Pramote Dechaumphai, S. Sucharitpwatskul (2018) - Finite Element Analysis With ANSYS Workbench [PDF]

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Finite Element Analysis with ANSYS Workbench

Pramote Dechaumphai Sedthawat Sucharitpwatskul

α Alpha Science International Ltd. Oxford, U.K.

Finite Element Analysis with ANSYS Workbench 280 pgs.

Pramote Dechaumphai Department of Mechanical Engineering Chulalongkorn University Pathumwan, Bangkok Thailand Sedthawat Sucharitpwatskul National Metal and Materials Technology Center National Science and Technology Development Agency KlongLuang, Pathumthani Thailand Copyright © 2018 ALPHA SCIENCE INTERNATIONAL LTD. 7200 The Quorum, Oxford Business Park North Garsington Road, Oxford OX4 2JZ, U.K. www.alphasci.com ISBN 978-1-78332-369-2 E-ISBN 978-1-78332-433-0 Printed from the camera-ready copy provided by the Authors.

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without prior written permission of the publisher.

PREFACE The book “Finite Element Analysis with ANSYSWorkbench” is written for students who want to use the software while learning the finite element method. The book is also suitable for designers and engineers before using the software to analyze realistic problems. The book contains twelve chapters describing different analyses of engineering problems. These problems are in the fields of solid mechanics, heat transfer and fluid flows. In each chapter, the governing differential equations and the finite element method are presented. An academic example is used to demonstrate the ANSYS procedure for solving it in detail. An application example is also included at the end of each chapter to highlight the software capability for analyzing realistic problems. The ANSYS files for application problems can be downloaded from the book website: https://goo.gl/BDSBRQ These files can be modified to increase understanding on how to use the software. The authors would like to thank the ANSYS, Inc., USA for providing the software to prepare this book and the CAD-IT Consultants (Asia) Pte Ltd for the book distribution. The authors appreciate Dr. Edward Warute Dechaumphai for proof-reading the book manuscript. Pramote Dechaumphai Sedthawat Sucharitpwatskul

CONTENTS Preface iii

Chapter 1 Introduction

1

1.1

Solving Engineering Problems   1 1.1.1 Problem Ingredients  2 1.1.2 Solution Methods  3 1.2 Finite Element Method   4 1.2.1 What is the Finite Element Method?   4 1.2.2 Finite Element Method Procedure   5 1.3 ANSYS Software  6 1.3.1 ANSYS Workbench  7 1.3.2 Screen and Tool Bars   7 1.3.3 Analyzing steps  9 1.4 Advantages of Finite Element Method   11

Chapter 2 Truss Analysis

13

2.1

Basic Equations   13 2.1.1 Differential Equation   13 2.1.2 Related Equations   14 2.2 Finite Element Method   15 2.2.1 Finite Element Equations   15 2.2.2 Element Types  15 2.3 Academic Example  17 2.3.1 Two Truss Members in One Dimension   17 2.3.2 Two Truss Members in Two Dimensions   29 2.4 Application  39 2.4.1 Twenty-one Truss Members in Two Dimensions   39

Chapter 3 Beam Analysis 3.1

3.2

Basic Equations   42 3.1.1 Differential Equation   42 3.1.2 Related Equations   42 Finite Element Method   43 3.2.1 Finite Element Equations   43 3.2.2 Element Types  43

41

vi 

Contents

3.3 Academic Example  45 3.3.1 Two Beam Members in Two Dimensions   45 3.3.2 Twenty-one Beam Members in Two Dimensions   55 3.4 Application  56 3.4.1 Racing Car Frame Structure   56

Chapter 4 Plane Stress Analysis

61

4.1

Basic Equations   61 4.1.1 Differential Equations    61 4.1.2 Related Equations    62 4.2 Finite Element Method    63 4.2.1 Finite Element Equations    63 4.2.2 Element Types   63 4.3 Academic Example    66 4.3.1 Plate with Circular Cut-out    66 4.4 Application   77 4.4.1 Stress in Motorcycle Chain Wheel    77

Chapter 5 Plate Bending Analysis

81

5.1

Basic Equations   82 5.1.1 Differential Equation   82 5.1.2 Related Equations   83 5.2 Finite Element Method   84 5.2.1 Finite Element Equations   84 5.2.2 Element Types  84 5.3 Academic Example  86 5.3.1 Simply-supported Plate under Uniform Loading   86 5.4 Application  96 5.4.1 Stress in Shelf Angle Bracket   96

Chapter 6 Three-Dimensional Solid Analysis 6.1

Basic Equations   100 6.1.1 Differential Equations   100 6.1.2 Related Equations   100 6.2 Finite Element Method   101 6.2.1 Finite Element Equations   101 6.2.2 Element Types  102 6.3 Academic Example  105 6.3.1 Simple 3D Solid Problem   105 6.4 Application  114 6.4.1 Stress in Aircraft Structural Component   114

99

Contents

Chapter 7 Vibration Analysis

vii  117

7.1

Basic Equations   118 7.1.1 Differential Equations   118 7.1.2 Related Equations   121 7.2 Finite Element Method   123 7.2.1 Finite Element Equations   123 7.2.2 Element Types  123 7.3 Academic Example  124 7.3.1 Vibration of Thin Plate   124 7.4 Application  132 7.4.1 Modal Analysis of Passenger Car Frame   132

Chapter 8 Failure Analysis

135

8.1 Buckling  136 8.1.1 Fundamentals  136 8.1.2 Academic Example  138 8.1.3 Application  147 8.2 Fatigue and Life Prediction   150 8.2.1 Fundamentals  150 8.2.2 Academic Example  153 8.2.3 Application  162

Chapter 9 Heat Transfer Analysis

165

9.1

Basic Equations   166 9.1.1 Differential Equation   166 9.1.2 Related Equations   167 9.2 Finite Element Method   168 9.2.1 Finite Element Equations   168 9.2.2 Element Types  168 9.3 Academic Example  171 9.3.1 Plate with Specified Edge Temperatures   171 9.4 Application  179 9.4.1 Three-dimensional Heat Transfer Through Fins   179

Chapter 10 Thermal Stress Analysis 10.1 Basic Equations   186 10.1.1 Differential Equations   186 10.1.2 Related Equations   187 10.2 Finite Element Method   188 10.2.1 Finite Element Equations   188 10.2.2 Element Types  189 10.3 Academic Example  190

185

viii 

Contents

10.3.1 Thermal Stress Analysis of Thin Plate   190 10.4 Application  202 10.4.1 Thermal Stress in Combustion Engine Cylinder   202

Chapter 11 Incompressible Flow Analysis

207

11.1 Basic Equations   208 11.1.1 Differential Equations   209 11.1.2 Solution Approach  209 11.2 Finite Volume Method  209 11.2.1 Finite Volume Equations   210 11.2.2 SIMPLE Method  211 11.3 Academic Example  212 11.3.1 Lid-Driven Cavity Flow   212 11.3.2 Flow past Cylinder in Channel   222 11.4 Application  227 11.4.1 Flow in Piping System   227

Chapter 12 Compressible Flow Analysis

231

12.1 Basic Equations   232 12.1.1 Differential Equations   232 12.1.2 Related Equations   233 12.2 Finite Volume Method  234 12.2.1 Finite Volume Equations   234 12.2.2 Computational Procedure  236 12.3 Academic Example  238 12.3.1 Mach 3 Flow over Inclined Plane   238 12.3.2 Mach 3 Flow over Cylinder   251 12.4 Application  255 12.4.1 Flow over Shuttle Nose and Cockpit   255 Bibliography  259 Index 263

Chapter 1 Introduction

1.1 Solving Engineering Problems Computer-Aided Engineering (CAE) has played an important role in engineering design and analysis. Designers and engineers nowadays use CAE software packages to improve their product quality. The software packages help reducing designed time and material consumption while increasing the product strength and life time. Trial-and-error process, based solely on intuition of designers and engineers, is minimized or eliminated. Most of CAE software packages employ the finite element and finite volume methods to provide design and analysis solutions. These methods are based on engineering mathematics together with the application of numerical methods. The output numerical solutions are converted and displayed graphically so that

2

Chapter 1 Introduction

the simulated results can be understood easily. Without knowing how the software solves the problem, it is difficult for new users to be confident with the validity of output solutions. Mathematics and engineering governing equations embedded in these CAE software packages represent the nature of the problem being considered. As an example of fluid flow problem, mass and momentums must be conserved at any location in the flow domain. Such conservations are expressed in form of partial differential equations that are taught in fluid flow courses. This means users should have some background in mathematics together with the understanding of their physical meanings. By employing the finite volume method, these partial differential equations are transformed into a large set of algebraic equations. A computer program is developed to solve these algebraic equations for the flow solutions. The computed solutions are displayed as color graphics on computer screen. Similarly, users need to understand the equilibrium equations before analyzing a structural problem. These equilibrium equations are again in form of the partial differential equations as seen in many solid mechanics textbooks. The finite element method transforms these differential equations into their corresponding algebraic equations. A computer program is developed to solve such algebraic equations for the deformed shape and stresses that occur in the structure. The explanation above indicates that users should have backgrounds in mathematics and physics of the problem being solved. Users are also needed to understand the finite element/volume method prior to use any CAE software package. They can then convince themselves on the solutions generated by the software. This is one of the main reasons that most universities are offering the finite element/volume method courses to engineering students. 1.1.1 Problem Ingredients Solutions to an engineering problem depend on the three components:

1.1 Solving Engineering Problems

3

(a) Differential Equations. The differential equations interpret and model physical behavior of the problem into mathematical functions. For example, if we would like to determine temperature distribution of a ceramic cup containing hot coffee, we need to solve the differential equation that describes the conservation of energy at any location on the cup. The differential equation contains partial derivative terms representing conduction heat transfer inside the cup material. Such differential equation is not easy to solve using analytical approaches. (b) Boundary Conditions. The temperature distribution on the cup depends on the coffee temperature inside the cup and the surrounding ambient temperature outside the cup surface. Different boundary conditions thus affect the cup temperature solution. (c) Geometry. Cup shapes also affect their temperature distribution, even though they are made from the same material and placed under the same boundary conditions. The cup temperature changes if the cup is larger or thicker. The three components above always affect the solutions of the problem being solved. In undergraduate classes, we learned how to solve simplified forms of differential equations subjected to simple boundary conditions on plain geometries to obtain exact or analytical solutions. For real-life practical problems, they are governed by coupled differential equations which are quite sophisticated. Their boundary conditions and geometries are complicated. Numerical methods such as the finite element and finite volume methods are employed to provide accurate approximated solutions. 1.1.2 Solution Methods Methods for finding solutions can be categorized into two types: (a) Analytical Method. The analytical method herein refers to a mathematical technique used to find an exact or analytical solution for a given problem. The technique can provide

4

Chapter 1 Introduction

solutions only for simple problems as taught in undergraduate courses where differential equations, boundary conditions and geometries are not complicated. Most problems are limited to one dimensional problems so that their governing equations can be simplified from partial to ordinary differential equations. (b) Numerical Method. If the differential equations, boundary conditions and geometry of a given problem are complicated, solving with analytical method is not feasible. We need to find an approximate solution from a numerical method. There are many numerical techniques for finding solutions to complex problems. The popular techniques widely used are the finite element and finite volume methods. This is mainly because both techniques can handle problems with complex geometry effectively. Both the finite element and finite volume methods transform the governing differential equations into algebraic equations. In the process, many numerical techniques are needed. The techniques include solving a large set of algebraic equations, understanding concepts of the interpolation functions, determining derivatives and integrations of functions numerically, etc. Details of these techniques are taught in undergraduate numerical method courses and can be found in many introductory numerical method textbooks.

1.2 Finite Element Method Because most of CAE commercial software packages employ the finite element method to solve for solutions, we will introduce the method in this section. 1.2.1 What is the Finite Element Method? The finite element method is a numerical technique for finding approximated solutions of problems in science and engineering. These problems are governed by the three components including differential equations, boundary conditions and geometries.

1.3 ANSYS Software

5

The method starts by dividing the problem domain or geometry into a number of small elements. These elements are connected via nodes where the unknowns are to be determined. The finite element equations for each element are derived from the governing differential equations describing physics. These finite element equations are assembled into a large set of algebraic equations. The boundary conditions are then imposed to the set of algebraic equations to solve for solutions at each node. We will understand the procedure of the finite element method in details in the following section. 1.2.2 Finite Element Method Procedure The finite element method procedure generally consists of 6 steps: Step 1: The first step is to construct the domain geometry of the given problem. The geometry may consist of straight lines, curves, circles, surfaces or solid shapes in three dimensions. Different software packages have their unique ways to create geometry. Users may have to spend some times to familiarize with the software. A finite element mesh is then generated on the constructed geometry. Depending on the complexity of the geometry, a mesh may consist of various element types such as line, triangular or brick element. These elements are connected at nodes for which the problem unknowns are located. Step 2: The second step is to select the element types. For examples, a line element may consist of two or three nodes, or a triangular element may have three or six nodes. The number of element nodes affects the interpolation functions used in that element. Selecting an element with more nodes will increase the number of unknowns and thus the computational time. However, the solution accuracy can also increase when a more complicated interpolation function is used. Step 3: The third step is the most important step of the finite element method. This step is the derivation of the finite element

6

Chapter 1 Introduction

equations from the governing differential equations. The derived finite element equations are in the form of algebraic equations that can be computed numerically. The transformation process must be carried out correctly so that the derived algebraic equations can yield accurate solutions. Step 4: The finite element equations from all elements are then assembled to become a large set of algebraic equations. Assembling element equations must be done properly. This is similar to placing jigsaw pieces at appropriate locations to yield the complete picture. Step 5: The boundary conditions of the problem are then imposed on the set of algebraic equations before solving for the nodal unknowns. The nodal unknowns are the displacements for structural problem and are the temperatures for heat transfer problem. Step 6: Other quantities of interest can then be solved. For structural problem, stresses in the structure can be determined after the displacements are known. For heat transfer problem, heat fluxes can be computed once the nodal temperatures are obtained. The six steps above indicate that the method is quite general and suitable for a large class of problems in science and engineering. The three problem ingredients which are the differential equations, boundary conditions and geometry are handled in the third, fifth and first step of the method, respectively.

1.3 ANSYS Software ANSYS software was first developed in 1970 by John Swanson who was an engineer at Westinghouse Astronuclear Laboratory. The software was originally for stress analysis of nuclear reactor components. He later founded Swanson Analysis System, which was named as ANSYS, Inc. His ANSYS software then became an industry leading finite element program for analyzing engineering problems and optimizing products. At the

1.1 Solving Engineering Problems

7

same time, NASTRAN (NAsa STRuctural ANalysis program) was also popular and being used by NASA Engineers. I remembered Dr. Swanson came to NASA Langley Research Center, Hampton, Virginia to promote his software while I was an engineer there. He gave coffee cups with the early yellow/black ANSYS logo to NASA engineers working in the CAE department. Nowadays, ANSYS is a software widely used all over the world for analyzing a large class of problems in many fields. This is mainly because the software is easy to learn and use. Various problems can be solved conveniently while solutions are displayed graphically on the computer screen. 1.3.1 ANSYS Workbench In the early days of ANSYS development, the Disk Operating System (DOS) was the most widely used operating system on computers. ANSYS users needed to type long and specific commands through keyboards. These commands were required to construct model geometry, such as lines, arcs, surfaces, volumes, etc. Various commands were also needed to create meshes, apply boundary conditions and execute the problem for solutions. Using the software for analyzing a problem at that time was not convenient at all. Development of Windows environment has provided the ease of using the software. With mouse and keyboard, users can interact with the software graphically. Lately, ANSYS has introduced the Workbench function which further simplifies the use of the software via Graphic User Interface (GUI). The ANSYS Workbench is employed to solve various types of problems presented as examples in this book. 1.3.2 Screen and Tool Bars The starting workbench window consists of the menu and tool bars at the top. The large two areas below are the Toolbox and Project Schematic regions as shown in the figure.

8

Chapter 1 Introduction

File View Tools Units Help

The frequently used menu items are: Create a new file, open an existing file, save the current file, import existing model, etc. Arrange the window layout, customize the toolbox, etc. Set the license preference, select options of appearance, languages, graphics interaction, etc. Select unit systems, define user’s units, etc. Get Help from ANSYS. The toolbox region on the left side of the screen contains numerous systems. These include the analysis, component, custom and external connection systems with design exploration. The analysis system consists of several tools for solving different classes of problems such as static and transient structural analyses, buckling and modal analyses, steady-state and transient thermal analyses, fluid flow analysis. These analytical tools are shown in the figure.

1.3 ANSYS Software

9

The project schematic region on the right side of the screen is the working area. This larger region is for the user to view what is happening at different stages starting from creating geometry domain, discretizing domain into a number of small elements, applying boundary conditions, solving for solutions and displaying results. 1.3.3 Analyzing steps The analyzing steps via the Workbench follow the standard finite element procedure. As an example of analyzing a static structural problem, we double-click at the Static Structural under the Analysis Systems in the Toolbox window. A small cell of Static Structural will appear in the Project Schematic window as shown in the figure.

The cell consists of seven items as follows: 1. Static Structural Perform static analysis of a structure. 2. Engineering Data Provide engineering data associated with the problem, such as the material modulus of elasticity, Poisson’s ratio, coefficient of thermal expansion, etc.

10

Chapter 1 Introduction

3. Geometry

Create model geometry of the problem by constructing lines, arcs, circles, surfaces, etc. This step is normally time consuming especially for complex configuration. An imported CAD model file could help reducing the effort.

4. Model

Assign materials and generate a mesh by discretizing the model into a number of small elements. The process is performed automatically.

5. Setup

Specify boundary conditions such as the constraints and loadings, as well as some specific analysis settings.

6. Solution

Solve the problem for solutions. This step is executed automatically if the information provided in the preceding steps is complete.

7. Results

Display solutions in different forms, such as color contours, vectors and surface plots.

The check mark symbol ( ) will appear on the right side of the step if that step has been carried out correctly. ANSYS Workbench uses different symbols to explain status of the step as follows: Nothing is done because upstream data is not available. Refresh is needed since upstream data has changed. Attention is required. User interaction is needed. Update is required because upstream data was modified. Everything is OK. Solution is interrupted. Need correction to resume action. Solution is in progress.

1.4 Advantages of Finite Element Method

11

We will follow the above procedure, step by step, to analyze different types of problems in the following chapters. These include structural, heat transfer and fluid flow problems using one-, two- and three-dimensional finite element models. We will find that, if we performed each step correctly, the ANSYS Workbench will show the check mark symbol ( ) on the right side of the step. But if we see other symbols, we need to go back and fix that step before moving on. The process thus ensures us that everything has been done appropriately before obtaining the final solutions.

1.4 Advantages of Finite Element Method The finite element method is popular and widely used by scientists and engineers all over the world for analyzing various types of problems. Examples of problems are as follows. (a) Stress analyses of large-scale structures such as bridges, ships, trains, aircrafts, automobiles and buildings. Structural analysis for small-scale products are such as automotive and electronic parts, furniture, machine equipment, etc. (b) Vibration and dynamic analyses of high-voltage power transmission towers, expressway signs under strong wind, crash simulation of automobiles, turbine blades operating under high pressure and temperature, etc. (c) Fluid analyses of air flows over cities, air ventilation in large halls, inside offices, cleanrooms, computer cases, etc. (d) Electromagnetic analyses around power transmission lines, electric motors, sensitive electronic devices, etc. (e) Bio-mechanic analyses of blood flow in human hearts and veins, artificial joints and bones, etc.

12

Chapter 1 Introduction (f) Analyses of other problems in which their experiments are dangerous to human or too costly for multiple tests, such as hazardous chemical reaction in gas chambers, prediction of bomb explosion phenomena, flow field around hypersonic aerospace plane, etc.

Advantages of the finite element method as highlighted above have led to many commercial software packages. Users of these software packages must have good background and understanding of the method prior to using them. Basic mathematical theories and the finite element method for structural, heat transfer and fluid flow analyses will be presented in the following chapters with examples. Understanding materials in these chapters is encouraged before using the ANSYS Workbench with confidence.

Chapter 2 Truss Analysis

Analysis of truss structures is normally used as the first step toward understanding the finite element method. The analysis is simple because the truss (rod or spring) element contains only a displacement unknown in its axial direction at each node. The finite element equations are easy to derive and problems with few elements can be solved by hands.

2.1 Basic Equations 2.1.1 Differential Equation A one-dimensional equilibrium equation, in the xdirection of a truss member without the inclusion of its body force, is governed by the equilibrium equation,

14

Chapter 2 Truss Analysis

 x x

 0

where  x is the truss axial stress. 2.1.2 Related Equations The truss stress varies with the strain  x by the Hook’s law,

 x  E x where E is the modulus of elasticity or Young’s modulus. The strain  x is related to the displacement according to the small deformation theory as,

x 

u x

where u  u ( x) is the displacement that varies with the distance x along the length of the truss member. Thus, the stress can be written in form of the displacement as,

x  E

u x

The governing differential equation, for the case of constant Young’s modulus, becomes,

E

 2u x2

 0

For a truss member that lies only in the x-direction, the displacement distribution u  u ( x) can be derived from the differential equation above. This is done by performing integrations twice and applying the problem boundary conditions. The stress of the truss member can be then determined. However, if the problem contains many truss members oriented in three dimensions, it is not easy to determine their deformed shape and member stresses. The finite element method offers a convenient way to find the solution as explained in the following section.

2.2 Finite Element Method

15

2.2 Finite Element Method 2.2.1 Finite Element Equations Finite element equations can be derived from the governing differential equation by using the Method of Weighted Residuals (MWR). The idea of the method is to transform the differential equation into the corresponding algebraic equations by requiring that the error is minimum. These algebraic equations consist of numerical operations of addition, subtraction, multiplication and division. Such operations allow the use of calculators to determine solutions for small problems. For larger problems, a computer program must be developed and employed. The derived finite element equations are normally written in matrix form so that they can be used in computer programming easily. The finite element equations for the truss element are,

 K u  F  where  K  is the element stiffness matrix; u is the column matrix or vector that consists the nodal displacement unknowns; and  F  is the column matrix or vector that contains the nodal loads. These matrices depend on the element types used as explained in the following section. 2.2.2 Element Types The standard two-node truss element is shown in the figure. The element lies in the x-coordinate direction and consists of a node at each end. The element length is L with the crosssectional area of A and made from a material that has the Young’s modulus of E . At an equilibrium condition, the forces at node 1 and 2 are F1 and F2 , causing the displacements of u1 and u 2 in its axial direction, respectively.

16

Chapter 2 Truss Analysis

u2

u1 F1

1

2

A, E

F2

x L The displacement distribution is assumed to vary linearly along the element axial x-direction in the form, u  u ( x)  N1 ( x) u1  N2 ( x) u2   N1 ( x) N2 ( x)   1  u2    N ( x )  u (1 2)

u ( x)

u2

u1

where N1 ( x ) and N2 ( x) are the element interpolation functions. For this two-node element, they are,

N1 ( x )  1 

2

1 x

L

(21)

and

N2 ( x) 

x L

x L

A truss element may contain more than two nodes. As an example, the three-node truss element, as shown in the figure, assumes the displacement distribution in the form, u ( x)

u3

u ( x)  N1 ( x) u1  N2 ( x) u2  N3 ( x) u3

u1

1

u2

2

3

 u1    N1 ( x) N2 ( x) N3 ( x)  u2    u3    N ( x )  u 

xL L (13) (31) 2 2 where N1 ( x ) , N2 ( x) และ N3 ( x) are the interpolation functions expressed by,

17

2.3 Academic Example

N1 ( x )  1 

3 x 2 x2  2 L L

;

N2 ( x) 

4 x 4 x2  2 L L

;

x 2 x2  L L2 The assumed displacement distribution of the threenode element is more complicated than that of the two-node element. Thus, the three-node element can provide higher solution accuracy. However, the element requires more computational time because it contains more nodal unknowns. The finite element equations for the two-node element are, AE  1 1  u1   F1         L  1 1  u2   F2  If we have a finite element model consisting of 10 elements, we need to establish 10 sets of finite element equations. These element equations are then assembled to form up a system of equations. The problem boundary conditions are applied before solving for the displacement unknowns at nodes. If a finite element model containing many truss elements is in two or three dimensions, the finite element equations above are needed to transform into to two or three dimensions accordingly. The transformation causes the finite element matrices to become larger leading to a larger set of algebraic equations. Such the larger set of algebraic equations requires more computer memory and computational time. However, these requirements do not pose any difficulty to current computers. Commercial software packages today have been developed to analyze complex truss structures containing a large number of elements effectively.

N3 ( x )  

2.3 Academic Example 2.3.1 Two Truss Members in One Dimension A model with two truss members connected together in one dimension is shown in the figure. The two truss members have

Chapter 2 Truss Analysis

18

the lengths of 0.5 and 1.0 m, cross-sectional areas of .002 and .001 m2, and made from materials with Young’s modulus of 5 107 and 10 107 N m 2 , respectively. The left end of the model is fixed at a wall while the connecting point and the right end are subjected to the forces of 300 and 500 N, respectively. By using only one twonode element to represent each truss member, determine the deformed configuration and the truss member stresses.

A1  .002 m 2

A2  .001 m 2

E1  5 10 7 N m 2

E2  10  10 7 N m 2

2

1 1 0.5 m

3

300 N

500 N

x

2 1.0 m

We will employ the ANSYS Workbench to analyze this problem by going through the steps in details as follows.

(a) Starting ANSYS Workbench    



Open the ANSYS Workbench, set the Units menu on the upper tab to Metric (kg,m,s,C,A,N,V). On the Analysis Systems window, click twice on the Static Structural item. A new small box will appear on the Project Schematic window. Retype the name in the lower blue tab as the desired project name, e.g., 1D Truss Problem, and hit Enter. Right click on the Engineering Data tab and select the Edit… option, the A2: Engineering Data window will appear. Double click on Click here to add a new material and type in a new material name, e.g., “Material 1”, and hit Enter. Click at the Isotropic Elasticity under Linear Elastic and drag it to the Property list at the bottom of the window. Enter the Young’s Modulus value as 5e7 and hit Enter,

2.3 Academic Example

 

19

enter the Poisson Ratio value as 0.3 and hit Enter, and close this window. Repeat the same process to provide data of the second material with the Young’s Modulus of 10e7 and Poisson Ratio of 0.3 and assign the name as “Material 2”. Then, close the Engineering Data tab and click at the Project tab on the upper menu, it will bring back to the main Project Schematic window.

(b) Creating Geometry



  

Right click on the Geometry tab and select the Properties option, the Properties of Schematic window will open. Select the Line Bodies under the Basic Geometry Options. Then, close this small window. Right click on the Geometry tab and select the New Geometry…. This will launch the ANSYS Design Modeler (green logo DM). On DM window, set unit in the Units menu on the upper tab to Meter. On the Tree Outline window, right click on XYPlane and select Look at. The X-Y-Z coordinates on the Model View in 3D view will become X-Y coordinates in 2D view.

20

Chapter 2 Truss Analysis

 





Select the Sketching tab below the Tree Outline window, the Sketching Toolboxes will appear in the same place. Select the Settings tab and then Grid, activate the buttons Show in 2D and Snap. The grid will appear on the main window. Grid snapping provides convenience for drawing model geometry. Change the Major Grid Spacing to 1 m and hit Enter, Minor-Steps per Major to 2 and hit Enter, and Snaps per Minor to 1 and hit Enter.

Enlarge the scale by clicking at the Box Zoom icon on the upper part of the screen (icon with plus sign on the magnifying glass) and draw a box with appropriate size to

2.3 Academic Example

21

zoom in. Click it again after appropriate scale is shown on the window. The model can be moved around using the Pan icon, the four arrows icon on the upper part of the screen.



 

  

Click on Modeling tab, and then click the New Sketch icon (a small blue geometry symbol with * on the upper part of the DM window) to create Sketch1 which will appear under XYPlane. Note that this name Sketch1 can be deleted or renamed by right clicking on it and selecting an option. Click on Sketch1 to start drawing the first line for the left truss element. Click the Sketching tab and select Draw. Choose Line to create the first line with the end coordinates of (0,0) and (0.5,0). This is done by first clicking at the coordinates of (0,0), move the cursor to the coordinates of (0.5,0), and click the mouse again. Click on Generate (the icon with yellow lightning on the upper-left part of the screen). The first line will become dark green. If the model contains many lines that have same material property and cross-sectional area, the same process can be used to create the additional lines. The next important step is to click the Concept tab on top of the screen and select Lines From Sketches. Then, select Sketch1, this line will become yellow.

