or less has been the goal.

These tests were run on an Intel Xeon processor using Microsoft Windows 7 Enterprise 64-bit . These results are reported in the test documentation. Slightly different results may be obtained when different processor types or operating systems are used.

The tests contained in this manual are a partial subset of the full set of tests that are run by ANSYS developers to ensure a high degree of quality for the Workbench product. The verification of the Workbench product is conducted in accordance with the written procedures that form a part of an overall Quality Assurance program at ANSYS, Inc.

You are encouraged to use these tests as starting points when exploring new Workbench features. Geometries, material properties, loads, and output results can easily be changed and the solution 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 Options Analysis Type

Element Type Test Case Number

Linear Static Structural Solid VMMECH001 Linear Static Structural Solid VMMECH002 Free Vibration Modal Solid VMMECH003 Nonlinear, Visco-plastic Materials Structural Solid VMMECH004 Linear Static Thermal Solid VMMECH005 Nonlinear Static Thermal Solid VMMECH006 Nonlinear Thermal Stress Static Structural Solid VMMECH007 Linear Static Thermal Solid VMMECH008 Linear Static Structural Solid VMMECH009 Free Vibration Modal Shell VMMECH010 Nonlinear, Large Deformation Static Structural Shell VMMECH011 Buckling Solid VMMECH012 Buckling Shell VMMECH013 Harmonic Solid VMMECH014 Harmonic Solid VMMECH015 Fatigue Static Structural Solid VMMECH016 Linear Thermal Stress Static Structural Solid VMMECH017

Linear, Inertia relief Static Structural Solid VMMECH018 Linear Static Structural Beam VMMECH019 Shell Modal Beam VMMECH020 Buckling Beam VMMECH021 Nonlinear, Contact Static Structural Solid VMMECH022 Linear Static Structural Beam VMMECH023

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Static Structural 2-D Solid, Plane Strain VMMECH030 Static Structural 2-D Solid, Plane Stress VMMECH031 Linear Thermal Stress Static Structural 2-D Solid, Axisym-metric VMMECH032 Electromagnetic Static Structural Solid VMMECH033 Nonlinear, Large Deformation Static Structural Solid VMMECH034 Coupled (Static Thermal and Static Stress) Solid VMMECH035 Sequence Loading Static Structural Solid VMMECH036 Transient Thermal 2-D Solid, Axisym-metric VMMECH037 Flexible Dynamic Transient Structural Solid VMMECH038 Flexible Dynamic Transient Structural Solid VMMECH039 Spring Static Structural Beam VMMECH040 Electromagnetic Static Structural Solid VMMECH041 Hydrostatic Fluid Static Structural Solid VMMECH042 Modal Beam VMMECH043 Linear Thermal Stress Static Structural Beam VMMECH044 Static Structural Shell VMMECH045 Static Structural Shell VMMECH046 Nonlinear, Plastic Materials Static Structural 2-D Solid, Axisym-metric VMMECH047 Static Structural Beam VMMECH048 Static Structural Beam VMMECH049 Static Structural Axisymmetric Shell VMMECH050 Static Structural Axisymmetric Shell VMMECH051 Rigid Dynamic Multipoint Con-straint VMMECH052 Rigid Dynamic Multipoint Con-straint VMMECH042 Rigid Dynamic Multipoint Con-straint VMMECH054 Rigid Dynamic Multipoint Con-straint VMMECH055

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Rigid Dynamic Multipoint Con-straint VMMECH056 Rigid Dynamic Multipoint Con-straint VMMECH057 Rigid Dynamic Multipoint Con-straint VMMECH058 Static Structural 2-D Plane Stress Shell VMMECH059 Flexible Dynamic Transient Structural Solid VMMECH060 Multipoint Con-straint Static Structural Beam VMMECH061 Static Structural Axisymmetric Shell VMMECH062 Nonlinear, Large Deformation Static Structural Shell VMMECH063 Static Structural Shell VMMECH064 Linear Thermal Stress Static Structural Solid Shell VMMECH065 Static Structural Shell VMMECH066 Static Structural Solid VMMECH067 Nonlinear, Plastic Materials Static Structural 2-D Solid, Plane Strain VMMECH068 Static Structural Shell VMMECH069 Nonlinear, Large Deformation Static Structural 2-D Solid VMMECH070 Static Thermal 2-D Thermal Solid VMMECH071 Linear Thermal Stress Static Structural 2-D Thermal Solid VMMECH072 Modal Solid VMMECH073