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Chapter 2 Truss Analysis











Click Apply icon on the right side of the Base Objects tab in the Details View window at the lower left of the screen. The line will become cyan, then click on Generate. The right side of the Base Objects tab will show 1 Sketch. The 1 Part, 1 Body item will appear in the Tree Outline window. To draw the second line representing the right element, select New Sketch, the item Sketch 2 will pop up beneath Sketch 1. The second line that connects between the coordinates of (0.5,0) and (1.5,0) can be drawn using the same process. Then, select Concept and Lines From Sketches. Select Sketch 2, this second line will become yellow. Before clicking on Apply button, be sure to change Add Material on the right side of Operation in the Details View window to Add Frozen. Without doing this, by default, the two lines would become a single line and only one material property is allowed. Click on Base Objects again, the Apply button appears. Select Sketch 2 and click Apply, the second line will become cyan. The 2 Parts, 2 Bodies will appear in the Tree Outline window.

Double-click on 2 Parts, 2 Bodies, the two items under the same name of Line Body appear beneath it. Note that these two line bodies are not connected yet. To connect them

2.3 Academic Example

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

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23

together, hold Ctrl key and select both Line Body. Then, right click and select Form New Part. The two line bodies will be connected to become 1 Part, 2 Bodies. At this point, we now have a model consisting of two lines. Next step is to create the cross sections of the two lines. For the left line, select Rectangular item in Cross Section under the Concept tab. In the Details of Rect1 window, change the base value B to 0.05 and hit Enter, the height value H to 0.04 and hit Enter. A blue rectangular cross section will appear on the main Graphic window. Then, click Generate.

Repeat the same process to create the cross section of the right line by selecting Rectangular in Cross Section under Concept tab. In the Details of Rect2 window, change the base B value to 0.04 and the height H value to 0.025. Next, assign the cross sections Rect1 and Rect2 to the two line bodies. Double click at 1 Part, 2 Bodies and select the first Line Body, assign Rect1 to the Cross section selection in the Details of Line Body window. Similarly, select the second Line Body, assign Rect2 to the Cross section selection in the Details of Line Body window. Save file as 1D Truss Problem through the File button at the upper left of the screen, and close the DM window.

24

Chapter 2 Truss Analysis

(c) Assigning Material Properties and Creating Mesh

 

On the main Project Schematic window, double click on Model, the truss model will appear back on the main window. Double click on Geometry cell, the Part 1 item will pop up. Click on this Part 1, two lines of Line Body will pop up. Click on the first Line Body and select Material 1 (the name assigned earlier containing material properties of left element) which is on the right-hand-side of Assignment under Material in the Details of “Line Body” window. The left line will become green.

2.3 Academic Example  



25

Repeat the same process to assign Material 2 containing material properties of the right element to the second Line Body. Select Mesh under Model section, the Details of “Mesh” window will appear on the lower left of the screen. Select Element Size under Sizing and change the value on the right column to 10 and hit Enter. This input value of 10 m is to ensure that each line body is modelled by using only 1 element. Right click at the Mesh again and select Generate Mesh. A finite element mesh with only two truss elements will appear.

Save the project and close the DM window.

(d) Applying Boundary Conditions, Solving for and Displaying Solutions  On the main Project Schematic window, double click on Setup, the truss model will appear back on the main window.  We first apply the fixed boundary condition at the left end by selecting Analysis Settings under Static Structural. Click on the Support tab on the upper menu bar with Fixed Support option, then select Vertex icon (box with arrow and green dot). Move the cursor to the center of the left end

26

Chapter 2 Truss Analysis

edge and click at it, a small green spot will appear. Then, click Apply button next to the Geometry button under the Details of “Fixed Supports” window.





Repeat the same process to constrain the displacements in the Y- and Z-directions of the middle and right nodes. This is done by selecting Analysis Setting under Static Structural. Select Displacement under Supports tab, hold Ctrl key and click at the middle and right nodes. Then, click Apply and change the value of Displacement to 0 only for Y and Z Component. To apply the boundary condition of the force at the middle node, click on the Analysis Settings, select the Loads tab on the upper menu bar with Force option, and select Vertex icon. Move the cursor to the middle node and click at it. Select Components option on the right side of Define By in the Details of “Force” window. Click Apply button, and input X Component as 300 and hit Enter.

2.3 Academic Example

27



Similarly, the boundary condition for applying force of 500 N at the right node can be performed in the same way.



For truss analysis, all the nodes must be the hinge type which is free to rotate. Click Tools at the upper tab and select Options…. Select Connections and change the right side of Fixed Joints to No and click OK. The problem is now ready to solve for solution. Right click on the Solution item under Static Structural and select the Solve tab. Click the Solution item, the Deformation tab will appear on the lower menu bar. Click on this Deformation tab and Select the Total option, the Total Deformation item will pop-up beneath the Solution item. Right click at the Total Deformation item and select Evaluate All Results, the displacement solution in form of color fringe plot will appear. Note that the node and element numbers can be displayed by clicking at the Preferences tab on the upper part of the screen. Then, select Node Numbers and Element Numbers under Mesh Display and click Enter. Nodal displacement values can also be exported as a text file by right clicking on Total Deformation or Directional Deformation. Then, select Export… and Export Text File, respectively.

 





28

Chapter 2 Truss Analysis



 

Solution of displacement values can be displayed by selecting the Probe tab on the upper bar of the screen. Move the mouse to the desired location and click at it, the corresponding displacement value will appear. To display element stresses, right click on Solution and select Insert, Beam Tool and Beam Tool, respectively, a Beam Tool item will appear beneath Solution item. Right click on this Beam Tool item, select Insert, Beam Tool, Stress and Direct Stress. Select Unaveraged on the right side of Display Option in the Details of “Direct Stress 2” box. Right click on Direct Stress 2 item and select Evaluate All Results, the uniform element stresses will be displayed.

29

2.3 Academic Example 2.3.2 Two Truss Members in Two Dimensions

A two-dimensional truss model consisting of only two members is shown in the figure. The two members have the same cross-sectional area of .0004 m2 and made from the same material with the Young’s modulus of 7 1010 N m 2 . The lower right hinge is subjected to a horizontal force of 500 N pulling to the right and a downward force of 2500 N. Each member is modelled by a two-node truss element. We will use ANSYS Workbench to solve for the deformed shape of the truss structure and the stresses that occur in each member. Y A  .02  .02  .0004 m 2

E  7 1010 N m 2 1m 2500 N 500 N X 1m When using ANSYS to solve a truss structure, it is important to keep in mind that the truss element allows only axial loading. The connection between truss members must be hinge type which is free to rotate. The truss element is different from the beam element that allows bending. We will use the beam elements to analyze frame structures in the following chapter. (a) Starting ANSYS Workbench



Open the ANSYS Workbench, set the Units menu on the upper tab to Metric (kg,m,s,C,A,N,V).

30

Chapter 2 Truss Analysis

  



On the Analysis Systems window, click twice on the Static Structural item. A new small box will appear on the Project Schematic window. Retype the name in the lower blue tab as the desired project name, e.g., Two Truss Element Problem, and hit Enter. Right click on the Engineering Data tab and select the Edit… option, the A2: Engineering Data window will pop-up. Double click on Click here to add a new material and type in a new material name, e.g., “My Aluminum Material”, and hit Enter. Click at the Isotropic Elasticity under Linear Elastic and drag it to the Property list at the bottom of the window. Enter the Young’s Modulus value as 7e10 and hit Enter, Enter the Poisson Ratio value as 0.3 and hit Enter, and close this window. Then, close the Engineering Data tab and click at the Project tab on the upper menu, it will bring back to the main Project Schematic window.

(b) Creating Geometry



Right click on the Geometry tab and select the Properties option, the Properties of Schematic window will open. Select the Line Bodies under the Basic Geometry Options. Then, close this small window.

2.3 Academic Example

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31

Right click on the Geometry tab and select the New This will launch the ANSYS Design Geometry…. Modeler (green logo DM). On DM window, set unit in the Units menu on the upper tab to Meter.

On the Tree Outline window, right click on XYPlane and select Look at. The X-Y-Z coordinates on the Model View in 3D view will become X-Y coordinates in 2D view. Select the Sketching tab below the Tree Outline window, the Sketching Toolboxes will pop-up in the same place. Select the Settings tab and then Grid, activate the buttons Show in 2D and Snap. The grid will appear on the main window. Grid snapping provides convenience for drawing model. Change the Major Grid Spacing to 1 m and hit Enter, Minor-Steps per Major to 2 and hit Enter, and Snaps per Minor to 1 and hit Enter. Enlarge the scale by clicking at the Box Zoom icon on the upper part of the screen (icon with plus sign on the magnifying glass) and draw a box with appropriate size to zoom in. Click it again after appropriate scale is shown on the window. The model can be moved around using the Pan icon, the four arrows icon on the upper part of the screen.

32

Chapter 2 Truss Analysis



 

  



Click on Modeling tab, and then click the New Sketch icon (a small blue geometry symbol with * on the upper part of the DM window) to create Sketch1 which will appear under XYPlane. Note that this name Sketch1 can be deleted or renamed by right clicking on it and selecting an option. Click on Sketch1 to start drawing the two truss elements. Click the Sketching tab and select Draw. Choose Line to create the first line with the end coordinates of (0,0) and (1,0). This is done by first clicking at the coordinates of (0,0) on the model, move the cursor to the coordinates of (1,0), and click the mouse again. Then, follow the same procedure to create the second line with the end coordinates of (0,1) and (1,0). Click on Generate (the icon with yellow lightning on the upper-left part of the screen). The desired two lines will become dark green. The next important step is to go to the Concept tab on top of the screen and select Lines From Sketches. Select the Sketch1, the two lines will become yellow. Click Apply icon on the right side of the Base Objects tab in the Details View at the lower left of the screen. Both two lines will become cyan. Then, click on Generate. The right side of the Base Objects tab will show 1 Sketch. The 1 Part, 1 Body item will appear in the Tree Outline window. We now have a model consisting of two lines.

2.3 Academic Example

33



Next step is to create the truss cross section. Select Rectangular item in Cross Section under the Concept tab. In the Details of Rect1 window, change the base value B to 0.02 and hit Enter, the height value H to 0.02 and hit Enter. A blue rectangular cross section will appear on the main Graphic window. Then, click Generate.



Next, assign this cross section to the Line Body. Double click at 1 Part, 1 Body and select the Line Body, assign Rect1 to the Cross section selection in the Details of Line Body window. Save file as Two Truss Element Problem, and close the DM window.



34

Chapter 2 Truss Analysis

(c) Assigning Material Properties and Creating Mesh



On the main Project Schematic window, double click on Model, the truss model will appear back on the main window. Click on the Z arrow head to display the model in 2D.



Double click on Geometry item, the Line Body item will pop-up. Select the Line Body item and select “My Aluminum Material” (the name assigned earlier containing material properties of this problem) which is on the righthand-side of Assignment under Material in Details of “Line Body” window. The truss model will become green.



Select Mesh under Model section, the Details of “Mesh” window will appear on the lower left of the screen. Select at Element Size under Sizing and change the value on the right column to 10 and hit Enter. This input value of 10 m. is to ensure that each truss is modelled by only 1 element. Right click at the Mesh again and select Generate Mesh. A finite element mesh with only two truss elements will appear.



Save the project and close the DM window.

2.3 Academic Example

35

(d) Applying Boundary Conditions, Solving for and Displaying Solutions



 

On the main Project Schematic window, double click on Setup, the truss model will appear back on the main window. Click on Show Mesh tab on the upper tool bar to display the mesh. Next, the boundary conditions of displacement constraint at the two left ends can be applied. This will be done, one at a time, starting from the lower left end. Select Analysis Settings under Static Structural. Select the Supports tab on the upper menu bar with Displacement option, then select Vertex icon (box with arrow and green dot). Move the cursor to the lower left end and click at it, the left end will become green. Click Apply button next to the Geometry button under the Details of “Displacement” window. Change X, Y and Z Component to Constant as 0.

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Chapter 2 Truss Analysis

 







Repeat the same process to apply boundary condition of zero displacement for X, Y and Z Components at the upper left end. Repeat the same process to apply boundary condition of zero displacement only for the Z Component at the lower right end. Note that the Z displacement must be zero for all nodes of the 2D planar truss problems. Repeat the similar process to apply boundary condition of the two applied forces at the right hinge by first selecting the Analysis Settings, select the Loads tab on the upper menu bar with Force option, and select Vertex icon. Move the cursor to the right hinge and click at it. Select Components option on the right side of Define By in the Details of Force window. Click Apply button, and input X Component as 500 and hit Enter, Y component as -2500 and hit Enter. Note that, mesh can be shown by clicking the Show Mesh icon on the upper menu bar.

This problem is a truss analysis, thus all the nodes must be the hinge type which are free to rotate. Click Tools at the upper tab and select Options…. Select Connections and change the right side of Fixed Joints to No and click OK. The problem is now ready to solve for solution. Right click the Solution item and under Static Structural and select the Solve tab.

2.3 Academic Example

37



Click the Solution item, the Deformation tab will appear on the lower menu bar. Click on this Deformation tab and Select the Total option, the Total Deformation item will pop-up beneath the Solution item. Right click at the Total Deformation item and select Evaluate All Results, the solution in form of color fringe plot will appear as shown in the figures as isometric view.



Note that the node and element numbers can be displayed by clicking at the Preferences tab on the upper part of the screen. Then, select Node Numbers and Element Numbers under Mesh Display and click Enter. To show displacement components such as the Y-displacement, click Deformation tab and select Directional. The Directional Deformation will appear beneath the Solution item. Right click at Directional Deformation and change X Axis on right side of Orientation in the Details of “Directional Deformation” box to Y Axis. Then, select Evaluate All Results, model deformation in the Ydirection will appear. Nodal displacement values can also be exported as a text file by right clicking on Total Deformation or Directional Deformation. Then, select Export… and Export Text File, respectively.





38

Chapter 2 Truss Analysis



 

Solution of Displacement values can be displayed on the screen by selecting the Probe tab on the upper bar of the screen. Move the mouse to the desired location and click at it, the corresponding displacement value will appear. To display element stresses, right click on Solution and select Insert, Beam Tool and Beam Tool, respectively, a Beam Tool item will appear beneath Solution item. Right click on this Beam Tool item, select Insert, Beam Tool, Stress and Direct Stress. Select Unaveraged on right side of Display Option in the Details of “Direct Stress 2” box. Right click on Direct Stress 2 item and select Evaluate All Results, the element stresses will be displayed.

39

2.4 Application

2.4 Application 2.4.1 Twenty-one Truss Members in Two Dimensions The same process used for analyzing the two truss members in the preceding example can be employed to solve a more complicated truss structure. A truss structure, as shown in the figure, consists of 21 members. All members have the same crosssectional area of .01 m2 and made from a structural steel material with the Young’s modulus of 2  1011 N m 2 . The structure is supported by the hinges at the left and right ends, and is subjected to different vertical loadings as shown in the figure. A  .1 .1  .01 m 2 E 

2 1011

N

Y

m2

5m

1m

5m X

1000 N

2000 N

2000 N

1000 N

4000 N The finite element model consisting of 21 two-node truss elements can be constructed easily by using the Line command. The boundary conditions of fixed x- and y-displacements at the left and right end hinges and the concentrated forces at lower nodes of the model can be applied conveniently as shown in the figure.

40

Chapter 2 Truss Analysis

Once the problem has been set up, solutions of the deformed shape and stresses in the truss members can be obtained without difficulty as shown in the figures. The figures show the deformed shape and identify the truss members that have high stresses. Users can change the member cross-sectional areas or apply other types of boundary conditions to obtain different solutions. These solutions will increase understanding of the problem behaviors that change with the geometry and boundary conditions.

Chapter 3 Beam Analysis

Frame structures such as high-voltage power transmission towers, large bridges and tall buildings use beams to provide high strength at low weight. Most of the beam cross sections are in I, L, U, C, O and rectangular shapes with different areas and moments of inertia. In an undergraduate strength of materials course, analytical solutions of the deflection and stress of a single beam under simple loadings and boundary conditions are normally derived by using conventional approach. However, for a complicated frame structure with many beam members oriented in three dimensions, it is difficult to obtain solutions by following such approach. The finite element method through the use of a software can provide solutions effectively as explained in this chapter.

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Chapter 3 Beam Analysis

3.1 Basic Equations 3.1.1 Differential Equation A beam that lies in the x-direction with its cross section in the y-z plane is shown in the figure. The beam is subjected to a distributed load p( x) causing the deflection of w in the z-direction and the displacement of u in the x-direction. z

p ( x)

z

w u

x

y

If beam deflection is small, the small deformation theory stating that the plane sections before and after deflection remain plane is applied. This lead to the relation such that the displacement u can be written in form of the deflection w as u   z w x . In addition, if the beam is long and slender, the deflection w may be assumed to vary with x only, i.e., w  w( x) . These two assumptions yield to the equilibrium equation of the beam deflection as, 2  2 w  E I    p x 2  x 2  where E is the beam Young’s modulus, and I is the moment of inertia of the cross-sectional area. As an example, the moment of inertia of the rectangular cross section is I  b h3 12 where b and h is the width and height of the cross section, respectively. 3.1.2 Related Equations The stress  x along the axial x-coordinate of the beam varies with the strain  x according to the Hook’s law as,  x  E x

43

3.2 Finite Element Method Since the strain is related to the displacement and deflection as, u 2 w  z 2 x x Then, the stress can be determined from the deflection as,

x



2 w x2 For a typical beam in a three-dimensional frame structure, its deflection may be in a direction other than the zcoordinate. In addition, the beam may be twisted by torsion caused by the applied loads or affected by other members. These influences must be considered and included for the analysis of three dimensional beam structures.

x

 E z

3.2 Finite Element Method 3.2.1 Finite Element Equations Finite element equations can be derived directly from the beam governing differential equation by using the method of weighted residuals. Detailed derivation can be found in many finite element textbooks including the one written by the same author. The derived finite element equations are in the form,

 K     F  where  K  is the element stiffness matrix;   is the element vector containing nodal unknowns of deflections and slopes; and F  is the element vector containing nodal forces and moments. These element matrices depend on the selected beam element types as explained in the following section. 3.2.2 Element Types The basic beam bending element with two nodes is shown in the figure. Each node has two unknowns of the deflection w and slope  .

44

Chapter 3 Beam Analysis

1

w1 E, I

1 x

2

Distribution of the deflection w is assumed in the form,

w2 2

L

 w1      w( x)   N1 N2 N3 N4   1    N ( x )    (1 4) (41)  w2  2  where the element interpolation functions are, x 2 x 3 N1  1  3    2   L L

;

N2 

2 x x   1  L 

x 2 x N3     3  2  L L 

;

N4 

x2  x    1 L L 

These interpolation functions lead to the finite element equations as,  F1   6 3L 6 3L   w1   1          2 EI 3L 2 L2 3L L2   1   M1  p0 L  L 6         2  1  L3  6 3L 6 3L   w2   F2     L 6  3L L2 3L 2 L2  2   M 2  where F1 and F2 are the forces, while M1 and M2 are the moments, at node 1 and 2, respectively. The last vector contains the nodal forces and moments from the distributed load p0 which is uniform along the element length. The finite element equations above can be used to determine beam bending behavior. If a problem contains only few beam elements, we can use a calculator to solve for the solution. However, if a problem consists of many beam elements, we need to develop a computer program to solve for the solution instead. For a frame structure containing a large number of beam elements oriented in three dimensions, the element matrices as shown above

45

3.3 Academic Example

must be transformed into three dimensions too. The sizes of element matrices increase and the element equations now contain more unknowns. Developing a computer program is thus a must in order to analyze practical problems. We will use ANSYS through the Workbench to solve for beam solution behaviors. We will start with simple academic type example containing only few elements before analyzing a more realistic problem in three dimensions.

3.3 Academic Example 3.3.1 Two Beam Members in Two Dimensions A two-dimensional frame structure consisting of two members is shown in the figure. The two members have the same cross-sectional area of .0004 m2 and made from the same material with the Young’s modulus of 7 1010 N m 2 . The lower right end is subjected to a horizontal force of 500 N pulling to the right and a downward force of 2500 N. Each member is modelled by a twonode beam element. We will use ANSYS Workbench to solve for the deformation shape and the stresses that occur in the members.

Y

A  .02  .02  .0004 m 2 E  7  1010 N m 2

1m 2500 N 500 N 1m

X

46

Chapter 3 Beam Analysis

(a) Starting ANSYS Workbench 

Open the ANSYS Workbench, set the Units menu on the upper tab to Metric (kg,m,s,C,A,N,V).



On the Analysis Systems window, click twice on the Static Structural item. A new small box will appear on the Project Schematic window.

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Retype the name in the lower blue tab as the desired project name, e.g., Two Beam Element Problem, and hit Enter.

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Right click on the Engineering Data tab and select the Edit… option, the A2: Engineering Data window will pop-up. Double click on Click here to add a new material and type in a new material name, e.g., “My Aluminum Material”, and hit Enter.



Click at the Isotropic Elasticity under Linear Elastic and drag it to the Property list at the bottom of the window. Enter the Young’s Modulus value as 7e10 and hit Enter, enter the Poisson Ratio value as 0.3 and hit Enter, and close this window.



Close the Engineering Data tab and click at the Project tab on the upper menu, it will bring back to the main Project Schematic window.

3.3 Academic Example

47

(b) Creating Geometry 

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   

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Right click on the Geometry tab and select the Properties option, the Properties of Schematic window will open. Select the Line Bodies under the Basic Geometry Options. Then, close this small window. Right click on the Geometry tab and select the New Geometry…. This will launch the ANSYS Design Modeler (green logo DM).

On DM window, set unit in the Units menu on the upper tab to Meter. On the Tree Outline window, right click on XYPlane and select Look at. The X-Y-Z coordinates on the Model View in 3D view will become X-Y coordinates in 2D view. Select the Sketching tab below the Tree Outline window, the Sketching Toolboxes will pop-up in the same place. Select the Settings tab and then Grid, activate the buttons Show in 2D and Snap. The grid will appear on the main window. Grid snapping provides convenience when drawing model. Make sure that the Major Grid Spacing is set to 1 m, Minor-Steps per Major is 2, and Snaps per Minor is 1.

48

Chapter 3 Beam Analysis



Enlarge scale by clicking at Box Zoom icon on the upper part of the screen (icon with plus sign on the magnifying glass) and draw a box with appropriate size to zoom in. Click it again after appropriate scale is showing on the window.



Click on Modeling tab, and then click the New Sketch icon (a small blue geometry symbol with * on the upper part of the DM window) to create Sketch1 which will appear under XYPlane. Note that this name Sketch1 can be deleted or renamed by right clicking on it and selecting an option. Next, we draw the two beam elements. Click on Sketch1. Click the Sketching tab and select Draw. Choose Line to create the first line with the end coordinates of (0,0) and (1,0). This is done by clicking at the coordinates of (0,0) on

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3.3 Academic Example

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49

the model, move the cursor to the coordinates of (1,0), and click the mouse again. Then, follow the same procedure to create the second line with the end coordinates of (0,1) and (1,0). Click on Generate (the icon with yellow lightning on the upper-left part of the screen). The desired two lines will become dark green. The next important step is to go to the Concept tab on top of the screen and select Lines From Sketches. Select the Sketch1, the two lines will become yellow. Click Apply icon on the right side of the Base Objects tab in the Details View at the lower left of the screen. Both two lines will become cyan. Then, click on Generate. The right side of the Base Objects tab will show 1 Sketch. The 1 Part, 1 Body item will appear in the Tree Outline window. We now have a model consisting of two lines.

The next step is to create the beam cross section. Select Rectangular item in Cross Section under the Concept tab. In the Details of Rect1 window, change the base value B to 0.02 m and hit Enter, the height value H to 0.02 m and hit Enter. A blue rectangular cross section will appear on the main Graphic window.

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Chapter 3 Beam Analysis



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Next, assign this cross section to the Line Body. Double click at 1 Part, 1 Body and select the Line Body, assign Rect1 to the Cross section selection in the Details of Line Body window. Save file as Two Beam Element Problem, and close the DM window.

(c) Assigning Material Properties and Creating Mesh  



On the main Project Schematic window, double click on Model, the beam model will appear back on the main window. Double click on Geometry item, the Line Body item will pop-up. Select the Line Body item and select “My Aluminum Material” (the name assigned earlier containing material properties of this problem) which is on the righthand-side of Assignment under Material in Details of “Line Body” window. The beam model will become green. Select Mesh under Model section, the Details of “Mesh” window will appear on the lower left of the screen. Select Element Size under Sizing and change the value on the right column to 0.2 and hit Return so that the generated element length is approximately 0.2 m. Right click at the Mesh again and select Generate Mesh. A finite element

3.3 Academic Example



51

mesh with the 2-node beam elements will appear as shown in the figure. Save the project and close the DM window.

(d) Applying Boundary Conditions, Solving for and Displaying Solutions   

On the main Project Schematic window, double click on Setup, the beam model will appear back on the main window. Next, the boundary conditions of constraints on the two left ends can be applied. This will be done, one at a time, starting from the lower left end. Select Analysis Settings under Static Structural. Select the Supports tab on the upper menu bar with Fixed Support option, then select Vertex icon (box with arrow and green dot). Move the cursor to the lower left end and click at it, the left end will become green. Click Apply button next to the Geometry button under the Fixed Support window.

52

Chapter 3 Beam Analysis





Repeat the same process to apply boundary condition of fixed support at the upper left end. This is done by selecting the Analysis Settings. Select the Supports tab on the upper menu bar with Fixed Support option, then select Vertex icon (box with arrow and green edge). Move the cursor to the upper left and click at it, the upper left end will become green. Click Apply button next to the Geometry button under the Fixed Support window. Repeat the similar process to apply boundary condition of the two applied forces along the right connection by first selecting the Analysis Settings, select the Loads tab on the upper menu bar with Force option, and select Vertex icon. Move the cursor to the right connection and click at it. Select Components option on the right side of Define By

3.3 Academic Example



 

 

53

in the Details of “Force” window. Click Apply button, and input X Component as 500 and hit Enter, Y component as -2500 and hit Enter. Note that, mesh can be shown by clicking the Show Mesh icon on the upper menu bar. It is noted that these beam elements are connected as fixed joints. This can be verified by clicking the Tools button on the top menu. Select Options… and follow by Connections. The right-hand-side of the Fixed Joints item must be Yes. The problem is now ready to solve for solution. Right click the Solution item and under Static Structural and select the Solve tab. Click the Solution item, the Deformation tab will appear on the lower menu bar. Click on this Deformation tab and Select the Total option, the Total Deformation item will pop-up beneath the Solution item.

Right click on Solution, select Insert, select Beam Tool, and Beam Tool. The Beam Tool will pop-up beneath the Solution item. Right click on Solution and select Solve, the program will start to solve the model.

54

Chapter 3 Beam Analysis    



After completion, click Total Deformation beneath the Solution item, the deformed model with show on the main window. Click the Direct Stress to show the axial stress results that occur in beams. Double click on Beam Tool item beneath the Solution item, the Direct Stress, Minimum Combined Stress and Maximum Combined Stress items will pop-up. Click the Minimum Combined Stress to show the combination of the direct stress and minimum bending stress results that occur in beams.

Similarly, click on the Minimum Combined Stress and Maximum Combined Stress to display these stresses, respectively, on the two beams as shown in the figures.

55

3.3 Academic Example 3.3.2 Twenty-one Beam Members in Two Dimensions

The same process used for analyzing the two beam members in the preceding example can be employed to solve a more complicated frame structure. A frame structure, as shown in the figure, consists of 21 members. All members have the same cross-sectional area of .01 m2 and made from a structural steel material with the Young’s modulus of 2  1011 N m 2 . The frame structure is fixed at the lower left and right ends and subjected to five vertical loadings as shown in the figure. A  .1 .1  .01 m 2 E 

2 1011

N

Y

m2

5m

1m 5m

X 1000 N 2000 N

2000 N

1000 N

4000 N

The finite element model consisting of 21 two-node beam elements can be constructed easily by using the Line command. The boundary conditions of the complete constraint at the left and

56

Chapter 3 Beam Analysis

right ends and the concentrated forces at lower nodes of the model can be applied conveniently as shown in the figure. Once the problem has been set up, solutions of the deformed shape and stresses in the beam members can be obtained without difficulty as shown in the figures. The figures show the deformation shape and stresses that occur in these beam members.

3.4 Application 3.4.1 Racing Car Frame Structure A racing car frame structure is made from structural steel material that has the density of 7,850 Kg m 3 , the Young’s modulus of 2 1011 N m 2 and the Poisson’s ratio of 0.3. The engine and driver weights, as shown in the figure by the spheres at the front and middle of the frame structure, are 10 and 75 Kg, respectively. The deformed configuration and member stresses are to be determined when the forces from the weights are triple, simulating while the structure drops into a road pit-hole. We will use ANSYS through the Workbench to analyze the problem.

3.4 Application

57

We start from importing the CAD model of the frame structure which consists of straight and curved lines as shown in the figure. These lines are then assigned as circular pipe with the diameter and thickness of 25.4 and 1.5 mm, respectively.

We then create a finite element model which contains many beam elements as shown in the figure.

58

Chapter 3 Beam Analysis

We apply the simply-supported boundary conditions at the eight locations denoted by small square symbols as shown in the figure. The applied forces from the engine and driver are transferred to the rectangular frames surrounding them. The additional forces when the frame structure drops into a pit-hole are embedded through the vertical acceleration with the magnitude of 34.445 m sec 2 .

The computed member stresses are displayed on the deformed frame structure as shown in the figure. The solution indicates the locations where high stresses occur. Such solution

3.4 Application

59

helps engineers to understand the frame structure behavior. They can further improve the design by modifying the frame structure configuration, changing member diameters, selecting different materials, etc. The analysis is performed and the process is repeated until engineers satisfy with their design.

Chapter 4 Plane Stress Analysis Determination of deformation and stresses for twodimensional elasticity problems is rather difficult by using the conventional approach. The finite element method alleviates such difficulty especially for problems with complicated geometry. This chapter begins with the governing differential equations of the plane stress problem. Finite element equations and their element matrices are derived for a simple triangular element. ANSYS with its Workbench are employed to solve for solutions of examples in both academic and application problems.

4.1 Basic Equations 4.1.1 Differential Equations The equilibrium conditions at any point of a membrane that lies in x-y plane, under in-plane forces with exclusion of body forces, are governed by the two partial differential equations,

62

and

Chapter 4 Plan Stress Analysis

 xy  x   0 x y  xy  y   0 x y

where  x and  y is the normal stress in the x and y direction, respectively, while  xy is the shearing stress. The basic unknowns of the two equations above are the u and v displacement in the xand y-direction, respectively. 4.1.2 Related Equations The normal stresses  x and  y together with the shearing stress  xy can be written in forms of the strain components according to the Hook’s law as,  x  0   x  1  E      0    y   C     1     y   2  1   (31) (33) (31)    0 0 (1  ) 2   xy   xy  where  x and  y is the normal strain in the x- and y-direction, respectively, while  xy is the shearing strain. The elasticity matrix

C  depends on the material Young’s modulus E and the Poisson’s ratio  . For small deformation theory, these strain components varies with the displacement u and v in the x- and ydirection as, u v u v x  ; ; y   xy   x y y x The stress-strain relations and strain-displacement relations above lead to the two partial differential equations in the forms,

  E  u v     E  u v             0 x 1  2  x y   y  2(1  )  y x     

4.2 Finite Element Method

63

  E  u v     E  u v       0       x  2(1  )  y x   y 1  2  x y       which could be solved for the two displacement components u and v in the x- and y-direction, respectively. Determination of the strain components  x ,  y ,  xy and stress components  x ,  y ,  xy can then be followed.

4.2 Finite Element Method 4.2.1 Finite Element Equations Finite element equations can be derived by applying the method of weighted residuals to the partial differential equations. Detailed derivation can be found in many finite element textbooks including the one written by the same author. The derived finite element equations are written in matrix form as,  K     F  where  K  is the element stiffness matrix;   is the element vector containing the nodal displacement unknowns of u and v ; and  F  is the element vector containing the nodal forces in the xand y-direction. Number of equations and sizes of these element matrices depend on the element type selected. These element equations are formed up for every element before assembling them together to become a large set of simultaneous equations. Boundary conditions of the problem are then imposed before solving them for the displacement solutions of u and v at nodes. 4.2.2 Element Types Triangular and quadrilateral elements are the two popular element types used in the plane stress analysis. The triangular element may contain three or six nodes, while the quadrilateral element may consist of four or eight nodes as shown in the figures.

64

Chapter 4 Plan Stress Analysis 3

3

4

5

2

2 1

6

1

3

4

3

7

4

6 8

2

2 5

1

1

The three-node triangular element is the simplest twodimensional element to understand. Both u and v displacements are assumed to vary as flat planes over the element as shown in the figure. u ( x, y )

v ( x, y )

3

u2 2

Displacement

v2

t

y

1 x

u ( x, y )  v ( x, y ) 

N1 ( x , y )u1  N 2 ( x , y )u2  N 3 ( x , y )u3 N1 ( x , y ) v1  N 2 ( x , y ) v2  N 3 ( x , y ) v3

where N1 , N 2 , N3 are the element interpolation functions, 1 i  1, 2, 3 (ai  bi x  ci y ) Ni ( x, y )  2A and A is the element area,

4.2 Finite Element Method

65

1  x2 ( y3  y1 )  x1 ( y2  y3 )  x3 ( y1  y2 )  2 In the above equations, xi , yi ; i  1, 2, 3 are coordinates of the three nodes. The parameters ai , bi , ci ; i  1, 2, 3 depend on the nodal coordinates as follows, a1  x2 y3  x3 y2 b1  y2  y3 c1  x3  x2

A 



x3 y1  x1 y3

b2



y3  y1

c2



x1  x3

a3 

x1 y2  x2 y1

b3 

y1  y2

c3 

x2  x1

a2

The finite element equations corresponding to the three-node triangular element above are,  K     F  (66) (61)

(61)

where the element stiffness matrix  K  can be determined from,

 K    B T  C   B  t A (66)

where

 B 

(36)

(63)

 b1 1  0 2A   c1

(33) (36)

0

b2

0

b3

c1 0 c2 0 b1 c2 b2 c3

0 c3   b3 

 T  u1 v1 u2 v2 u3 v3  (61)

and

 F T   F1x F1 y F2 x F2 y F3x F3 y  (61)

Similarly, the finite element equations for the four-node quadrilateral element can be determined in the same way, except the process is more complicated. These element matrices suggest that it is nearly impossible to solve plane stress problems by hands even though they contain only few elements. Developing a finite element computer program is thus required. A model with few hundred elements can be solved easily by using a computer program. We will employ ANSYS through its Workbench to analyze plane stress problems in the following sections. We will find that the software can provide solutions conveniently and effectively for model containing a large number of elements.

66

Chapter 4 Plan Stress Analysis

4.3 Academic Example 4.3.1 Plate with Circular Cut-out A rectangular plate with dimensions of 1.4  1 m is shown in the figure. The plate has a circular cut-out with radius of .2 m at its center. The plate is made from a material that has the Young’s modulus of 7  1010 N m 2 and Poisson’s ratio of 0.3. The plate is subjected to the uniform loadings of 7 107 N m2 along the left and right edges. y  0  7 107 N/m 2

1m

x

.2 m 1.4 m Since the problem has its solution symmetry, the upper right quarter of the plate can be used for analysis. Model of the upper right quarter is shown in the figure with symmetrical boundary condition along the left and bottom edges.

y

 0  7 107 N/m 2

.5 m .2 m 

x .7 m

Steps for analyzing this problem by employing the ANSYS Workbench are as follows.

4.3 Academic Example

67

(a) Starting ANSYS Workbench    



Open the ANSYS Workbench, set the Units menu on the upper tab to Metric (kg,m,s,C,A,N,V). On the Analysis Systems window, click twice on the Static Structural item. A new small box will appear on the Project Schematic window. Retype the name in the lower blue tab as the desired project name, e.g., Plane Stress Problem, and hit Enter. Right click on the Engineering Data tab and select the Edit… option, the A2: Engineering Data window will pop-up. Double click on Click here to add a new material and type in a new material name, e.g., “My Material”, and hit Enter. Click at the Isotropic Elasticity under Linear Elastic and drag it to the Property list at the bottom of the window. Enter the Young’s Modulus value as 7e10 and hit Enter, enter the Poisson Ratio value as 0.3 and hit Enter, and close this window. Then, close the Engineering Data tab and click at the Project tab on the upper menu, it will bring back to the main Project Schematic window.

(b) Creating Geometry 

Right click on the Geometry tab and select the Properties option, the Properties of Schematic window will open.

Chapter 4 Plan Stress Analysis

68



 

 

 

Change the Analysis Type under the Advanced Geometry Options from 3D to 2D. Then, close this small window. Right click on the Geometry tab and select the New Geometry…. This will launch the ANSYS Design Modeler (green logo DM).

On DM window, set unit in the Units menu on the upper tab to Meter. On the Tree Outline window, right click on XYPlane and select Look at. The X-Y-Z coordinates on the Model View in 3D view will become X-Y coordinates in 2D view. Select the Sketching tab below the Tree Outline window, the Sketching Toolboxes will pop-up in the same place. Select the Settings tab and then Grid, activate the buttons Show in 2D and Snap. The grid will appear on the main window. Grid snapping provides convenience when drawing model. Make sure that the Major Grid Spacing is set to 1 m, Minor-Steps per Major is 10, and Snaps per Minor is 1. Enlarge the scale by clicking at the Box Zoom icon on the upper part of the screen (icon with plus sign on the magnifying glass) and draw a box with appropriate size to zoom in. Click it again after appropriate scale is showing on the window.

4.3 Academic Example



  

 



69

Click on Modeling tab, and then click the New Sketch icon (a small blue geometry symbol with * on the upper part of the DM window) to create Sketch1 which will appear under XYPlane. Note that this name Sketch1 can be deleted or renamed by right clicking on it and selecting an option. Click again on the same New Sketch icon to create Sketch2. Next we draw the rectangle. Click on Sketch1. Click the Sketching tab and select Draw. Choose Rectangle to create a rectangle with the vertices of (0,0) and (.7,.5). This is done by clicking at the coordinates of (0,0) on the model, move the cursor to the coordinates of (.7,.5), and click the mouse again. Click on Generate (the icon with yellow lightning on the upper-left part of the screen). The desired rectangle will pop up in dark green. Next we draw the circle. Click the Modeling tab and select Sketch2. Then click the Sketching tab. Select the Draw tab and choose Circle. Draw a circle with center at the coordinates of (0,0). Do not worry about the size of the circle, it will be taken care later. Then, click the Generate button. Select the Dimensions tab and choose Radius. Left click on the circle that just drew, drag the mouse outward without releasing the mouse until seeing an arrow with notation R1, then release the mouse. The desired circle will pop up in dark green.

Chapter 4 Plan Stress Analysis

70



On the Details View window that just appears on the lower left of the DM screen, adjust the radius to 0.2 and hit Enter. Click on Generate, the circle will radius of 0.2 will appear.



The next important step is to go to the Concept tab on top of the screen and select Surfaces From Sketches. Press the Ctrl key to select both the Sketch1 and Sketch2, the rectangle and circle will become yellow. Click Apply icon on the right side of the Base Objects tab in the Details View at the lower left of the screen. Both rectangle and circle will become cyan. The right side of the Base Objects tab will show 2 Sketches. Then, click on Generate. We now have both rectangular and circular surfaces.

 



4.3 Academic Example  

 

 

 

71

The next step is to subtract the circle from rectangle. Click on the Create tab located at the upper part of the screen, and select Boolean. In the Details window, select the Operation tab and choose Subtract. Note that, under the Operation tab (shown as Not Selected at present), the Target Bodies will become rectangle (which is Sketch1) while the Tool Bodies will become the circle (in Sketch2). Click on the Not Selected tab next to the Target Bodies tab, the Apply and Cancel tabs will appear. Go to Tree Outline window, click twice at the 2 Parts, 2 Bodies and select the first Surface Body tab that represents the rectangle (which is Sketch1), the rectangle area become yellow. Click Apply tab, the right side tab will say 1 Body while the rectangle will become dark green. Click on the Not Selected tab next to the Tool Bodies tab, the Apply and Cancel tabs will appear. Go to Tree Outline window again, select the second Surface Body tab that represents the circle (which is Sketch2), the circle area become yellow. Click Apply tab, the right side tab will say 1 Body while the circle will become dark green. The, click Generate, a quarter of the plate with circular cutout is displayed. Save file as Plane Stress Problem, and close the DM window.

Chapter 4 Plan Stress Analysis

72

(c) Assigning Material Properties and Creating Mesh  





On the main Project Schematic window, double click on Model, the solid plate model will appear back on the main window. Double click on Geometry item, the Surface Body item will pop-up. Select the Surface Body item and select “My Material” (the name assigned earlier containing material properties of this problem) which is on the right-hand-side of Assignment under Material in Details of “Surface Body” window. The plate model will become green.

Select Mesh under Model section, the Details of “Mesh” window will appear on the lower left of the screen. Click at Relevance with the value of 100. Right click at the Mesh again and select Generate Mesh. A finite element mesh with most of 4-node quadrilateral elements will appear as shown in the figure. Save the project and close the DM window.

(d) Applying Boundary Conditions, Solving for and Displaying Solutions 

Next, the boundary conditions of constraints along the bottom and left edges can be applied. This will be done, one at a time, starting from the bottom edge.

4.3 Academic Example

73



On the main Project Schematic window, double click on Setup. Select Analysis Settings under Static Structural. Select the Supports tab on the upper menu bar with Frictionless Support option, then select Edge icon (box with arrow and green edge). Move the cursor to the bottom edge and click at it, the edge will become green. Click Apply button next to the Geometry button under the Frictionless Support window.



Repeat the same process to apply boundary condition of frictionless support along the left edge. This is done by selecting the Analysis Settings. Select the Supports tab on the upper menu bar with Frictionless Support option, then select Edge icon (box with arrow and green edge). Move the cursor to the left edge and click at it, the left edge will become green. Click Apply button next to the Geometry button under the Frictionless Support window. Note that the mesh can be displayed by clicking on the Show Mesh button on the upper menu bar.



Chapter 4 Plan Stress Analysis

74



Repeat the same process to apply boundary condition of uniform loading along the right edge by first selecting the Analysis Settings, select the Loads tab on the upper menu bar with Pressure option, and select Edge icon. Move the cursor to the right edge and click at it. Click Apply button, and change Magnitude value to -7e7, and hit Enter. If preferred, mesh can be shown by clicking the Show Mesh icon on the upper menu bar.



The problem is now ready to solve for solution. Right click the Solution item and under Static Structural and select the Solve tab.

4.3 Academic Example

75



Click the Solution item, the Deformation tab will appear on the lower menu bar. Click on this Deformation tab and Select the Total option, the Total Deformation item will pop-up beneath the Solution item. Right click at the Total Deformation item and select Evaluate All Results, the solution in form of color fringe plot will appear as shown in the figure.



Click the Solution item, the Stress tab will appear on the lower menu bar. Click on this Stress tab and Select the Equivalent (von-Misses) option, the Equivalent Stress item will pop-up beneath the Solution item. Right click at the Equivalent Stress item and select Evaluate All Results, the solution in form of color fringe plot will appear as shown in the figure.

Chapter 4 Plan Stress Analysis

76



The normal stress in X-direction is displayed by selecting the Solution item and then the Stress tab with Normal option and select X Axis under the Orientation in the Details of Normal Stress window.



The normal stress in Y-direction is displayed by selecting the Solution item and then the Stress tab with Normal option and select Y Axis under the Orientation in the Details of Normal Stress window.



The shear stress is displayed by selecting the Solution item and then the Stress tab with Shear option.

4.4 Application

77

4.4 Application 4.4.1 Stress in Motorcycle Chain Wheel A motorcycle chain wheel as shown in the figure is made from steel. The steel material has the Young’s modulus of 2  1011 N m 2 and Poisson’s ratio of 0.3. The chain wheel is subjected to an applied torque at the center and the chain resistant force along one side of the outer rim. The ANSYS file of this problem can be downloaded from the book website.

78

Chapter 4 Plan Stress Analysis

The analysis starts from importing the CAD file of the chain wheel. The two-dimensional mesh is then generated on the mid-plane of the wheel by using the Mid-Surface command. The finite element model contains mostly the four-node quadrilateral elements with very few three-node triangular elements as shown in the figure.

The boundary conditions consist of constraining the right half of the outer rim and applying the torque of 39 N-m at the wheel center as highlighted in the figure.

4.4 Application

79

The predicted von-Mises stress on the deformed wheel is shown in the figure. Detailed stresses in the two inserts indicate the locations of high stress that should be concerned. Such solutions also provide insight into the problem. Engineers may alter the design configuration to reduce the stress or to further increase the strength.

Chapter 5 Plate Bending Analysis

Plate bending analysis is needed for design of new products and corrugated structures today. Examples are drinking water bottles, soda cans, high pressure gas containers, automotive bodies, airplane fuselage, etc. Analysis solutions of these problems are difficult to obtain in the past by using the classical method. At present, the finite element method has played important role for providing detailed solutions. The method has become an essential tool to engineers for designing new products and analyzing complicated structures. This chapter begins with the differential equation and related equations for solving plate bending problems. The finite element formulation is described by using a simple plate bending element for the ease of understanding. An academic plate bending problem is then analyzed by employing ANSYS Workbench. A shelf angle bracket that we have seen in our everyday life is used as

82

Chapter 5 Plate Bending analysis

an application example at the end of the chapter to demonstrate the advantages of the finite element method.

5.1 Basic Equations 5.1.1 Differential Equation Derivation of the differential equation for plate bending is similar to that for the beam bending as explained in Chapter 3. A thin plate with its thickness of h that lies in the x-y plane is shown in the figure. The plate is subjected to the pressure of p ( x, y ) on the upper surface causing the deflection of w in the z-direction and the in-plane displacement of u and v in the x- and y-direction, respectively.

p ( x, y )

u v

y

x

w

z

h

The basic assumption of plane section remains plane before and after deflection leads to the relations of u   z  w  x and v   z  w  y . Together with the additional assumption of the deflection w that varies with x and y only, w  w( x, y ) , the governing differential equation representing equilibrium condition of plate bending can be derived in the form,

2 Mxy 2 M y 2 Mx 2   x2 x y y2 where the bending moment components are,

 p

83

5.1 Basic Equation

 2 w 2 w  M x   D  2  2  y   x 2 w M xy  D(1  ) x y

;

 2 w 2 w  M y   D  2  2  y   x D 

;

E h3 12(1  2 )

In these bending moment component equations, D represents the plate flexural rigidity that depends on the Young’s modulus E, the Poisson’s ratio  , and the plate thickness h . Substituting these bending moment component equations into the governing differential equation above yields the final form of the plate differential equation. The final form is of fourth-order differential equation containing only one unknown of the deflection w. 5.1.2 Related Equations From the relations of the in-plane displacements u and v with the deflection w, the strain components become,

x  y 

u x v y

u v  y x Then, the stress components are,

 xy 

2 w x2 2 w  z 2 y  z

  2z

2 w x y

x 

E ( x  y ) 1  2

 

2 w  E  2 w    z y 2  1  2  x2

y 

E ( x   y ) 1  2

 

E  2 w 2 w    z 1  2  x2 y 2 

 xy 

E  xy 2(1  )

 

E 1 

 2 w    z  x y 

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5.2 Finite Element Method 5.2.1 Finite Element Equations Finite element equations can be derived by applying the method of weighted residuals to the governing differential equation. Detailed derivation can be found in many finite element textbooks including the one written by the same author. The application leads to the finite element equations in matrix form as,

 K   

FQ   FM   Fp 

where  K  is the element stiffness matrix;   is the element vector containing nodal deflections in the z-direction and rotations about the x- and y-coordinates;  FQ  is the element vector of the nodal shearing forces;  FM  is the element vector of the nodal

bending moments; and  Fp  is the element vector containing nodal loads from the applied pressure p ( x, y ) . 5.2.2 Element Types

Size of the matrices in the finite element equations above depends on the element type selected. Element types could be in triangular or quadrilateral shapes as shown in the figures. These elements may consist of only corner nodes as well as additional nodes on their edges. 3

4 4

5

6

2 1

3

7

6

8 2 1

5

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5.2 Finite Element Method

We will consider the four-node rectangular element as shown in the figure because it is one of the simplest element type. The element has dimensions of a  b with the thickness of h . Each node contains three unknowns which are the deflection w in the zdirection and the rotations  x and  y about the x- and ycoordinates, respectively. unknowns.

Thus, the element has a total of 12 y

3

4 z

 y1

w1 1

 x1

b h 2

x

a

Distribution of the deflection is assumed in the form,

w( x, y)   N1 N2 N3 N4 N5 N6 N7 N8 N9 N10 N11 N12    where

 T   w1 x1  y1 w2 x 2  y 2 w3 x3  y3 w4 x 4  y 4 

The element interpolation functions, Ni , i  1 to 12 , are rather complicated. As an example,

N7  

xy x2 y xy2 x3 y xy3 3 2 3 2 2 3  2 3 ab ab ab ab ab

This leads to a complicated element stiffness matrix with lengthy coefficients, such as,

K77 

D  a2 b2   60 2  60 2  30  42(1  )  15ab  b a 

Derivation of element matrices must be performed carefully. Symbolic manipulation software can help alleviating such task.

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5.3 Academic Example 5.3.1 Simply-supported Plate under Uniform Loading A square plate with the dimensions of 2  2 m and thickness of 0.01 m is shown in the figure. The plate is made from a material that has the Young’s modulus of 7.2  1010 N/m 2 and Poisson’s ratio of 0.25. The plate is simply supported along its four edges and is subjected to a uniform pressure of 1, 200 N/m 2 . We will employ ANSYS through the Workbench to solve for the plate deflection and stresses. z

7.2 1010

E    0.25

N

m2

y

x 2m p  1200 N /m

2

2m

h  0.01 m

Steps for analyzing this plate bending problem by employing the ANSYS Workbench are as follows. (a) Starting ANSYS Workbench   

Open the ANSYS Workbench, set the Units menu on the upper tab to Metric (kg,m,s,C,A,N,V). On the Analysis Systems window, click twice on the Static Structural item. A new small box will appear on the Project Schematic window. Retype the name in the lower blue tab as the desired project name, e.g., Simply-supported Plate, and hit Enter.

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Right click on the Engineering Data tab and select the Edit… option, the A2: Engineering Data window will pop-up. Double click on Click here to add a new material and type in a new material name, e.g., “My Plate Material”, and hit Enter. Click at the Isotropic Elasticity under Linear Elastic and drag it to the Property list at the bottom of the window. Enter the Young’s Modulus value as 7.2e10 and hit Enter, enter the Poisson Ratio value as 0.25 and hit Enter, and close this window. Close the Engineering Data tab and click at the Project tab on the upper menu, it will bring back to the main Project Schematic window.

(b) Creating Geometry

  

Right click on the Geometry tab and select the New Geometry…. This will launch the ANSYS Design Modeler (green logo DM). On DM window, set unit in the Units menu on the upper tab to Meter. On the Tree Outline window, right click on XYPlane and select Look at. The X-Y-Z coordinates on the Model View in 3D view will become X-Y coordinates in 2D view.

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 

 

Select the Sketching tab below the Tree Outline window, the Sketching Toolboxes will pop-up in the same place. Select the Settings tab and then Grid, activate the buttons Show in 2D and Snap. The grid will appear on the main window. Grid snapping provides convenience when drawing model. Make sure that the Major Grid Spacing is set to 1 m, Minor-Steps per Major is 5, and Snaps per Minor is 1. Enlarge the scale by clicking at the Box Zoom icon on the upper part of the screen (icon with plus sign on the magnifying glass) and draw a box with appropriate size to zoom in. Click it again after appropriate scale is showing on the window.

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Click on Modeling tab, and then click the New Sketch icon (a small blue geometry symbol with * on the upper part of the DM window) to create Sketch1 which will appear under XYPlane. Note that this name Sketch1 can be deleted or renamed by right clicking on it and selecting an option. Next we draw the square. Click on Sketch1. Click the Sketching tab and select Draw. Choose Rectangle to create a square with the vertices of (-1,-1) and (1,1). This can be done by clicking at the coordinates of (1,-1) on the model, move the cursor to the coordinates of (1,1) and click it again. Click on Generate (the icon with yellow lightning on the upper-left part of the screen). The desired square will pop up in dark green.

The next important step is to go to the Concept tab on top of the screen and select Surfaces From Sketches. Select the Sketch1, the square will become yellow. Click Apply icon on the right side of the Base Objects tab in the Details View at the lower left of the screen. The square will become cyan. The right side of the Base Objects tab will show 1 Sketch. Then, click on Generate. We now have the desired square surface.

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Chapter 5 Plate Bending analysis

  

Click SurfaceSk1 and change the value of Thickness in the Details View window to 0.01 and hit Enter. Click ISO tab and Zoom tab on the upper menu bar so that the model is displayed in three dimensions. Save file as Simply-supported Plate, and close the DM window.

(c) Assigning Material Properties and Creating Mesh

 



On the main Project Schematic window, double click on Model, the thin plate model will appear back on the main window. Double click on Geometry item, the Surface Body item will pop-up. Select the Surface Body item and select “My Plate Material” (the name assigned earlier containing material properties of this problem) which is on the righthand-side of Assignment under Material in Details of “Surface Body” window. The plate model will become green. Select Mesh under Model section, the Details of “Mesh” window will appear on the lower left of the screen. Click at Relevance with the value of 100. Right click at the Mesh again and select Generate Mesh. A finite element mesh with most of 4-node quadrilateral elements will appear as shown in the figure.

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91

Save the project and close the DM window.

(d) Applying Boundary Conditions, Solving for and Displaying Solutions

 

Next, the boundary conditions of simply support along the four edges can be applied. These will be done, one at a time, starting from the top edge. Select Analysis Settings under Static Structural. Select the Supports tab on the upper menu bar with Simply Supported option, then select Edge icon (box with arrow and green edge). Move the cursor to the top edge and click at it, the edge will become green. Click Apply button next to the Geometry button under the Details of Simply Support window.

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Chapter 5 Plate Bending analysis



Repeat the same process to apply simply-supported boundary condition along the other three edges.



Repeat the same process to apply boundary condition of uniform loading on the plate surface by first selecting the Analysis Settings, select the Loads tab on the upper menu bar with Pressure option, and select Face icon. Move the

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cursor to the plate face and click at it. Click Apply button, and change Magnitude value to 1200, and hit Enter. If preferred, mesh can be shown by clicking the Show Mesh icon on the upper menu bar.







The problem is now ready to solve for solution. Right click the Solution item and under Static Structural and select the Solve tab. Click the Solution item, the Deformation tab will appear on the lower menu bar. Click on this Deformation tab and Select the Total option, the Total Deformation item will pop-up beneath the Solution item. Right click at the Total Deformation item and select Evaluate All Results, the solution in form of color fringe plot will appear. Click the Solution item, the Stress tab will appear on the lower menu bar. Click on this Stress tab and Select the Equivalent (von-Mises) option, the Equivalent Stress item will pop-up beneath the Solution item. Right click at the Equivalent Stress item and select Evaluate All Results, the solution in form of color fringe plot will appear as shown in the figure.

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Chapter 5 Plate Bending analysis



The normal stress on the top surface in X-direction is displayed by selecting the Solution item and then the Stress tab with Normal option. Select X Axis under the Orientation and follow by Top under Position in the Details of Normal Stress window. Right click at the Normal Stress item and select Evaluate All Results, the solution in form of color fringe plot will appear as shown in the figure.

 The normal stress on the bottom surface in Y-direction is displayed by selecting the Solution item and then the Stress

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95

tab with Normal option. Select Y Axis under the Orientation, and then select Bottom under Position in the Details of Normal Stress window. Right click at the Normal Stress item and select Evaluate All Results, the solution in form of color fringe plot will appear as shown in the figure.



The shear stress on the top or bottom surface is displayed by selecting the Solution item and then the Stress tab with Shear option. Right click at the Shear Stress item and select Evaluate All Results, the solution in form of color fringe plot will appear as shown in the figure.