Rigid Body Dynam-ics Solid Spring VMMECH074 Harmonic Solid VMMECH075 Static Structural Shell VMMECH076

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Modal Pipe VMMECH081 Spectral Mass Fracture Mechanics Static Structural Solid VMMECH082 Mode Superposi-tion Transient Dynamic Spring, Mass VMMECH083 Nonlinear, Hypere-leastic Static Structural Solid VMMECH084 Composite Material Static Structural Solid VMMECH085 Static Structural Solid VMMECH086 Submodeling (2D-2D) Modal Line Body VMMECH087 Point Mass Bearing Connection Linear Perturbation Static Structural Beam VMMECH088 Modal Harmonic Contact-Based De-bonding Static Structural Solid VMMECH089 Interface Delamina-tion Static Structural Solid VMMECH090

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VMDM001: Extrude, Chamfer, and Blend Features

Overview

Extrude, Chamfer, and Blend

Feature:

Millimeter

Drawing 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 rectrect-angular 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.

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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 Target Results 0 189561.95 189561.95 Volume (mm3) -0.0135 44433.3 44439.29 Surface Area (mm2) 0 52 52 Number of Faces 0 1 1 Number of Bodies

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

Overview

Revolve, Sweep, Extrude, and Skin-Loft

Feature:

Millimeter

Drawing Units:

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.

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Results Comparison

Error (%) Design-Modeler Target Results 0 1130973.3 1130973.3 Volume (mm3) 0 81053.1 81053.1 Surface Area (mm2) 0 3 3 Number of Faces 0 1 1 Number of Bodies

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

Overview

Extrude, Revolve, Skin-Loft, and Sweep

Feature:

Millimeter

Drawing Units:

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:

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Vc = 47123.9 mm3 and Vd = 28352.8 mm3. •

• And total volume that gets subtracted from box due to Sweep Feature = 75476.7 mm3. • Hence Net volume of box, Va = 814260.3 - 75476.7 = 738783.6 mm3.

• Sum of volumes of all four bodies = Va+Vb+Vc+Vd = 738783.6 + 11134.15 + 47123.9 +28352.8 = 825394.4 mm3.

Results Comparison

Error (%) Design-Modeler Target Results 0 825394.5 825394.4 Volume (mm3) 0 101719.95 101719.47 Surface Area (mm2) 0 22 22 Number of Faces 0 4 4 Number of Bodies

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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 Analysis

Analysis Type(s): Solid Element Type(s):

Test Case

An assembly of three prismatic bars is supported at both end faces and is axially loaded with forces F₁ and F₂. Force F₁ is applied on the face between Parts 2 and 3 and F₂ is applied on the face between Parts 1 and 2. Apply advanced mesh control with element size of 0.5”.

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

Figure 4: Schematic Loading Geometric Properties Material 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 F_{(Y direction)}2 = -500 Length of Part 2 = 3" Length of Part 3 = 3”

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Results Comparison

Error (%) Mechanical Target Results 0.127 901.14 900 Y Reaction Force at Top

Fixed Support (lbf )

-0.190 598.86

600 Y Reaction Force at Bottom

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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:

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 Loading Geometric Properties Material Properties Pressure = -100 Pa Length = 15 m E = 1000 Pa Width = 5 m ν = 0 Thickness = 1 m Hole radius = 0.5 m

Results Comparison

Error (%) Mechanical Target Results 0.864 315.2 312.5 Maximum Normal X Stress

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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 Analysis

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 Loading Geometric Properties Material Properties 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"

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Loading Geometric Properties Material Properties Thickness of all plates = 1"

Results Comparison

Error (%) Mechanical Target Results -0.23 310.21 310.911 1st Frequency Mode (Hz) -0.78 315.6 318.086 2nd Frequency Mode (Hz) -0.77 315.64 318.086 3rd Frequency Mode (Hz) -1.38 346.73 351.569 4th Frequency Mode (Hz) -1.27 347.11 351.569 5th Frequency Mode (Hz) -1.22 437.06 442.451 6th Frequency Mode (Hz)

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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 Analysis

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

Loading Geometric Properties Material Properties Temp = 400°C = 673°K h = 1 cm E_x (Young's Modulus) = 60.6 GPa thickness = 1 cm Velocity (x-direc-tion) = 0.01 (Poisson's Ratio) = 0.4999 cm/sec @ y = 1 cm S_o = 29.7 MPa Q/R = 21.08999E3 K Time = 20 sec A = 1.91E7 s-1 = 7.0