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Chapter 5 Plate Bending analysis

5.4 Application 5.4.1 Stress in Shelf Angle Bracket A shelf angle bracket with the thickness of 1 mm, as shown in the figure, is made from a material that has the Young’s modulus of 2 1011 N m 2 and Poisson’s ratio of 0.3. The vertical side of the bracket is fixed on a wall by screws at the three holes as shown in the figure. The horizontal side supports a vertical load of 100 N. We will use ANSYS through the Workbench to solve for the deformed shape and the stress that occurs in the bracket.

Since the bracket is made from a thin metal sheet, we use the plate bending elements to model it. We first import the CAD model and discretize it into many small elements as shown in the figure.

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97

The next step is to impose the boundary condition of fixed support along rims of the three holes. On the upper surface, we impose the boundary condition of the frictionless support with the vertical load of 100 N. These imposed boundary conditions are shown in the figure.

The computed von-Mises stress distribution on the deformed shape is displayed in the figure. As expected, the maximum stress occurs at the inner corner of the bracket. Since the ANSYS files of this problem can be downloaded from the book website, users can modify the problem to obtain different solutions. As an example, the stress is reduced if the bracket thickness is increased. Changing problem geometry and boundary conditions can increase understanding of the solution behaviors. This often leads to improvement of the design and efficiency of the products.

Chapter 6 Three-Dimensional Solid Analysis

Determination of stresses in a three-dimensional elastic solid is difficult in the past even when its shape is uncomplicated. This is because their solutions must be solved from the three coupled partial differential equations. The finite element method helps finding these solutions effectively. This chapter begins with the governing differential equations of the three-dimensional elastic solid and the related equations. The finite element equations are presented and different finite element types are highlighted. ANSYS software through its Workbench is then employed to analyze simple and application problems.

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Chapter 6 Three-Dimensional Solid Analysis

6.1 Basic Equations 6.1.1 Differential Equations The equilibrium conditions at any location of a threedimensional elastic solid in x-y-z coordinate system, with exclusion of body forces, are governed by the three partial differential equations,  xy  x  xz  0   x y z  xy  y  yz  0   x y z  yz  xz  z  0   x y z where  x ,  y , z are the normal stress components in the x, y,

z coordinate directions, and  xy ,  xz ,  yz are the shearing stress components. The three differential equations of the problem suggest that there must be three basic unknowns. These unknowns are the displacement components u ( x, y, z ), v ( x, y , z ) and w( x, y, z ) in the x, y and z coordinate directions, respectively. Thus, the stress components in the differential equations must be written in forms of the three displacement components prior to solving them. 6.1.2 Related Equations The six stress components can be written in forms of the six strain components according to the Hooke’s law as,    C    (61)

where and

(66) (61)

    x  y  z  xy  xz  yz   T   x  y  z  xy  xz  yz  T

The matrix C  is the elasticity matrix which depends on the Young’s modulus and Poisson’s ratio.

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101

The six strain components are written in terms of the three displacement components based on the small deformation theory as, u v u   xy  x  ; y x x v u w y  ;  xz   y z x v w w   yz  z  ; z y z The six strain-displacement relations are substituted into the six stress-strain relations, so that the stress components can be written in terms of the displacement components. These stress components are then further substituted into the three governing differential equations. The final three governing differential equations are now in forms of the three displacement components. The three displacement components thus can be solved from the three differential equations.

6.2 Finite Element Method 6.2.1 Finite Element Equations Finite element equations can be derived by applying the method of weighted residuals to the three partial differential equations. Detailed derivation can be found in many finite element textbooks including the one written by the same author. It is noted that the finite element equations can also be derived by using the variational method. The method is based on the minimum total potential energy principle. This later method was often used to derive the finite element equations for solid problems in the past. The derived finite element equations are written in matrix form as,  K     F  where  K  is the element stiffness matrix;   is the element vector containing the nodal displacements u, v and w in the x, y

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Chapter 6 Three-Dimensional Solid Analysis

and z coordinates, respectively; and  F  is the element vector containing the nodal forces in the x, y and z coordinates. Sizes of the element matrices and the number of element equations depend on the element types selected. Popular element types are described in the following section. 6.2.2 Element Types Tetrahedral and hexahedral elements are often used in the analysis. The tetrahedral element contains four or 10 nodes while the hexahedral element may consist of eight or 20 nodes as shown in the figures. Elements with more number of nodes provide higher solution accuracy but require extra computational time. For complicated three-dimensional solid models, the tetrahedral elements are normally used because the mesh is easier to generate. For model with simple geometry, the hexahedral elements are preferred because they can provide a more accurate solution accuracy.

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6.2 Finite Element Method

Since there are 3 displacement unknowns at each node, a problem containing only few elements is almost impossible to solve by hands. A computer program is needed to carry out the analysis for solutions. For ease of understanding, the four-node tetrahedral element is explained herein. The element contains 12 displacement unknowns as shown in the figure.

w4 4 v4

u4

w3 w1 z

x

u1 1 y

3

v3

u3

v1 w2 u2 2

v2

Distribution of the u displacement component over the element is assumed in the form, u ( x, y , z )   N1 N 2 N 3 N 4  u where the interpolation functions are, 1 i  1, 2,3, 4 Ni   ai  bi x  ci y  di z  6V In the equation above, V is the element volume, the parameters a i , bi , c i , di depend on the nodal coordinates xi , yi , z i . The element vector u contains the nodal displacements in the xcoordinate direction,

uT  u1 u 2 u3 u 4  Distribution of the v and w displacement components over the element are in the same form as,

v( x, y, z )   N1 N 2 N3 N 4  v

104

where

Chapter 6 Three-Dimensional Solid Analysis

w( x, y, z )   N1 N 2 N 3 N 4  w vT  v1 v2 v3 v4 

wT   w1 w2 w3 w4  Thus, the element vector of nodal unknowns contains the total of 12 unknowns as,

 T  u1 v1 w1 u 2 v2 w2 u3 v3 w3 u 4 v4 w4  The element vector containing the six strain components can be determined from,     B    (61)

(612) (121)

where the matrix  B  is called the strain-displacement matrix which relates the six strain components with the 12 nodal displacements. The element stiffness matrix  K  can then be determined from,

K 

(1212)

  B C   B V T

(126)

(66) (612)

These element matrices are determined for all elements before assembling them to become a large stiffness matrix of the system equations. Boundary conditions are then applied and the system equations are solved for all nodal displacement solutions u i , v i , wi . Once all nodal solutions u i , v i , wi are obtained, the element stresses are determined from,    C   B    (61)

(66) (612) (121)

The same process is applied for the hexahedral element but the number of equations is larger. For example, the 8-node hexahedral element contains 24 equations while the 20-node element consists of 60 equations. Developing a computer program is thus a must for solving a problem. We will use the ANSYS software through its Workbench to analyze three-dimensional solid problems in the following sections.

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6.3 Academic Example 6.3.1 Simple 3D Solid Problem A three-dimensional solid with simple geometry is shown in the figure. The solid is made from a material that has the Young’s modulus of 107 psi and Poisson’s ratio of 0.3. The left end face of the cylinder is fixed to a wall while the upper right face of the solid block is subjected to the applied pressure of 2,000 psi. We will employ ANSYS software with its Workbench to analyze the problem for the deformed shape and stresses that occur in the model.

1 p E  psi   0.3 p  2, 000 psi

4

10 7

3 Y

Fixed

4 X

1 6

2

Z

(a) Starting ANSYS Workbench  

Open the ANSYS Workbench, set the Units menu on the upper tab to U.S.Customary (lbm,in,s,F,A,lbf,V). On the Analysis Systems window, click twice on the Static Structural item. A new small box will appear on the Project Schematic window.

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Retype the name in the lower blue tab as the desired project name, e.g., Solid Stress Problem, and hit Enter. Right click on the Engineering Data tab and select the Edit… option, the A2: Engineering Data window will pop-up. Double click on Click here to add a new material and type in a new material name, e.g., “My Aluminum Material”, and hit Enter. Click at the Isotropic Elasticity under Linear Elastic and drag it to the Property list at the bottom of the window. Change the unit of Young’s Modulus to psi and enter the value of 1e7 and hit Enter, enter the Poisson Ratio value as 0.3 and hit Enter, and close this window. Close the Engineering Data tab and click at the Project tab on the upper menu, it will bring back to the main Project Schematic window.

(b) Creating Geometry

  

Right click on the Geometry tab and select the New DesignModeler Geometry…. This will launch the ANSYS Design Modeler (green logo DM). On DM window, set unit in the Units menu on the upper tab to Inch. On the Tree Outline window, right click on XYPlane and select Look at. The X-Y-Z coordinates on the Model View in 3D view will become X-Y coordinates in 2D view.

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On the Tree Outline window, select on XYPlane. Select the Sketching tab below the Tree Outline window, the Sketching Toolboxes will pop-up in the same place. Select the Settings tab and then Grid, activate the buttons Show in 2D and Snap. The grid will appear on the main window in two dimensions. Grid snapping provides convenience when drawing model. Make sure that the Major Grid Spacing is set to 1 in, Minor-Steps per Major is 1, and Snaps per Minor is 1. Enlarge the scale by clicking at the Box Zoom icon on the upper part of the screen (icon with plus sign on the magnifying glass) and draw a box with appropriate size to zoom in. Click it again after appropriate scale is showing on the window. Also pan the drawing frame by clicking at the Pan icon on the upper part of the screen (icon with four opposite arrows) so that the model will be fitted inside the window. Click it again after appropriate frame is obtained. Click on Modeling tab, and then click the New Sketch icon (a small blue geometry symbol with * on the upper part of the DM window) to create Sketch1 which will appear under XYPlane representing the lower square. Note that this name Sketch1 can be deleted or renamed by right clicking on it and selecting an option.

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Chapter 6 Three-Dimensional Solid Analysis

Click again on the same New Sketch icon to create Sketch2 and Sketch3 to represent the upper rectangle and the circle, respectively. Next, we draw the lower square with the size of 4×4 in. Click on Sketch1 then click the Sketching tab and select Draw. Choose Rectangle to create the lower square with the vertices of (-2,-2) and (2,2). This is done by clicking at the coordinates of (-2,-2) on the model, move the cursor to the coordinates of (2,2), and click it again. Click on Generate (the icon with yellow lightning on the upper-left part of the screen). The lower square will pop up in dark green. Next, we draw the upper rectangle. Click on Sketch2 then click the Sketching tab and select Draw. Choose Rectangle to create this upper rectangle with the vertices of (-2,2) and (-1,6). This is done by clicking at the coordinates of (-2,2) on the model, move the cursor to the coordinates of (-1,6), and click it again. Click on Generate (the icon with yellow lightning on the upper-left part of the screen). The upper rectangle will pop up in dark green. Then, we draw the circle. Click the Modeling tab and select Sketch3. Then click the Sketching tab. Select the Draw tab and choose Circle. Draw a circle with center at the coordinates of (0,0). Do not worry about the

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size of the circle, it will be taken care later. Then, click the Generate button. Select the Dimensions tab and choose Radius. Left click on the circle that just drew, drag the mouse outward without releasing the mouse until seeing an arrow with notation R1, then release the mouse. The desired circle will pop up in dark green. On the Details View window that just appears on the lower left of the DM screen, adjust the radius to 1 and hit Enter. Click on Generate, the circle will radius of 1” will appear. Click on Modeling tab, and then click Generate.

The next step is to extrude the lower square and upper rectangles for 2” and the cylinder for 8” into the z-direction.

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Chapter 6 Three-Dimensional Solid Analysis

Click on the ISO tab on the upper menu bar so that the model is shown in isometric view Select Sketch1 and click Extrude to make a solid body of the lower square with thickness of 2”. Click Apply next to the Geometry under the Details View window, and change the FD1 value under the Details of Extrude1 window to be 2, and hit Enter. Then, click Generate so that the lower square becomes a 3D solid in dark grey as shown in the Figure. Select Sketch2 and click Extrude to make a solid body of the upper rectangle with thickness of 2”. Click Apply next to the Geometry under the Details View window, and change the FD1 value under the Details of Extrude2 window to be 2, and hit Enter. Then, click Generate so that the upper rectangle becomes a 3D solid in dark grey. Select Sketch3 and click Extrude to make a solid body of the circle with thickness of 8”. Click Apply next to the Geometry under the Details View window, and change the FD1 value under the Details of Extrude3 window to be 8, and hit Enter. Then, click Generate so that the circle becomes a solid cylinder in dark grey. Save the project as 3D Solid Stress and close the DM window. The model is ready for meshing, but before that, we will specify the boundary conditions on model geometry first.

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(c) Assigning Material Properties and Creating Mesh

 



On the main Project Schematic window, double click on Model, the 3D solid model will appear back on the main window. Double click on Geometry item, the Solid item will popup. Select the Solid item and select “My Aluminum Material” (the name assigned earlier containing material properties of this problem) which is on the right-hand-side of Assignment under Material in Details of “Solid” window. The 3D solid model will become green.

Select Mesh under Model section, the Details of “Mesh” window will appear on the lower left of the screen. Click at Relevance with the value of 80. Right click at the Mesh

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again and select Generate Mesh. A finite element mesh with three-dimensional elements will appear as shown in the figure. Save the project and close the DM window.

(d) Applying Boundary Conditions, Solving for and Displaying Solutions

 

Next, the boundary conditions of the fixed left end and the pressure load can be applied. This will be done, one at a time, starting from the left end face. On the main Project Schematic window, double click on Setup. Select Analysis Settings under Static Structural. Select the Supports tab on the upper menu bar and select Fixed Support item, then select Face icon (box with arrow and green face). Move the cursor to the left end face and click at it, this face will become green. Click Apply button next to the Geometry button.

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

Select Analysis Settings under Static Structural again. Select the Loads tab on the upper menu bar and select Pressure item, then select Face icon (box with arrow and green face). Move the cursor to the upper right face and click at it, this face will become green. Change the Magnitude value to 2000 psi and click Apply button next to the Geometry button.



The problem is now ready to solve for solution. Right click the Solution item and under Static Structural and select the Solve tab.



Click the Solution item, the Deformation tab will appear on the lower menu bar. Click on this Deformation tab and

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Chapter 6 Three-Dimensional Solid Analysis



Select the Total option, the Total Deformation item will pop-up beneath the Solution item. Right click at the Total Deformation item and select Evaluate All Results, the solution in form of color fringe plot will appear as shown in the figure. Click the Solution item, the Stress tab will appear on the lower menu bar. Click on this Stress tab and Select the Equivalent (von-Mises) option, the Equivalent Stress item will pop-up beneath the Solution item. Right click at the Equivalent Stress item and select Evaluate All Results, the solution in form of color fringe plot will appear as shown in the figure.

6.4 Application 6.4.1 Stress in Aircraft Structural Component A structural component in an aircraft wing is made from aluminum material that has the Young’s modulus of 7 1010 N m2 and Poisson’s ratio of 0.3. The component is held by internal screws at the two holes on the left portion of the figure. The component is subjected to an offset loading of 100 N at the other end. Since the ANSYS files of this problem can be downloaded from the book website, users can study the model in

6.4 Application

115

details. Users can further understand the solution behaviors by modifying the boundary conditions.

With the imported CAD model, a finite element mesh is generated as shown in the figure. The mesh consists of 5,221 tetrahedral elements and 9,881 nodes. Since there are three displacement unknowns of u , v, w at each node, the problem thus contains the total of 29,643 equations before applying the boundary conditions.

The model is constraint as fixed support at the two holes as shown by the symbol A in the figure. An offset loading is applied at the other end as indicated by the symbol B.

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Chapter 6 Three-Dimensional Solid Analysis

With the mesh and applied boundary conditions, the analysis can be performed. The computed von-Mises stress is displayed on the deformed model as shown in the figure. The stress distribution suggests that the design is appropriate. This can be seen by a relatively uniform stress on the right structural portion of the model.

Chapter 7 Vibration Analysis

Vibration and dynamics analyses play important role in structural and machine design. A stay-cable bridge and a computer hard disk drive under certain conditions may vibrate with large amplitude. Inappropriate design can cause structural failure and machine breakdown. Study of vibration and dynamics behaviors is thus important to new structural and machine design today. Structures and machines are normally comprised of truss, beam, plate and solid components. The finite element method is a widely used tool to provide detailed solution behaviors effectively. In this chapter, we start by reviewing a standard vibrational problem that we have learnt in our undergraduate courses. We then look at the differential equations that govern the vibration behaviors of the truss, beam, plate and solid components. Simple finite elements and their element equations are introduced

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Chapter 7 Vibration Analysis

before using ANSYS software to solve an academic type problem. Benefits of the method and software are demonstrated by analyzing a practical application of an automobile frame structure.

7.1 Basic Equations 7.1.1 Differential Equations A classical example that we have learnt in the vibration course is the harmonic oscillation of a mass-spring system as shown in the figure. By using the Newton’s second law, the differential equation that describes the mass movement u in the xdirection with time t can be derived as,

k

k m

x

t u

u0

m

u

x

T

d 2u  ku  0 dt 2 d 2u  2 u  0 or, dt 2 where  2  k m represents the square of the circular frequency, i.e., k   m In the above equation, m is the mass and k is the spring stiffness. The general solution of the governing differential equation is, m

119

7.1 Basic Equations u (t )



A sin  t  B cos  t

where A and B are constants that can be determined from the initial conditions. As an example, if the initial displacement and velocity are u0 and zero, respectively, the mass movement behavior is as shown in the figure. For an oscillating cycle of T, the frequency f that represents the cycles per second, or Hertz, is, 

f

1 T

Thus, the circular frequency  and the frequency f are related by,



   (rad sec)



f 2   (2 rad cycle) (cycles sec)

The value of  above is also known as the natural circular frequency. The oscillation in this classical example is called free vibration. If the mass is subjected to an external force F (t ) in the form, F (t )  F0 sin  f t

k

m u (t )

then, the governing differential equation of the mass-spring system becomes,

m F (t )

d 2u  ku dt 2



F0 sin  f t

The general solution of this differential equation is,

u (t )  A sin  t  B cos  t 

( F0 sin  f t ) k 1  ( f  )2

The last term in the solution above suggests that the oscillating magnitude u (t ) becomes very large if the applied forcing frequency f is closed to the natural frequency  of the system. Knowing the natural frequency  of the system is thus important to avoid an uncontrollable vibration caused by the external force.

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The idea above could be applied to problem with complicated geometry. The frequency f of the external forces is often difficult to control while the natural frequency  of the system is controllable. Since we know that the natural frequency of the system depends on the overall stiffness and mass, we can alter either the system stiffness or mass. The system mass is not easy to change in general but its stiffness can be altered by modifying, (a) the model geometry (b) the material (c) the boundary conditions The finite element method can provide natural frequency solutions conveniently for different model configuration, materials and boundary conditions. The method is thus suitable for vibration analysis of complicated structures. Since the structures often consist of the truss, beam, plate and solid components, we will look at the differential equations that govern their vibration behaviors as follows. 1D Truss

A

 2u t 2

 EA

 2u x 2

The displacement u  u ( x , t ) varies with the axial coordinate x of the truss and time t,  is the material density, A is the crosssectional area, and E is the material Young’s modulus. 1D Beam

A

2w t 2



EI

4w x 4

The deflection w  w( x , t ) varies with the axial coordinate x of the beam and time t,  is the material density, A is the cross-sectional area, E is the material Young’s modulus and I is the moment of inertia of area.

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7.1 Basic Equations

2D Plate 2w 4w 4w  Eh 3   4 w   h 2  2 t x 2 y 2 y 4  12(1   2 )  x 4 The deflection w  w ( x , y , t ) varies with the coordinates x, y and time t,  is the material density, h is the plate thickness, E is the material Young’s modulus and  is the Poisson’s ratio. 3D Solid  2u  2  t

 xy  x  xz   x y z  xy  y  yz  2v    2  x y z t 2    xz  z  w yz    2  x y z t The displacement components u  u ( x , y , z , t ), v  v ( x , y , z , t ), w  w ( x , y , z , t ) vary with the coordinates x, y , z and time t . The quantities  x ,  y ,  z are the normal stress components while  xy ,

xz ,  yz are the shearing stress components. 7.1.2 Related Equations After the displacement unknowns are solved from the above differential equations, the stresses can be determined by using the related equations as follows. 1D Truss The stress  is determined from the computed displacement u as, 



E

u x

1D Beam The stress  of the beam is determined at any z coordinate from the computed deflection w as,

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Chapter 7 Vibration Analysis



  Ez

2w x 2

2D Plate The stress components  x ,  y ,  xy of the plate are determined at any z coordinate from the computed deflection w as,

x



y



E  2w 2w   z  y 2  1   2  x 2 E  2w 2w    z  1   2  x 2 y 2 



 xy  

E 1 

 2 w   x y  z  

3D Solid The three normal stress components  x ,  y ,  z and three shearing stress components  xy , xz ,  yz are determined from the computed displacement components u , v, w as, E u v w     (1   ) x y z  (1   )(1  2 )  E v w   u  (1   )     y z  (1   )(1  2 )  x

x



y



z



E v w   u   (1   )   y z  (1   )(1  2 )  x

 xy



E  u v   2(1   )  y x 

 xz



E  u w     2(1   )  z x 

 yz



E  v w   2(1   )  z y 

7.2 Finite Element Method

123

7.2 Finite Element Method 7.2.1 Finite Element Equations The finite element equations for the truss, beam, plate and solid elements can be derived by applying the method of weighted residuals to their differential equations. Detailed derivation can be found in many finite element textbooks including the one written by the same author. The derived finite element equations are in the same form of,  M    K    F  where  M  is the element mass matrix;  K  is the element stiffness matrix;  F is the element load vector;   is the element vector containing nodal unknowns; and  is the element vector containing nodal accelerations. After assembling all element equations together and applying the boundary conditions, solutions of the element nodal unknowns   at different times can be determined using the method of: (a) modal superposition, and (b) recurrence relations. The modal superposition method involves determination of the eigenvalues and eigenvectors as the first step. The recurrence relations method employs the finite difference approximation to transform the acceleration vector  into the nodal unknown vector   . Details of these two methods are omitted herein for brevity. They can be found in many advanced finite element method books including the book written by the author. 7.2.2 Element Types Truss, beam, plate and solid elements are presented in the preceding chapters. With their element interpolation functions, the corresponding element stiffness matrix  K  and element load vector  F can be derived. The mass matrix  M  that arises in this chapter for analysis of vibration problems is in the form of an integral over element domain  as,

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Chapter 7 Vibration Analysis

M 



T    N   N  d 

where  N  is the element interpolation function matrix of the element type selected. As an example, the mass matrix corresponding to the two-node truss element with the length of L and crosssectional area of A in chapter 2 is,  AL  2 1  M   6  1 2  Similarly, the mass matrix of the two-node beam element with the length of L and cross-sectional area of A in chapter 3 is, 22 L 54 13L   156  3L2  4 L2 13L  AL  22 L  M   22 L  13L 156 420  54   4 L2   13L 3L2 22 L The mass matrices of other element types can also be determined in the same way. It is noted that many symbolic manipulation software packages, such as MATLAB, Mathematica, Maple, Maxima, etc., can be used to derive the mass matrices for many element types in closed-form expressions. The element matrices in closed-form expressions can help reducing the computational time. Several finite element computer programs have been developed to analyze structural vibration and dynamics problems in the past. We will employ ANSYS through its Workbench to analyze an academic type and realistic problems in the following sections.

7.3 Academic Example 7.3.1 Vibration of Thin Plate A square plate with the dimensions of 1  1 m and thickness of 0.01 m is shown in the figure. The plate is made from a material that has the Young’s modulus of 10.92  106 N m 2 and Poisson’s ratio of 0.3. The plate is clamped along its four edges. We will employ ANSYS through the Workbench to determine its

125

7.3 Academic Example

frequencies and the corresponding mode shapes. It is noted that this example is a classical problem often used to study plate vibration because their exact solutions are available. Understanding their solution behaviors will provide confidence prior to analyzing problems with more complicated geometry. Y

  1 Kg/m 3 E  10.92  10 6 N/m 2 1m

v  0.3

h  0.01 m

X 1m

(a) Starting ANSYS Workbench    



Open the ANSYS Workbench, set the Units menu on the upper tab to Metric (kg,m,s,C,A,N,V). On the Analysis Systems window, click twice on the Modal item. A new small box will appear on the Project Schematic window. Retype the name in the lower blue tab as the desired project name, e.g., Plate Modal Analysis, and hit Enter. Right click on the Engineering Data tab and select the Edit… option, the A2: Engineering Data window will pop-up. Double click on Click here to add a new material and type in a new material name, e.g., “My Plate Material”, and hit Enter. Click at the Density under Physical Properties and drag it to the Property list at the bottom of the window. Enter the Density value as 1 and hit Enter.

126 

Chapter 7 Vibration Analysis

Click at the Isotropic Elasticity under Linear Elastic and drag it to the Property list at the bottom of the window. Enter the Young’s Modulus value as 10.92e6 and hit Enter, enter the Poisson’s Ratio value as 0.3 and hit Enter. Then, close the Engineering Data tab and click at the Project tab on the upper menu, it will bring back to the main Project Schematic window.

(b) Creating Geometry 

Right click on the Geometry tab and select the New Geometry…. This will launch the ANSYS Design Modeler (green logo DM).

7.3 Academic Example    

 





127

On DM window, set unit in the Units menu on the upper tab to Meter. On the Tree Outline window, right click on XYPlane and select Look at. The X-Y-Z coordinates on the Model View in 3D view will become X-Y coordinates in 2D view. Select the Sketching tab below the Tree Outline window, the Sketching Toolboxes will pop-up in the same place. Select the Settings tab and then Grid, activate the buttons Show in 2D and Snap. The grid will appear on the main window in two dimensions. Grid snapping provides convenience when drawing model. Make sure that the Major Grid Spacing is set to 1 m, Minor-Steps per Major is 1, and Snaps per Minor is 1. Enlarge the scale by clicking at the Box Zoom icon on the upper part of the screen (icon with plus sign on the magnifying glass) and draw a box with appropriate size to zoom in. Click it again after appropriate scale is showing on the window.

Click on Modeling tab, and then click the New Sketch icon (a small blue geometry symbol with * on the upper part of the DM window) to create Sketch1 which will appear under XYPlane. Note that this name Sketch1 can be deleted or renamed by right clicking on it and selecting an option. Next we draw the square. Click on Sketch1.

128 

  

  

Chapter 7 Vibration Analysis

Click the Sketching tab and select Draw. Choose Rectangle to create the square with the vertices of (0,0) and (1,1). This is done by clicking at the coordinates of (0,0) on the model, move the cursor to the coordinates of (1,1) and click the mouse again, the square domain will appear. Click on Generate (the icon with yellow lightning on the upper-left part of the screen). The desired square will pop up in dark green. The next important step is to go to the Concept tab on top of the screen and select Surfaces From Sketches. Select the Sketch1, the square will become yellow. Click Apply icon on the right side of the Base Objects tab in the Details View at the lower left of the screen. The square will become cyan. The right side of the Base Objects tab will show 1 Sketch. Then, click on Generate. We now have the desired square domain. Click SurfaceSk1 and change the value of Thickness in the Detail View window to 0.01 and hit Enter. Click ISO tab and save file as Plate Modal Analysis, then close the DM window.

(c) Assigning Material Properties and Creating Mesh 

On the main Project Schematic window, double click on Model, the thin plate model will appear back on the main window.

7.3 Academic Example

129



Double click on Geometry item, the Surface Body item will pop-up. Select the Surface Body item and select “My Plate Material” (the name assigned earlier containing material properties of this problem) which is on the righthand-side of Assignment under Material in Details of “Surface Body” window. The plate model will become green.



Select Mesh under Model section, the Details of “Mesh” window will appear on the lower left of the screen. Click at Relevance with the value of 100. Right click at the Mesh again and select Generate Mesh. A finite element mesh will appear as shown in the figure. Save the project and close the DM window.



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Chapter 7 Vibration Analysis

(d) Applying Boundary Conditions, Solving for and Displaying Solutions  

  

Next, the boundary conditions of the four clamped edges can be applied. These will be done, one at a time, starting from the top edge. Select Analysis Settings under Modal. Select the Supports tab on the upper menu bar with Fixed Support option, then select Edge icon (box with arrow and green edge). Move the cursor to the top edge and click at it, the edge will become green. Click Apply button next to the Geometry button under the Details of “Fixed Support” window.