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Loading Geometric Properties Material Properties m = 0.23348 h_o = 1115.6 MPa

µ

= 18.92 MPa = 0.07049 a = 1.3

Results Comparison

Error (%) Mechanical Target Results -6.3 -791.76 845.00 F_x, N

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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 Analysis

Test Case

A furnace wall consists of two layers: fire brick and insulating brick. The temperature inside the furnace is 3000°F (T_f) and the inner surface convection coefficient is 3.333e-3 BTU/s ft2°F (h_f). The ambient temperature is 80°F (T_a) and the outer surface convection coefficient is 5.556e-4 BTU/s ft2°F (h_a). Find the Temperature Distribution.

Figure 8: Schematic Loading Geometric Properties Material Properties 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

Error (%) Mechanical Target Results 0.202 336.68 336 Minimum Temperature (°F) 0.007 2957.2 2957 Maximum Temperature (°F)

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VMMECH006: Heater with Nonlinear Conductivity

Overview

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

Reference:

Nonlinear Static Thermal Analysis

Analysis Type(s):

Solid

Element 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.

Loading Geometric Properties

Material Properties

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-002 32 2.75e-002 1000

Results Comparison

Error (%) Mechanical Target Results 0.96 480.58 476 Maximum Temperature (°F) -0.003 9.9997 10 Maximum Total Heat Flux

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VMMECH007:Thermal Stress in a Bar with Temperature Dependent Conductivity

Overview

Any basic Heat Transfer book

Reference:

Nonlinear Thermal Stress Analysis

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

Rear face tem-perature T = 100°C Length = 20 m E = 2e11 Pa Width = 2 m ν = 0 Breadth = 2 m α = 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°C 3.91e-002 5 Reference tem-perature = 5°C 0.215 800

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Analysis

Temperature at a distance "z" from rear face is given by:

z = − + −

Thermal strain in the z direction in the bar is given by:

ε T = × × −

−5

Deformation in the z direction is given by:

=

_∫

× −× − 0

Heat flux in the z direction is given by:

= × −

Results Comparison

Error (%) Mechanical Target Results -0.016 38.014 38.02 Minimum Temperature (°C) 0.042 0.00049521 0.000495 Maximum Thermal strain (z

= 20) (m/m)

0.000 0.001425

0.001425 Maximum Thermal strain (z

= 0) (m/m) 0.905 0.002341 0.00232 Maximum Z Deformation (m) 0.042 0.16507 0.165 Maximum Z Heat Flux (z =

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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 Analysis

Test Case

A steel cooling spine of cross-sectional area A and length L extend from a wall that is maintained at temperature T_w. The surface convection coefficient between the spine and the surrounding air is h, the air temper is T_a, and the tip of the spine is insulated. Apply advanced mesh control with element size of 0.025'.

Find the heat conducted by the spine and the temperature of the tip.

Figure 11: Schematic Loading Geometric Properties Material Properties Loading Geometric Properties Material Properties T_w = 100°F E = 4.177e9 psf Cross section = 1.2” x 1.2” ν = 0.3 _T_a_{= 0°F} Thermal conductiv-ity k = 9.71e-3 BTU/s ft °F h = 2.778e-4 BTU/s ft2 °F L = 8”

Results Comparison

Error (%) Mechanical Target Results 0.055 79.078 79.0344 Temperature of the Tip (°F)

-0.041 6.3614e-3

6.364e-3 Heat Conducted by the

Spine (Heat Reaction) (BTU/s)

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VMMECH009: Stress Tool for Long Bar with Compressive Load

Overview

Any basic Strength of Materials book

Reference:

Test Case

A multibody of four bars connected end to end has one of the end faces fixed and a pressure is applied to the opposite face as given below. The multibody is used to nullify the numerical noise near the contact regions.

Find the maximum equivalent stress for the whole multibody and the safety factor for each part using the maximum equivalent stress theory with tensile yield limit.