Repeat the same process to apply clamped-supported boundary condition along the other three edges. The problem is now ready to solve for solution. Right click the Solution item and under Modal and select the Solve tab. Click the Solution item, the Deformation tab will appear on the lower menu bar. Click on this Deformation tab and Select the Total option, the Total Deformation item will

7.3 Academic Example

131

pop-up beneath the Solution item. Right click at the Total Deformation item and select Evaluate All Results, the first mode shape with the frequency of 57.789 Hz in form of color fringe plot will appear as shown in the figure.



Change Mode item under Definition in the Details of “Total Deformation” Window to 2, right click at the Total Deformation item and select Evaluate All Results, the second mode shape with the frequency of 119.72 Hz in form of color fringe plot will appear as shown in the figure.



Change Mode item under Definition in the Details of “Total Deformation” Window to 4, right click at the Total Deformation item and select Evaluate All Results, the

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Chapter 7 Vibration Analysis

fourth mode shape with the frequency of 176.54 Hz in form of color fringe plot will appear as shown in the figure.

7.4 Application 7.4.1 Modal Analysis of Passenger Car Frame

A passenger car frame as shown in the figure composes of thin metal sheets. The metal sheets have the Young’s modulus of 2  1010 N m 2 and Poisson’s ratio of 0.3. We will use ANSYS

7.4 Application

133

through its Workbench to solve for the natural frequencies and mode shapes. We import a CAD file containing the car frame geometry and construct a finite element model for it. We select plate bending elements that allow bending together with plane stress elements for in-plane movement. By using the element size of approximately 5 cm, the finite element model consists of 10,147 elements and 10,068 nodes as shown in the figure. Since there are 6 unknowns of the three displacements and three rotations at each node, the problem thus contains the total of 60,408 equations.

Before analyzing the problem, we change the value of Max Modes to Find under Options in the Details of “Analysis Setting” window to 10. The software will determine the solutions up to the first ten modes. Solutions of the first and tenth modes are shown in the figures.

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Chapter 7 Vibration Analysis

Results from the modal analysis provide important information to designers on the natural frequencies and their mode shapes. Designers can modify the geometry to avoid a large amplitude that may occur when the natural and forcing frequencies are closed to each other. Since the ANSYS files of this problem are available from the book website, readers are can explore the solution behaviors by changing the problem geometry and boundary conditions to increase understanding.

Chapter 8 Failure Analysis Failure analysis is important in structural and machine design. Large structures and machine components may fail under repeated loading. This can occur even though the magnitude of the repeated load is much less than the critical static load. In this chapter, we will employ the finite element method via ANSYS Workbench to predict life of structural components caused by buckling and fatigue. We will use academic examples as well as a practical application to demonstrate capability of the software for failure analysis. For static loading, the popular failure theory of a ductile material is the maximum shear stress theory. Based on the Tresca criterion, the theory states that,

max   yield 2 where max is the maximum shear stress and  yield is the yield stress.

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Chapter 8 Failure Analysis

The distortion energy theory is another popular theory. With the von-Mises criterion, the theory states that,

 max

  yield

where max is the maximum von-Mises stress. It is noted that the maximum stresses above are reduced by the safety factor n for problems in application.

8.1 Buckling 8.1.1 Fundamentals Buckling is a common failure of frame structure caused by too high compressive loading. Beam buckling is an academic example often shown in undergraduate class because it is easy to understand. A beam with the length of L and moment of inertia of I is made from material that has the Young’s modulus of E. The beam is constrained at the left end while the right end is subjected to a compressive load of P as shown in the figure.

E, I P L

The critical bucking load Pcr according to the Euler’s formula is,

Pcr   2 EI k L2 where k is the factor depending on the end boundary conditions. For examples, k  1 when both ends are pinned or hinged, and k  0.7071 when the left end is clamped. The lowest critical buckling load, sometimes called the Euler’s critical load, causes the beam to bend in one direction as shown in the preceding figure. The lowest critical bucking load is,

137

8.1 Buckling

Pcr ( mode 1)   2 E I L2

This critical bucking load corresponds to the first mode shape. For the second mode, the beam shape is similar to an Scurve as shown in the following figure. The corresponding buckling load is, Pcr ( mode 2)  4 2 E I L2 E, I

P L For higher modes, the beam shapes behave in the same fashion but are more complicated. The example above contains only a single beam, determination of its mode shapes and critical buckling loads is not difficult. For a complicated structure with many beams and plates, the classical method cannot provide solution effectively. The finite element method offers a convenient way to yield the mode shapes with critical buckling loads. The method starts from deriving finite element equations for all elements in the structural model. These element equations are in the algebraic form of,

 M    K    0 where  M  is the mass matrix;  K  is the stiffness matrix;   is

the vector containing nodal unknowns; and  is the vector con-

taining nodal accelerations. Then, the eigenvalue problem is solved from,

 K    2 M 

 0

where  denotes the natural frequency. The equations above lead to the eigenvalues i and corresponding eigenvectors. Details for finding the eigenvalues and eigenvectors can be found in advanced finite element textbooks, including the book written by the author.

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Chapter 8 Failure Analysis

We will use ANSYS through its Workbench to find the eigenvalues, eigenvectors together with the critical buckling loads and mode shapes by using an academic example of a single beam in the following section. 8.1.2 Academic Example A rectangular shape beam with the length of 1 m and cross-sectional dimensions of 0.02  0.01 m is shown in the figure. The beam material is the structural steel that has the Young’s modulus of 2  1011 N m2 . The left end is clamped into a wall while the right end is simply supported so that it can move only in its axial direction. The right end is subjected to a compressive force of P = 1 N. We will employ ANSYS to determine the critical buckling loads at different mode shapes.

y h

P b

x L

; P  1N L  1 m ; E  2  1011 N m2 b  .02 m  1 3 b h  1.666667 109 m4  I  12 h  .01 m  (a) Starting ANSYS Workbench   

Open the ANSYS Workbench, set the Units menu on the upper tab to Metric (kg,m,s,C,A,N,V). On the Analysis Systems window, click twice on the Static Structural item. A new small box will appear on the Project Schematic window. Retype the name in the lower blue tab as the desired project name, e.g., Beam Buckling Analysis, and hit Enter.

8.1 Buckling 



139

Right click on the Engineering Data tab and select the Edit… option, the A2: Engineering Data window will pop-up. Double click on Click here to add a new material and type in a new material name, e.g., “My Beam Material”, and hit Enter. Click at the Isotropic Elasticity under Linear Elastic and drag it to the Property list at the bottom of the window. Enter the Young’s Modulus value as 2e11 and hit Enter, enter the Poisson’s Ratio value as 0.3 and hit Enter, and close this window. Then, close the Engineering Data tab and click at the Project tab on the upper menu, it will bring back to the main Project Schematic window.

(b) Creating Geometry 

  

Right click on the Geometry tab and select the Properties, the Properties of Schematic window will appear. Activate the Line Bodies under the Basic Geometry Options and close this small window. Right click on the Geometry tab and select the New Geometry…. This will launch the ANSYS Design Modeler (green logo DM). On DM window, set unit in the Units menu on the upper tab to Meter. On the Tree Outline window, right click on XYPlane and select Look at. The X-Y-Z coordinates on the Model View in 3D view will become X-Y coordinates in 2D view.

Chapter 8 Failure Analysis

140

 

 

Select the Sketching tab below the Tree Outline window, the Sketching Toolboxes will pop-up in the same place. Select the Settings tab and then Grid, activate the buttons Show in 2D and Snap. The grid will appear on the main window in two dimensions. Grid snapping provides convenience when drawing model. Make sure that the Major Grid Spacing is set to 1 m, Minor-Steps per Major is 1, and Snaps per Minor is 1. Enlarge the scale by clicking at the Box Zoom icon on the upper part of the screen (icon with plus sign on the magnifying glass) and draw a box with appropriate size to zoom in. Click it again after appropriate scale is showing on the window.

8.1 Buckling 

 

   

141

Click on Modeling tab, and then click the New Sketch icon (a small blue geometry symbol with * on the upper part of the DM window) to create Sketch1 which will appear under XYPlane. Note that this name Sketch1 can be deleted or renamed by right clicking on it and selecting an option. Next we draw the beam model. Click on Sketch1. Click the Sketching tab and select Draw. Choose Line to create the first line with the end coordinates of (0,0) and (1,0). This is done by clicking at the coordinates of (0,0) on the model, move the cursor to the coordinates of (1,0) and click the mouse again. Click on Generate (the icon with yellow lightning on the upper-left part of the screen). The desired line will become dark green. If the model is too small to see, it can be enlarged by clicking at the Box Zoom icon on the upper part of the screen. Click it again after finishing. The next important step is to go to the Concept tab on top of the screen and select Lines From Sketches. Select the Sketch1, the line will become yellow. Click Apply icon on the right side of the Details of Line1 tab in the Details View at the lower left of the screen. The line will become cyan. Then, click on Generate. The right side of the Base Objects tab will show 1 Sketch. The 1 Part, 1 Body item will appear in the Tree Outline window.

Chapter 8 Failure Analysis

142

 

We now have the required beam model. The next step is to create the beam cross section. Select Rectangular item in Cross Section under the Concept tab. In the Details of Rect1 window, change the base value B to 0.02 m and hit Enter, the height value H to 0.01 m and hit Enter. A blue rectangular cross section will appear on the main Graphic window.



Next, assign this cross section to the Line Body. Double click at 1 Part, 1 Body and select the Line Body, assign Rect1 to the Cross section selection in the Details of Line Body window. Save file as Beam Buckling Analysis, and close the DM window.



(c) Assigning Material Properties and Creating Mesh  

On the main Project Schematic window, double click on Model, the beam model will appear back on the main window. Double click on Geometry item, the Line Body item will pop-up. Select the Line Body item and select “My Beam Material” (the name assigned earlier containing material properties of this problem) which is on the right-hand-side of Assignment under Material in Details of “Line Body” window. The beam model will become green.

8.1 Buckling 



143

Select Mesh under Model section, the Details of “Mesh” window will appear on the lower left of the screen. Select at Element Size under Sizing and change the value on the right column to 0.1 and hit Enter so that the generated element length is approximately 0.1 m. Right click at the Mesh again and select Generate Mesh. A finite element mesh with the 2-node beam elements will appear as shown in the figure. Save the project and close the DM window.

(d) Applying Boundary Conditions, Solving for and Displaying Solutions   

On the main Project Schematic window, double click on Setup, the beam model will appear back on the main window. Next, the boundary conditions on both ends can be applied. This will be done, one at a time, starting from the left end. Select Analysis Settings under Static Structural. Select the Supports tab on the upper menu bar with Fixed Supported option, then select Vertex icon (box with arrow and green dot). Move the cursor to the left end and click at it, the left end will become green. Click Apply button next to the Geometry button under the Details of “Fixed Support” window.

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



Repeat the similar process to apply the constraints of no translation in the y- and z-direction at the right end. Select the Analysis Settings and select the Supports tab on the upper menu bar with Displacement option, then select Vertex icon (box with arrow and green dot). Move the cursor to the right end and click at it, the right end will become green. Click Apply button next to the Geometry button under the Details of Displacement window. Change Y Component and Z Component to Constant with the value of 0 and hit Enter. Repeat the similar process to apply the boundary condition of axial compressive force at the right end by selecting the Analysis Settings, select the Loads tab on the upper menu bar with Force option, and select Vertex icon. Move the cursor to the right end and click at it. Click Apply button and change Vector on the right-hand-side of Define By to Components. Then, input X Component as -1 and hit Enter. Note that, mesh can be shown by clicking the Show Mesh icon on the upper menu bar.

8.1 Buckling

    

145

The problem is now ready to solve for solution. Right click the Solution item and under Static Structural and select the Solve tab. Click the Solution item, then click on Deformation tab on the upper menu bar and select the Total option, the Total Deformation item will pop-up beneath the Solution item. Right click on Solution and select Evaluate All Results, the program will start to solve the model. After completion, the computed displacement will be shown on the main window. Save file and close the DM window.

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(e) Eigenvalue Buckling Part 











The eigenvalue buckling analysis can be now performed. The data from the static structural analysis (model, mesh, solution, etc.) can be transferred directly to the buckling analysis. Drag the Eigenvalue Buckling icon from the Analysis Systems Toolbox window and drop it on to the Solution cell of the highlighted Static Structural in the Project Schematic window. Retype the name in the lower blue tab as the desired project name, e.g., Eigenvalue Buckling, and hit Enter.

Double click on Setup under Eigenvalue Buckling, a new window of Multiple System appears. Right click on Solution under Eigenvalue Buckling and select Solve, the buckling analysis is now performed. Click on Solution and select Total Deformation under Deformation tab. Right click on Total Deformation and select Evaluate All results, the fundamental mode shape will appear as shown in the figure with the computed Load Multiplier of 6727.2. Because the input force P is 1 N, this means the critical load is 6727.2 N.

8.1 Buckling



147

Right click on Mode button in Total Deformation window and change 1 to 2. Right click on Total Deformation and select Evaluate All results, the second mode shape will appear as shown in the figure with the computed Load Multiplier of 19,881.

8.1.3 Application

Buckling of Detergent Bottle A detergent bottle, lying down in the horizontal zdirection as shown in the figure, is made from polyethylene material that has the Young’s modulus of 1.1109 N m 2 and the Poisson’s ratio of 0.42. The bottle is subjected to an external compressive loading in the z-direction of 5 Kg. In addition, the

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Chapter 8 Failure Analysis

bottle has to support hydrostatic pressure of the liquid detergent with the density of 106 Kg mm3 . We will employ ANSYS through its Workbench to analyze the possibility of buckling when the bottle thickness is 0.5 mm.

Using the imported CAD file of the bottle, we begin by discretizing the model into a number of small plate elements with their element sizes of approximately 5 mm. The generated finite element mesh consists mostly of the quadrilateral elements with few triangular elements. The mesh contains a total of 4,385 elements and 4,193 nodes. Users may change the element sizes, if preferred, by downloading the ANSYS files from the book website.

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149

The next step is to apply the problem boundary conditions. As shown in the figure, we fix the bottle base (A symbol) and apply the compressive load of 49.05 N in the zdirection (B symbol). The hydrostatic pressure from the liquid detergent inside the bottle is applied in the z-direction (C symbol).

The computed von-Mises stress is displayed on the bottle deformed shape. The maximum stress occurs at the bottle neck as shown in the figure.

The critical buckling load is then determined by following the same procedure as explained in the preceding academic example. The computed load multiplier as shown in the

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150

figure is 3.3892. This means the critical bucking load of the bottle is 166.24 N because the applied load is 49.05 N. In another word, the bottle won’t buckle if the applied compressive load is less than 166.24 N.

This example highlights benefits of the finite element method to provide information necessary for the design of complicated model. The ANSYS software though the use of its Workbench helps the analysis process to proceed with ease.

8.2 Fatigue and Life Prediction 8.2.1 Fundamentals Fatigue is one of the common problems that causes structural failure. The failure may occur even though the stress is less than the yield or ultimate stress if the structure is under cyclic loading. As shown in the figure, the cyclic stress varies up and down with cycles, where max is the maximum stress, mim is the minimum stress, m is the mean stress, and  a is the alternating stress.

151

8.2 Fatigue and Life Prediction Stress

a  max  min

m Cycles

The stress variation behavior as shown in the figure creates fatigue leading structural failure. Life of a structural part or machine component is usually predicted by using the S-N curve of the test specimen as shown in the figure. Stress

S-N curve

e Endurance limit

Se

Ne

Cycles

If the computed effective stress  e is less than the endurance limit stress Se , the structural part or machine component is safe. In the opposite way, if the computed effective stress  e is larger than the endurance limit stress Se , we can determine the number of cycles before the structural part or machine component will fail. The criteria to estimate that a structural part or machine component may fail are suggested by: (a) Soderberg, (b) Goodman, and (c) Gerber, as follows.

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(a) Soderberg criterion states that a structural part or machine component is safe if, a  1  m  y Se n

where Se is the endurance limit stress,  y is the yield stress and n is the design safety factor. (b) Goodman criterion states that a structural part or machine component is safe if, a  1  m  Se n u where  u is the ultimate stress.

(c) Gerber criterion states that a structural part or machine component is safe if, 2

a

  1   m  Se n  u  The three criteria when n  1 are plotted as shown in the figure. The Soderberg criterion is the most conservative measure while the Goodman and Gerber criteria are the lesser ones, respectively.

a Gerber

Se

Goodman Soderberg

y

u

m

If the computed stress does not meet one of the criteria above, the structural part or machine component may fail at a limited time. Its limited life is normally estimated in form of the

153

8.2 Fatigue and Life Prediction

stress cycles. As an example, if we follow the Goodman criterion, the effective stress e is determined from,

a   m e u



1 n

The computed effective stress e is used to further determine the number of cycles Ne from the S-N curve as shown earlier. Thus, the life of the structural part or machine component can be predicted. We will employ an academic example and an application problem to demonstrate life prediction of a structural part and a machine component in the following sections. 8.2.2 Academic Example A U-shape plate with its thickness of 0.02 m is shown in the figure. The plate is made of structural steel material with the Young’s modulus of 2  1011 N m2 and Poisson’s ratio of 0.3. The plate is clamped along the left edge while the right edge is subjected to a cyclic pressure loading of 108 Pa. We will use ANSYS software through the Workbench to estimate the plate life in form of the pressure cycles. y

E  2 1011 N/m 2 ,   0.3, t  .02 m .2



R  .05



p  108 Pa

.2 .2

x

.6

.2

(a) Starting ANSYS Workbench



Open the ANSYS Workbench, set the Units menu on the upper tab to Metric (kg,m,s,C,A,N,V).

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Chapter 8 Failure Analysis

On the Analysis Systems window, click twice on the Static Structural item. A new small box will appear on the Project Schematic window. Retype the name in the lower blue tab as the desired project name, e.g., Life Prediction, and hit Enter. Right click on the Engineering Data tab and select the Edit… option, the A2: Engineering Data window will pop-up. Use the default material of Structural Steel with the Young’s Modulus value as 2e11 and Poisson’s Ratio value as 0.3, and close this window. Then, close the Engineering Data tab and click at the Project tab on the upper menu, it will bring back to the main Project Schematic window.

(b) Creating Geometry

   

Right click on the Geometry tab and select the New Geometry…. This will launch the ANSYS Design Modeler (green logo DM). On DM window, set unit in the Units menu on the upper tab to Meter. On the Tree Outline window, right click on XYPlane and select Look at. The X-Y-Z coordinates on the Model View in 3D view will become X-Y coordinates in 2D view. Select the Sketching tab below the Tree Outline window, the Sketching Toolboxes will pop-up in the same place.

8.2 Fatigue and Life Prediction



 



155

Select the Settings tab and then Grid, activate the buttons Show in 2D and Snap, the grid will appear on the main window. Grid snapping provides convenience when drawing model. Change the Major Grid Spacing to 1 m, Minor-Steps per Major is 5, and Snaps per Minor is 1. Enlarge the scale by clicking at the Box Zoom icon on the upper part of the screen (icon with plus sign on the magnifying glass) and draw a box with appropriate size to zoom in. Click it again after appropriate scale is showing on the window.

Click on Modeling tab, and then click the New Sketch icon (a small blue geometry symbol with * on the upper part of

156

 







Chapter 8 Failure Analysis

the DM window) to create Sketch1 which will appear under XYPlane. Note that this name Sketch1 can be deleted or renamed by right clicking on it and selecting an option. Click on Sketch1, then click the Sketching tab and select Draw. Choose Line to create the lower line with the end coordinates of (0,0) and (1,0). This is done by clicking the mouse at the coordinates of (0,0) on the model, move the cursor to the coordinates of (1,0), and click the mouse again. Then, follow the same procedure to create all other lines, and click Generate. The left fillet is created by selecting Modify tab, then change the Radius to 0.05 and hit Enter. Click at the corner to create the fillet. Follow the same procedure for the right fillet.

Next, click Extrude to make a solid body of the plate with thickness of 0.02. Click Apply next to the Geometry under the Details View window, and change the FD1 value under the Details of Extrude1 window to be 0.02, and hit Enter. Then, click Generate so that the plate becomes a 3D solid in dark grey. Click ISO tab to display model in 3D and save the file as Life Prediction, then close the DM window.

8.2 Fatigue and Life Prediction

157

(c) Generating Mesh

   

On the Project Workbench window under Project Schematic, click twice on Model. On the pop-up Outline window, select Mesh. Change the value on the right-hand-side of Relevance under the Details of Mesh window to 100. Click Update on the menu bar above the Outline window, a mesh will be generated.

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(d) Applying Boundary Conditions, Solving for and Displaying Solutions

 



The boundary conditions of the edge constraints and loading can now be applied. These will be done, one at a time, starting from the fixed left edge. Select Analysis Settings under Static Structural. Select the Supports tab on the upper menu bar with Fixed Support option, then select Face icon (box with arrow and green face). Move the cursor to the left edge and click at it, the edge will become green. Click Apply button next to the Geometry button under the Details of “Fixed Support” window.

Next, apply the loading on the right edge. This is done by selecting the Loads tab on the upper menu bar with Pressure option, then select Face icon (box with arrow and green face). Move the cursor to the right edge and click at it, the edge will become green. Click Apply button next to the Geometry button under the Details of “Pressure”

8.2 Fatigue and Life Prediction

159

window and input the value of 1e8 and hit Enter. Click Show Mesh button on the upper tab to display mesh on the model.



Note that the right edge is constrained so that it can’t move in the z-direction. This can be done by selecting the Supports tab on the upper menu bar with Displacement option, then select Face icon (box with arrow and green face). Move the cursor to the right edge and click at it, the edge will become green. Click Apply button next to the Geometry button under the Details of “Displacement” window and change the Z Component to be Constant as 0.



The problem is now ready to solve for solution. Right click the Solution item and under Static Structural and select the Solve tab.

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

Click the Solution item, the Deformation tab will appear on the lower menu bar. Click on this Deformation tab and Select the Total option, the Total Deformation item will pop-up beneath the Solution item. Right click at the Total Deformation item and select Evaluate All Results, the deformed shape will appear as shown in the figure.



To show the maximum principal stress, click the Solution item, then Stress tab and select the Maximum Principal option, the Maximum Principal Stress item will pop-up beneath the Solution item. Right click at the Maximum Principal Stress item and select Evaluate All Results, the maximum principal stress will be plotted on the deformed shape as shown in the figure.

8.2 Fatigue and Life Prediction

161

(e) Fatigue Analysis Part



To perform fatigue analysis, right-click on Solution under Static Structural. Select Insert, Fatigue, and Fatigue Tool. In the Detail of “Fatigue Tool”, set the Mean Stress Theory to None.



Right-click on Fatigue Tool in the Outline, and select Insert, then Life. Right click on Life and select Evaluate All results, the life in form of cycles will appear as shown in the figure.



Follow the same procedure by right-clicking on Fatigue Tool, and selecting Insert, then Safety Factor. Right click

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on Safety Factor and select Evaluate All results, the life in form of cycles will appear as shown in the figure.

8.2.3 Application Fatigue and Life Prediction of Piston Rod A piston rod lying in the x-y-z coordinates as shown in the figure is made from a material that has the Young’s modulus of 2 1011 N m2 and Poisson’s ratio of 0.3. The rod is subjected to a force from the piston pin with the magnitude of 20,000 N in the negative y-direction. We will use ANSYS through its Workbench to estimate the life span of this piston rod.

8.2 Fatigue and Life Prediction

163

We start by importing the CAD file of the piston rod. A finite element mesh is constructed from the CAD model by assigning the element size of 1 mm. The generated mesh, as shown in the figure, consists of 93,575 tetrahedral elements.

We apply the boundary condition of the compressive force from piston pin (A symbol) along the inner surface of the pin tube as shown in the figure. The applied force has magnitude of 20,000 N in the negative y-direction. We also apply the fixed boundary condition along the inner surface of the crankshaft tube (B symbol) at the other end of the rod.

The analysis is performed to determine the deformation shape and maximum principal stress. The figure shows that the

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maximum principal stress occurs at the outer surface connecting between the pin tube and axial bar of the rod.

The computed stress is the used to estimate the life span of the rod. The predicted life in form of the loading cycles is shown in the figure. Result indicates that the location of maximum stress has the shortest life span.

Since the ANSYS files are available from the book website, users may alter the boundary conditions by changing magnitude and direction of the applied force. This will increase understanding of the solution behaviors. Such understanding may lead to shape modification of the piston rod. The modified rod shape with lower stress will increase its life span.

Chapter 9 Heat Transfer Analysis Heat transfer problem is one of the simplest problems normally used to study the finite element method. This is mainly because the heat transfer problem contains only one basic unknown of the temperature. The temperature has a clear physical meaning which is easy to understand. Solving a heat transfer problem by using analytical method in the past was difficult. Exact solution is not available if the problem has complicated boundary conditions. The finite element method helps alleviating such difficulty, especially when the geometry of the problem is complicated. In this chapter, we begins by reviewing the governing differential and related equations of heat transfer problem. The finite element method for analyzing the heat transfer problem is described. Typical element equations and popular element types are presented. ANSYS through its Workbench is then employed to solve academic example and application problem.

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Chapter 9 Heat Transfer Analysis

9.1 Basic Equations 9.1.1 Differential Equation The conservation of energy at any location in an isotropic three-dimensional solid is described by the differential equation,

   T    T    T   T   Q  0 k   k  k   x  x  y  y  z  z   t   where  is mass density of the solid material, c is the specific heat, k is the thermal conductivity coefficient, Q is the internal heat generation rate per unit volume, and T is the temperature that varies with the coordinates x, y, z and time t . For steady-state heat transfer, the differential equation above becomes,

c

  T    T    T   k  + k  + k  +Q  0  x  x  y  y  z  z  If heat transfer occurs only in the two-dimensional x-y plane with constant thermal conductivity coefficient k , the differential equation reduces to,

2T 2T + x2 y2

 

Q k

which is in form of the Poisson’s equation. In addition, if there is no internal heat generation, the governing differential equation reduces further to,

2T 2T + x2 y2

 0

which is called the Laplace’s equation. Even though the Laplace’s equation above looks very simple, its exact solution T ( x, y ) is still difficult to derive especially when the problem geometry is complicated.

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9.1 Basic Equations 9.1.2 Related Equations

The governing differential equations in the preceding section are derived by using the Fourier’s law. The law relates the conduction heat flux components with the temperature gradients. For isotropic material, the conduction heat flux in the x-direction is,

T x Boundary conditions of heat transfer at the domain surface may consist of,  k

qx

(a) Specified temperature,

T  Ts where Ts may be constant or varies with x-, y-, z-coordinates and time t. (b) Specified heat flux,

q   qs where qs is the specified heat flux which must be in equilibrium with the conduction heat flux q at the surface. (c) Convection heat transfer,

q  h(Ts  T ) where h is the surface convection coefficient and T is the surrounding medium temperature. (d) Radiation heat transfer,

q   (Ts4  T4 ) where  is the surface emissivity and  is the Stefan-Boltzmann constant. For transient heat transfer, an initial condition is needed,

T ( x, y, z,0)  T0 ( x, y, z) where T0 is the initial temperature of the solid.

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Chapter 9 Heat Transfer Analysis

9.2 Finite Element Method 9.2.1 Finite Element Equations Finite element equations can be derived by applying the method of weighted residuals to the governing differential equation. Details of the derivation can be found in many finite element textbooks including the one written by the author. The derived finite element equations in matrix form are, C T   K    K    K T  c

h

r

 Qc   QQ   Qq   Qh   Qr  where C  is the capacitance matrix;  K c  is the conduction

matrix;  K h  is the convection matrix;  K r  is the radiation matrix; T is the vector containing rate of change of nodal temperatures;

T  is the vector containing nodal temperatures; Qc  is the

conduction load vector; QQ  is the heat generation load vector;

Qq  is the specified heating load vector; Qh  is the convection

load vector; and Qr  is the radiation load vector. These element matrices and load vectors depend on element types as described in the following section. 9.2.2 Element Types The one-dimensional two-node rod element is shown in the figure. The finite element matrices and load vectors can be derived in closed form, such as, qs h(T  T )  (T 4  T4 )

Q

p

2

1 x

L

A

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9.2 Finite Element Method

1 1  6 1 q pL 1 hT pL 1 QAL 1 QQ   2 1 ; Qq   s2 1 ; Qh   2 1    These closed-form matrices and vectors can be used to develop a finite element computer program directly. The three-node triangular element is a simple element type for learning the finite element method in two dimensions. The element consists of a node at each corner as shown in the figure. The finite element matrices and load vectors can be derived in closed form. Examples of these matrices and load vectors are, h(T  T ) qs 3  (T 4  T4 )

 cAL  2

C  

1 2

;

 Kc  

Q y

kA  1 L  1

2

t x

1

2 1 1 1 2 1 C    12  1 1 2  2 1 1 hA  1 2 1  Kh     12 1 1 2  b1b1  c1c1 kt  b1b2  c1c2  Kc   4A   b1b3  c1c3

 cAt 

1 QAt   ; QQ   1 3  1 1 hT A   ; Qh   1 3  1 b1b2  c1c2 b1b3  c1c3  b2b2  c2c2 b2b3  c2c3   b2b3  c2c3 b3b3  c3c3 

where bi , ci ; i =1, 2, 3 are the coefficients that depend on the nodal coordinates xi , yi and A is the element area. Details for determining these coefficients and area are given in chapter 4.

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The four-node quadrilateral element, as shown in the figure, is a popular two-dimensional finite element. This is because the quadrilateral element can provide a more accurate solution as compared to two triangular elements. However, numerical integration is needed to compute the finite element matrices and load vectors.

qs

h(T  T ) 4

 (T 4  T4 )

Q

y

3

1

2

t

x

The four-node tetrahedral element is a simple element type. The element contains four faces with a node at each corner as shown in the figure. The element matrices and load vectors can be derived in closed form ready for computer programming. Examples of an element matrix and a load vector are,

qs

4

h(T  T )

Q

 (T 4  T4 )

y

x

3

1

z

2

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9.3 Academic Example

C  

2  cV 1 20 1  1

1 2

1 1

1 1

2 1

1 1  1  2

;

QQ 

1 QV 1   4 1 1

where V is the element volume. The hexahedral element is a widely used element for analyzing three-dimensional problems. The element consists of eight nodes and six faces as shown in the figure. The element can provide higher solution accuracy as compared to the tetrahedral element. Because the element employs more complicated interpolation functions, its element matrices and load vectors must be determined by using numerical integration. qs

5

8

6 7

h(T  T )

 (T 4  T4 )

1 2

z

4

y x

3

9.3 Academic Example 9.3.1 Plate with Specified Edge Temperatures A rectangular plate with the dimensions of 2  1 m and thickness of 0.01 m has specified temperatures along the four edges as shown in the figure. The plate is made from a material that has the thermal conductivity coefficient of 1 W m-C . We

Chapter 9 Heat Transfer Analysis

172

will employ ANSYS through its Workbench to determine the temperature distribution in the plate. This example is an academic example for which the exact temperature solution is in the simple form of, T ( x, y ) 

Y

1T 0

sin( x 2) sinh( y 2) sinh( 2)

T  sin( x 2)

k 1

T 0

X T 0 2

(a) Starting ANSYS Workbench    

Open the ANSYS Workbench, set the Units menu on the upper tab to Metric (kg,m,s,C,A,N,V). On the Analysis Systems window, click twice on the Steady-State Thermal item. A new small box will appear on the Project Schematic window. Retype the name in the lower blue tab as the desired project name, e.g., Plate with Specified Edge Temperatures, and hit Enter. Right click on the Engineering Data tab and select the Edit… option, the A2: Engineering Data window will pop-up. Double click on Click here to add a new material and type in a new material name, e.g., “My Ideal Material”, and hit Enter.

9.3 Academic Example 







173

Click on the Isotropic Thermal Conductivity under Thermal and drag it to the Property list at the bottom of the window. Enter the Isotropic Thermal Conductivity value as 1 and hit Enter. Then, close the Engineering Data tab and click on the Project tab on the upper menu, it will bring back to the main Project Schematic window.

Right click on the Geometry tab and select the Properties option, the Properties of Schematic window will open. Change the Analysis Type under the Advanced Geometry Options from 3D to 2D. Then, close this small window. Right click on the Geometry tab and select the New Geometry…. This will launch the ANSYS Design Modeler (green logo DM).

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(b) Creating Geometry

   

 



On DM window, set unit in the Units menu on the upper tab to Meter. On the Tree Outline window, right click on XYPlane and select Look at. The X-Y-Z coordinates on the Model View in 3D view will become X-Y coordinates in 2D view. On the Tree Outline window, select on XYPlane and Select the Sketching tab below the Tree Outline window, the Sketching Toolboxes will pop-up in the same place. Select the Settings tab and then Grid, activate the buttons Show in 2D and Snap. The grid will appear on the main window in two dimensions. Grid snapping provides convenience when drawing model. Make sure that the Major Grid Spacing is set to 1 m, Minor-Steps per Major is 1, and Snap per Minor is 1. Enlarge the scale by clicking at the Box Zoom icon on the upper part of the screen (icon with plus sign on the magnifying glass) and draw a box with appropriate size to zoom in. Click it again after appropriate scale is showing on the window.

Click on Modeling tab, and then click on the New Sketch icon (a small blue geometry symbol with * on the upper part of the DM window) to create Sketch1 which will appear under XYPlane. Note that this name Sketch1 can

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be deleted or renamed by right clicking on it and selecting an option. Next, to draw the rectangle with the size of 2×1, click on Sketch1. Click on the Sketching tab and select Draw. Choose Rectangle to create a rectangle with the vertices of (0,0) and (2,1). This is done by clicking at the coordinates of (0,0) on the model, move the cursor to the coordinates of (2,1), and click the mouse again. Click on Generate (the icon with yellow lightning on the upper-left part of the screen). The desired rectangle will pop-up in dark green.

The next important step is to go to the Concept tab on top of the screen and select Surfaces From Sketches. Select Sketch1, the rectangle will become yellow. Click Apply icon on the right side of the Base Objects tab in the Details View at the lower left of the screen. The rectangle will become cyan. The right side of the Base Objects tab will show 1 Sketch. Then, click on Generate. We now have rectangular domain of the plate. Save file as Plate with Specified Edge Temperatures, and close the DM window.

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(c) Assigning Material Properties and Creating Mesh

 

On the main Project Schematic window, double click on Model, the plate model will appear back on the main window. Double click on Geometry item, the Surface Body item will pop-up. Select the Surface Body item and change the Thickness to 0.01 and hit Enter. Then, change select the material name as “My Ideal Material” (the name assigned earlier containing material properties of this problem) which is on the right-hand-side of Assignment under Material in Details of “Surface Body” window. The plate model will become green.

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Select Mesh under Model section, the Details of “Mesh” window will appear on the lower left of the screen. Change the Relevance value under the Details of “Mesh” window to 100. Right click at the Mesh again and select Generate Mesh. A finite element mesh with the 4-node quadrilateral elements will appear as shown in the figure. Save the project and close the DM window.

(d) Applying Boundary Conditions, Solving for and Displaying Solutions





Next, the boundary conditions of specified temperatures along the 4 edges can be applied. We will apply zero temperature to the right, bottom and left edges at the same time before applying the temperature along the top edge. On the main Project Schematic window, double click on Setup. Select Analysis Settings under Steady-State Thermal. Select the Temperature tab on the upper menu bar, then select Edge icon (box with arrow and green edge). Hold the Ctrl button and click at the right, bottom and left edges. These edges will become green. Click Apply button next to the Geometry button under the Details of Temperature window, and change temperature value next to the Magnitude button to 0, and hit Enter. Note that the righthand-side of Geometry in the Details of “Temperature” window will become 3 Edges.

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Repeat the same process to apply boundary condition along the top edge. This is done by first selecting the Analysis Settings. Select the Temperature tab on the upper menu bar, then select Edge icon. Move the cursor to the top edge and click at it, the bottom edge will become green. Click Apply button next to the Geometry button, select Function next to the Magnitude button and enter sin(90*x), and hit Enter. Note that the Angular Measure of the Function beneath the Magnitude button herein is set to Degrees, not in radians.

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The problem is now ready to solve for solution. Click on the Solution tab and under Steady-State Thermal and right click on the Solve icon, the analysis will be performed. Click on the Solution item, the Thermal tab will appear on the lower menu bar. Click on this Thermal tab and select the Temperature option, the Temperature item will popup beneath the Solution item. Right click at the Temperature item and select Evaluate All Results, the solution in form of color fringe plot will appear as shown in the figure.

9.4 Application 9.4.1 Three-dimensional Heat Transfer through Fins A three-dimensional heat sink, consisting of fins as shown in the figure, is made from a material that has the thermal conductivity coefficient of k  237 W m-  C . The heat sink transfers heat from the bottom surface to the surrounding air by fins. The bottom surface of the heat sink is subjected to a specified heating of q  1500 W m2 . The convection coefficient of fin surface is h  40 W m2 -  C and the surrounding air temperature is

T  30  C . We will use ANSYS through its Workbench to determine the temperature distribution of this heat sink.

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Chapter 9 Heat Transfer Analysis

Y k  237 W/m -  C

1 8

2

h  40 W/m 2 -  C T  30  C

2

Z 2 q  1500 W/m 2

15

X

Model of the heat sink can be constructed easily by using the Line command under the Draw tab. Each line can be drawn to yield the cross section of the model on X-Y plane as shown in the figure.

The Extrude command is used to create the complete model in three dimensions as shown in the figure.

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A finite element mesh is then constructed by using the Element Size of approximately 1 mm. The mesh consists of 1,320 hexahedral elements as shown in the figure.

The boundary condition of convection heat transfer is applied to all surfaces except the bottom one as shown in the figure.

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Chapter 9 Heat Transfer Analysis

The heating on the bottom surface of the model is applied in the same fashion as shown in the figure.

With the mesh and boundary conditions, the problem is then analyzed. The computed temperature distribution is displayed on the heat sink model as shown in the figure.

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Since the ANSYS files are available from the book website, users can modify the boundary conditions to obtain different temperature solutions. This will increase understanding on how to solve heat transfer problems and interpret their solutions. Analyzing heat transfer problems is simpler than other problems because the temperature is only the basic unknown.

Chapter 10 Thermal Stress Analysis Analyzing practical problems sometimes requires knowledge more than one engineering discipline. Thermal stress problem is one of these problems that need the knowledge of heat transfer and solid stress disciplines. Heat transfer analysis is performed firstly to obtain temperature solution. The computed temperature solution is then used as input data to determine the deformation and thermal stresses of the solid. Examples of these problems are automotive engines, electric motors, computer microchips, as well as ceramic cups after pouring hot coffee into them. We will study on how to analyze thermal stress problems in this chapter. The chapter starts from presenting the differential equations that govern heat transfer and equilibrium equations in solids. Corresponding finite element equations are derived for both analysis disciplines. ANSYS is then employed to solve both academic and application problems. We will see that the current finite element software can analyze interdisciplinary problems, such as the thermal stress problem, effectively.

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10.1 Basic Equations Since the differential equations and related equations for heat transfer and stress analyses were presented in details in the preceding chapters, this chapter will review essential equations and show additional equations that relate the two disciplines together. 10.1.1 Differential Equations The conservation of energy at any location in an isotropic three-dimensional solid is represented by the differential equation,

   T    T    T   T   Q  0 k   k  k   x  x  y  y  z  z   t   where  is the mass density, c is the specific heat, k is the thermal conductivity coefficient, and Q is the internal heat generation rate. In the above differential equation, T is the temperature that varies with the x-, y-, z-coordinates and time t. If the temperature change does not significantly alter the solid strain rate, the quasi-static analysis may be used for thermal stress solutions. The condition simplifies the analysis procedure and reduces overall computational time. The computed temperature at a given time is input into the stress analysis to determine the corresponding thermal stress solution. The thermal stress solution is solved from the governing differential equations of the solid,  xy  x  + + xz  0 x y z  xy  y  yz + +  0 x y z  yz  xz  z + +  0 x y z where x ,  y ,  z are the normal stress components in the x, y, z

c

directions, respectively, and  xy ,  xz ,  yz are the shearing stress components.

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10.1 Basic Equations

These six stress components depend on the six strain components and temperature. The six strain components are functions of the three displacement components u , v, w in the x, y, z directions, respectively. The three displacement components are solved from the three governing differential equations above. 10.1.2 Related Equations Boundary conditions for heat transfer problem are: (a) specified temperature, (b) specified surface heating, (c) surface convection, and (4) surface radiation. Details of these boundary conditions are provided in chapter 9. Boundary conditions for stress analysis of solid problem are: (a) specified displacements, and (b) specified tractions on the solid surface. Details of these boundary conditions are described in chapter 6. The basic unknowns of the solid problem are the three displacement components u , v, w which are solved from the three governing differential equations. Since the differential equations are written in forms of the stress components, the relations between the stress and displacement components must be provided. The six stress components can be written in forms of the six strain components as,    C    0  (61)

where

(66)

(61)

    x  y  z  xy  xz  yz  T

The matrix C  is the material elasticity matrix. The total and thermal strain components are,

 T  x  y z  xy  xz  yz  0 T   T  T  T 0 0 0  where  is the coefficient of thermal expansion and T is the difference between the temperature and reference temperature Tref for zero stress, T

 T ( x , y , z )  Tref

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188

For small deformation theory, the strain components are written in forms of the displacement components u , v, w as,

x y

z

u u  v ;  xy   x y  x  u w v ;  xz    z x y w v w ;  yz    z z y 

As mentioned earlier, the heat transfer problem is firstly analyzed for temperature solution. The computed temperature is input into the solid problem for stress analysis. Once the stress analysis is performed and the displacement components are obtained, the six strain components can then be computed. Determination of the six stress components is followed to complete the analysis of thermal stress problem.

10.2 Finite Element Method 10.2.1 Finite Element Equations Finite element equations for heat transfer problem can be derived by applying the method of weighted residuals to the governing differential equation as described in chapter 9. The finite element equations are in the form, C T   K    K    K T  c

h

r

 Qc   QQ   Qq   Qh   Qr  The element matrices on the left-hand side of the equations are the capacitance, conduction, convection and radiation matrices, respectively. The vectors on the right-hand side of the equations are associated with conduction, internal heat generation, specified heating, convection and radiation, respectively. Forms and sizes of these element matrices and vectors depend on the element types. The unknowns of the finite element equations above are the nodal temperatures.

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10.2 Finite Element Method

Similarly, the finite element equations for solid problem can be derived by applying the method of weighted residuals to the governing differential equations as described in chapter 6. The finite element equations are in the form,  K    F   F0  where  K  is the element stiffness matrix;  F  is the element vector containing nodal forces, and  F0  is the element vector containing nodal forces from temperature change. In the finite element equations, the unknowns are the displacement components u , v, w at nodes which are contained in the element vector   . 10.2.2 Element Types A common finite element mesh should be employed for both heat transfer and solid stress analyses. Nodal temperatures obtained from heat transfer analysis can be transferred directly to the same nodes of the solid stress analysis. The overall thermal stress analysis thus can be performed conveniently. The finite element equations of the solid stress problem include the load vector  F0  from the temperature change. This load vector affects the solid solutions of the deformation and stresses. As an example, the load vector  F0  due to temperature change for the two-node truss element as shown in the figure is,  1 F0   A E  (T  Tref )   1 where A is the truss cross-sectional area, E is the material Young’s modulus,  is the coefficient of thermal expansion, T is the average element temperature, and Tref is the reference temperature for zero stress. u2

u1

F0

1

A, E ,  x

L

2

F0

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Chapter 10 Thermal Stress Analysis

The vector  F0  for two- and three-dimensional element types can be derived without difficulty. The finite element equations for both heat transfer and solid stress problems suggest that the process for solving thermal stress problem is straight forward. Again, to avoid difficulty of transferring nodal temperatures from the heat transfer analysis to the solid stress analysis, a common finite element mesh should be used. We will use ANSYS through the Workbench to carry out the thermal stress analysis for both academic and application problems as demonstrated in the following section.

10.3 Academic Example 10.3.1 Thermal Stress Analysis of Thin Plate A rectangular plate with the dimensions of 3  2 ft and thickness of 0.01 ft is made from aluminum material that has the properties as shown in the figure. The plate is subjected to a rooflike temperature distribution with the temperature of 245F and 95F along the X-direction at Y = 0 and 1 ft, respectively. 245F 95F Y

1' X

95F 1.5'

1'

1.5'

k  137 Btu ft-hr- o F,

  0.29,

E  1.5  109 lb ft 2

  12.7  106

o F,

Tref  80 o F

Due to symmetry, we will use only the upper right quarter of the plate as shown in the figure for the analysis. We will

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employ ANSYS to provide roof-like temperature distribution. The computed nodal temperatures will be transferred to the stress analysis to determine the plate deformation and thermal stresses. It is noted that this problem is a classical thermal stress problem for which the analytical solution and experiment data are available. Y

95  F

1'

245  F

X

1.5' (a) Starting ANSYS Workbench    



Open the ANSYS Workbench, set the Units menu on the upper tab to U.S.Customary (lbm,in,s,F,A,lbf,V). On the Analysis Systems window, click twice on the Steady-State Thermal item. A new small box will appear on the Project Schematic window. Retype the name in the lower blue tab as the desired project name, e.g., Heat Transfer Part, and hit Enter. Right click on the Engineering Data tab and select the Edit… option, the A2: Engineering Data window will pop-up. Double click on Click here to add a new material and type in a new material name, e.g., “Aluminum Material”, and hit Enter. Click at the Isotropic Thermal Conductivity and drag it to the Property list at the bottom of the window. Enter the Isotropic Thermal Conductivity value as 137 BTU/(ft^2 hr (F/ft)) and hit Enter, and close this window.

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Close the Engineering Data tab and click at the Project tab on the upper menu, it will bring back to the main Project Schematic window.

(b) Creating Geometry 

Right click on the Geometry tab and select the Properties option, the Properties of Schematic window will open. Change the Analysis Type under the Advanced Geometry Options from 3D to 2D. Then, close this small window.

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Right click on the Geometry tab and select the New Geometry…. This will launch the ANSYS Design Modeler (green logo DM). On DM window, set unit in the Units menu on the upper tab to Foot. On the Tree Outline window, select on XYPlane and select the Sketching tab below the Tree Outline window, the Sketching Toolboxes will pop-up in the same place. Select the Settings tab and then Grid, activate the buttons Show in 2D and Snap. The grid will appear on the main window in two dimensions. Grid snapping provides convenience when drawing model. Make sure that the Major Grid Spacing is set to 1 ft, Minor-Steps per Major is 2, and Snap per Minor is 1. Enlarge the scale by clicking at the Box Zoom icon on the upper part of the screen (icon with plus sign on the magnifying glass) and draw a box with appropriate size to zoom in. Click it again after appropriate scale is showing on the window.

Click on Modeling tab, and then click the New Sketch icon (a small blue geometry symbol with * on the upper part of the DM window) to create Sketch1 which will appear under XYPlane. Note that this name Sketch1 can be deleted or renamed by right clicking on it and selecting an option.

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Next, to draw the rectangle with the size of 1.5  1 ft, click on Sketch1. Click the Sketching tab and select Draw. Choose Rectangle to create a rectangle with the vertices of (0,0) and (1.5,1). This is done by clicking at the coordinates of (0,0) on the model, move the cursor to the coordinates of (1.5,1), then click the mouse again. Click on Generate (the icon with yellow lightning on the upper-left part of the screen). The desired rectangle will pop-up in dark green. The next important step is to go to the Concept tab on top of the screen and select Surfaces From Sketches. Select Sketch1, the rectangle will become yellow. Click Apply icon on the right side of the Base Objects tab in the Details View at the lower left of the screen. The rectangle will become cyan. The right side of the Base Objects tab will show 1 Sketch, with 1 Part, 1 Body appears in the Tree Outline window. Then, click on Generate. We now have rectangular domain representing the upper right quarter of the plate.

Save file as Thermal Stress Problem, and close the DM window.

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(c) Assigning Material Properties and Creating Mesh    



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On the main Project Schematic window, double click on Model, the plate model will appear back on the main window. Select Units on the upper menu bar to U.S.Customary (ft,lbm,lbf,F,s,V,A). Double click on Geometry item, the Surface Body item will pop-up. Select the Surface Body item and change the Thickness of the plate to .01 ft and hit Enter. Select “Aluminum Material” (the name assigned earlier containing material properties of this problem) which is on the right-hand-side of Assignment under Material in Details of “Surface Body” window. The plate model will become green. Select Mesh under Model section, the Details of “Mesh” window will appear on the lower left of the screen. Change the Relevance value under the Details of Mesh window to 100. Right click at the Mesh again and select Generate Mesh. A finite element mesh with the 4-node quadrilateral elements will appear as shown in the figure. Save the project and close the DM window.

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(d) Applying Boundary Conditions, Solving for and Displaying Solutions  



Next, the boundary conditions of specified temperatures along the top and bottom edges can be applied. This will be done, one at a time, starting from the bottom edge. On the main Project Schematic window, double click on Setup. Select Analysis Settings under Steady-State Thermal. Select the Temperature tab on the upper menu bar, then select Edge icon (box with arrow and green edge). Move the cursor to the bottom edge and click at it, the bottom edge will become green. Click Apply button next to the Geometry button under the Details of Temperature window, and change temperature value next to the Magnitude button to 245, and hit Enter.

Repeat the same process to apply boundary condition along the top edge. This is done by first selecting the Analysis

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Settings. Select the Temperature tab on the bottom menu bar, then select Edge icon. Move the cursor to the top edge and click at it, the top edge will become green. Click Apply button next to the Geometry button, and change temperature value next to the Magnitude button to 95, and hit Enter. The problem is now ready to solve for temperature solution. Click the Solution tab and under Steady-State Thermal and click the Solve icon (the icon with yellow lightning on the top menu bar). Click the Solution item, the Thermal tab will appear on the lower menu bar. Click on this Thermal tab and select the Temperature option, the Temperature item will pop-up beneath the Solution item. Right click at the Temperature item and select Evaluate All Results, the solution in form of color fringe plot will appear as shown in the figure.

(e) Stress Analysis Part 

The thermal stress analysis can be now performed. The data from the thermal analysis (model, mesh, temperature solution, etc.) can be transferred directly to the stress analysis on the same mesh.

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Drag the Static Structural icon from the Analysis Systems Toolbox window and drop it on to the Solution cell of the highlighted Heat Transfer Part in the Project Schematic window. Retype the name in the lower blue tab as the desired project name, e.g., Thermal Stress Part, and hit Enter.

Right-click at the Engineering Data under the Heat Transfer Part project window, the A2: Engineering Data window will appear again. Click at the Isotropic Elasticity under Linear Elastic and drag it to the Property list at the bottom of the window. Enter the Young’s Modulus of 1.5e9 psf and hit Enter, the Poisson’s Ratio of 0.29 and hit Enter. Click at My Aluminum Material item assigned earlier and expand the Physical Properties tab in the Toolbox window. Click at the Isotropic Instantaneous Coefficient of Thermal Expansion item and drag to Property item, then enter the value of 12.7e-6 /F. Also enter Temperature value of 80F and hit Enter. Then, close this A2,B2:Engineering Data window.

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(f) Applying Boundary Conditions and Solve for Solution of Stress Analysis Part  



The next step is to apply the boundary condition of frictionless support along the bottom and left edges. On the main Project Schematic window under Static Structural, double click on Setup. Select Analysis Settings under Static Structural. Select the Supports tab on the upper menu bar with Frictionless Support option, then select Edge icon (box with arrow and green edge). Move the cursor to the bottom edge and click at it, the edge will become green. Click Apply button next to the Geometry button under the Frictionless Support window. Repeat the same process to apply boundary condition of frictionless support along the left edge. This is done by selecting the Analysis Settings. Select the Supports tab on the upper menu bar with Frictionless Support option, then select Edge icon (box with arrow and green edge). Move the cursor to the left edge and click at it, the left edge will become green. Click Apply button next to the Geometry button under the Frictionless Support window. The mesh can be shown by clicking at the Show Mesh button on the upper mane bar.

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The problem is now ready to solve for solution. Right click the Solution item and under Static Structural and select the Solve tab. Click the Solution item, the Deformation tab will appear on the lower menu bar. Click on this Deformation tab and Select the Total option, the Total Deformation item will pop-up beneath the Solution item. Right click at the Total Deformation item and select Evaluate All Results, the solution in form of color fringe plot will appear as shown in the figure.

Click the Solution item, the Stress tab will appear on the lower menu bar. Click on this Stress tab and Select the Equivalent (von-Mises) option, the Equivalent Stress

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201

item will pop-up beneath the Solution item. Right click at the Equivalent Stress item and select Evaluate All Results, the solution in form of color fringe plot will appear as shown in the figure.



The normal stress in X-direction is displayed by selecting the Solution item and then the Stress tab with Normal option and select X Axis under the Orientation in the Details of “Normal Stress” window.



Similarly, the normal stress in Y-direction is displayed by selecting the Solution item and then the Stress tab with Normal option and select Y Axis under the Orientation in the Details of “Normal Stress” window.

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Chapter 10 Thermal Stress Analysis



The shear stress is displayed by selecting the Solution item and then the Stress tab with Shear option.

10.4 Application 10.4.1 Thermal Stress in Combustion Engine Cylinder A combustion engine cylinder as shown in the figure is made from a material that has the thermal conductivity coefficient of 237 W m-  C , the Young’s modulus of 7.11010 N m2 , the Poisson’s ratio of 0.3, and the thermal expansion coefficient of 2.3 105  C . The top surface of the cylinder is subjected to an internal pressure of 106 N m2 and heat flux of 100,000 W m2 .

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10.4 Application

Convection heat transfer is assumed to occur on all other surfaces to the surrounding medium temperature of 30  C with the convection coefficient of 40 W m2 -  C . We will employ ANSYS through its Workbench to determine the temperature distribution, deformation and thermal stresses in the cylinder.

k



237

W m-  C

h



40

W m2 - C

T



30

C

q



100, 000

W m2

E



7 10 10

N m2





0.3





2.3  10 5

C

We start from heat transfer analysis to determine the cylinder temperature distribution. A finite element mesh is constructed with a large number of tetrahedral elements as shown in the figure.

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Chapter 10 Thermal Stress Analysis

The heat flux of 100,000 W m2 is applied on top of the cylinder surface. The boundary condition of convection heat transfer is applied to all other surfaces using the convection coefficient of 40 W m2 -  C and the surrounding medium temperature of 30  C . Application of these boundary conditions through the Workbench is highlighted in the figure.

The heat transfer analysis is performed and the computed temperature distribution is displayed as shown in the figure. High temperature occurs on the cylinder top surface where the heat flux is applied. The cylinder temperature decreases gradually from top to bottom due to convection heat transfer to the surrounding medium.

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205

The cylinder temperature obtained from the heat transfer analysis is transferred to the stress analysis to predict deformation and stress. The combustion pressure with a magnitude of 106 N m2 is applied on the cylinder top surface while the cylinder pin surface is constrained as shown in the figure.

The predicted von-Mises stress is displayed on the cylinder deformed shape as shown in the figure. The figure indicates high stress occurs in the region above the cylinder pin hole. Such solution helps designer to understand cylinder behavior under both the mechanical and thermal loadings.

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Chapter 10 Thermal Stress Analysis

Since ANSYS files of this problem are available from the book website, users may modify boundary conditions in the heat transfer and stress analyses. Such modification will lead to different solution and increase understanding of the thermal stress behavior. Users will also realize benefits of the finite element method that can analyze the multidisciplinary problems effectively.

Chapter 11 Incompressible Flow Analysis Computational Fluid Dynamics (CFD) has played important role for the flow analysis recently. CFD provides detailed flow behaviors over complicated configuration, such as flow over an automobile body, flow over a city, flow circulation inside an office, etc. CFD also provides insight into some flow behaviors that might be harmful to human and reduces cost of performing experiments. Most of CFD software packages employ the finite element and finite volume methods to solve for flow solutions. The finite volume method is popular because it can provide accurate flow solutions at reasonable cost. ANSYS includes Fluent software which can perform different classes of flow analyses effectively. This chapter demonstrates the use and capability of Fluent to analyze both academic and application problems.