Figure 12: Schematic Material Properties Tensile Yield (Pa) ν E (Pa) Mater-ial 2.07e8 0 1.93e11 Part 1 2.8e8 0 7.1e10 Part 2 2.5e8 0 2e11 Part 3 2.8e8 0 1.1e11 Part 4 Loading Geometric 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

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Part 4: 2 m x 2 m x 2 m

Results Comparison

Error (%) Mechanical Target Results 0.000 2.5e8 2.5e8 Maximum Equivalent Stress

(Pa)

0.000 0.828

0.828 Safety Factor for Part 1

0.000 1.12

0.000 1

1 Safety Factor for Part 3

0.000 1.12

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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 Analysis

Analysis Type(s): Shell Element Type(s):

Test Case

A rectangular plate is simply supported on both the smaller edges and fixed on one of the longer edges as shown below. Sizing mesh control with element size of 6.5 mm is applied on all the edges to get accurate results.

Find the first five modes of natural frequency.

Figure 13: Schematic Loading Geometric Properties Material Properties Length = 0.25 m E = 2e11 Pa ν = 0.3 Width = 0.1 m ρ = 7850 kg/m3 Thickness = 0.005 m

Results Comparison

Error (%) Mechanical Target Results -0.952 590.03 595.7 1st Frequency Mode (Hz) -0.987 1118.4 1129.55 2nd Frequency Mode (Hz) -0.667 2038.1 2051.79 3rd Frequency Mode (Hz) -0.994 2879.3 2906.73 4th Frequency Mode (Hz)

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Error (%) Mechanical Target Results -0.489 3350 3366.48 5th Frequency Mode (Hz)

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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)

Test Case

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

Figure 14: Schematic Loading Geometric Properties Material Properties Pressure = 6585.18 Pa Radius = 0.25 m E = 2e11 Pa ν = 0.3 Thickness = 0.0025 m

Results Comparison

Error (%) Mechanical Target Results -1.008 0.0012374 0.00125 Total deformation (m)

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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 Analysis Analysis Type(s): Solid Element 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.

Figure 15: Schematic Loading Geometric Properties Material Properties 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-tion Length of smal-ler diameter = 0.1 m

Results Comparison

Error (%) Mechanical Target Results 2.0356 22.958 22.5 Load Multiplier

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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: Buckling Analysis Analysis Type(s): Shell Element 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.

Figure 16: Schematic Loading Geometric Properties Material Properties 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

Error (%) Mechanical Target Results 0.4 546.07 544 Load Multiplier

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VMMECH014: Harmonic Response of a Single Degree of Freedom System

Overview

Any basic Vibration Analysis book

Reference: Harmonic Analysis Analysis Type(s): Solid Element 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

Figure 17: Schematic Material Properties ρ (kg/m3) ν E (Pa) Material 1e-8 0.34 1.1e11 Shaft 1 1e-8 0.34 1.1e11 Shaft 2 1e-8 0.35 4.5e10 Shaft 3 1e-8 0.35 4.5e10 Shaft 4 Loading Geometric Properties Force = 1e7 N (Z-direction) Each cylinder: Diameter = 20 mm Point Mass = 3.1044 Kg Length = 50 mm

(50)

Results Comparison

Error (%) Mechanical Target Results 0.591 0.14123 0.1404 Maximum Amplitude without damping (m) 0.000 180 180 Phase angle without

damp-ing (degrees)

0.577 0.1408

0.14 Maximum Amplitude with

damping (m)

0.000 175.58

175.6 Phase angle with damping

(51)

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: Harmonic Analysis Analysis Type(s): Solid Element 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. Figure 18: Schematic Material Properties ρ (kg/m3) ν E (Pa) Material 7850 0.3 2e18 Block 2 1e-8 0.35 4.5e10 Shaft 2 15700 0.3 2e18 Block 1 1e-8 0.35 9e10 Shaft 1

(52)

Loading Geometric

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

Error (%) Mechanical Target Results 1.5 0.21174 0.20853 Maximum Amplitude for

vertex A (m)

1.2 0.075838

0.074902 Maximum Amplitude for

(53)

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

Overview

Any basic Machine Design book

Reference: Fatigue Analysis Analysis Type(s): Solid Element 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

Figure 20: Scenario 2

E = 2e11 Pa ν = 0.3

(54)

Yield Tensile Strength = 3.5e8 Pa Endurance Strength = 2.2998e6 Pa

Alternating Stress (Pa) Number of Cycles 4.6e8 1000 2.2998e6 1e6 Loading Geometric

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 Target Results -0.156 3329.9 3335.1049 Life

Stress Component - Component

X _Damage _299.8406 _300.31 _0.157 0.132 0.019025 0.019 Safety Factor -0.764 14653 14765.7874 Life

Y _Damage _67.724 _68.247 _0.772 -0.683 0.045378 0.04569 Safety Factor 0.001 14766 14765.7874 Life

Z _Damage _67.724 _67.725 _0.001 0.013 0.045696 0.04569 Safety Factor

(55)

VMMECH017: Thermal Stress Analysis with Remote Force and Thermal Loading

Overview

Reference:

Linear Thermal Stress Analysis

Test Case

A cylindrical rod assembly of four cylinders connected end to end has frictionless support applied on all the cylindrical surfaces and both the flat end faces are fixed. Other thermal and structural loads are as shown below.

Find the Deformation in the x direction of the contact surface on which the remote force is applied. To get accurate results apply a global element size of 1.5 m.

Figure 21: Schematic Loading Geometric Properties Material Properties Given temperature (End A) = 1000°C Diameter = 2 m E = 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

Error (%) Mechanical Target Results -1.5 0.10025 0.101815 Maximum X Deformation (m)

(56)

(57)

VMMECH018: A Bar Subjected to Tensile Load with Inertia Relief

Overview

Reference:

Linear Static Structural Analysis (Inertia Relief On)

Test Case

A long bar assembly is fixed at one end and subjected to a tensile force at the other end as shown below. Turn on Inertia Relief.

Find the deformation in the z direction

Figure 22: Schematic Loading Geometric Properties Material Properties 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 from ρ = 7850 kg/m3 End A: 2 m, 5 m, 10 m, and 3 m.

Analysis

δz = − ρ 2 where:

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

(58)

Results Comparison

Error (%) Mechanical Target Results 0.172 2.5043E-06 2.5e-6 Maximum Z Deformation (m)

(59)

VMMECH019: Mixed Model Subjected to Bending Loads with Solution Combination

Overview

Reference:

Analysis Type(s):

Beam and Shell

Element 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. Figure 23: Scenario 1 Figure 24: Scenario 2 Loading Geometric Properties Material Properties Force F = -10 N (y direction) Shell = 160 mm x 500 mm x 10 mm E = 2e5 Pa ν = 0

(60)

Loading Geometric Properties Material Properties Moment M = -4035 Nmm @ z-ax-is Beam rectangu-lar cross section = 10 mm x 10 mm Beam length = 500 mm

Analysis

δy = + 3 2 l l where:

I = total bending length of the mixed model

I = moment of inertia of the beam cross-section

Results Comparison

Error (%) Mechanical Target Results 0.929 -7.2542 -7.18742 Maximum Y-Deformation (mm)

(61)

VMMECH020: Modal Analysis for Beams

Overview

Reference: Modal Analysis Analysis Type(s): Beam Element 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

Figure 26: Schematic Material Properties ρ (kg/m3) ν E (Pa) Material 1e-8 0.34 1.1e11 Spring 7.85e5 0 2e11 Mass

(62)

Spring beam length = 500 mm

Mass beam length = 5 mm

Results Comparison

Error (%) Mechanical Target Results 0.160 1190.5 1188.6 Natural Frequency of Axial

(63)

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: Buckling Analysis Analysis Type(s): Beam Element 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.

Figure 27: Schematic Loading Geometric Properties Material Properties Force on L₁ = -1000 N (x direc-tion) L₁ = 50 mm E = 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

Error (%) Mechanical Target Results -0.407 10.198 10.2397 Load Multiplier

(64)

(65)

VMMECH022: Structural Analysis with Advanced Contact Options

Overview

Any basic Strength of Material book

Reference:

Nonlinear Static Structural Analysis

Test Case

An assembly of two parts with a gap has a Frictionless Contact defined between the two parts. The end faces of both the parts are fixed and a given displacement is applied on the contact surface of Part 1 as shown below.

Find the Normal stress and Directional deformation - both in the z direction for each part for the following scenarios:

• Scenario 1: Interface treatment - adjust to touch.

• Scenario 2: Interface treatment - add offset. Offset = 0 m. • Scenario 3: Interface treatment - add offset. Offset = 0.001 m. • Scenario 4: Interface treatment - add offset. Offset = -0.001 m.

Validate all of the above scenarios for Augmented Lagrange and Pure Penalty formulations.