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Chapter 11 Incompressible Flow Analysis

11.1 Basic Equations 11.1.1 Differential Equations The flow behavior in three dimensions is governed by the full Navier-Stokes equations consisting of the conservation of mass, momentums and energy. There are five differential equations which are coupled in complicated form. Solving the full Navier-Stokes equations requires extensive computational effort. The equations are thus normally reduced into simplified forms according to different classes of flow behaviors. In this chapter, we will concentrate on the steady-state incompressible laminar flow analysis in two-dimensional Cartesian coordinates. The Navier-Stokes equations, in this case, consist of only three differential equations. These equations are: (a) conservation of mass, (b) conservation of momentum in the x-direction, and (c) conservation of momentum in the y-direction as follows. u v  x y

 0

  p   u    u  (  uu )  (  vu )         x y x x  x  y  y    p   v    v  (  uv)  (  vv)         x y y x  x  y  y  where  is the density, u and v are the velocity components in the x- and y-directions, respectively, p is the pressure, and  is the dynamic viscosity. The three basic unknowns of the three differential equations above are the velocity components u ( x, y), v( x, y ) and pressure p( x, y) . It is noted that the differential equations form a set of coupled nonlinear differential equations. Such the set of differential equations is more difficult to solve as compared to the differential equations in the preceding chapters.

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209

11.1.2 Solution Approach By observing the three differential equations above, the velocity components u and v should be determined from the x- and y-momentum equations. This means the pressure p should be obtained from the mass equation. However, the mass equation does not contain the pressure p at all. Thus, the pressure p must be determined together with the velocity components u and v in the momentum equations such that the mass is also satisfied. The idea above suggests that the solution process should be an iteration process. The process is continued until the converged solutions of u, v and p are achieved. It is noted that the x- and y-momentum equations contain the convection terms which are in form of the first-order partial derivative. These convective terms are nonlinear and require additional effort for solutions. These terms may yield oscillated solutions if the mesh is not fine enough. Fine mesh is thus normally needed which requires more computational time. These factors must be realized prior to solving flow problems using any CFD software.

11.2 Finite Volume Method The finite volume method is a popular method for analyzing CFD problems. The method provides accurate flow solution with reasonable computational effort. Details of the method can be found in many CFD textbooks including the one written by the author. The method starts from dividing the computational domain into a number of cells as shown in the figure. Herein, we use rectangular cells to simplify explanation of the method. Each cell consists of the three unknowns which are the velocity components u, v and the pressure p. The cell is surrounded by the north cell N, the east cell E, the south cell S, and the west cell W. The concept of staggered grids is applied to reduce error that might occur during the computation.

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210

p cell

u cell j2

N

J 1

j 1 y y 2

W

E

P

J

staggered grid original grid

j J 1

S

j 1

v cell

J 2

j2 I 2 I 1 i2 i i 1 x 2

I

i 1

I 1 i2

x

11.2.1 Finite Volume Equations With the concept of staggered grids, the u cell is moved to the left of the p cell as shown in the figure. The new N, E, S and W cells corresponding to this u cell are established. The velocity u is then determined from,

aPu u P 

 anbu unb

 ( pI 1, J  pI , J ) Au

where Au is the flow area on the left and right edges of the cell. The subscript nb means the neighbor cells of the u cell. Similarly, the velocity v is determined from,

aPv vP 

 anbv vnb

 ( pI , J 1  pI , J ) Av

where Av is the flow area on the top and bottom edges of the cell. The coefficients aP and anb in the equations above consist of the convection and diffusion terms of the p cell. These

211

11.2 Finite Volume Method

coefficients are derived depending on the algorithms selected. Common algorithms are the central differencing, upwinding, hybrid differencing, power-law, and QUICK algorithms. Understanding these algorithms and applying them appropriately can improve the solution accuracy. From the equations for determining the velocity components u and v of any p cell above, the computational approach to find the solutions should be an iteration process. The process starts from an initial guess of the flow solution for the entire domain. The iteration process is performed and terminated when a converged solution is obtained. One of the efficient processes is the SIMPLE (Semi-Implicit Method for Pressure-Linked Equation) method. The method is briefly explained in the following section. 11.2.2 SIMPLE Method The SIMPLE method consists of the three main steps as follows. Step1 Assume the velocity components u  , v  and the pressure p  for all cells in the flow domain. Then, determine the new velocity components u  and v  from,

and

p u A x p  v  A y

aPu u P 

 anbu unb 

aPv vP

 anbv vnb



Step 2 Assign u , v and p as the corrections which are the differences between the correct solutions and assumed solutions in step 1, i.e., u  u  u  ;

v   v  v ;

p 

p  p

Then, determine the velocity components u and v from, uP  uP 

Au p aPu x

212

and

Chapter 11 Incompressible Flow Analysis

vP  vP 

Av p aPv y

while p is obtained by solving the differential equation,  u   v     Au p    Av p    u    v      x  aP x  y  a P y  y   x

which is in the form of the Poisson’s equation. Step 3 Check whether the solutions u, v and p converge to the correct solutions. This is equivalent to the values of u, v and p are closed to zero, so that u  u  , v  v and p  p . If the u, v  and p are not converged to the specified tolerance, reset u   u ,

v  v and p   p , then repeat the iteration process. The process continues until u, v and p for all cells converge to the correct solutions. During the iteration process, many software packages show plot of the solutions u, v and p that change with the iteration numbers. The plot provides good information to ensure convergence of the solutions.

11.3 Academic Example We will use Fluent which is embedded in ANSYS through the Workbench to analyze flow circulation in a cavity and flow past a cylinder in a channel. 11.3.1 Lid-Driven Cavity Flow A unit square cavity filled with a fluid is shown in the figure. The specified velocity along the top edge induces flow circulation in the cavity. The flow behavior depends on the Reynold’s number defined by,

213

11.3 Academic Example

Re 

uL 

y u 1

1

Fluid

 1   0.01

x

1 By employing Fluent in ANSYS software, the steps for analyzing the problem are as follows. (a) Starting ANSYS Workbench 

Open the ANSYS Workbench, set the Units menu on the upper tab to Metric (kg,m,s,C,A,N,V).

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Chapter 11 Incompressible Flow Analysis

On the Analysis Systems window, click twice on the Fluid Flow (Fluent) item. A new small box will appear on the Project Schematic window. Replace the name Fluid Flow (Fluent) in the lower blue tab by retyping the desired project name, e.g., Lid Driven Cavity Flow, and hit Enter. Right click on the Geometry tab and select the Properties option, the Properties of Schematic window will open. Change the Analysis Type under the Advanced Geometry Options from 3D to 2D. Then, close this window.

(b) Creating Geometry

    

 

On the Project Schematic window, right click on the Geometry tab and select the New Geometry…. This will launch the ANSYS Design Modeler (green logo DM). On DM window, set unit in the Units menu on the upper tab to Meter. On the Tree Outline window, right click on XYPlane and select Look at. The X-Y-Z coordinates on the Model View in 3D view will become X-Y coordinates in 2D view. Select the Sketching tab below the Tree Outline window, the Sketching Toolboxes will pop-up in the same place. Select the Settings tab and then Grid, activate the buttons Show in 2D and Snap. The grid will appear on the main window. Grid snapping provides convenience when drawing model. Make sure that the Major Grid Spacing is set to 1 m, Minor-Steps per Major is 1, and Snaps per Minor is 1. Enlarge the scale by clicking at the Box Zoom icon on the upper part of the screen (icon with plus sign on the magnifying glass) and draw a box with appropriate size to zoom in. Click it again after finishing.

11.3 Academic Example



 

215

Click on Modeling tab, and then click the New Sketch icon (a small blue geometry symbol with * on the upper part of the DM window) to create Sketch1 which will appear under XYPlane. Note that this name Sketch1 can be deleted or renamed by right clicking on it and selecting options. Next, to draw the unit square, click on Sketch1. Click the Sketching tab and select Draw. Choose Rectangle to create a square with the vertices of (0,0) and (1,1). This is done by clicking at the coordinates of (0,0), move the cursor to the coordinates of (1,1) and click the mouse again. Click on Generate (the icon with yellow lightning on the upper-left part of the screen). The desired square will pop up in dark green.

216   



 



Chapter 11 Incompressible Flow Analysis

The next important step is to go to the Concept tab on top of the screen and select Surfaces From Sketches. Select the Sketch1, the square will become yellow. Click Apply icon on the right side of the Base Objects tab in the Details View at the lower left of the screen. The square will become cyan. The right side of the Base Objects tab will show 1 Sketch with 1 Part, 1 Body appears in the Tree Outline window. Then, click on Generate. We now have a unit square domain.

The domain is ready for meshing, but before that, we will specify boundary conditions on the domain first. We will give the name of the upper edge as Lid. On the upper tools bar, select the Selection Filter: Edges icon (box with arrow and green edge)

Place the cursor near the top edge of the square and click, the top edge will become green. Then, right click to select

11.3 Academic Example

 



217

Named Selection, and click Apply tab next to the Geometry tab in the lower left of Details View window. The right tab will become 1 Edge. Then, click on Generate. Right click in the NameSel1 in the Tree Outline window and choose Rename. Type Lid and hit Enter. Then, click on Generate. It is noted that all other boundaries, by default, are viscous walls with zero velocity components, i.e., we don’t have to specify anything.

Click on Save Project icon (diskette icon on top of the screen) to save the work under the file name Lid Driven Cavity Flow and close the DM window.

(c) Generating Mesh

   

On the Project Workbench window under Project Schematic, click twice on Mesh. On the pop-up Outline window, select Mesh. In the Details of “Mesh” window, click the plus sign (+) next to Sizing to expand it. Change Relevance Center to Fine.

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Chapter 11 Incompressible Flow Analysis

Click Update on the menu bar above the Outline Window. A mesh will be generated. Close the window and return to the Project Workbench window. Click on Save Project icon to save the work.

(d) Setting up Fluid Properties and Boundary Conditions





On the Project Workbench window under Project Schematic, click twice on Setup. Click OK on the Fluent Launcher window. Wait for few seconds, the mesh that just created will appear on the central window. The left side of the screen is the Tree window consisting of the three main sections: Setup, Solution, and Results.

11.3 Academic Example

 

 

219

Click on Models under Setup section. Make sure that everything is off except the third option must be Viscous – Laminar. Next, click on Materials under Setup section, and double click on Fluid. Change the Density value to 1.0 and the Viscosity value to 0.01. Then click Change/Create button and Close button. Now, specifying the boundary conditions. Select Boundary Conditions, click at Lid, and Edit…. Select Moving Wall and input the Speed (m/s) as 1.0 and click OK. On the wall-surface_body zone, make sure that it is No Slip Condition, and click OK. Note that Fluent assumes any other edges as viscous wall.

(e) Solving for Solution



Under Solution section in the Tree window, select Run Calculation, set Number of Iterations to 1000, Reporting Interval to 10, and click on Calculate button. If it asks for initial condition, click on Yes button.



If it works properly, residual curves of the continuity and momentum equations that decrease with the number of iterations will be plotted on the main window

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Chapter 11 Incompressible Flow Analysis

(f) Displaying Results



Under Results section in the Tree window, select Graphics. Choose and click twice on Contours and select Velocity… in the Contours of box. Select the Filled button and click Display button.



Under Result section in the Tree window, select Graphics. Choose and click twice on Vectors and select Velocity… in the Vectors box, and click Display button.

11.3 Academic Example

221



Under Result section in the Tree window, select Graphics. Choose and click twice on Contours and select Velocity… with Stream Function in the Contours box under the Contour of. Click Display button to show the streamlines.



The analysis can be repeated for higher Reynolds number, such as when Re=1,000. In the Materials option under Setup section, change the Viscosity value to 0.001 and reanalyze the problem. Result of the path lines for Re=1,000 can be displayed in the same fashion. Flow circulations appear clearly near both lower corners of the cavity.

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Chapter 11 Incompressible Flow Analysis



The analysis is repeated for Re=5,000. Result of the path lines is shown below. Flow circulation now appear near the upper left corner in addition to the lower two corners of the cavity.

11.3.2 Flow past Cylinder in Channel We will use Fluent in ANSYS to analyze the flow past a cylinder in a channel. The fluid properties, flow domain geometry and boundary conditions are shown in the figure. y

3





x

The flow domain can be constructed by creating a rectangle and a circle with the given dimensions. The Subtract

11.3 Academic Example

223

command is used to take away the circular region from the rectangular region. The procedure to subtract a region from another is the same as constructing a plate with a circular cutout as explained in chapter 4.



Next, the boundary names of the flow inlet, flow outlet and cylinder edge are assigned as Inlet, Outlet and Cylinder, respectively. This will provide the convenience in applying boundary conditions later.



A mesh representing the flow domain is then constructed as shown in the figure. The mesh consists mostly of quadrilateral elements with few triangular elements.

224

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Chapter 11 Incompressible Flow Analysis

To perform the flow analysis, click on the Fluent Launcher and enter the Density as 1.0 and the Viscosity as 0.1 in the same way as in the preceding example. The boundary conditions are then applied by clicking on Inlet (the name assigned earlier) and enter the Velocity Magnitude as 1.0. Click on Outlet and enter the Pressure as 0.0. On the Cylinder, the boundary condition is selected as No Slip. We follow the same procedure as explained in the preceding example to execute the problem for solutions. The solution residuals associated with the continuity and the two momentum equations decrease with the number of iterations are shown in the figure.

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225



The computed solutions such as the velocity contours, velocity vectors and flow path lines are shown in the figures.



It is noted that if there is an additional square inside the channel as shown in the figure, the same procedure is applied for the solution.

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Chapter 11 Incompressible Flow Analysis

y uv0

(0,10)

u 1 v0

 1   0.1

3

 10

p0

x

6 20

10

uv0 

10

6

10 (40, 10)

The computed flow solutions in forms of the velocity contours and path lines for this latter case are shown in the figures. Such solutions highlight benefits and capabilities of the software to handle complicated flow domain effectively. The computed solutions provide insight into the flow field to increase understanding of the flow behaviors.

11.4 Application

227

11.4 Application 11.4.1 Flow in Piping System A piping system as shown in the figure consists of pipes with different diameters, reservoir, reducing adapter and elbow. Water flows into the larger pipe on the left side of the figure at the speed of 1 m/s. The water leaves the smaller pipe on the right side of the figure at the atmospheric pressure. The water density is 998.2 kg/m3 and its viscosity is 0.001003 kg/m-s. We will use Fluent in ANSYS to analyze the flow behavior in this piping system.

We start from importing the CAD file of the threedimensional piping system. The flow domain is highlighted as shown in the figure.

Discretizing the flow domain in three dimensions leads to a large number of elements and hence the flow unknowns. In order to understand the flow behavior, this particular problem may

228

Chapter 11 Incompressible Flow Analysis

be analyzed firstly by using a two-dimensional domain. This can reduce the number of unknowns and provide adequate information of the flow behavior. The two-dimensional model is shown in the figure.

We apply the boundary conditions of water inlet velocity at 1 m/s as shown by symbol A in the figure. The water exits from the piping system at the atmospheric pressure denoted by symbol B.

With the flow domain, the mesh can be constructed easily. By assigning the cell size of approximately 10 mm, the mesh consists of 3,419 cells as shown in the figure.

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229

Since the flow velocity is relatively high, we may turn on the standard k-e model which is under the Setup>Models> Viscous item. The water properties can be obtained by using the data base inside the software by selecting Setup>Materials> Fluid>Water-liquid(h2o). The analysis the then performed to yield the flow velocity and pressure as shown in the figures.

The computed solutions help designers to understand flow behavior in details. The solutions show the effect of the reservoir, reducing adapter and elbow to the flow field. Since the ANSYS files of this problem can be downloaded from the book website, users may want to change the inlet and outlet boundary conditions to increase understanding of the flow behavior.

Chapter 12 Compressible Flow Analysis Compressible flow occurs in many applications such as flow in turbine engines, flow over supersonic aircrafts and rockets. The flow behaviors consist of shock wave, expansion wave and shock-shock interaction phenomena. These phenomena are complicated and difficult to predict by numerical methods in the past. Fluent in ANSYS contains analysis capability that can provide solutions representing such complicated phenomena effectively. In this chapter, we begin with the conservative equations of the compressible flow. Theoretical background of the cellcentered method for compressible flow analysis is presented. Capability of Fluent is demonstrated by analyzing academic problems that have analytical solutions. Application example is also presented to highlight complicated compressible flow behaviors.

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12.1 Basic Equations 12.1.1 Differential Equations In order to reduce the complexity of mathematics and increase understanding of the formulation, we will consider the compressible flow in two-dimensional Cartesian coordinates. The flow is governed by the conservation of mass, x- and y-momentums and energy equations. These four equations are written in the conservative form as,    U   E I  EV   FI  FV   0 t x y where U  is the vector containing the conservative variables,

U  

 u       v    

The vectors EI  and FI  contain the inviscid fluxes in the x- and y-directions as,  u   v    uv   u 2  p    EI    FI    2  ;   v  p    uv    v  pv    u  pu  The vectors EV  and FV  contain the viscous fluxes in the x- and y-directions as, 0 0         x  xy    ;  EV      F    V   xy y     u x  v xy  q x  u xy  v y  q y  In the above equations,  is the fluid density, u and v are the velocity components in the x- and y-directions, p is the

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233

pressure,  is the total energy,  x and  y are the normal stresses,  xy is the shearing stress, q x and q y are the heat fluxes in the xand y-directions. 12.1.2 Related Equations The total energy consists of the internal energy e and the kinetic energy as, 1  = e  u 2  v 2  2 The internal energy e can be written in forms of the temperature T or the pressure p as, e = cvT = p (  1)  where  is ratio of the specific heats at constant pressure and volume,  = c p cv The pressure p can also be written in form of the total energy  and velocity components u, v as, p =

1 2

  1     u 2  v 2  



The internal energy e is used to determine the enthalpy h from,

1 h =  e =     u 2  v 2  2   and the total enthalpy H from, H = h

  1 2 2 1 2 2 u  v  =    u  v  2 2

The speed of sound a is determined from the pressure and density,

a

=

 p 

In the above differential equations, the normal stress components and shearing stress are written on forms of the velocity components u, v as,

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Chapter 12 Compressible Flow Analysis

2  u v  2   v u   2   ; μ 2   ; y = 3  x y  3  y x   u v   xy =      y x  The heat fluxes qx and q y vary with the temperature T according to

x =

the Fourier’s law, qx =

k

T x

;

qy =

k

T y

The fluid thermal conductivity k is determined from, k = cp  Pr where Pr is the Prandtl number and  is the dynamic viscosity that can be determined from the Sutherland’s law.

12.2 Finite Volume Method 12.2.1 Finite Volume Equations For simplicity in understanding the derivation of the finite volume equations, we will concentrate on the inviscid flow analysis. A typical equation representing any one of the four Navier-Stokes equations can be written in the form, U EI FI   = 0 y t x If we consider the mass equation, then U   ; EI  u ; FI  v . Similarly, if we consider the x-momentum equation, then U  u ; EI  u 2  p ; FI  uv . To derive the finite volume equations, the method of weighted residuals is employed with unit weighting function to yield, F  U  E  t dA    xI  yI  dA = 0 A A The Gauss’s theorem is applied to introduce the boundary integral term so that the equations become,

12.2 Finite Volume Method

235

U dA   ( E I n x  FI n y ) dS = 0  t A S



where nx and ny are directions cosines of the unit vector normal to the cell edge. The integrand in the second integral term represents the flux Fn normal to the cell edge,

Fn = E I n x  FI n y So that the finite volume equations reduce to, U  t dA =   Fn dS S A The fluxes normal to the cell edge for the four Navier-Stokes equations are, u n x   v n y Vn         2 u  p  nx  uv n y   uVn  p nx  Fn =  =    uv nx   v 2  p  n y  vVn  p n y     V  pV   u  pu nx   v  pv  n y  n n   In the above equation, Vn is the velocity normal to the cell edge. As an example of a triangular cell in the figure, the normal velocity to the cell edge is, Vn = u nx  v n y while the tangential velocity to the cell edge is,  u n y  v nx

Vt = 3

S3

1

x

Vn

Vt

nx 2

v

3

nˆ

S1

S2 y

ny

S2

y

S3

1 x

u

S1

2

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Chapter 12 Compressible Flow Analysis

12.2.2 Computational Procedure The finite volume equations are discretized by applying the forward difference approximation to the integral term associated with time, U U m 1  U m dA = A  t t A

where superscript m refers to the mth step and t is the time step. For the integral term associated with the flux across the cell edge, we replace it by the numerical flux,

 Fn dS S

=

  Fn dS S

So that the finite volume equations become, t U m1  U m =    Fn dS A S

~

The numerical flux Fn from the left cell L to the right cell R with the common edge of length S is determined using the Roe’s averaging method, ~ 1 1 Fn = A (UL  UR ) ( FnL  FnR )  2 2 where FnL and FnR are the fluxes of the left cell L and the right cell R, respectively. The determinant A is computed from the Jaco-bian matrix which will be shown later. The quantities U L and U R represent the conservation variables of the left cell L and the right cell R, respectively. The final form of finite volume equations becomes, t m m U m1  U m =   FnR  A m (ULm  URm )  S  FnL  2A The computational procedure starts from using U m at time step m to determine U m1 at time step m+1. The procedure is performed for all the cells in the flow domain for transient analysis. For steady-state analysis, the computation is terminated when the

12.2 Finite Volume Method

237

result on the left-hand-side of the equation for every cell is less than the specified tolerance. The Jacobian matrix is determined from,

 A = R 1  R  where

R

1

 1  c2   u  c2 =  v  2  c    2  c

 Vn  0  =  0   0

R 

1 2c 2 u  cx 2c 2 v  cy

0  ny nx Vt

1 2c 2 u  cx 2c 2 v  cy

2c 2   Vn c 1  2 2c 2

0

0

Vn

0

0

Vn  c

0

0

       2c 2    Vn c 1    2  2c 2

  0  0   Vn  c  0

   c 2  u   Vt  ny =    Vn c c x   u    Vn c  c x   u

 v



 0 c y  v     c y   v   nx

These three matrices contain coefficients which are,

cx = cnx

;

cy = c n y

;

p =

 =

;

 =  1

Vn

;

Vt =

c =

and

p 

1 2 (u  v 2 ) 2 = u nx  v n y

H =     (  1)

 ( 1) (H   ) 

 u n y  v nx

238

Chapter 12 Compressible Flow Analysis

Values in these equations are average between the left and right cell values, L H L  R H R L R ; H =  = L  R

u =

 L uL   R uR L  R

;

v =

 L vL   R vR L  R

where the subscripts L and R refer to the left and right cells, respectively.

12.3 Academic Example 12.3.1 Mach 3 Flow over Inclined Plane y Shock wave

.454

M 3

1 .0

Symmetry 0 .5

  1.4 M  3



  20

Inviscid wall

.546 x

1 .5

; R  287 m 2 sec 2  K ; T  300 K ; a  347.19 m sec ; u  1041.57 m sec 3   1.18 kg m ; p  101598 N m 2 c p  1004.5 m 2 sec 2  K

A Mach 3 inviscid flow over an inclined plane is probably one of the simplest examples for understanding the compressible flow behavior. The problem statement is shown in

12.3 Academic Example

239

the figure with the fluid properties and flow conditions. The problem has analytical solution so that the computed solution can be compared. The shock wave angle  is determined from the transcendental function,  M 2 sin 2   1  2 cot   2  = tan   M   cos 2   2  We will employ Fluent in ANSYS through its Workbench to solve for the flow behavior. (a) Starting ANSYS Workbench    



Open the ANSYS Workbench, set the Units menu on the upper tab to Metric (kg,m,s,C,A,N,V). On the Analysis Systems window, click twice on the Fluid Flow (Fluent) item. A new small box will appear on the Project Schematic window. Retype the name in the lower blue tab as the desired project name, e.g., Mach 3 Flow over Wedge, and hit Enter. Right click on the Geometry tab and select the Properties option, the Properties of Schematic window will open. Change the Analysis Type under the Advanced Geometry Options from 3D to 2D. Then, close this window. Back to the Project Schematic window, right click on the Geometry tab and select the New Geometry…. This will launch the ANSYS Design Modeler (green logo DM).

240

Chapter 12 Compressible Flow Analysis

(b) Creating Geometry    

 





On DM window, set unit in the Units menu on the upper tab to Meter. On the Tree Outline window, right click on XYPlane and select Look at. The X-Y-Z coordinates on the Model View in 3D view will become X-Y coordinates in 2D view. Select the Sketching tab below the Tree Outline window, the Sketching Toolboxes will pop-up in the same place. Select the Settings tab and then Grid, activate the buttons Show in 2D and Snap. The grid will appear on the main window. Grid snapping provides convenience when drawing model. Make sure that the Major Grid Spacing is set to 1 m, Minor-Steps per Major is 4, and Snaps per Minor is 1. Enlarge the scale by clicking at the Box Zoom icon on the upper part of the screen (icon with plus sign on the magnifying glass) and draw a box with appropriate size to zoom in. Click it again after finishing.

Click on Modeling tab, and then click the New Sketch icon (a small blue geometry symbol with * on the upper part of the DM window) to create Sketch1 which will appear under XYPlane. Note that this name Sketch1 can be deleted or renamed by right clicking on it and selecting options. Next, to draw the flow domain, click on Sketch1.

12.3 Academic Example 

 

  

241

Click the Sketching tab and select Draw. Choose Line to create a lower horizontal line with the vertices of (0,0) and (0.5,0). This is done by clicking at the coordinates of (0,0) of the model, drag the cursor to the coordinates of (0.5,0), and click the mouse again. Click on Generate (the icon with yellow lightning on the upper-left part of the screen). The lower horizontal line will pop up in dark green. Follow the same procedure to create the left vertical line, as well as the upper horizontal line. Then, create the right vertical line in the same manner with the vertices of (2,1) and (2,0.5). This length can be shortened to 0.454 by selecting the Dimensions tab followed by General. The exact length is obtained by clicking at the line and drag the cursor slightly to the right, then change the value of V1 in the Details window to 0.454, hit Enter and click on Generate.

The last inclined line can now be drawn by selecting Draw and choose Line. Then, click at the lower left and upper right vertices, respectively, followed by Generate. The next important step is to go to the Concept tab on top of the screen and select Surfaces From Sketches. Select the Sketch1, the domain boundary will become yellow.

242





 

Chapter 12 Compressible Flow Analysis

Click Apply icon on the right side of the Base Objects tab in the Details View at the lower left of the screen. The domain boundary will become cyan. The right side of the Base Objects tab will show 1 Sketches. Then, click on Generate. We now have the desired flow domain.

The domain is ready for meshing, but before that, we will specify the boundary conditions on the domain first. We will assign the name for the left, top and right boundaries as Farfield. Similarly, we will assign the names for the bottom and inclined boundaries as Symmetry and Wedge, respectively.

12.3 Academic Example

243



Farfield: On the upper tools bar, select the Selection Filter: Edges icon (box with arrow and green edge)



Hold the Ctrl key while clicking the mouse at the left, top and right edges, these edges will become green. Then, right click to select Named Selection, and click Apply tab next to the Geometry tab in the lower left of Details View window. The right tab will become 3 Edges. Then, click on Generate. Right click at the NameSel1 in the Tree Outline window and choose Rename. Type Farfield and hit Enter. Then, click on Generate. Symmetry: On the upper tools bar, select the Selection Filter: Edges icon again. Click the mouse at the bottom edge, this edge will become green. Then, right click to select Named Selection, and click Apply tab next to the Geometry tab in the lower left of Details View window. The right tab will become 1 Edge. Then, click on Generate. Right click at the NameSel2 in the Tree Outline window and choose Rename. Type Symmetry and hit Enter. Then, click on Generate. Wedge: On the upper tools bar, select the Selection Filter: Edges icon again. Click the mouse at the inclined edge, this edge will become green. Then, right click to select Named Selection, and click Apply tab next to the Geometry tab in the lower left of Details View window. The right tab will become 1 Edge. Then, click on Generate. Right click at the NameSel3 in the Tree Outline window and choose Rename. Type Wedge and hit Enter. Then, click on Generate.

  

  



244

 

Chapter 12 Compressible Flow Analysis

Click on the Save Project icon (diskette icon on top of the screen) to save the work as Mach 3 Flow over Wedge. Close the DM and go back to the Workbench.

(c) Generating Mesh      

On the Project Workbench window under Project Schematic, click twice on Mesh. On the pop-up Outline window, select Mesh. In the Details of “Mesh” window, click the plus sign (+) next to Sizing to expand it. Change Relevance Center to Fine. Click Update on the menu bar above the Outline Window. A mesh will be generated. Close the window, and return to the Project Workbench window.