Figure 28: Schematic Loading Geometric Properties Material Properties Given displace-ment = (0, 0, 0.0006) m Gap = 0.0005 m E = 2e11 Pa Dimensions for each part: 0.1 m x 0.1 m x 0.5m ν = 0

Results Comparison

(66)

Er-ror (%) Mechan-ical Tar-get Results 0.000 6e-4 6e-4 Maximum directional z

de-formation Part 1 (m) Adjust To Touch -0.357 5.9786e-4 6e-4

Maximum directional z de-formation Part 2 (m)

0.000 2.4e8

2.4e8 Maximum normal z stress

Part 1 (Pa)

-0.354 -2.3915e8

-2.4e8 Maximum normal z stress

Part 2 (Pa)

0.000 6e-4

6e-4 Maximum directional z

de-formation Part 1 (m) Add Offset. Offset = 0 m

-0.356

0.99644e-4 1e-4

Maximum directional z de-formation Part 2 (m)

0.000 2.4e8

Part 1 (Pa)

-0.355 -3.9858e7

-4e7 Maximum normal z stress

Part 2 (Pa)

0.000 6e-4

de-formation Part 1 (m) Add Offset. Offset =

0.001 m -0.355 1.0961e-3 1.1e-3 Maximum directional z de-formation Part 2 (m)

0.000 2.4e8

Part 1 (Pa)

-0.357 -4.3843e8

-4.4e8 Maximum normal z stress

Part 2 (Pa)

0.000 6e-4

de-formation Part 1 (m) Add Offset. Offset =

-0.001 m

0.000 0

0 Maximum directional z de-formation Part 2 (m)

0.000 2.4e8

Part 1 (Pa)

0 0

0 Maximum normal z stress

(67)

VMMECH023: Curved Beam Assembly with Multiple Loads

Overview

Reference:

Analysis Type(s): Beam Element 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.

Figure 29: Schematic Equivalent Loading: Loading Geometric Properties Material Properties 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

(68)

Loading Geometric Properties Material Properties UDL w = -5 N/mm (y direction) on both beams Included angle = 45° Beam 2: E₂ = 2e5 MPa ν₂ = 0

This UDL is applied as an edge force ρ₂ = 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

Error (%) Mechanical Target Results 0.619 -8.4688 -8.416664 Minimum Y Deformation (mm)

(69)

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

Beams

Overview

Reference: Harmonic Analysis Analysis Type(s): Beam Element 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.

Figure 30: Schematic Material Properties ρ (kg/m3) ν E (Pa) Mater-ial 1e-8 0.34 1.1e11 Spring 7.85e5 0 2e11 Mass Loading Geometric Properties Harmonic force F = 1 e6 N (z-direc-tion) Cross-section of each beam: Outer radius = 10 mm

(70)

Loading Geometric Properties Inner radius = 5 mm Length of longer beam = 100 mm Length of shorter beam = 5 mm

Results Comparison

Er-ror (%) Mechan-ical Target Results -0.859 4.078e-3 4.11332e-3 Maximum z directional deforma-tion without damping (m) Mode Superposi-tion -0.876 4.0765e-3 4.11252e-3 Maximum z directional deforma-tion with damping (m)

-0.003

4.1132e-3

4.11332e-3 Maximum z directional deforma-tion without damping (m) Full Method -1.046 4.0695e-3 4.11252e-3 Maximum z directional deforma-tion with damping (m)

(71)

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:

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. Figure 31: Schematic Loading Geometric Properties Material Properties P = 30000 psi Inner Cylinder: Both cylinders are made of r_i = 4” the same

mater-ial

r_o = 6.005” R_i = 6” E = 3e7psi

(72)

Loading Geometric Properties Material Properties R_o = 8” ν = 0 ρ = 0.28383 lbm/in3 Length of both cylinders = 5”

Results Comparison

Error (%) Mechanical Target Results 1.0 35738 35396.67 Maximum normal y stress, inner

cylinder (psi)

0.0 42279

42281.09 Maximum normal y stress, outer

cylinder (psi)

Note

(73)

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

Overview

Any standard Machine Design and Strength of Materials book

Reference: Fatigue Analysis Analysis Type(s): Shell Element 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.

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 1.08e9 1000 1.38e8 1e6 Loading Geometric Properties Moment M = 0.15 Nm (counterclock-wise @ z-axis) Length L = 12e-3 m Width W = 1e-3 m

(74)

Loading Geometric Properties Thickness T = 1 e-3 m

Results Comparison

Er-ror (%) Mechan-ical Target Results 0.000 9e8 9e8 Maximum normal x-stress (Pa)

0.279

6.4981e-4 6.48e-4

Maximum total deformation (m)

0.020 0.15333 0.1533 Safety factor SN-Goodman 0.005 1844.4 1844.3 Life 0.020 0.15333 0.1533 Safety factor SN-Soderberg 0.005 1844.4 1844.3 Life 0.020 0.15333 0.1533 Safety Factor SN-Gerber 0.005 1844.4 1844.3 Life

(75)

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

Overview

Any standard Thermal Analysis book

Reference:

Thermal Stress Analysis

Test Case

A plate of length (L), width (W), and thickness (T) is fixed along the width on one edge and heat flow (Q) is applied on the same edge. The opposite edge is subjected to a temperature of 20 °C. Ambient temperature is 20 °C. To get accurate results, apply a sizing control with element size = 2.5e-2 m.

Find the maximum temperature, maximum total heat flux, maximum total deformation, and heat reaction at the given temperature.

Figure 33: Schematic Loading Geometric Properties Material Properties Heat flow Q = 5 W Length 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

(76)

where:

T_h = maximum temperature T₁ = given temperature Total heat flux is:

=

Temperature at a variable distance z from the fixed support is given by:

z h h = − − ×      1

Thermal deformation in the z-direction is given by:

δ α l =

_∫

− 0

Results Comparison

Er-ror (%) Mechan-ical Target Results 0.000 86.116 86.1157 Maximum Temperature (°C) 0.000 2e4 2e4 Maximum Total Heat Flux (W/m2)

0.781

7.9958e-5

7.93386e-5 Maximum Total Deformation (m)

0.000 -5

-5 Heat Reaction (W)

(77)

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

Overview

Any standard Strength of Materials book

Reference:

Static Structural Analysis

Test Case

A semi-cylinder is fixed at both the end faces. The longitudinal faces have frictionless support. A bolt pretension load is applied on the semi-cylindrical face. To get accurate results, apply sizing control with element size of 0.01 m.

Find the Z directional deformation and the adjustment reaction due to the bolt pretension load.

Figure 34: Schematic Loading Geometric Properties Material Properties Pretension as pre-load = 19.635 N Length L = 1 m E = 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:

δP re t en si o n = ×

(78)

Results Comparison

Er-ror (%) Mechan-ical Target Results 0.004 -5.0002E-08 -5.00E-08 Minimum z-directional deformation

(m)

-0.996

4.9502E-08 5.00E-08

Maximum z-directional deformation (m)

0.000 1.00E-07

1.00E-07 Adjustment Reaction (m)

(79)

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 Analysis

Test Case

A rectangular beam is loaded in pure bending. For an elastic-perfectly-plastic stress-strain behavior, show that the beam remains elastic at M = Myp = σ_ypbh2 / 6 and becomes completely plastic at M = Mult = 1.5 Myp. To get accurate results, set the advanced mesh control element size to 0.5 inches.

Figure 35: Stress-Strain Curve

Figure 36: Schematic Loading Geometric Properties Material Properties M = 1.0 M_yp to 1.5 M_yp Length L = 10” E = 3e7 psi Width b = 1” ν = 0.0 Height h = 2” σ_yp = 36000 psi _(M yp = 24000 lbf -in)

(80)

Analysis

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

Results Comparison

Error (%) Mechanical Target M/M_yp Equivalent Stress (psi) State Equivalent Stress (psi) State 0.164 36059 fully elastic 36000 fully elastic 1 0.800 36288 elastic-plastic 36000 elastic-plastic 1.25 -solution not converged plastic solution not converged plastic 1.5

(81)

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

Overview

Reference:

Plane Strain Analysis

Analysis Type(s): 2D Structural Solid Element 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. Figure 37: Schematic Loading Geometric Properties Material Properties 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.

(82)

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

Results Comparison

Error (%) Mechanical Target Results 0.000 30 30 Normal Stress 0.000 30 30 Maximum Normal Stress in

the X-Direction (psi)

0.018 0.16553e-2

0.1655e-2 Maximum Total Deformation

(in)

0.000 -5

-5 Reaction Moment (lbf-in)

(83)

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

Overview

Reference:

Plane Stress Analysis

Analysis Type(s):

2D Structural Solid

Element 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.

Tensile Yield (Pa) ν E (Pa) Material 2.07e8 0 1.93e11 Part 1 2.8e8 0 7.1e10 Part 2 2.5e8 0 2e11 Part 3 2.8e8 0 1.1e11 Part 4 Loading Geometric Properties 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

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