12.3 3 Academic Example

245

(d)) Setting up u for Analyysis 

On thee Project Workbench window w under Project Schemaatic, click tw wice on Settup. Click OK on thee Fluent Launch her window w



Wait forr few secon nds, the messh that just created c willl appear on the central c winddow. Select Density-Ba D ased for Soolver in thhe Task Page for compresssible flow analysis.





The leftt side of thee screen is thhe Tree win ndow consiisting of the threee main sectiions: Setup p, Solution, and Resultts.

246

Chapter 12 Compressible Flow Analysis



Click on Models under Setup section. Double-click on Energy to turn it On, so that the energy equation will be solved together. Double-click on Viscous and select Inviscid because inviscid analysis will be performed.



Next, click on Materials under Setup section, and double click on Fluid. Change Density from Constant to Idealgas. Change the Cp (Specific Heat) value to 1004.5. Then click Change/Create button and Close button.



Now, specify the boundary conditions. Select Boundary Conditions, double-click at farfield, select Type and change wall to pressure-far-field. Then, input Gauge Pressure as 101598 and Mach Number as 3 and click OK.

12.3 Academic Example

  



247

Next, double-click at symmetry, select Type as symmetry and click OK. Then, double-click at wedge, select Type as wall and click OK. We also need to provide the free-stream values. Select Reference Values. Then, input the Density as 1.18, the Pressure as 101598, the Temperature as 300, and the Velocity as 1041.57. We also need to provide the free-stream values. Select Reference Values, then change the Density to as 1.18, the Pressure to 101598, the Temperature to 300 and the Velocity to 1041.57, then hit Enter button.

248

Chapter 12 Compressible Flow Analysis

(e) Solving for Solution 

In the Solution Methods under Solution, be sure to select Second Order Upwind under Spatial Discretization.



In the Solution Controls under Solution, ensure that the Courant Number is set to 5. In the Monitors under Solution, select Residuals - Print, Plot and double-click on Edit button. Then, change the Convergence Absolute Criteria to 1e-6, and click OK.





In the Solution Initialization under Solution, select Standard Initialization. Also select farfield from the dropdown box under Compute from, and click Initialize button.

12.3 3 Academic Example



249

We aree now readdy to run for solutioons. Selecct Run Calculaation, changge Number of Iterattions to 10 000 and Reportiing Intervaal to 10. Then, T click k Calculatee button twice. If it wo orks properrly, residu ual curves of the d continuiity, momenntum and ennergy equattions that decrease with thee number of o iterationss will be pllotted on thhe main window w

(f) Displayin ng Results 

Under Result R section in the Tree window w, select Grraphics. Choose and doublee-click on Contours C an nd select Deensity…

250

Chap pter 12 Coompressiblee Flow Ana alysis C o Select the of. t Filled button b and click under the Contours Display buutton.



Choose andd double-cliick on Conttours and select s Temp perature… undder the Con ntours of. Select S the Filled F buttonn and click Displlay button.



Choose annd double-click on Coontours andd select Veelocity… to dispplay the Veelocity Maggnitude.

12.3 3 Academic Example



251

Choose and double-click on Contours and select Velocity… under the Con ntours of with w Stream Function option. o

12.3.2 Macch 3 Flow over o Cylinder Inviiscid flow over a cylinder c is another classical c exam mple normaally used to study com mpressible fllow behavioor. The flow w field connsists of a bow shocck wave with w varyinng flow prop perties behiind it. Wee will use Fluent F in ANSYS A to provide solu utions of a Mach 3 flo ow over 0.3 meter diaameter cyliinder as show wn in the figure. f Thee problem statement of o the symm metrical flow w domain, boundary b c conditions a and fluid properties p are a also show wn in the figgure.

252

Chapter 12 Compressible Flow Analysis

Computational domain

 .45

Bow shock M=3

R T M a u

 Cylinder

Symmetry .15

.15

p cp

        

1.4 287 m2/sec2-K 300 K 3 347.19 m/sec 1041.57 m/sec 1.18 kg/m3 101598 N/m2 1004.5 m2/sec2-K

.15

(a) Creating Geometry 

We start from constructing the flow domain which consists of the lower horizontal line, the right vertical line, the upper left curvature and the lower right cylinder edge. These lines can be constructed easily by using the Draw command to form up the flow domain is shown in the figure.

12.3 Academic Example 

253

Next, we assign the name for the upper left curvature and the right vertical line as Farfield. We also assign the names for the symmetrical line and cylinder edge as Symmetry and Cylinder, respectively. These assigned names aid application of the boundary conditions later.

(b) Generating Mesh 

With the constructed flow domain, a mesh is generated. A fine mesh is used to capture detailed flow solution. The mesh contains a total of 11,877 cells for which most of them are in quadrilateral shape. Detail of the mesh in front of the cylinder above the symmetrical line is shown in the figure.

(c) Setting up for Analysis  

After loading the Fluent Launcher, enter the fluid Density as 1.18 and the Cp (Specific Heat) as 1004.5. Apply the boundary conditions by clicking on Farfield (the name assigned earlier) and select Type as pressure-farfield. Enter the Gauge Pressure as 101598 and the Mach Number as 3.

254  

Chapter 12 Compressible Flow Analysis

Click on Symmetry and select Type as symmetry. Finally, click on Cylinder and select Type as wall.

(d) Solving for Solution 

We follow the same procedure as in the preceding example to execute the problem. The residuals corresponding to the continuity, momentums and energy equations reduce with the number of iterations are shown in the figure.

(e) Displaying Results 

The converged flow solutions can then be displayed. The figures from left to right and top to bottom show the flow density, temperature, Mach number and pressure, respectively.

12.4 Application

255

12.4 Application 11.4.1 Flow over Shuttle Nose and Cockpit During reentry at high-speed, the Shuttle is subjected to high aerodynamic heating and pressure. Inviscid flow analysis is normally used as the first step to provide flow behavior around the vehicle. The analysis also provides good estimation of the aerodynamic pressure on the vehicle body. As shown in the figure, the Shuttle geometry is rather complicated. The flow domain surrounding the vehicle is thus huge and complicated too. Such the flow domain requires a large amount of small cells to capture detailed flow behavior.

In order to demonstrate the software capability for predicting complicated flow behavior, we reduce the problem size by concentrating only the two-dimensional domain in front of the nose and cockpit as highlighted in the figure. The flow condition is at Mach 3 and five degrees angle of attack. The flow boundary conditions include the specified horizontal velocity of 1,041 m/s. The air density is 1.18 kg/m3 at the temperature of 300 K.

256

Chap pter 12 Coompressiblee Flow Ana alysis

It is nooted that thee quality off the flow solution stroongly m Herrein, a strucctured messh consistin ng of dependds on the mesh. 271,5399 cells is ussed in the analysis. a Thhe cell shappes are closeed to rectang gle and aliggned with the nose and a cockpit edges. Such structurred mesh caan improve the solutionn accuracy. The coomputationaal proceduree follows th he steps presented in thhe precedinng exampless. The preddicted densiity, temperaature, Mach number n andd pressure are a shown in the figu ures from leeft to right, reespectively..

12.4 Application

257

Pressure and temperature distributions are shown in more details in the figures. By using the fine mesh, the nose bow shock is quite sharp and the flow behaviors behind it change smoothly. The nose bow shock hits the shock from the cockpit creating the shock-shock interaction phenomenon.

258

Chapter 12 Compressible Flow Analysis

Since the ANSYS files can be downloaded from the book website, users can exercise more on this problem. Users may change the flow Mach number and angle of attack to obtain different flow solutions. Such practice provides good experience to realize that solving compressible flows always requires considerable effort. The effort is from the fact that the governing differential equations are strongly coupled and nonlinear. A large number of small cells are thus needed to provide accurate flow solutions. A large amount of small cells requires excessive computational time and computer memory.

Bibliography Carslaw, H. S. and Jaeger, J. C., Conduction of Heat in Solids, Second Edition, Oxford University Press, Oxford, 1995. Cook, R. D., Malkus, S. D., Plesha, M. E. and Witt, R. J., Concepts and Applications of Finite Element Analysis, Fourth Edition, John Wiley & Sons, New York, 2002. Dechaumphai, P. and Wansophark N., Numerical Methods in Engineering: Theories with MATLAB, Fortran, C and Pascal Programs, Alpha Science International, Oxford, 2011. Dechaumphai, P. and Phongthanapanich, S. Easy Finite Element Method with Software. Oxford: Alpha Science International, 2009. Dechaumphai, P., “Adaptive Finite Element Technique for Thermal Stress Analysis of Built-Up Structures”, JSME International Journal, Vol. 39, No. 2, 1996, pp. 223-230. Dechaumphai, P., “Improvement of Plane Stress Solutions Using Adaptive Finite Elements”, Journal of Chinese Institute for Engineers, Vol. 19, No. 3, 1996, pp. 375-380. Dechaumphai, P., “Progress in Integrated Analysis with Adaptive Unstructured Meshing”, NASA CP-3142, May 1992, pp. 5979. Dechaumphai, P., Calculus and Differential Equations with Mathematica, Alpha Science International, Oxford, 2016. Dechaumphai, P., Calculus and Differential Equations with MATLAB, Alpha Science International, Oxford, 2016. Dechaumphai, P., Computational Fluid Dynamics by Finite Element and Finite Volume Methods, Third Edition, Chulalongkorn University Press, Bangkok, 2017. Dechaumphai, P., Finite Element Method: Fundamentals and Applications, Alpha Science International, Oxford, 2010.

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Finite Element Analysis with ANSYS Workbench

Ferreira, A. J. M., MATLAB Codes for Finite Element Analysis, Springer, New York, 2009. Finlayson, B. A. and Seriven, L. E., “The Method of Weighted Residuals - A Review”, Applied Mechanics Review, Vol. 19, No. 9, 1966, pp. 735-748. Heldenfels, R. R. and Roberts, W. M., “Experimental and Theoretical Determination of Thermal Stress in a Plate”, NASA TN-2769. 1952. Huebner, K. H., Thornton, E. A. and Byrom, T. G. The Finite Element Method for Engineers. Third Edition, John Wiley & Sons, New York, 1995. Hughes, T. J. R. The Finite Element Method, Linear Static and Dynamic Analysis. New York : Dover, 2000. Kaplan, W., Advanced Calculus, Fifth Edition, Addison-Wesley, Massachusetts, 2003. Kreith, F. and Bohn, M. S., Principles of Heat Transfer, Sixth Edition, Thomson, New York, 2006. Kreyszig, E., Advanced Engineering Mathematics, Tenth Edition, John Wiley and Sons, New York, 2011. Munson, B. R., Young, D. F., Okiishi, T. H. and Huebsch, W. W., Fundamentals of Fluid Mechanics, Fifth Edition, John Wiley & Sons, New York, 2009. O’Neil, P. V., Advanced Engineering Mathematics, Seventh Edition, Cengage Learning, Stamford, 2012. Patankar, S. V., Numerical Heat Transfer and Fluid Flow, Hemisphere, Taylor & Francis, New York, 1980. Roe, P. L., “Approximate Reimann Solvers, Parameter Vector, and Difference Schemes”, Journal of Computational Physics, Vol. 43, 1981, pp. 357-372. Timoshenko, S. and Goodier, J. N. Theory of Elasticity. Third Edition, New York: McGraw-Hill, 1970. Ugural, A. C. and Fenster, S. K. Advanced Strength and Applied Elasticity. Fourth Edition, New York: Prentice Hall, 2003.

Bibliography

261

Versteeg, H. K. and Malalasekera, W., An Introduction to Computational Fluid Dynamics: The Finite Volume Method, Second Edition, Pearson Education, Essex, 2007. White, F. M. Viscous Fluid Flow. Third Edition, New York: McGraw-Hill, 2005. Zienkiewicz, O. C., Taylor, R. L. and Nithiarasu, P. The Finite Element for Fluid Dynamics. Sixth Edition, Oxford: Elsevier, 2005. Zienkiewicz, O. C., Taylor, R. L. and Zhu, J. Z. The Finite Element Method: Its Basis and Fundamentals. Sixth Edition, Oxford: Elsevier, 2005.

Index Aircraft component, 114 Analysis, 3D solid, 99 Beam, 41 compressible flow, 231 failure, 135 heat transfer, 165 incompressible flow, 207 plane stress, 61 plate bending, 81 thermal stress, 185 truss, 13 vibration, 117 bracket, 96 ANSYS, 6 Workbench, 7 Area, cross-sectional, 120 triangle, 64 Beam, 41 cross section, 42 Beam structure, two-dimensional, 45, 55 Bending moments, 82 Boundary conditions, 3 clamped, 125, 138, 153 fixed, 18, 45, 105 hinged, 29, 136 pinned, 136 simply-supported, 86, 138

symmetry, 66, 191 Buckling, 135 beam, 136 Buckling load, critical, 136 Euler, 136 factor, 136 lowest critical, 136 Button, Display, 220, 250 File, 23 Filled, 220, 250 Mode, 147 Show in 2D, 20, 31, 47, 68, 88, 107, 127, 140, 155, 174, 193, 214, 240 Show Mesh, 159 Snap, 20, 31, 47, 68, 88, 107, 127, 140, 155, 174, 193, 214, 240 Stream Function, 251 Subtract, 222 Temperature, 250 Velocity Magnitude, 250 Cell, edge, 235 Engineering Data, 9 Fluid Flow (Fluent), 214, 239 Geometry, 10, 176

264

Finite Element Analysis with ANSYS Workbench

Modal, 125 Model, 10 Results, 10 Setup, 10 Solution, 10, 146 Static Structural, 9, 18, 30, 46, 67, 86, 105, 138, 154, 198 Steady-State Thermal, 172, 191 triangle, 235 Chain wheel, 77 Computational Fluid Dynamics, 207 Computer-Aided Engineering, 1 Conservation, energy, 232 mass, 208, 232 momentums, 208, 232 Convection coefficient, 167, 179, 204 Convergence, 212 Criterion, Tresca, 135 von-Mises, 136 Cycles, 119, 151 stress, 153 Deflection, 42, 82, 121 Density, material, 120 Detergent bottle, 147 Differential equations, 3 3D solid, 100, 186 beam, 42 truss, 13 compressible flow, 232 heat transfer, 166, 186

incompressible flow, 208 Laplace, 166 mass-spring, 118 Navier-Stokes, 208, 234 plane stress, 61 plate bending, 82 Poisson, 166 Displacement, 14, 42, 118 components, 188 initial, 119 Domain, 5 Eigenvalues, 137 problem, 137 Eigenvectors, 137 Element equations, 3D solid, 101, 189 beam, 43 heat transfer, 168, 188 plane stress, 123 plate, 84 truss, 15 vibration, 123, 137 Element, area, 15, 64 assembling, 6 beam, 44 domain, 123 hexahedral, 102, 171 length, 15, 44 quadrilateral, 64, 84, 170 rod, 13 spring, 13 tetrahedral, 102, 115, 170 three-node truss, 16 triangular, 64, 84, 169 two-node, 16, 168, 189 type, 5 volume, 103

265

Index Emissivity, 167 Endurance limit, 151 Engine cylinder, 202 temperature, 203 thermal stress, 204 Equations, algebraic, 6 finite element, 5 finite volume, 234 Failure criterion, 151 Gerber, 152 Goodman, 152 Soderberg, 152 Fatigue, 150 File, imported, 57, 133, 148, 163, 227 Fin heat transfer, 179 Finite element, advantages, 11 equations, 5, 15, 43, 63, 84, 101, 123, 137, 168, 188, 189 procedure, 5 Finite volume, cell, 209, 228, 235 equations, 210, 234, 236 method, 209, 236 Flexural rigidity, 83 Flow, compressible, 231 conservative variables, 234 convection, 210 diffusion, 210 enthalpy, 233 flux, 235 heat fluxes, 233 incompressible, 207

internal energy, 233 inviscid, 238 lid-driven cavity, 212 normal flux, 235 normal stresses, 233 normal velocity, 235 pressure, 208, 211, 232 shear stress, 233 sound speed, 233 specific heat ratio, 233 steady-state, 236 tangential velocity, 235 temperature, 234 total energy, 233 total enthalpy, 233 transient, 236 velocities, 208, 211, 232 Fluid, density, 208, 232 viscosity, 208, 234 thermal conductivity, 234 Frame structure, 56 Frequency, 119 circular, 118 forcing, 119 natural, 119, 137 Geometry, 3 Graphic User Interface, 7 Heat flux, 167 specified, 167, 204 Heat transfer, 165, 186 conduction, 167, 186 convection, 167, 204 radiation, 167 steady-state, 166 transient, 166, 186 Hertz, 119, 131

266

Finite Element Analysis with ANSYS Workbench

Icon, Apply, 22, 26, 36, 49, 52, 70, 74, 89, 128, 141, 175, 194, 199, 216, 242 Box Zoom, 20, 31, 48, 68, 88, 107, 127, 140, 155, 174, 193, 214, 240 Edge, 73, 91, 130, 177, 196, 199 Eigenvalue Buckling, 146 Face, 92, 113, 159 Generate, 21, 32, 49, 70, 89, 108, 128, 141, 155, 175, 194, 215, 241 New Sketch, 21, 32, 48, 69, 89, 107, 127, 141, 155, 174, 193, 215, 240 Pan, 21 Show Mesh, 74, 93, 199 Vertex, 25, 36, 51, 143 Internal heat generation, 166, 186 Interpolation functions, 3D solid, 103 beam, 44 matrix, 124 plane, 64 plate, 85 three-node truss, 16 two-node truss, 16 Item, Analysis Setting, 26, 52, 74, 92, 113, 130, 144, 158, 177, 196, 199 Assignment, 176, 195 Boundary Conditions, 219, 246 Calculate, 219, 249 Contours, 220, 249

Convergence Absolute Criteria, 248 Cross Section, 142 Density, 219, 224, 246 Energy, 246 Equivalent Stress, 160 Gauge Pressure, 246 Geometry, 25, 34 Graphics, 220, 249 Isotropic Instantaneous Coefficient of Thermal Expansion, 198 Line Body, 25, 34, 39 Mach Number, 246 Mean Stress Theory, 161 Mesh, 25, 34, 50, 177, 195, 217, 244 Modal, 130 Model, 220, 246 Monitor, 248 Name Selection, 217, 243 Number of Iterations, 219, 249 Pressure, 224, 247 Rectangle, 142 Reference Values, 247 Reporting Interval, 219, 249 Results, 220, 245 Run Calculation, 219, 249 Setup, 146, 218, 221, 245 Solution Controls, 248 Solution Initialization, 248 Solution Methods, 248 Solution, 27, 36, 53, 74, 93, 113, 130, 144, 146, 179, 197, 200, 218, 245

Index Solve, 27, 36, 53, 74, 93, 113, 179, 200 Solver, 245 Specific Heat, 246 Stream Function, 221 Surface Body, 176 Temperature, 247 Vectors, 220 Velocity, 247 Viscosity, 219, 221 Viscous, 229, 246 Jacobian matrix, 237 determinant, 236 Law, Hooke, 14, 42, 62, 100 Newton’s second, 118 Sutherland, 234 Life prediction, 150 Load, compressive, 136 critical static, 135 cycles, 164 cyclic, 153 offset, 115 repeated, 135 static, 135 Mass, 118 density, 166, 186 Mass-spring system, 118 Material, ductile, 135 Matrix, capacitance, 168, 169, 188 conduction, 168, 169, 188 convection, 168, 188

267 elasticity, 62, 187 Jacobian, 236 mass, 123, 124, 137 radiation, 168, 188 stiffness, 15, 43, 65, 84, 101, 123, 137, 189 strain-displacement, 104 Menu, File, 8 Help, 8 Tools, 8 Units, 8, 18, 29, 46, 67, 86, 105, 125, 138, 153, 172, 191, 213, 239 Update, 157, 218, 244 View, 8 Method, analytical, 3 finite element, 4 finite volume, 209 numerical, 4 Roe’s averaging, 236 SIMPLE, 211 variational, 101 weighted residuals, 15, 43, 63, 84, 168, 234 Modal superposition, 123 Mode shape, 131, 137 Model, k-epsilon, 229 Modulus, elasticity, 14 Young, 14, 42, 62, 83, 120 Moment of inertia of area, 42, 120, 136 Number, Prandtl, 234 Reynold, 212, 221

268

Finite Element Analysis with ANSYS Workbench

Option, Add Frozen, 22 Analysis Type, 214, 239 Basic Geometry, 19, 30 Beam Tool, 28, 38, 53 Boolean, 71 Circle, 69, 108 Combined Stress, 54 Connections, 27, 36 Cross Section, 23, 33, 49 Density-Based, 245 Direct Stress, 28, 38, 54 Displacement, 26, 35, 159 Element Numbers, 27, 37 Equivalent (von-Mises), 75, 93, 114, 160, 200 Evaluate All Results, 27, 37, 131, 146, 160 Export Text File, 27, 37 Fatigue Tool, 161 Fatigue, 161 Fixed Joints, 53 Fixed Support, 25, 51, 112, 130, 143 Force, 26, 36, 52, 144 Frictionless Support, 73, 199 Ideal-gas, 246 Insert, 28, 38, 53, 161 Inviscid, 246 Life, 161 Line, 141, 156, 241 Line Bodies, 19 Lines From Sketches, 21, 32, 49, 141 Magnitude, 74, 93, 113, 177, 196 Max Modes to Find, 133 Mesh Display, 27

Metric, 18, 29, 46, 67, 86, 125, 138, 153, 172, 213, 239 Mid-Surface, 77 Mode, 131 Moving Wall, 219 No Slip, 219, 224 Node Numbers, 27, 37 Normal, 76, 93, 201 Pressure, 113, 158 Radius, 69, 109 Rectangle, 69, 89, 108, 128, 175, 194, 215 Rectangular, 23, 33, 49 Safety Factor, 161 Second Order Upwind, 248 Shear, 202 Simply Support, 91 Standard Initialization, 248 Stress, 28, 38, 54, 75, 93, 114, 200 Surfaces From Sketches, 70, 89, 128, 175, 194, 215, 241 Temperature, 179, 197 Total, 27, 37, 53, 75, 93, 114, 130, 144, 160, 200 U.S.Customary, 105, 191 Viscous-Laminar, 219 Oscillation, cycle, 119 harmonic, 118 magnitude, 119 Passenger car frame, 132 Piping system, 227 Piston rod, 162

269

Index Plane stress, 61 Plate, bending, 81, 124 with hole, 66 simply-supported, 86 Poisson’s ratio, 62, 83, 120 Pressure, 42, 82 Racing car, 56 Recurrence relations, 123 Residual curves, 219, 249 Safety factor, 136 Screen, 7 Shock wave, 238 angle, 239 oblique, 238 bow, 252 Shock-shock interaction, 257 Shuttle, 255 bow shock, 256 cockpit, 255 nose, 255 Slope, 43 S-N curve, 151 Software, ANSYS, 6 Fluent, 207, 231 Maple, 124 Mathematica, 124 MATLAB, 124 Maxima, 124 NASTRAN, 6 package, 1, 5 symbolic manipulation, 124 Solid stress, 99 Solution, converged, 212

Specific heat, 166, 179, 186 Spring stiffness, 118 Steel, 77 structural, 138, 153 Stefan-Boltzmann constant, 167 Stiffness matrix, 3D solid, 104, 189 beam, 44 plane, 65, 85 two-node truss, 17 Strain, axial, 14 bending, 42 normal, 62, 83, 100, 187 shearing, 62, 83, 100, 187 Stress, 151 alternating, 150 axial, 13 bending, 42 effective, 151, 153 maximum, 150 maximum shear, 135 maximum von-Mises, 136 mean, 150 minimum, 150 normal, 62, 83, 100, 122, 186 shearing, 62, 83, 100, 122, 186 yield, 135 Support, frictionless, 97 Symmetry, 66, 86 Tab, Add Material, 22 Advanced Geometry Options, 214, 239

270

Finite Element Analysis with ANSYS Workbench

Angular Measure, 178 Base Objects, 22 Concept, 21, 32, 49, 70, 89, 128, 175, 194, 215, 241 Create, 71 Deformation, 27, 37, 53, 75, 93, 113, 130, 144, 146, 200 Density, 125 Dimensions, 241 Draw, 21, 32, 48, 69, 89, 108, 128, 141, 156, 175, 180, 194, 215, 241, 252 Element Size, 25, 50, 143, 181 Engineering Data, 18, 30, 46, 67, 87, 105, 125, 139, 154, 172, 191, 198 Evaluate All Results, 179, 197 Extrude, 110, 155, 180 Fixed Supports, 158 Function, 178 Generate Mesh, 72, 90, 112, 129, 143, 177, 195 Geometry, 19, 30, 139, 154, 173, 192, 214, 239 Grid, 20, 31, 47, 68, 88, 107, 127, 140, 155, 174, 193, 214, 240 ISO, 90, 108 Isotropic Elasticity, 18, 30, 46, 67, 87, 105, 126, 139, 154, 198 Isotropic Thermal Conductivity, 173, 191

Line, 48, 180 Linear Elastic, 18, 30, 46, 67, 87, 105, 126, 139, 154, 198 Loads, 26, 36, 52, 74, 92, 113, 144, 158 Major Grid Spacing, 20, 31, 47, 68, 88, 107, 127, 140, 155, 174, 93, 214, 240 Meter, 174, 214, 240 Minor-Steps per Major, 20, 31, 47, 68, 88, 107, 127, 140, 155, 174, 193, 214, 240 Modeling, 21, 32, 48, 69, 89, 107, 127, 141, 155, 174, 193, 215, 240 Modify, 155 Operation, 71 Physical Properties, 198 Poisson Ratio, 19, 30, 46, 67, 86, 87, 105, 198 Preference, 27, 37 Probe, 28, 38 Properties, 18, 30, 46, 67, 87, 105, 126, 139, 154, 173, 191, 198, 214, 239 Radius, 155 Relevance, 72, 90, 111, 129, 157, 177, 195, 217, 244 Settings, 20, 31, 47, 68, 88, 107, 127 Sizing, 143 Sketching, 20, 31, 47, 69, 89, 107, 127, 141, 155, 174, 214, 240

271

Index Snap per Minor, 20, 31, 47, 68, 88, 107, 127, 140, 155, 174, 193, 214, 240 Supports, 26, 35, 52, 144, 158 Target Bodies, 71 Temperature, 46, 177, 196, 198 Thermal, 179, 197 Thickness, 90, 128, 176, 195 Tool Bodies, 71 Tools, 27, 36, 53 Young’s Modulus, 18, 30, 46, 67, 86, 87, 105, 139, 154, 198 Zoom, 90 Temperature, 166, 183, 186 initial, 167 specified, 167 surrounding medium, 167, 179, 204 zero stress, 187, 190 Theorem, Gauss, 234 Theory, distortion energy, 136 failure, 135 maximum shear stress, 135 Thermal conductivity, 166, 171, 179, 186 Thickness, plate, 83, 121 Time, 118, 166, 186 Time step, 236 Tool bar, 7

Truss structure, one-dimensional, 17 two-dimensional, 29, 39 Vector, acceleration, 123, 137 conduction, 168, 188 convection, 168, 169, 188 deflection, 43, 84 displacement, 15, 65, 101, 189 force, 15, 43, 65, 84, 101, 189 heat generation, 168, 169, 188 load, 123 moment, 84 radiation, 168, 188 slope, 43 specified heating, 168, 182, 188 stress, 104 temperature, 168, 188 thermal load, 189 unknown, 123, 137 Velocity, initial, 119 Vibration, 3D solid, 121 beam, 120 force, 119 free, 119 plate, 121 truss, 120 Water, 227 density, 227 viscosity, 227

272

Finite Element Analysis with ANSYS Workbench

Window, Analysis Systems, 18, 46, 67, 86, 105, 125, 138, 154, 172, 191, 198, 214, 239 Details of Displacement, 35 Details of Fatigue Tool, 161 Details of Fixed Supports, 26, 51, 158 Details of Force, 26, 36, 53 Details of Line Body, 24, 33, 50, 142 Details of Mesh, 25, 34, 50, 72, 90, 111, 129, 143, 157, 177, 195, 217, 244 Details of Normal Stress, 75, 93, 201

Details of Pressure, 158 Details of Solid, 111 Details of Surface Body, 72, 90, 129, 176 Details of Temperature, 177, 196 Details of Total Deformation, 131 Details of View, 90, 109, 217, 243 Details View, 22, 49 Fluent Launcher, 218, 245, 253 Multiple System, 146 Project Schematic, 18, 30, 46 Toolbox, 198 Tree Outline, 19 Workbench, 7 symbols, 10